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An Introduction to Quantum Physics

An Introduction to Quantum Physics

A First Course for Physicists, Chemists,Materials Scientists, and Engineers

Stefanos Trachanas

Authors

Stefanos TrachanasFoundation for Research & Technology–Hellas (FORTH)Crete University Press100 Nikolaou PlastiraVassilika Vouton70013 HeraklionGreece

and

University of CreteDepartment of PhysicsP.O. Box 220871003 HeraklionGreece

Manolis AntonoyiannakisThe American Physical SocietyEditorial Office1 Research RoadRidge, NY 11961United States

and

Columbia UniversityDepartment of Applied Physics & AppliedMathematics500 W. 120th StreetNew York, NY 10027United States

Leonidas TsetserisNational Technical University of AthensDepartment of PhysicsZografou Campus15780 AthensGreece

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vii

Contents

Foreword xixPreface xxiiiEditors’ Note xxvii

Part I Fundamental Principles 1

1 The Principle of Wave–Particle Duality: An Overview 31.1 Introduction 31.2 The Principle of Wave–Particle Duality of Light 41.2.1 The Photoelectric Effect 41.2.2 The Compton Effect 71.2.3 A Note on Units 101.3 The Principle of Wave–Particle Duality of Matter 111.3.1 From Frequency Quantization in Classical Waves to Energy

Quantization in Matter Waves: The Most Important GeneralConsequence of Wave–Particle Duality of Matter 12

1.3.2 The Problem of Atomic Stability under Collisions 131.3.3 The Problem of Energy Scales: Why Are Atomic Energies on the Order

of eV, While Nuclear Energies Are on the Order of MeV? 151.3.4 The Stability of Atoms and Molecules Against External

Electromagnetic Radiation 171.3.5 The Problem of Length Scales: Why Are Atomic Sizes on the Order of

Angstroms, While Nuclear Sizes Are on the Order of Fermis? 191.3.6 The Stability of Atoms Against Their Own Radiation: Probabilistic

Interpretation of Matter Waves 211.3.7 How Do Atoms Radiate after All? Quantum Jumps from Higher to

Lower Energy States and Atomic Spectra 221.3.8 Quantized Energies and Atomic Spectra: The Case of Hydrogen 251.3.9 Correct and Incorrect Pictures for the Motion of Electrons in Atoms:

Revisiting the Case of Hydrogen 251.3.10 The Fine Structure Constant and Numerical Calculations in Bohr’s

Theory 29

viii Contents

1.3.11 Numerical Calculations with Matter Waves: Practical Formulas andPhysical Applications 31

1.3.12 A Direct Confirmation of the Existence of Matter Waves: TheDavisson–Germer Experiment 33

1.3.13 The Double-Slit Experiment: Collapse of the Wavefunction UponMeasurement 34

1.4 Dimensional Analysis and Quantum Physics 411.4.1 The Fundamental Theorem and a Simple Application 411.4.2 Blackbody Radiation Using Dimensional Analysis 441.4.3 The Hydrogen Atom Using Dimensional Analysis 47

2 The Schrödinger Equation and Its Statistical Interpretation 532.1 Introduction 532.2 The Schrödinger Equation 532.2.1 The Schrödinger Equation for Free Particles 542.2.2 The Schrödinger Equation in an External Potential 572.2.3 Mathematical Intermission I: Linear Operators 582.3 Statistical Interpretation of Quantum Mechanics 602.3.1 The “Particle–Wave” Contradiction in Classical Mechanics 602.3.2 Statistical Interpretation 612.3.3 Why Did We Choose P(x) = |𝜓(x)|2 as the Probability Density? 622.3.4 Mathematical Intermission II: Basic Statistical Concepts 632.3.4.1 Mean Value 632.3.4.2 Standard Deviation (or Uncertainty) 652.3.5 Position Measurements: Mean Value and Uncertainty 672.4 Further Development of the Statistical Interpretation: The Mean-Value

Formula 712.4.1 The General Formula for the Mean Value 712.4.2 The General Formula for Uncertainty 732.5 Time Evolution of Wavefunctions and Superposition States 772.5.1 Setting the Stage 772.5.2 Solving the Schrödinger Equation. Separation of Variables 782.5.3 The Time-Independent Schrödinger Equation as an Eigenvalue

Equation: Zero-Uncertainty States and Superposition States 812.5.4 Energy Quantization for Confined Motion: A Fundamental General

Consequence of Schrödinger’s Equation 852.5.5 The Role of Measurement in Quantum Mechanics: Collapse of the

Wavefunction Upon Measurement 862.5.6 Measurable Consequences of Time Evolution: Stationary and

Nonstationary States 912.6 Self-Consistency of the Statistical Interpretation and the Mathematical

Structure of Quantum Mechanics 952.6.1 Hermitian Operators 952.6.2 Conservation of Probability 982.6.3 Inner Product and Orthogonality 992.6.4 Matrix Representation of Quantum Mechanical Operators 1012.7 Summary: Quantum Mechanics in a Nutshell 103

Contents ix

3 The Uncertainty Principle 1073.1 Introduction 1073.2 The Position–Momentum Uncertainty Principle 1083.2.1 Mathematical Explanation of the Principle 1083.2.2 Physical Explanation of the Principle 1093.2.3 Quantum Resistance to Confinement. A Fundamental Consequence of

the Position–Momentum Uncertainty Principle 1123.3 The Time–Energy Uncertainty Principle 1143.4 The Uncertainty Principle in the Classical Limit 1183.5 General Investigation of the Uncertainty Principle 1193.5.1 Compatible and Incompatible Physical Quantities and the Generalized

Uncertainty Relation 1193.5.2 Angular Momentum: A Different Kind of Vector 122

Part II Simple Quantum Systems 127

4 Square Potentials. I: Discrete Spectrum—Bound States 1294.1 Introduction 1294.2 Particle in a One-Dimensional Box: The Infinite Potential Well 1324.2.1 Solution of the Schrödinger Equation 1324.2.2 Discussion of the Results 1344.2.2.1 Dimensional Analysis of the Formula En = (ℏ2𝜋2∕2mL2)n2.

Do We Need an Exact Solution to Predict the Energy Dependence onℏ, m, and L? 135

4.2.2.2 Dependence of the Ground-State Energy on ℏ, m, and L : The ClassicalLimit 136

4.2.2.3 The Limit of Large Quantum Numbers and QuantumDiscontinuities 137

4.2.2.4 The Classical Limit of the Position Probability Density 1384.2.2.5 Eigenfunction Features: Mirror Symmetry and the Node

Theorem 1394.2.2.6 Numerical Calculations in Practical Units 1394.3 The Square Potential Well 1404.3.1 Solution of the Schrödinger Equation 1404.3.2 Discussion of the Results 1434.3.2.1 Penetration into Classically Forbidden Regions 1434.3.2.2 Penetration in the Classical Limit 1444.3.2.3 The Physics and “Numerics” of the Parameter 𝜆 145

5 Square Potentials. II: Continuous Spectrum—ScatteringStates 149

5.1 Introduction 1495.2 The Square Potential Step: Reflection and Transmission 1505.2.1 Solution of the Schrödinger Equation and Calculation of the Reflection

Coefficient 1505.2.2 Discussion of the Results 153

x Contents

5.2.2.1 The Phenomenon of Classically Forbidden Reflection 1535.2.2.2 Transmission Coefficient in the “Classical Limit” of High

Energies 1545.2.2.3 The Reflection Coefficient Depends neither on Planck’s Constant nor

on the Mass of the Particle: Analysis of a Paradox 1545.2.2.4 An Argument from Dimensional Analysis 1555.3 Rectangular Potential Barrier: Tunneling Effect 1565.3.1 Solution of the Schrödinger Equation 1565.3.2 Discussion of the Results 1585.3.2.1 Crossing a Classically Forbidden Region: The Tunneling Effect 1585.3.2.2 Exponential Sensitivity of the Tunneling Effect to the Energy of the

Particle 1595.3.2.3 A Simple Approximate Expression for the Transmission

Coefficient 1605.3.2.4 Exponential Sensitivity of the Tunneling Effect to the Mass of the

Particle 1625.3.2.5 A Practical Formula for T 163

6 The Harmonic Oscillator 1676.1 Introduction 1676.2 Solution of the Schrödinger Equation 1696.3 Discussion of the Results 1776.3.1 Shape of Wavefunctions. Mirror Symmetry and the Node

Theorem 1786.3.2 Shape of Eigenfunctions for Large n: The Classical Limit 1796.3.3 The Extreme Anticlassical Limit of the Ground State 1806.3.4 Penetration into Classically Forbidden Regions: What Fraction of Its

“Lifetime” Does the Particle “Spend” in the Classically ForbiddenRegion? 181

6.3.5 A Quantum Oscillator Never Rests: Zero-Point Energy 1826.3.6 Equidistant Eigenvalues and Emission of Radiation from a Quantum

Harmonic Oscillator 1846.4 A Plausible Question: Can We Use the Polynomial Method to Solve

Potentials Other than the Harmonic Oscillator? 187

7 The Polynomial Method: Systematic Theory andApplications 191

7.1 Introduction: The Power-Series Method 1917.2 Sufficient Conditions for the Existence of Polynomial Solutions:

Bidimensional Equations 1947.3 The Polynomial Method in Action: Exact Solution of the Kratzer and

Morse Potentials 1977.4 Mathematical Afterword 202

8 The Hydrogen Atom. I: Spherically Symmetric Solutions 2078.1 Introduction 207

Contents xi

8.2 Solving the Schrödinger Equation for the Spherically SymmetricEigenfunctions 209

8.2.1 A Final Comment: The System of Atomic Units 2168.3 Discussion of the Results 2178.3.1 Checking the Classical Limit ℏ→ 0 or m → ∞ for the Ground State of

the Hydrogen Atom 2178.3.2 Energy Quantization and Atomic Stability 2178.3.3 The Size of the Atom and the Uncertainty Principle: The Mystery of

Atomic Stability from Another Perspective 2188.3.4 Atomic Incompressibility and the Uncertainty Principle 2218.3.5 More on the Ground State of the Atom. Mean and Most Probable

Distance of the Electron from the Nucleus 2218.3.6 Revisiting the Notion of “Atomic Radius”: How Probable is It to Find

the Electron Within the “Volume” that the Atom SupposedlyOccupies? 222

8.3.7 An Apparent Paradox: After All, Where Is It Most Likely to Find theElectron? Near the Nucleus or One Bohr Radius Away from It? 223

8.3.8 What Fraction of Its Time Does the Electron Spend in the ClassicallyForbidden Region of the Atom? 223

8.3.9 Is the Bohr Theory for the Hydrogen Atom Really Wrong? Comparisonwith Quantum Mechanics 225

8.4 What Is the Electron Doing in the Hydrogen Atom after All? A FirstDiscussion on the Basic Questions of Quantum Mechanics 226

9 The Hydrogen Atom. II: Solutions with AngularDependence 231

9.1 Introduction 2319.2 The Schrödinger Equation in an Arbitrary Central Potential:

Separation of Variables 2329.2.1 Separation of Radial from Angular Variables 2329.2.2 The Radial Schrödinger Equation: Physical Interpretation of the

Centrifugal Term and Connection to the Angular Equation 2359.2.3 Solution of the Angular Equation: Eigenvalues and Eigenfunctions of

Angular Momentum 2379.2.3.1 Solving the Equation for Φ 2389.2.3.2 Solving the Equation for Θ 2399.2.4 Summary of Results for an Arbitrary Central Potential 2439.3 The Hydrogen Atom 2469.3.1 Solution of the Radial Equation for the Coulomb Potential 2469.3.2 Explicit Construction of the First Few Eigenfunctions 2499.3.2.1 n = 1 : The Ground State 2509.3.2.2 n = 2 : The First Excited States 2509.3.3 Discussion of the Results 2549.3.3.1 The Energy-Level Diagram 2549.3.3.2 Degeneracy of the Energy Spectrum for a Coulomb Potential:

Rotational and Accidental Degeneracy 2559.3.3.3 Removal of Rotational and Hydrogenic Degeneracy 257

xii Contents

9.3.3.4 The Ground State is Always Nondegenerate and Has the FullSymmetry of the Problem 257

9.3.3.5 Spectroscopic Notation for Atomic States 2589.3.3.6 The “Concept” of the Orbital: s and p Orbitals 2589.3.3.7 Quantum Angular Momentum: A Rather Strange Vector 2619.3.3.8 Allowed and Forbidden Transitions in the Hydrogen Atom:

Conservation of Angular Momentum and Selection Rules 263

10 Atoms in a Magnetic Field and the Emergence of Spin 26710.1 Introduction 26710.2 Atomic Electrons as Microscopic Magnets: Magnetic Moment and

Angular Momentum 27010.3 The Zeeman Effect and the Evidence for the Existence of Spin 27410.4 The Stern–Gerlach Experiment: Unequivocal Experimental

Confirmation of the Existence of Spin 27810.4.1 Preliminary Investigation: A Plausible Theoretical Description

of Spin 27810.4.2 The Experiment and Its Results 28010.5 What is Spin? 28410.5.1 Spin is No Self-Rotation 28410.5.2 How is Spin Described Quantum Mechanically? 28510.5.3 What Spin Really Is 29110.6 Time Evolution of Spin in a Magnetic Field 29210.7 Total Angular Momentum of Atoms: Addition of Angular

Momenta 29510.7.1 The Eigenvalues 29510.7.2 The Eigenfunctions 300

11 Identical Particles and the Pauli Principle 30511.1 Introduction 30511.2 The Principle of Indistinguishability of Identical Particles in Quantum

Mechanics 30511.3 Indistinguishability of Identical Particles and the Pauli Principle 30611.4 The Role of Spin: Complete Formulation of the Pauli Principle 30711.5 The Pauli Exclusion Principle 31011.6 Which Particles Are Fermions and Which Are Bosons 31411.7 Exchange Degeneracy: The Problem and Its Solution 317

Part III Quantum Mechanics in Action: The Structureof Matter 321

12 Atoms: The Periodic Table of the Elements 32312.1 Introduction 32312.2 Arrangement of Energy Levels in Many-Electron Atoms:

The Screening Effect 324

Contents xiii

12.3 Quantum Mechanical Explanation of the Periodic Table:The “Small Periodic Table” 327

12.3.1 Populating the Energy Levels: The Shell Model 32812.3.2 An Interesting “Detail”: The Pauli Principle and Atomic

Magnetism 32912.3.3 Quantum Mechanical Explanation of Valence and Directionality of

Chemical Bonds 33112.3.4 Quantum Mechanical Explanation of Chemical Periodicity: The Third

Row of the Periodic Table 33212.3.5 Ionization Energy and Its Role in Chemical Behavior 33412.3.6 Examples 33812.4 Approximate Calculations in Atoms: Perturbation Theory and the

Variational Method 34112.4.1 Perturbation Theory 34212.4.2 Variational Method 346

13 Molecules. I: Elementary Theory of the Chemical Bond 35113.1 Introduction 35113.2 The Double-Well Model of Chemical Bonding 35213.2.1 The Symmetric Double Well 35213.2.2 The Asymmetric Double Well 35613.3 Examples of Simple Molecules 36013.3.1 The Hydrogen Molecule H2 36013.3.2 The Helium “Molecule” He2 36313.3.3 The Lithium Molecule Li2 36413.3.4 The Oxygen Molecule O2 36413.3.5 The Nitrogen Molecule N2 36613.3.6 The Water Molecule H2O 36713.3.7 Hydrogen Bonds: From the Water Molecule to Biomolecules 37013.3.8 The Ammonia Molecule NH3 37313.4 Molecular Spectra 37713.4.1 Rotational Spectrum 37813.4.2 Vibrational Spectrum 38213.4.3 The Vibrational–Rotational Spectrum 385

14 Molecules. II: The Chemistry of Carbon 39314.1 Introduction 39314.2 Hybridization: The First Basic Deviation from the Elementary Theory

of the Chemical Bond 39314.2.1 The CH4 Molecule According to the Elementary Theory: An

Erroneous Prediction 39314.2.2 Hybridized Orbitals and the CH4 Molecule 39514.2.3 Total and Partial Hybridization 40114.2.4 The Need for Partial Hybridization: The Molecules C2H4, C2H2, and

C2H6 40414.2.5 Application of Hybridization Theory to Conjugated

Hydrocarbons 408

xiv Contents

14.2.6 Energy Balance of Hybridization and Application to InorganicMolecules 409

14.3 Delocalization: The Second Basic Deviation from the ElementaryTheory of the Chemical Bond 414

14.3.1 A Closer Look at the Benzene Molecule 41414.3.2 An Elementary Theory of Delocalization: The Free-Electron

Model 41714.3.3 LCAO Theory for Conjugated Hydrocarbons. I: Cyclic Chains 41814.3.4 LCAO Theory for Conjugated Hydrocarbons. II: Linear Chains 42414.3.5 Delocalization on Carbon Chains: General Remarks 42714.3.6 Delocalization in Two-dimensional Arrays of p Orbitals: Graphene and

Fullerenes 429

15 Solids: Conductors, Semiconductors, Insulators 43915.1 Introduction 43915.2 Periodicity and Band Structure 43915.3 Band Structure and the “Mystery of Conductivity.” Conductors,

Semiconductors, Insulators 44115.3.1 Failure of the Classical Theory 44115.3.2 The Quantum Explanation 44315.4 Crystal Momentum, Effective Mass, and Electron Mobility 44715.5 Fermi Energy and Density of States 45315.5.1 Fermi Energy in the Free-Electron Model 45315.5.2 Density of States in the Free-Electron Model 45715.5.3 Discussion of the Results: Sharing of Available Space by the Particles of

a Fermi Gas 46015.5.4 A Classic Application: The “Anomaly” of the Electronic Specific Heat

of Metals 463

16 Matter and Light: The Interaction of Atoms withElectromagnetic Radiation 469

16.1 Introduction 46916.2 The Four Fundamental Processes: Resonance, Scattering, Ionization,

and Spontaneous Emission 47116.3 Quantitative Description of the Fundamental Processes: Transition

Rate, Effective Cross Section, Mean Free Path 47316.3.1 Transition Rate: The Fundamental Concept 47316.3.2 Effective Cross Section and Mean Free Path 47516.3.3 Scattering Cross Section: An Instructive Example 47616.4 Matter and Light in Resonance. I: Theory 47816.4.1 Calculation of the Effective Cross Section: Fermi’s Rule 47816.4.2 Discussion of the Result: Order-of-Magnitude Estimates and Selection

Rules 48116.4.3 Selection Rules: Allowed and Forbidden Transitions 48316.5 Matter and Light in Resonance. II: The Laser 48716.5.1 The Operation Principle: Population Inversion and the Threshold

Condition 487

Contents xv

16.5.2 Main Properties of Laser Light 49116.5.2.1 Phase Coherence 49116.5.2.2 Directionality 49116.5.2.3 Intensity 49116.5.2.4 Monochromaticity 49216.6 Spontaneous Emission 49416.7 Theory of Time-dependent Perturbations: Fermi’s Rule 49916.7.1 Approximate Calculation of Transition Probabilities Pn→m(t) for an

Arbitrary “Transient” Perturbation V (t) 49916.7.2 The Atom Under the Influence of a Sinusoidal Perturbation:

Fermi’s Rule for Resonance Transitions 50316.8 The Light Itself: Polarized Photons and Their Quantum Mechanical

Description 51116.8.1 States of Linear and Circular Polarization for Photons 51116.8.2 Linear and Circular Polarizers 51216.8.3 Quantum Mechanical Description of Polarized Photons 513

Online Supplement

1 The Principle of Wave–Particle Duality: An OverviewOS1.1 Review QuizOS1.1 Determining Planck’s Constant from Everyday Observations

2 The Schrödinger Equation and Its Statistical InterpretationOS2.1 Review QuizOS2.2 Further Study of Hermitian Operators: The Concept of the

Adjoint OperatorOS2.3 Local Conservation of Probability: The Probability Current

3 The Uncertainty PrincipleOS3.1 Review QuizOS3.2 Commutator Algebra: Calculational TechniquesOS3.3 The Generalized Uncertainty PrincipleOS3.4 Ehrenfest’s Theorem: Time Evolution of Mean Values and the

Classical Limit

4 Square Potentials. I: Discrete Spectrum—Bound StatesOS4.1 Review QuizOS4.2 Square Well: A More Elegant Graphical Solution for Its EigenvaluesOS4.3 Deep and Shallow Wells: Approximate Analytic Expressions for Their

Eigenvalues

5 Square Potentials. II: Continuous Spectrum—ScatteringStates

OS5.1 Review QuizOS5.2 Quantum Mechanical Theory of Alpha Decay

xvi Contents

6 The Harmonic OscillatorOS6.1 Review QuizOS6.2 Algebraic Solution of the Harmonic Oscillator: Creation and

Annihilation Operators

7 The Polynomial Method: Systematic Theory and ApplicationsOS7.1 Review QuizOS7.2 An Elementary Method for Discovering Exactly Solvable PotentialsOS7.3 Classic Examples of Exactly Solvable Potentials: A Comprehensive

List

8 The Hydrogen Atom. I: Spherically Symmetric SolutionsOS8.1 Review Quiz

9 The Hydrogen Atom. II: Solutions with Angular DependenceOS9.1 Review QuizOS9.2 Conservation of Angular Momentum in Central Potentials, and Its

ConsequencesOS9.3 Solving the Associated Legendre Equation on Our Own

10 Atoms in a Magnetic Field and the Emergence of SpinOS10.1 Review QuizOS10.2 Algebraic Theory of Angular Momentum and Spin

11 Identical Particles and the Pauli PrincipleOS11.1 Review QuizOS11.2 Dirac’s Formalism: A Brief Introduction

12 Atoms: The Periodic Table of the ElementsOS12.1 Review QuizOS12.2 Systematic Perturbation Theory: Application to the Stark Effect and

Atomic Polarizability

13 Molecules. I: Elementary Theory of the Chemical BondOS13.1 Review Quiz

14 Molecules. II: The Chemistry of CarbonOS14.1 Review QuizOS14.2 The LCAO Method and Matrix MechanicsOS14.3 Extension of the LCAO Method for Nonzero Overlap

15 Solids: Conductors, Semiconductors, InsulatorsOS15.1 Review QuizOS15.2 Floquet’s Theorem: Mathematical Study of the Band Structure for an

Arbitrary Periodic Potential V(x)OS15.3 Compressibility of Condensed Matter: The Bulk ModulusOS15.4 The Pauli Principle and Gravitational Collapse: The Chandrasekhar

Limit

Contents xvii

16 Matter and Light: The Interaction of Atoms withElectromagnetic Radiation

OS16.1 Review QuizOS16.2 Resonance Transitions Beyond Fermi’s Rule: Rabi OscillationsOS16.3 Resonance Transitions at Radio Frequencies: Nuclear Magnetic

Resonance (NMR)

Appendix 519Bibliography 523Index 527

xix

Foreword

As fate would have it, or perhaps due to some form of quantum interference,I encountered Stefanos Trachanas’ book on Quantum Physics in its prenatalform. In the late 1970s, while a graduate student at Harvard, Trachanas wasworking on a set of notes on quantum physics, written in his native language.He occasionally lent his handwritten notes (the file-sharing mode of that era) tofriends who appreciated his fascination with Nature’s wonders. At the time, I wasan undergraduate student at the technical school down the river, struggling tolearn quantum physics, and was very grateful to have access to Trachanas’ notes.I still remember the delight and amazement I felt when reading his notes, fortheir clarity and freshness, and for the wonderful insights, not to be found in anyof the classic physics texts available at the time (our common native languagealso helped). It is a great pleasure to see that in the latest version of his book onQuantum Physics, this freshness is intact, enriched from decades of teachingexperience. This latest version is of course a long way from his original set ofnotes; it is a thorough account of the theory of quantum mechanics, expertlytranslated by Manolis Antonoyiannakis and Leonidas Tsetseris, in the form of acomprehensive and mature textbook.

It is an unusual book. All the formulas and numbers and tables that you findin any other textbook on the subject are there. This level of systematic detail isimportant; one does expect a textbook to contain a complete treatment of thesubject and to serve as a reference for key results and expressions. But there arealso many wonderful insights that I have not found elsewhere, and numerouselaborate discussions and explanations of the meaning of the formulas, a crucialingredient for developing an understanding of quantum physics.

The detailed examples, constantly contrasting the quantum and the classicalpictures for model systems, are the hallmark of the book. Another key charac-teristic is the use of dimensional analysis, through which many of the secrets ofquantum behavior can be elucidated. Finally, the application of key concepts torealistic problems, including atoms, molecules, and solids, makes the treatmentof the subject not only pedagogically insightful but also of great practical value.

The book is nicely laid out in three parts: In Part I, the student is intro-duced to “the language of quantum mechanics” (the author’s astute definitionof the subject, as mentioned in the Preface), including all the “cool” (myquotes) concepts of the quantum realm, such as wave–particle duality and theuncertainty principle. Then, in Part II, the language is used to describe the

xx Foreword

standard simple problems, the square well, the harmonic oscillator, and theCoulomb potential. It is also applied to the hydrogen atom, illustrating howthis language can capture the behavior of Nature at the level of fundamentalparticles—electrons and protons. Finally, in Part III, the student is given athorough training in the use of the quantum language to address problemsrelevant to real applications in modern life, which is dominated by quantumdevices, for better or for worse. Many everyday activities, from using a cellphone to call friends to employing photovoltaics for powering your house, aredirectly related to quintessentially quantum phenomena, that is, the physicsof semiconductors, conductors, and insulators, and their interaction withlight. All these phenomena are explained thoroughly and clearly in Trachanas’book. The reader of the book will certainly develop a deep appreciation of theprinciples on which many everyday devices are based. There is also a lovelydiscussion of the properties of molecules and the nature of the chemical bond.The treatment ranges from the closed sixfold hydrocarbon ring (“benzene”) tothe truncated icosahedron formed by 60 carbon atoms (“fullerene”), with severalother important structures in between. This discussion touches upon the originof chemical complexity, including many aspects related to carbon, the “elementof life” (again my quotes), and occupies, deservedly, a whole chapter.

For the demanding reader, there are several chapters of higher mathematicaland physical sophistication. The two cases that stand out are Chapter 7 onthe polynomial method and Chapters 10 and 11 on the nature of spin and onidentical particles. The treatment of the polynomial method is quite unusualfor an introductory text on quantum physics, but it is beautifully explained insimple steps. Although the author suggests that this chapter can be skipped atfirst reading, in my view, it is not to be missed. Anyone who wondered why allbooks deal with just three standard problems (square well, harmonic oscillator,Coulomb potential), will find here some very enticing answers, and a wonderfuldiscussion of which types of problem yield closed analytical solutions. For thepractitioners of numerical simulations, this approach provides elegant insightsto the well-known Kratzer and Morse potentials. It is satisfying to see that thesefamiliar tools for simulating the properties of complex systems have simpleanalytical solutions. Finally, Trachanas argues that the nature of the electron’s“spin” is related to the essence of quantum measurement, and this is nicelyconnected to the character of elementary particles, “fermions” or “bosons,”and to their interaction with magnetic fields. The concepts are deep, yet theirexplanation is elegant and convincing. It is presented through a playful set ofquestions and answers, with no recourse to technical jargon. The effect of thisapproach is powerful and empowering: The reader is left with the impressionthat even the most puzzling concepts of quantum physics can actually be graspedin simple, intuitive terms.

Foreword xxi

A famous joke among physicists is that “One does not really understandquantum mechanics, but simply gets used to it.” To an undergraduate studentbeing exposed to quantum physics for the first time, this phrase may come veryclose to how it feels to speak Nature’s language of the atomic scale. Trachanas’Quantum Physics aims to remove this feeling and in my opinion it succeedsbrilliantly.

Cambridge, MassachusettsMarch, 2017

Efthimios KaxirasHarvard University

xxiii

Preface

Learning quantum mechanics is like learning a foreign language. To speak it wellone needs to relocate to the country where it is spoken, and settle there for awhile—to make it one’s day-to-day language. This book has been designed so thatthe teaching of quantum mechanics as a “foreign language” satisfies this residencerequirement. Once the readers become familiar with the fundamental principles(Part I) and study some simple quantum systems (Part II), they are invited to “set-tle” in the atomic world (Part III) and learn quantum mechanics in action. To talkthe language of quantum mechanics in its natural habitat. So, in a way, this is adouble book: Quantum Mechanics and Structure of Matter. It includes a com-plete introduction to the basic structures of nonnuclear matter—not simply as“applications” but as a necessary final step toward understanding the theory itself.

This is an introductory book, aimed at undergraduate students with no priorexposure to quantum theory, except perhaps from a general physics course.From a mathematical perspective, all that is required from the readers is to havetaken a Calculus I course and to be simply familiar with matrix diagonalizationin linear algebra.

Those readers with some previous exposure to quantum mechanicalconcepts—say, the wave–particle duality principle—can readily proceed toChapter 2. But a quick browse through Chapter 1 may prove useful for them also,since this is a quite conceptual chapter that prepares the ground for acceptanceof the rather bizarre quantum mechanical concepts, which are so alien to oureveryday experience.

An integral part of the book is the online supplement. It contains reviewquizzes, theory supplements that cover the few additional topics taught in amore formal course on quantum mechanics, and also some further applications.Installed in an open-source platform—Open edX—designed for massive openonline courses (MOOCs), the online supplement offers an interactive onlinelearning environment that may become an integral part of academic textbooksin the future. It can be freely accessed at http://www.mathesis.org.

Had it not been for the generous decision of colleagues Manolis Antonoyian-nakis and Leonidas Tsetseris to undertake the translation and editing of theoriginal Greek edition, this book would not have seen the light of day. Mydeepest thanks therefore go to them. They tirelessly plowed through the originaltext and my own continuous—and extensive—revisions, as well as two entirely

http://www.mathesis.org

xxiv Preface

new chapters (Chapters 1 and 7). Throughout this process, their comments andfeedback were critical in helping me finalize the text.

In the last stages of the editing process, I have benefited from a particularlyfruitful collaboration with Manolis Antonoyiannakis, and also with our youngercolleague Tasos Epitropakis, who copyedited the last version of the text, andundertook the translation and editing of the online supplement.

I am also grateful to Manolis for managing the whole project, from the bookproposal, to peer review, to negotiating and liaising with Wiley-VCH, to marketresearch and outreach.

The book was fortunate to have Jerry Icban Dadap (Columbia University)as the very first critical reader of its English edition. He read the manuscriptfrom cover to cover and made numerous comments that helped improve thetext considerably. Vassilis Charmandaris (University of Crete), Petros Ditsas(University of Crete), Eleftherios Economou (University of Crete, and Founda-tion for Research & Technology–Hellas), Themis Lazaridis (City College of NewYork), Nikos Kopidakis (Macquarie University), Daniel Esteve (CEA-Saclay),Che Ting Chan (Hong Kong University of Science and Technology), and Pak WoLeung (Hong Kong University of Science and Technology) also made varioususeful comments and recommendations, as did Jessica Thomas (AmericanPhysical Society) on two early chapters. I also thank Dimitrios Psaltis (Universityof Arizona) and Demetrios Christodoulides (University of Central Florida) fortheir encouragement.

I am indebted to—and humbled by—Efthimios Kaxiras (Harvard University)for his artfully crafted Foreword. I could not have hoped for a better introductionto the book! I am also grateful to Nader Engheta (University of Pennsylvania) forhis constant encouragement, support, and endorsement.

Prior to this English edition, the book has benefited enormously from its wideuse as the main textbook for quantum mechanics courses in most universitiesand polytechnics in Greece and Cyprus. I owe a lot to the readers and instructorswho supported the book and provided feedback throughout these years. But mygreatest debt is to John Iliopoulos (Ecole Normale Superieure) for his invaluableadvice and comments during the first writing of this book, and his generoussupport of its original Greek version.

At a technical level, the skillful typesetting by Sofia Vlachou (Crete UniversityPress), the design of the figures by Iakovos Ouranos, and the installation of theonline supplement on Open edX by Nick Gikopoulos (Crete University Press)have contributed critically to the quality of the final product. At the proofs stage,the assistance of Katerina Ligovanli (Crete University Press) was invaluable.Finally, I am grateful to Wiley’s production team, and especially Sujisha KunchiParambathu, for a smooth and productive collaboration.

Heraklion, Crete Stefanos TrachanasMarch 2017 Foundation for Research & Technology–Hellas (FORTH)

Preface xxv

The Teaching Philosophy of the Book

The long—and ever-increasing—list of quantum mechanics textbooks tells usthat there are wide-ranging views about how to teach the subject and what topicsto include in a course. Aside from our topic selection, the pedagogic approachthat perhaps sets this book apart from existing textbooks can be summed up inthe following six themes:

1. Extensive Discussions of Results: The Physics Behind the FormulasThroughout the book—and especially in Part II—we discuss in detail the resultof every solved problem, to highlight its physical meaning and understandits plausibility, how it behaves in various limits, and what are the broaderconclusions that can be derived from it. These extensive discussions aim atgradually familiarizing students with the quantum mechanical concepts, anddeveloping their intuition.

2. Dimensional Analysis: A Valuable ToolUnlike most textbooks in the field, dimensional analysis is a basic tool forus. It not only helps us simplify the solution of many quantum mechanicalproblems but also extract important results, even when the underlying theoryis not known in detail or is quite cumbersome. For example, to be able toshow—on purely dimensional grounds, and in one line of algebra—thatthe ultraviolet catastrophe is a consequence of the universality of thermalradiation, is, in our opinion, more interesting and profound than the detailedcalculation of the corresponding classical formula.

3. Numerical Calculations and Order-of-Magnitude Estimates: Numbers MatterThe ability to make order-of-magnitude estimates and execute transparentnumerical calculations—in appropriate units—in order to understand aphysical result and decide whether it makes sense is of primary importance inthis book. This is a kind of an art—with much physics involved—that needs tobe taught, too. In this spirit, the order-of-magnitude estimates of basic quan-tities involved in a problem, the use of suitable practical formulas in helpfulunits, the construction of dimensionless combinations of quantities that allownumerical calculations to have universal validity, or even the use of a naturalsystem of units (such as the atomic system), are some of the tools widelyemployed in the book. In contrast to classical theories, quantum mechanicscannot be properly understood via its equations alone. Numbers matter here.

4. Exact Solutions: Is There a Method?The exactly solvable quantum mechanical problems are importantbecause—thanks to the explicit form of their solutions—they allow studentsto develop familiarity with the quantum mechanical concepts and methods.It is therefore important for students to be able to solve these problems ontheir own, and thus reach the level of self-confidence that is necessary for ademanding course such as quantum mechanics.

But the traditional way of presenting methods of exact solution, especiallythe power-series method, does not serve this purpose in our opinion. Themethod itself has never been popular among students (and many teachersalso), while the application—in the hydrogen atom, for example—of finely

xxvi Preface

tuned transformations in order to arrive at a specific eponymous equationthat has known solutions, dispels any hope that one might ever be able tosolve the problem on one’s own.

On this topic, at least, this book can promise something different: Namely,that the exact solution and the calculation of eigenvalues for problems suchas (but not only) the harmonic oscillator or the hydrogen atom will not beharder than for the infinite-well potential. And even further, that readers willbe able to execute such calculations on their own, in a few minutes, with noprior knowledge of any eponymous differential equation or the correspondingspecial function. The pertinent ideas and techniques are presented, at a verybasic level, in Chapters 6, 8, and 9, while those interested in a systematicpresentation can consult Chapter 7.

5. The Weirdness of Quantum Mechanics: Discussion of Conceptual ProblemsQuantum mechanics is not just a foreign language. It is a very strangelanguage, often at complete odds with the language of our classical world.Therefore, it cannot be taught in the same manner as any classical theory.Aside from its equations and calculational rules, quantum mechanics alsorequires a radical change in how we perceive physical reality and the kindof knowledge we can draw from it. Thus, innocuous questions like “what isthe electron really doing in the ground state of the hydrogen atom” or “whatexactly is the spin of an electron” cannot be properly answered without theappropriate conceptual gear. The development of the pertinent conceptsbegins in Chapter 1 and continues in Chapters 8 and 10, where we discussquestions such as those mentioned and discover the central role of themeasurement process in quantum mechanics—both for the very definition ofphysical quantities (e.g., spin) and for the fundamental distinction to be madebetween questions that are valid in the quantum mechanical context (i.e.,experimentally testable) and those that are not.

6. Online Quizzes: Student Engagement and Self-LearningThe interactive self-examination of students is another pedagogical featureof the book, drawing on the author’s growing experience with Massive OpenOnline Courses (MOOCs) and his experimentation with various forms ofblended learning. In contrast to the conventional textbook (or take-home)problems, the online quizzes allow and encourage a much greater varietyof targeted questions—many of them of a conceptual character—as well assuitable multi-step problems that make it easier for students (thanks alsoto immediate access to their answers) to identify their own weaknesses andproceed to further study as they deem necessary. The online quizzes will be aliving—and evolving—element of the book.

At any rate, the fundamental teaching philosophy of this book is what weearlier called the residence requirement: the residence, for some time, in the“country” where quantum mechanics is the spoken language. Only in that“place” can we come to terms with the weirdness of quantum mechanics.And, if, having fulfilled this residence requirement, there remain lingeringobjections, they may actually turn out to be legitimate, leading some oftoday’s students to a fundamental revision of the quantum mechanical theorytomorrow. If such a revision is indeed to come.

xxvii

Editors’ Note

This book is a labor of love. Officially, the translation project began in 2006. Bythen, Stefanos had completely rewritten and recast his original, three-volumetextbook into one comprehensive volume. But the idea of bringing Stefanos’swork to a global audience was born in the early 1990s, when, as undergradu-ate students, we experienced firsthand—through his books and lectures at aUniversity of Crete summer school—his original style of explaining quantummechanics, combining a high command of the material with an eagerness todemystify and connect with the students. Since then, we have often mused withfellow physicists and chemists that Stefanos’s work ought to be translated toEnglish one day. After all, his books are taught in most departments of physics,chemistry, materials science, and engineering in Greece. By 2006, we felt thatit was time to break the language barrier. And what better place to start thanStefanos’s signature book on quantum physics?

The translation and editing of this book has been a challenge. Both of us havehad demanding full-time jobs, so this had to be a part-time project on our “free”time. Translating into a target language other than one’s mother tongue is tricky.And Stefanos’s native prose is highly elaborate, with rich syntax, long sentences,and a playfulness that is challenging to translate. Progress has thus been slowand intermittent. Over time, we developed a methodology for how to collaborateeffectively, utilize online tools, resolve translation issues, and calibrate our prose.And we revised the text relentlessly: Each chapter has been edited at least adozen times.

But it was not all work and no play. While editing the book, we were able toexpand our understanding, particularly with the new material accompanying thisEnglish edition, or on historical aspects we had previously overlooked. And dur-ing a series of marathon phone calls with Stefanos, we were often able to digressfrom editorial issues and discuss physics, as if we were, once again, young stu-dents at a summer school in Crete, our whole life ahead of us and time on our side.

Needless to say, there are many people who helped us complete the project.First and foremost, we are indebted to Stefanos for being an inspiring teacher,a dear friend, and an extraordinary colleague. His contribution goes wellbeyond having authored the original book: He oversaw our translation, gave usvaluable feedback, and took the opportunity to add two new chapters, reviseextensively the rest of the text, redesign dozens of problems, and expand theonline supplement. We are fortunate to have Wiley-VCH as our publisher. We

xxviii Editors’ Note

are grateful to Valerie Moliere, formerly consultant senior commissioning editorat Wiley-VCH and currently at the Institution of Engineering and Technology,for her professionalism, support, and enthusiasm during the early stages ofthe book. Special thanks to: Nina Stadhaus, our project editor at Wiley-VCH,for a smooth collaboration and for her patience during the preparation of themanuscript; Claudia Nussbeck, for critical assistance in the design of the cover;and Sujisha Kunchi Parambathu, our production editor, for the skillful andefficient processing of our manuscript.

Manolis: I am indebted to Richard M. Osgood Jr., for hosting me at theDepartment of Applied Physics & Applied Mathematics at Columbia Universityfor the duration of this project, and for his constant encouragement and support.The Columbia community has provided me with invaluable access to peopleand resources (library access, online tools, etc.) that critically affected thebook. I am deeply grateful to Jerry Icban Dadap (Columbia University) forsubstantive feedback, encouragement, and his inspiring friendship. My editorialposition at the journals of the American Physical Society (APS) has aided myprofessional development in numerous ways since 2003, and I am grateful to mycolleagues—especially from Physical Review B and Physical Review Letters—fora productive collaboration throughout this time, and to the APS in general forthe privilege of working in this historic organization. My writing style owes alot to the influence of Fotis Kafatos, whom I had the honor to advise from 2008to 2010 in his capacity as President of the European Research Council. I amthankful to my math mentor Manolis Maragakis for guidance and for instillingin me a sense of urgency about this project. I must also thank Nader Engheta(University of Pennsylvania), Daniel Esteve (CEA-Saclay), Dimitrios Psaltis(University of Arizona), Francisco-Jose Garcia-Vidal (Universidad Autonoma deMadrid), Miles Blencowe (Dartmouth College), and Che Ting Chan (Hong KongUniversity of Science and Technology) for their encouragement. But my deepestgratitude goes to my family: my parents, Yannis and Chrysoula, for their love,support, and wise counsel; and, of course, my wife Katerina and our daughterNefeli, without whom none of this would be possible—they gave me the sweetestmotive to complete the work, while graciously accepting that I had to spend toomany evenings, weekends, and vacations away from them.

Leonidas: I wish to thank Sokrates Pantelides of Vanderbilt University for host-ing me as a research associate and research assistant professor in the period thatoverlapped with the first years of this project. But, my deepest thanks go to mywife Nektaria and our son Ioannis for their understanding and patience duringall these countless, multihour sessions I had to “borrow” from our “free” time.

While this translation project has been a labor of love for us, the ultimatejudges of our work are the readers, of course. We hope they will enjoy readingStefanos’s book as much as we have.

New York Manolis Antonoyiannakis(1) American Physical Society

(2) Columbia University

Athens, Greece Leonidas TsetserisNational Technical University of Athens

1

Part I

Fundamental Principles

3

1

The Principle of Wave–Particle Duality: An Overview

1.1 Introduction

In the year 1900, physics entered a period of deep crisis as a number of peculiarphenomena, for which no classical explanation was possible, began to appear oneafter the other, starting with the famous problem of blackbody radiation. By 1923,when the “dust had settled,” it became apparent that these peculiarities had acommon explanation. They revealed a novel fundamental principle of nature thatwas completely at odds with the framework of classical physics: the celebratedprinciple of wave–particle duality, which can be phrased as follows.

The principle of wave–particle duality: All physical entities have a dualcharacter; they are waves and particles at the same time. Everything we used toregard as being exclusively a wave has, at the same time, a corpuscular character,while everything we thought of as strictly a particle behaves also as a wave. Therelations between these two classically irreconcilable points of view—particleversus wave—are

E = h f , p = h𝜆, (1.1)

or, equivalently,

f = Eh, 𝜆 = h

p. (1.2)

In expressions (1.1) we start off with what we traditionally considered to be solelya wave—an electromagnetic (EM) wave, for example—and we associate its wavecharacteristics f and 𝜆 (frequency and wavelength) with the corpuscular charac-teristics E and p (energy and momentum) of the corresponding particle. Conversely,in expressions (1.2), we begin with what we once regarded as purely a particle—say,an electron—and we associate its corpuscular characteristics E and p with thewave characteristics f and 𝜆 of the corresponding wave. Planck’s constant h, whichprovides the link between these two aspects of all physical entities, is equal to

h = 6.62 × 10−27 erg s = 6.62 × 10−34 J s.

Actually, the aim here is not to retrace the historical process that led to this fun-damental discovery, but precisely the opposite: Taking wave–particle duality as

An Introduction to Quantum Physics: A First Course for Physicists, Chemists, Materials Scientists, and Engineers,First Edition. Stefanos Trachanas, Manolis Antonoyiannakis and Leonidas Tsetseris.© 2018 Wiley-VCH Verlag GmbH & Co. KGaA. Published 2018 by Wiley-VCH Verlag GmbH & Co. KGaA.

4 1 The Principle of Wave–Particle Duality: An Overview

granted, we aim to show how effortlessly the peculiar phenomena we mentionedearlier can be explained. Incidentally, these phenomena merit discussion notonly for their historical role in the discovery of a new physical principle butalso because of their continuing significance as fundamental quantum effects.Furthermore, we show that the principle of wave–particle duality should berecognized as the only sensible explanation to fundamental “mysteries” of theatomic world—such as the extraordinary stability of its structures (e.g., atomsand molecules) and the uniqueness of their form—and not as some whim ofnature, which we are supposed to accept merely as an empirical fact.

From its very name, it is clear that the principle of wave–particle duality can benaturally split in two partial principles: (i) the principle of wave–particle dualityof light and (ii) the principle of wave–particle duality of matter. We proceed toexamine both these principles, in relation to the peculiar phenomena and prob-lems that led to their postulation.

1.2 The Principle of Wave–Particle Duality of Light

According to the preceding discussion, the wave–particle duality says thatlight—which in classical physics is purely an EM wave—has also a corpuscularcharacter. The associated particle is the celebrated quantum of light, thephoton. The wavelike features f and 𝜆 of the corresponding EM wave, and theparticle-like features E and p of the associated particle, the photon, are relatedthrough expressions (1.1). We will now see how this principle can explain twokey physical phenomena—the photoelectric effect and the Compton effect—thatare completely inexplicable in the context of classical physics.

1.2.1 The Photoelectric Effect

With this term we refer today to the general effect of light-induced removalof electrons from physical systems where they are bound. Such systems canbe atoms and molecules—in which case we call the effect ionization—or ametal, in which case we have the standard photoelectric effect studied at theend of the nineteenth and the beginning of twentieth century. What makes theeffect peculiar from a classical perspective is the failure of classical physics toexplain the following empirical fact: The photoelectric effect (i.e., the removalof electrons) is possible only if the frequency f of the incident EM radiation isgreater than (or at least equal to) a value f0 that depends on the system fromwhich the removal occurs (atom, molecule, metal, etc.). We thus have

f ≥ f0. (1.3)

In classical physics, a “threshold condition” of the type (1.3) has no physicaljustification. Whatever the frequency of the incident EM wave, its electric fieldwill always produce work on the electrons, and when this exceeds the workfunction of the metal—the minimum energy required for extraction—electronswill be ejected from it. In other words, in classical physics, the frequency plays nocrucial role in the energy exchanges between light and matter, while the intensity

1.2 The Principle of Wave–Particle Duality of Light 5

of the electric field of light is the decisive factor. Clearly, the very existence ofa threshold frequency in the photoelectric effect leaves no room for a classicalexplanation. In contrast, the phenomenon is easily understood in quantummechanics. A light beam of frequency f is also a stream of photons with energy𝜖 = h f ; therefore, when quantized light—a “rain of light quanta”—impingeson a metal, only one of two things can happen: Since the light quantum is bydefinition indivisible, when it “encounters” an electron it will either be absorbedby it or “pass by” without interacting with it.1 In the first case (absorption), theoutcome depends on the relative size of 𝜖 = h f and the work function, W , ofthe metal. If the energy of the light quantum (i.e., the photon) is greater thanthe work function, the photoelectric effect occurs; if it is lower, there is no sucheffect. Therefore, the quantum nature of light points naturally to the existence ofa threshold frequency in the photoelectric effect, based on the condition

h f ≥ W ⇒ f ≥ Wh

= f0, (1.4)

which also determines the value of the threshold frequency f0 = W∕h. Forh f >W , the energy of the absorbed photon is only partially spent to extract theelectron, while the remainder is turned into kinetic energy K (= m𝑣2∕2) of theelectron. We thus have

h f = W + K = W + 12

m𝑣2, (1.5)

which is known as Einstein’s photoelectric equation. Written in the form

K = h f − W ( f ≥ f0), (1.6)

Equation (1.5) predicts a linear dependence of the photoelectrons’ kinetic energyon the light frequency f , as represented by the straight line in Figure 1.1.

Therefore, by measuring K for various values of f we can fully confirm—ordisprove—Einstein’s photoelectric equation and, concomitantly, the quantumnature of light, as manifested via the photoelectric effect. In addition, we candeduce the value of Planck’s constant from the slope of the experimental line.

The discussion becomes clearer if in the basic relation 𝜖 = h f = hc∕𝜆 weexpress energy in electron volts and length in angstroms—the “practical units”of the atomic world (1 Å = 10−10 m, 1 eV = 1.6 × 10−19 J = 1.6 × 10−12 erg). Theproduct hc, which has dimensions of energy times length (since h has dimensionsof energy times time), then takes the value hc = 12 400 eV Å, and the formula forthe energy of the photon is written as

𝜖(eV) = 12 400𝜆(Å)

≈ 12 000𝜆(Å)

. (1.7)

1 For completeness, let us also mention the possibility of scattering. Here, the photon “collides”with an electron, transfers to it part of its energy and momentum, and scatters in another directionas a photon of different frequency (i.e., a different photon). This is the Compton effect, which weexamine in the coming section. But let us note right away that Compton scattering has negligibleprobability to occur for low-energy photons like those used in the photoelectric effect.


0.5

4

3

2

1

1 1.5 2

f (1015 Hz)

f0 = 0.72 × 1015 Hz

W = 3 eV

f0

K (eV)

Figure 1.1 The kinetic energy K of electrons as a function of photon frequency f .The experimental curve is a straight line whose slope is equal to Planck’s constant.

f Light

Vacuum tube

V

AI

Figure 1.2 The standard experimental setup for studying the photoelectric effect. Thephotoelectric current occurs only when f > f0 and vanishes when f gets smaller than thethreshold frequency f0. The kinetic energy of the extracted electrons is measured by reversingthe polarity of the source up to a value V0—known as the cutoff potential—for which thephotoelectric current vanishes and we get K = eV0.

The last expression is often used in this book, since it gives simple numericalresults for typical wavelength values. For example, for a photon with𝜆 = 6000 Å—at about the middle of the visible spectrum—we have 𝜖 = 2 eV. Weremind the readers that the electron volt (eV) is defined as the kinetic energyattained by an electron when it is accelerated by a potential difference of 1 V.Figure 1.2 shows a typical setup for the experimental study of the photoelectriceffect. Indeed, Einstein’s photoelectric equation is validated by experiment,thus confirming directly that light is quantized, as predicted by the principle ofwave–particle duality of light.


Example 1.1 A beam of radiation of wavelength 𝜆 = 2000 Å impinges on ametal. If the work function of the metal is W = 2 eV, calculate: (i) the kineticenergy K and the speed 𝑣 of the photoelectrons, (ii) the cutoff potential V0.

Solution: If we set 𝜆 = 2000 Å in the relation 𝜖(eV) = 12 000∕𝜆(Å), we obtain𝜖 = 6 eV. So if we subtract the work function 2 eV, we obtain 4 eV for the kineticenergy of the outgoing electrons. The speed of the photoelectrons can then becalculated by the relation

K = 12

m𝑣2 = 12

mc2(𝑣

c

)2⇒

𝑣

c=√

2Kmc2

=√

2 × 4 eV12× 106 eV

= 4 × 10−3 ⇒𝑣

c= 4 × 10−3 ⇒ 𝑣 = 1.2 × 108 cm∕s.

Here we wrote 12m𝑣2 as 1

2mc2(𝑣∕c)2 in order to express mc2 in eV (mc2 = 0.5 MeV

for an electron) and the electronic speed as a fraction of the speed of light(which is useful in several ways: for example, it helps us assess the validity ofour nonrelativistic treatment of the problem). As for the cutoff potential V0, it isequal to V0 = 4 V, since K = 4 eV and K = e ⋅V0.

We should pause here to remark how much simpler and more transparent ourcalculations become when, instead of using the macroscopic units of one systemor another (cgs or SI), we use the “natural” units defined by the very phenomenawe study. For example, we use eV for energy, which also comes in handywhen we express the rest mass of particles in terms of their equivalent energyrather than in g or kg. In this spirit, it is worthwhile to memorize the numbersmec2 ≈ 0.5 MeV and mpc2 ≈ 1836 mec2 = 960 MeV ≈ 1 GeV for electrons andprotons, respectively. We will revisit the topic of units later (Section 1.2.3).

1.2.2 The Compton Effect

According to expressions (1.1), a photon carries energy 𝜖 = h f and momentump = h∕𝜆. And because it carries momentum, the photon can be regarded as aparticle in the full sense of the term. But how can we verify that a photon hasnot only energy but also momentum? Clearly, we need an experiment wherebyphotons collide with very light particles—we will shortly see why. We can thenapply the conservation laws of energy and momentum during the collision tocheck whether photons satisfy a relation of the type p = h∕𝜆.

Why do we need the target particles to be as light as possible—that is, electrons?It is well known that when small moving spheres collide with considerably largerstationary ones, they simply recoil with no significant change in their energy,while the large spheres stay practically still during the collision. Conversely, if thetarget spheres are also small (or even smaller than the projectile particles), thenupon collision they will move, taking some of the kinetic energy of the impingingspheres, which then scatter in various directions with lower kinetic energy.Therefore, if photons are particles in the full sense of the term, they will behaveas such when scattered by light particles, like the electrons of a material: They will


transfer part of their momentum and energy to the target electrons and end upwith lower energy than they had before the collision. In other words, we will have

𝜖′ = h f ′< 𝜖 = h f ⇒ f ′< f ⇒ 𝜆′ >𝜆, (1.8)

where the primes refer to the scattered photons. This shift of the wavelength togreater values when photons collide with electrons is known as the Comptoneffect. It was confirmed experimentally by Arthur H. Compton in 1923, whenan x-ray beam was scattered off by the electrons of a target material. Whywere x-rays used to study the effect? (Today we actually prefer 𝛾 rays for thispurpose.) Because x- (and 𝛾) rays have very short wavelength, the momentump = h∕𝜆 of the impinging photons is large enough to ensure large momentumand energy transfer to the practically stationary target electrons (whereby thescattered photons suffer a great loss of momentum and energy). In a Comptonexperiment we measure the wavelength 𝜆′ of the scattered photon as a functionof the scattering angle 𝜃 between the directions of the impinging and scatteredphoton. By applying the principles of energy and momentum conservation wecan calculate the dependence 𝜆′ = 𝜆′(𝜃) in a typical collision event such as theone depicted in Figure 1.3.

Indeed, if we use the conservation equations—see Example 1.2—to eliminatethe parameters E, p, and 𝜙 (which we do not observe in the experiment, as theypertain to the electron), we eventually obtain

Δ𝜆 = 𝜆′ − 𝜆 = hmc

(1 − cos 𝜃) = 𝜆C(1 − cos 𝜃), (1.9)

where

𝜆C = hmc

= 0.02427 Å ≈ 24 × 10−3 Å (1.10)

is the so-called Compton wavelength of the electron. It follows from (1.9) thatthe fractional shift in the wavelength, Δ𝜆∕𝜆, is on the order of 𝜆C∕𝜆, so it isconsiderable in size only when 𝜆 is comparable to or smaller than the Comptonwavelength. This condition is met in part for hard x-rays and in full for 𝛾 rays.

Compton’s experiment fully confirmed the prediction (1.9) and, concomi-tantly, the relation p = h∕𝜆 on which it was based. The wave–particle duality oflight is thus an indisputable experimental fact. Light—and, more generally, EMradiation—has a wavelike and a corpuscular nature at the same time.

f, λ

y

x

f′, λ′

E, p

θ

ϕ

Figure 1.3 A photon colliding witha stationary electron. The photonis scattered at an angle 𝜃 with awavelength 𝜆′ that is greater thanits initial wavelength 𝜆. Theelectron recoils at an angle 𝜙withenergy E and momentum p.


Example 1.2 In a Compton experiment the impinging photons have wave-length 𝜆 = 12 × 10−3Å = 𝜆C∕2 and some of them are detected at an angle of 60∘with respect to the direction of the incident beam. Calculate (i) the wavelength,momentum, and energy of the scattered photons and (ii) the momentum, energy,and scattering angle of the recoiling electrons. Express your results as a functionof the electron mass and fundamental physical constants.

Solution: For 𝜆 = 𝜆C∕2 and 𝜃 = 60∘ (⇒ cos 𝜃 = 1∕2), the formula Δ𝜆 = 𝜆′ − 𝜆 =𝜆C(1 − cos 𝜃) yields 𝜆′ = 𝜆C, which is twice the initial wavelength. The momen-tum and energy of the photon before and after scattering are

p𝛾 =h𝜆= h𝜆C∕2

= h(h∕mc)∕2

= 2mc, p′𝛾 =

h𝜆′

= h𝜆C

= mc

and

𝜖 = h f = hc𝜆

= hc𝜆C∕2

= 2mc2, 𝜖′ = hc𝜆′

= hc𝜆C

= mc2,

where the index “𝛾” in the momentum symbol p denotes the photon (in custom-ary reference to “𝛾 rays”) to disambiguate it from the symbol p of the electronicmomentum. We can now write the conservation laws of energy and momentumas follows:

• Conservation of energy

𝜖 + mc2 = 𝜖′ + E ⇒ 2mc2 + mc2 = mc2 + E ⇒ E = 2mc2.

• Conservation of momentum along the x-axis (Figure 1.3 with 𝜃 = 60∘)

p𝛾 + 0 = p′𝛾 cos 𝜃 + p cos𝜙 ⇒ 2mc + 0 = mc 1

2+ p cos𝜙

⇒ p cos𝜙 = 32

mc. (1)

• Conservation of momentum along the y-axis

0 + 0 = p′𝛾 sin 𝜃 − p sin𝜙 ⇒ 0 = mc

√3

2− p sin𝜙

⇒ p sin𝜙 = mc√

32. (2)

If we now take the square of (1) and (2) and add them, we get

p2 = 3m2c2 ⇒ p =√

3 mc

and, based on (1), we find that√

3 mc cos𝜙 = (3∕2)mc ⇒ cos𝜙 =√

3∕2 ⇒𝜙 = 30∘.

Now that p and E for the electron (p =√

3 mc, E = 2mc2) have been calculated,one may wonder whether they satisfy the relativistic energy–momentum relationE2 = c2p2 + m2c4.2 Indeed they do, as the readers can readily verify.

2 The use of relativistic formulas here is necessary because the speeds of the recoiling electrons areindeed relativistic.


1.2.3 A Note on Units

At this point we should pause to make some remarks on the system of units. Wehave already suggested (see Example 1.1) that both the cgs and SI system of unitsare equally unsuited for the atomic world, since it would be quite unreasonableto measure, for example, the energy in joules (SI) or erg (cgs). The natural scaleof energies in atoms is the electron volt, a unit that is 19 orders of magnitudesmaller than the joule and 12 orders smaller than the erg! Likewise, the naturallength unit in the atomic world is the angstrom (= 10−10 m), since it is the typicalsize of atoms. In this spirit, it is inconvenient to express, say, Planck’s constantin erg s or J s, and hc—another useful constant—in erg cm or J m; instead, it iseasier to use the corresponding practical units eV s for h and eV Å for hc. Thereis, however, one instance in atomic physics where we cannot avoid choosingone system over another: The basic force law governing atomic and molecularstructure—Coulomb’s law—has a much more convenient form in cgs than SIunits, namely,

F = 14𝜋𝜖0

q1q2

r2 (SI) F =q1q2

r2 (cgs), (1.11)

whence we immediately see why the cgs system is preferable over SI in atomicphysics. In SI units all mathematical expressions of the basic quantum resultsfor atoms appear much less elegant because they carry the cumbersome factor1∕4𝜋𝜖0. For example, in cgs units the quantum formula for the ionization energyof the hydrogen atom has the simple form WI = me4∕2ℏ2, while in SI units itbecomes WI = me4∕32𝜋2𝜖2

0ℏ2! Therefore, our choice is to go with the cgs system

for the mathematical expression of Coulomb’s law, but to make all calculations inthe practical units eV and Å, or even in the so-called atomic system of units, whichwe will introduce later. As you will soon find out, the practical unit of energy (i.e.,the eV) is much better suited than the joule, even for calculations concerningphysical quantities, like voltage or electric field intensities in atoms, where theSI units (V or V∕m) are certainly preferable. The reason is that the energy uniteV is directly related both to the fundamental unit of charge e and the SI unitof volt. A pertinent example was the calculation—without much effort!—of thecutoff potential in Example 1.1. The same holds true for electric field intensitiesin atoms, where the SI unit V∕m (or V∕cm) arises naturally from the energyunit eV.

So, even readers who are adherents of the SI system will find that the practicalenergy unit, eV, is much closer to the SI system than the joule itself.

As for the cgs system, we remind the readers that its basic units—length, mass,and time—are the centimeter (cm), the gram (g), and the second (s), while forderivative quantities such as force, energy, and charge, the cgs units are the dyn,the erg, and the esu-q (electrostatic unit of charge), respectively. These units arerelated to their SI counterparts as follows:

Quantity cgs SI ConversionForce dyn N (newton) 1 N = 105 dynEnergy erg J (joule) 1 J = 107 ergCharge esu-q C (coulomb) 1 C = 3 × 109 esu-q

1.3 The Principle of Wave–Particle Duality of Matter 11

Another advantage of the cgs system is that the charge q has mechanicalunits (due to the form of Coulomb’s law), while in the SI system charge is anindependent physical quantity whose unit is not related to the mechanical unitsof the system. Therefore, in the SI system, dimensional analysis—which we useextensively in this book—becomes unnecessarily cumbersome, since there arenow four fundamental quantities instead of three. Conversely, in the cgs system,the electric charge q—or rather, its square, via the relation F = q2∕r2—hasmechanical units, namely,

[q2] = [F ⋅ r2] = [F ⋅ r ⋅ r] = E ⋅ L ⇒ [q2] = erg cm.

Incidentally, another quantity with dimensions of energy times length, like q2, isthe product hc. The ratio q2∕hc is thus a dimensionless quantity, which we shallencounter later on in this chapter.

Problems

1.1 The ionization energy of the hydrogen atom is WI = 13.6 eV. Will there be aphotoelectric effect (i.e., ionization of the atom) if it is exposed to ultraviolet(UV) light of wavelength 480Å? What is the speed of the extracted electron?

1.2 Besides the threshold frequency, another remarkable feature of the pho-toelectric effect is the practically vanishing time between the incidence ofthe light beam on the photocathode and the extraction of electrons. Evenfor a very weak beam, photoelectrons are produced almost instantaneously(𝜏 < 10−9 s). To see how the classical theory fails here also, estimate thetime needed to extract an electron from an atom exposed to a light beamof the same intensity as, say, a light bulb of 100 W at a distance of 1 m.Treat the atom as a light collector that absorbs all EM energy incident onits cross-sectional area.

1.3 A photon of 𝜆 = 𝜆C impinges on a stationary electron and scatters at anangle of 180∘ (𝜃 = 𝜋). Calculate the momentum and energy of the electronafter the collision and confirm your results by performing an appropriatetest. What could this test be?

1.4 A photon of 𝜆 = 𝜆C∕2 is Compton scattered by an initially stationaryelectron. (a) Calculate (in terms of h,m, and c) the wavelength, momentum,and energy of the photon scattered at a 120∘ angle. (b) Calculate thescattering angle, momentum, and energy of the electron after the collision.

1.3 The Principle of Wave–Particle Duality of Matter

As emphasized in the introduction, relations (1.2) of the wave–particle duality ofmatter are similar to those of light—relations (1.1)—but they have to be viewedin reverse order. In case (1.2), we are talking about entities (e.g., electrons) we


used to recognize as particles in classical physics (so they are described by theirenergy E and momentum p), but we now learn they are also waves. Their wavefeatures f and 𝜆 are connected to the corpuscular attributes E and p via relations(1.2). The electron—the most fundamental particle of nonnuclear matter—is thusa particle and a wave at the same time. We were already aware of its corpuscularnature; after all, we first came across the electron as a particle. So we just need toexamine if it is also a wave with 𝜆 = h∕p, as Louis de Broglie first hypothesized in1923. Let us examine how we can infer the existence of these waves.

1.3.1 From Frequency Quantization in Classical Waves to EnergyQuantization in Matter Waves: The Most Important General Consequenceof Wave–Particle Duality of Matter

To experimentally verify the wavelike nature of electrons, the obvious test is tolook for interference phenomena between electronic waves, just as in classicalwaves. This would be a direct confirmation. But there is also an indirect confir-mation, invoking a characteristic feature of standing waves, namely, frequencyquantization. A standing classical wave—localized on a finite object—can onlyexist if its frequency takes a discrete sequence of values known as the eigenfre-quencies of the system. The most representative examples are the standing wavesof definite frequency—the so-called normal modes—on a string. As it followsfrom Figure 1.4, the allowed frequencies of the string’s vibrations—f = c∕𝜆,where c is the speed of wave propagation—are given by

L = n𝜆2⇒ 𝜆 = 2L

n⇒ f = c

𝜆= c

(2L∕n)= c

2L⋅ n, (1.12)

which means that the only possible vibrations of the string are those with integermultiples of the fundamental frequency f1 = c∕2L.

But if the frequency is quantized in classical systems, so too will be theparticle’s energy, since the wave–particle duality of particles—namely, therelation E = hf —provides a direct link between their energy and the frequencyof the corresponding wave. So if a quantum particle, say an electron, is trappedsomewhere in space (e.g., in an atom or a molecule), the associated de Brogliewave will be a standing wave with quantized frequency, and therefore the energyE = hf of the electron will also be quantized. As we will see shortly, energyquantization for particles that are trapped in some region of space (and thusperform confined motion) is the deepest consequence of the wave–particleduality of matter.

n = 1

L = λ/2

0 L

L = 3λ/2L = λ

n = 2 n = 3

Figure 1.4 Standing classical waves on a string. A standing wave of this kind can only beformed when an integer number of half-waves fit on the string. That is, L = n 𝜆

2(n = 1, 2,…).


1.3.2 The Problem of Atomic Stability under Collisions

We will now see that energy quantization of atomic electrons provides the onlyreasonable explanation for the mystery of atomic stability. Why is this a mystery?Because atoms remain completely unscathed, even though they continuouslyundergo violent collisions with each other. If we picture the electrons in atomsorbiting around the nucleus, like planets around a sun, then it is as if their orbitsdo not change at all upon innumerable collisions with other “solar systems.”But there is more to it. Even if we took apart an atom—by removing all itselectrons—and let it “reconstruct” itself, it would reemerge in identical formand shape. The evidence for these statements is that atoms always emit the samecharacteristic frequencies—the same spectrum—while their chemical behavioralso remains unaltered. In fact, chemical stability is an essential prerequisite forour very existence. Note, however, that in the discussion we only consideredatomic stability against collisions. We have completely ignored the stability ofatoms against the radiation emitted by their electrons, which, being chargedparticles in accelerated motion, ought to radiate and lose energy until they fallinto the nucleus. We discuss this problem in Section 1.3.6. Until then, let ussimply accept that, for some reason, the classical laws of EM radiation do nothold in the atomic world. Let us then see how the problem of atomic stabilityagainst collisions can be explained naturally by assuming that the energy ofatomic electrons is quantized. In the hydrogen atom, for example, if the electron,which has quantized energy, occupies the lowest possible state—the so-calledground state—then the smallest possible change for the atom is a transfer of theelectron from the ground state to the next available state, namely, the first excitedstate. In other words, the electron can only make a discontinuous transition—aquantum jump (or leap)—toward an excited atomic state. Now, if the environ-ment offers less energy to the atom than the energy required for such a leap, asis the case for thermal collisions at room temperature, then no transition canoccur at all. Indeed, the energy difference between the ground and first excitedstates of any atom—or molecule—is on the order of a few eV, while the averagethermal energy at room temperature is about a 100th of this value. As a result,thermal collisions at room temperature do not provide sufficient energy to excitethe atoms, which thus behave as stable and invariant entities during collisions.We have to reject the classically allowed small, gradual changes in energy, andconsider only those quantum jumps for which the minimum required energy isavailable. Hence, energy quantization arises as the only conceivable explanationof the mystery of atomic stability. The “equation”

Quantization = stability

emerges thus as the fundamental conceptual equation of quantum physics.And since the only natural mechanism of quantization we are aware of involvesstanding waves, the following reasoning also applies:

Stability → quantization → wavelike behavior.

This explanation of the central mystery of the atomic world—the remarkablestability of its structures—demonstrates that the notion of wavelike behavior


for particles is not so “crazy” after all. In hindsight, we can regard it as the onlynatural explanation of the most fundamental problem put forward by the studyof matter at the atomic level.

Example 1.3 To appreciate the extreme conditions under which atoms man-age to retain their structural stability, calculate the typical frequency of collisionsbetween air molecules. Treat air particles as spheres of 1 Å diameter and assumean approximate value for the particle density n of air equal to 1020molecules∕cm3,which is about 1000 times less than the density of solid matter.

Solution: Our approach is to first estimate the mean free path 𝓁 of airmolecules—that is, the mean path traversed by a particle between collisions—andthen divide it by the mean thermal speed 𝑣 to obtain the average time 𝜏 betweentwo consecutive collisions. The frequency of collisions will then be equal to 1∕𝜏 .We can easily calculate the mean free path if we realize that a molecule collideswith another when it travels far enough to cover the whole volume available toit. This volume is equal to V∕N = 1∕(N∕V ) = 1∕n, where n is the number ofparticles per cm3. Therefore, the quantities 𝓁 (mean free path), 𝜎 (cross sectionof molecules), and n (particle density) are all related via the expression

𝓁 ⋅ 𝜎 = volume covered by a molecule that travels a distance𝓁= volume of the space available per molecule = 1∕n

⇒ 𝓁 = 1n𝜎. (1)

For n ≈ 1020 cm−3 and 𝜎 ≈ (10−8 cm)2 = 10−16 cm2, we find

𝓁 ≈ 10−4 cm. (2)

We can also obtain the mean thermal speed 𝑣—actually, the root-mean-square(rms) speed—of the air molecules (of mass M) as follows:

12

M𝑣2 = 32

kT ⇒ Mc2(𝑣

c

)2= 3kT ⇒

𝑣

c=√

3kTMc2

⇒𝑣

c≈

√

3 ⋅ 140

eV

30 × 109 eV⇒ 𝑣 ≈ 105 cm∕s. (3)

Here, we expressed the kinetic energy as 12Mc2(𝑣∕c)2 in order to use familiar

numbers such as the proton’s rest energy mpc2 ≈ 2000 ⋅ mec2 ≈ 2000 ⋅ 0.5 MeV ≈1 GeV = 109 eV. As a typical molecule of air we take the nitrogen molecule N2with mass 28 (≈ 30) times that of a proton. For kT at room temperature weused the rounded value kT ≈ 1∕40 eV, which results from (kT)T≈12 000 K ≈ 1 eV.Naturally, we have rounded the numbers significantly, as we are only making anorder-of-magnitude estimate.

From expressions (2) and (3), we obtain the time between collisions 𝜏 = 𝓁∕𝑣 ≈10−9 s, so the frequency of collisions is f = 1∕𝜏 ≈ 109 s−1. Each air molecule thusundergoes approximately one billion collisions with other molecules per second.And yet it remains intact. Surely, molecules are very robust structures!


1.3.3 The Problem of Energy Scales: Why Are Atomic Energies on the Orderof eV, While Nuclear Energies Are on the Order of MeV?

The main idea of the previous discussion—namely, that microscopic particles inconfined motion inside a structure (such as an atom or a molecule) are repre-sented by standing matter waves—helps us understand another central mysteryof the atomic world: The smaller the region inside which a particle resides, thegreater the energy of that particle. The most typical examples of this mystery arethe atom and the nucleus. Atomic electrons (of the outer shell, for heavy atoms)have energies on the order of a few eV, while the corresponding energies forprotons and neutrons inside the nucleus are one million times greater—that is,on the order of a few MeV! Again, the explanation lies in the wave–particle dual-ity expression 𝜆 = h∕p and the realization that the first (fundamental) standingwave in a region of space—recall the example of the string—has a wavelength𝜆 on the order of the linear size of the region. The wavelengths of the higherstanding waves are even smaller. So we can say that the largest wavelength—theone that corresponds to the ground state—will be about the size of

𝜆max ≈ 2L,where L is the linear size of the region within which the standing wave is formed.In this case, the relation 𝜆 = h∕p ⇒ p = h∕𝜆 shows that the momentum of thetrapped wave–particle cannot be smaller than

pmin = h𝜆max

≈ h2L.

And, if we are interested in the state of lowest energy—which is certainly themost important state—then p ≈ pmin, and the formula

p ≈ h2L

provides a good estimate of the momenta of particles trapped inside a quantumsystem of linear dimension L. For the corresponding kinetic energy, p2∕2m, ofthese particles, we have

K ≈ h2

8mL2 . (1.13)

The conclusion is now clear: The smaller the region inside which a particle ismoving, the smaller the wavelength (in the first standing wave) of that particleand, consequently, the greater its momentum and energy. Figure 1.5 should helpvisualize the physics of this key fact.

If we now apply formula (1.13) on a nucleon—where m = mN =mass of aproton or neutron and L ≈ 2R (R =nuclear radius)—or an electron of an outeratomic shell—where m = me and L = 2a (a =atomic radius)—we obtain

KN = h2

32mNR2 = h2

32mea2a2

R2

me

mN= Ke

(aR

)2 me

mN. (1.14)

Given now that a ≈ 1 Å ≈ 10−10 m, R ≈ 10−15 m, and mN ≈ 1836 me, expression(1.14) yields

KN ≈ (106 − 107)Ke, (1.15)


L

λ/2

2= L

λ

λ = 2L

2L

h

h2

8mL2

p ≈

K ≈

Figure 1.5 A standing matter wave ofspherical shape. A particle trappedinside a bounded region—a sphericalvolume in our case—of lineardimension L, is described (in the stateof lowest energy) by a sphericalstanding wave that vanishes only atthe boundary of this region. For itswavelength we thus have𝜆∕2 ≈ L ⇒ 𝜆 ≈ 2L.

which tells us that the kinetic energies KN of protons and neutrons inside thenucleus are a few million times greater than the kinetic energies Ke of theouter-shell electrons in atoms. (Inner-shell electrons have greater energies thanelectrons in the outer shells, since they are moving in a smaller region of space.)

Furthermore, we can use the formula Ke = h2∕32mea2 to obtain a typical valuefor the kinetic energy of the outer electrons

Ke ≈ a few eV. (1.16)

We combine Eq. (1.15) with Eq. (1.16) to obtain

KN ≈ a few MeV. (1.17)

If we now take the next logical step, namely, that the energies released inchemical and nuclear reactions should be on the same order of magnitude as theenergies of outer-shell atomic electrons and nucleons, respectively, then we candeduce another fundamental feature of our world: Energies released in chemicalreactions can only be on the order of an eV, while energies released in nuclearreactions must be on the order of an MeV per reaction. We can thus say that eVand MeV define the chemical and nuclear energy scales, respectively.

We can now reexamine the problem of atomic stability. If the energy scaleof electrons in atoms—in the hydrogen atom, for simplicity—is on the orderof an eV, then differences between adjacent energy levels (remember, they arequantized) should be of the same order, that is, a few eV. Note, for example, thatthe first excited state of the atom will correspond to a standing matter wave with𝜆 = L (one-half that of the ground state), so the electronic momentum p = h∕𝜆will double and the kinetic energy will quadruple compared to the ground state.(Provided, of course, that all standing waves of the atom occur within the samevolume in space, which is not exactly true; in its excited states the atom is bigger.)The energy difference between the ground and first excited states of an atom,such as hydrogen, would thus also be on the order of an eV; this energy differencedetermines the atom’s stability against collisions, as we noted earlier. We remindthe readers that the factor kT that determines the average magnitude of thermalenergies at temperature T via the relation

K = 32

kT (1.18)


takes (at room temperature) the approximate value

kT|T≈300 K ≈ 140

eV. (1.19)

We can thus see that thermal collisions at room temperature—but also atmuch higher temperatures, say, a few thousand degrees—cannot cause atomicexcitations. Atoms emerge from their incessant collisions—roughly one billioncollisions per second, as we saw—completely intact. In reality, not all atoms ofa gas have the same thermal kinetic energy—(1.18) is merely a mean value—butobey a Maxwell–Boltzmann distribution, so some of them are much moreenergetic than others and able to cause mutual excitations when they collidewith each other. So the exact picture is this: Even at room temperature, a smallfraction of atoms in a gas are excited, but the overwhelming majority remainsintact in their ground state.

In the case of an atomic nucleus, where the energy difference between theground and first excited states is on the order of an MeV, a similar reasoningleads us to conclude that nuclear stability against collisions is a million timesgreater than atomic stability. The critical temperature for the stability of anucleus is thus a few billion degrees kelvin, compared to a few thousand degreesfor an atom. Therefore, for thermonuclear reactions to occur, as in the interiorof a star, the temperature needs to rise to billions of degrees! And yet ther-monuclear reactions inside stars occur—for without such reactions, we wouldnot exist—even though the typical temperature in their interior is no greaterthan 10–20 million degrees! The resolution of this mystery has a quantumorigin also and is discussed in the online supplement of Chapter 5.

1.3.4 The Stability of Atoms and Molecules Against ExternalElectromagnetic Radiation

There are two types of external “perturbations” that atoms and molecules areoften subjected to, and which could threaten their structural stability. The firstperturbation is thermal collisions—actually, electric forces between electrons ofapproaching atoms—which we have already examined. The second type of per-turbation is the ubiquitous electromagnetic radiation—visible light, infrared (IR),UV, x-rays, radio waves, and so on—that hits atoms continuously. Does EM radi-ation change the structure of atoms? If the atoms were classical systems, thenthe answer would surely be in the affirmative, since they would have to “respond”to any external perturbation, however small, by changing their structure accord-ingly; for example, by slightly adjusting their electronic orbits. However, atomsare not classical but quantum systems and therefore their states have quantizedenergies that can only change via specific quantum jumps. In other words, atomscannot absorb an arbitrary amount of energy, but only the amount required for atransition from the ground state—if this is where they start from—to any one oftheir excited states. Now, due to the wave–particle duality of light, the incidentEM radiation on an atom is also quantized with an energy quantum equal to

𝜖 = hf = hc𝜆

⇒ 𝜖(eV) ≈ 12 000𝜆(Å)

. (1.20)


For example, for visible light, where

4000 Å < 𝜆 < 7400 Å (visible light), (1.21)

the energies of optical photons span the range

1.6 eV < 𝜖 < 3 eV (visible light) (1.22)

with a typical value—for 𝜆 ≈ 6000 Å—equal to 2 eV. Thus, UV photons—beingmore energetic, and hence more chemically potent, than optical photons—haveenergies greater than 3 eV, while IR photons have energies less than1.6 eV. In other words, UV- and IR-light photon energies lie to the rightand left, respectively, of the visible range (1.22). For radiowaves—wheref ≈ 100 MHz ⇒ 𝜆 = c∕f = 3 m = 3 × 1010 Å—we have

𝜖 (radiowaves) = 12 0003 × 1010 eV = 0.4 × 10−6 eV ≈ 1μeV.

What happens when one of the aforementioned kinds of radiation impinges onan atom? Take, for example, the hydrogen atom, for which the first excitationenergy—equal to the energy difference between its first excited and groundstates—is 10.2 eV. Clearly, any radiation whose photons have energies lessthan 10.2 eV cannot induce any changes to the hydrogen atom. The photonsof the impinging radiation “bounce off” the atom to another direction; theyare scattered, as we say, leaving the atom intact. Hence, we can conclude thatenergy quantization of atomic electrons—and the corresponding energy scale onthe order of an eV—combined with light quantization, ensures atomic stabilityagainst not only collisions but also all of EM radiation with energy below theUV: visible, IR, microwaves, radio waves, and so on. No matter how long atomsor molecules are bombarded by such radiation—provided its intensity is not toohigh—they remain completely unaffected. Similarly, radiation at such frequenciescannot cause chemical reactions. The reason is that for chemical reactions thereis also an energy threshold, a minimum energy barrier the light quantum has tosurpass for a reaction to occur. And just like the typical energies in atoms and(small) molecules are on the order of a few eV, this threshold energy is also onthe order of a few eV—typically greater than 3 eV. So the only kinds of radiationthat are chemically potent are those in the UV range and beyond (x-rays,𝛾 rays, etc.). This means, among other things, that visible light is not chemicallydangerous—for if it were, we would not be here (since our planet is awashwith it)!

We thus come to the realization that the crucial feature of the photoelectriceffect—the existence of a threshold frequency (and energy) for the phenomenonto occur—is completely general: It holds for chemical reactions, excitations,dissociations of molecules, and so on. As a consequence, all radiation withphoton energies below the energy threshold is “harmless,” in the sense thatit cannot cause the abovementioned effects. Given also that all “threshold”energies are on the order of a few eV, atoms and molecules are completely“safe” against all radiation from the visible range and below (in energy). (Visiblelight can actually cause some reactions, but these belong to a very specificcategory.)


1.3.5 The Problem of Length Scales: Why Are Atomic Sizes on the Orderof Angstroms, While Nuclear Sizes Are on the Order of Fermis?

So far, we considered the problem of energy scales in the microscopic world. Wewondered if there is a simple explanation to a seemingly paradoxical feature ofnature: The energies trapped in a nucleus—in spite of its minuscule size—are amillion times greater than in an atom, even though the latter is roughly 100 000times larger. We saw that a nucleus is an energy giant precisely because it is sosmall. Let us recall the reason for this: Since the “fundamental” standing wave ina region of space of linear size L has a wavelength on the order of L (𝜆 ∼ 2L), themomentum p = h∕𝜆 of the particle trapped there will be inversely proportionalto L and will thus increase as the region shrinks in size. So the fact that thenucleus is an energy giant should no longer surprise us, but should instead beviewed as a direct consequence of its size in conjunction with the principle ofwave–particle duality. The tinier a structure, the more energetic the particles thatlie inside it. But to determine the specific energy scales—eV in atoms and MeVin nuclei—we also have to know the length scales of these structures, namely,that they are on the order of an angstrom (= 10−10 m) for atoms, and a fermi(= 10−15 m) for nuclei. We took these characteristic lengths as given. So our nextquestion is: Can we explain the characteristic length scales of atoms and nuclei?Why should the size of atoms, for example, be on the order of an angstrom andnot much smaller or much larger? Let us try to answer this question, startingwith the self-evident idea that the ground state of an atom—which is essentiallywhat we are after—has to be such that the total energy (kinetic + potential) isminimized. We will take the hydrogen atom as a representative example andthink first of the qualitative mechanism that possibly determines its size. Thereare two energy terms: potential and kinetic. The former favors a short distancebetween the electron and the nucleus, ideally with the electron right on thenucleus and at rest. In that case, the total classical energy3

E = K + V = 12

m𝑣2 − e2

r(1.23)

tends to minus infinity for r = 0 and is thus clearly minimized. But the possibilityof an electron at rest on the nucleus exists only in classical physics. In the contextof the wave–particle duality of matter, it is not possible to have a particle at restat a specific point. In fact, the exact opposite is true: If the electron is “squeezed”in such a tiny region like that of the nucleus, or even smaller, it will develop ahuge momentum—on the order of h∕4R, where R is the nuclear radius—anda corresponding kinetic energy K = h2∕32mR2 that will hurl it away from thenucleus. We thus see that the notion of an electron confined in the vicinity ofthe nucleus does not minimize the total energy of the atom. While the potentialenergy is then minimized, the kinetic energy grows with no bound. To obtainthe “correct” size of the atom, we need to find its radius a for which the totalenergy is minimized. For the kinetic energy, we use the quantum expression

3 Recall that we are using the cgs system precisely because the expression for the electrostaticpotential energy, V = q1q2∕r, does not contain the cumbersome factor 1∕4𝜋𝜖0. Note also that hereq1 = proton charge = e and q2 = electron charge = −e.


E(a)

E(a) ≈ h2 e2

32ma2 a

a0 a

Figure 1.6 Total energy of the hydrogen atom as a function of its size, a. The real radius of theatom is the one that minimizes its total energy.

K = h2∕32ma2, while for the potential energy we can use the approximationV ≈ −e2∕a, even though the electron is not located at this exact distance a,since it is now a wave that extends throughout the spherical volume of radiusa. (This is the so-called probability cloud as we will shortly see.) But for crude,order-of-magnitude estimates we can still use the (approximate) expression ofthe total energy

E ≈ h2

32ma2 − e2

a. (1.24)

Figure 1.6 shows the total energy of the atom as a function of its possible size a.The function has a minimum—obtained from the condition dE∕da = 0—at

a0 = h2

16me2 ≈ 1 Å, (1.25)

which indeed corresponds to the correct order of magnitude of atomic radii.The general conclusion about the mechanism that determines the characteristic

length scales of various atomic-scale structures is now clear. No matter howstrong the mutual attraction between particles that form a microscopic structure,it will never be able to compress them to an infinitesimally small volume, because,in that case, the particles would develop an exceedingly large kinetic energy(due to their extreme localization) that would offset the energy gained by thereduction of their potential energy. The minimum total energy is thus achievedat an optimum size, which is determined by the balance between the attractivepotential energy term (that pulls particles together) and the repulsive kineticenergy term (that resists their extreme localization). In this regard, the muchsmaller size of the nucleus compared to the atom must be accounted for by themuch stronger nuclear forces exerted between the nucleons, and also by theirbigger masses that weaken their resistance to localization. In fact, if we considerthat the strength of nuclear forces—as measured by the relevant couplingconstant g, which is the analog of e in (1.25)—is about 100 times greater than


that of the EM force,4 and also that the nucleons’ mass is about 2000 timesgreater than an electron’s mass, then formula (1.25), applied to the nucleus, givesa nuclear radius five orders of magnitude smaller than the atomic radius. This isin full agreement with observation.

Let us also add here that all atoms have roughly the same size—on the orderof an Å—because their outer electrons are subject to (approximately) the sameelectric attraction from the nucleus as the single electron in the hydrogen atom.The reason is that inner electrons screen a large fraction of the nuclear chargefrom the outer electrons. In other words, from the vantage point of the outerelectrons, heavier atoms resemble hydrogen and must therefore have roughly thesame size.

1.3.6 The Stability of Atoms Against Their Own Radiation: ProbabilisticInterpretation of Matter Waves

In 1911, the Rutherford experiment showed that the atom consists of a tinynucleus with the electrons orbiting far away, like planets around the sun. No clas-sical model could explain how such an atom may last more than a few tenths ofa nanosecond! Whatever the classical orbit of the electrons, their motion wouldsurely be an accelerated one (with linear or centripetal acceleration). As a result,electrons would emit EM radiation continuously, lose energy, and ultimatelyfall—in an infinitesimally small amount of time—on the nucleus. Conclusion:A truly classical atom cannot exist. But can the quantum atom—based on theprinciple of wave–particle duality of electrons—also solve the mystery of theatoms’ stability against their own radiation, as it solved the two previous mysteries(stability against collisions and stability against external radiation)? Here theanswer is not a resounding yes, as it was for the other two questions. At this point,the mystery of the stability of atoms against their own radiation cannot be solveddirectly, because the quantum theory has not been “set up” yet. Nevertheless, thisproblem can be at least bypassed with the following reasoning: If the principleof wave–particle duality is correct, then the orbital motion of electrons (whichis where radiation comes from) has no physical meaning. Let us elaborate:Orbital motion means that the electron at any given time is found at a specificlocation (i.e., localized in space), whereas the very concept of a wave postulatesa physical entity that is spread out in space. Moreover, a particle is by definition“indivisible”—within certain limits—while a wave can always be divided, forexample, by letting a part of it be transmitted through one slit and another partthrough another. A wave is thus always extended and divisible; a particle is alwayslocalized and indivisible. At this point, the reader would be justified to ask: Howcan the principle of wave–particle duality of matter then be true? How can we saythat a particle is at the same time a wave? How can we fit within the same physicalentity two mutually exclusive properties, such as “localized and indivisible” onthe one hand, and “extended and divisible” on the other? We have just arrived atthe most critical question of quantum theory—a question that, as we see in the

4 Measured in dimensionless units, the strength of (strong) nuclear interactions, 𝛼s = g2∕ℏc, is oforder 1, while the corresponding parameter 𝛼 = e2∕ℏc for electromagnetic forces has the knownvalue of 1∕137.


next chapter, leads to the celebrated statistical (or probabilistic) interpretationof matter waves. Here is what this interpretation says (Max Born, 1927):

The function 𝜓 = 𝜓(r) that describes a matter wave (its so-called wave-function) does not represent a measurable physical quantity. It is rathera mathematical wave—a probability wave—whose squared amplitude|𝜓(r)|2 yields the probability per unit volume to locate the particle in thevicinity of an arbitrary point r.

We thus have

P(r) = |𝜓(r)|2, (1.26)

where P(r) is the probability per unit volume—the probability density—oflocating the particle in the vicinity of an arbitrary point in space. The totalprobability of finding the particle anywhere in space is given by the integral overall space

∫ |𝜓(r)|2 dV = 1, (1.27)

which clearly equals unity.Given this interpretation, the wavefunction 𝜓 has no immediate physical

meaning—as it does not represent some sort of a physical wave—so it can takecomplex values in general. This is why absolute values are necessary in (1.26)or (1.27). According to (1.26), quantum particles frequent locations where theirwave is strong—“stormy” areas—and avoid “calm” places where their waveis weak. In the context of such an interpretation, the contradiction betweenparticles and waves is removed at once, since the particle need not cease being aparticle and does not have to physically “disperse” throughout the volume of thewave. The wave simply describes the probability of detecting the particle here orthere, but never here and there at the same time. When we do locate the particle,our detectors always record an integral and indivisible entity. No experimenthas ever “captured” half an electron or a quarter of a proton. To give readers anidea of how we describe quantum particles, we depict in Figure 1.7 two simpleexamples of one-dimensional wavefunctions.

This short detour in our discussion helped us arrive at the following basicconclusion: The correct interpretation of the principle of wave–particle dualitystrips the concept of electronic orbits in atoms of any physical meaning. Asa result, it makes no sense to speak of accelerated motion of electrons, nor,therefore, of emission of radiation from them. In other words, we do not havea solution to the problem mentioned earlier—but we do not have a problemeither! However, a new question pops up naturally at this point: How do atomsradiate after all? We address this question in the following section.

1.3.7 How Do Atoms Radiate after All? Quantum Jumps from Higherto Lower Energy States and Atomic Spectra

Let us make some “impromptu” thoughts on this question, using again thehydrogen atom as an example. Like any standing wave, a standing electron


x = 0

(a) (b)

x = 0

Figure 1.7 Typical one-dimensional wavefunctions. (a) An extended wavefunction: Theposition of the particle is known with very low precision. There is a significant probability oflocating the particle in regions away from the “most frequented” location at x = 0.(b) A localized wavefunction: The position of the particle is known with very high precision.In the vast majority of the measurements, we would detect the particle in the immediatevicinity of x = 0.

wave around the nucleus can exist in a number of possible forms—the so-callednormal modes. The first form corresponds to the state of lowest energy and thenext ones correspond to excited atomic states. The corresponding energies arequantized according to some discrete sequence E1,E2,… ,En,… Since thesesuccessive standing waves around the nucleus represent the only possible energystates of the electron, we have the following two scenarios:(a) If the electron is in the ground state, then it obviously cannot radiate; for if

it did, then it would lose energy and would have to move to a lower energystate, which, however, does not exist.

(b) If the electron is in an excited state—say, the first excited state—it can bede-excited,5 but only according to the basic quantum rules described earlier.First of all, a gradual de-excitation is impossible because the electron wouldthen be able to gradually shed its excess energy in the form of radiation andtransit to states with gradually decreasing energy, which, again, do not exist.The only available state to go to is the ground state, which, however, is located(in the hydrogen atom) 10.2 eV below the first excited state. So what can theexcited electron do to “shed” its excess energy and return to the ground state?Very simply, a quantum jump: an abrupt transition from the excited to itsground state via emission of the energy difference 10.2 eV in the form of aUV photon.

Atoms, therefore—and, likewise, molecules, and all other quantumsystems—emit light only when they undergo a transition from one of theirexcited states to a lower state. When this happens, a photon is emitted withenergy hf , equal to the energy difference between the initial and final states ofthe transition. We thus have

En − Em = h fnm (n > m), (1.28)

5 As we shall later see (e.g., Chapters 9 and 16), excited states are always unstable and getde-excited by emission of electromagnetic radiation from the atom.


where En (n > 1) is the energy of the initial excited state of the atom and Em is theenergy of the final state (which may or may not be its ground state).

The frequencies fnm are what we observe in the so-called line emission orabsorption spectrum of a gas made of the atoms or molecules we wish tostudy. We thus realize that the quantization of electronic energies in atoms ormolecules is reflected in the line spectra of the corresponding substances ingas form. In turn, these spectra are our best “tool” for measuring the allowedenergies in a quantum system.

The frequencies fnm that correspond to the transitions n → m are known asBohr frequencies. Theoretical physicists, however, prefer to use the same term forthe corresponding angular frequencies, 𝜔nm = 2𝜋 fnm, because 𝜔 is better suitedthan f (= 𝜔∕2𝜋) for the mathematical description of harmonic oscillationsor waves. Note, for example, that the mathematical expression of a harmonicoscillation x(t) = A sin(2𝜋t∕T)—where T is the period—takes the much simplerform x(t) = A sin𝜔t if we introduce the angular frequency 𝜔, via the relation

𝜔 = 2𝜋T

= 2𝜋 f . (1.29)

In the same spirit, theoretical physicists prefer to write the fundamental relation𝜖 = h f in the equivalent form

𝜖 = h f = h 𝜔2𝜋

= h2𝜋𝜔 = ℏ𝜔,

where

ℏ = h2𝜋

(1.30)

is the so-called reduced Planck’s constant. As we will see later, the mathematicalexpressions of basic quantum results are considerably simplified when writtenin terms of ℏ instead of h. Thus, the use of ℏ instead of h is now common inquantum physics, while one can always revert to the older symbol wheneverthere is a need to use quantities closer to what is experimentally measured, suchas the frequency f , or the wavelength 𝜆.

Having thus opted to use ℏ over h, we can rewrite the second expression—p =h∕𝜆—of the wave–particle duality as

p = h𝜆= 2𝜋ℏ

𝜆= ℏ

2𝜋𝜆

= ℏk, (1.31)

where

k = 2𝜋𝜆

(1.32)

is the so-called wavenumber of the wave. Clearly, k is the spatial equivalentof 𝜔, with 𝜆 in place of T , as we should have expected, since 𝜆 is the spatialand T is the temporal period of a sinusoidal wave. The modern version of thewave–particle duality is thus written as

E = ℏ𝜔, p = ℏk, (1.33)

which is clearly more elegant than the older form.


1.3.8 Quantized Energies and Atomic Spectra: The Case of Hydrogen

With hydrogen being the simplest and most abundant element, it is no accidentthat the spectrum of its atoms in gaseous form has been studied extensivelyin the visible and its neighboring regions (IR and UV) of the EM spectrum.A remarkable result of those studies is the famous Balmer’s formula for theemitted frequencies

fnm = R( 1

m2 − 1n2

)

, (1.34)

where n and m are positive integers (n > m) and R = 3.27 × 1015 s−1 is theso-called Rydberg constant, which has dimensions of frequency.

A comparison of formulas (1.28) and (1.34) gives the following expression forthe quantized energies of the hydrogen atom:

En = −hRn2 = −13.6

n2 eV, (1.35)

where the negative sign appears because we are talking about bound states. Here,the number 13.6 eV is simply the numerical value of hR expressed in units of eV.

According to (1.35), the ground-state energy of the atom is E1 = −13.6 eV. Itsopposite, WI = 13.6 eV, is the ionization energy, as confirmed by chemical data.

It also follows from (1.35) that the energy of the first excited state of the atom isE2 = −3.4 eV, and, therefore, the first excitation energy E2 − E1 is indeed equal tothe value 10.2 eV we already mentioned. This unusually large value explains theremarkable stability of the hydrogen atom against external influences (thermalcollisions or EM radiation).

The conclusion from the preceding discussion should be clear. Spectral data forthe hydrogen atom fully confirm all the general atomic properties we identifiedearlier based solely on the principle of wave–particle duality for electrons. Theelectronic energies in the atom are indeed quantized—they only take the discretevalues of (1.35)—and the energy scale is indeed on the order of a few eV, as wepredicted. And given that the energy scale stems from the length scale, the sizeof the hydrogen atom must be on the order of an Å. The same holds true for thesize of all heavier atoms, for reasons we already mentioned (Section 1.3.5).

Based on this, we also need to introduce a suitable terminology to reflect thefundamental role in quantum physics of the allowed—or quantized—energies ofa quantum system. Thus, we now speak of energy levels of the system and depictthem in the so-called energy-level diagram, as in Figure 1.8.

1.3.9 Correct and Incorrect Pictures for the Motion of Electrons in Atoms:Revisiting the Case of Hydrogen

After our discussion so far, how can we picture the hydrogen atom, at least inits ground state? Which description would be consistent with the wave natureof the electron and at the same time devoid of concepts with no experimentalmeaning, like the electronic orbit? The answer is plain to see. The picture weseek should be the analog of a classical standing wave with a similar geometry;for example, a classical sound wave in the interior of a hollow sphere filled withair. The simplest form of such a wave—corresponding to the “fundamental”


n = 1

n = 2

E1 = –13.6 eV

E2 = –3.4 eV

E3 = –1.5 eV

Min

imu

m e

xcita

tio

n e

ne

rgy

Ion

iza

tio

n e

ne

rgy

ΔE =

10

.2 e

V

WI =

13

.6 e

V

E4 = –0.85 eV

n = 3

n = 4

Figure 1.8 The energy-level diagram for the hydrogen atom and the two basic quantitiesassociated with it. The ionization energy WI is the minimum energy needed to remove theelectron from the atom. The minimum excitation energy ΔE is the energy required to affectthe smallest possible change to the atom in its ground state.

frequency—is a spherically symmetric compression (high pressure) that period-ically becomes a rarefaction (low pressure), while retaining its spherical shape.Such a wave has a time dependence of the form

p(r, t) = p(r) cos𝜔t, (1.36)

where the pressure p is measured with respect to atmospheric pressure—thatis, a positive sign refers to higher-than-atmospheric and a negative sign tolower-than-atmospheric pressure.

Note that, just like in the case of a string, the following is also true for thenormal oscillation modes of two- or three-dimensional objects: The shape of thefundamental oscillation has no nodes (i.e., no nodal lines for two-dimensionalobjects and no nodal surfaces for three-dimensional ones), while it also hasthe full symmetry of the problem. In a string, the fundamental oscillationis symmetric with respect to its midpoint; in a two-dimensional object withcircular geometry, such as a drum, it is rotationally symmetric with respect toits center; and likewise for three-dimensional objects with spherical symmetry.It follows that the “fundamental” standing electron wave around the nucleus isalso spherically symmetric with no nodal surfaces.

The answer, therefore, to the question “how are we to picture the hydrogenatom in its ground state,” is depicted in Figure 1.9: a spherically symmetriccloud of probability density that engulfs the nucleus. Such a cloud roughly


Figure 1.9 The correct quantum picture for the ground state of thehydrogen atom. The wave nature of the electron is incompatiblewith motion along some classical orbit. Instead, we are forced tothink of the electron (in its ground state) as a spherically symmetricprobability cloud about the nucleus.

Figure 1.10 A false picture that should bediscarded. Here the electron supposedly formssomething like sinusoidal standing matterwaves along a circle of radius r. But this pictureis a flawed projection to three-dimensionalspace of the classical picture for a wave on astring. Three-dimensional waves—quantum orclassical—typically fill the space and surely donot look like standing sound waves in acircular tube.

r

λ

represents—according to the probabilistic interpretation of matter waves wementioned earlier—the region of space where it is highly likely to find theelectron. (In reality, the wave extends outside the shaded region, but withexponentially diminishing amplitude.)

But unlike pressure fluctuations in a classical gas (that can vanish periodicallyas compressions become rarefactions), a probability “compression,” as inFigure 1.9, cannot periodically disappear, since the electron it describes wouldthen also disappear! It follows that quantum waves—precisely because of theirinterpretation as probability waves—cannot evolve temporally as in (1.36), butmust instead have a different time dependence, which we shall unravel later.6For the time being, let us retain the notion that quantum waves are similar toclassical waves in their spatial form, but distinctly different with respect to theirphysical interpretation and time evolution.

The preceding discussion focused on a rudimentary description of the mostbasic quantum system, namely, the electron in the ground state of the hydrogenatom. But it also serves another purpose in helping us eliminate the false pictureof Figure 1.10 for the hydrogen atom.

6 Instead of cos𝜔t in (1.36), the time dependence has the complex form exp (−i𝜔t), where𝜔 = E∕ℏ. But since |e−i𝜔t| = 1, such a time dependence implies that the probability distribution ofthe electron around the nucleus (in a state of a given energy and frequency) remains unchanged intime. For the ground state in particular, this means that the physical and chemical properties of theatom remain invariant in time, as one should expect. (By the way, this is the reason atoms in theirground state do not radiate.) The complex form of the time evolution is thus a crucial differencebetween quantum and classical waves, without which the physical interpretation of the formerwould be impossible.


According to this false picture—used extensively in many textbooks, due to itssuccess in explaining Bohr’s quantization condition (see subsequent text)—thestanding electronic waves of the hydrogen atom are formed along a circle whoseradius r satisfies the relation

2𝜋r = n𝜆. (1.37)

In other words, electronic waves are formed if an integer number of wavelengthsfits on the said circle. But since 𝜆 = h∕p, expression (1.37) can be written asrp = nh∕2𝜋, or, equivalently,

𝓁 = nℏ, (𝓁 = rp = m𝑣r), (1.38)

which is the celebrated Bohr’s quantization condition (Bohr, 1913): The electronin the hydrogen atom can only move along specific quantized circular orbitsfor which its angular momentum, 𝓁 = m𝑣r, is an integer multiple of Planck’sconstant, ℏ. Equation (1.38) together with Newton’s law for a circular orbit,

m𝑣2

r= e2

r2 (cgs), (1.39)

form a system of two equations with two unknowns, 𝑣 and r. Its solution yields

rn = n2 ℏ2

me2 = n2a0 (allowed radii) (1.40)

𝑣n = e2

ℏ

1n

(allowed speeds) (1.41)

En = −me4

2ℏ21n2 (allowed energies), (1.42)

where a0 = ℏ2∕me2 = 0.529 Å ≈ 0.5 Å is the so-called Bohr radius, whichcorrectly predicts the size of the atom in its ground state (n = 1). Likewise, thequantity WI = me4∕2ℏ2 in the energy formula gives the ionization energy ofthe atom and its numerical value (13.6 eV) agrees with our earlier finding usingexperimental data.

But despite the empirical success of Bohr’s theory—and its theoreticaljustification based on de Broglie waves—we should not give in to the temptingthought that Bohr’s theory describes something real. There are no quantizedorbits—since no orbits exist at all—and no standing de Broglie waves of thetype shown in Figure 1.10. As we have seen, this picture is flawed; therefore, thesooner we put it aside, the better.7

7 It is curious that this flawed picture appears to go back all the way to the time of Schrödinger andde Broglie! (Which may partly explain its endurance, despite its falseness.) Indeed, as recounted byFelix Bloch (Section 2.2), Schrödinger himself had used this picture to obtain Bohr’s quantizationrules, an approach for which he was chided by Debye who characterized this way of thinking as“childish,” thus prodding Schrödinger to delve deeper and come up with his eponymous waveequation. So, to a small extent, we may owe the discovery of the Schrödinger equation to this flawedpicture of waves fitted along a stationary orbit. Actually, waves of this form can also appear inmodern quantum mechanics, but only in the, so-called, classical (or semiclassical) limit of thetheory.


Let us now see how we can use the correct three-dimensional picture of thespherical probability cloud for the hydrogen atom (Figure 1.9), in order to viewin the right context the approximate condition 𝜆 ∼ 2L we frequently invokedto explain the basic features of the atomic world. For a spherical probabilitycloud we can say the following. Although the concept of the wavelength hasmeaning only for sinusoidal waves, we can introduce a sort of wavelengthfor three-dimensional standing waves (which are anything but sinusoidal),by defining the distance between adjacent nodes as half-wavelength, or thedistance between a peak and the nearest node as quarter-wavelength. Therefore,for a spherical probability cloud with a peak at r = 0 and the node at r = a,we obtain the approximate expression 𝜆∕4 ≈ a ⇒ 𝜆 = 4a. We could arrive atthe same result by saying that a diameter of the sphere is half a wavelength(𝜆∕2 = 2a ⇒ 𝜆 ≈ 4a) because the wave vanishes at the endpoints of the diameter.

1.3.10 The Fine Structure Constant and Numerical Calculations in Bohr’sTheory

Although we emphasized that Bohr’s theory of quantized orbits does not providea correct picture of the atom—since the wave nature of electrons excludes theexistence of orbits—the theory is nevertheless useful for quick calculations thatgive us a first quantitative description of the essential features of the hydrogenatom. These calculations are further simplified if we use the so-called finestructure constant, which is defined, in cgs units, as

𝛼 = e2

ℏc≈ 1

137(fine structure constant), (1.43)

and is a dimensionless quantity, as already mentioned in Section 1.2.3. Using(1.43) we can rewrite formula (1.41) as

𝑣n = e2

ℏccn= 𝛼c

n(n = 1, 2,…), (1.44)

which tells us that the speed of the electron in the first Bohr orbit (n = 1) isapproximately 137 times smaller than the speed of light. Even though this smallvalue justifies, to first order, the nonrelativistic treatment of the problem, it alsoshows that relativistic effects are not negligible to second order. Such effects aremanifested particularly as small shifts in the energy levels of the atom—and thecorresponding spectral lines—and thus produce the so-called fine structure ofthe spectrum. The name of the quantity (1.43) refers exactly to this fine structure.

The constant 𝛼 is also quite useful for quick numerical calculations, since itallows us to eliminate from the pertinent expressions the square of the electriccharge e2 by setting

e2 = 𝛼ℏc, (1.45)

so that e2 is given in terms of more familiar physical constants. Actually, thesubstitution (1.45) in various expressions allows us to perform the calculation ina quick and elegant manner. In the case of ionization energy, for example, we have

WI =me4

2ℏ2 = 12

m(𝛼ℏc)2

ℏ2 = 12𝛼2mc2 = 1

2

( 1137

)20.5 MeV = 13.6 eV.


As you see, we are able to perform the calculation promptly—and in theappropriate energy unit of eV—by simply using the rest energy of the electron (avalue worth remembering, since it is customary to refer to particle masses notin units of g or kg, but in terms of their equivalent energy). We also note that thefactor 𝛼2 before mc2 alerts us about the order of magnitude of atomic energies:They are some five orders of magnitude smaller than an MeV, that is, on theorder of 10 eV.

In practice, it also pays to remember the numerical values of some equivalentcombinations of parameters of the hydrogen atom (e.g., m, e, and ℏ) with thedimension of energy. Three such equivalent combinations that arise often areme4∕ℏ2, ℏ2∕ma2

0, and e2∕a0, where a0 (= ℏ2∕me2) is the Bohr radius. They areequal to each other and have twice the value of the ionization energy of the atom,namely, 27.2 eV. As we shall see later—when we introduce the so-called atomicunits—this value is the natural unit of energy in the atomic world; it is known asone Hartree or the atomic unit of energy.

In the following example, we describe how to calculate in a quick and trans-parent way another important property of Bohr’s theory: the intensity of theelectric field acting on the electron in its ground state.

Example 1.4 Calculate, in SI units, the electric potential and the electric fieldintensity at a distance of one Bohr radius from the nucleus of the hydrogen atom.

Solution: The electric potential is obviously V = −27.2 V, since the correspond-ing potential energy of the electron is V = −e2∕a0 and the absolute value of thisquantity is 27.2 eV, that is, 27.2 e ⋅ V. For the intensity of the electric field weshould note that the pertinent unit in the SI system is the volt per meter (V∕m),given that the product × distance yields the potential difference between twopoints. In the present case we have

= ea2

0≡ e2

a0

1e

1a0

= 27.2 eV ⋅1e⋅

10.5 × 10−10 m

= 5.44 × 1011 V∕m.

Here we simply rewrote the initial expression e∕a20 to form the energy combina-

tion e2∕a0 = 27.2 eV and then divided by the electric charge e to isolate the volt.Finally, we divided by the Bohr radius a0 = 0.5 × 10−10 m to obtain V∕m, whichis the unit of the electric field in the SI system.

It should be evident by now how cumbersome the conventional systems ofunits—SI or cgs—are for calculations in the atomic world. If there is any shred ofdoubt remaining, we encourage the readers to attempt the preceding calculationin any one of those systems. Good luck!

In the following example we expand Bohr’s theory to the so-called hydrogen-likeatoms (or ions).

Example 1.5 Calculate (in eV) the ionization energies of the first threehydrogen-like atoms—He+, Li++, and Be+++—together with their correspondingradii.


Solution: Let us first clarify that the so-called hydrogen-like atoms are simplyions of heavier elements that have been stripped of all but one of their electrons.A hydrogen-like atom is thus identical to hydrogen, except that it has Z protonsin its nucleus. Since the nuclear charge is then Ze, the Coulomb force on theelectron is equal to F = (Ze) ⋅ e∕r2 ≡ Ze2∕r2, instead of e2∕r2 for hydrogen.Clearly, applying Bohr’s theory to any given hydrogen-like atom with atomicnumber Z yields the “same” results as for hydrogen, albeit with Ze2 instead of e2.For example, for the ionization energy WI = me4∕2ℏ2 we obtain

WI(Z) = WI(H)|e2→Ze2 = mZ2e4

2ℏ2 = Z2WI(H) = Z2 ⋅ 13.6 eV,

while for the corresponding radii we have

a0(Z) = a0(H)|e2→Ze2 = 1Z

ℏ2

me2 = 1Z

⋅ 0.5 Å.

In particular, for the ionization energies of He+ (singly ionized helium), Li++(doubly ionized lithium), and Be+++ (triply ionized beryllium), we find

WI(He+) = Z2 ⋅ 13.6 eV|Z=2 = 54.4 eV,WI(Li++) = 122.4 eV, WI(Be+++) = 217.6 eV,

in excellent agreement with experimental data. Moreover, the complete set ofallowed energies of a hydrogen-like atom is given by the formula

En(Z) = En(H)|e2→Ze2 = Z2En(H) = −Z2 ⋅ 13.6n2 eV.

For instance, in the case of He+ we obtain

En(He+) = −54.4n2 eV.

This prediction is in spectacular agreement with the “dark spectral lines” in thesun’s absorption spectrum, which are therefore attributed to the existence ofHe ions in the solar atmosphere. Actually, the successful explanation of theselines was one of the early triumphs of Bohr’s theory and played a decisive rolein its adoption by the scientific community. Nevertheless, a few years later,the wave–particle duality of matter was discovered, leading to the realizationthat Bohr’s theory, although a successful calculational model for hydrogen-likeatoms, was based on notions (such as quantized orbits) that have no place in amodern quantum mechanical context.

1.3.11 Numerical Calculations with Matter Waves: Practical Formulasand Physical Applications

To facilitate calculations with matter waves, we first need to rewrite formula𝜆 = h∕p so that it gives us the wavelength of a particle directly in Å, once we haveits energy in eV. Indeed, what we typically know is not the speed or momentumof, say, an electron, but its energy in eV, since we normally deal with a beam ofsuch particles that have been accelerated by some potential difference expressed


in volts. So, for this purpose, we rewrite formula 𝜆 = h∕p as

𝜆 = hp= h

√2mE

= hc√

2mc2 ⋅ E= 12 400 eV Å

√2 ⋅ 0.5 × 106 eV ⋅ E(eV)

,

where we invoked the nonrelativistic energy–momentum formula E = p2∕2m ⇒

p =√

2mE and multiplied the numerator and denominator by the speed of lightc. We thus obtained mc2 in the denominator and hc in the numerator. The latteris a constant whose value, in eV Å, is already known to us. (It appears in therelation 𝜖 = hf = hc∕𝜆 for photons, where we found that 𝜖(eV) = 12 400∕𝜆(Å)and therefore hc = 12 400 eV Å.) We thus obtain the following practical formula:

𝜆e(Å) = 12.4√

E(eV)≈ 12

√E(eV)

, (1.46)

while for the proton and the neutron (i.e., the nucleons) we have

𝜆N(Å) = 𝜆e(Å)√

me

mp=

𝜆e(Å)√

1836≈𝜆e(Å)

43. (1.47)

Here we took into account that mp ≈ mn ≈ 1836 me, and that the dependenceof the wavelength on the mass is 1∕

√m. According to (1.47), the de Broglie

wavelength of a nucleon is about 43 times smaller than the correspondingwavelength of an electron of the same energy. Taking (1.46) into account, thecorresponding formula for nucleons becomes

𝜆N(Å) = 0.289√

E(eV)≈ 0.3

√E(eV)

. (1.48)

A simple order-of-magnitude test of formula (1.46) is the following: For electronsin the ground state of the hydrogen atom, whose kinetic energy is 13.6 eV, thewavelength should be roughly twice the atomic diameter—that is, on the orderof a few angstroms. Indeed, (1.46) confirms this expectation.

An interesting practical conclusion emerges from the comparison of formula(1.46) and the corresponding one for photons; namely, 𝜆𝛾 (Å) = 12 400∕𝜖(eV): Toattain a wavelength on the order of a few angstroms, an electron needs an energyof a few eV, while a photon needs a few thousand eV, respectively. A directimplication of this fact is the concept of the electron microscope. As is knownfrom optics, the resolution of a microscope is limited by the wavelength of thelight being used. For a given 𝜆, we cannot see any details of the object if theirsize d is less than or equal to 𝜆. To obtain a resolution of a few angstroms withan “optical” microscope, we would need to employ photons of a few thousandeV—that is, x-rays—which are hard to manipulate, for example, focus.

However, if our microscope uses electrons instead of photons—that is, matterwaves instead of EM ones—then a resolution of a few angstroms can be achievedwith electrons of very low energy (a few eV), which can readily be used in manyapplications. In fact, if our desired resolution is 100 Å—which is good enoughfor most biological uses of the electron microscope—then the required energydecreases to a few hundredths of an eV, a nondestructive energy for the sampleexamined. (It does not cause ionizations or dissociations of chemical bonds,


etc.) At the same time, the manipulation of an electron beam is feasible using anappropriate system of electric and magnetic lenses.

Another important application of matter waves pertains to the so-calledthermal neutrons. These are produced in large quantities in nuclear reactors wheninitially fast-moving neutrons impinge on a material containing light nuclei8 (e.g.,graphite). In this way, neutrons slow down through collisions with the graphitenuclei and eventually reach thermal equilibrium. Their kinetic energies are thenapproximately equal to E(eV) ≈ (kT)T≈300 K ≈ 1∕40 eV, so formula (1.48) yields

𝜆n(Å) ≈ 0.3√

E(eV)≈ 0.3 ⋅

√40 ≈ 2 Å.

This result means that thermal neutrons are the prototypical matter wave forcrystallographic studies, because, first, they interact with lattice nuclei (and aretherefore sensitive to their positions) and, second, internuclear distances areon the same order of magnitude as the wavelength of these neutrons (so therequired resolution is available).

Going to the other extreme—the macroscopic world—let us now calculate thewavelength of a dust particle, which has a mass of 1 mg and is moving at a speedof 1 cm/s. In this case, we have (in cgs units)

𝜆 = hm𝑣

≈ 6 × 10−27

10−3 ⋅ 1= 6 × 10−24 cm.

To observe the wave nature of a “particle” that has such a tiny wavelength, we needinterference or diffraction experiments with slits or obstacles of a size similarto that of the wavelength. But neither of these exists in nature, since even thesmallest “object” we know of—the atomic nucleus—has dimensions of 10−13 cm.

We therefore conclude that, even though the principle of wave–particle dualityof matter is in theory also applicable to the motion of macroscopic bodies, it hasno measurable consequences for them and can thus be ignored in practice.

1.3.12 A Direct Confirmation of the Existence of Matter Waves: TheDavisson–Germer Experiment

As we noted at the beginning of this section, the direct confirmation of theexistence of matter waves requires an interference experiment. The simplest suchexperiment was conducted for the first time—somewhat inadvertently—in 1927by Clinton Davisson and Lester Germer. When they bombarded a nickel crystalwith a monoenergetic beam of electrons with E = 54 eV, they were surprised todiscover that apart from the normal reflection, the beam had a preferential angleof oblique reflection equal to 51∘ with respect to the direction of incidence. Thecrucial thing to note here is that—due to their low energy—electrons do notpenetrate the crystal, and reflection, therefore, takes place only at the surface.Figure 1.11 depicts the main idea of the experiment.

If we adopt the de Broglie hypothesis, then the theoretical analysis of theexperiment is simple. The incident electron wave is reflected at the locationsof the Ni atoms in the form of secondary spherical wavelets with the same

8 Because only then there is a sizable energy transfer from the neutrons to the nuclei.


d

Toward detector

A

Bθ

Figure 1.11 Theoretical analysis of the Davisson–Germer experiment. The electrons arescattered preferentially toward those directions 𝜃 that satisfy the condition of constructiveinterference d sin 𝜃 = n𝜆 (n = 0, 1, 2,…).

wavelength as the initial wave. These wavelets interfere constructively onlyin those directions 𝜃 for which the path difference AB (= d sin 𝜃) betweentwo adjacent reflected “rays” is an integer multiple of the wavelength 𝜆 of theelectrons. In other words, constructive interference occurs when

d sin 𝜃 = n𝜆. (1.49)

In the case at hand, we have d = 2.15 Å (this was already known from earlierdiffraction measurements of crystalline nickel using x-rays) and therefore,according to formula (1.46) for E = 54 eV,

𝜆(Å) = 12.4√

54= 1.68 Å.

Thus, condition (1.49) yields

sin 𝜃n = n 𝜆d= n 1.68

2.15= 0.78n,

whence we see that, apart from the trivial case of normal reflection(n = 0 ⇒ 𝜃0 = 0), there is only one more scattering direction at an angle

sin 𝜃1 = 0.78 ⇒ 𝜃1 = 51∘,

just like the experiment revealed! Thus, from 1927 onward de Broglie waveswere no longer a theoretical conjecture, but an irrefutable experimental fact.

1.3.13 The Double-Slit Experiment: Collapse of the Wavefunction UponMeasurement

The evidence we presented so far in support of the principle of wave–particleduality of matter is so compelling that one may wonder why there would be aneed at all for yet another pertinent experiment. Let us therefore stress rightaway that the double-slit experiment—originally a thought experiment that wassubsequently conducted numerous times—is not included here as just anotherconfirmation of the dual nature of particles. It serves, rather, as an ideal “tool” for


investigating the central concept of quantum theory, namely, the probabilisticinterpretation of matter waves. Such an interpretation is compulsory for reasonswe already explained (Section 1.3.6). It provides the only conceivable way tocombine, in the same physical entity, the mutually exclusive properties of beinglocalized and indivisible (akin to particles) on the one hand, and extended anddivisible (akin to waves) on the other. By interpreting the wave associated witha particle as a probability wave, we no longer deny the corpuscular nature of theparticle. The wave here simply describes the probability of finding the particlehere or there, but never here and there at the same time. This implies thatquantum particles (e.g., electrons) are always detected as integral and indivisibleentities. They leave, for example, point-like traces on a photographic plate.

However, the “abstract” nature of quantum waves—we could also call theminformation waves or even waves of knowledge (of probabilistic nature) pertinentto a particle’s state—has as an inescapable consequence the so-called collapse ofthe wavefunction upon measurement. This effect underlies all major paradoxesof quantum mechanics, such as those arising from the double-slit experimentwe shall discuss shortly. The collapse of the wavefunction highlights the centralrole of the measurement process in the quantum world. What this term means issomewhat self-evident: If a measurement yields some information about the par-ticle, then its wavefunction immediately after the measurement must reflect whatwe just measured, and must “incorporate” the information obtained from themeasurement. That is, if we were to repeat the measurement a second time—onthe same particle that was just observed—we should always confirm the firstmeasurement. For example, if we measure the position of a particle—say, in onedimension, for simplicity—and locate the particle in the vicinity of a point x0,then its wavefunction after the measurement can only be highly localized aboutthe point x0. The said wavefunction thus represents a particle whose location isnow known to us with high precision—or at least with as much precision as wewere able to measure it. The result of the measuring process is thus an abrupt—infact, instantaneous—“shrinking” of the initial wavefunction to its new form thatis dictated by the result of the measurement. We depict all this in Figure 1.12.

The “collapse” of the wavefunction so that it “adapts” to experimental datais thus an inescapable logical consequence of the probabilistic interpretation.Without this collapse the probabilistic interpretation would make no sense.

We are now ready to discuss the double-slit experiment in the abovementionedcontext.

Two alternative pictures of the experiment are shown in Figure 1.13. Inthe first picture, a beam of particles with prescribed momentum—and hence,wavelength—impinges on a plate with a double slit. The beam is depicted as aplane wave that emerges on the other side of the plate as two circular waveletscentered on the two slits. An image of interference fringes appears on thescreen—a sort of photographic plate that records the particles arriving at itsvarious regions—depending on whether the two circular wavelets emanatingfrom the slits arrive in the specific region in phase or out of phase. In the secondpicture, the impinging beam is shown as a “stream of particles.” The particlesfall on the plate, go through one or the other slit, and form interference fringes


ψ(x)

ψ0(x)

x0

x

Figure 1.12 The “collapse” of the wavefunction in a position measurement: When ameasurement detects a particle at position x0, its wavefunction collapses immediately to ahighly localized form around x0 and instantaneously vanishes elsewhere. It is as if themeasurement “sucks” the wavefunction, only to concentrate it suddenly at the point wherethe particle was located. Clearly, such an instantaneous collapse has a nonlocal character; thatis, it seems to imply some sort of action at a distance. But since the wavefunction is amathematical entity—not a physical wave with energy and momentum distributed inspace—this instantaneous collapse does not imply a corresponding instantaneous transfer ofenergy or momentum and therefore it does not violate the theory of relativity. A measurementsimply “removes” all possibilities to locate the particle anywhere else than the position it wasfound to be.

ψ

ψ1

ψ2

(a) (b)

|ψ1 + ψ2|2

Figure 1.13 Two alternative pictures for the double-slit experiment: (a) The wave picture. (b)The particle picture. Both pictures are legitimate. But only the wave picture—with theunderstanding of the wave as a probability wave—provides a qualitatively and quantitativelycorrect understanding of this experiment. The particle picture is just to remind us that thereare only particles “behind” the wave that describes how they move in space.

as before, by appearing in various regions in the screen in smaller or greaternumbers. After all, it is the same experiment.

But in contrast to the wave picture—which predicts naturally the interferencefringes of the actual experiment—the particle picture can never lead us to thecorrect description of the phenomenon. The reason is that the particle pictureimplies from the outset the false notion that the particles move in classical orbits.On such a flawed basis, no valid predictions can be drawn, especially aboutinterference fringes. The conclusion is completely general: The only basis forthe proper description of quantum phenomena is always the wave picture, with


the additional clarification that the corresponding waves are to be interpretedas probability amplitude waves,9 not as classical waves. We thus arrive at thekey point for the double-slit experiment. Let us view it from the perspectiveof a person who is skeptical of the statistical interpretation, and whose line ofthinking is the following: “There is no doubt that the incident beam behaves likea wave, since it produces the expected interference fringes on the screen. Buthow do we know that this is a probability amplitude wave and not a truly classicalwave? Can we experimentally distinguish these two possibilities, since they bothlead to the same interference fringes?” This is the real conceptual question thatneeds to be clarified by the double-slit experiment.

Let us begin then. A probability wave—if this is what is going on here—doesnot represent a measurable physical disturbance. It takes physical meaning onlythrough the particle it describes: An experiment can only detect the particle andnothing else. The experimental question pertinent to the particle is clearly thefollowing: Which slit did it go through? Clearly, the particle can pass througheither one slit or the other (but not through both) because in the context ofwave–particle duality (and its probabilistic interpretation) particles are integral,indivisible entities, and are always detected as such.

Therefore, if we were to place two detectors near the exit of each slit—so thatthey register an event when a particle passes through—then we expect the follow-ing experimental outcome (provided that the incident beam is so dilute that onlyone particle arrives at the plate at any one time): (i) Only one of the two detectorswill register an event every time. This means that it is only through one slit that“something” goes through at that instant—obviously, this is the particle—whilenothing at all goes through the other slit. (ii) If we repeat this experiment manytimes and measure how many times a particle passes through one slit or the other,the two numbers will tend to be equal. The reason for this is that the incomingplane wave has the same amplitude at the entry point of each slit and therefore theprobabilities of locating the particle at the entry of one or the other slit are equal.

But if the waves in our experiments were classical waves of some kind—thatis, if they represented some measurable physical disturbance—then the twodetectors would continuously register an event, since in this case, a physicalwave (not a probability wave) would be transmitted continuously through bothslits. It is clear, therefore, that if we confirm the prediction of the probabilisticinterpretation—that something passes through one slit and nothing passesthrough the other—then the classical picture of a continuous passing of the wavethrough both slits fails and is thus rejected. The experiment actually confirms thequantum mechanical prediction. It is only through one slit at a time that some-thing passes, while the transmission numbers are indeed equalized eventually.

But the most fundamental difference between the two types of waves—theprobability waves and the classical waves—is what we are about to describe now.If the waves in our experiment are indeed probability waves, then the placementof detectors immediately behind the slits would have a dramatic consequence:

9 The term amplitude is necessary because the wave nature is represented by the wavefunction 𝜓 ,that is, by the wave amplitude. But for brevity we will often use the term probability wave instead ofprobability amplitude wave.


Interference fringes disappear! By contrast, if the waves are classical then thepresence of the detectors should not affect the interference fringes!

For classical waves, the abovementioned statement is self-evident. In theclassical world the act of observation can always be made so as not to affect theobserved phenomenon in any significant way.

It is therefore impossible to explain in classical physics the disappearance ofthe interference fringes due to the presence of the two detectors. And yet thisdisappearance is an experimental fact. All pertinent experiments confirm itbeyond any doubt. As soon as we activate the detectors to alert us as to whichslit the particle passed through, the interference fringes vanish!

A straightforward explanation of the disappearance of interference fringes isprovided by the collapse of the wavefunction due to measurement. We remind thereaders that in the context of the statistical interpretation, the wave describing theparticle is not a physical wave but an “information wave,” and, as such, it shouldalways respond instantaneously to the “new knowledge” about the particle that isobtained via measurement. In other words, the probability (or information) waveought to express what we know at every instance.

Here is the direct implication of this thinking on our experiment (Figure 1.14).As soon as the detector observes the particle passing through slit #1, theprobability to detect it simultaneously at slit #2 vanishes and the correspondingwavelet (in the vicinity of slit #2) disappears at once. The probability waveimmediately after the measurement—where transmission through slit #1 wasrecorded—contains only the circular wavelets centered on that slit. A similarstatement can be made when the other detector records transmission throughslit #2: We would have circular wavelets around slit #2, and nothing aroundslit #1. If we were dealing with classical waves, the only way to obtain such anoutcome would be to shut one or the other slit at a time. As a result, there wouldbe no interference fringes on the screen (since the two slits would never be openat the same time), but a mere merging of the two diffraction patterns10 aroundthe projections of the slits onto the screen. Indeed, this is what we obtain in ourexperiment when the detectors near the slits are in operation, informing us fromwhich slit the particle passed through every time. But as soon as we switch thedetectors off, interference fringes reappear in all their glory!

Let us also note that the quantum nature of the incident wave is revealed clearlyfrom the way the interference fringes form in the course of the experiment. Thefringes appear gradually as the “spots”—that is, the traces of the particles imping-ing on the screen—accumulate. And while all such particles are described by thesame quantum wave—and are thus in no way different from one another—weobserve that each particle “lands” at a different spot on the screen. Each suchevent is fundamentally unpredictable and nobody can say why a particle fallson a particular spot. Yet all these events together must form the probabilitydistribution described by the intensity of the quantum wave in each region of the

10 In a typical diffraction experiment, a wave beam that passes through a tiny slit emerges in wideangles if the size of the slit is comparable to or narrower than the wavelength of the beam. Becauseof the beam’s angular opening, the trace of the diffracted beam on a screen is not only centered atthe projection of the slit but also extends away from it with decreasing intensity.


ψ

ψ1

ψ2 = 0

or

ψ

ψ1 = 0

ψ2

Figure 1.14 Quantum mechanical explanation of the disappearance of interference fringes.Because of the measurement, the wavefunction collapses into the wave that passes throughthe slit where detection occurs. Simultaneous “emission” of probability waves from both slits isno longer possible, and interference fringes disappear.

screen. This intensity is high in areas where the two circular wavelets interfereconstructively and low in areas where the two wavelets interfere destructively.

We conclude the section by examining how we got here. We initially stated ourtwofold goal: to empirically check the principle of wave–particle duality—thisfundamental principle of quantum theory—and to familiarize our readers withit. Our aim was to make them view the principle not as a whim of nature(something we have to accept simply because it is empirically correct), but as theonly natural explanation of the most challenging mysteries of the atomic world:the inexplicable, from a classical viewpoint, stability of its structures (say, atomsand molecules) and the uniqueness of their form: The fact that no matter howmany times we “break up” an atom or a molecule and allow it to form anew, italways emerges in identical form. In other words, the microscopic constituentsof nature—atoms, molecules, nuclei, and so on—have no history. Their formis predetermined—like that of normal oscillation modes in classical standingwaves—and does not depend on how and when they were created. In the wordsof Niels Bohr:

I had best begin by telling you a little about the history of this theory.My starting point was not at all the idea that an atom is a small-scaleplanetary system and as such governed by the laws of astronomy. I nevertook things as literally as that. My starting point was rather the stability ofmatter, a pure miracle when considered from the standpoint of classicalphysics. By ‘stability’ I mean that the same substances always have thesame properties, that the same crystals recur, the same chemical com-pounds, etc. In other words, even after a host of changes due to externalinfluences, an iron atom will always remain an iron atom, with exactlythe same properties as before. This cannot be explained by the principles


of classical mechanics, certainly not if the atom resembles a planetarysystem. Nature clearly has a tendency to produce certain forms—I use theword ‘forms’ in the most general sense—and to recreate these forms evenwhen they are disturbed or destroyed.[Werner Heisenberg, Physics and Beyond: Encounters and Conversations,Translated by Arnold J. Pomerans, Harper & Row (New York, 1971).]

We encouraged the readers earlier, and emphatically continue to do so now, toreturn to the “mystery of the atomic stability” every time they feel intimidatedby the paradoxical features of quantum theory. When they realize time and againthe scandalous failure of classical physics to explain this mystery, they can retracethe chain of thought we put forward in Section 1.3.2 and summarize again hereas follows:

Stability → Quantization → Wavelike behavior → Wave= Probability wave → Collapse of the quantum wave upon measurement.

We leave it to the readers to traverse this chain of thought—as a kind ofconceptual exercise—by bringing forward the arguments we developed earlier.There is not much room for alternatives. Based on our knowledge today, it isclear that the fundamental principles of quantum mechanics arise readily as theonly natural explanation of the mystery of atomic stability and the uniqueness ofatomic structures.

Actually, it is classical physics—not quantum mechanics—that ought to shock uswhen we try to explain the phenomena of the atomic world.

Problems

1.5 To familiarize yourselves with the “practical formulas” of wave–particleduality, calculate the following quantities:(a) The de Broglie wavelength of an alpha particle accelerated by a poten-

tial difference of 50 V.(b) The energy of the incident electron beam in a Davisson–Germer exper-

iment, where the crystal has d = 2.48Å and the angle of maximumoblique reflection is 𝜃 = 30∘.

(c) The de Broglie wavelength of protons in a CERN experiment (CERNis the European Organization for Nuclear Research near Geneva,Switzerland), where their energy is on the order of 10 TeV = 1013 eV.Show first that for ultrarelativistic particles, whose rest energy ispractically negligible compared to their kinetic energy, the formula forthe wavelength is the same as for photons. Why is this to be expected?

1.6 (Particle in a tubule.) For some linear organic molecules the valenceelectrons can practically move freely along the molecule’s main axis,without being able to leave the molecule. We can therefore approximate

1.4 Dimensional Analysis and Quantum Physics 41

their motion with that of a free-moving particle—of mass m—inside atubule of length L (equal to the molecule’s length) from which it cannotexit. The particle is permanently trapped there. Apply the condition forthe formation of standing waves to calculate the allowed energies of theparticle trapped in the tubule.

1.7 What do you expect will happen to the Bohr radius (a0 = ℏ2∕me2) andthe ground-state energy of the hydrogen atom (E1 = −me4∕2ℏ2) in thefollowing limits:

(a) ℏ → 0, (b) m → ∞?

First, state your prediction—taking care to explain your rationale—andthen check whether it is correct.

1.8 Apply Bohr’s quantization condition to obtain the allowed energies of aparticle in a central force field, like the one in hydrogen, but with a forcelaw F = −kr, known as a three-dimensional harmonic oscillator. Do thesame for F = −gr3.

1.9 In a Davisson–Germer experiment—where d = 2.15Å (nickel crystal)—theelectrons of the incident beam have energy 64 eV. What is the angle ofoblique reflection in this case?

1.10 In the (hypothetical) double-slit experiment of the figure, you are asked to

υ = 7 cm/s

1

2

A

r 1=2 cm

r 2=2.

5 cm

decide whether point A on the screen willbe a local maximum or a local minimumof the expected interference pattern. Usethe rounded values h ≈ 7 × 10−27erg s,me ≈ 10−27g for simplicity. What happenswhen the speed of the electrons becomes(a) two times greater and (b) four timesgreater?

1.4 Dimensional Analysis and Quantum Physics

1.4.1 The Fundamental Theorem and a Simple Application

In its simplest version, dimensional analysis is merely a test of the dimensionalcorrectness of a physics formula (i.e., whether both sides of the formula havethe same physical dimension). Let us call this the passive use of the dimensionalmethod: Given a formula, we simply check whether it is dimensionally correct.But here we wish to speak of the active use of the method: how to use dimensionalanalysis to find a formula that describes a physical phenomenon without evenknowing its underlying theory! The conditions under which we can do this areexpressed in the following fundamental theorem of dimensional analysis.


Theorem 1.1 If a physical quantity—say, D —depends only on three others—letus call them A,B,C —then the dependence of D on A,B,C is determined on purelydimensional grounds up to a dimensionless multiplicative constant. In particular,we will have

D = 𝜎A𝛼B𝛽C𝛾 , (1.50)

where 𝜎 is an arbitrary dimensionless constant and 𝛼, 𝛽, and 𝛾 are suitable expo-nents that are determined by equating the physical dimensions of length, mass,and time of both sides of the equation.

Let us see how the method works in a simple problem from elementary physics.

Example 1.6 Use dimensional analysis to find the formula for the angularfrequency 𝜔 of a harmonic oscillation as a function of the parameters of theproblem.

Solution: The crucial step appears at the very end of the statement of theproblem. On which parameters of the problem do we expect the quantity𝜔 to depend? Two obvious candidates are k and m: the spring constant thatdetermines the restoring force via the known relation F = −kx and the mass mof the oscillating body. Is there a third parameter? Come to think of it, this oughtto be the maximum displacement a of the body from its equilibrium position;that is, its oscillation amplitude. At this point, the reader may object that theamplitude does not appear in the familiar formula 𝜔 =

√k∕m. But recall that we

are not supposed to know anything about this formula, so we should considerall reasonable options. Based now on this theorem and the dependence of thefrequency 𝜔 on the three quantities k,m, and a, we have

𝜔 = 𝜎k𝛼m𝛽a𝛾 . (1)

To determine 𝛼, 𝛽, and 𝛾 , we start from the expressions[𝜔] = T−1, [m] = M, [a] = L, (2)

while for k the definition F = −kx yields

[k] = [F]L

= [M ⋅ acceleration]L−1 = (M ⋅ LT−2)L−1 = MT−2, (3)

where the brackets denote the physical dimension of a physical quantity withrespect to length, mass, and time—L,M, and T—which are the basic units in thecgs system. By substituting now (2) and (3) into (1) we obtain

T−1 = (MT−2)𝛼M𝛽L𝛾 = L𝛾M𝛼+𝛽T−2𝛼. (4)

Upon equating the dimensions of length, mass, and time of both sides in (4),we get

𝛾 = 0, 𝛼 + 𝛽 = 0, −2𝛼 = −1 ⇒ 𝛼 = 1∕2, 𝛽 = −1∕2, 𝛾 = 0,

which means that the desired formula has the form

𝜔 = 𝜎

√km,


where 𝜎 is the anticipated dimensionless constant. The latter cannot bedetermined on dimensional grounds alone, but we can calculate it with a singleexperiment for two convenient values of k and m. It should also be stressed thatthe independence of 𝜔 from the oscillation amplitude a—a very special featureof the harmonic oscillator—emerged through purely dimensional arguments, soit does not depend on the details of the theoretical description of the problem.It is a purely dimensional result.

An equally useful exercise for the readers is to prove, using dimensionalarguments again, that for a nonlinear power law of the form F = −kx3, theformula for 𝜔 becomes

𝜔 = 𝜎

√km

a

whose characteristic feature is the linear dependence of 𝜔 on the amplitude ofoscillation. The greater the amplitude, the greater the frequency of oscillationand thus the smaller its period. Can you explain why?

We hope the given example has also clarified the reason the fundamentaltheorem of dimensional analysis holds. It does so because, upon equatingthe dimensions L,M,T of both sides of (1.50), we obtain a system of threeequations with three unknowns from which the desired exponents 𝛼, 𝛽, and 𝛾can be uniquely determined. It can also be easily shown that, among all possiblefunctional forms D = f (A,B,C), only the form (1.50)—that is, a product ofpowers—can be compatible with the requirement that both sides of the equationhave the same dimension.11 Also, it goes without saying that the three quantitiesA,B, and C in (1.50) are dimensionally independent. That is, none of themcan be expressed in terms of the other two. It is equally clear that when thedesired quantity D depends, not on three, but on four or more quantities, thenthe dimensional method cannot by itself determine the desired formula, noteven up to a dimensionless multiplicative constant. And yet, even in thosecases, a suitable use of the dimensional method can lead to remarkable results,depending on whether certain dimensionless parameters of the problem can beconsidered small.

Closing, we cannot fail to note how advantageous the cgs system of units isover SI, from the perspective of dimensional analysis. In cgs units, the basicphysical quantities are only three—L,M,T—while in SI units we should alsoadd the electric charge as an independent unit. Since this increase of the basicquantities in SI units is completely artificial (as artificial as measuring tempera-ture in nonmechanical units), it can easily be remedied. All it takes is to realizethat the two systems differ mainly with respect to Coulomb’s law, which—in

11 For example, if the function f (A,B,C) could be expanded in a Taylor series—normally this is notpossible—then we would have

D = f (A,B,C) =∑

𝜇,𝜈,𝜆

𝜎𝜇𝜈𝜆

A𝜇B𝜈C𝜆 (𝜆, 𝜇, 𝜈 are positive integers or zero).

However, given that of all the products of the series, only one can have the correct dimensions, theform (1.50) emerges as the only possible choice.


the hydrogen atom, for example—has the form e2∕r2 in cgs as opposed to(1∕4𝜋𝜖0)e2∕r2 in SI units. Clearly, the (more elegant) results of the cgs systemare converted to their counterparts in the SI system upon the substitution

e2 → kCe2 (kC = 1∕4𝜋𝜖0).

A practical question remains. Is there a considerable number of physicalphenomena that depend only on three parameters so that one can use thedimensional method to predict their behavior? The answer is yes. The mostfundamental physical phenomena have in almost all cases few—and quite oftenonly three—parameters, because by their nature they relate to the simplestpossible manifestations of fundamental laws. Two important examples wediscuss here pertain to this category.

1.4.2 Blackbody Radiation Using Dimensional Analysis

The thermal radiation of bodies—also known as blackbody radiation—is a fun-damental physical phenomenon for a simple reason: It is completely independentof the material the radiating body consists of. After all, it is for this reason that allincandescent bodies “look” exactly the same; they are visually indistinguishable.The physical explanation of this remarkable fact lies in the thermal natureof their light. Namely, before being emitted by any hot body, light interactsrepeatedly with its material and eventually reaches thermal equilibrium withthe body. By the time light is emitted, it is thermal light. Therefore, the spectraldistribution of its intensity has a universal character, just like the distribution ofmolecular speeds for a gas in thermal equilibrium with the walls of its container.The experimental data for the thermal radiation of bodies—and the definitionsof the quantities needed for its description—are given in Figure 1.15.

As expected, experimental efforts to investigate thermal radiation focused,right from the start, on its two most prominent features: the total intensity(i.e., the surface area below the corresponding curve) and the location of its

0.25

1.2

1

0.8

0.6

0.4

0.2 T2 = 300 K

T1 = 600 K

f (1014 Hz)

J (10–10 W/m2 Hz)

0.5 0.75 1 1.25 1.5

Figure 1.15 The spectral distribution curve of blackbody radiation. J = spectral intensitydef=

radiated energy, per unit time Δt, per unit frequency Δf , and per unit surface ΔS of the

radiating body = ΔE

ΔtΔfΔS. I = total intensity

def= ∫ ∞

0 J(f , T)df .


maximum value, as well as the dependence of these features on the temperatureof the radiating body. The following empirical laws summarize the major findingson these quantities:

Stefan–Boltzmann law Wien’s law

I = 𝜎T4 𝜆max = bT

The total intensity of the thermal radiationemitted by a blackbody depends on thefourth power of its absolute temperature

The wavelength at which maximumemission occurs is inversely proportional tothe absolute temperature of the body

𝜎 = 5.67 × 10−8 W∕m2∕K4 b = 0.3 cm K ⇒ 𝜆max(cm) = 0.3T(K)

By the end of 1899, Planck had succeeded—utilizing all prior knowledge—tofind the full mathematical formula for the spectral distribution J( f ,T):

J( f ,T) = 2𝜋hc2

f 3

ehf ∕kT − 1(Planck’s general empirical formula), (1.51)

where k is Boltzmann’s constant and c the speed of light. Immediately thereafter,Planck also concluded that the only theoretical assumption that could explain hisformula was the quantization of light. Namely, that the energy of light is quantizedin integer multiples of 𝜖 = h f , where h is our familiar Planck’s constant.

In the following examples, we examine what conclusions we can draw onthermal radiation using purely dimensional arguments.

Example 1.7 Use dimensional analysis to make a prediction within the contextof classical physics for the spectral distribution J( f ,T) of blackbody radiation andcomment on the result.

Solution: Our first step is to find the quantities that J depends on. Obviously,two of these are the variables of the problem—the frequency f and temperatureT—but also the physical constants of the laws that govern the phenomenon. Suchconstants are the speed of light c, since we are dealing with an EM phenomenon,and Boltzmann’s constant k, since we are also dealing with thermodynamics. Onthe other hand, J cannot depend on any properties of the light-emitting materialbecause thermal radiation does not depend on these either; it has a universalcharacter. It follows that J cannot depend on any parameters such as the mass andcharge of the electron or the nuclear masses, which determine atomic structureand, concomitantly, all properties of macroscopic matter. We thus have

J = J( f ,T , k, c). (1)

It appears therefore that J depends on four parameters (not three), so thedimensional method is not sufficient to determine the sought dependence.But recall that k is not a truly fundamental constant, but more the result of ahistorical accident: the fact that we discovered the concept of temperature beforefiguring out its physical meaning as a measure of the thermal kinetic energy ofatoms or molecules in a gas. We thus had to devise, after the fact, the constant k


to ensure that the product kT assigns to T its correct physical meaning and thecorrect units. It is for this reason that T never appears on its own12 but alwaystogether with k as the product kT .

In this spirit, we can equivalently write (1) asJ = J( f , kT , c) ⇒ J = 𝜎f 𝛼(kT)𝛽c𝛾 ,

with exponents to be calculated in the manner we presented earlier. We thusobtain

𝛼 = 2, 𝛽 = 1, 𝛾 = −2 ⇒ J ∼ f 2kT∕c2, (2)where we have used the proportionality symbol to avoid a repeated reference tothe dimensionless multiplicative constant 𝜎 that is always present in formulasderived using dimensional analysis.

Relation (2)—known as the Rayleigh-Jeans law—is a truly profound result.It tells us that applying classical physics to the problem of thermal radiationinescapably leads us to the so-called ultraviolet catastrophe: the boundlessincrease of radiated EM energy at high frequencies. Such an increase wouldmake the total radiated intensity I diverge. So classical physics is not simplyunable to explain the phenomenon: It produces an irrational prediction; acatastrophe. But there is more. The fact that this prediction resulted solelyfrom dimensional requirements—and not from a detailed calculation that couldentail some revisable assumptions—should leave no doubt in our mind thatthere is really no “cure” for this catastrophe. At least in the problem of thermalradiation, classical physics is fundamentally wrong. And now we know why. Inthe context of quantum theory, the UV catastrophe is avoided, because at highfrequencies the energy h f of the light quanta is so high (h f ≫ kT) that theirthermal excitation is impossible. (Again, quantization saves the day.)

Let us note, finally, that an alternative—and much simpler—way to arriveat relation (2) is to combine the quantities f , kT , and c, to produce the unitsof the desired quantity J . These units emerge directly from the definition ofJ = ΔE∕(Δt ⋅ Δf ⋅ ΔS), whence we obtain [ J] = E∕L2. Of the given quantities, kThas dimensions of energy, while a combination of the other two—c and f —thathas dimensions of length is c∕f , which, as we know, is the wavelength of theradiation. At this point, we are only interested in the fact that c∕f has dimensionsof length (i.e., [c∕f ] = L), so the right combination of kT , f and c that yields thecorrect dimensions E∕L2 for J is kT∕(c∕f )2 = f 2kT∕c2, as given. For the skepticalreaders—to whom this whole process may seem a bit arbitrary—we would stressthat, since there is one and only correct combination (according to the theorem),then no matter how we arrived at it, it is bound to be the correct one.

Example 1.8 Use dimensional analysis to see if it can lead us to the abovemen-tioned empirical laws of thermal radiation. That is, the laws of Stefan–Boltzmannand Wien. Assume that we are dealing with a quantum phenomenon, so thatPlanck’s constant ℏ appears in the formulas, as does the speed of light c.

12 Apart from obvious exceptions (e.g., the specific heat cV = (dQ∕dT)V ) where the temperature T ,not kT , appears in the definition of the quantity. In this case, all we have to do is a trivial substitutionof T with kT and proceed as before.


Solution: The first thing to note is this: Both desired quantities, I and 𝜆max,do not depend on f , because I relates to the total radiated power—that is, theintegral of J over all f —while 𝜆max relates only to the position of maximumradiation. But I and 𝜆max both depend on the temperature T—always in the formof the product kT—and on c and ℏ as we saw earlier. We thus have

I = I(ℏ, c, kT), 𝜆max = 𝜆max(ℏ, c, kT)

so the conditions for the fundamental theorem are met, namely, the desiredquantity depends only on three physical parameters. Applying the theorem (inthe usual systematic manner) yields

I ∼ (kT)4

c2ℏ3 , 𝜆max ∼ℏckT, (1)

which are indeed the correct empirical laws—as far as dependence ontemperature is concerned—but also with realistic numerical values for thecoefficients, as the reader can verify. We can thus confirm the practical rule thatthe dimensionless multiplicative constant in the dimensional method is nevera “very large” or a “very small” number. For order-of-magnitude estimates, theundetermined multiplicative constant can safely be regarded as a number of theorder of unity.

Note, finally, that formulas (1) can also be derived in the nonsystematic way wesketched earlier. Since [I] = E∕L2T , [𝜆max] = L, and kT provides an energy term,we need combinations of kT , c, and ℏ with dimensions of length and time. Suchcombinations can be obtained easily if we realize that [ℏ] = ET and [ℏc] = EL, sowe obtain

[ℏ

kT

]

= ETE

= T ,[ℏckT

]

= ELE

= L

and then

[I] = EL2T

⇒ I ∼ kT(ℏckT

)2⋅(

ℏ

kT

) = (kT)4

c2ℏ3 , [𝜆max] = L ⇒ 𝜆max =ℏckT.

1.4.3 The Hydrogen Atom Using Dimensional Analysis

It would be interesting to check whether dimensional analysis can tell us some-thing about the two basic empirical quantities regarding the hydrogen atom: thesize of the atom—on the order of Å, as we know—and its ionization potential (orenergy) whose empirical value is 13.6 eV. We will take it as given that classicalphysics cannot describe this or any other atom, since the very existence ofthe atom would not be possible within the framework of classical physics: Theelectron would have collapsed onto the nucleus. And even if such a collapse wereavoided somehow, classical physics cannot provide a mechanism for the atom tohave a definite size. Thus, our study will proceed within the quantum context, sothat the sought quantities a (atomic radius) and WI (ionization energy) will befunctions of the form

a = a(ℏ,m, e), WI = WI(ℏ,m, e) (1.52)


that is, functions of Planck’s constant, the electron mass, and the electroncharge. Why not the proton mass? Because the proton, being much heavier thanthe electron, is practically fixed at the center of the atom and its mass can beregarded as infinite to a first approximation. But to the second approximation,the proton mass does play a role, and to account for it we should substitutethe electron mass with the reduced mass of the electron–proton system. Butwhy would the speed of light not appear in formulas (1.52)? Since the quantumatom—in its ground state—does not radiate, c need not appear and play a rolein determining this state. Therefore, the quantities we are interested in dependonly on the triplet ℏ,m, and e. We thus determine their dependence usingdimensional arguments as usual to obtain

a ∼ ℏ2

me2 , WI ∼me4

ℏ2 . (1.53)

Since the combination e2∕cℏ—the fine structure constant—is dimensionless,it is clear that e2∕ℏ has dimensions of speed. So the combination m(e2∕ℏ)2 =m × (speed)2 has dimensions of energy and is thus the desired unique expressionfor WI. Moreover, the ratio e2∕a has energy dimensions (it is the potentialenergy of two electron charges separated by a distance a); if we equate it withthe expression for WI, we obtain the dimensionally correct formula for a. Asfor the numerical values of the quantities in (1.53), we recall from previouscalculations that 0.5 Å is the value for a and 27.2 eV is the value for WI, both inthe correct order of magnitude for these quantities. Thus, the mere introductionof Planck’s constant, aided by dimensional analysis, can give us plausible resultsfor the hydrogen atom (even without a detailed theory). Surely this is a strongindication that in the correct theory for the atom, Planck’s constant will play akey role. Note also that the expression me4∕ℏ2, apart from being correct as anorder of magnitude, is also exactly twice the ionization potential. Therefore, wecan comfortably assume that the formula

WI =me4

2ℏ2 (1.54)

is exact and that the empirical relation for energies, En = −13.6 eV∕n2, can alsobe written as

En = −me4

2ℏ21n2 , (n = 1, 2,… ,∞) (1.55)

which is indeed the correct mathematical expression for the allowed energies ofthe atom, as we shall see later.

If you now combine what we said here and our discussion in Section 1.3.9,you will arrive at a plausible explanation of how Bohr was led to his theory.As we shall see again and again in the book, dimensional analysis can be apowerful tool.

Problems

1.11 Use dimensional analysis to predict—up to a multiplicative dimensionlessconstant—the formula for the angular frequency𝜔 of a nonlinear oscillatorwith a force law of the form F = −kx3. What is the most interesting featureof your result?

Further Problems 49

1.12 Use dimensional analysis to show that in a world of d dimensions,the formulas for the quantities Jcl (classical prediction for the spectraldistribution of blackbody radiation), I, and 𝜆max will have the form

(a) Jcl ∼ kT(

fc

)d−1

,

(b) I ∼ (kT)d+1

ℏdcd−1 ,

(c) 𝜆max ∼ℏckT.

1.13 The Stefan–Boltzman law is, of course, expected to be a special conse-quence of Planck’s general formula (1.51) for the spectral distributionof blackbody radiation. Show that this is indeed the case and that thetheoretical prediction for 𝜎 is

𝜎 = 2𝜋5

15k4

c2ℏ3 .

Does this prediction agree with the experimental value of 𝜎?

1.14 You suspect that you may have made an error in copying 𝜎 (i.e.,the Stefan–Boltzmann constant) from some book as 𝜎 = 5.67 ×10−4 W∕m2∕K4. Can you argue, using facts from everyday experience,why this numerical value is completely wrong?

1.15 Consider a hypothetical universe where the value of Planck’s constant is10 times lower compared to ours. Would the radiative intensity of a hotbody be different in such a universe compared to ours? If yes, then by howmuch?

1.16 As the universe expands, the wavelength of cosmic microwave background(CMB) photons—this wonderful thermal afterglow of the Big Bang—gets“stretched” by the same factor. This is because, as space itself expands, thedistance between two successive crests (or troughs) of a propagating EMwave increases. Given that the CMB is presently observed to have a tem-perature of T ≈ 3 K, calculate the following: (a) The present intensity ofthe CMB and (b) the intensity and peak-emission wavelength of the CMBwhen the universe was 10 times smaller than its present size (i.e., when thedistance between two distant galaxies was 10 times smaller than what ispresently observed).

1.17 The energy of the photons corresponding to the peak emission of a hotbody is equal to 4 eV. Calculate the total emitted intensity from that body,in units of W∕m2.

Further Problems

1.18 Consider the general case of Compton scattering, whereby the wavelength𝜆 of the impinging photon is 𝜆 = k𝜆C, where k is an arbitrary dimensionlessnumber, and the scattering angle 𝜃 is also arbitrary. Use momentum andenergy conservation to calculate the quantities 𝜆′, p′

𝛾 , 𝜖′𝛾 (= wavelength,


momentum, and energy of the scattered photon) as well as 𝜙, p, andE (= scattering angle, momentum, and energy of the scattered electron).Specifically, show that the following relations hold.For the photon:

𝜆′ = 𝜆C(1 + k − cos 𝜃), p′𝛾 =

mc1 + k − cos 𝜃

, 𝜖′𝛾 =mc2

1 + k − cos 𝜃.

For the electron:

tan𝜙 = k1 + k

1tan(𝜃∕2)

, K ≡ E − mc2 = mc2 1 − cos 𝜃k(1 + k − cos 𝜃)

,

p = mc 1 − cos 𝜃k(1 + k − cos 𝜃)

√

k2

tan2(𝜃∕2)+ (k + 1)2,

while for the photon before the “collision” we have

𝜆 = k𝜆C, p𝛾 =mck, 𝜖𝛾 =

mc2

k.

Once you confirm that these general formulas reproduce the results ofExample 1.2, apply them to obtain the results in the following two specialcases:(a) k = 1, 𝜃 = 𝜋∕2,(b) k = 2, 𝜃 = 2𝜋∕3.

1.19 Aside from its historical role in the development of quantum theory,Bohr’s quantization condition—that only those circular orbits are allowedfor which the angular momentum 𝓁 = m𝑣r of the electron is an integermultiple of Planck’s constant ℏ—is still useful, as it provides a quick wayto approximately calculate the allowed energies for various central forcefields. In fact, these calculated energies—with a possible exception of thefirst few of them—reflect the essential features of the energy spectrum. Inthis spirit, apply Bohr’s condition 𝓁 = nℏ to show that for an attractivecentral force of the form F = ±gr𝜈 (the sign depends on 𝜈 being positiveor negative, respectively, while g is assumed positive), the radii of theallowed orbits, the speeds, and the energies of the particle on these orbitsare given by the formulas

rn = (ℏn)2∕(𝜈+3)

(mg)(𝜈+3) , 𝑣n = 1m(mg)1∕(𝜈+3)(nℏ)(𝜈+1)∕(𝜈+3), (1)

En = ±g2𝜈 + 3𝜈 + 1

(ℏn)2(𝜈+1)∕(𝜈+3)

(mg)(𝜈+1)∕(𝜈+3) . (2)

Do these formulas reproduce the known results for hydrogen? What aboutthe three-dimensional harmonic oscillator, where F = −gr?

1.20 It may sound hard to believe, but the temperature at the surface of thesun—a very distant object—can actually be deduced with reasonableaccuracy on the basis of the following observations from everyday

Further Problems 51

experience, in conjunction with the law I = 𝜎T4 and the known value for𝜎 (𝜎 = 5.67 × 10−8 W∕m2∕K4).Observation #1: A solar panel works! That is, it produces roughly thesame work (albeit a bit slower) as an electrical device (e.g., water heater)that consumes power on the order of 2–3 kW, which is typical of allenergy-intensive home appliances. Needless to say, the light-collectingsurface of a solar panel is on the order of 1 m2.Observation #2: The solar disk—whose apparent size is roughly equal tothat of the moon’s disk—can be blocked out (as you can verify yourselves)by an object (e.g., part of our finger) of size 1 cm at the end of our stretchedarm, which is roughly 1 m away from our eyes.Utilize these two observations to calculate—or, at least, estimate—thetemperature at the sun’s surface.

1.21 Given that life on earth is the outcome of a primordial evolutionaryadaptation to environment, a key element of which is sunlight (a sourceof both energy and “information” about the world around us), it is notunreasonable to assume that terrestrial living beings gradually “tuned in”to the sun, to utilize the energetically richer region of its spectrum. Afterall, this is the only way to explain the remarkable fact that the eyes of allliving beings “see” roughly in the same spectral range. Actually, the sameholds true for all light-collecting molecules (e.g., chlorophyll) of plants.

Use the given reasoning to estimate—in conjunction with the formula𝜆max (cm) = 0.3∕T(K)—the temperature at the sun’s surface. Does yourresult roughly agree with what you obtained in the previous problem?

1.22 Use dimensional analysis to predict (without referring to Bohr’s theory) thedependence of the radius of the ground state orbit, and the correspondingenergy, on the parameters ℏ, m, and g for an arbitrary central potential ofthe form V (r) = ±gr𝜈 for a given 𝜈. Do your results agree with those ofProblem 1.19?

53

2

The Schrödinger Equation and Its Statistical Interpretation

2.1 Introduction

As we saw in Section 1.3, wave–particle duality implies that all knownparticles—electrons, protons, neutrons, and also more complex structures,such as atoms or molecules—have a wavelike character that is governed by thefundamental relations

E = ℏ𝜔, p = ℏk,

where E, p are the energy and momentum of the particle, and 𝜔, k are the angu-lar frequency and wavenumber of the corresponding matter wave, respectively.But if particles—for example, electrons—are also waves, then to build a theoryof quantum particles, we need to find the wave equation they satisfy. This is thetopic of the following section.

Classical waves have a clear physical meaning. But attributing physical meaningto quantum waves is not so straightforward. Indeed, this is our main objective inthis chapter. We present a self-contained and systematic approach to the subject,based only on wave–particle duality as explained in the introduction of Chapter1, and reformulated in its modern form in Section 1.3.7, formulas (1.33).

We hope that Chapter 1 has familiarized readers with the basic quantummechanical concepts, thus providing a useful background for what follows.

2.2 The Schrödinger Equation

Once at the end of a colloquium, I heard Debye saying something like:‘Schrödinger, you are not working right now on very important problemsanyway. Why don’t you tell us some time about that thesis of de Broglie,which seems to have attracted some attention?’ So, in one of the nextcolloquia, Schrödinger gave a beautifully clear account of how de Broglieassociated a wave with a particle and how he could obtain the quantizationrules of Niels Bohr and Sommerfeld by demanding that an integer numberof waves should be fitted along a stationary orbit. When he had finished,Debye casually remarked that he thought this way of talking was rather


54 2 The Schrödinger Equation and Its Statistical Interpretation

childish.1 […] [To] deal properly with waves, one had to have a waveequation. […]

Just a few weeks later [Schrödinger] gave another talk in the colloquiumwhich he started by saying:

‘My colleague, Debye, suggested that one has to have a wave equation;well, I found one.’

Felix Bloch

From a talk given on 26 April, 1976, at the Washington, DC meeting of theAmerican Physical Society. Source: Physics Today 29, 23 (1976).

2.2.1 The Schrödinger Equation for Free Particles

We start off with the simplest possible case, free motion in one dimension.Wave–particle duality tells us that if E and p are the energy and momentum ofthe particle, respectively, then the frequency 𝜔 and the wavenumber k of thecorresponding matter wave will be

𝜔 = E∕ℏ, k = p∕ℏ.

In classical wave theory, a wave with a well-defined frequency and wavenumberis necessarily a sinusoidal wave and is thus described by the function

u(x, t) = cos (kx − 𝜔t) or sin (kx − 𝜔t),

which in complex form can be written as

u(x, t) = ei(kx−𝜔t).

In analogy with classical waves, we assume that a matter wave of definite momen-tum and energy will be described by the wavefunction

𝜓(x, t) = ei(px−Et)∕ℏ, (2.1)

where the use of the complex form is now necessary for reasons that will soonbecome clear.

Our aim is to find the wave equation satisfied by 𝜓(x, t). It is reasonable toexpect that the equation we seek should meet the following criteria:

1. It must be linear and homogeneous: This would ensure that the superpositionof two solutions is also a solution, as is the case for the classical wave equation.

2. It must have constant coefficients: This follows from the requirement that forV = 0 all points in space are equivalent (principle of homogeneity of space).Likewise, all points in time are equivalent (homogeneity of time). Also, thecoefficients of the equation should be independent of any varying particlecharacteristics (like energy and momentum), and should only depend on

1 For a discussion of this picture, see also Section 1.3.9 and Figure 1.10.

2.2 The Schrödinger Equation 55

fixed parameters of the particle (like its mass), and on the necessary physicalconstants (Planck’s constant).

3. It must reproduce the classical nonrelativistic energy–momentum relation(E = p2∕2m).

The linearity requirement 1 means that any wavefunction 𝜓(x, t) will alwaysbe a superposition of plane waves “with suitable momenta”—recall the Fouriertheorem—which in turn implies that a plane wave will satisfy the same waveequation satisfied by any wavefunction 𝜓 . So all we need to do is figure out whatkind of wave equation the wavefunction (2.1) of a plane wave satisfies.

Differentiating (2.1) with respect to t and x, we obtain𝜕𝜓

𝜕t= − iE

ℏ𝜓,

𝜕𝜓

𝜕x=

ipℏ𝜓. (2.2)

Let us now introduce new operators in place of 𝜕∕𝜕t and 𝜕∕𝜕x,

E = iℏ 𝜕𝜕t, p = −iℏ 𝜕

𝜕x, (2.3)

so that (2.2) take the elegant form

E𝜓 = E𝜓, p𝜓 = p𝜓, (2.4)

which says that the action of the operators E and p on the wavefunction of a planematter wave generates the same wavefunction multiplied by E and p, respectively.

Equations (2.4) tell us that in order to reproduce the energy–momentum rela-tion E = p2∕2m, the wavefunction 𝜓(x, t) must satisfy the symbolic equation

E𝜓 =p2

2m𝜓, (2.5)

where

p2 = p ⋅ p =(

−iℏ 𝜕

𝜕x

)(

−iℏ 𝜕

𝜕x

)

= −ℏ2 𝜕2

𝜕x2 .

In this way, we can rewrite (2.5) as

iℏ 𝜕𝜓𝜕t

= − ℏ2

2m𝜕2𝜓

𝜕x2 , (2.6)

which is called the Schrödinger equation for free particles.Equation (2.5) recovers the well-known energy–momentum relation for the

plane wave (2.1), as we intended. Indeed, every time the operator p acts on 𝜓of (2.1), it multiplies it by the momentum p, so that

p2𝜓 ≡ p(p𝜓) = p(p𝜓) = p(p𝜓) = p(p𝜓) = p2𝜓.

In the left-hand side of (2.5), the operator E yields E𝜓 = E𝜓 , so we can rewrite(2.5) for a plane wave as

E𝜓 =p2

2m𝜓

and obtain E = p2∕2m. We state once again that the Schrödinger equation forfree particles, (2.5) or (2.6), will hold for any wavefunction 𝜓(x, t), since the lattercan always be written as a superposition of plane waves.


The Schrödinger equation for free particles is often written in the equivalentsymbolic form

iℏ 𝜕𝜓𝜕t

= H0𝜓, (2.7)

where

H0 =p2

2m= − ℏ2

2m𝜕2

𝜕x2

is the so-called free Hamiltonian operator. Let us remind the readers here that inclassical mechanics, the Hamiltonian H is defined as the function

H = H(x, p) =p2

2m+ V (x),

which gives the total energy (kinetic + potential) of the particle. We use the sym-bol E when we refer to a specific energy value, and H (the Hamiltonian) when wespeak of the total energy H = p2∕2m + V (x) as a function of p and x. For a vanish-ing potential V , only the kinetic term p2∕2m survives in the Hamiltonian, whichis then written as H0 and called the free Hamiltonian. Note also that in the form(2.7) of the Schrödinger equation for free particles, the operator H0 is obtainedby substituting p → p = −iℏ(𝜕∕𝜕x) in the classical expression H0 = p2∕2m.We will make use of this fact later.

We now apply the above-mentioned procedure to construct the freeSchrödinger equation in a situation where the energy–momentum relationdiffers from E = p2∕2m.

Example 2.1 In relativity theory, the energy–momentum relation for a free par-ticle of rest mass m is

E2 = c2p2 + m2c4.

Write down the one-dimensional quantum wave equation in this case.

Solution: As before, all we need to do is substitute (2.3) in the energy–momentumrelation, which leads to the symbolic equation

E2𝜓 = (c2p2 + m2c4)𝜓,

that is,

−ℏ2 𝜕2

𝜕t2𝜓 =(

−c2ℏ2 𝜕2

𝜕x2 + m2c4)

𝜓,

or, equivalently,(𝜕2

𝜕x2 − 1c2

𝜕2

𝜕t2 − m2c2

ℏ2

)

𝜓 = 0.

This is the so-called Klein–Gordon equation, which is of second order withrespect to time, as we might have expected, since the energy is squared in therelativistic energy–momentum relation.

2.2 The Schrödinger Equation 57

2.2.2 The Schrödinger Equation in an External Potential

The symbolic form (2.7) of the free Schrödinger equation is suggestive of how wemight go about generalizing it in the presence of an external potential. Indeed,looking at the form (2.7), we can immediately guess that in the case of a particlemoving in a potential V (x), the Schrödinger equation would be

iℏ 𝜕𝜓𝜕t

= H𝜓. (2.8)

Here, we replaced the operator H0 of the free Hamiltonian with the operator Hof the full Hamiltonian

H =p2

2m+ V (x) = − ℏ2

2m𝜕2

𝜕x2 + V (x),

which is derived from the expression of the classical Hamiltonian of the particle,by substituting

x → x, p → −iℏ 𝜕

𝜕x,

so that (2.8) takes the form

iℏ 𝜕𝜓𝜕t

= − ℏ2

2m𝜕2𝜓

𝜕x2 + V (x)𝜓. (2.9)

The extension to three-dimensional (3D) motion in an arbitrary poten-tial V (r) can now be seen clearly. Once again, the symbolic form is (2.8)with 𝜓 = 𝜓(x, y, z, t), but the operator H derives now from the 3D classicalHamiltonian

H =p2

2m+ V (r) =

p2x + p2

y + p2z

2m+ V (x, y, z)

by substituting

x → x, y → y, z → z,

px → −iℏ 𝜕𝜕x, py → −iℏ 𝜕

𝜕y, pz → −iℏ 𝜕

𝜕z.

In a more compact form, we have

r → r, p → −iℏ∇,

where

∇ = x 𝜕

𝜕x+ y 𝜕

𝜕y+ z 𝜕

𝜕z

is the well-known del operator. The Hamiltonian operator H now becomes

H = − ℏ2

2m

(𝜕2

𝜕x2 + 𝜕2

𝜕y2 + 𝜕2

𝜕z2

)

+ V (x, y, z),

or, equivalently,

H = − ℏ2

2m∇2 + V (r),


where

∇2 = 𝜕2

𝜕x2 + 𝜕2

𝜕y2 + 𝜕2

𝜕z2

is the Laplacian operator (or Laplacian, for simplicity). We are finally ready towrite the 3D Schrödinger equation as

iℏ 𝜕𝜓𝜕t

= − ℏ2

2m∇2𝜓 + V (r)𝜓.

It should be clear to the readers that this discussion is not a proof of theSchrödinger equation, which, like any other fundamental physical law, cannotbe proved mathematically. Our aim has simply been to show that the form ofthis equation is not at all arbitrary, but rather, it is the most plausible choice wecan make based on the empirical facts we have. The ultimate “judge” will be theexperiment, of course—and we will later invoke its “verdict.”

2.2.3 Mathematical Intermission I: Linear Operators

We used the term operator earlier without defining it properly. Basically it is avery general and rather trivial concept. An operator is any mapping of one set ofmathematical objects into another (which usually is the same as the initial set).

To be more specific, consider the set of functions 𝜓(x) defined on the interval−∞ < x < +∞, and assume that these functions have all the “nice” properties(i.e., they are continuous, differentiable many times, etc.) so that we can do allsorts of things (i.e., “actions” or “operations”) on them. In this set of functions, theact of multiplying 𝜓(x) by the variable x defines a mapping of the set into itself.We can then think of x as an operator whose effect on a function is to multiplyit by x. Another example is the act of differentiating with respect to x, namely,D = d∕dx: This is an operator that produces the derivative of the initial function.Another example: Take any function, square it, add 3, and subtract its derivative!This whole process constitutes an operator!

So an operator is nothing to be afraid of. Even further, the operators used inquantum mechanics are linear, which means that

A(c1𝜓1 + c2𝜓2) = c1(A𝜓1) + c2(A𝜓2),

that is, their action on a linear combination of functions produces the same linearcombination of their respective images. (By image we mean the function resultingfrom the action of an operator on an initial function.) Clearly, both operatorsA = x and B = D = d∕dx are linear, since

x(c1𝜓1 + c2𝜓2) = c1(x𝜓1) + c2(x𝜓2), andd

dx(c1𝜓1 + c2𝜓2) = c1

d𝜓1

dx+ c2

d𝜓2

dx.

Throughout this book, we only deal with linear operators. Therefore, for simplic-ity, we will omit the term “linear” from now on.

The addition or multiplication of operators acting on a common set of functionsis self-explanatory:

Addition of operators: (A + B)𝜓def= A𝜓 + B𝜓

Problems 59

Multiplication of operators: (AB)𝜓def= A(B𝜓)

Algebraic operations between operators have the same properties as algebraicoperations between numbers, except one: the commutative property of multipli-cation. In other words, while for numbers we can always write

𝛼𝛽 = 𝛽𝛼,

for operators this is not generally true: The order in which two operators appearin a product matters. For example, for A = x and B = D = d∕dx, if we form theproduct AB = x(d∕dx) and act on a function 𝜓(x), we obtain

(AB)𝜓(x) =(

x ddx

)

𝜓(x) = x d𝜓dx

≡ x𝜓 ′.

If we now reverse the order of the two operators in the product, by forming BA =(d∕dx)x and acting on the same function, we obtain

(AB)𝜓 =(

ddx

x)

𝜓 = ddx

(x𝜓) = 𝜓 + x𝜓 ′.

Clearly, we have AB ≠ BA in general. The fact that we cannot “commute”—that is,switch the order of—two operators in a product has far-reaching consequencesin quantum mechanics, as we will later see (Chapter 3).

Before going further, let us clarify that the caret symbol is sometimes used todenote not operators but unit vectors. For example, we wrote previously for thedel operator:

∇ = x 𝜕

𝜕x+ y 𝜕

𝜕y+ z 𝜕

𝜕z.

In this instance, x, y, z are not operators but unit vectors along the directionsx, y, z of the Cartesian coordinate system. We hope that this double usage in nota-tion will not confuse the readers—depending on the context, it should be clear inevery instance what we mean.

Problems

2.1 Find a classical “wave equation” whose solutions are plane waves u(x, t) =exp(i(kx − 𝜔t)) and whose relation between frequency and wavenumber(the so-called dispersion relation) has the form 𝜔 = 𝛼k3, where 𝛼 is a givenconstant.

2.2 Write down the Hamiltonian operator for the following physical systems:(a) A free particle in one and three dimensions.(b) A particle in one dimension and in a force field of the form: (i) F = −kx,

(ii) F = −kx3.(c) The electron in a hydrogen atom.


2.3 Statistical Interpretation of Quantum Mechanics

2.3.1 The “Particle–Wave” Contradiction in Classical Mechanics

Wave–particle duality (which we used as the empirical basis of the previous dis-cussion) says that particles have wavelike behavior, yet they retain their corpus-cular nature, that is, they always exist as indivisible pointlike entities. Clearly,this last feature is incompatible with any classical interpretation of these waves.For instance, let us suppose that the wavefunction 𝜓(x) describes some mea-surable physical disturbance, like the sound field. It would then follow that thecorresponding particle physically extends throughout the volume occupied bythe wave, and has a distribution density that would most probably be proportionalto the “intensity” |𝜓(x)|2 of this “matter field.” But such a classical interpretationof a matter wave is in direct contradiction with the well-known experimental factthat each particle is always detected as an indivisible pointlike entity.

The failure of a classical interpretation of matter waves is also apparent fromthe fact that the wavefunction 𝜓(x, t) necessarily takes complex values. Indeed, if𝜓(x, t) were any real-valued function, then in the Schrödinger equation (2.9) theright-hand side would be necessarily real, while the left-hand side would be nec-essarily imaginary, which cannot be. But if 𝜓(x, t) is a complex-valued function,it cannot describe any measurable (i.e., physically meaningful) physical distur-bance. Contrast this with the classical wave equation

∇2u − 1c2

𝜕2u𝜕t2 = 0, (2.10)

which has only real coefficients and therefore always assumes real-valued solu-tions. Just to make things clear: While we often use complex expressions for thesolutions of (2.10), for example, u(x, t) = exp (i(kx − 𝜔t) for a plane wave,2 thiscomplex form is used purely for convenience, since an exponential can be handledmore easily than the cosine or sine functions. But at the end of such calculations,we always have to extract the real part of u, which is also a solution of the waveequation.

However, the situation is totally different when we are dealing with theSchrödinger equation: Now the equation itself has complex coefficients, so thewavefunction can never be purely real.

For example, a plane wave 𝜓 = exp(i(px − Et)∕ℏ) is a solution of the freeSchrödinger equation, but its real or imaginary parts alone are not solutions, aswe argued before and readers can readily check.

It is therefore evident that a classical interpretation of matter waves is impos-sible. We cannot resolve the particle-versus-wave contradiction in a classicalcontext.

2 Although we use the terms plane wave and sinusoidal wave as synonyms, strictly speaking theyare not: For a plane wave the wavefronts always lie on a plane. This does not necessarily mean thatthe wave is sinusoidal, since a suitably chosen collection of sinusoidal waves can also form a planewave. But since quantum mechanical waves (of well-defined frequency and wavenumber) havenecessarily complex form, it would perhaps be misleading to call them “sinusoidal”; we prefer theterm “plane wave” instead.

2.3 Statistical Interpretation of Quantum Mechanics 61

2.3.2 Statistical Interpretation

Faced with the manifest inability of classical physics to provide an acceptablephysical interpretation of matter waves, M. Born3 formulated in 1926 the sta-tistical interpretation, which can be stated as follows:

The statistical interpretation of the wavefunction: The wavefunction does notrepresent a physically observable classical wave, but rather a “probability wave.”The probability density—that is, the probability per unit length (or volume)—oflocating the particle in any region in space is given by the square of the absolutevalue of the wavefunction.

According to this statistical interpretation, the probability density P(x) is4

P(x) = |𝜓(x)|2 = 𝜓∗(x)𝜓(x).

Note that in the given expression, we did not explicitly refer to the time variable inthe wavefunction. And the reason is simple. Within the context of the statisticalinterpretation, the time t is a mere parameter with no particular importance. Soit suffices to write 𝜓(x) and refer to just one frame of the wavefunction 𝜓(x, t),that is, to its form at the particular time t.

The probability of locating the particle in the region between x and x + dx isthen

P(x)dx = |𝜓(x)|2dx,

whereas the total probability of locating the particle anywhere in space(−∞ < x < +∞) is

∫+∞

−∞P(x)dx = ∫

+∞

−∞|𝜓(x)|2 dx.

For the statistical interpretation to make sense, the total probability must beunity:

∫+∞

−∞|𝜓(x)|2 dx = 1. (2.11)

Equation (2.11) is called the normalization condition, and any wavefunctionthat satisfies it is called a normalized wavefunction. Clearly, for a wavefunction tobe “normalizable,” it has to obey the inequality

∫+∞

−∞|𝜓(x)|2 dx < ∞,

which is another way to say that the integral converges. A wavefunction 𝜓(x)with this property is called square integrable. We can actually take any square

3 German–British physicist, who was awarded the Nobel Prize in 1954.4 The complex conjugate of any complex number z = x + iy is defined as z∗ = x − iy. The square ofa complex number’s absolute value (also called the square of its modulus) is then

|z|2 = z∗z = x2 + y2.

Writing z in the so-called polar form z = 𝜌ei𝜙, we get z∗ = 𝜌e−i𝜙 and |z|2 = z∗z = 𝜌ei𝜙 ⋅ 𝜌e−i𝜙 = 𝜌2

⇒ |z| = 𝜌.


integrable wavefunction and multiply it by a suitable normalization factor suchthat the total probability comes out to be unity. Therefore, to test whether a wave-function describes a physically realizable state of the particle, we should checkwhether it is square integrable. An obvious necessary condition for square inte-grability is that the wavefunction vanishes at ±∞,5

𝜓(−∞) = 𝜓(+∞) = 0.

Later on, we will turn our attention to wavefunctions that are not squareintegrable and yet are finite everywhere, for example, the plane wave𝜓p(x) = exp(ipx∕ℏ). Actually, such “unphysical” wavefunctions constitutevery useful idealizations of physically realizable states, as we will see.

Let us also add that the statistical interpretation of |𝜓(x)|2 as probability perunit length implies that

[𝜓2] = L−1 ⇒ [𝜓] = L−1∕2.

To recap, by interpreting 𝜓 as a probability wave, the classical contradiction of“particle” versus “wave” is resolved, since the particle is no longer forced to denyits corpuscular nature and to physically “spread” itself throughout the volumewhere 𝜓 differs from zero. The wavefunction 𝜓 represents the probabilityamplitude to locate the particle “here or there,” but never “here and there”simultaneously.

2.3.3 Why Did We Choose P(x) = |𝝍(x)|2 as the Probability Density?

To begin with, the probability density P(x) can only be positive. This observationdoes not take us very far, however, since we could construct an infinite numberof positive expressions from 𝜓(x); for instance, we could have chosen

P(x) = |𝜓(x)|, (2.12)

which is as simple as

P(x) = |𝜓(x)|2. (2.13)

So the question arises: Is there another physical requirement for P(x) (otherthan mere positivity) that could possibly dictate the choice (2.13) uniquely? Sucha requirement indeed exists, and it is the conservation of probability. As the wave-function evolves in time—that is, 𝜓(x) → 𝜓(x, t)—the total probability

∫+∞

−∞P(x, t)dx

must stay independent of t: Once we have normalized the wavefunction so thatthe total probability is unity at a given time, say, t = 0, then this property shouldnot change in time. The total probability must stay equal to unity forever.

If this requirement were not met, we would end up with probability “surpluses”or “deficits” as time evolved, and the statistical interpretation would not make

5 The sufficient condition for square integrability is that the wavefunction vanishes at ±∞ fasterthan 1∕

√x. Why?


sense. This is what sets (2.12) apart from (2.13): the latter expression conservesthe total probability, while the former does not. In other words, the integral

∫+∞

−∞|𝜓(x, t)|2 dx

is independent of time, whereas

∫+∞

−∞|𝜓(x, t)|dx

is not, as we can easily show. (We prefer nevertheless to postpone the proof untilwe have defined some necessary mathematical concepts; see the last section ofthis chapter.) But even on purely physical grounds we can see why the properchoice for the probability density is the square of the wavefunction |𝜓(x, t)|2, andnot any other positive quantity. Here is why. We know from classical wave theorythat the energy density of a wave is given by the square of the “waving” quantity(i.e., the square of the wave amplitude). For example, in an electromagnetic wave,where the wave amplitude is given by the electric and magnetic fields E and B,the energy density in cgs units is

𝑈 = 18𝜋

(E2 + B2).

Integrating this expression throughout all space gives the total energy of the wave,which is a constant of motion; it is independent of time.

In matter waves, the “waving” quantity is the wavefunction 𝜓(x, t), even if itdoes not describe a measurable physical disturbance in itself. Thus we can inferthat, to ensure conservation of the total probability,6 which seems to be the analogof the total energy, we should rather choose the square of the wavefunction forthe probability density.

2.3.4 Mathematical Intermission II: Basic Statistical Concepts

It should be evident by now that the new theory we are building—quantummechanics—is really a statistical theory. To proceed further, we shall thereforeneed some basic statistical concepts, such as the mean (or expectation) valueand the standard deviation or uncertainty. We define the mean value first.

2.3.4.1 Mean ValueConsider a statistical quantity A that can take distinct values a1, a2,… , an,….Suppose we make N measurements of A, yielding N1 times the value a1, N2 timesthe value a2, and so on. The mean value ⟨A⟩ ≡ A is then

⟨A⟩ =N1a1 + · · · + Nnan + · · ·

N=∑

nan fn,

6 Given that the total probability should equal unity, you may wonder whether we have beenmaking much ado about the conservation of unity! This is not the case: While we are free tonormalize the wavefunction initially so that the total probability is one at that moment in time, thereis no guarantee it will remain so at later times; this depends on whether a conservation law exists.


where fn = Nn∕N are the frequencies at which the possible values an appear inthe measurements. When N → ∞ the frequencies fn tend to the probabilities Pnfor the values an. The mean value in this limit is then

⟨A⟩ =∑

nanPn. (2.14)

So the mean value of a statistical quantity is equal to the sum of its possible valuesmultiplied by the corresponding probabilities.

Any function G(A) of a statistical quantity A is also a statistical quantity withpossible values gn = G(an). Its mean value ⟨G(A)⟩ will therefore be

⟨G(A)⟩ =∑

nG(an)Pn. (2.15)

We often deal with statistical quantities whose possible values span a continu-ous range, usually from −∞ to +∞. In this case, the probability of a specific valuehas no meaning—and if it did, its value would be zero—so we only talk aboutthe probability of occurrence of a continuous range of values. We thus introducethe probability density P(a), defined such that the product

P(a)dais equal to the probability of occurrence of values in the infinitesimal intervalfrom a to a + da. So, P(a) is the probability per unit interval of the continuousstatistical (or stochastic) variable a.

The mean value ⟨A⟩ of a continuous statistical quantity is given by an expressionsimilar to (2.14), but with an integral instead of a discrete sum. That is,

⟨A⟩ =∑

aaP(a) ≡ ∫

+∞

−∞aP(a) da,

where we have used the “continuous sum” symbol∑

a—which is equivalent tothe integration symbol ∫ da—to stress even further the similarity between thediscrete and continuous cases. We thus obtain the following: The mean value ofa statistical quantity with a continuous range of values is equal to the integralof the continuous statistical variable multiplied by the corresponding probabilitydensity.

The mean value of any function G(A) of the statistical quantity A, is analogousto (2.15) for the discrete case:

⟨G(A)⟩ = ∫+∞

−∞G(a)P(a) da.

Let us note also that the sum of probabilities of all possible outcomes mustequal unity, for both the discrete and continuous cases. Thus, the following nor-malization conditions must be obeyed,

∑

nPn = 1 ∫

+∞

−∞P(a) da = 1

(Discrete case) (Continuous case)whereas there is also the possibility of a mixed spectrum, with both discrete andcontinuous parts.


2.3.4.2 Standard Deviation (or Uncertainty)Of course, there is more to life than the mean value of the statistical distribu-tion of a quantity. We often wish to know how “concentrated” the possible valuesare about the mean. The degree of concentration of a statistical distribution iscalled dispersion or standard deviation and is denoted by ΔA or 𝜎A, where A isthe statistical quantity of interest.

In quantum mechanics the symbol ΔA is most often used (and goes also by thename uncertainty of A), whereas elsewhere in statistics the symbol 𝜎A (and thename standard deviation) is customarily preferred.

We need to define the uncertainty ΔA such that it has the desired meaning. Itshould clearly be a measure of the mean distance of all possible values from themean. One definition of such a measure is

(ΔA)2 = ⟨(A − ⟨A⟩)2⟩, (2.16)

that is, the square of the uncertainty is defined as the mean square deviation fromthe mean value.

The reason we use the square of the deviation, (A − ⟨A⟩)2, rather than the devi-ation itself, A − ⟨A⟩, is simple: When we are trying to determine how dispersed astatistical distribution is, we do not really care whether one value lies to the rightor to the left of the mean, but only how far it lies from it; we are only interestedin the size (not the sign) of the deviation A − ⟨A⟩. From a practical point of view,the “recipe” for computing ΔA is this: For a discrete distribution, we calculatethe squared difference of each possible value from the mean, multiply it by thecorresponding probability, and sum up all these products. Here is an example.

Example 2.2 A statistical quantity A has two possible values, a1 = 4 and a2 = 8,with corresponding probabilities

P1 = 1∕4 and P2 = 3∕4.

Find the uncertainty ΔA of this quantity.

Solution: The mean value ⟨A⟩ is

⟨A⟩ = a1P1 + a2P2 = 4 ⋅14+ 8 ⋅

34= 7,

so that

(ΔA)2 = (a1 − ⟨A⟩)2P1 + (a2 − ⟨A⟩)2P2 = (4 − 7)2 ⋅14+ (8 − 7)2 ⋅

34= 3,

that is,

ΔA =√

3.

We will now show that we can express the uncertainty in an equivalent, simplerform

(ΔA)2 = ⟨A2⟩ − ⟨A⟩2, (2.17)


which tells us that the square of the uncertainty equals the mean value of thesquare, minus the square of the mean value.

Proof : The square of the deviation from the mean can be written explicitly as

(A − ⟨A⟩)2 = A2 − 2⟨A⟩A + ⟨A⟩2 ≡ A2 − 2𝜆A + 𝜆2, (2.18)

where 𝜆 = ⟨A⟩ is some constant. By taking the mean value of (2.18) we obtain

(ΔA)2 = ⟨A2 − 2𝜆A + 𝜆2⟩ = ⟨A2⟩ − 2𝜆⟨A⟩ + 𝜆2

= ⟨A2⟩ − 2⟨A⟩⟨A⟩ + ⟨A⟩2 = ⟨A2⟩ − ⟨A⟩2.

Here, we have used the well-known properties of the mean value:

1∶ ⟨A + B + · · · ⟩ = ⟨A⟩ + ⟨B⟩ + · · ·2∶ ⟨cA⟩ = c⟨A⟩3∶ ⟨c⟩ = c,

where A,B,… are any statistical quantities and c a given constant. □

Let us also stress that the mean value of the square of a statistical quantity Ais calculated in the same way as we previously suggested for a general functionG(A). That is,

⟨A2⟩ =∑

na2

nPn ⟨A2⟩ = ∫ a2P(a) da

(Discrete distribution) (Continuous distribution)

Readers can verify the following fact (which one should expect on simplequalitative grounds): If the uncertainty of a statistical distribution is zero, thenthe distribution consists of a single possible value, with 100% probability ofoccurrence. This is another way of saying that we know the statistical quantitywith full certainty. (This is the literal meaning of the phrase “zero uncertainty”after all.)

Finally, let us introduce another concept from statistics, the statistical moment,which is defined as follows: The statistical moment In of order n is the mean valueof the nth power of the statistical variable. That is, In = ⟨An⟩.

In this context, the mean value ⟨A⟩ is nothing but the statistical moment of firstorder, that is, ⟨A⟩ = I1, while for the squared uncertainty we have (ΔA)2 = I2 − I2

1 .Equivalently, we can see from the definition (ΔA)2 = ⟨(A − ⟨A⟩)2⟩ that the squareof the uncertainty is the statistical moment of second order of the deviation fromthe mean value ⟨A⟩.

The importance of statistical moments of all orders can be seen from the fol-lowing theorem, which is stated without proof.

Theorem 2.1 Knowledge of the statistical moments of all orders determines astatistical distribution uniquely.


Problems

2.3 The possible values of a discrete statistical distribution are a1 = 2, a2 = 3,and a3 = 4, with corresponding probabilities P1 = 1∕4, P2 = 1∕4, and P3 =1∕2. Without doing any computation, select the correct answer for ⟨A⟩ andΔA from the list of values proposed here:

(a) ⟨A⟩ = 3, 74

, 3, 134

, 174

.

(b) ΔA =√

114

, 32

, 2, 52

.

2.4 Explain why the following inequalities for the mean value ⟨A⟩ and the uncer-tainty ΔA of an arbitrary statistical distribution are expected to hold

amin ≤ ⟨A⟩ ≤ amax, ΔA ≤ |amax − amin|

2and then proceed to prove them. (amin and amax are the minimum and max-imum possible values of the distribution, respectively.)

2.5 The probability density P(a) of a continuous statistical variable a, whichtakes values in the interval 0 ≤ a ≤ 1, has the explicit form

P(a) = Na(1 − a). (1)

Compute the normalization coefficient N , the mean value ⟨A⟩, and theuncertainty ΔA of the distribution.

2.3.5 Position Measurements: Mean Value and Uncertainty

Suppose now we wish to calculate the mean value of the position ⟨x⟩ of a quantumparticle, given its wavefunction𝜓(x) at a given moment in time. According to theearlier discussion, the mean value is given by the expression

⟨x⟩ = ∫+∞

−∞xP(x)dx = ∫

+∞

−∞x|𝜓(x)|2 dx,

whereas the mean value F(x) of any function of x is

⟨F(x)⟩ = ∫+∞

−∞F(x)P(x)dx = ∫

+∞

−∞F(x)|𝜓(x)|2 dx.

More specifically, for F(x) = x2, we get

⟨x2⟩ = ∫+∞

−∞x2|𝜓(x)|2 dx,

so that when we know the wavefunction 𝜓(x), we can also calculate the positionuncertainty Δx from (2.17), which in this case can be written as

(Δx)2 = ⟨x2⟩ − ⟨x⟩2.


According to the discussion in the previous section, the position uncertaintyΔx provides a quantitative measure of the space interval where it is highly likelyto find the particle. We should also stress here that the position uncertainty Δxis not some nebulous or ill-defined quantity, but is exactly determined once weknow the wavefunction 𝜓(x) of the particle. Qualitatively speaking, the form ofthe wavefunction tells us immediately whether Δx is large or small. If 𝜓(x) isspread out over an extended region, then the position uncertainty of the particlewill clearly be high. It is also clear that if the distribution P(x) is symmetrical aboutsome point x0, then ⟨x⟩ = x0. A pertinent example is the following.

Example 2.3 Calculate the mean position ⟨x⟩ and the position uncertainty Δxof a particle whose wavefunction at a given moment in time is

𝜓(x) = Ne−𝜆x2∕2, (1)

where the numerical coefficient in the exponential has been set to 1∕2 for conve-nience (so that it cancels upon squaring 𝜓 to obtain the probability density).

Solution: The first step is to calculate the normalization coefficient N from therequirement that the total probability of locating the particle from −∞ to +∞ beunity:

∫+∞

−∞|𝜓(x)|2 dx = N2 ∫

+∞

−∞e−𝜆x2 dx = 1. (2)

The definite integral I = ∫ +∞−∞ exp (−𝜆x2)dx—also known as the Gaussian inte-

gral—can be calculated via the double integral

J = ∫+∞

−∞ ∫+∞

−∞e−𝜆(x2+y2) dx dy = ∫

+∞

−∞e−𝜆x2 dx ⋅∫

+∞

−∞e−𝜆y2 dy = I ⋅ I = I2,

which is easily computed once we switch from Cartesian to polar coordinates,𝜌, 𝜃,

x2 + y2 = 𝜌2, dx dy = 𝜌 d𝜌 d𝜃.

We thus find

J = ∫+∞

0 ∫2𝜋

0e−𝜆𝜌2

𝜌 d𝜌 d𝜃 = ∫+∞

0e−𝜆𝜌2

𝜌 d𝜌 ⋅ ∫2𝜋

0d𝜃

= 12𝜆 ∫

+∞

0e−𝜆𝜌2 d(𝜆𝜌2) ⋅ 2𝜋 (𝜆𝜌2 = 𝜉)

⇒ J = 𝜋

𝜆 ∫∞

0e−𝜉 d𝜉 = 𝜋

𝜆(−e−𝜉)

||||

∞

0= 𝜋

𝜆

and (since J = I2)

I = ∫+∞

−∞e−𝜆x2 dx =

√𝜋

𝜆.

Going back to (2) we have

N2 ∫+∞

−∞e−𝜆x2 dx = N2

√𝜋

𝜆= 1 ⇒ N = 4

√𝜆

𝜋,


where we have selected by convention the positive value for N , since the sign of𝜓(x) has no physical significance. (Can you explain why?) Thus, the normalizedwavefunction is

𝜓(x) = 4

√𝜆

𝜋e−𝜆x2∕2 (3)

and is plotted in Figure 2.1. Given now that𝜓(x) is clearly symmetric about x = 0,we have

⟨x⟩ = 0.

To calculate the position uncertainty (Δx)2 = ⟨x2⟩ − ⟨x⟩2 = ⟨x2⟩, we need theintegral

⟨x2⟩ = ∫+∞

−∞x2|𝜓(x)|2 dx =

√𝜆

𝜋 ∫+∞

−∞x2e−𝜆x2 dx, (4)

which is a special case of the general integral

I2n = ∫+∞

−∞x2ne−𝜆x2 dx.

The latter is easily computed by successive differentiations, with respect to 𝜆, ofour familiar Gaussian integral

I0(𝜆) = ∫+∞


√𝜋

𝜆.

Indeed, each differentiation of e−𝜆x2 with respect to 𝜆 produces a factor −x2. So,after n successive differentiations, the desired term x2n will appear. We thus arriveat the general formula

I2n = (−1)n dn

d𝜆n

√𝜋

𝜆= 1 ⋅ 3 · · · (2n − 1)

(2𝜆)n

√𝜋

𝜆, (5)

ψ(x)

x

Figure 2.1 The Gaussian wavefunction 𝜓(x) = 4√𝜆∕𝜋e−𝜆x2∕2. The mean position of the particle

is zero, ⟨x⟩ = 0, and the corresponding uncertainty is Δx = 1∕√

2𝜆. As we should expect(why?), the position uncertainty decreases as the parameter 𝜆 increases.


which, in our case of n = 1, gives

I2 = ∫+∞

−∞x2e−𝜆x2 dx = 1

2𝜆

√𝜋

𝜆.

Combining this result with (4) we obtain

⟨x2⟩ = (Δx)2 = 12𝜆

⇒ Δx = 1√

2𝜆. (6)

Given that the term 𝜆x2 in the exponential of 𝜓(x) ought to be dimensionless—otherwise its series expansion would contain terms of different dimension—thedimensions of 𝜆 should be [𝜆] = L−2. Thus, the result (6) is dimensionally correct.It is also physically plausible: As 𝜆 increases, the exponential e−𝜆x2 decays rapidlyas we move away from the origin, and so Δx ought to decrease also, since thewavefunction then becomes “narrower” and “taller” because of the normalizationfactor in (3). We ask readers to check for themselves that this normalization factoris also dimensionally correct.

Some readers may wonder by now: How come the wavefunction 𝜓(x) =4√𝜆∕𝜋 e−𝜆x2∕2 of this problem is real, when we have previously argued that

wavefunctions are necessarily complex? Let us clarify then, that our earlierstatement was not that “wavefunctions are necessarily complex.” What weactually said is that the solutions to the Schrödinger equation have this property.In other words, it is the form 𝜓(x, t) of the solution at each time t that mustnecessarily be complex. But a single frame 𝜓(x, t0) of such a solution (i.e., theform of the solution at a given moment in time) can surely be real. And that isexactly the case with the wavefunction of the present problem and of similarinstances later.

Problems

2.6 From the following list of one-dimensional wavefunctions, select those thatare square integrable. The parameter 𝜆 is positive everywhere.

(a) 𝜓(x) = N tanh 𝜆x ≡ N e𝜆x − e−𝜆x

e𝜆x + e−𝜆x

(b) 𝜓(x) = N x√

x2 + a2

(c) 𝜓(x) = N sin kx ⋅ e−𝜆x2∕2

(d) 𝜓(x) = Ne𝜆x + e−𝜆x

(e) 𝜓(x) = Nxe−𝜆|x|

(f ) 𝜓(x) = Neikx−𝜆|x|

2.7 Plot the wavefunction

𝜓(x) = Nxe−𝜆x2∕2 (𝜆 > 0)

2.4 Further Development of the Statistical Interpretation: The Mean-Value Formula 71

and then calculate the mean position ⟨x⟩ and the position uncertainty Δx interms of the given parameter 𝜆. Check your results for dimensional consis-tency.

2.8 Do the same as before for the wavefunction

𝜓(x) = Ne−𝜆|x| (𝜆 > 0).

2.4 Further Development of the StatisticalInterpretation: The Mean-Value Formula

2.4.1 The General Formula for the Mean Value

Our development of the statistical interpretation of the wavefunction so farallows us to predict the statistical distribution of position measurements only. Butbesides position, there are surely other interesting measurable quantities of a par-ticle. Therefore, to complete the statistical interpretation of quantum mechanics,we should find a general way of calculating the statistical distribution of mea-surements for any physical quantity. We will accomplish this task in two steps.

First, we will seek a general formula for the mean value of measurements of anyphysical quantity. This search will yield the famous mean-value formula. Second,we will conclude that the mean-value formula “says it all.” As we will later show,the mean-value formula for a physical quantity allows us to calculate not only themean value but also the uncertainty, as well as all statistical moments of the dis-tribution of measurements of that quantity. In other words, this formula providescomplete knowledge of the statistical distribution of the measured quantity.

Here is how the argument goes. Since we are looking for a mean-value formulawith the same general form for all physical quantities, we might as well start fromour previously acquired knowledge of the full statistical distribution of one suchquantity: position. We will write down the expression for its mean value and seewhether (and how) we could generalize it to other physical quantities as well. Themean value ⟨x⟩ can be written successively as

⟨x⟩ = ∫+∞

−∞xP(x)dx = ∫

+∞

−∞x𝜓∗ 𝜓 dx = ∫

+∞

−∞𝜓∗ x𝜓 dx,

where the last expression

⟨x⟩ = ∫+∞

−∞𝜓∗ (x𝜓)dx (2.19)

suggests that a possible generalization for any physical quantity A might be7

⟨A⟩ = ∫+∞

−∞𝜓∗ (A𝜓)dx, (2.20)

7 For a three-dimensional problem, this expression would be ⟨A⟩ = ∫ 𝜓∗(A𝜓)dV . For notationalsimplicity, we are using here one-dimensional expressions, since their generalization to threedimensions is usually obvious. For the same reason we will also omit the limits of integration, withthe implication that they span (−∞,+∞) in one dimension, and all space in three dimensions.


where A is some suitable operator for the physical quantity A. Equation (2.20) tellsus that the operator A acts upon the function to its right, and hence the need touse the parenthesis for clarity.

Equation (2.19) says immediately that for position, x, the appropriateoperator is

x = x,

which results in a simple multiplication of the wavefunctions by the variable x.Let us now proceed to obtain a plausible choice of operators for other physical

quantities. To begin with, for one-dimensional problems, any physical quantityA will always be a function of the two basic quantities, position and momentum.That is,

A = A(x, p). (2.21)

With the operators x for position and p for momentum at hand, we can thenconstruct the operator A by substituting x → x, p → p in the classical expression(2.21). In a way, this substitution takes care of the so-called classical limit of quan-tum mechanics: The new theory incorporates the “memory” of the old one, andperhaps even reproduces it as a suitable limiting case.

We shall then have

A = A(x, p).

So, all we need to do is to find the suitable quantum mechanical operator pfor the momentum variable p. A useful clue along this direction is obtainedby recalling that in producing the symbolic form of the Schrödinger equation,iℏ(𝜕𝜓∕𝜕t) = H𝜓 , we introduced the Hamiltonian operator H by substitutingx → x, p → −iℏd∕dx in the classical expression H = p2∕2m + V (x).

The substitution x → x coincides with the choice x = x we made earlier for theposition operator. So we are strongly encouraged to choose for the momentumoperator the expression

p = −iℏ ddx,

whence the Hamiltonian operator, corresponding to the physical quantity“energy,” becomes

H =p2

2m+ V (x) = − ℏ2

2md2

dx2 + V (x),

and is thus identical to the operator appearing in the Schrödinger equation, whichis what we set out to do.

The generalization to three dimensions is straightforward. For the basic quan-tities of position and momentum, the quantum mechanical operators are

x = x, y = y, z = z,

px = −iℏ 𝜕

𝜕x, py = −iℏ 𝜕

𝜕y, pz = −iℏ 𝜕

𝜕z


and in vector notation,

r = r, p = −iℏ∇.

Thus, for any physical quantity A = A(r,p), we can write

A = A(r,−iℏ∇).

For angular momentum, for example,

𝓵 = r × p,

whose components along the Cartesian axes are

𝓁x = ypz − zpy, 𝓁y = zpx − xpz, 𝓁z = xpy − ypx,

the respective quantum mechanical operators become

𝓁x = −iℏ(

y 𝜕

𝜕z− z 𝜕

𝜕y

)

, 𝓁y = −iℏ(

z 𝜕

𝜕x− x 𝜕

𝜕z

)

𝓁z = −iℏ(

x 𝜕

𝜕y− y 𝜕

𝜕x

)

and, in vector form,

𝓵 = r × p = −iℏ r × ∇.

Before taking the next step, namely, to calculate the uncertainty ΔA for anyphysical quantity, let us make one more notational simplification. So far wehave used the symbol A for the quantum mechanical operator corresponding tothe physical quantity A. But since it will be self-evident from now on that whenwe write down the symbol of a physical quantity we actually refer to its quantummechanical operator, we will drop the “hat” and write simply A instead of Athroughout, except for special circumstances where the distinction needs to bemade. Thus, we will avoid using a cumbersome notation, and, as you will see,there will be practically no room for confusion.

2.4.2 The General Formula for Uncertainty

From the general statistical formula for the uncertaintyΔA of any statistical quan-tity A, we have

(ΔA)2 = ⟨A2⟩ − ⟨A⟩2.

For a quantum mechanical quantity A, we have seen that the mean value of itsstatistical distribution is given by the expression

⟨A⟩ = ∫ 𝜓∗(A𝜓)dx,

where A denotes now the quantum mechanical operator of the quantity A.Clearly, the square of a physical quantity is also a physical quantity; so theoperator of the former can only be the square of the operator of the latter. The


formula for the mean value holds for any physical quantity, therefore it holds forA2, and so we can use this formula to find ⟨A2⟩:

⟨A2⟩ = ∫ 𝜓∗ (A2𝜓)dx.

Once we know ⟨A⟩ and ⟨A2⟩, the uncertaintyΔA is immediately found from theexpression (ΔA)2 = ⟨A2⟩ − ⟨A⟩2. So we now have all the tools to calculate meanvalues and uncertainties of all physical quantities. In particular, we can calculatethe uncertainties Δx and Δp and confirm with concrete examples the renowneduncertainty principle

(Δx)(Δp) ≥ ℏ

2,

which becomes now a well-defined and mathematically verifiable (or falsifiable)inequality, since the quantitiesΔx andΔp can be calculated exactly once we knowthe wavefunction 𝜓(x). Before proceeding to prove the position–momentumuncertainty relation, we can test its correctness with as many examples as wewish. The following examples should help us apply these concepts and drawsome useful general conclusions.

Example 2.4 Show that the mean value ⟨p⟩ of momentum vanishes for each ofthe following cases: (a) The wavefunction 𝜓(x) is real. (b) 𝜓(x) is an even or anodd function.

Solution: (a) The general expression for the mean value ⟨p⟩ is

⟨p⟩ = ∫+∞

−∞𝜓∗ (p𝜓)dx = −iℏ∫

+∞

−∞𝜓∗ (x) 𝜓 ′(x)dx.

(a) If 𝜓(x) is a real function, then 𝜓∗(x) = 𝜓(x) and the integrand becomes

F(x) = 𝜓(x)𝜓 ′(x) = 12

ddx

(𝜓2),

so that

⟨p⟩ = − iℏ2 ∫

+∞

−∞

ddx

(𝜓2)dx = − iℏ2𝜓2

|||||

+∞

−∞

= 0.

In the last step we used the fact that a square-integrable function vanishes at ±∞,that is, 𝜓(±∞) = 0.

(b) We remind the readers that a function 𝜓(x) is even if 𝜓(−x) = 𝜓(x) andodd if 𝜓(−x) = −𝜓(x). From these definitions we can immediately see that thederivative of an even function is odd and vice versa.

Whether𝜓(x) is even or odd, the integrand F(x) = 𝜓∗(x)𝜓 ′(x) is odd, and, sincethe range of integration (−∞,+∞) is symmetric about the origin, the integralvanishes.

Both cases (a) and (b) appear frequently in quantum mechanics, so it is good forreaders to keep them in mind and spare themselves unnecessary computations.


Example 2.5 Show that the mean value of the squared momentum in aone-dimensional problem can always be written in the form

⟨p2⟩ = ℏ2 ∫+∞

−∞|𝜓 ′(x)|2 dx.

This expression is useful, since it saves us the trouble of calculating the secondderivative, and elucidates the expected positivity of ⟨p2⟩. The three-dimensionalgeneralization is

⟨p2⟩ = ℏ2 ∫ |∇𝜓(r)|2 dV .

Why should we have anticipated that ⟨p2⟩ is positive?

Solution: We begin with the expression for the mean value ⟨p2⟩

⟨p2⟩ = ∫+∞

−∞𝜓∗(p2𝜓)dx = −ℏ2 ∫

+∞

−∞𝜓∗𝜓 ′′ dx.

Using integration by parts,8 we obtain

⟨p2⟩ = −ℏ2𝜓∗𝜓 ′||+∞−∞ + ℏ2 ∫

+∞

−∞𝜓

′∗𝜓 ′ dx = 0 + ℏ2 ∫+∞

−∞|𝜓 ′(x)|2 dx,

which is what we intended to show. The three-dimensional case is proved in asimilar fashion, as the readers can verify.

The fact that ⟨p2⟩ ≥ 0 is not at all surprising. Since there may be circumstanceswhere ⟨p⟩ = 0, the expected positivity of the expression

(Δp)2 = ⟨p2⟩ − ⟨p⟩2

can only be guaranteed if ⟨p2⟩ ≥ 0 for every wavefunction 𝜓(x). The sameinequality holds for any other physical quantity A; that is,

⟨A2⟩ ≥ 0.

In fact the positivity of ⟨A2⟩ is nothing to write home about. If a physical quantity,such as A2, is positive in classical mechanics, it is quite reasonable to expect thatthe corresponding quantum mechanical mean value will also be positive.

Example 2.6 Calculate the uncertainty product (Δx)(Δp) for the Gaussianwavefunction

𝜓(x) = 4

√𝜆

𝜋e−𝜆x2∕2

and confirm that it satisfies the Heisenberg inequality, Δx ⋅ Δp ≥ ℏ∕2.

Solution: Let us note first that in an earlier example for the same wavefunction(Example 2.3) we found that Δx = 1∕

√2𝜆. So all we need to do now is calculate

8 We integrate the equality (fg)′ = f ′g + f g′ from a to b to obtain ∫ ba fdg = fg|ba − ∫ b

a (df )g, or,equivalently, ∫ b

a f g′dx = fg|ba − ∫ ba f ′gdx.


Δp. Given that𝜓(x) is both a real and an even function, the mean momentum ⟨p⟩vanishes (Example 2.4). We thus have

(Δp)2 = ⟨p2⟩ = ℏ2 ∫+∞

−∞|𝜓 ′(x)|2 dx

= ℏ2

√𝜆

𝜋⋅ 𝜆2 ∫

+∞

−∞x2e−𝜆x2 dx = ℏ2

√𝜆

𝜋⋅ 𝜆2 ⋅

12𝜆

√𝜋

𝜆= ℏ2𝜆

2and therefore

Δp =√⟨p2⟩ = ℏ

√𝜆

2⇒ (Δx)(Δp) = 1

√2𝜆

⋅ ℏ

√𝜆

2= ℏ

2.

Thus, the Heisenberg inequality, Δx ⋅ Δp ≥ ℏ∕2, is satisfied in this case as anequality. This feature (i.e., the equality) is particular to the Gaussian form of thewavefunction and should not be expected to hold in general.

Example 2.7 Show that if two wavefunctions 𝜓 and 𝜓 ′ differ by a constantphase, that is, 𝜓 ′ = eia𝜓 , they describe the same physical state. Does this holdwhen the phase is not constant but depends on position? Note that the symbol𝜓 ′ does not denote now the derivative of 𝜓 , but the new wavefunction eia𝜓 .

Solution: What we need to show here is that

⟨A⟩′ = ∫ (𝜓 ′)∗(A𝜓 ′)dx = ⟨A⟩,

which says that both wavefunctions 𝜓 and 𝜓 ′ produce the same mean value forany physical quantity A. Indeed,

⟨A⟩′ = ∫ (𝜓eia)∗(Aeia𝜓)dx =∫ 𝜓∗e−iaeia(A𝜓)dx =∫ 𝜓∗(A𝜓)dx = ⟨A⟩.

The crucial step in the given proof was the relation A(eia𝜓) = eia(A𝜓), whichholds because A is a linear operator. We can thus move the phase factor eia tothe left of A where it cancels with e−ia.

But things change when the phase is not constant, that is, when a = a(x). SinceA is a differential operator in general, it also acts on the phase factor. For example,for A = p = −iℏd∕dx we have

p(𝜓eia(x)) = eia(x)((p𝜓) + ℏa′(x)𝜓)⇒ ⟨p⟩′ = ⟨p⟩ + ℏ⟨a′(x)⟩.

As you see, for states that differ by a spatially dependent phase, the mean valuesof the momentum are different.

Problems

2.9 At a given time, the state of a particle is described by the wavefunction

𝜓(x) = Nxe−𝜆x2∕2.

2.5 Time Evolution of Wavefunctions and Superposition States 77

(a) Calculate the quantities ⟨x⟩, Δx, ⟨p⟩, Δp, and make sure that the uncer-tainty product Δx ⋅ Δp does not violate Heisenberg’s principle.

(b) In the vicinity of which points is it most likely to locate the particle ina position measurement?

2.10 Show that the mean value of momentum for every wavefunction of theform

Ψ(x) = 𝜓(x)eikx

is ℏk for every real and square-integrable wavefunction 𝜓(x).

2.11 Show that if the wavefunction describing the particle’s state is purely real,then the mean values of all three components of its angular momentumvanish.

2.12 Show that when the wavefunction describing the particle’s three-dimensional motion depends only on r (≡ √

x2 + y2 + z2 )—that is, when𝜓 = 𝜓(r)—then all three components of its angular momentum have awell-defined value that is equal to zero.

2.13 The electron’s wavefunction in the ground state of the hydrogen atom hasthe form

𝜓 = Ne−r∕a0 ,

where a0 = ℏ2∕me2 is the so-called Bohr radius. Calculate:(a) The mean distance ⟨r⟩ of the electron from the nucleus (the latter is

assumed stationary at the origin).(b) The distance (irrespective of direction) where the probability of locat-

ing the electron is maximum.(c) The mean values ⟨px⟩, ⟨py⟩, ⟨pz⟩ of the momentum components.

2.5 Time Evolution of Wavefunctions and SuperpositionStates

2.5.1 Setting the Stage

In the two preceding sections we were preoccupied with understanding thephysical meaning of a wavefunction at a fixed moment in time, and we thereforedid not pay any attention to the issue of time evolution. But the question “howdo things evolve in time?” remains central to any physical theory, and it isappropriate to address it now from a quantum mechanical perspective. Recallthat in one-dimensional Newtonian mechanics, we predict the future based onour knowledge of the initial position x(0) and the initial velocity 𝑣(0)—that is,x(0)—of a body at a certain time t = 0. This requirement stems from the factthat Newton’s equation mx = F(x) is a second-order differential equation with


respect to time, and so both x(0) and x(0) are needed to uniquely determine itssolution.

In quantum mechanics, Newton’s equation is replaced by the Schrödingerequation

iℏ𝜕𝜓𝜕t

= H𝜓, (2.22)

which is of first order with respect to time. It is reasonable, then, to expect thatthe necessary initial condition for the complete determination of its solutionwill be

𝜓(x, 0) = 𝜓(x), (2.23)

that is, the requirement is that we know the wavefunction at an initial time t = 0.So our mathematical task now is to solve (2.22) given (2.23); that is, to find theunique solution 𝜓(x, t) of (2.22), which for t = 0 is identical to the given initialwavefunction 𝜓(x).

From a physical perspective, once we know this solution we can predict, say,the mean position or the mean momentum of the particle after the lapse of timet. We can thus observe its “motion” in space, subject, of course, to the limitationsstemming from the statistical interpretation of 𝜓 (i.e., the uncertainty principle).

2.5.2 Solving the Schrödinger Equation. Separation of Variables

We can solve (2.22) easily with a method known as separation of variables, whichis in fact the only general method of exact solution for the partial differentialequations of mathematical physics. In this instance, the method consists ofsearching for special solutions with the separable form

𝜓(x, t) = 𝜓(x)T(t), (2.24)

that is, solutions that are products of one function depending only on x, andanother depending only on t. To check whether such solutions indeed exist, weplug (2.24) in (2.22) and obtain

iℏ𝜓(x)T(t) = (H𝜓)T , (2.25)

where we assumed that the Hamiltonian operator

H = − ℏ2

2m𝜕2

𝜕x2 + V (x) (2.26)

acts only on the function 𝜓(x), whereas T(t) is a constant with regard to thatoperator. By dividing both sides of (2.25) by the product 𝜓(x) ⋅ T(t) (= 𝜓(x, t)),we find

iℏ TT

= H𝜓𝜓. (2.27)

The left-hand side of (2.27) is a function of t only, whereas the right-hand sideis a function of x only. As one can easily verify, two functions of unrelatedvariables—like x and t in our case—can only be equal to each other if they are


equal to a common constant, usually called the separation constant. So in thecase at hand, we have

iℏ TT

= H𝜓𝜓

= E, (2.28)

where we have denoted the separation constant with the letter E, since it rep-resents the particle’s energy, as we will shortly see. From (2.28) two equationsemerge, namely,

iℏT = ET (2.29)

and

H𝜓 = E𝜓, (2.30)

which are both ordinary differential equations, in contrast to the originalequation that was a partial differential equation. And this is precisely the objec-tive of the method of separation of variables: to transform a partial differentialequation into a set of ordinary differential equations.

Looking back at (2.29) and (2.30), we immediately observe that the formerequation can be readily solved to find9

T(t) = e−iEt∕ℏ. (2.31)

The latter equation can be expanded and rewritten in the form(

− ℏ2

2md2

dx2 + V (x))

𝜓(x) = E𝜓(x)

or

− ℏ2

2m𝜓 ′′ + V (x)𝜓 = E𝜓 ,

or even in the standard form

𝜓 ′′ + 2mℏ2 (E − V (x))𝜓 = 0, (2.32)

whereby all terms have been moved over to the left side and the coefficient of thesecond derivative is unity, as is customary in the theory of ordinary differentialequations.

Equation (2.30), or its equivalent form (2.32), is the so-called time-independentSchrödinger equation, in contrast to the original equation (2.22), which is knownas the time-dependent Schrödinger equation. Note that the solution (2.31) of thetemporal equation (2.29) is always the same, and is independent of the poten-tial V (x), which is assumed as time-independent throughout. The precedingdiscussion suggests that the process of solving the time-dependent Schrödingerequation reduces fully to solving its time-independent version (2.30) or (2.32),which will therefore be our main concern from here on. In recognition of the factthat (2.32) is the equation to be solved for in each specific problem henceforth, it

9 There is a multiplicative constant in the general solution, which we omit for the moment bysetting it equal to unity. We will recover it shortly.


is also called the Schrödinger equation, without the qualifier time-independent,since the absence of t creates no confusion here.

We later show that (2.30) has solutions that vanish at ±∞, only if the parame-ter E (the energy of the particle, as we promised to show) takes distinct valuesE1,… ,En,…, with corresponding solutions 𝜓1,… , 𝜓n,…. We call the formereigenvalues and the latter eigenfunctions of the problem; or, more precisely, energyeigenvalues and energy eigenfunctions.

According to this discussion, there is an infinite sequence of separable solutionsof (2.22) in the form

𝜓n(x, t) = 𝜓n(x)e−iEnt∕ℏ (n = 1,… ,∞), (2.33)

whereby En and 𝜓n(x) are the eigenvalues and eigenfunctions, respectively, thatwe find by solving the time-independent Schrödinger equation (2.30) or (2.32).Given now that the original equation (2.22) is linear and homogeneous, everylinear combination of its solutions will also be a solution. Therefore, the function

𝜓(x, t) =∞∑

n=1cn𝜓n(x, t) =

∞∑

n=1cn𝜓n(x)e−iEnt∕ℏ (2.34)

will also be a solution of the time-dependent Schrödinger equation, for any val-ues of the constants c1,… , cn,…. Since there is an infinite number of arbitraryconstants in this solution, it is reasonable to go a step further and assume thatthe infinite series (2.34) represents the general solution of the time-dependentSchrödinger equation. This will indeed hold provided the constants c1,… , cn,…can always be chosen such that the infinite series (2.34) also satisfies the initialcondition (2.23) for any 𝜓(x). That is,

𝜓(x, 0) =

( ∞∑

n=1cn𝜓n(x)e−iEnt∕ℏ

)||||||t=0

=∞∑

n=1cn𝜓n(x) = 𝜓(x),

where 𝜓(x) is, of course, the initial wavefunction of the system. We are thus ledto a new and interesting mathematical question: Can an infinite series expansionof the form

𝜓(x) =∞∑

n=1cn𝜓n(x) (2.35)

reproduce any given wavefunction𝜓(x)? (Subject, of course, to the condition that𝜓(x) is square integrable, so that it may represent a physically realizable state.)

It turns out that it can. As we will shortly see, the corresponding coefficients cnfor the wavefunction 𝜓(x) are given by

cn = ∫+∞

−∞𝜓∗

n (x)𝜓(x)dx. (2.36)

Let us take this result as given for now. Then the solution of (2.22), subject to thecondition (2.23), becomes

𝜓(x, t) =∞∑

n=1cn𝜓n(x)e−iEnt∕ℏ, (2.37)


where En and 𝜓n(x) are the solutions of the time-independent Schrödingerequation, H𝜓 = E𝜓 , and cn are given by Eq. (2.36), where 𝜓(x) is the initial formof the wavefunction. Solving the time-dependent Schrödinger equation is thusequivalent to solving the time-independent equation (2.30) or (2.32). Our nextstep is to investigate the solutions of the latter equation.

2.5.3 The Time-Independent Schrödinger Equation as an EigenvalueEquation: Zero-Uncertainty States and Superposition States

The first thing to note is that Eq. (2.30) is remarkably simple: It consists of a linearoperator H that acts on a wavefunction 𝜓(x) to produce the same wavefunction𝜓 multiplied by a number (the eigenvalue E). Equations of this form are known aseigenvalue equations, in direct analogy to the eigenvalue equations for matrices,

AX = aX,

where the action of the matrix A on the column vector X has a similar effect: Itsimply multiplies it with a number (the eigenvalue a) leaving it otherwise intact.

The fact that the Schrödinger equation is an eigenvalue equation is of funda-mental physical importance, as can be deduced from the following theorem:

Theorem 2.2 The solutions of the Schrödinger equation represent physical stateswith well-defined energy, equal to the eigenvalue E.

Proof : We will show that for a wavefunction𝜓 satisfying the equation H𝜓 = E𝜓 ,the mean energy of the particle is equal to E, and the corresponding uncertaintyvanishes. Indeed, given that the quantum mechanical operator of energy is theHamiltonian, H , we have

⟨E⟩ ≡ ⟨H⟩ = ∫ 𝜓∗(H𝜓) dx = ∫ 𝜓∗(E𝜓) dx

= E∫ 𝜓∗𝜓 dx = E∫ |𝜓|2 dx = E.

For the squared uncertainty, (ΔE)2, given by

(ΔE)2 ≡ (ΔH)2 = ⟨H2⟩ − ⟨H⟩2,

we have already shown that ⟨H⟩ = E. All we have to do now is find ⟨H2⟩ byemploying again the mean-value formula

⟨H2⟩ = ∫ 𝜓∗(H2𝜓)dx = ∫ 𝜓∗(H(H𝜓))dx

= ∫ 𝜓∗(HE𝜓)dx = E∫ 𝜓∗(H𝜓)dx = E∫ 𝜓∗(E𝜓)dx

= E2∫ 𝜓∗𝜓dx = E2∫ |𝜓|2dx = E2.

Therefore,

(ΔE)2 = ⟨H2⟩ − ⟨H⟩2 = E2 − E2 = 0,


as we promised to show. But if the uncertainty of some statistical quantity van-ishes, then the only possible value for that quantity is its mean value. Therefore,in this case, where ⟨E⟩ = E, the only possible outcome of an energy measurementis the eigenvalue E corresponding to the eigenfunction 𝜓(x). □

If we now assume that the Schrödinger equation has physically acceptablesolutions only for a discrete set of values E1,… ,En,…, we immediately realizethat these are in fact the only possible values an energy measurement could yield.We say that the energy is a quantized quantity or that it has a discrete spectrum.But this is not always the case, as we shall see later in the book.

The next question is this: What would a measurement of the energy yield whenthe wavefunction of our particle is not an eigenfunction 𝜓n, but some arbitrarywavefunction 𝜓? In this case, we have seen that we can write 𝜓 as an infinite—ingeneral—superposition of eigenfunctions of the form

𝜓 =∞∑

n=1cn𝜓n. (2.38)

What if we attempt to measure the particle’s energy in a superposition state like(2.38)? To answer this crucial question, we need one basic property of the eigen-functions 𝜓n of the Schrödinger equation, H𝜓n = En𝜓n. This property, known asthe orthogonality relation of the eigenfunctions, states that

∫+∞

−∞𝜓∗

n (x)𝜓m(x)dx = 0 (n ≠ m), (2.39)

that is, the integral of the product10 of two eigenfunctions corresponding to dif-ferent eigenvalues vanishes. In the following section we prove this property andexamine its meaning further.

Equation (2.39) allows us to readily prove Eq. (2.36) above. Starting with

𝜓 =∑

mcm𝜓m

we multiply both sides by 𝜓∗n (x), and integrate from −∞ to +∞:

∫+∞

−∞𝜓∗

n𝜓 dx = ∫+∞

−∞𝜓∗

n

(∑

mcm𝜓m

)

dx =∑

mcm ∫

+∞

−∞𝜓∗

n𝜓m dx. (2.40)

The last integral vanishes for m ≠ n and is unity for m = n, since it then becomesthe normalization integral of 𝜓n. So the last term of (2.40) is equal to cn, andtherefore cn = ∫ +∞

−∞ 𝜓∗n𝜓ndx, as we wanted to show.

We can now answer the question we posed earlier about the allowed values ofE that a measurement could yield. For this we need to calculate the mean energy

10 We clearly imply here that the first term in this product is always the complex conjugate of theeigenfunction. This is in direct analogy to the formula of the mean value ⟨A⟩ = ∫ 𝜓∗(A𝜓)dx, or evenof the normalization integral ∫ |𝜓|2dx = ∫ 𝜓∗𝜓dx = 1, whereby we always take the complexconjugate of the function appearing first, so that the end result is guaranteed to be a real number.


⟨E⟩ = ⟨H⟩ for a superposition state of the form (2.38):

⟨E⟩ = ⟨H⟩ = ∫ 𝜓∗(H𝜓) dx = ∫(∑

nc∗n𝜓∗

n

)(

H

(∑

mcm𝜓m

))

dx

= ∫(∑

nc∗n𝜓∗

n

)(∑

mcm(H𝜓m)

)

dx =∑

n,mc∗ncm ∫ 𝜓∗

n (H𝜓m) dx

=∑


n Em𝜓m dx =∑

n,mEmc∗ncm ∫ 𝜓∗

n𝜓m dx.

Since the operator H is linear, it commutes with the summation symbol, which inturn commutes with the integration symbol, provided both processes (summa-tion and integration) converge. Note also that when we multiply two sums, thesummation index used should be different for each sum. (Why?)

Based on the orthogonality relation (2.39) in this double sum, only terms withm = n survive. We thus obtain

⟨E⟩ =∑

nEn|cn|

2. (2.41)

This expression is identical to the mean-value formula of an arbitrary (discrete)statistical quantity A:

⟨A⟩ =∑

nanPn, (2.42)

whose possible values are an, and the respective probabilities Pn. By comparing(2.41) and (2.42) we immediately conclude the following.

Theorem 2.3 (i) The only values that an energy measurement could possiblyyield are the eigenvalues En of the corresponding quantum mechanical operator H,that is, the values resulting from the solution of the Schrödinger equation. (ii) Fora given wavefunction 𝜓 , the probability that a measurement yields the eigenvalueEn is

Pn = |cn|2

(

cn = ∫ 𝜓∗n𝜓 dx

)

.

The probability is thus equal to the square of the absolute value of the coefficientcn of the eigenfunction𝜓n that appears in the expansion of𝜓 in terms of the energyeigenfunctions.

Example 2.8 The state of a particle at a moment in time is given by the wave-function

𝜓 = 1√

3𝜓1 +

√23𝜓2, (1)

where 𝜓1, 𝜓2 are the energy eigenfunctions with eigenvalues E1,E2, respectively.

(a) What would be the possible results of an energy measurement in state (1),and what are the respective probabilities?


(b) Calculate the mean energy and the energy uncertainty of the particle in state(1). To obtain numerical results, assume that E1 = 3 and E2 = 6 in some sys-tem of units.

Solution: As we discussed, (1) is a superposition state containing only theeigenfunctions 𝜓1, 𝜓2. Therefore, the only possible values that a measurement ofenergy could yield, are the eigenvalues E1 and E2. The corresponding probabili-ties, P1 and P2, are given by the squares of the absolute values of the coefficientsc1 = 1∕

√3 and c2 =

√2∕3:

P1 = |c1|2 = 1

3, P2 = |c2|

2 = 23.

Note that P1 + P2 = 1, as it should. (There are only two possible outcomes for anenergy measurement, E1 or E2, so the sum of the respective probabilities mustequal unity.)

Once we know the possible values for the energy and the corresponding prob-abilities, it is trivial to calculate the mean value ⟨E⟩ and the uncertainty ΔE, usingstandard statistical formulas:

⟨E⟩ = P1E1 + P2E2 = 13⋅ 3 + 2

3⋅ 6 = 5

and

⟨E2⟩ = P1E21 + P2E2

2 = 13⋅ 32 + 2

3⋅ 62 = 27,

so that

(ΔE)2 = ⟨E2⟩ − ⟨E⟩2 = 27 − 25 = 2 ⇒ ΔE =√

2.

Here is an extra question for the readers: What tests would you run to make surethat these results, ⟨E⟩ = 5 and ΔE =

√2, are free from some “grave” error? For

example, if you had found ⟨E⟩ = 4 and ΔE =√

10, would you be suspicious thatsomething went wrong? More generally, what can you say about the maximumpossible value of the uncertainty ΔA of a statistical quantity having a discrete,say, distribution? Can you show—or, even better, can you make it qualitativelyevident—that it will always be

ΔA ≤ amax − amin

2, (2)

where amin and amax are the minimum and maximum values of the statisticalquantity at hand? When does the equality in (2) hold?

Problems

2.14 You are given the general superposition state at t = 0

𝜓 = c1𝜓1 + c2𝜓2, (1)

where 𝜓1 and 𝜓2 are energy eigenstates with eigenvalues E1 and E2 respec-tively.


(a) Show that the energy uncertainty in state (1) is equal toΔE = |c1c2| |E1 − E2|.

For what values of c1 and c2 is the energy uncertainty maximized?(b) What is the value of ΔE after time t?

2.15 Somebody claims that the two superposition states

𝜓 = 1√

3𝜓1 +

√23𝜓2 and 𝜙 = 1

√3𝜓1 + i

√23𝜓2

are physically equivalent, since the presence of i in the second state doesnot change the probabilities to find the particle in one or the other energyeigenstate. So both ⟨E⟩ and ΔE are the same in the two states. Do youagree?

2.5.4 Energy Quantization for Confined Motion: A Fundamental GeneralConsequence of Schrödinger’s Equation

One last issue we need to discuss here is energy quantization. We claimedearlier that the Schrödinger equation has physically acceptable solutions onlyfor a discrete sequence of values E1,E2,… ,En,…. Actually, this is true only forsolutions that describe confined motion of the particle (i.e., solutions that vanishat ±∞). We will prove this quantization phenomenon for the slightly simpler casewhereby the solutions of the Schrödinger equation are required to vanish at twofinite points a and b, not at ±∞. The problem at hand now is to solve the equation

𝜓 ′′ + 2mℏ2 (E − V (x))𝜓 = 0 (2.43)

in the interval a ≤ x ≤ b, subject to the constraint that the solutions vanish atthe endpoints of this interval, so that the following boundary conditions are met:

𝜓(a) = 0, 𝜓(b) = 0. (2.44)We will then prove the following:

Theorem 2.4 For the Schrödinger equation to have vanishing solutions at theendpoints of a finite interval, the energy of the particle can only take a discretesequence of values.

Proof : Given that (2.43) is a second-order linear (and homogeneous) equation, ithas two linearly independent solutions 𝜓1(x,E) and 𝜓2(x,E). These solutions alsodepend on the particle’s energy, since it appears as a parameter in the Schrödingerequation. The general solution of (2.43) can then be written in the form

𝜓(x,E) = c1𝜓1(x,E) + c2𝜓2(x,E),and the arbitrary constants c1, c2 must now be chosen so that the boundary con-ditions (2.44) are satisfied. So, we require that

𝜓(a,E) = c1𝜓1(a,E) + c2𝜓2(a,E) = 0,𝜓(b,E) = c1𝜓1(b,E) + c2𝜓2(b,E) = 0.


We thus obtain the following homogeneous system of equations(𝜓1(a,E) 𝜓2(a,E)𝜓1(b,E) 𝜓2(b,E)

) (c1c2

)

= 0,

which (as is well known) has a nonvanishing solution (c1, c2) only when the deter-minant of the matrix of its coefficients vanishes. That is,

Δ(E) = det(𝜓1(a,E) 𝜓2(a,E)𝜓1(b,E) 𝜓2(b,E)

)

= 𝜓1(a,E)𝜓2(b,E) − 𝜓2(a,E)𝜓1(b,E) = 0. □

The roots E1,E2,… ,En,… of the function Δ(E) form a discrete set because ifthey did not, then Δ(E) would be identically zero. The conclusion that the energyspectrum is discrete needs to be modified when we are dealing with an infiniteinterval. In that case, the existence of vanishing solutions at±∞depends criticallyboth on the form of the potential and on the energy range we are looking at. Theenergy spectrum can then be discrete, continuous, or even mixed,11 dependingon whether there exist vanishing solutions at ±∞. We will study this situationextensively in Chapters 4 and 5. As we shall see there, if the particle’s energy issuch that the corresponding classical motion is confined, the quantum “motion”will also be confined (i.e., the particle’s wavefunction vanishes at infinity) andthe energy will be quantized. As we emphasized in the previous chapter, energyquantization—like frequency quantization in classical standing waves—is alwayspresent when the wave is localized in a finite region of space.

2.5.5 The Role of Measurement in Quantum Mechanics: Collapse of theWavefunction Upon Measurement

The preceding discussion raises a fundamental question regarding the role ofmeasurement in a quantum system. The question becomes rather simple for thecase of the superposition state of Example 2.8. We claimed that in one-third ofthe measurements we would obtain the value E1, and in the other two-thirds wewould get E2. What would be the state (i.e., wavefunction) of the particle aftermeasuring one or the other value? We can easily answer this question with the fol-lowing reasoning that we already presented in Chapter 1: A second measurementought to confirm the first one with 100% certainty. But this can only happen if thewavefunction resulting from the first measurement is identical to the eigenfunctionof the measured eigenvalue. In this case—see Example 2.8—this implies that oncea measurement yields the value E1, the wavefunction of the particle afterwardsmust be 𝜓1; when a measurement yields E2, the wavefunction after that measure-ment must be 𝜓2. This is the famous collapse of the wavefunction that has beendiscussed in Chapter 1, and is now shown schematically in Figure 2.2.

It transpires from this discussion that the process of measurement plays a dras-tically different role in quantum mechanics than in classical physics. In classicalphysics, we can always decouple the influence of the measuring device on the

11 A mixed spectrum is discrete in one energy range, and continuous in another.


ψ = ∑ cnψnnEn Enψn ψn

Pn = |cn|2 Pn = 1

Figure 2.2 Collapse of the wavefunction upon measurement. When the particle is in thesuperposition state 𝜓 =

∑ncn𝜓n, the first measurement can yield any one of the possible

outcomes En(n = 1, 2,…) with a priori probabilities Pn = |cn|2. Once a measurement yields a

particular value, however, the wavefunction “produced” by the measuring device is theeigenfunction of the eigenvalue that was just measured. And because of this fact, a seconddevice that performs the measurement right after the first one will confirm its outcome with100% probability.

measured system, either by making it negligibly small, or by calculating and cor-recting for it afterwards. (As a standard example, you can think of the processof measuring the temperature of a solid body using a thermometer whose heatcapacity is known.) In contrast, in quantum mechanics, we can neither makethe influence of the measuring device negligibly small, nor can we predict it andaccount for it after the measurement. Indeed, according to the discussion, theinfluence of the measuring device on the measured system will always take theform

𝜓

(

=∑

ncn𝜓n

)

→ 𝜓n.

Clearly, such an influence is neither negligible nor predictable. When the state𝜓 =

∑cn𝜓n enters the measuring device, nobody can predict which one of the

eigenfunctions 𝜓n will be selected, and therefore which one of the possible out-comes will appear on the “screen” of our measuring device. The outcome of anyparticular measurement in quantum mechanics is fundamentally unpredictable.

If we were to examine any particular measurement process from a physicalperspective, we would arrive at the same conclusions regarding the nature ofquantum measurement. A typical example is measuring the position of a particleusing an appropriate light beam (the so-called Bohr’s microscope). To measurethe particle’s position with a desirable accuracy Δx we must use light with acorrespondingly small wavelength: specifically, it should be 𝜆 ≤ Δx. What doesthis mean? It means that the photons of the light beam will be energetic enough(𝜖 = hf = hc∕𝜆) to “kick” the particle far away from its original position once theyimpinge on it. Thus, the influence of this measurement process on the particleis neither negligible, nor predictable. In fact, the quantum measurement deter-mines the state of the quantum system after the measurement, not before. That isthe essence of the measurement principle. After a measurement yields the resultEn the system’s state will surely be described by the eigenfunction 𝜓n. And since𝜓n is an eigenfunction of the measured quantity, each subsequent measurementcan only confirm the outcome of the first measurement, with 100% probability.

Some readers may have suspected that most results of this and the previoussection hold for any physical quantity A, not simply for the energy. The grounds


for this generalization are rather obvious: Nowhere in the proofs (wherever aproof was provided) did we specify that the operator H need be the Hamiltonianoperator—it could be the operator of any physical quantity for that matter. Wecan therefore conclude that this discussion holds for other physical quantities,too. We provide more evidence for this generalization in the following section.The examples given below serve not only as a useful application of the ideas wehave presented so far but also as an opportunity for suggesting some necessarygeneralizations.

Example 2.9 In classical wave theory, the instantaneous form of a wave with awavelength 𝜆 (and wavenumber k = 2𝜋∕𝜆) is given by

u(x) = A sin kx or A cos kx.

One would expect, therefore, that in quantum mechanics, the instantaneous formof the wavefunction of a particle with a well-defined momentum p would be

𝜓(x) = A sin kx or A cos kx, (1)

where k = p∕ℏ due to wave–particle duality (p = ℏk).Show that the correct form of the particle’s wavefunction is not (1), but rather

𝜓(x) = Aeikx, (2)

while the well-defined momentum of the particle is p = ℏk.

Solution: As we discussed before, a physical quantity can only have a well-definedvalue if the wavefunction describing the particle’s state is an eigenfunction ofthe corresponding quantum mechanical operator. Given that the momentumoperator is

p = −iℏ ddx

we can immediately guess that neither the sine nor the cosine function couldever be its eigenfunctions. The differentiation would turn one function into theother, contrary to the requirement that an eigenfunction remain “intact” (up toa multiplicative constant) when the relevant operator acts on it. On the otherhand, the complex expression (2) is indeed a momentum eigenfunction, sincethe differentiation leaves the exponential “intact,” apart from multiplying it by afactor ik. In particular, for the wavefunction (2) we have

p𝜓 = ℏk𝜓,

which means that (2) is indeed a state of well-defined momentum with eigenvaluep = ℏk.

You may have noticed, however, that (2) does not actually vanish at infinity, andthus cannot be a physically acceptable wavefunction according to what we havesaid so far. Besides, this wavefunction corresponds to infinite total probability,that is, ∫ |𝜓|2dx = ∞, since 𝜓 = |A| = constant. This is a result we should haveactually expected. Even in classical physics, a wave with a well-defined wavelength(or wavenumber) is a sine wave extending throughout space; it has infinite total


energy, which is the classical analog of infinite total probability. Despite this prob-lem, we use sine waves extensively in classical physics. They serve as mathemat-ically useful idealizations of physically realizable waves. Besides, we can alwaysconstruct physical waves by a suitable superposition of a number of plane (sine)waves. For the same reason, in quantum mechanics, we do not reject wavefunc-tions of the form (2). In fact, we use them extensively as useful idealizations ofphysically realizable states. In this spirit it is reasonable to relax the requirementthat only square-integrable wavefunctions be allowed, and accept also wavefunc-tions that do not vanish at infinity but are, nevertheless, finite everywhere.

Example 2.10 The momentum of a particle (in one dimension) was measuredto be p0. In which state was the particle before the measurement? After the mea-surement? The same question if the measurement yielded the particle’s positionto be x = a.

Solution: According to the measurement principle (collapse of wavefunction),the wavefunction of the particle, after the measurement of its momentum yieldedp0, is the corresponding eigenfunction

𝜓p0= Aeip0x∕ℏ, (1)

where A is a constant. But the result of this measurement does not allow us tosay anything at all about the wavefunction before the measurement. The particlecould have any wavefunction 𝜓(x), which can always be written as

𝜓(x) =∑

pc(p)𝜓p(x) = ∫ c(p)eipx∕ℏ dp,

namely, as a continuous superposition of momentum eigenstates. The mea-surement process selects each time one of these eigenstates in an unpredictablefashion.

If instead of momentum we measured the particle’s position yielding the valuex = a, then the wavefunction 𝜓a(x) immediately after the measurement shouldsatisfy the eigenvalue equation of the position operator x = x, where a is theeigenvalue. We thus have,

x𝜓a(x) = a𝜓a(x) ⇒ (x − a)𝜓a(x) = 0,

which means that 𝜓a(x) vanishes for each x ≠ a, and probably diverges for x = a,since the probability of locating the particle at any other position is then zero.This type of function, vanishing everywhere except at one point where it becomesinfinite, is unusual from a mathematical point of view, yet it arises quite naturallyin quantum mechanics.

To describe such singular wavefunctions mathematically, P. Dirac introducedin 1927 the so-called Dirac delta function; it is denoted as 𝛿(x − a) and sketchedin Figure 2.3.

The standard way to define 𝛿(x)mathematically is to take the limit of a sequenceof ordinary functions as these become gradually narrower and taller, so that theirintegral from −∞ to +∞, that is, the area under the curve, remains constant, and


δ(x) δ(x – a)

x

x = 0 x = a

(a) (b)

Figure 2.3 Dirac delta function. (a) The function 𝛿(x). (b) The function 𝛿(x − a). Both (a) and (b)describe the same function, but are centered at different points. The function 𝛿(x) is centeredat x = 0, so it becomes infinite there and vanishes everywhere else, while 𝛿(x − a) is centeredat x = a.

equal to unity. The limit of that sequence is the generalized function 𝛿(x), whichis thus defined formally from the condition

∫+∞

−∞𝛿(x)dx = 1

and likewise for 𝛿(x − a); it is again ∫ 𝛿(x − a)dx = 1. More generally,

∫+∞

−∞𝛿(x)f (x)dx = f (0),

where f (x) is an arbitrary ordinary function. (Can you explain why?)To return to our question: What is the wavefunction of a particle after its posi-

tion was measured to be x = a? The answer is

𝜓a(x) = 𝛿(x − a), (2)

where 𝛿(x − a) is the Dirac delta function, centered at x = a.A direct consequence of our discussion so far is that in quantum mechan-

ics, it is absolutely impossible to determine simultaneously the exact positionand the exact momentum of a particle. If we attempted to measure the parti-cle’s momentum with perfect accuracy, such that Δp = 0, then its wavefunctionafter the measurement would have the form (1). The corresponding probabilitydensity for the particle’s position would then be

P(x) = |𝜓p0(x)|2 = |A|2,

which is constant throughout space. Thus, the particle can be found anywherein space, and its position is completely undetermined (Δx = ∞). By measuringexactly the particle’s momentum we lost all hope of knowing where it might be.If, on the other hand, we attempted to measure its position with perfect accuracy,then its wavefunction after the measurement would take the form (2). The wave-functions (2) and (1) are exact opposites: While (2) is completely localized at one


point, (1) is spread out throughout space. What would be the momentum uncer-tainty for the particle having the wavefunction (2)? To begin with, since this isnot a momentum eigenfunction, the corresponding momentum uncertainty Δpwould surely be nonzero. In fact, Δp = ∞, given that the position uncertaintyfor (2) vanishes (i.e., Δx = 0), since that wavefunction is a position eigenfunctionwith eigenvalue a.

The last two examples allowed us to discuss the uncertainty principle, but theyalso prompt us to comment on another important element of the theory: the exis-tence of physical quantities, such as the position and momentum, whose spec-trum is continuous (as opposed to discrete). For example, we saw that the eigen-value equations for these quantities (position and momentum) can have solutionsfor any value of the eigenvalue parameters, p0 or a, in the range (−∞,+∞). Thisis something we should expect, at least for position: A hypothetical quantizationof position would imply that the particle could only be located in some discretepoints in space and not in others. Such an occurrence has never been observed(and would actually have some rather bizarre consequences for physics). Positionis thus a continuous variable in quantum mechanics as in classical mechanics. Thesame holds true for momentum.

2.5.6 Measurable Consequences of Time Evolution: Stationaryand Nonstationary States

We are now ready to discuss the physical meaning of the solutions we obtained.We begin with the separable solutions,

𝜓n(x, t) = 𝜓n(x)e−iEnt∕ℏ, (2.45)

for which we can immediately see that the probability distribution of locating theparticle in space is

Pn = |𝜓n(x, t)|2 = |𝜓n(x)|2|e−iEnt∕ℏ|2 = |𝜓n(x)|2.

Clearly, Pn is independent of time, since the time dependence has the form ofa phase factor12 e−iEnt∕ℏ, whose absolute value is unity. The mean value of anyphysical quantity is also independent of time,

⟨A⟩ = ∫ 𝜓∗n (x, t)(A𝜓n(x, t)) dx = ∫ 𝜓∗

n (x)eiEnt∕ℏ(A𝜓n(x))e−iEnt∕ℏ dx

= ∫ 𝜓∗n (x)(A𝜓n(x)) dx,

as the temporal phase factor cancels out because of the complex conjugate formof the wavefunction to the left of A in the given integral.

It is evident, then, that the time evolution of states described by the separablewavefunctions (2.45) has no physical consequence whatsoever. Such states aretherefore called stationary, since, basically, nothing changes with time for them!

This conclusion follows naturally from our discussion in the previoussection. Given that the eigenfunctions 𝜓n(x) have well-defined energy En,

12 The term phase factor refers to any complex number of the form ei𝜙 whose absolute value (in thecomplex sense) is always one.


they must also have a well-defined frequency 𝜔n = En∕ℏ, since E = ℏ𝜔 fromwave–particle duality. The time evolution of such states would then have theform 𝜓n(x, t) = 𝜓n(x)e±i𝜔nt , as in classical waves, but with the crucial differencethat in the quantum case the time evolution necessarily has a complex form.Recall that in classical waves, we use the complex exponential exp(±i𝜔nt) forconvenience, but we always take its real or imaginary part in the end, that is, thecosine or the sine, which is in itself a solution of the classical wave equation. Inthe case of “quantum waves,” a time evolution of the classical form

𝜓n(x, t) = 𝜓n(x) sin𝜔t (or cos𝜔t) (2.46)

is impossible, for two equally fundamental reasons: (i) The form (2.46) is a realfunction and, as such, it cannot be a solution of the time-dependent Schrödingerequation as we have repeatedly stressed. (ii) A sinusoidal time evolution, such asin (2.46), implies that the wavefunction 𝜓n(x, t) becomes identically zero at vari-ous moments in time, namely, at t = 0, 𝜋

𝜔,

2𝜋𝜔,… This is physically unacceptable,

of course, since it would mean that the particle actually disappears from timeto time!

So we should have expected a time evolution of the form of (2.45) over (2.46) forthe eigenfunctions of the Schrödinger equation corresponding to a well-definedenergy of the particle, and, therefore, to a well-defined frequency of the quan-tum wave.

The fact that the passage of time bears no physical consequence on the statesdescribed by (2.45)—the stationary states as we called them—is certainly some-thing we are not used to in classical physics where there is no analogous behavior.But let us clarify that this “stationary” behavior is the exception rather than therule. For the general solution of the Schrödinger equation

𝜓(x, t) =∑

ncn𝜓n(x)e−iEnt∕ℏ,

the time evolution no longer has the form of a single phase factor that, neces-sarily, cancels out with its own complex conjugate in the quantum expressionsof measurable quantities, such as the position probability density, mean values,uncertainties, and so on. For example, the mean value (at time t) of some physicalquantity A is13

⟨A⟩t = ∫ 𝜓∗(x, t)(A𝜓(x, t)) dx

= ∫(∑

ncn𝜓ne−iEnt∕ℏ

)∗ (

A

(∑

mcm𝜓me−iEmt∕ℏ

))

dx

=∑

n,mc∗ncmei(En−Em)t∕ℏ ∫ 𝜓∗

n (A𝜓m) dx

≡ ∑

n,mc∗ncmAnmei𝜔nmt , (2.47)

13 As we have already noted, when we omit the integration limits we imply that the integration isfrom −∞ to +∞. It should also be clear that we have only limited the discussion to one dimensionfor simplicity; the results obtained here are equally valid for the 2D and 3D cases as well.


where Anm = ∫ 𝜓∗n (A𝜓m)dx are the so-called matrix elements of the operator A,14

and 𝜔nm = (En − Em)∕ℏ are the so-called Bohr frequencies of the problem. (Thereason for this name is that these are the frequencies of photons emitted whenthe particle jumps from the state with energy En to the state with energy Em.)

Expression (2.47) means that, for an arbitrary superposition state 𝜓(x, t), thereare measurable consequences of time evolution, since the mean values of thephysical quantities are changing in time. For instance, if A = x, then the meanposition of the particle at time t would be written as

⟨x⟩t =∑

n,mc∗ncmxnmei𝜔nmt. (2.48)

In particular, if the initial wavefunction is a superposition of only two eigenfunc-tions of the form

𝜓 = c1𝜓1 + c2𝜓2 (2.49)

then only the terms with n and m equal to 1 and 2 survive in the double infinitesum (2.48). The mean position of the particle is then

⟨x⟩t = |c1|2x11 + |c2|

2x22 + c∗1c2x12ei𝜔t + c1c∗2x21e−i𝜔t, (2.50)

where, for simplicity, we have set 𝜔12 = 𝜔. And since

x21 = ∫ 𝜓∗2 (x𝜓1)dx =

(

∫ 𝜓∗1 (x𝜓2)dx

)∗

= x∗12,

the coefficients of the last two terms in (2.50) are complex conjugates of eachother, so we can use the substitution

z = c∗1c2x12 = 𝜌ei𝜙, z∗ = c1c∗2x21 = 𝜌e−i𝜙

to rewrite (2.50) in the form

⟨x⟩t = 𝛼 + 𝛽 cos (𝜔t + 𝜙), (2.51)

where we have also set

𝛼 = |c1|2x11 + |c2|

2x22, 𝛽 = 2𝜌

in order to make the essential feature of (2.51) stand out better: The mean posi-tion of the particle in a superposition state, such as (2.49), oscillates in time with aBohr frequency𝜔 corresponding to the energy difference between the two super-position states, that is, 𝜔 = (E1 − E2)∕ℏ. Hence, as soon as the particle lies ina superposition of energy eigenstates, the time evolution has nontrivial conse-quences in the measurable quantities of our system.

A simple application of this discussion is given in the following example.

14 This terminology will become more meaningful at the end of this chapter, where we prove thatany quantum mechanical operator can be represented by a matrix with elements given by thisformula. For the time being, we just use this odd name “matrix elements” to refer to the numbersAnm = ∫ 𝜓∗

n (A𝜓m)dx.


Example 2.11 At time t = 0, a particle is in a state described by the superposi-tion of the energy eigenstates 𝜓1 and 𝜓2,

𝜓 = 12𝜓1 +

√3

2𝜓2 (1)

with eigenvalues E1 and E2, respectively.

(a) What is the time evolution of state (1)?(b) What is the mean energy and the corresponding uncertainty of the particle

at t = 0, and after time t?(c) How does time evolution affect the mean position of the particle?

Solution: (a) Since, as we saw earlier, each eigenfunction 𝜓1, 𝜓2 evolves in timewith a phase factor e−iEt∕ℏ, the time evolution of a superposition state such as (1)will itself be a superposition of the time-evolved forms of 𝜓1 and 𝜓2. So we have

𝜓(x, t) = 12𝜓1(x)e−iE1t∕ℏ +

√3

2𝜓2(x)e−iE2t∕ℏ. (2)

(b) Since (2) is a superposition state of the form

𝜓(x, t) = c1(t)𝜓1(x) + c2(t)𝜓2(x)

with

c1(t) =12

e−iE1t∕ℏ, c2(t) =√

32

e−iE2t∕ℏ

the probabilities of measuring the eigenvalues E1 and E2 at time t will be

P1 = |c1(t)|2 = 14|e−iE1t∕ℏ|2 = 1

4, P2 = |c2(t)|2 = 3

4|e−iE2t∕ℏ|2 = 3

4.

These values are independent of time, as we should expect, since energy is a con-served quantity of the system. For a statistical theory such as quantum mechanics,this can only mean that the statistical distribution of the possible results of energymeasurements will be constant in time. The mean energy ⟨E⟩ and the correspond-ing uncertainty ΔE are, indeed, independent of time

⟨E⟩ = 14

E1 +34

E2, ΔE =√

34

|E1 − E2|

as we just argued.(c) Contrary to energy, the position measurements of the particle are time

dependent, as is their mean value, ⟨x⟩t , for the reasons we discussed above. Themean position of the particle oscillates about an initial value with frequency𝜔 = |E1 − E2|∕ℏ.

Problems

2.16 Decide whether the energy spectrum is discrete, continuous, or mixed foreach of the following one-dimensional potentials. In each case give theenergy range where the spectrum is discrete or continuous.

2.6 Self-Consistency of the Statistical Interpretation 95

(a) V = 12kx2,

(b) V = − V0

cosh 𝜆x,

(c) V = V0

cosh 𝜆xThe parameters k, V0, and 𝜆 are positive.

2.17 A position measurement in one dimension has located the particle at thepoint x = a. What is the wavefunction of the particle just after the mea-surement? If we wait some time and repeat the measurement on the sameparticle we measured before, should we expect to find it at x = a again?

2.18 Show that the momentum eigenstates in one dimension

𝜓p(x) = Aeipx∕ℏ (1)

are also energy eigenstates if the particle is moving with no force acting onit (free motion). Use this fact to write down the time evolution of state (1).

2.6 Self-Consistency of the Statistical Interpretationand the Mathematical Structure of Quantum Mechanics

2.6.1 Hermitian Operators

Before closing the chapter, it would be appropriate to pause for a while and reflecton whether the statistical interpretation of quantum mechanics, which we havejust presented, is self-consistent. The first question to be examined is this: Giventhat the wavefunctions 𝜓(x, t) are necessarily complex and that the quantumoperators A(x, p) = A(x,−iℏd∕dx) also include the imaginary unit i, how can weguarantee that the mean value

⟨A⟩ = ∫+∞

−∞𝜓∗(x, t)(A𝜓(x, t))dx

will always be real, as is required for any measurable physical quantity?As we will shortly see, the fact that all mean values are real is warranted by

the following common property of quantum mechanical operators, known as thehermitian property.

Definition 2.1 A (linear) operator A, acting on a space of complex-valued func-tions, is called hermitian if for each pair of such functions 𝜓(x) and 𝜙(x), thefollowing relation holds:

∫ 𝜓∗(A𝜙)dx = ∫ (A𝜓)∗𝜙dx, (2.52)

that is, the action of the operator can be transferred, without a change in the endresult, from one function of an integral like (2.52) to the other, in the manner shown.


If a quantum mechanical operator satisfies (2.52) (i.e., it is hermitian), we caneasily show that its mean value ⟨A⟩ is always real. Indeed, by applying (2.52) for𝜙 = 𝜓 , we obtain

∫ 𝜓∗(A𝜓)dx = ∫ (A𝜓)∗𝜓 dx =(

∫ (A𝜓)𝜓∗ dx)∗

≡(

∫ 𝜓∗(A𝜓)dx)∗

.

From the definition of the mean value, ⟨A⟩ = ∫ 𝜓∗(A𝜓)dx, we then have⟨A⟩ = ⟨A⟩∗,

which surely means that ⟨A⟩ is a real number, since it is equal to its own complexconjugate.15 The question now becomes whether quantum mechanical operatorshave this property or not. For the position operator, A = x, (2.52) is clearly satis-fied, while the mean value

⟨x⟩ = ∫ 𝜓∗(x𝜓)dx = ∫ x|𝜓|2 dx

is real for any wavefunction 𝜓(x). The hermitian property can easily beshown for the momentum operator also. The formal proof goes as follows.We want to show that for p = −iℏd∕dx we have

∫+∞

−∞𝜓∗(p𝜙)dx = ∫

+∞

−∞(p𝜓)∗𝜙dx.

Beginning with the left-hand side—and performing an integration by parts—weobtain successively,

∫+∞

−∞𝜓∗(p𝜙)dx ≡ ∫

+∞

−∞𝜓∗(−iℏ𝜙′)dx = −iℏ∫

+∞

−∞𝜓∗𝜙′ dx

= −iℏ⎛⎜⎜⎜⎝

−∫+∞

−∞𝜓

′∗𝜙dx + 𝜓∗𝜙|+∞−∞⏟⏟⏟

0

⎞⎟⎟⎟⎠

= iℏ∫+∞

−∞𝜓

′∗𝜙dx

= ∫+∞

−∞(−iℏ𝜓 ′)∗𝜙dx = ∫

+∞

−∞(p𝜓)∗𝜙dx.

What happened is rather obvious. The integration by parts introduced a minussign, which in turn cancels out with another minus sign introduced by the com-plex conjugation in the first function, which turns i to −i. So we now realize whythe momentum operator has the complex i in its expression: to make p hermitian.

The position and momentum operators are thus hermitian, and the same isexpected to hold in general for operators of all other physical quantities, sincethese are functions of position and momentum. (With some qualifications thatneed not be discussed here. See OS2.2.)

The fact that the mean value ⟨A⟩ is always real guarantees that the eigenvaluesa of the corresponding operator A are also real, since for A𝜓 = a𝜓 we will have⟨A⟩ = a. Thus, even though the expressions of many quantum mechanical oper-ators have a complex form, the hermitian property of these operators guarantees

15 We remind the readers that for z = x + iy we have z∗ = x − iy. The equalityz = z∗ ⇒ x + iy = x − iy immediately implies then that y = 0, and z = x = a real number.


that all measurable quantities (mean values, eigenvalues, etc.) will always comeout real in the end.

Another general property that derives from the hermitian nature of an operator(and which we presented earlier without proof) is the so-called orthogonality ofthe eigenfunctions, namely, that

∫ 𝜓∗1𝜓2 dx = 0, (2.53)

where 𝜓1, 𝜓2 are any two eigenfunctions of a quantum mechanical operator Awith distinct eigenvalues a1 and a2. The proof of (2.53) is quite simple. We beginwith the definition of a hermitian operator,

∫ 𝜓∗(A𝜙)dx = ∫ (A𝜓)∗𝜙dx (2.54)

and apply it for 𝜓 = 𝜓1 and 𝜙 = 𝜓2, where 𝜓1 and 𝜓2 are eigenfunctions obeyingthe eigenvalue equations

A𝜓1 = a1𝜓1 and A𝜓2 = a2𝜓2.

Thus, (2.54) can be rewritten as

∫ 𝜓∗1 (A𝜓2) dx = ∫ (A𝜓1)∗𝜓2 dx

⇒ ∫ 𝜓∗1 (a2𝜓2) dx = ∫ (a1𝜓1)∗𝜓2 dx

⇒ a2 ∫ 𝜓∗1𝜓2 dx = a∗

1 ∫ 𝜓∗1𝜓2 dx = a1 ∫ 𝜓∗

1𝜓2 dx

⇒ (a2 − a1)∫ 𝜓∗1𝜓2 dx = 0 ⇒ ∫ 𝜓∗

1𝜓2 dx = 0,

where we have taken into account the previously shown fact that the eigenvaluesof A are real and also that they are distinct from each other.

Suppose, now, that the wavefunction𝜓 is not an eigenfunction of A but a super-position state of the general form

𝜓 =∑

ncn𝜓n. (2.55)

By the orthogonality relation for eigenfunctions, that is,

∫ 𝜓∗n𝜓m dx = 0 (n ≠ m),

we can write the mean value ⟨A⟩ as

⟨A⟩ = ∫ 𝜓∗(A𝜓) dx = ∫(∑

ncn𝜓n

)∗ (

A

(∑

mcm𝜓m

))

dx

=∑


n (A𝜓m) dx =∑


n (am𝜓m) dx

=∑

n,mc∗ncmam ∫ 𝜓∗

n𝜓m dx =∑

nc∗ncnan,


where in the last line we took into account the fact that the integral ∫ 𝜓∗n𝜓mdx

is nonzero only when m = n, in which case it is equal to the normalization inte-gral of the eigenfunction 𝜓n, which is equal to one. This calculation is, of course,the same as that presented earlier for the energy; we only included it here forcompleteness. Surely enough, the end result is also identical:

⟨A⟩ =∑

nan|cn|

2, (2.56)

which allows us again to interpret the quantities

Pn = |cn|2 (2.57)

as the probabilities of appearance of the eigenvalues an. Equation (2.56) can thenbe viewed as the usual expression for the mean value of a statistical quantity Awith possible values an and corresponding probabilities Pn = |cn|

2.In view of this, the requirement for completeness of the statistical interpreta-

tion of quantum mechanics raises one more concern: whether each physicallyrealizable wavefunction 𝜓 , that is, every square-integrable function 𝜓(x), can bewritten as an infinite superposition of eigenfunctions of an operator describing aphysical quantity. In mathematics, this property is called the completeness of aneigenfunction set and can be shown to hold for all quantum mechanical operators(although the relevant proofs are far from trivial).

The conclusion is simple: The self-consistency and completeness of the statis-tical interpretation of quantum mechanics are warranted by a common mathe-matical property of all quantum mechanical operators: the hermitian property(or hermiticity). The relevant theorem, which we have already proved up to thepoint of the completeness of the eigenfunctions, is this:

Theorem 2.5 Every16 hermitian operator has real eigenvalues and a completeset of orthogonal eigenfunctions.

2.6.2 Conservation of Probability

Before we go any further, let us pause for one moment and ask: What about prob-ability conservation? As you may recall from the beginning of this chapter, thisconservation was also required for the statistical interpretation to make sense (inaddition to the mean values being real). In other words, it is necessary that thequantity

I = ∫+∞

−∞𝜓∗(x, t)𝜓(x, t)dx = ∫

+∞

−∞|𝜓(x, t)|2 dx (2.58)

be independent of time to ensure that once we normalize the wavefunction𝜓(x, 0) at a certain moment t = 0 and make the total probability equal to one,

16 To be precise, the completeness of the set of the eigenfunctions raises some additionalrequirements to be satisfied by hermitian operators for the theorem to hold. However, quantummechanical operators typically meet these requirements, save for a few special cases of theHamiltonian operator for highly singular potentials.


then this property is conserved in time: The total probability should alwaysremain unity.

It is interesting that this crucial property is warranted also by the hermitianproperty of the Hamiltonian operator H , which determines the time evolution ofthe system via the Schrödinger equation

iℏ 𝜕𝜓𝜕t

= H𝜓. (2.59)

We will show that (2.58) is independent of time (i.e., dI∕dt = 0), by using (2.59)and the fact that H is a hermitian operator whose mean value is always real.That is,

∫ 𝜓∗(H𝜓)dx = ∫ (H𝜓)∗𝜓 dx. (2.60)

Proof : We differentiate (2.58) with respect to t to getdIdt

= ddt ∫ 𝜓∗𝜓 dx = ∫

𝜕𝜓∗

𝜕t𝜓 dx + ∫ 𝜓∗ 𝜕𝜓

𝜕tdx. (2.61)

We now invoke (2.59) to substitute for the time derivatives,𝜕𝜓

𝜕t= 1

iℏ(H𝜓) ⇒ 𝜕𝜓∗

𝜕t= − 1

iℏ(H𝜓)∗,

so that (2.61) can be recast asdIdt

= − 1iℏ∫ (H𝜓)∗𝜓 dx + 1

iℏ∫ 𝜓∗(H𝜓) dx

= 1iℏ

(

∫ 𝜓∗(H𝜓)dx − ∫ (H𝜓)∗𝜓 dx)

,

which clearly vanishes due to (2.60), that is, due to the hermitian property ofoperator H . □

But there is more to the conservation of probability. We also need to considerthe so-called local conservation: how local concentrations of probability movefrom one place to another as time goes on. We discuss this topic in the onlinesupplement of this chapter.

2.6.3 Inner Product and Orthogonality

We conclude this section by introducing the concept of inner product, whichallows us to present the previous definitions and proofs in a more elegant way.We define the inner product of two wavefunctions 𝜓(x) and 𝜙(x) as

(𝜓, 𝜙) = ∫+∞

−∞𝜓∗(x)𝜙(x)dx. (2.62)

Using (2.62), the definition of a hermitian operator (2.52) can be written as

(𝜓,A𝜙) = (A𝜓, 𝜙), (2.63)

which can be worded as follows:


Definition 2.2 An operator is called hermitian if it can be transferredunchanged from one “vector” of the inner product to the other.

The term “vector” implies, of course, that the functions𝜓(x) and𝜙(x) belong toa vector space.17 Indeed, this is legitimate, since quantum mechanical wavefunc-tions possess the property required for elements of such a space: Every linearcombination of them belongs to the same set, that is, it is a square-integrablefunction.

In the spirit of this geometrical picture—treating wavefunctions as vectors in afunction space—it would make sense to introduce the notion of orthogonality oftwo wavefunctions via the (self-evident) definition:

Definition 2.3 Two wavefunctions 𝜓(x) and 𝜙(x) are called orthogonal if theirinner product vanishes. That is,

(𝜓, 𝜙) ≡ ∫ 𝜓∗(x)𝜙(x)dx = 0,

which is the same definition we gave earlier, except that now it carries a nice geo-metric meaning.

We should also note that the inner product (2.62) shares all the essential prop-erties of the ordinary inner product A ⋅B between two 3D vectors, apart from theobvious difference that our vectors can now take complex values. Specifically:

I: (𝜓, 𝜙) = (𝜙, 𝜓)∗

II: (𝜓,𝜓) = 0 ⇔ 𝜓 = 0

III:(

𝜓,∑

cn𝜙n

)

=∑

cn(𝜓, 𝜙n)(∑

cn𝜓n, 𝜙)

=∑

c∗n(𝜓n, 𝜙)

whereby the two versions of property III derive from each other via property I,but we list them separately, to emphasize that:

The inner product (𝜓, 𝜙) is linear with respect to its second vector, andanti-linear with respect to the first one.

This means that when the second vector is a linear combination of vectors(i.e., wavefunctions), the summation symbol and the coefficients of that combi-nation can “come out” of the inner product with no change; while for a linearcombination in the first vector, its coefficients can only “come out” of the innerproduct by turning into their complex conjugates. Hence the term “anti-linear,”which implies a property with all the features of linearity, except that thecoefficients of the linear combination become their own complex conjugates.The origin of this “small” difference between the first and second vector, is, of

17 We call vector space a set of objects if every linear combination of them belongs to the same set.The definition implies that the set is such that we can define the addition between any two of itselements (vector addition), as well as the multiplication of any one of its elements with a (complexor real) number (scalar multiplication), such that the result of these two operations is also anelement of the set.


course, the symbol of complex conjugation in the first function of the integral∫ 𝜓∗𝜙dx that defines the inner product.

Complex conjugation is necessary, since our functions are complex valued andthe inner product (𝜓,𝜓) =

∑𝜓∗𝜓dx—which represents the “length” of vector𝜓

(see subsequent text)—should be real and positive for any 𝜓 .The inner product notation allows us to recast some of the previous expressions

in a more elegant form that highlights their geometrical significance. For instance,expression (2.36) for the coefficients cn of the expansion of 𝜓 into eigenfunctions𝜓n can be rewritten as

cn = (𝜓n, 𝜓), (2.64)

which is essentially identical to the expressions

Ax = x ⋅A, Ay = y ⋅A, Az = z ⋅A

that give the components of the vector A as projections onto the unit vectors x, y, zof three-dimensional space. In this spirit, because the eigenfunctions 𝜓n can bethought of as a complete basis of orthonormal vectors in wavefunction space, thecoefficients cn are also the components of the vector𝜓 in this basis. We could thenexpect that the inner product of the two vectors,

𝜓 =∑

cn𝜓n, 𝜙 =∑

dn𝜓n

can be written (as in three-dimensional space) in the form

(𝜓, 𝜙) =∑

c∗ndn, (2.65)

that is, as a sum of the products of their respective components with the complexconjugation symbol in the components of the first vector, since these componentsare now complex numbers. In particular, for 𝜙 = 𝜓 , Eq. (2.65) becomes

(𝜓, 𝜙) ≡∥𝜓∥2 =∑

|cn|2, (2.66)

where we used the definition for the length of a vector—denoted as ∥𝜓 ∥—as thesquare root of its inner product with itself. That is,

∥𝜓 ∥= (𝜓,𝜓)1∕2 ∶ length of the vector 𝜓.

Given now that quantum mechanical states are normalized (i.e., ∫ |𝜓|2dx =∫ 𝜓∗𝜓dx = (𝜓,𝜓) = 1), expression (2.66) confirms that, for a properly normal-ized quantum state, the sum of probabilities for all possible outcomes shouldequal unity. Formula (2.66) has also a clear geometrical meaning: The square ofthe length of a vector equals the sum of the squares of the absolute values of itscomponents. This is the Euclidean formula of length (or Pythagoras’ theorem, ifyou prefer) seen in a new light!

2.6.4 Matrix Representation of Quantum Mechanical Operators

In the context of the abovementioned geometrical interpretation, we could pic-ture the quantum mechanical states 𝜓 as column vectors—of infinite dimension,


in general—with their components being the “coordinates” cn in some basis ofeigenfunctions. That is,

𝜓 =

⎛⎜⎜⎜⎜⎜⎜⎝

c1

c2

⋮

cn

⋮

⎞⎟⎟⎟⎟⎟⎟⎠

. (2.67)

This representation of 𝜓 paves the way for the notion that quantum mechanicaloperators can also be represented as suitable square matrices—but of infinitedimension—that act on the vectors in the usual fashion (multiplication of amatrix by a column). Indeed, if 𝜓 ′ is the vector resulting from the action ofan operator A onto an initial state vector 𝜓—that is, if 𝜓 ′ = A𝜓—then thecoordinates c′n = (𝜓n, 𝜓

′) and cm = (𝜓m, 𝜓) of these two vectors will be relatedas follows:

c′n = (𝜓n, 𝜓′) = (𝜓n,A𝜓) =

(

𝜓n,A

(∑

mcm𝜓m

))

=

(

𝜓n,∑

mcm(A𝜓m)

)

=∑

mcm(𝜓n,A𝜓m) ≡

∑

m(𝜓n,A𝜓m)cm,

where we invoked both the linearity of operator A (to “move” it inside the sum)and the linearity of the inner product with respect to its second vector (to “move”the summation symbol and the coefficients cn outside the inner product).

The end result,

c′n =∑

m(𝜓n,A𝜓m)cm ≡ ∑

mAnmcm, (2.68)

where

Anm = (𝜓n,A𝜓m), (2.69)

is in agreement with our earlier announcement. If A is thought of as a matrixwith elements given by (2.69), then expression (2.68) can only be the result ofthe action of that matrix onto the column vector (2.67) describing the quantumstate of the system of interest. The term “matrix elements” we previously used forthe quantities Anm = ∫ 𝜓∗

n (A𝜓m)dx is thus completely justified. Indeed, these arethe elements of a matrix

A =⎛⎜⎜⎜⎝

A11 A12 · · · A1m · · ·⋮ ⋮ · · · ⋮ · · ·

An1 An2 · · · Anm · · ·⋮ ⋮ ⋮

⎞⎟⎟⎟⎠

(2.70)

that represents the physical quantity A. Given, now, that the operator A ishermitian—and therefore, (𝜓,A𝜙) = (A𝜓, 𝜙)—we can conclude for its matrixelements Anm that

Anm = (𝜓n,A𝜓m) = (A𝜓n, 𝜓m) = (𝜓m,A𝜓n)∗ = A∗mn,

2.7 Summary: Quantum Mechanics in a Nutshell 103

which means that the corresponding matrix will be hermitian, since it has thecharacteristic hermitian property of matrix theory: Elements that are symmetricwith respect to the diagonal are complex conjugates of each other; and elementson the diagonal are real numbers.

The conclusion of this discussion is clear: Not only is quantum theoryself-consistent (i.e., its statistical interpretation does not contradict itself ), but italso has an amazing mathematical structure reflecting its physical meaning in amost transparent manner. The reader who is receptive to mathematical beautywill thus find one more reason to appreciate quantum theory.

Problems

2.19 You are given the superposition state

𝜓 = N(𝜓1 + 2i𝜓2 + 𝜓3),

where 𝜓1, 𝜓2, 𝜓3 are normalized eigenfunctions of some physical quantityA with eigenvalues a1 = −1, a2 = 0, a3 = 1. Calculate the mean value ⟨A⟩and the uncertainty ΔA of the quantity A.

2.20 You are given the superposition states

𝜓 = N(2𝜓1 + i𝜓2 + 2𝜓3), 𝜙 = N(𝜓1 + 2i𝜓2 − 2𝜓3)

where 𝜓1, 𝜓2, and 𝜓3 are eigenstates of some physical quantity A with dif-ferent eigenvalues.(a) Normalize the states 𝜓 and 𝜙.(b) Calculate the “lengths” ∥𝜓 ∥ and ∥𝜙∥ of the vectors 𝜓 and 𝜙 after

normalization. What do you expect to find?(c) Calculate the inner product (𝜓, 𝜙) to check whether the vectors𝜓 and

𝜙 are orthogonal. Are they?

2.7 Summary: Quantum Mechanics in a Nutshell

Can we state the principles of quantum mechanics in the form of a few fundamen-tal statements—or postulates if you prefer—from which everything else followsas a mathematical consequence? The answer is yes. Below we present these pos-tulates for a quantum system of one particle, for simplicity.

Postulate 1: The law of motion: The Schrödinger equation.The state of a quantum system is fully described by a wavefunction 𝜓(r, t) whosetime evolution is determined by the Schrödinger equation

iℏ 𝜕𝜓𝜕t

= H𝜓. (2.71)


Here, H is the Hamiltonian operator

H = − ℏ2

2m∇2 + V (r)

that results from the classical expression for the energy of the problem (i.e., theHamiltonian) by substituting the classical quantities position and momentum ofthe particle, with the operators

xi = xi, pi = −iℏ 𝜕

𝜕xi(xi ≡ x, y, z). (2.72)

Postulate 2: The statistical interpretation.For an arbitrary wavefunction𝜓 , the mean value of the measurements of a physicalquantity A is given by the expression

⟨A⟩ = (𝜓, A𝜓) = ∫ 𝜓∗(A𝜓)dV , (2.73)

where A is the quantum mechanical operator of that quantity, which resultsby performing the substitutions (2.72) in the classical expression A = A(xi, pj).That is,

A = A

(

xi,−iℏ 𝜕

𝜕xj

)

. (2.74)

Postulate 3: The measurement principle: Collapse of the wavefunction.Following the measurement of the quantity A that yielded the eigenvalue an,the state of the system is described by the eigenfunction 𝜓n of the measuredeigenvalue.

Readers are justified at this point to ask: Why did we not include the very statisti-cal interpretation of the wavefunction𝜓? Namely, the fact that P(r, t) = |𝜓(r, t)|2measures the probability, per unit volume, to find the particle near the pointr? Actually, we did not include it because it follows from the mean-value for-mula and the stated form of the position operators xi = xi. For a one-dimensionalproblem, for example—where x = x, 𝜓 = 𝜓(x)—the mean-value formula (2.73)becomes

⟨x⟩ = ∫ 𝜓∗(x, 𝜓)dx = ∫ x|𝜓(x)|2 dx,

whence it is evident that P(x) = |𝜓(x)|2. Actually, the deeper reason why themean-value formula (2.73) allows us to infer the full statistical distribution ofany physical quantity is that it is valid for all powers of A, since they are physicalquantities as well. This means that we can use the formula

In(A) = ⟨An⟩ = (𝜓, An𝜓)

Further Problems 105

to calculate all statistical moments of a quantity, and thus infer its full statisti-cal distribution, according to the basic theorem of Section 2.3.4. This statisticaldistribution is inferred readily—in the manner described in Section 2.5.3—and issummarized in the following statement, which almost completes the list of fun-damentals one has to keep in mind before proceeding to applications. (Absentfrom the list is the famous uncertainty principle, since it is not an independentphysical principle, as we discuss in the next chapter.)

For an arbitrary quantum state𝜓 , the probability amplitude cn for a measurementof the quantity A, to yield the eigenvalue an, is given by the formula

cn = (𝜓n, 𝜓),

that is, by the coefficient of the eigenfunction 𝜓n in the expansion

𝜓 =∑

cn𝜓n

of the wavefunction 𝜓 into eigenfunctions of the quantity A.

However awkward quantum mechanics may seem, it has all the features of afundamental physical theory. Just like classical mechanics or electromagnetictheory, quantum mechanics has been put forward with such clarity and axiomaticcompleteness,18 that there is absolutely no escape from its consequences: noroom for an alternative interpretation within the theory if experiment failed toconfirm it. No such thing has happened to date.

Further Problems

2.21 A quantum particle of mass m performs a one-dimensional motion (not inthe classical sense, obviously) under the influence of the force

F = −kx (k > 0)

and its state, at a given time, is described by the wavefunction

𝜓(x) = Ne−𝜆x2∕2. (1)

(a) Does the particle have a well-defined energy? If not, is there a suitablevalue of 𝜆 for which this happens?

(b) For any given 𝜆, calculate the mean energy of the particle in state (1)and sketch its dependence on 𝜆. Is there anything noteworthy aboutyour result?

2.22 At a given time, the wavefunction of a particle in three-dimensional spacehas the form

𝜓(x, y, z) = 𝛿(x − a) ei(ky+qz).

18 This will be strictly true later on, when the Pauli principle for a system of identical particles willbe added to the list as the fourth fundamental postulate.


Which physical quantities have well-defined values in this state? What aretheir values?

2.23 At a given time, the wavefunction of a particle has the form

𝜓 = 1√

2(𝜓1 + 𝜓2), (1)

where 𝜓1 and 𝜓2 are normalized eigenfunctions of the energy with eigen-values E1 and E2, respectively.(a) Calculate the mean value and uncertainty of the energy for state (1).(b) Assuming that 𝜓1(x) is an even function and 𝜓2(x) is odd (which is

often the case, in practice), calculate the mean position of the parti-cle after time t. For greater simplicity you may also assume that thefunctions 𝜓1, 𝜓2 are real.

2.24 Prove the following properties of hermitian operators:(a) The sum of two hermitian operators is always a hermitian operator.(b) The product of two hermitian operators is a hermitian operator,

provided the two operators commute.

107

3

The Uncertainty Principle

The true meaning of Planck’s constant is the following: It is a universalmeasure of the indeterminacy that is an intrinsic characteristic of physicallaws because of the principle of wave–particle duality.

M. Born – W. Heisenberg(Joint announcement at the Solvay Conference)

3.1 Introduction

We now arrive at the “core” of the physical interpretation of the quantummechanical formalism: the uncertainty principle, one of the greatest scientificdiscoveries of the twentieth century, in our view.

As we will shortly see, the uncertainty principle (or indeterminacy principle, asit is sometimes called) is not an independent physical principle but a necessaryconsequence of the wave–particle duality and its statistical interpretation, pre-sented in Chapters 1 and 2. This fact will become clear when we actually provethe position–momentum uncertainty principle,

Δx ⋅ Δp ≥ ℏ∕2, (3.1)

as a purely mathematical inequality between two well-defined quantities (theuncertainties Δx and Δp) determined from the wavefunction 𝜓(x) of the system.

Inequality (3.1) tells us that regardless of the particular wavefunction 𝜓(x)describing the state of a physical system, the product of the uncertainties in theposition and momentum of the particle cannot be smaller than half of Planck’sconstant. However, our main goal in this chapter is not to prove (3.1) but tounderstand it. That is, we wish to understand its physical and mathematical ori-gin and unravel its physical consequences. After all, what makes the uncertaintyprinciple so fundamental—indeed, the “trademark” of quantum theory—is notso much whether it is an independent or a derivative principle, but its centralrole in understanding the theory itself and, more importantly, the world around


108 3 The Uncertainty Principle

us. We discuss first the position–momentum uncertainty principle and thenproceed to examine its time–energy counterpart.

3.2 The Position–Momentum Uncertainty Principle

3.2.1 Mathematical Explanation of the Principle

It is not difficult to bring to light the mathematical mechanism behind (3.1). First,we recall that the quantities Δx and Δp for a particular wavefunction 𝜓(x) can becalculated from the familiar statistical expressions

(Δx)2 = ⟨x2⟩ − ⟨x⟩2, (Δp)2 = ⟨p2⟩ − ⟨p⟩2.

The values on the right-hand side can be obtained by applying the quantummechanical formula ⟨A⟩ = ∫ 𝜓∗(A𝜓) dx for the mean values of the position(x = x) and momentum (p = −iℏ d∕dx) operators. In the following, we willsimplify our discussion by assuming that the wavefunction 𝜓(x) is real. We thenhave ⟨p⟩ = 0 and

(Δp)2 = ℏ2 ∫ |𝜓 ′(x)|2 dx (3.2)

as we already showed in Example 2.5. Equation (3.2) tells us that the momen-tum uncertainty Δp is a rough measure of the slope values of the wavefunction𝜓(x). The steeper the wavefunction, the greater the uncertainty in the particle’smomentum. On the other hand, we already know that the position uncertaintyΔx is a measure of how narrow or how broad a wavefunction is. The narrower thewavefunction, the smaller Δx is, and vice versa: The uncertainty in the particle’sposition is greater for a broader wavefunction.

The mathematical mechanism behind the uncertainty principle begins toemerge. If we wish to lower the uncertainty in the position of a particle, weneed to “employ” a narrow and (consequently) tall wavefunction 𝜓 , since thearea under the square of 𝜓 must always be unity. But a “thin” wavefunctionhas necessarily high slope values—it goes up and down very abruptly—so theuncertainty in momentum is also high. As we attempt, therefore, to lowerthe uncertainty in the particle’s position, we increase the uncertainty in itsmomentum. And vice versa: If we wish to reduce Δp, the wavefunction of theparticle must become broad and short to allow for relatively small slope values.But a broad wavefunction necessarily has large Δx, that is, a large uncertainty inthe position of the particle. Evidently, Δx and Δp cannot be simultaneously small,so we cannot know with high accuracy both the position and momentum of theparticle. If we wish to know the position of the particle with high accuracy (smallΔx), then we must tolerate a large indeterminacy in knowing its momentum(large Δp). Conversely, if we are interested in knowing the momentum with highaccuracy (small Δp), then we must accept a limited knowledge of where theparticle is (large Δx). The simultaneous precise knowledge of the position andmomentum of a particle is thus fundamentally impossible. This fact is not theresult of imperfections in our measuring tools, but “an intrinsic characteristic ofphysical laws,” as the founders of quantum theory have categorically stated.

3.2 The Position–Momentum Uncertainty Principle 109

This “anticorrelation” between the uncertainties Δx and Δp became alreadyevident in Examples 2.3 and 2.6. As we saw then, the Gaussian wavefunction𝜓(x) = 4

√𝜆∕𝜋 exp (−𝜆x2∕2) has uncertainties Δx and Δp

Δx = 1√

2𝜆, Δp = ℏ

√𝜆

2. (3.3)

This dependence of Δx and Δp on the parameter 𝜆 is testimony to the mecha-nism just described. If 𝜆 is large, the exponential exp (−𝜆x2∕2) decays very rapidlyfor large x, and the Gaussian wavefunction becomes narrow and tall,1 so thatΔx is small and Δp is large. The exact opposite trend is observed for small 𝜆,whereby the wavefunction “decays” very slowly as x increases. In other words,the wavefunction becomes broad and short and has thus large Δx and small Δp,in agreement with formulas (3.3). Figure 3.1 depicts this “inversely proportionalcorrelation” between Δx and Δp.

3.2.2 Physical Explanation of the Principle

We will now consider the uncertainty principle from a purely physical perspec-tive. Specifically, we will show that it is an inescapable consequence of the princi-ple of wave–particle duality of matter. This should come as no surprise, given thatwave–particle duality is really the cornerstone of the whole quantum mechanicaledifice. What we will do in this section is highlight in a purely physical mannerthe anticipated connection between these two principles.

(a) (b)

Figure 3.1 The mathematical mechanism behind the uncertainty principle. (a) A narrow andtall wavefunction. Δx is small, but Δp is large because the wavefunction goes up and downvery abruptly (it has high slope values). (b) A broad and short wavefunction. Δp is clearlysmaller now, since the wavefunction goes up and down more smoothly, but Δx is large.Conclusion: Δx and Δp cannot simultaneously become small, or, even more so, vanish. Inquantum mechanics, it is impossible to have a concurrent precise knowledge of the positionand momentum of a particle.

1 Note that the “height” of the wavefunction 𝜓(x) = 4√𝜆∕𝜋 exp (−𝜆x2∕2) is determined by its value

at x = 0, 𝜓(0) = 4√𝜆∕𝜋, and therefore it increases with 𝜆: The wavefunction is “short” for small 𝜆 and

“tall” for large 𝜆.


Let us start with a simple statement. According to the principle ofwave–particle duality, a particle with a completely determined momentum(i.e., with Δp = 0) is, at the same time, a wave with a definite wavelength𝜆 = h∕p. But a wave with definite 𝜆 is necessarily a plane wave2 that extendsunabated throughout space and which, therefore, leaves the position of the par-ticle completely undetermined. Thus, Δx = ∞, in agreement with Heisenberg’sinequality for Δp = 0.

Readers with basic knowledge of classical wave theory may recognize some-thing familiar in this statement. For example, they may recall that if we wishto create a wave that is localized in space, we need to “bring together” manysinusoidal waves with different wavelengths or wavenumbers.3 In fact, thesmaller the localization region of the composite wave, the broader the spectrumof wavelengths (or wavenumbers) we need to add. Let us recall how localizationworks: The individual sinusoidal waves that make up the composite wave(called a wavepacket) interfere constructively inside the localization region anddestructively outside it. After all, the whole point about sinusoidal waves isprecisely that we can always use a suitable superposition of them to describeany kind of wave disturbance. But our claim here is more specific, namely, thatin order to form a wavepacket with spatial extent Δx, we need to use sinusoidalwaves that span a wavenumber range Δk such that

Δx ⋅ Δk ≈ 1. (3.4)Thus, the narrower the wavepacket we wish to construct, the broader the spec-trum of wavelengths (or wavenumbers) we need to use. For example, the for-mation of a disturbance with spatial extent Δx requires all wavelengths up toΔx, that is, 𝜆 ≤ Δx. Only then can we reproduce the abrupt slope values (on theorder of, at least, 1∕Δx) exhibited by the disturbance in this spatial region. (Theslope values of a wave with wavelength 𝜆 are on the order of 1∕𝜆.) And sinceall 𝜆 values extend from zero to Δx, the corresponding wavenumber range isΔk ≈ 1∕𝜆 ≥ 1∕Δx. Hence, Δx ⋅ Δk ≥ 1, or, roughly speaking, Δx ⋅ Δk ≈ 1.

If we now take relation (3.4) as known, or at least plausible, we recognize theuncertainty principle as its direct consequence. Indeed, since p = ℏk, we findΔp = ℏΔk ⇒ Δk = Δp∕ℏ, and (3.4) can be rewritten as

Δx ⋅ Δp ≈ ℏ, (3.5)which is how we usually express the uncertainty principle, namely, as an approx-imate equality to facilitate the approximate calculations we will perform shortly.

These arguments provide an alternative view of the wavefunction, as awavepacket that comprises waves of many different wavenumbers and thusmomenta. Because the width Δk of the wavenumbers is inversely proportionalto the width Δx of the wavefunction, the same relation must be true also for theexperimentally detectable dispersion Δp of the associated momenta.

To recap, the uncertainty principle emerges as a property familiar from classicalwaves, but its meaning is different in quantum mechanics due to the probabilistic

2 It is a wave of the form 𝜓 = Aeikx, as we saw in Example 2.9.3 We recall that the wavenumber k of a sinusoidal wave is defined as k = 2𝜋∕𝜆. It is the spatialanalog of the angular frequency 𝜔 = 2𝜋∕T .


nature of the associated waves. In hindsight, we can think of relation (3.4) as akind of classical uncertainty principle. Indeed, this is how we view it today, sinceit was the quantum uncertainty principle that made relation (3.4) famous andhighlighted its significance.

We will now take a different approach with regard to the uncertainty principle,where our focus will be on the role of measurement and the limitations posedby the wave behavior of particles on our ability to achieve a simultaneous, exactknowledge of their position and momentum. We will employ a classic examplefrom the celebrated collection of gedanken (or thought) experiments of quantummechanics. These experiments were invented by the protagonists of quantumtheory, mainly in the critical years of 1924–1927, to help clarify the conceptualfoundations of the then nascent theory. The term thought experiment refers to anidealized experimental setup that is not meant for actual use, but serves mainlyas a conceptual tool to elucidate some fundamental aspects of a physical theory.

In this case, our thought experiment for the measurement of a particle’s posi-tion is based on the setup in Figure 3.2. The idea behind the experiment is thefollowing. Particles are forced to pass (either one by one, or as a beam) throughthe opening of a slit with width D. As they pass through the slit, particles havea position determined with accuracy Δy ≈ D along the y-axis. Of course, as the“opening” of the slit D becomes smaller, the accuracy in the position of the par-ticles along the y-axis increases. However, because particles are also waves withwavelengths 𝜆 = h∕p (where p is their momentum along the x-axis), they undergodiffraction as they pass through the slit.

To calculate the angular opening 𝜃 of the diffracted beam, we have to applyHuygens’ principle. That is, we assume that every infinitesimal part of the prop-agating wavefront reaching the slit is a source of circular wavelets, and that thesuperposition of these wavelets produces the diffracted wave. The angle 𝜃 is the

λ

Dθ θ

pΔpy

Figure 3.2 Wave–particle duality and the uncertainty principle in a position measurement.Because of diffraction, any attempt to increase the accuracy of a measurement by decreasingthe slit opening leads to even stronger diffraction, and, concomitantly, to a greater range ofvalues for the particles’ transverse momenta. The wave nature of the particles rendersfundamentally impossible the simultaneous, exact knowledge of their position andmomentum along any axis. As the accuracy in position increases (D ≈ Δy ↘), the uncertaintyin momentum increases (Δpy ↗), and vice versa (Δy ↗⇒ Δpy ↘).


direction where destructive interference occurs, which, in turn, defines the angu-lar region within which most of the diffracted wave is confined. It is, therefore,the diffraction angle.

But for order-of-magnitude estimates—and in particular for D ≪ 𝜆—we candivide the slit in just two “infinitesimal” parts, say, the top and bottom half ofthe slit (see the pertinent figure). In the same approximation, the center of eachpart will generate circular wavelets, which interfere destructively when the pathdifference, D sin 𝜃∕2, between the two “rays” depicted in the figure is equal tohalf a wavelength. That is, when

D D/2 θ

sin θ2

D

D2

sin 𝜃 = 𝜆

2⇒ 𝜆 = D sin 𝜃,

whence for the momentum uncertainty Δpy(see the relevant triangle in Figure 3.2) wehave, successively,

Δpy = p sin 𝜃 = h𝜆⋅ sin 𝜃 = h

D sin 𝜃⋅ sin 𝜃

= hD

≈ hΔy

⇒ Δy ⋅ Δpy ≈ h

in complete agreement with the uncertaintyprinciple.

3.2.3 Quantum Resistance to Confinement. A Fundamental Consequenceof the Position–Momentum Uncertainty Principle

From our discussion so far, it may seem as if nature is whimsical in denying usaccess to its secrets! Indeed, this is how the uncertainty principle is perceived bythose who do not understand it properly, or approach the subject with a superfi-cial philosophical disposition that leads them to exclaim that “everything is uncer-tain according to quantum mechanics!”

It is thus appropriate to present now another aspect of the uncertainty princi-ple, which highlights its fundamental role in the structure and stability of matter,properties that underlie our very existence.

The main point here is simple. If the position of a particle is known with veryhigh precision (say, Δx ≈ a, with a being very small), then there is very highuncertainty in knowing its momentum, according to the approximate version(3.5) of the uncertainty principle. Note that in using the approximate version ofthe uncertainty principle (3.5), it is implied that Δx and Δp are no longer strictlydefined quantities that are calculated rigorously from a wavefunction, but onlyorder-of-magnitude estimates.

From (3.5) we find

Δp ≈ ℏ

Δx≈ ℏ

a,

whence we conclude that a particle whose position is known with very high accu-racy (small a) must have a huge spread in its momentum and a correspondingly


large spread in its kinetic energy. We can thus say that such a particle must haveon average a very large momentum and a correspondingly large kinetic energy.

But the position of a particle may be known (with a certain accuracy Δx ≈ a)not only because it was measured but also because the particle happens to be“trapped” in a physical system with known position and size. For example, theaccuracy in the position of electrons bound in an atom is on the order of anangstrom, while the position of nuclear particles (protons and neutrons) is knownwith an accuracy on the order of a fermi (1 fermi ≡ 1 F = 10−13 cm = 10−15 m).Actually, the very entrapment of a particle in a physical system—an atom,molecule, or nucleus—constitutes an act of measurement, where the physicalsystem that enforces the trapping becomes itself a measuring apparatus ofsome sort.

Therefore, if we know that a particle is enclosed within a physical system with alinear dimension a,4 we have Δx ≈ a ⇒ Δp ≈ ℏ∕a. So, its mean kinetic energy is

⟨K⟩ ≡ K =⟨p2⟩

2m=

(Δp)2

2m≈ ℏ2

2ma2 ,

where we took into account the fact that⟨p⟩ = 0 ⇒ (Δp)2 = ⟨p2⟩ − ⟨p⟩2 = ⟨p2⟩,

since the average momentum of a particle trapped in a finite space must be zero.(Otherwise, if ⟨p⟩ ≠ 0, the particle would have an average momentum in a certaindirection and would eventually escape from the trap.)

Thus, we arrived at the conclusion that the mean kinetic energy of a particletrapped in a region of dimension a is equal to

K ≈ ℏ2

2ma2 .

This result is sensational. Just because a quantum particle is trapped in a finiteregion, it has to have a minimum kinetic energy equal to ℏ2∕2ma2! In fact, thesmaller this region, the higher the energy of the particle. In other words: Thenarrower its “prison,” the more “vigorous” the particle! This resistance to con-finement is arguably the most important of all quantum phenomena in nature.

The stability and incompressibility of atoms (what we called the central mysteryof the atomic world in Chapter 1) is a direct manifestation of this phenomenon.Despite the huge intra-atomic vacuum—that is, the empty space between theelectrons and the nucleus—the electrons of an atom do not fall onto the nucleus,but resist strongly against any attempt to reduce the atomic volume via externalpressure. The underlying reason for this effect is the resistance to confinement,namely, the fact that the kinetic energy of the electrons increases excessively whenwe try to further reduce the extent of their spatial motion.

We can follow the same line of reasoning to explain the magnitude of nuclearenergies. Being trapped in a remarkably tiny enclave that is a hundred thousand

4 In the case of atoms and nuclei, we adopt the convention that a refers to their radius and do notbother much about the accuracy of the formulation, since we are only talking about rough estimates.For the same reason, we are not particularly interested whether the physical system isone-dimensional or three-dimensional, so we retain the symbol Δx, even though Δr would be moreappropriate for three-dimensional systems.


times smaller than an atom, nuclear particles must be more energetic than atomicelectrons by many orders of magnitude. Indeed, as we saw in Chapter 1 usingwave–particle duality (which is where the uncertainty principle originates from),the nucleus is an energy giant exactly because it is a dwarf in size.

The conclusion is unambiguous. Far from undermining our ability to study thephysical world, the uncertainty principle lays the foundation for the stability ofits basic constituents, namely, atoms and molecules. We owe our very existenceto the uncertainty principle.

3.3 The Time–Energy Uncertainty Principle

We will now discuss a different kind of an uncertainty principle, which isexpressed in a similar to (3.5) approximate form as

Δt ⋅ ΔE ≈ ℏ, (3.6)

where ΔE is the uncertainty in knowing the energy of a system, and Δt is a kind of“time uncertainty” that is yet to be defined. Surely,Δt cannot have the same phys-ical meaning as the uncertainties of other physical quantities, since time is not adynamic quantity but a parameter that accompanies our measurements. The dis-tinction becomes clear once we realize that, while it is meaningful to enquire, forexample, about the position of a particle, it makes no sense to ask “what is thetime of a particle?” A valid question to ask is, for example, “what is the positionof the particle at this or that moment?”

Time is therefore an external parameter for our physical system, which is whywe have not introduced a quantum mechanical operator of time, as we did forother physical quantities.

It follows that the quantityΔt in (3.6) is inherently different from the uncertain-ties of other physical quantities. What could its meaning be? As we will discussshortly, Δt should be interpreted as the characteristic time of evolution for thephysical system of interest. In other words, Δt is the time it takes to produce anoticeable change in the properties of the system. In the context of this interpre-tation, it makes sense to use the symbol 𝜏 (the typical symbol for the characteristictime of a phenomenon) instead of Δt, and rewrite Eq. (3.6) in the form

𝜏 ⋅ ΔE ≈ ℏ, (3.7)

whose physical meaning can now be described as follows:The slower the rate of change of a physical system (large 𝜏), the more well-defined

its energy is (small ΔE). And vice versa: The faster the rate of change (small 𝜏), thehigher the uncertainty in its energy (large ΔE).

As a first example of this new principle, let us examine what happens forΔE = 0, that is, when the physical system is in an energy eigenstate. Accordingto (3.7) we then have 𝜏 = ∞, so the system does not evolve at all, since it takesan infinite amount of time to notice a change in its state. Indeed, recall that wecalled the energy eigenstates “stationary” precisely because their time evolutionproduces no change to the measurable quantities of the system.

3.3 The Time–Energy Uncertainty Principle 115

Finite evolution times 𝜏 can only occur for states with unspecified energy, thatis, when ΔE ≠ 0. For example, the superposition state

𝜓 = c1𝜓1 + c2𝜓2 (3.8)

has ΔE ≠ 0, and specifically,

ΔE ≈ |E1 − E2|, (3.9)

where E1 and E2 are the two possible energy values that a measurement can yield.The uncertaintyΔE can be calculated rigorously as a function of E1, E2, c1, and c2,5but for order-of-magnitude estimates we can always set it equal to the differencebetween the extreme values of E, as in (3.9). As for the time evolution of (3.8) wehave

𝜓(x, t) = c1𝜓1e−iE1t∕ℏ + c2𝜓2e−iE2t∕ℏ,

whence the mean value for an arbitrary physical quantity A at time t is readilycomputed (see Section 2.5.6) and found to be

⟨A⟩t = 𝛼 + 𝛽 cos 𝜔t,

where 𝜔 = |E1 − E2|∕ℏ and 𝛼, 𝛽 are constants that depend on c1 and c2.We thus realize that, as soon as the system ceases to have a well-defined energy,

it acquires a nontrivial time evolution as depicted in Figure 3.3. If we now mon-itor the evolution of the system by measuring at every instance the mean value⟨A⟩t of quantity A, then the time it takes to observe a noticeable change in thisvalue is on the order of a quarter or a half of the period T = 2𝜋∕𝜔. Therefore, thecharacteristic evolution time of the system is roughly

𝜏 ≈ T2

= 𝜋

𝜔= ℏ𝜋

E1 − E2≈ h

ΔE≈ ℏ

ΔE,

which evidently agrees with the time–energy uncertainty relation, if we ignore thenumerical coefficients that have no significance in order-of-magnitude estimates.

To recap, as soon as our physical system ceases to have a definite energy, sothat ΔE ≠ 0, it acquires also a nontrivial time evolution with a characteristic time𝜏 ≈ ℏ∕ΔE. Conversely, if we know nothing about the energy of the system, butnotice that its measurable properties change appreciably in a time interval 𝜏 , wecan immediately infer that the system does not have a well-defined energy andthe corresponding uncertainty ΔE is on the order of ℏ∕𝜏 .

The best-known manifestation of all these ideas is the correlation betweenmean lifetime and line width of excited states of an atom. As we know, an excited

5 Specifically, we have

(ΔE)2 = ⟨E2⟩ − ⟨E⟩2 = P1E21 + P2E2

2 − (P1E1 + P2E2)2 (Pi = |ci|

2, i = 1, 2)

⇒ ΔE = |c1c2| |E1 − E2| (readers are encouraged to verify this), (1)

where, as we noted in Section 2.5.3, and can be easily shown from (1), (ΔE)max = |E1 − E2|∕2. Fororder-of-magnitude estimates we often set ΔE ≈ |E1 − E2|.


α + β

α – β

⟨A⟩t = α + β cos ωt

T / 4 T / 2 3T / 4 T

t

α

ω = (E1 – E2) / h

Figure 3.3 Time evolution of the mean value of a physical quantity in a superposition ofenergy eigenstates with energies E1 and E2 (E1 > E2). The system evolves with a characteristictime on the order of a quarter or a half of the period T = 2𝜋∕𝜔. That is,𝜏 ≈ T∕2 = 𝜋∕𝜔 = ℏ𝜋∕(E1 − E2) ≈ ℏ∕ΔE, in agreement with the time–energy uncertaintyrelation (𝜏 ⋅ ΔE ≈ ℏ ≈ h).

state is not strictly stationary, as we have assumed so far,6 because an electron insuch a state can jump to a lower energy level with the simultaneous emission ofa photon. Just like other quantum quantities, the time it takes for this process tohappen cannot be predicted exactly. Therefore, what we call the mean lifetime ofan excited state is actually the statistical average of the de-excitation times of apopulation of identically excited atoms.

The mean lifetime is thus the average time it takes the atom to de-excite; aprocess that represents a noticeable change in its state. The mean lifetime of anenergy level corresponds thus to what we called the characteristic evolution timeof a quantum system. Therefore, an unstable excited level cannot have a rigor-ously defined energy; instead, its uncertainty ΔE is approximately given by therelation

ΔE ≈ ℏ

𝜏, (3.10)

where 𝜏 is the mean lifetime of the level. So, in the familiar energy-level diagram,excited levels must be represented as broadened lines of width ΔE, whereasfor the ground state we have ΔE = 0, since its mean lifetime is infinite. Theenergy-level diagram should look as in Figure 3.4.

Given now that the energy difference between these levels has also a spreadΔE,the energy of the emitted photon will be uncertain by the same amount. We thus

6 This is an unambiguous prediction of the Schrödinger equation in its usual form, where the fullelectromagnetic field has been ignored and only the electrostatic potential between electrons andthe nucleus has been included in the Hamiltonian. This is, of course, an approximation, adequate forpredicting atomic structure, but unable to account for the spontaneous decay of excited states. To dothat we need the full force, not only of the classical electromagnetic theory but of its quantumversion as well. Nevertheless, for a semiclassical treatment of this process, see Section 16.6.

3.3 The Time–Energy Uncertainty Principle 117

ΔE ≠ 0

ΔE = 0

Excited state level(τ = finite)

Ground state level(τ = ∞)

Figure 3.4 Two typical energy levels and their energy broadenings. The excited levels havefinite lifetimes, hence a finite width, while the ground state has an infinite lifetime andvanishing width.

predict the frequency spread of the emitted photon to be

Δf = ΔEh

≈ 1𝜏, (3.11)

which is experimentally manifested as a broadening of the corresponding spec-tral line.

However, let us note that it is difficult to directly confirm this prediction, sincespectral lines are actually much broader for two reasons: First, the Doppler effectdue to the thermal motion of atoms in a gas, and, second, thermal collisionsbetween atoms. Therefore, to detect the so-called natural line width in (3.10)or (3.11) we need special experimental techniques. In any case, pertinent experi-ments (which are readily performed nowadays using laser light) completely con-firm the theoretical prediction.

Let us also add that the order of magnitude for the mean lifetime of excitedatomic states is roughly

𝜏 ≈ 10−8 s.

Hence, the corresponding energy uncertainties ΔE are on the order of

ΔE ≈ ℏ∕𝜏 ≈ 10−7 eV,

and, as expected, are much smaller than the distances between energy levels,which are on the order of a few eV.

One last point worth discussing here is the classical analog of the time–energyuncertainty principle

Δt ⋅ Δ𝜔 ≈ 1, (3.12)

which can be easily derived from (3.6) with the substitution E = ℏ𝜔⇒ ΔE =ℏΔ𝜔. Equation (3.12) is the time version of relation (3.4) with the correspondencex → t and k → 𝜔. The physical meaning of (3.12) is that it relates the width Δt ofa time pulse to the spectral width of the frequencies included in the pulse. It tellsus that the narrower the pulse, the broader its frequency spectrum.

A demonstration from everyday life of the classical uncertainty relation (3.12) isthe following: While your radio is on, plug a desk lamp in power. You will observethat the radio generates some noise the moment the light switch is turned on.Use (3.12) to explain this effect and you will thus obtain a palpable experimentaldemonstration of the time–energy uncertainty relation!


3.4 The Uncertainty Principle in the Classical Limit

If quantum mechanics is indeed a fundamental physical theory (as we haveclaimed), then it must hold true not only in the atomic world (for which it wasinvented) but also in the macroscopic realm where classical theory reigns. Butfor the two theories not to be in conflict, the application of quantum mechanicsat the macroscopic level must reproduce classical results. This implies that theapplication of the position–momentum uncertainty principle should have nomeasurable effect for macroscopic bodies (large mass m), or for tiny particlesmoving on trajectories with macroscopic dimensions.

Let us examine, as a first example, a macroscopically heavy particle with a massof 1 mg, say, a dust particle. A measurement of its position with accuracy on theorder of 10−6 m = 10−4 cm is certainly satisfactory. In this case (Δx ≈ 10−4 cmand m = 10−3 g) the uncertainty relation Δx ⋅ Δp ≈ ℏ⇒ Δx ⋅ mΔ𝑣 ≈ ℏ gives thefollowing uncertainty for the speed of the dust particle:

Δ𝑣 ≈ ℏ

m Δx= 10−27

10−3 × 10−4 = 10−20 cm∕s,

which is clearly beyond any possibility of experimental detection. We thus realizethat, even though the uncertainty relation applies, in principle, also to macro-scopic objects, it does not set any serious limitation on our ability to simultane-ously know their positions and velocities. And because it is possible to know theposition and velocity simultaneously, it makes sense to speak of the trajectory ofa macroscopic body and apply the laws of classical mechanics to calculate thistrajectory.

In contrast, the concept of a trajectory is completely meaningless for micro-scopic particles that move in regions of microscopic dimensions. A good casein point is provided by atomic electrons. If it were possible to define a quan-tum mechanical trajectory for these electrons, this would surely have the formof an “orbital torus” (a kind of “bicycle inner tube”) whose cross-section diam-eter would be much smaller than its major radius. We would then know thedistance of the electron from the nucleus with uncertainty much smaller thana Bohr radius, say Δr ≈ 10−2 Å. The uncertainty principle along the radial direc-tion (Δr ⋅ Δpr ≈ ℏ) would then give an uncertainty in the radial velocity 𝑣r equalto

Δ𝑣r ≈Δpr

m≈ ℏ

mΔr≈ 1010 cm∕s,

which is comparable to the speed of light! It is thus clear that the concept of atrajectory for atomic electrons is completely untenable. The motion of electronsinside the atom must be described solely in quantum mechanical terms, namely,via a wavefunction that provides the probability of locating the electron at differ-ent regions of space around the nucleus.

But when the electron follows a macroscopic trajectory (e.g., inside an acceler-ator), then it is possible to know its radial position with accuracy on the order of,say, a millimeter. In this case, the uncertainty in the radial velocity is on the orderof 10 cm∕s, a tiny fraction of the electron’s orbital velocity in the accelerator,

3.5 General Investigation of the Uncertainty Principle 119

which is close to the speed of light. Hence, the concept of a trajectory (and,therefore, the classical laws of motion) can be applied not only to macroscopicobjects but also to microscopic particles that move in a path of macroscopicdimensions.

Also negligible are the ramifications of the time–energy uncertainty relation(𝜏 ⋅ ΔE ≈ ℏ) for macroscopic systems, where it is rather difficult to have charac-teristic times of change smaller than a nanosecond (𝜏 ≈ 10−9 s). For such systemswe have

ΔE ≈ ℏ

𝜏= 10−27

10−9 = 10−18 erg = 10−25 J,

which is a negligible indeterminacy in energy, certainly beyond the detectioncapability of any measuring apparatus.

Let us now revisit the position–momentum uncertainty relation and note thatthe gradual weakening of its implications as we move from the microscopicto the macroscopic world is a direct result of the corresponding weaken-ing of the wave manifestations of matter. In other words, the application ofthe wave–particle duality principle on macroscopic objects (such as a dustparticle with a mass of 1 mg) has no detectable consequences because thecorresponding wavelengths are so small that wave phenomena are beyondexperimental observation. For example, even if the dust particle moves so slowlythat its speed is barely measurable, for example, 𝑣 ≈ 10−6 cm∕s, its wavelengthwill be

𝜆 ≈ ℏ

m ⋅ 𝑣≈ 10−27

10−3 × 10−6 ≈ 10−18 cm.

To observe the wave nature of such a particle requires “slits” or obstacleswhose dimensions are several orders of magnitude smaller than what existsin nature. Just like the wave nature of physical objects is not detectable inthe macroscopic world, the same is true for the position–momentum uncer-tainty, since the latter is a direct consequence of the wave–particle duality ofmatter.

3.5 General Investigation of the Uncertainty Principle

3.5.1 Compatible and Incompatible Physical Quantities and the GeneralizedUncertainty Relation

The position–momentum uncertainty relation we discussed earlier summarizesthe fundamental finding that the corresponding uncertainties cannot vanishsimultaneously, that is, for the same physical state. Since the concurrent exactknowledge of these quantities is impossible, they are, in a way, incompatible.Precise knowledge of one of them is incompatible with precise knowledge of theother. A natural question to ask at this point is the following: Are position andmomentum the only incompatible physical quantities, or are there more suchpairs? And if so, then how do we recognize them?


Consider an arbitrary pair of physical quantities A and B. Our question is: Canthe uncertainties ΔA and ΔB of these quantities vanish simultaneously? First, letus assume that ΔA = 0. We know already that this condition of zero uncertaintyis fulfilled for quantum mechanical states that are solutions of the eigenvalueequation

A𝜓n = an𝜓n, (3.13)that is, for the eigenfunctions of quantity A. If we now also require that ΔB = 0,then the eigenfunctions 𝜓n of A must also be eigenfunctions of quantity B, withdifferent eigenvalues bn, in general. Thus the following eigenvalue equation mustbe satisfied,

B𝜓n = bn𝜓n. (3.14)However, we can easily show that (3.13) and (3.14) cannot always be simultane-ously true. To that purpose, we act on both sides of (3.13) with operator B and onboth sides of (3.14) with operator A, and we find

B ⋅ (3.13) ⇒ BA𝜓n = B(an𝜓n) = anB𝜓n = anbn𝜓n (3.15)A ⋅ (3.14) ⇒ AB𝜓n = A(bn𝜓n) = bnA𝜓n = bnan𝜓n. (3.16)

Since anbn = bnan, the right-hand sides of (3.15) and (3.16) are equal, so theleft-hand sides must also be equal. Thus, we have

BA𝜓n = AB𝜓n

for all eigenfunctions 𝜓n. It must then be true thatBA𝜓 = AB𝜓 (3.17)

for any wavefunction 𝜓 , since 𝜓 can always be written as a superposition of theeigenfunctions 𝜓n. But from the definition of operator equality,7 (3.17) impliesthat

AB = BA, (3.18)which tells us that the operators commute with each other: The order in whichthey are multiplied does not matter. Note that (3.18) is not a trivial condition,because quantum mechanical operators do not commute in general, as we alreadystressed in Section 2.2.3. The position and momentum operators provide themost famous example of noncommuting operators. If we form the product xpand act on an arbitrary wavefunction 𝜓(x), we obtain

xp ⋅ 𝜓(x) ≡ x(p𝜓(x)) = x(

−iℏ ddx

⋅ 𝜓(x))

= −iℏ x𝜓 ′, (3.19)

while if we reverse their order and act again on 𝜓(x), we find

px ⋅ 𝜓(x) ≡ p(x𝜓) = −i ℏ ddx

(x𝜓) = −iℏ 𝜓 − iℏ x𝜓 ′. (3.20)

7 Two operators are equal if their action on an arbitrary wavefunction 𝜓 yields the same result.That is,

A = B ⇔ A𝜓 = B𝜓, for every 𝜓.


It is thus evident that xp 𝜓(x) ≠ px 𝜓(x), and, therefore, xp ≠ px. If we now sub-tract (3.20) from (3.19), we get

xp 𝜓 − px 𝜓 ≡ (xp − px)𝜓 = iℏ 𝜓.

Based on the definition for the equality of two operators, we then obtain

xp − px = iℏ,

which can also be cast in the form

[x, p] = iℏ, (3.21)

where the symbol

[A,B]def= AB − BA

is the so-called commutator of the A and B operators. The commutator is a usefulquantity, since it is zero when the quantities A and B commute (and are thuscompatible), and vice versa.

We can summarize our findings as follows:

Theorem 3.1 Two quantum mechanical quantities A and B can be simultane-ously measured with absolute precision if and only if their operators commutewith each other, that is, if [A,B] = 0. Such quantities are termed compatible. Con-versely, if their operators do not commute, that is, if [A,B] ≠ 0, then the quantitiesare called incompatible and cannot be measured simultaneously with absoluteprecision.

We can directly apply this discussion to position and momentum in one dimen-sion. The impossibility of a simultaneous measurement of these two quantitieswith absolute precision is now explained as a consequence of (3.21), the noncom-mutativity of the corresponding operators: The physical quantities of position andmomentum are incompatible.

Let us now examine three-dimensional motion. Position and momentum arenow vectors with three components, r = (x, y, z), and p = (px, py, pz). The firstthing to notice is that the three components of position are mutually compati-ble, because the order in which they multiply a wavefunction does not matter. Inparticular, we have

xy ⋅ 𝜓(x) = yx ⋅ 𝜓(x) ⇒ xy = yx ⇒ [x, y] = 0

and similarly for the other two combinations, x, z and y, z. The momentum com-ponents px, py, and pz are also compatible, since they represent differentiationwith respect to x, y, and z, the order of which does not matter: Clearly, we may dif-ferentiate first with respect to x and then with respect to y, or first with respect to yand then with respect to x, but the result will be the same in both cases (assumingthat the function has a continuous derivative). The physical conclusion is evi-dent. We can measure simultaneously the three spatial coordinates of a particleand pinpoint its position in 3D space with whatever accuracy we wish. Likewise,we can measure simultaneously all three momentum components and obtain full


knowledge of the magnitude and direction of the velocity after the measurement.8However, we cannot have it both ways: simultaneous exact knowledge of r andp is impossible. For it is clear that, in analogy to the one-dimensional case, thefollowing commutation relations hold

[x, px] = iℏ, [y, py] = iℏ, [z, pz] = iℏ, (3.22)

while for the components of the two vectors along different axes, we have

[x, py] = 0, [y, pz] = 0, [z, py] = 0, etc., (3.23)

since x is a constant with respect to y-differentiation and can thus be interchangedwith it, and likewise for the other components.

In the online supplement of this chapter, readers are asked to show that theposition-momentum uncertainty relation Δx ⋅ Δp ≥ ℏ

2is simply a special case of

the following more general relation for any pair of incompatible physical quanti-ties.

Generalized uncertainty relation: The product of the uncertainties of two incom-patible physical quantities can never be smaller than half of the absolute meanvalue of their commutator.

We thus have

ΔA ⋅ ΔB ≥ 12|⟨[A,B]⟩|. (3.24)

For position and momentum (in one dimension), we have A = x, B = p, [A,B] =[x, p] = iℏ, so that (3.24) is written as

Δx ⋅ Δp ≥ 12|⟨[x, p]⟩| = 1

2|⟨iℏ⟩| = 1

2|iℏ| = ℏ

2or

Δx ⋅ Δp ≥ ℏ

2,

which is our familiar uncertainty relation for these physical quantities.Similarly, applying (3.24) in the cases of (3.22) yields

Δx ⋅ Δpx ≥ ℏ

2, Δy ⋅ Δpy ≥ ℏ

2, Δz ⋅ Δpz ≥ ℏ

2. (3.25)

Combined with (3.23), expressions (3.25) tell us that in three dimensions theposition–momentum uncertainty principle applies only for the same-axis com-ponents of the corresponding vectors, while there is no restriction whatsoeverfor the simultaneous measurement of components along different axes.

3.5.2 Angular Momentum: A Different Kind of Vector

Having discussed position and momentum, which are two of the three funda-mental vector quantities of mechanics, it is now time to consider the third one,namely, angular momentum. Actually, in the context of quantum mechanics,

8 We recall that a quantum measurement determines the state of the quantum system after themeasurement.


angular momentum is more significant than position and momentum, as willbecome apparent in later chapters. So, can we simultaneously measure all threecomponents of angular momentum, the same way we can for position andmomentum? The answer depends on whether the commutators [𝓁x,𝓁y], [𝓁y,𝓁z],and [𝓁z,𝓁x] vanish or not. It suffices to calculate the first commutator, since wecan then easily obtain the other two via a cyclic permutation9 of x, y, and z. Astraightforward way to calculate [𝓁x,𝓁y] is to have the products 𝓁x𝓁y and 𝓁y𝓁xact on a three-dimensional wavefunction 𝜓(x, y, z), find their difference, andexamine what happens.10 First, we have11

𝓁x = ypz − zpy = y(

−iℏ 𝜕𝜕z

)

− z(

−iℏ 𝜕𝜕y

)

= −iℏ(

y 𝜕𝜕z

− z 𝜕𝜕y

)

and likewise

𝓁y = zpx − xpz = −iℏ(

z 𝜕𝜕x

− x 𝜕𝜕z

)

.

We thus obtain12

𝓁x𝓁y𝜓 = (−iℏ)2(

y 𝜕𝜕z

− z 𝜕𝜕y

) (

z 𝜕𝜕x

− x 𝜕𝜕z

)

𝜓

= −ℏ2(

y 𝜕𝜕z

− z 𝜕𝜕y

)

(z𝜓x − x𝜓z)

= −ℏ2(

y 𝜕𝜕z

(z𝜓x) − yx 𝜓zz − z2𝜓xy + xz𝜓yz

)

= −ℏ2(

y𝜓x + yz𝜓xz − yx𝜓zz − z2𝜓xy + xz𝜓yz

)

(3.26)

and

𝓁y𝓁x𝜓 = −ℏ2(

z 𝜕𝜕x

− x 𝜕𝜕z

) (

y 𝜕𝜕z

− z 𝜕𝜕y

)

𝜓

= −ℏ2(

z 𝜕𝜕x

− x 𝜕𝜕z

)

(y𝜓z − z𝜓y)

= −ℏ2(

zy𝜓xz − z2𝜓xy − xy𝜓zz + x 𝜕𝜕z

(z𝜓y))

= −ℏ2(

zy𝜓xz − z2𝜓xy − xy𝜓zz + x𝜓y + xz𝜓yz

)

. (3.27)

9 The cyclic permutation of an ordered set of objects—in our case, x, y, and z—is defined as follows:We switch from the first object to the second one, then to the third one, and so on, and from the lastobject back to the first one, as if the objects were arranged in a circle. A cyclic permutation of x, y,and z is therefore the “motion” x → y (x “goes to” y), y → z, z → x, and so on.10 A more elegant way to calculate commutators is described in the online supplement of thischapter.11 We recall that

𝓵 = r × p =|||||||

x y zx y zpx py pz

|||||||

= x(ypz − zpy) + y(zpx − xpz) + z(xpy − ypx), and so on.

12 We use the symbols 𝜓x ≡ 𝜕𝜓∕𝜕x, 𝜓xy ≡ 𝜕2𝜓∕𝜕x𝜕y = 𝜓yx, and so on, for simplicity.


If we now subtract (3.26) from (3.27), all second-derivative terms cancel, and onlyfirst-derivative terms (shown in the boxes) survive. We thus obtain

(𝓁x𝓁y − 𝓁y𝓁x)𝜓 ≡ [𝓁x,𝓁y]𝜓 = −ℏ2(y𝜓x − x𝜓y) ≡ ℏ2(

x 𝜕𝜕y

− y 𝜕𝜕x

)

𝜓

≡ iℏ(

x(

−iℏ 𝜕𝜕y

)

− y(

−iℏ 𝜕𝜕x

))

𝜓 = iℏ(xpy − ypx)𝜓

= iℏ 𝓁z 𝜓,

which can be recast as an operator equality, namely,[𝓁x,𝓁y] = iℏ 𝓁z (3.28)

and similarly (via cyclic permutations of x, y, and z)[𝓁y,𝓁z] = iℏ 𝓁x , [𝓁z,𝓁x] = iℏ 𝓁y . (3.29)

The conclusion is impressive. In contrast to the position and momentum vec-tors, whose components commute with each other and can be measured simul-taneously with absolute accuracy, the components of angular momentum do notcommute, so their concurrent exact knowledge is impossible. In other words, inquantum mechanics it is impossible to fully know the angular momentum vector.The maximum knowledge we can attain for the angular momentum of a parti-cle is limited to one of its components and the magnitude of the vector, or themagnitude squared, which is represented in quantum mechanics by the operator

𝓵 2 = 𝓁2x + 𝓁2

y + 𝓁2z . (3.30)

The possibility of knowing the magnitude of the vector 𝓵 in addition to one of itscomponents is a consequence of the commutation relations

[𝓁i,𝓵2] = 0, i = x, y, z, (3.31)

which can be readily shown from Eqs. (3.28)–(3.30). Relations (3.31), inconjunction with (3.28) and (3.29), tell us that any one—but only one at atime!—component of the angular momentum can be measured simultaneouslywith its magnitude—that is, with 𝓵 2. The final conclusion is worth repeating:For the angular momentum vector 𝓵, quantum mechanics allows us to simulta-neously know only its magnitude and one of its components.

As we will see later (Chapters 9 and 10), the key role of the angular momentumvector in quantum mechanics stems directly from these properties, and specif-ically from the commutation relations (3.28) and (3.29) satisfied by its compo-nents. Angular momentum is a premier quantum quantity. (As we could havesuspected, perhaps, given that angular momentum has the same physical dimen-sions as Planck’s constant.)

Problems

3.1 Excited rotational states of diatomic molecules have lifetimes on the orderof a few seconds. Calculate the uncertainty in the frequency of the photonsemitted when these states are de-excited.

Problems 125

3.2 Certain nuclear particles produced in high-energy experiments (these par-ticles are known as “resonances”) have lifetimes on the order of 10−23 s andmasses on the order of a few GeV (say, 1 GeV). What is the maximum accu-racy in knowing their mass?

3.3 Take a spectral line in the visible spectrum and estimate first its naturalwidth ΔfN (N=natural) due to the uncertainty principle (𝜏 ≈ 10−8 s). Then,for the same spectral line, estimate the room-temperature Doppler broad-ening ΔfD due to the thermal motion of the emitting atoms of the gas. Whatdo you conclude from the comparison of these two values?

3.4 Someone claims that the z coordinate as well as the z component of momen-tum, pz, of a particle can be measured (with arbitrary accuracy) simultane-ously with the corresponding z component of angular momentum, 𝓁z. Doyou agree with this claim?

3.5 Consider the following pairs of physical quantities:

(x,H), (p,H)(

H =p2

2m+ V (x)

)

∶ 1D

(x,𝓁z), (x,𝓁x), (x, py), (x, px), (px,𝓁y) ∶ 3D

Which of these pairs can be simultaneously measured with arbitrary accu-racy? Write down the generalized uncertainty relation in all other cases.

3.6 Show that the commutator [A,B] = AB − BA satisfies the following generalproperties:

[A,B] = −[B,A], [A, 𝜆B + 𝜇C] = 𝜆[A,B] + 𝜇[A,C],[A,BC] = [A,B]C + C[A,B].

Now use these properties—in conjunction with the basic commutation rela-tion [x, p] = iℏ—to show that

[x, p2] = 2iℏp, [x, p3] = 3iℏp2, … , [x, pn] = niℏpn−1

and then,

[p, xn] = −niℏxn−1,

from which the following (reasonable) generalizations can be derived:

[x,A(x, p)] = iℏ𝜕A𝜕p, [p,A(x, p)] = −iℏ𝜕A

𝜕x.

3.7 Apply the results you obtained in the previous problem to prove the follow-ing uncertainty relations:

Δx ⋅ ΔE ≥ ℏ

2|⟨𝑣⟩|, Δp ⋅ ΔE ≥ ℏ

2|⟨F(x)⟩|,

where 𝑣 = p∕m is the velocity operator of the particle, and F(x) =−dV (x)∕dx is the force the particle experiences.

127

Part II

Simple Quantum Systems

129

4

Square Potentials. I: Discrete Spectrum—Bound States

4.1 Introduction

In this and the following chapter, we study a class of problems with potentialsthat are piecewise constant functions, like the one in Figure 4.1. Because of theirgraphical representation, we call them square potentials.

The convenience of dealing with square potentials is that the Schrödingerequation, being a linear differential equation with constant coefficients, canbe solved exactly in each separate interval where the potential has a constantvalue. We can thus obtain explicit solutions and directly confirm all the generalprinciples of quantum mechanics we presented in previous chapters.

We begin with the one-dimensional Schrödinger equation

𝜓 ′′ + 2mℏ2 (E − V (x))𝜓 = 0 (4.1)

and cast it in the more compact form

𝜓 ′′ + (𝜖 − 𝑈 (x))𝜓 = 0, (4.2)

where

𝑈 (x) = 2m V (x)ℏ2 , 𝜖 = 2m E

ℏ2 , (4.3)

while—as one can verify from (4.2)—the physical dimensions of 𝑈 and 𝜖 are

[𝑈 ] = [𝜖] = L−2.

Since the quantities 𝑈 (x) and 𝜖 are related to V (x) and E via a common mul-tiplicative factor, we will occasionally refer to them as “potential” and “energy,”respectively. We trust that the use of distinct symbols𝑈 and 𝜖 leaves no room forconfusion with the actual potential and particle energy.

Let us now examine the mathematical conditions that the desired solution𝜓(x)of Eqs (4.1) or (4.2) ought to satisfy. First, since 𝜓(x) is the solution of a linearsecond-order differential equation, its second derivative must exist. Therefore,𝜓 and 𝜓 ′ must also exist and be continuous functions everywhere, including thepoints of discontinuity of the potential V (x). If x0 is a discontinuity point (i.e., the


130 4 Square Potentials. I: Discrete Spectrum—Bound States

V2V(x)

– V0

xa

V1V (x) =

V1; −∞< x < 0− 0 < x < aV0;

2V ; a < x < ∞

Figure 4.1 An example of a square potential.

potential V (x) has a finite jump there), the following continuity conditions shouldapply

𝜓L(x0) = 𝜓R(x0), 𝜓 ′L(x0) = 𝜓 ′

R(x0), (4.4)

where 𝜓L(x) and 𝜓R(x) are the respective solutions to the left and right of thediscontinuity point x0.

Relations (4.4) are also known as matching conditions, precisely because theytell us how to “match” the particular solutions of (4.1) in the regions to the leftand to the right of the discontinuity point x0. Matching is of course necessaryhere because the potential takes different values on either side of x0, and thecorresponding solutions have different forms in these two regions.

This discussion holds only if the discontinuity of the potential at x0 is finite. Ifthe discontinuity is infinite, it follows from (4.1) that the second derivative alsodiverges, and therefore the first derivative cannot be continuous at x0. If this state-ment is not completely clear to you, consider that for 𝜓 ′′ to diverge somewhere,𝜓 ′ must have a vertical jump at this point, which implies that𝜓 ′ is a discontinuousfunction there.

It is also clear from the discussion that the conditions (4.4) are automaticallysatisfied when the potential V (x) is a continuous function, or, even better, whenit is a function given in analytic form throughout the range −∞ < x < +∞.

Another general observation worth making pertains to the form of the energyspectrum (whether it is discrete or continuous) and the physical meaning of thecorresponding solutions. For this purpose, let us examine the typical “potentialwell” shown in Figure 4.2.

We will see in more detail later that the physical meaning of the quantummechanical solutions in these two energy regions is completely analogous to thatof the classical problem. In particular:

(a) In the energy region −V0 < E < 0, the quantum mechanical solutions vanishexponentially at ±∞ and so they are square integrable. Particles describedby such solutions have a vanishing probability to escape to infinity, so theirmotion is confined around the bottom of the potential. The physical behaviorassociated with these solutions is evidently analogous to the finite classicaloscillations in the same energy range.

(b) In the region E > 0, the solutions of the Schrödinger equation are finiteeverywhere, but do not vanish at ±∞ and are thus not square integrable.

4.1 Introduction 131

V(x)

E > 0

E < 0

xx1 x2

– V0

Figure 4.2 The energy spectrum of a one-dimensional potential well. In the range of negativeenergies (E < 0), where the classical motion is confined within the interval x1 ≤ x ≤ x2, theenergy spectrum is discrete and the corresponding solutions represent the bound states ofthe particle. Conversely, in the range of positive energies (E > 0), where the classical motion isunrestricted, the energy spectrum is continuous and solutions describe the scattering of theparticle by the potential.

This feature is actually not surprising, because in the corresponding classicalcase, the particle can also reach infinity with a nonvanishing speed 𝑣∞determined by the relation E = m𝑣2

∞∕2, since the potential energy vanishesat infinity. It is thus reasonable to expect that the quantum mechanical solu-tions in the energy range E > 0 describe particles that perform unconfinedmotion in the whole range from −∞ to +∞. Such solutions are suitable forthe description of particle scattering by the potential of interest.

Indeed, for a potential like the one in Figure 4.2, V (x) → 0 as x → +∞, and theSchrödinger equation for large x takes the asymptotic form

𝜓 ′′∞ + 𝜖𝜓∞ = 0 (𝜖 = 2mE∕ℏ2),

whence we easily conclude that for 𝜖 < 0 we can put 𝜖 = −𝛾2, and the asymptoticsolution for 𝜓∞ can be chosen to be a decreasing exponential, since the otherexponential (e𝛾x for x → +∞ or e−𝛾x for x → −∞) blows up at infinity. Similarly,for 𝜖 > 0 we put 𝜖 = k2, and the asymptotic solution in this case is 𝜓∞ = exp(±ikx), which is finite everywhere and, therefore, physically acceptable. We stressagain that the general requirement for physically acceptable solutions—validboth for bound and unbound states—is that they are finite everywhere, includinginfinity.

Nevertheless, there is a fundamental feature of the quantum mechanicalsolutions of case (a) that distinguishes them clearly from the correspondingclassical solutions. In the energy range of bound states, the Schrödinger equationhas solutions with the desired features (to vanish at ±∞) only when the energyE takes values from a discrete sequence (discrete spectrum). On the other hand,in the region E > 0, there are physically allowed solutions for any value of E,so the spectrum is continuous for E > 0. From the perspective of wave theorythis property ought to be expected. Indeed, as we have discussed repeatedlyin Chapter 1, the frequency of classical waves is also quantized—that is, the


frequency spectrum is discrete—if the corresponding solution is spatiallyconfined. In effect, we have the following two-way relations:

localized solutions ↔ discrete spectrumnonlocalized solutions ↔ continuous spectrum.

In view of this, it is reasonable to divide the study of square potentials in twochapters. In this chapter we discuss problems with bound states. In Chapter 5, weexamine cases where the spectrum is continuous and the corresponding solutionscan be used to describe scattering problems and the corresponding scatteringstates.

4.2 Particle in a One-Dimensional Box: The InfinitePotential Well

4.2.1 Solution of the Schrödinger Equation

The potential well in this case is schematically shown in Figure 4.3.The fact that the particle cannot exit the box means that the probability of

finding it outside the interval 0 < x < L is zero. Therefore, the wavefunctionmust vanish everywhere outside this interval and be nonzero only inside. Thewavefunction 𝜓(x) will thus have the general shape of Figure 4.4 and satisfy the

V(x)

L

x

0

∞ ∞

m

V (x) =∞;

< <

<xx L

L>x0;

0;

0

Figure 4.3 The infinite potential well. The potential function V(x) for a particle that movesfreely inside a one-dimensional box (≡tubule) but cannot escape from it. The potential is zeroinside the box and infinite outside.

L

x

0

Figure 4.4 General shape of a wavefunction in an infinite potential well. The wavefunction iszero everywhere outside the box and nonzero only inside.

4.2 Particle in a One-Dimensional Box: The Infinite Potential Well 133

boundary conditions

𝜓(0) = 0, 𝜓(L) = 0, (4.5)

so that it remains continuous as we cross the box boundaries. (Remember thatwe cannot require that the derivative of the wavefunction be also continuous,because the potential V (x) has infinite discontinuities at the points x = 0 andx = L.)

It follows from the discussion that to study the quantum motion of the particleinside the box, we have to solve the one-dimensional Schrödinger equation

𝜓 ′′ + 2mℏ2 (E − V (x))𝜓 = 0 (4.6)

in the interval 0 < x < L, where V (x) = 0, and with the boundary conditions (4.5)at the edge points. We thus find

𝜓 ′′ + 2m Eℏ2 𝜓 = 0 ⇒ 𝜓 ′′ + k2𝜓 = 0, (4.7)

where the quantity

k2 = 2m Eℏ2 (4.8)

is clearly positive because, even in quantum mechanics, the total energy of theparticle inside the box—where the potential energy vanishes—cannot becomenegative.

As we know, the general solution of (4.7) is written as

𝜓(x) = Asin kx + Bcos kx,

and we now have to apply the boundary conditions (4.5) to proceed. The first ofthese conditions gives

𝜓(0) = 0 ⇒ A ⋅ 0 + B ⋅ 1 = 0 ⇒ B = 0,

which means that the general solution is limited to 𝜓 = A sin kx. Applying nowthe second boundary condition we obtain

𝜓(L) = 0 ⇒ Asin kL = 0 ⇒ sin kL = 0,

where, in the second step, we required that A cannot be zero, since we would thenhave 𝜓(x) ≡ 0 and there would be no particle inside the box in the first place! Thecondition sin kL = 0 is satisfied only when

kL = n𝜋, n = 1, 2,… ,∞, (4.9)

where we have excluded negative values of n because the substitution n → −nmerely flips the sign of the solutions

𝜓(x) = Asin kx = Asin n𝜋xL, (4.10)

and such a sign change has no physical significance, as we know. Combining nowthe relations k = n𝜋∕L from Eq. (4.9) and k2 = 2mE∕ℏ2 from Eq. (4.8), we obtain

E = En = ℏ2𝜋2

2mL2 n2, n = 1, 2,… (4.11)


Hence, the energy of the particle inside the box is quantized and the allowedenergy values are given by (4.11). In particular, the minimum allowed energy, alsoknown as the ground-state energy, for the particle inside the box is

E1 = ℏ2𝜋2

2mL2 . (4.12)

As for the undetermined constant A in the eigenfunctions (4.10), it is calculatedvia the normalization condition

∫L

0

||𝜓n(x)||

2dx = 1, (4.13)

which expresses the self-evident requirement that the total probability of findingthe particle inside the box must be unity. The pertinent integral can be readilycalculated by substituting sin2 kx = (1 − cos 2kx)∕2 and taking into account thefact that kL = n𝜋, to obtain

∫L

0sin2kx dx = L

2− 1

4ksin 2kL |

||k=n𝜋∕L

= L2.

Equation (4.13) now yields

|A|2 ∫L

0sin2kx dx = |A|2 L

2= 1 ⇒ A = ±

√2L,

so the normalized eigenfunctions are

𝜓n =√

2L

sin n𝜋xL. (4.14)

Here we chose the positive value for A, since the sign of the wavefunction (or,more generally, any multiplicative factor of the form eia) has no physical signifi-cance, as we noted earlier.

Note that the appearance of L in the denominator of the square root in (4.14)is no accident. Because of the statistical interpretation of 𝜓2 as a probability perunit length, we always have1

[𝜓2] = L−1 ⇒ [𝜓] = L−1∕2,

which implies that every normalized one-dimensional wavefunction must havedimensions of L−1∕2, as in (4.14).

As is customary in quantum physics, we depict in a joint diagram (seeFigure 4.5) the potential function, the allowed energies (shown as horizontal linesegments on the vertical axis of energy), and the corresponding eigenfunctions(sketched on these line segments).

4.2.2 Discussion of the Results

Some features of these results—namely, energy quantization and the existenceof a nonzero minimum energy for the particle in the box—were expected and

1 We remind the readers that the symbol [A] denotes the physical dimensions of quantity A.


Figure 4.5 The first three eigenvalues andeigenfunctions of an infinite potential well.

∞ ∞

E3

E2

E1

E1

x

Vmin

Figure 4.6 The energy of the ground state in an arbitrary potential well always lies above thebottom of the well. Falling to the bottom of a potential well is impossible in quantummechanics. Due to the uncertainty principle, quantum particles move like “crazy” even in thestate of maximal rest: the state of minimum total energy.

need no further discussion here. Energy quantization is always expected forconfined motion, while the nonzero ground-state energy is a consequence ofthe uncertainty principle. We suggest that the readers perform the relevantapproximate calculation and extend the argument to justify the statement in thecaption of Figure 4.6.

4.2.2.1 Dimensional Analysis of the Formula En = (ℏ2𝝅

2∕2mL2)n2. Do WeNeed an Exact Solution to Predict the Energy Dependence on ℏ, m, and L?The answer is definitely no. Since we have only three parameters, and basedon the fundamental theorem of dimensional analysis (Section 1.4.1), there is aunique way of combining them to obtain a quantity with physical dimensionsof energy, namely, 𝜖 = ℏ2∕mL2. We can confirm this result either by using thesystematic approach of Chapter 1 or by combining known formulas that involveenergy and the parameters ℏ,m, and L. For example, by combining the formulasE = p2∕2m and p = h∕𝜆⇒ p ≈ ℏ∕L, we see immediately that E ≈ ℏ2∕mL2. Thus,there is a purely dimensional reason why the combination ℏ2∕mL2 appears inthe expression for the energy eigenvalues and we need not solve the problem tofind it.


4.2.2.2 Dependence of the Ground-State Energy on ℏ, m, and L : TheClassical LimitA good way to check our physical understanding of a certain problem is whetherwe can qualitatively explain the main features of the dependence of a physicalquantity on the given parameters. So let us put ourselves to the test, with theformula (4.12) for the ground-state energy of a particle in a box. First, the fact thatL appears in the denominator is plausible. As we know well by now, the smallerthe physical dimension of a confining region, the greater the kinetic energy of theparticle, and the further away from the bottom of the potential its ground-stateenergy is expected to be.

Let us now examine why ℏ appears in the numerator, and the mass m in thedenominator. First, recall that ℏ is the characteristic parameter of quantum phe-nomena. Therefore, in the limit ℏ → 0, all typical quantum effects must disap-pear and the classically expected behavior must be recovered. The elevation ofthe ground state above the bottom of the potential is one such typical quan-tum effect. Therefore, in the classical limit ℏ→ 0, the ground-state energy mustapproach the lowest possible classical value, which is the minimum of the poten-tial curve. To put it differently: In the classical limit ℏ→ 0, the particle must fallto the bottom of the potential. Indeed, in this case we have E1 → 0 for ℏ → 0, asexpected.

The appearance of the mass in the denominator has a similar explanation. Here,the classical limit refers to large masses. The heavier a particle, the more classicalits behavior. In the limit m → ∞, the energy E1 of the ground state must thentend toward the bottom of the potential, which indeed is the case here, since mappears in the denominator of the formula E1 = ℏ2𝜋2∕2mL2.

The classical limit ℏ→ 0, m → ∞ can be also called the weak quantum limit,because it points to the direction where quantum features fade away and classicalbehavior is gradually restored. The opposite direction (ℏ→ ∞, m → 0) corre-sponds to the strong quantum limit, where quantum behavior manifests itselfwith great intensity, that is, the ground-state energy goes further away from thebottom of the potential, the distance between energy levels increases, and so on.

We can also arrive at these conclusions in the most general way via the funda-mental relation

𝜆 = hp= h

m𝑣,

which tells us, for instance, that the heavier a particle, the smaller its wave-length. But as the wavelength gets smaller, the wavelike behavior becomes lesspronounced, as do all quantum effects, since they are mere consequences ofthe wave character of motion. The exact opposite happens in the limit m → 0.The wavelength becomes huge, so the wavelike behavior and all associatedquantum features appear with great intensity. So the lighter a particle is, themore quantum mechanical its behavior. Indeed, as we will see later, the factthat the electron has such a minuscule mass has profound implications for thestructure of matter. The electron is the premier quantum particle of our world.

Let us make one more observation. We saw earlier that, as the confinementlength L decreases, quantum effects are enhanced. Conversely, when the particle


is allowed to move freely in a large area, its behavior is roughly classical.Therefore, the frequently made statement that quantum laws apply to particlesof the microscopic world is not generally valid and warrants clarification. Thecorrect statement is that quantum laws are manifest (strongly) in microscopicparticles that are confined in regions of microscopic dimensions.

4.2.2.3 The Limit of Large Quantum Numbers and Quantum DiscontinuitiesClassical behavior is recovered also for large quantum numbers. Why is that?Quite simply, large quantum numbers correspond to high energies where themomentum of the particle is also high, so the corresponding wavelength 𝜆 = h∕pis small enough that the wavelike behavior of the particle is negligible. Therefore,in the limit of large quantum numbers, quantum mechanical features should fadeaway and classical behavior should be gradually recovered. A typical quantumeffect is the quantization of the energy spectrum. We expect, therefore, that forlarge n, quantum discontinuities in the allowed energies gradually disappear andthe classical continuum is restored. In this context, if we now examine the for-mula En = n2E1 for the energy spectrum of an infinite potential well, we may besurprised to realize that in the limit of large n, the distance between successiveenergy levels

ΔEn = En+1 − En = (2n + 1)E1

does not decrease, but instead grows indefinitely. In other words, quantum dis-continuities are being amplified in the limit of large quantum numbers insteadof disappearing! What is going on? Well, it is wrong to expect the values of thediscontinuities themselves to gradually decrease and approach zero as n → ∞.Instead, it is the relative discontinuities (ΔEn)∕En that should gradually diminish.Indeed, we have

ΔEn

En= 2n + 1

n2 −−→n→∞

0.

As we now realize, even though the distance between successive eigenvaluesincreases as we go higher in the spectrum, the energies themselves increase muchfaster. Therefore at some point, for large enough n, the energy discontinuitiesbecome insignificant compared to the actual energy values. In fact, as we willsee shortly, the deep classical regime corresponds to values of n on the order of1027, so the relative discontinuity ΔEn∕En is then

ΔEn

En≈ 2

n≈ 2 × 10−27.

Put differently, quantum discontinuities are about 1027 times smaller than thevalue of the energy itself! As you may suspect, such values are beyond the capa-bilities of any experimental measurement. From a practical (experimental) pointof view, the energy spectrum can thus be regarded as essentially continuous.

What about the value n ≈ 1027 that corresponds to what we earlier calleddeep classical regime? This rough, order-of-magnitude, estimate for n emergeswhen we consider a typical classical system of a macroscopic particle in aone-dimensional box of macroscopic size. So let us assume we have a particle of


mass m = 1 g inside a box of width L = 1 cm and that its energy has also a typicalclassical value of, say, E = 1 erg. If we apply the formula En = ℏ2𝜋2n2∕2mL2 andignore (for order-of-magnitude estimates) the factor 𝜋2∕2, we obtain

ℏn ≈ 1 erg s

and, therefore, n ≈ 1027.

4.2.2.4 The Classical Limit of the Position Probability DensityIn addition to the disappearance of relative discontinuities, classical behavior isalso recovered for other characteristics of motion in the limit n → ∞. Take, forexample, the probability density P(x) of finding the particle in the vicinity of somepoint x. Classically, this quantity is proportional to the fraction of time (comparedto the full period) the particle spends inside the interval [x, x + dx]. If we considerthat in a single period the particle passes twice through the interval dx, we find

Pcl(x) dx = 2 dtT

⇒ Pcl(x) =2

T𝑣(x),

where 𝑣(x) (= dx∕dt) is the classical velocity of the particle at point x. Evidently, inregions where the velocity is large, the classical probability of finding the particleis small, and vice versa. In the case of an infinite potential well, the particle moveslike a ball that goes back and forth with constant speed between two completelyelastic walls. The period T is then T = 2L∕𝑣, so that Pcl(x) = 1∕L =constant. Inquantum mechanics, on the other hand, the position probability density is

P(x) = ||𝜓n(x)||

2 = 2L

sin2(n𝜋x

L

)

,

which for large n (e.g., n = 20) is plotted in Figure 4.7 (the figure also depicts thecorresponding classical distribution).

As we see from the figure, the quantum mechanical probability distributionPqu(x) oscillates symmetrically around the constant classical value Pcl = 1∕L. Ifwe now consider that in the classical limit, a huge number of such oscillationscan “fit” in any given region, no matter how small it is by macroscopic standards,we realize that what we actually measure is just a mean value of the oscillating

2

1.5

1

0.5

0.2 0.4 0.6 0.8 1

n = 20, L = 1

Pcl (x) = 1/L

x

Pqu (x) = (x)n2ψ

Figure 4.7 Quantum and classical position probability densities.


probability in the interval of interest. And this mean value is equal to the constantclassical value Pcl = 1∕L.

4.2.2.5 Eigenfunction Features: Mirror Symmetry and the Node TheoremTake a careful look at Figure 4.5 and you will notice that the eigenfunctionsare, in turn, even and odd with respect to the center of the box. This is ageneral feature of all potentials V (x) that are symmetric with respect to somepoint. If this point is x = 0, then the potential function satisfies the relationV (−x) = V (x), that is, it is an even function.

In this case, it is reasonable to expect that the eigenfunctions are even orodd, because only then do we have equal probabilities of finding the particle inpositive versus negative x, as required by the mirror symmetry of the problem.The fact that the eigenfunctions are alternately even and odd relates closely tothe so-called node theorem. According to this theorem, the number of nodes—thepoints where the wavefunction vanishes—increases by one as we move from theground state (with zero nodes) to higher states. In other words, the eigenfunc-tion of the first excited state has one node, of the second excited state has twonodes, and so on. From the point of view of wave theory, the node theorem isevident: Since quantum mechanical bound states are actually standing waves ofdefinite frequency—the so-called normal modes—their formation requires aninteger number of half-waves to fit within the given interval. The lowest statecorresponds to fitting one half-wave (with no intermediate nodes), the secondstate has two half-waves (and therefore one node), and so on. As we will findout later, this mechanism of eigenfunction formation is completely general, eventhough the various half-waves will not always have the simple sinusoidal formof the infinite potential well. It is now straightforward to see that, as the numberof nodes increases, the eigenfunctions will necessarily alternate between beingeven or odd, depending on whether their number of nodes is even or odd.

Finally, note that for each energy eigenvalue, there is only one eigenfunction,that is, the spectrum has no degeneracy. This lack of degeneracy is a property forbound states of all one-dimensional potentials, as the readers are asked to provein a problem at the end of the chapter.

4.2.2.6 Numerical Calculations in Practical UnitsFor real-world applications, where the particle is the electron and the proper unitsare the eV for energy and the Å for length, formula (4.11) for the allowed energiescan be rewritten as

En = ℏ2𝜋2

2meL2 n2 = ℏ2𝜋2

2mea20

(a0

L

)2n2 = 13.6 𝜋2

(a0

L

)2n2 eV, (4.15)

where a0 = ℏ2∕me2 ≈ 0.5 Å is the Bohr radius. We remind the readers (seeSection 1.3.10) that

ℏ2

2mea20= ℏ2

2me(ℏ2∕mee2)2 =mee4

2ℏ2 = WI = 13.6 eV.


Problems

4.1 Verify that the eigenfunctions (4.14) are orthogonal as they should. Whyshould they?

4.2 An electron is trapped in a one-dimensional box of length L = 2 Å. Cal-culate (in Å) the maximum wavelength of EM radiation the electron canabsorb while being in the ground state of the box.

4.3 The Square Potential Well


We will now examine what happens when the potential has finite depth, as inFigure 4.8. First of all, note that the potential has mirror symmetry. To take fulladvantage of this property, we shifted the origin of the axes at the center ofsymmetry of the problem, that is, at x = 0.

For the regions A, B, and C the Schrödinger equation is written as follows:A∶ 𝜓 ′′

A + (𝜖 − 𝑈0) 𝜓A = 𝜓 ′′A − 𝛾2𝜓A = 0

B∶ 𝜓 ′′B + 𝜖 𝜓B = 𝜓 ′′

B + k2 𝜓B = 0C∶ 𝜓 ′′

C + (𝜖 − 𝑈0) 𝜓 ′′C = 𝜓 ′′

C − 𝛾2𝜓C = 0,

where, as usual,

𝜖 = 2mEℏ2 , 𝑈0 =

2mV0

ℏ2

and

𝜖 = k2, 𝑈0 − 𝜖 = 𝛾2 (𝑈0 > 𝜖).

Let us now examine separately the even and odd solutions:

V0 V0

–a a

V(x)

A B C

x

E < V0

Figure 4.8 A square potential well of finite depth. The letters A, B, and C denote the threeregions of the x-axis where the potential has a given constant value. Since we are interested inbound states, the energy of the particle lies below the “edge” of the well (E < V0).

4.3 The Square Potential Well 141

Even SolutionsIn this case, the form of the wavefunction in each of the regions A, B, and C is

𝜓A = Ae𝛾x, 𝜓B = Bcos kx, 𝜓C = Ae−𝛾x.

As you can see, in regions A and C we kept only those exponentials that decay(i.e., die out) at infinity in each direction. In region A, for example, the generalsolution contains also the exponential exp (−𝛾x) that diverges (grows to infin-ity) in the limit x → −∞. So, to avoid having a divergent term in our solution forthis problem, we set to zero the constant multiplying the exponential exp (−𝛾x),which ensures the appropriate (for bound states) vanishing behavior of the wave-function at infinity. In region B, the solution is a linear combination of a sine anda cosine, but we kept only the cosine term, since we are here interested in evensolutions. Moreover, the constraint of having even solutions stipulates that theconstant C of 𝜓C = C exp (−𝛾x) has to be equal to A. In this way, the solutionsin regions A and C indeed satisfy the symmetry condition 𝜓A(x) = 𝜓C(−x).

Due to symmetry again, the continuity conditions for the wavefunction and itsderivative at x = a and x = −a are identical. Therefore, we need only consider theconditions for one of these two points (say, x = a). We thus have

𝜓B(a) = 𝜓C(a) ⇒ Bcoska = Ae−𝛾a (4.16)𝜓 ′

B(a) = 𝜓 ′C(a) ⇒ −B ksinka = −𝛾Ae−𝛾a, (4.17)

whence, if we divide (4.17) by (4.16), we obtain

tanka = 𝛾

k. (4.18)

Odd SolutionsIn this case, we have

𝜓A = Ae𝛾x, 𝜓B = Bsin kx, 𝜓C = −Ae−𝛾x,

where in regions A and C we retained again the “well-behaved” exponentials,while in region B we kept the sine, which is an odd function. Moreover, since wenow must have 𝜓C(x) = −𝜓A(−x), we chose C = −A. Note also that, for simplic-ity, we denoted the constants A and B using the same letter symbols as before,even though their specific values are different in the two cases (even versus oddsolutions). The continuity conditions at x = a are now written as

𝜓B(a) = 𝜓C(a) ⇒ Bsin ka = −Ae−𝛾a (4.19)𝜓 ′

B(a) = 𝜓 ′C(a) ⇒ B kcos ka = A𝛾e−𝛾a, (4.20)

and, if we divide again (4.19) by (4.20), we obtain

tan ka = −k𝛾. (4.21)

Graphical Solution of the Eigenvalue Equation Given that k2 = 𝜖, 𝛾2 = 𝑈0 − 𝜖, bothsides of Eqs (4.18) and (4.21) are functions of the unknown eigenvalue 𝜖, so theequations we need to solve have the general form f1(𝜖) = f2(𝜖). Thus, a graphicalsolution of these equations entails plotting the curves f1(𝜖) and f2(𝜖) as functions


of 𝜖 and finding their intersection points. This is the general idea, but in practiceone needs to get a little smarter. For example, it is not at all necessary to use 𝜖 itselfas a variable. We might just as well use as our variable a function 𝜉(𝜖), chosen sothat the curves are simpler to plot and their intersection points easier to inspect,thus resulting in a more accurate calculation of the eigenvalues. In addition, itmakes sense that the new variable 𝜉 is dimensionless, so that the graphical repre-sentations and our results have no dependence on the system of units we employ.An obvious choice of such a dimensionless parameter is the variable itself of thetangent function in (4.18) and (4.21), namely,

𝜉 = ak = a√𝜖 ⇒ a𝛾 = a

√𝑈0 − 𝜖 =

√𝜆2 − 𝜉2||

|𝜆=a√𝑈0

whereby Eqs (4.18) and (4.21) take the dimensionless form

tan 𝜉 =√𝜆2 − 𝜉2

𝜉(4.22)

and

tan 𝜉 = − 𝜉√𝜆2 − 𝜉2

, (4.23)

where

𝜆 = a√𝑈0 (4.24)

is a given dimensionless parameter of the problem. The graphical solution of(4.22) and (4.23) is shown in Figure 4.9 for a particular value of 𝜆 (𝜆 = 4) for whichthere are two intersection points 𝜉1 and 𝜉2 for (4.22) and only one for (4.23). Inother words, there are two bound states of even parity and only one of odd parity.

4

2

1 ξ1 ξ1

ξ2

ξ2π/2 3π/2π3 4

tan ξ

16 – ξ2

16 – ξ2

–2

–4

ξ

ξ

Figure 4.9 Graphical construction of the eigenvalues of the square well. For 𝜆 = 4 there aretwo bound states of even parity (𝜉1 and 𝜉2) and one of odd parity (𝜉1).


V0 V0V(x)

E3

E2

E1

–a ax

Figure 4.10 Eigenfunctions of a finite square well with 𝜆 = 4. This particular well has onlythree bound states.

In terms of 𝜉i and 𝜉i—the intersection points for even and odd solutions—theenergy eigenvalues of the well (Ei and Ei respectively) will be given by

Ei =ℏ2

2ma2 𝜉2i , Ei =

ℏ2

2ma2 𝜉2

i (i = 1, 2,…). (4.25)

Finally, let us remark that the eigenfunctions of the finite square well have thesame general shape as those of the infinite well, except that they do not vanish atthe edge points x = ±a. Instead, they have exponentially decaying “tails” insidethe classically forbidden region |x| > a. The wavefunction of the ground state iseven and node-less, the next state is odd with one node at x = 0, and so on. Thethree allowed eigenfunctions of the well for 𝜆 = 4 are depicted in Figure 4.10.


4.3.2.1 Penetration into Classically Forbidden RegionsA new quantum mechanical effect, which we have not encountered so far, is thepenetration of the particle inside a classically forbidden region, such as the inter-val |x| > a for a finite potential well. Now, from the viewpoint of wave theory,there is nothing paradoxical about this effect. Just like classical waves can entera region where they decay (e.g., electromagnetic waves inside a conductor), thesame effect occurs in quantum mechanics. In the region |x| > a, we have

k =√𝜖 − 𝑈0 = i

√𝑈0 − 𝜖 = i𝛾,

so the propagating wave exp (ikx) takes the exponentially decaying form exp(−𝛾x) with a decay constant 𝛾 . The fact that the particle has a nonzero probabilityto be in a region where the potential energy is greater than the total energy


leads often to the misunderstanding that, in quantum mechanics, the principleof energy conservation does not always hold. The error in this reasoning is thefollowing. In classical mechanics, both the kinetic and potential energy can beseparately measured and their sum gives the constant value of the particle’stotal energy. But in quantum mechanics this is no longer true, since the totalenergy is a quantity that characterizes the state of interest as a whole and cannotbe written down at each position as a sum of a kinetic and a potential term.In other words, it makes no sense to say that at a particular point x we haveE < V (x), because the quantities V (x) and E ≡ H = p2∕2m + V (x) cannot bemeasured simultaneously, since neither can x and p. So we cannot have a preciseknowledge of V (x) and E at the same time.

We can view this paradox of “penetration” in classically forbidden regions ina slightly different way: In region x > a, the wavefunction has the exponentiallydecaying form 𝜓(x) ∼ exp (−𝛾x), so the penetration is practically confinedwithin a distance l = 𝛾−1 outside the well. (Hereafter, we refer to l as the penetra-tion length.) To measure the position of the particle in the classically forbiddenregion x > a, we need to be able to locate it within a region that is at most equalto l, that is, Δx ⩽ l. Only then can we be certain of having detected the particleoutside the well. But then the particle’s kinetic energy after the measurementwill be at least equal to

(Δp)2

2m= ℏ2

2ml2 = ℏ2𝛾2

2m= ℏ2

2m(𝑈0 − 𝜖) = V0 − E.

Therefore, the very measurement that allows us to detect the particle in the classi-cally forbidden region perturbs its energy by at least the amount needed to “pull”it outside the well.

4.3.2.2 Penetration in the Classical LimitLet us now analyze qualitatively the dependence of the penetration length

l = 𝛾−1 =

√

ℏ2

2m(V0 − E)(4.26)

on the parameters of the problem. Note first that in the classical limit ℏ → 0,m → ∞, the penetration length vanishes as expected, since a classical particlecannot enter an energetically forbidden region. By contrast, in the strong quan-tum limit m → 0 (or ℏ→ ∞), the penetration length increases without bound.We are thus reminded once again that lighter particles have much more intensequantum mechanical behavior. Note also that l decreases as the height V0 of thewell increases, and vanishes in the limit V0 → ∞, whereby we obtain the infinitepotential well and the particle is indeed completely confined inside it. Now, thepenetration length also depends on the particle energy E, so that it is much greaterfor the higher excited energy states than for the ground state. And this is why weare justified to approximately treat a finite well as infinite, when only the groundstate or the first few excited states are concerned. But we cannot extend such anapproximation to much higher excited states. Intuitively, this result makes sense:When the particle is energetically near the bottom of a relatively deep well, it


cannot really “tell the difference” from an infinite well. Conversely, for states nearthe edges of the well, the finite depth of the well is directly noticeable.

4.3.2.3 The Physics and “Numerics” of the Parameter 𝝀Note finally that 𝜆—the parameter that determines the energy eigenvalues of thewell—is equivalently written in the more transparent form

𝜆 = a√𝑈0 = a

√2mV0

ℏ2 =

√V0

(ℏ2∕2ma2)=√

V0

,

(

= ℏ2

2ma2

)

(4.27)

that is, as the square root of the ratio of the two energy scales of the problem: thedepth of the well, V0, and the kinetic energy—due to the uncertainty principle—ofthe particle localized in an interval half the size of the well’s width, (= ℏ2∕2ma2).Now, if (V0∕)≫ 1, the well can be characterized as deep—its depth is muchgreater than the kinetic energy of a particle trapped in it—while if (V0∕)≪ 1,the well is shallow. And it is a simple exercise for the readers to show that thenumber N of bound states is determined solely by the value of 𝜆, through theformula

N =[𝜆

𝜋∕2

]

+ 1, (4.28)

where [x] is the integer part of the number x. To facilitate numerical calcula-tions in practical units, the quantity above can be rewritten—following a similartreatment in (4.15)—as

= ℏ2

2ma2 = ℏ2

2mea20

me

m

(a0

a

)2= 13.6

me

m

(a0

a

)2eV, (4.29)

whence the practical formula for 𝜆 will take the form

𝜆 =√

V0(eV)13.6

√mme

aa0. (4.30)

It is also useful to rewrite formulas (4.25) for the eigenvalues of the well as

Ei

= 𝜉2i ,

Ei

= 𝜉2

i , (4.31)

which means that their dimensionless values—or, equivalently, their values mea-sured in the appropriate unit —are the same for all wells. They are determinedsolely by the value of the parameter 𝜆.

As for the penetration length (4.26), we leave it to the readers to show that foran arbitrary eigenstate indexed by i, it is given by

𝓁i

a= 1

√

𝜆2 − 𝜉2i

, (4.32)

and that these values—measured in the appropriate unit of length—are again“universal” as before.

The general conclusion is clear: Dimensional analysis is also a powerful tool fornumerical calculations in quantum mechanical problems.


Problems

4.3 Use the graphical construction of Figure 4.9 to show that the ground-stateenergy of the square well, E1 = ℏ2𝜉2

1∕2ma2, is always lower than the energyof the corresponding state for the infinite potential well of the same size(that is, with L = 2a). Can you explain why this is so?

4.4 Decide which one of the following wells has more bound states. In bothcases, the particle in the well is an electron.(a) V0 = 3.4 eV, a = 3 Å,(b) V0 = 13.6 eV, a = 1.5 Å.

4.5 The ground-state energy of a square well is 1∕4 of its height, that is,E = V0∕4. Determine the number of bound states of this particular well.Hint: Show first that E∕V0 = 𝜉2∕𝜆2.

Further Problems

4.6 Calculate the uncertainties Δx and Δp for an arbitrary eigenfunction𝜓n(x) =

√2∕L sin (n𝜋x∕L) of an infinite potential well. In particular,

show that

Δx = L√

12

(

1 − 6n2𝜋2

)1∕2, Δp = ℏ𝜋

Ln.

Do these results satisfy Heisenberg’s uncertainty principle?

4.7 At a certain moment, the wavefunction of a particle inside an infinite wellof width L has the form

𝜓(x) = Nx(L − x). (1)

(a) Calculate the probability that an energy measurement yields the eigen-value E1 of the ground state. What is the wavefunction of the particleimmediately after such a measurement?

(b) Can you predict, without any calculation, the probability that anenergy measurement yields the energy of the first excited level of thewell for the above wavefunction?

4.8 At time t0 = 0 the state of a particle of mass m inside an infinite well isdescribed by the wavefunction

𝜓(x) = N sin3 𝜋xL. (1)

(a) What are the possible outcomes of energy measurements when theparticle is in this state, and what is the probability of each such out-come?


(b) Calculate the mean energy and the energy uncertainty of the particlein state (1). Express your results as functions only of the energy E1 ofthe ground state.

(c) What is the mean position of the particle at t0 = 0 and after the lapseof time t?

4.9 An electron is inside a square potential well with depth V0 = 4 eV andwidth L = 10 Å. Confirm that this well can be characterized as deep, anduse the analytic approximation described in the online supplement to cal-culate the energy eigenvalues in eV. For which of these eigenvalues canyou be confident that the approximation is very good (error< 1%), and forwhich ones do you expect a bigger error?

4.10 As we have already stressed, we characterize a particular well as deep orshallow for a certain particle (say, an electron) based not only on its depthV0 but also on its width L = 2a. In effect, the objective measure of thewell’s depth is how many bound states it can “accommodate.” This numberdepends on the dimensionless parameter 𝜆 that contains both the depthand the width of the well. In this context, decide which of the followingthree wells is the deepest and which the shallowest, assuming that the“guest” particle in the well is an electron.(a) L = 1 Å, V0 = 20 eV,(b) L = 10 Å, V0 = 2 eV,(c) L = 2 Å, V0 = 10 eV.For each of these cases, calculate the ground-state energy and any excitedstate energies you think can be calculated.

4.11 You are given the “semi-infinite” square well

V(x)

V0

a

x

V (x) =∞, −∞ < x < 00, 0 < x < aV0 x > a,

Solve the Schrödinger equation for a particle of mass m and show how tocalculate the allowed energies in this case. In particular, show that boundstates exist only when

V0 a2 >ℏ2𝜋2

8m.

Subsequently, show that we do not need to find the solutions of the prob-lem anew, as they can be derived directly from the solutions of the finitewell we treated in the text.


4.12 A particle of mass m is confined inside the two-dimensional rectangularbox shown in the figure. Find the allowed values of its energy.

y

x

L

L

Hint: One can solve the problem either fromscratch (with the method of separation ofvariables) or by treating the potential as a kind of“superposition” of two one-dimensional wells inthe x and y directions.

4.13 A particle of mass m is trapped inside the three-dimensional box shownin the figure. Calculate the allowed energy values. In particular, examine

a

b

c

z

y

x

the case of a cubic box with a = b = c = L andgive the expressions for the first five differenteigenvalues and the corresponding eigenfunc-tions. Pay attention to degeneracy, namely,whether certain eigenvalues are shared by morethan one eigenfunction.

4.14 For a very deep and narrow well, the potential function V (x) can bedescribed approximately by a delta function of the form

V (x) = −g𝛿(x) (g > 0).

Solve the Schrödinger equation for this potential and show that there isonly one bound state with eigenvalue E1 = −mg2∕2ℏ2 and correspondingeigenfunction 𝜓1(x) = Ne−𝛾|x|, where 𝛾 = mg∕ℏ2.Hint: Show first that the presence of the delta function in the Schrödingerequation implies the following matching conditions at x = 0:

𝜓L(0) = 𝜓R(0), 𝜓 ′L(0) − 𝜓

′R(0) = 𝜆𝜓(0),

where 𝜆 = 2mg∕ℏ2 and 𝜓L, 𝜓R are the solutions to the left and right ofx = 0, respectively.

4.15 Use any knowledge you have acquired from the theory of ordinary differ-ential equations (e.g., on the concept of a Wronskian) to prove that thebound states of the one-dimensional Schrödinger equation

𝜓 ′′ + (𝜖 − 𝑈 (x))𝜓 = 0

are not degenerate, that is, to each allowed energy value correspondsonly one eigenfunction. Subsequently, use this property to show thatfor a potential with mirror symmetry, that is, 𝑈 (x) = 𝑈 (−x), the energyeigenfunctions are either even or odd.

149

5

Square Potentials. II: Continuous Spectrum—ScatteringStates

5.1 Introduction

In the introduction of Chapter 4, we stated that quantum mechanical motionin a one-dimensional potential is confined, provided the corresponding classicalmotion in the same energy range is also confined. We also presented plausiblephysical arguments to justify our statement that the energy spectrum is discretefor confined and continuous for unconfined motion. The two problems we inves-tigated in Chapter 4 (namely, the infinite and finite potential wells) confirmedfully the first leg of this assertion (confined motion⇔ discrete spectrum). We nowexamine what happens when the motion is unconfined and whether the associ-ated spectrum is indeed continuous. Thus, in this chapter, we study two cases ofsquare potentials that allow only unconfined motion from the classical viewpoint,and which we expect to behave likewise in quantum mechanics.

Let us first recall (see Section 2.5.5) that the wavefunction

𝜓 = Aeikx (5.1)

is an eigenfunction of the momentum operator, p = −iℏd∕dx, with an eigenvaluep = ℏk. It therefore describes particles moving with definite momentum p = ℏkto the right or to the left for a positive or negative wavenumber k, respectively.In the language of waves we say that the wavefunction eikx (for k > 0) describesa wave that propagates rightward, while the wavefunction e−ikx describes a wavethat propagates leftward. If we now consider that, in a real experiment, we typi-cally have a very large number of particles that are all described by the wavefunc-tion (5.1), then it is clear that they form a beam whose current (or flux) density(defined as the number of particles crossing per second the unit area of a planeperpendicular to the direction of the beam) is given by the formula

J = P𝑣 = |𝜓|2𝑣 = |A|2 ℏkm. (5.2)

Relation (5.2) is a plausible quantum analog of the classical formula J = 𝜌𝑣, withthe probability density P = |𝜓|2 in place of the classical density 𝜌, and the speed𝑣 given by 𝑣 = p∕m = ℏk∕m.

For those readers who are unfamiliar with the classical hydrodynamic formulaJ = 𝜌𝒗, we note that it gives the current density J of any physical quantity (mass,


150 5 Square Potentials. II: Continuous Spectrum—Scattering States

charge, energy, etc.) distributed in space with density 𝜌 and moving with (local)velocity 𝒗. The magnitude of the vector J tells us how much of this physical quan-tity crosses per unit time a unit area that is perpendicular to the direction of theflow, which is the direction of the local velocity 𝒗. From a physical point of view,the formula J = 𝜌𝒗 is quite plausible. When we deal with the flow of an ordinaryliquid (a flow of mass), the current density must be proportional to the densityof the fluid and to the speed of its motion. We can also apply this hydrodynamicformula to a beam of quantum particles using the plausible analogy 𝜌→ P = |𝜓|2

and 𝑣→ p∕m = ℏk∕m. This can be shown rigorously using the quantum expres-sion for J (See OS2.3.). But for our purposes here, it is better to work with themore tangible semiclassical picture.

We begin with the simplest possible problem of a square potential that allowsonly unconfined motion: the so-called square potential step.

5.2 The Square Potential Step: Reflection andTransmission

5.2.1 Solution of the Schrödinger Equation and Calculation of theReflection Coefficient

As can be inferred from its name, the square potential step has the characteristicshape of Figure 5.1. Such a potential cannot, obviously, keep a particle in a boundstate, so we expect the spectrum to be continuous for all energies E > 0.

Since the solution in region B depends crucially on whether E > V0 or E < V0,we will study these two cases separately.Case I: E > V0The Schrödinger equation in each of the regions A and B is written as

A∶ 𝜓 ′′A + 𝜖𝜓A = 𝜓 ′′

A + k2𝜓A = 0 (k2 = 𝜖)B∶ 𝜓 ′′

B + (𝜖 − 𝑈0)𝜓B = 𝜓 ′′B + k′2𝜓B = 0 (k′2 = 𝜖 − 𝑈0 > 0)

and has a general solution𝜓A = A+eikx + A−e−ikx

𝜓B = B+eik′x + B−e−ik′x

}

. (5.3)

V0

BA

E

V(x)

Ae–ikx

eikx Beik′ x

x

Figure 5.1 Square potential step and classically forbidden reflection. The particle has a finiteprobability to be reflected when it encounters the potential step, even though its energy E isgreater than the step height V0.

5.2 The Square Potential Step: Reflection and Transmission 151

As in Chapter 4, we use the letter of each region as a subscript to denote thecorresponding wavefunction and the constants of the general solution. We alsouse the + index for the constant accompanying the plane wave exp (ikx) trav-eling to the right, and the − index for the wave exp (−ikx) traveling to the left.Finally, whenever the conditions of the problem allow us to eliminate one ofthe two constants, we will drop the ± index from the other constant also, forsimplicity.

As we discussed in the introductory comments of Chapter 4, the eigenfunctionsof the continuous spectrum are finite everywhere but are not square integrable,since they do not vanish at infinity. Their physical significance is that they aresuitable for the description of the scattering process. For a one-dimensional prob-lem, as in the present case, a scattering experiment consists of firing a beam ofparticles from the left and observing how many of them get reflected back, andhow many go over the step and get detected by a counter on the right. For thegeneral solution to simulate such an experiment, its form in region B must con-tain only the wave exp (ik′x) that describes rightward moving particles. Thus, thecoefficient B− of the wave exp (−ik′x) that comes from the right must vanish. Inregion A, the coefficients A+ and A− determine the densities of the incident andreflected beam, respectively, through the relation 𝜌 = P = |A|2, valid for a planewave𝜓 = A exp (ikx). And since we can adjust the characteristics of the incidentbeam at will, we can always set its density equal to unity, that is, A+ = 1.

Given the physical meaning of the conditions

A+ = 1, B− = 0, (5.4)

we refer to them as scattering boundary conditions or, more precisely, conditionsfor scattering from the left, since it is also possible to send the beam from the right,whence B− = 1 and A+ = 0.

The general solution (5.3) takes now the form

𝜓A = eikx + Ae−ikx, 𝜓B = Beik′x,

where we dropped the ± indices as there is now only one constant in each region.The experimentally interesting quantities in a one-dimensional scattering

experiment are the reflection (R) and transmission (T) coefficients, defined as

R =JR

JI, T =

JT

JI,

where JI, JR, and JT are the current densities of the incident, reflected, and trans-mitted beam, respectively. The coefficient R gives the probability of reflection ofthe particle, while T provides the probability of transmission. Since these are theonly available options (the particle either bounces back or goes through), the sumof these two probabilities must equal unity, that is,

R + T = 1. (5.5)

Indeed, if we consider the relation

JI = JR + JT,


which expresses the conservation of the number of particles at the reflectionpoint—the number of incident particles is equal to the sum of the numbers ofreflected and transmitted particles—then Eq. (5.5) follows directly from the defi-nitions of R and T . Now, for a plane wave 𝜓 = A exp (ikx), the current J is given,as we noted already, by the simple “hydrodynamic formula” (5.2). Indeed, in areal scattering experiment we use a beam of particles with the same momentum,so the probability density P = |𝜓|2 = |A|2 is equal to the actual beam density𝜌, which is the number of particles per unit volume (or per unit length, for aone-dimensional problem, as in this case).

For a potential step, the reflection and transmission coefficients are expressedin terms of the coefficients A and B as follows:

R =JR

JI=

|A|2 ℏkm

1 ⋅ ℏkm

= |A|2

and

T =JT

JI=

|B|2 ℏk′

m

1 ⋅ ℏkm

= |B|2 k′

k,

where the constants A and B will be computed from the continuity conditions atx = 0:

𝜓A(0) = 𝜓B(0) ⇒ 1 + A = B,𝜓 ′

A(0) = 𝜓 ′B(0) ⇒ ik − ikA = ik′B,

whereby

A = k − k′

k + k′ , B = 2kk + k′ . (5.6)

We thus find

R =(

k − k′

k + k′

)2

, T = 4kk′

(k + k′)2 , (5.7)

and the equality R + T = 1 is indeed satisfied, as can be easily checked.Case II: E < V0Our equations are now written as

𝜓 ′′A + 𝜖𝜓A = 𝜓 ′′

A + k2𝜓A = 0,𝜓 ′′

B + (𝜖 − 𝑈0)𝜓B = 𝜓 ′′B − 𝛾2𝜓B = 0 (𝛾2 = 𝑈0 − 𝜖)

and the physically acceptable solutions are𝜓A = eikx + Ae−ikx, 𝜓B = Be−𝛾x.

In region B, we retained only the exponential exp (−𝛾x) that vanishes at +∞,to ensure that the wavefunction remains finite everywhere (including infinity). Inregion A, we set the coefficient of the incident wave equal to unity, according tothe convention mentioned earlier. The continuity conditions at x = 0 now yieldthe constants A and B as

A = k − i𝛾k + i𝛾

, B = 2kk + i𝛾

. (5.8)


Unsurprisingly, expressions (5.8) are derived from (5.6) with the substitutionk′ = i𝛾 , because as we switch from E > V0 to E < V0, the quantity in the squareroot in the formula k′ =

√𝜖 − 𝑈0 becomes negative, and we get

k′ =√𝜖 − 𝑈0 = i

√𝑈0 − 𝜖 = i𝛾.

The reflection coefficient becomes then

R = |A|2 =||||

k − i𝛾k + i𝛾

||||

2= 1,

while the transmission coefficient vanishesT = 0,

which is another unsurprising result. Since the wavefunction in region B isa decaying exponential, the particle cannot reach +∞ to get detected by thecounter there. Thus, when the energy of the particle is smaller than the stepheight we have total reflection. Even in quantum mechanics, particles cannot gothrough a potential barrier of infinite width.

We summarize our results for both regimes (E < V0 and E > V0) as follows.

R =

E > V0−−−→ =k − k

k + k

2

=√

E −√E − V0√

E +√

E − V0

2

E < V0−−−→ = 1(5.9)

and

T =

E > V0−−−→ =4kk

(k + k )2=

4 E(E − V0)(√

E +√

E − V0)2

E < V0−−−→ = 0(5.10)

while in Figure 5.2 we sketch the coefficients R and T as functions of the energyE in the quantum and classical case.


5.2.2.1 The Phenomenon of Classically Forbidden ReflectionThe most notable consequence of the quantum mechanical solution wefound is, surely, the possibility of reflection by the potential step even whenthe incident particle has the energy required to go through (E > V0). Thisbehavior is the exact opposite of the penetration into classically forbiddenregions we encountered in Chapter 4. In that case, the particle enteredregions where it is not allowed to be, while now it sometimes avoids enteringregions where it has to enter! From the viewpoint of wave theory, how-ever, both these types of behavior are perfectly legitimate. For example,the potential jump at x = 0 in the present case is completely analogous tothe discontinuity of the refraction index n at the interface between twomedia, which causes an electromagnetic wave incident at the interface to bereflected.


Rqu

(a) (b)

(c) (d)

Rcl

Tqu

Tcl

1

1

1

1

E

V0

V0

V0

V0

E E

E

Figure 5.2 Reflection and transmission coefficients for the square potential step: quantumcase (a,b) and classical case (c,d).

5.2.2.2 Transmission Coefficient in the “Classical Limit” of High EnergiesAs is evident from relation (5.10), in the limit E → ∞, the transmission coefficienttends to unity, that is, it takes its classical value. There is no surprise here. As wenoted earlier, for very high energies (i.e., very small wavelengths), wave behavior isno longer noticeable and classical mechanics applies. The effect is actually similarto the reduction of wave optics to geometrical optics for small wavelengths.

5.2.2.3 The Reflection Coefficient Depends neither on Planck’s Constant noron the Mass of the Particle: Analysis of a ParadoxGiven that the reflection and transmission coefficients can also be defined in thecorresponding classical problem, it is reasonable to expect that in the limit ℏ→ 0or m → ∞, the functions R and T take their classical stepwise form (Figure 5.2).And yet, the expressions we obtained,

R =

(√E −

√E − V0

√E +

√E − V0

)2

, T =4√

E(E − V0)

(√

E +√

E − V0 )2,

contain neither ℏ nor m! Their absence is rather odd. Although we are dealingwith a purely quantum mechanical problem, the results appear to be independentfrom the characteristic quantum constant ℏ!

The origin of this paradox can be traced back to the physically unrealisticdiscontinuity of the potential at x = 0. Let us elaborate. As we said before, forsmall wavelengths the wave features gradually disappear and classical behavioris restored. But what does “small wavelengths” really mean? Small comparedto what? From the celebrated effect of diffraction through a slit, we know thatwavelike behavior appears when the wavelength is greater than, or at least onthe same order as, the slit size. If the wavelength is much smaller than the slit,then diffraction disappears and the straight-line propagation of geometrical


optics is restored. In other words, small wavelength here means small comparedto the size of the slit. We can readily generalize this conclusion by recalling thata slit is really just a change in the properties of the medium in which the wavepropagates. The medium is totally absorbing and opaque everywhere, except inthe region of the slit where it becomes transparent. The size of the slit is thusthe extent of the area where there is a noticeable change in the properties of themedium. Therefore, the general conclusion can be stated as follows:

When a wave propagates in an inhomogeneous medium, wave phenomenabecome apparent only if the wavelength is greater than the inhomogeneity length,that is, the linear size of the region wherein the properties of the medium exhibita noticeable change. To convince ourselves that this discussion applies also inquantum mechanics, let us first note that the time-independent Schrödingerequation

𝜓 ′′ + 2mℏ2 (E − V (x))𝜓 = 0 (5.11)

has the form of the classical (time-independent) wave equation

𝑈 ′′ + k2𝑈 = 0, (5.12)

which is derived directly from the time-dependent wave equation

uxx −1c2 utt = 0 (5.13)

upon setting u(x, t) = 𝑈 (x)ei𝜔t , or u(x, t) = 𝑈 (x)cos(𝜔t + 𝜙). (This is the substi-tution we make when we are looking for solutions of definite frequency 𝜔.) Withthis substitution, Eq. (5.13) transforms into (5.12) with k = 𝜔∕c. If the mediumof propagation is homogeneous, in which case c is a constant, then k is constantalso. If the medium is inhomogeneous, then both the propagation speed c andthe wavenumber k are functions of x.

By comparing (5.12) with (5.11), we see immediately that the Schrödingerequation itself is a wave equation with a varying wavenumber

k(x) =√

2mℏ2 (E − V (x)),

whose spatial variation is caused by the potential V (x) of the problem at hand.In the case of a potential step, the potential changes abruptly at one point,

so the inhomogeneity length is l = 0. Therefore, no matter how small ℏ is, thewavelength of the particle, 𝜆 = h∕p, is always infinitely greater that l = 0. Forthis reason, the wave character of the particle’s motion never diminishes and theresult does not depend on ℏ.

5.2.2.4 An Argument from Dimensional AnalysisActually, we could have predicted the independence of the reflection coefficientR from ℏ from the outset, using dimensional analysis. Since R is a dimensionlessquantity, it depends only on dimensionless combinations of the parameters ℏ, m,V0, and E of the problem. But the only such combination one can form is the ratio


E∕V0.1 Therefore, R can only depend on this ratio, that is, R = R(E∕V0), which isactually what we obtained earlier. Clearly, neither Planck’s constant nor the massof the particle can enter the formula for the reflection coefficient.

Once again, we appreciate dimensional analysis as a powerful tool for analyzingproblems with a small number of parameters.

Problems

5.1 The wavefunction of a particle incident from the left on a square potentialstep at x = 0 has the form (for x < 0)

𝜓A = 5eikx + 3e−ikx.

What is the probability for the particle to be reflected?

5.2 A beam of alpha particles with energy E = 50 eV impinges on a voltage step24 V high. What fraction of the particles will go through and what fractionwill be reflected?

5.3 After being accelerated by a voltage difference of 9 V, a beam of electronsimpinges on a square potential step, where 25% of them are reflected. Whatis the height of the potential step in eV?

5.3 Rectangular Potential Barrier: Tunneling Effect


The potential has now the shape of Figure 5.3. Evidently, the energy spectrum iscontinuous and spans the whole range from zero to infinity (0 < E <∞). For bothphysical and mathematical reasons we need to study separately the cases E > V0and E < V0.Case I: E > V0The Schrödinger equation in the three regions A, B, and C is now written as

𝜓 ′′A + 𝜖 𝜓A = 𝜓 ′′

A + k2 𝜓A = 0 (k2 = 𝜖)𝜓 ′′

B + (𝜖 − 𝑈0)𝜓B = 𝜓 ′′B + k′2𝜓B = 0 (k′2 = 𝜖 − 𝑈0)

𝜓 ′′C + 𝜖 𝜓C = 𝜓 ′′

C + k2𝜓C = 0.If we take into account the boundary conditions for scattering of particles comingfrom the left, the general solution is simplified to

𝜓A = eikx + Ae−ikx

𝜓B = B+ eik′x + B− e−ik′x

𝜓C = C eikx.

1 Here is another “uniqueness theorem” of dimensional analysis. Given any four physicalquantities—provided that any three of them are dimensionally independent—there is only onedimensionless combination of them.

5.3 Rectangular Potential Barrier: Tunneling Effect 157

E > V0

E < V0

V0

x = 0 x = L

xCBA

Figure 5.3 Scattering by a rectangular potential barrier. A quantum mechanical particle has afinite probability to be reflected even when its energy compels it classically to go through thebarrier (E > V0). Conversely, it has a finite probability to go through even when its energyforbids this within classical mechanics (E < V0).

Since the wavenumber k is the same in regions A and C, the reflection and trans-mission coefficients are now given by the expressions

R = |A|2, T = |C|2,

where, because of R + T = 1, we need only determine the constant A.The continuity conditions at points x = 0 and x = L yield

𝜓A(0) = 𝜓B(0) ⇒ 1 + A = B+ + B− (5.14)𝜓 ′

A(0) = 𝜓 ′B(0) ⇒ ik(1 − A) = ik′(B+ − B−) (5.15)

𝜓B(L) = 𝜓C(L) ⇒ B+ eik′L + B− e−ik′L = C eikL (5.16)𝜓 ′

B(L) = 𝜓 ′C(L) ⇒ ik′(B+ eik′L − B− e−ik′L) = ik C eikL. (5.17)

A quick way to calculate A from Eqs (5.14)–(5.17) is the following. From the firsttwo equations we can calculate the ratio B+∕B− as a function of A. Moreover, wecan divide both sides of the other two equations to show that the ratio B+∕B− isindependent of C. If we equate the two expressions for B+∕B−, we obtain a simpleequation for A as the only unknown constant. This procedure yields

Eqs (5.14), (5.15) ⇒B+

B−= (k′ + k) + (k′ − k)A

(k′ − k) + (k′ + k)A.

Eqs (5.16), (5.17) ⇒B+

B−= k′ + k

k′ − ke−2ik′L.

By equating the right-hand sides of these equations and solving for A we find

A = i(k′2 − k2) sin k′L2kk′ cos k′L − i(k′2 + k2) sin k′L

,

whence we obtain the reflection coefficient

R = |A|2 =𝑈 2

0 sin2 k′L𝑈 2

0 sin2 k′L + 4k2k′2

and the transmission coefficient

T = 1 − R = 4k2k′2

𝑈 20 sin2 k′L + 4k2k′2

.


Case II: E < V0Actually, we do not need to carry out the calculation anew, since the desiredresults can be obtained from the previous calculation with the simple substitutionk′ = i𝛾 . We recall that k′ =

√𝜖 − 𝑈0, so for 𝜖 < 𝑈0 we get

k′ =√𝜖 − 𝑈0 = i

√𝑈0 − 𝜖 = i𝛾 (𝛾 =

√𝑈0 − 𝜖).

The substitution k′ = i𝛾 in sin k′L gives

sin k L =12i

eik L − e−ik L k = iγ===

12i

e−γL − eγL

= i12

eγL − e−γL = i sinh γL,

where we used the definitions

cosh x = ex + e−x

2, sinh x = ex − e−x

2of the hyperbolic functions cosh x and sinh x. Thus, we arrive at the followingexpressions for R and T

R =𝑈 2

0 sinh2𝛾L

𝑈 20 sinh2

𝛾L + 4k2𝛾2

T = 4k2𝛾2

𝑈 20 sinh2

𝛾L + 4k2𝛾2.

Let us summarize the results for the transmission coefficient (the quantity we aremost interested in for this problem) in the energy ranges E > V0 and E < V0:

TE>V0(E) = 4k2k′2

𝑈 20 sin2k′L + 4k2k′2

=4𝜖(𝜖 − 𝑈0)

𝑈 20 sin2(L

√𝜖 − 𝑈0) + 4𝜖(𝜖 − 𝑈0)

(5.18)

TE<V0(E) = 4k2𝛾2

𝑈 20 sinh2

𝛾L + 4k2𝛾2=

4𝜖(𝑈0 − 𝜖)𝑈 2

0 sinh2(L√𝑈0 − 𝜖 ) + 4𝜖(𝑈0 − 𝜖)

.

(5.19)In Figure 5.4, we sketch the function T = T(E).


5.3.2.1 Crossing a Classically Forbidden Region: The Tunneling EffectThe fundamentally new quantum mechanical phenomenon that emerges fromthe solution of this problem is the tunneling effect: the ability of quantum par-ticles to cross classically forbidden regions of finite extent and continue theirmotion on the other side of the barrier. The term “tunneling” stems from theclassical analog of a ball that rolls up a hill, but its initial speed does not allowit to reach the hilltop and cross over to the other side. If the ball appears on theother side after all, we have no choice (in classical physics) but to assume thatthe hill has some sort of secret tunnel, which opens up when the ball reaches itsentrance and allows it to pass through to the other side! See Figure 5.5.


1

0.2

0.4

0.6

0.8

1

T(E / V0)

(E / V0)

2 3 4

Figure 5.4 The transmission coefficient as a function of the energy of the particle for a squarepotential barrier with 𝜆 = L

√𝑈0 = 5. The nonvanishing values of T for E < V0 confirm the

tunneling effect, while the values of T smaller than 1 for E > V0 show that “forbidden reflection”also takes place. The instances of T = 1 for E > V0 correspond to the so-called resonances.

Figure 5.5 Classical “analog” of the tunneling effect. The spherical ball does not have theenergy required to cross over the hill, and yet it does emerge on the other side using the“tunnel” on the hillside. The probabilistic nature of this quantum phenomenon—sometimesthe particle crosses the hill, sometimes it does not—is described classically with a randomopening or closing of the tunnel’s entrance!

Actually, we could have anticipated the possibility of particles crossing classi-cally forbidden regions earlier, when we encountered the possibility of quantumparticles penetrating into such regions. If a classically forbidden region has finiteextent (as in Figure 5.6), the exponential decay of the wavefunction inside it maynot always be sufficiently strong to eliminate the probability for the particle tocross the barrier and appear again in the classically allowed region.

5.3.2.2 Exponential Sensitivity of the Tunneling Effect to the Energyof the ParticleFrom a physical point of view, it is particularly interesting to study a barrier thatis sufficiently wide (large L) and high compared to the energy of the particle(V0 ≫ E). In this case, the variable 𝛾L =

√𝑈0 − 𝜖 L of the hyperbolic sine in

the transmission coefficient (5.19) is much greater than unity (𝛾L ≫ 1) and we


V0

EE

λλ

x

x = Lx = 0

Figure 5.6 General shape of the wavefunction of a particle that crosses a classically forbiddenregion. Even though the amplitude of the wavefunction decays exponentially inside thebarrier region, there is a finite probability for the particle to reach the other side and continueits motion as a wave with a significantly reduced amplitude.

can write

sinh 𝛾L = 12(e𝛾L − e−𝛾L) ≈ 1

2e𝛾L,

so that expression (5.19) takes the approximate form

T(E) ≈ 16 EV0

(

1 − EV0

)

exp (−2𝛾L)

= 16 EV0

(

1 − EV0

)

exp (−2L√

2m(V0 − E)∕ℏ2) .(5.20)

Of broader significance here is the exponential sensitivity of the transmissioncoefficient to variations of the particle energy and barrier width. That is, eventiny changes in the energy E or the width L result in huge changes in the transmis-sion coefficient. This exponential sensitivity of the tunneling effect with respect toenergy is seen in a spectacular way in the alpha decay of nuclei, where the lifetimeindeed varies exponentially with the energy of the emitted alpha particle. In par-ticular, while the energies of the emitted particles vary within a very narrow rangeof a few MeV (from 4 to 9 MeV), the lifetimes of radioactive nuclei range from10−7 s to 1010 years! We should stress here that in alpha decay, the exponentialenergy sensitivity is also enhanced by the shape of the barrier, which is not squarebut has the shape shown in Figure 5.7. As is evident from the figure, differencesin the particles’ energy also cause variations in the “active width” of the barrier,which in turn affect the probability of escape exponentially. We understand nowwhy the lifetimes of nuclei that undergo alpha decay show such gigantic varia-tions. The exponential sensitivity of the tunneling effect to the particle’s energyand the barrier’s width is the key to this fascinating phenomenon.

5.3.2.3 A Simple Approximate Expression for the Transmission CoefficientGiven that the energy of alpha particles does not vary much, the factor multi-plying the exponential in formula (5.20) varies very slowly compared to the hugevariation of the exponential in the same energy region. So we are justified to treatthis “prefactor” as a constant (we typically set it equal to unity) and write

T(E) ≈ exp (−2𝛾L), 𝛾 =√

2m(V0 − E)ℏ2 . (5.21)


V(r)

Vmax

V =r

2 Z e2

E < Vmax

R R′r

Figure 5.7 The potential experienced by an alpha particle in a radioactive nucleus. Inside thenucleus (r < R), the potential has the form of an attractive well due to the strong nuclearforces. Outside, it has the form of a repulsive Coulomb potential due to the electrostaticrepulsion between the escaping alpha particle and the rest of the nucleus. Since the alphaparticle stays for a long time inside the nucleus, we must have E < Vmax, so the only way for theparticle to escape the nucleus is via the tunneling effect.

This approximate form for the transmission coefficient can be understood easilybased on Figure 5.6. Indeed, inside the barrier, the magnitude of the wavefunctiondecreases exponentially by a factor exp (−𝛾L), so we have

𝜓(L)𝜓(0)

≈ e−𝛾L.

Since the transmission coefficient is given by the square of the ratio of amplitudesbefore and after the barrier, it is approximately equal to

T ≈||||

𝜓(L)𝜓(0)

||||

2≈ e−2𝛾L.

We can readily generalize this result for the arbitrarily shaped barrier ofFigure 5.8, as we explain here.

Inside the classically forbidden region x1 < x < x2 of this barrier, the decaycoefficient 𝛾 is no longer a constant (as in the square barrier, where 𝛾 =

√𝑈0 − 𝜖)

but depends on x as follows:

𝛾(x) =√𝑈 (x) − 𝜖 =

√2m(V (x) − E)

ℏ2 .

Therefore, a reasonable generalization of the formula T ≈ exp (−2𝛾L) is toreplace the constant decay coefficient 𝛾 with its mean value in the interval[x1, x2]. We remind the readers that the mean value of a variable quantity in theinterval of interest is given by its integral divided by the length of the interval.


E E

xx1 x2

Classically forbidden region

Figure 5.8 Tunneling through a potential barrier of arbitrary shape. The classically forbiddenregion is the interval between points x1 and x2.

Therefore, the mean decay coefficient 𝛾 is

𝛾 = 1L ∫

x2

x1

𝛾(x) dx (L = x2 − x1),

whence we obtain

T ≈ exp (−2𝛾 L) = exp(

−2∫x2

x1

𝛾(x) dx)

= exp

(

−2∫x2

x1

√2m(V (x) − E)

ℏ2 dx

)

.

(5.22)

This formula is the starting point for the quantum mechanical treatment ofalpha decay (see the online supplement of this chapter).

5.3.2.4 Exponential Sensitivity of the Tunneling Effect to the Massof the ParticleLet us now comment on the strong sensitivity of the tunneling effect to the massof the particle. The smaller the mass, the greater the probability for transmissionthrough a barrier. Conversely, as the mass increases, the transmission coefficientdecreases, until it vanishes altogether in the classical limit m → ∞. These state-ments are in agreement with our earlier observation that the lighter a particle, thestronger its quantum mechanical behavior. Tunneling through classically forbid-den regions is a purely quantum mechanical effect, so it is realized more readilyfor lighter particles. For example, let us compare the transmission coefficients foran electron and a proton of the same energy as they go through the same potentialbarrier V (x). We have

Te ≈ exp (−𝛽√

me ), Tp ≈ exp (−𝛽√

mp ),

where the factor 𝛽

𝛽 = 2∫x2

x1

√2(V (x) − E)

ℏ2 dx

is common in both cases, so its value is of no interest here.We can rewrite the transmission coefficient Tp of a proton as

Tp ≈ exp

(

−𝛽√

me

√mp

me

)

=(exp (−𝛽

√me )

)√ mp

me ,

Problems 163

to obtain

Tp = T√ mp

mee ≈ T43

e (mp ≈ 1836 me).If, say, the probability of electronic transmission through the barrier at a given

instance is 10%, that is, Te = 10−1, then we obtainTp ≈ 10−43,

which means that protons have practically zero probability to make it througha barrier that is almost “transparent” to electrons. Once again the conclusion isclear: The electron is nature’s premier quantum mechanical particle.

5.3.2.5 A Practical Formula for TAs it is customary by now, in order to facilitate numerical computations in prac-tical units (eV for V0 and E, and Å for L), we rewrite formula (5.21) as

T ≈ exp (−2𝛾L) = exp

(

−2√

V0 − E

)

, (5.23)

where = ℏ2∕2mL2, or, equivalently,

= ℏ2

2mea20

me

m

(a0

L

)2= 13.6

me

m

(a0

L

)2eV

and (5.23) is finally written as

T ≈ exp

(

−2√

V0 − E13.6

√mme

La0

)

, (5.24)

or—if we restrict ourselves to electrons—as

T ≈ exp

(

−2√

V0 − E13.6

La0

)

, (5.25)

where, of course, V0 and E are measured in eV.

Problems

5.4 Use formula (5.25) to calculate the probability for an electron to go througha square potential barrier 4 Å wide, whose top is 3.4 eV higher than theenergy of the electron.

5.5 Without repeating the calculation, predict the transmission probability forthe same barrier as before if the energy of the electron is raised so thatits “distance” from the top of the barrier becomes four times smaller thanbefore. The same question if that “distance” becomes four times greater.

5.6 Show that the following “exponential scaling laws” are valid for the tunnelingeffect:

T(m) = T(m0)√

m∕m0 , T(L) = T(L0)L∕L0 , T() = T(0)√

∕0 ,


where m0, L0, and 0 are some reference values for the mass of the transmittedparticle, the width of the barrier, and the energy distance = V0 − E from its top.

Further Problems

5.7 Solve the Schrödinger equation for the downward potential step of thefigure. Calculate the reflection probability as a function of the energy ofthe particle and the “step depth” V0. Discuss how we could have obtainedthis result from the corresponding upward potential step with a simple

E

– V0

V(x)

x

modification. Calculate the value ofthe reflection coefficient for E = 4 eV,V0 = 5 eV, and compare it with the corre-sponding value for an upward step withthe same height and an incident energyE = 9 eV. Is there a broader conclusionyou can draw?

5.8 Imagine that the square potential barrier we discussed in the textbecomes very narrow and very high, while its “area” V0L tends to aconstant value g. The barrier can then be approximated by a delta functionV = g𝛿(x), as shown in the figure. Solve the Schrödinger equation for thispotential—using the matching conditions we found in Problem 4.14—and

E

x

V = g δ(x)

show that the reflection and transmissioncoefficients are given by the expressions

R = 𝜆2

𝜆2 + 4k2 , T = 4k2

𝜆2 + 4k2 ,

where 𝜆 = 2mg∕ℏ2 and k2 = 2mE∕ℏ2. Notethat both R and T are functions only of theratio k∕𝜆 =

√ℏ2E∕2mg2. Could you have

predicted this feature?

5.9 The transmission coefficient for a square potential barrier can only be afunction of the five parameters ℏ, m, L, V0, and E of the problem; thatis, it must have the form T = f (ℏ,m, L,V0,E). However, you can employdimensional analysis to show that T is, in fact, a function of the formT = f (𝜆, 𝜇), where 𝜆 and 𝜇 are two dimensionless combinations of thefive original parameters. Show that this is indeed the case and find theexpressions for 𝜆 and 𝜇. Are these defined uniquely? Apply your findingsto formulas (5.18) and (5.19). Can these formulas be written as functions oftwo dimensionless combinations? Based on the discussion, what can youconclude about the possibility of “simulating” one square potential bar-rier with another? What would this possibility imply from an experimentalpoint of view? Have you heard of “simulation experiments” in hydrody-namics? If yes, comment on their possible relation to the discussion.


5.10 A particle with energy 1 eV impinges on a square potential barrier of width6 Å and height 3 eV. In another instance, the same particle with energy4 eV hits another square barrier of width 3 Å and height 12 eV. Show,without any calculation, that the transmission probability is the same inboth cases. What can you conclude regarding a more general condition ofequivalence between two such problems?

5.11 Outside the surface of a metal there exists a homogeneous electric fieldof intensity . In this case, the potential energy of the electrons insideand outside the metal has the shape shown in the figure below, wherewe assumed the potential inside the metal to be homogeneous, for sim-plicity. Show that for an electron of energy E (i.e., with a work functionW = −E = |E|) the probability to escape from the metal, due to the exter-nal electric field, is given by the formula

T(W ) ≈ exp

(

−4√

2m3ℏe W 3∕2

)

. (1)

The phenomenon of electron extraction from a metal by an external elec-tric field is known as cold emission. The term is used to distinguish it fromthe more familiar thermionic emission caused by heating the metal and

xx = l

V(x)

V(x) = – e Ɛ x

E W

– V0

thus increasing the kineticenergy of its electrons.Perform a numerical testof relation (1) to confirmthat the phenomenonof cold emission can beobserved under realisticconditions.

167

6

The Harmonic Oscillator

6.1 Introduction

The harmonic oscillator is, surely, the most “classic” problem of classicalmechanics. Its theory is simple. If a particle with mass m is subject to a restoringforce

F = −kx, (6.1)

which is proportional to the displacement from an attractive center at x = 0, thenNewton’s equation is written as

mx = −kx ⇒ x + 𝜔2x = 0, 𝜔 =√

km.

The solution of this equation takes the sinusoidal form

x(t) = A cos (𝜔t + 𝜑), (6.2)

which represents a harmonic oscillation of the particle with frequency𝜔 =

√k∕m and period

T = 2𝜋𝜔

= 2𝜋√

mk.

Note that the period is independent of the amplitude denoted by the constant Ain the solution (6.2). This unique feature of the classical harmonic oscillator hasits quantum analog, as we shall see soon.

The force (6.1) comes from a potential V (x) defined by the familiar relation

F = −kx = −dVdx

⇒ V = 12

kx2,

where V (x) is known as the parabolic potential, or the potential of a harmonicoscillator. Its characteristic shape is shown in Figure 6.1.

The significance of the parabolic potential stems from the fact that it is a verygood approximation to any potential in the vicinity of a stable equilibrium point.Indeed, consider an arbitrary one-dimensional potential V (x) with a local mini-mum (i.e., a stable equilibrium point), which we place at the origin of the x-axis.


168 6 The Harmonic Oscillator

V(x)

x

V =1

2k x2

Figure 6.1 The potential of a harmonic oscillator.

We can then expand V (x) in a Taylor series around x = 0 to find

V (x) = V (0) + V ′(0)x + 12

V ′′(0)x2 + · · · (6.3)

Since x = 0 is a stable equilibrium point, we have V ′(0) = 0 and V ′′(0) = k > 0.Furthermore, since we can choose the potential energy reference level at will, wecan set V (0) = 0, and thus write (6.3) as

V (x) = 12

kx2 + · · ·

The higher powers in this expansion can be neglected for small x (i.e., for smalloscillations around the equilibrium point), in which case only the “parabolicterm” survives; it provides a good approximation of the full potential in thevicinity of its minimum (see Figure 6.2).

As we see later in the book, the parabolic approximation—a plausible namefor the given process—finds a straightforward application in the study of thevibrational motion of diatomic molecules.

1

2k x2

V(x)

x

Figure 6.2 The parabolic approximation. Any potential V(x) can be approximated by aharmonic oscillator potential in the vicinity of a local minimum (x = 0 in this case).

6.2 Solution of the Schrödinger Equation 169

6.2 Solution of the Schrödinger Equation

A quantum mechanical treatment of the harmonic oscillator requires the replace-ment of Newton’s equation with the Schrödinger equation

𝜓 ′′ + 2mℏ2 (E − V (x))𝜓 = 0.

For V (x) = 12kx2 = 1

2m𝜔2x2, the equation takes the form

𝜓 ′′ + 2mℏ2

(

E − 12

m𝜔2x2)

𝜓 = 0 (6.4)

and we expect that its solution will lead to energy quantization, since the motionis always confined in the classical problem as well. We will now present theprocess of solving (6.4) in four main steps.STEP 1: Simplification of the equation with the use of dimensional analysis.The first step in solving (6.4) is to simplify its form using dimensional analy-sis. The basic idea is simple. Since the numerical values of the parameters ℏ, m,and 𝜔 depend on the system of the basic units L,M, and T we are using, we canalways—through a suitable choice of these units—give them any numerical valuewe wish. In particular, we can set all three equal to unity

ℏ = m = 𝜔 = 1,in the Schrödinger equation, and recover the dependence of our results on ℏ, m,and𝜔 at the end of the calculation using the fundamental theorem of dimensionalanalysis (Section 1.4). But we can put aside for now the issue of “dimensionalrestoration,” and solve the dimensionally simplified version of the Schrödingerequation, namely,

𝜓 ′′ + (2E − x2)𝜓 = 0, (6.5)whose form is much more “tractable” than the original one.STEP 2: Search for solutions that vanish at infinity: “Removal” of the asymptoticfactor.Since we seek solutions that vanish at ±∞, we need to study the behavior of (6.5)in the limit of large x, where it takes the asymptotic form

𝜓′′

∞ − x2𝜓∞ = 0, (6.6)because 2E is negligible compared to x2 in that regime (large x). The index ∞ in𝜓∞ serves as a reminder that we are looking for the solution at large x, not for thefull solution 𝜓 valid for all x.

It is reasonable to assume that the solution of (6.6) for large x is a decayingexponential, and the simplest one we can think of is

𝜓∞ = e−𝜆x2. (6.7)

The presence of x2 in the exponent guarantees that the solution vanishes at bothlimits ±∞. We could, of course, have chosen any even power of x in the exponent,but x2 is certainly the simplest such power, and, as we will shortly see, it is alsothe correct one. If we insert (6.7) into (6.6), we get

(4𝜆2x2 − 2𝜆)e−𝜆x2 − x2e−𝜆x2 = 0.


The equation need only be satisfied for large x values, as it is only there that werequire (6.7) to be a solution. For large x we have 4𝜆2x2 ≫ 2𝜆, so the equationbecomes

(4𝜆2x2 − x2)e−𝜆x2 = 0,

which is clearly satisfied for 𝜆 = ±1∕2. However, only for 𝜆 = +1∕2 do we obtaina physically acceptable solution that vanishes at infinity.

Therefore, the wavefunctions of the harmonic oscillator will behave atinfinity—that is, for large x—like the decaying exponential

𝜓∞(x) = e−x2∕2.

The next logical step is to write down the full solution 𝜓(x) as a product of theform

𝜓(x) = 𝜓∞(x)H(x) = e−x2∕2H(x), (6.8)

where the first factor (i.e., the asymptotic exponential) guarantees that the solu-tion vanishes at ±∞, while the second factor (i.e., the complementary functionH(x)) provides the expected wavelike form that befits the bound states (so thatthe ground state has zero nodes, the first excited state has one node, the secondexcited state two nodes, etc.; in general, the nth excited state must have n nodes).Moreover, the function H(x) should not increase at infinity faster than the decayrate of the exponential e−x2∕2, so that the required vanishing of the full solutionat infinity is not cancelled. The most obvious example of a function H(x) thatsatisfies these constraints—not to “cancel” the vanishing effect of the asymptoticexponential for large x and to have n real roots—is an nth-degree polynomial.Therefore, we assume that the complementary function H(x) in (6.8) is a polyno-mial Hn(x). Given the parabolic form of the potential, we expect that the problemhas an infinite number of bound states. So the degree n of Hn(x) is expected totake all values from n = 0 (ground state) to n = ∞.

From a mathematical point of view, (6.8) is merely a change of the dependentvariable—the old and new variables are 𝜓(x) and H(x), respectively—that has tobe inserted in the original Eq. (6.5) to convert it into an equation for the newunknown function H(x). Doing so we obtain

H′′ − 2xH′ + (2E − 1)H = 0. (6.9)

What remains now is to examine whether this equation has indeed polynomialsolutions, as we assumed.STEP 3: Search for polynomial solutions.A glance at an equation like (6.9) may generate mixed feelings to the readers whohave taken an introductory course in differential equations. First, we are dealingwith a linear (and homogeneous) equation, which is certainly a reassuring fact,since nonlinear equations can be exactly solved only in rare cases.

However, we are not dealing with an ideal situation either! Even though (6.9) islinear, it is not an equation with constant coefficients that can be exactly solvedwith the familiar—and beloved!—method of exponential substitution. Theequation has variable coefficients and, unfortunately, the only general method


of solving such equations is the power-series method. Why “unfortunately?”Because in practice, few of us feel comfortable using this method on our owninitiative.

A change of tactics is therefore rather necessary. We will ask the readers toattempt to solve (6.9) with the oldest of all methods; the “method of maximuminnocence!” The method used by people who do not know how to solve a problem,yet still need to solve it!

The equation at hand is (6.9) and the only thing we know about it is that it mayhave polynomial solutions, which is a plausible assumption from a physical per-spective. How would we go about exploring this possibility further? Quite simply,we would test whether (6.9) has indeed solutions of this form, starting with thesimplest solution of all: a polynomial of zeroth degree namely, a constant. Let usthen set H0 = c = constant in (6.9) and see what happens. We obtain

0 − 2x ⋅ 0 + (2E − 1) ⋅ c = 0,

and, since c ≠ 0, this equation is satisfied only if

E = E0 = 1∕2,

so that the corresponding solution can be written via (6.8) as

𝜓0(x) = ce−x2∕2.

The constant c remains undetermined because both Eq. (6.9) and the originalEq. (6.5) are homogeneous and, therefore, their solutions are defined up to anarbitrary multiplicative constant. The latter is determined by the normalizationcondition ∫ |𝜓(x)|2 dx = 1 and the result—already known to us from Chapter2—is c = 𝜋−1∕4. The normalized solution can thus be written as

𝜓0(x) =1

4√𝜋

e−x2∕2.

The readers may have not yet realized that we have just obtained a truly signifi-cant physical result with practically no effort. We actually found both the energy(E0 = 1∕2) and the wavefunction (𝜓0 = 𝜋−1∕4e−x2∕2) of the most important stateof the harmonic oscillator: its ground state!

Encouraged by our success we now proceed to the next step. We willtest whether (6.9) has a solution in the form of a first-degree polynomialH1(x) = ax + b, where a and b are arbitrary constants. Equation (6.9) now gives

0 − 2x ⋅ a + (2E − 1)(ax + b) = 0 (6.10)

⇒ (2E − 1)b + (2E − 3)ax = 0, (6.11)

which is satisfied only if1

(2E − 1)b = 0, (2E − 3)a = 0,

1 We recall that a polynomial is identically zero only when the coefficients of all its powers vanish.Therefore, to solve an equation like (6.10) we have to group all constant terms together, all x termstogether, and so on—as in (6.11)—and set their coefficients equal to zero.


whereby we deduce that E = 3∕2, b = 0, while a is arbitrary. The possibility thatb ≠ 0 ⇒ E = 1∕2 is easily rejected, because in that case, the second equationwould give a = 0 and the degree of the polynomial would not be one, as weassumed, but zero. In other words, we would have simply reproduced the pre-vious solution.

To recap, we found that if the energy of the particle takes the value

E1 = 32,

then Eq. (6.9) has the first-degree polynomial solution, H1(x) = ax and the cor-responding wavefunction is

𝜓1(x) =√

2√𝜋

xe−x2∕2,

where the arbitrary constant a is found from the appropriate normalization con-dition to be equal to a =

√

2∕√𝜋.

The fact that once we tried as a solution the first-degree polynomial H1 = ax +b we obtained b = 0 ⇒ H1(x) = ax, should not go unnoticed. We should actuallyhave anticipated this result, since the harmonic oscillator potential has reflec-tion (or mirror) symmetry, and, therefore, its solutions must be alternately evenand odd functions, as we already noted in Chapter 4. Given now that the expo-nential factor in the wavefunctions is even, it is the polynomials Hn(x) that mustbe responsible for this interchange of symmetry type. Therefore, Hn(x) must beeven or odd, depending on its degree. Specifically, polynomials of even (odd)degree must be even (odd) functions, which practically means that polynomialsof even degree (n = 0, 2, 4,…) contain only even powers of x, and polynomialsof odd degree (n = 1, 3, 5,…) only odd powers. Therefore, in this case (n = 1) wecould have set H1(x) = ax right away, since the polynomial can only contain oddpowers. We could have also set a = 1 for simplicity, since our equation is homo-geneous and cannot be expected to allow for the constant a to be determined.Setting H1(x) = x in (6.9) we obtain

0 − 2x ⋅ 1 + (2E − 1)x = 0 ⇒ (2E − 3)x = 0 ⇒ E = E1 = 3∕2,

whereby we see immediately that, by exploiting the symmetry of the problem, weobtain the desired second eigenvalue in less than one line of algebra.

It suffices to study one more case (n = 2) to eliminate any lingering doubt thatthe solutions we are seeking are of polynomial form. For n = 2 we have

H2(x) = x2 + a, (6.12)

where, apart from invoking the symmetry of the polynomial to retain only evenpowers of x, we used the fact that our equation is homogeneous, in order to setthe coefficient of x2 equal to one. Inserting (6.12) in (6.9), we obtain

2 − 2x(2x) + (2E − 1)(x2 + a) = 0⇒ (2E − 5)x2 + ((2E − 1)a + 2) = 0

⇒ 2E − 5 = 0, (2E − 1)a + 2 = 0 ⇒ E = 52, a = −1

2


⇒ H2(x) = x2 − 12∼ 2x2 − 1

⇒ 𝜓2(x) = N(2x2 − 1)e−x2∕2.

In the last two steps, we again took advantage of the fact that our equations arehomogeneous, to write the polynomial H2 in the simpler form 2x2 − 1, insteadof x2 − (1∕2), and we introduced a normalization factor N in the final expres-sion for𝜓2. As always, the factor N is determined by the normalization condition∫ |𝜓2(x)|2dx = 1.

A mere glance at the eigenvalues we found, namely,

E0 = 12, E1 = 3

2, E2 = 5

2,

leads us to suspect that for a polynomial solution of arbitrary degree n, thecorresponding eigenvalue is

En =odd integer

2= 2n + 1

2= n + 1

2, n = 0, 1, 2, · · · (6.13)

We can easily prove (6.13) as follows. Suppose that Eq. (6.9) is indeed satisfied byan nth degree polynomial of the general form

H(x) = a0 + · · · + anxn.

But if the equation is satisfied by the polynomial H(x) for any x, it must also besatisfied for large x, where

H(x) ∼ xn (large x),since in this regime, the lower powers of x are negligible compared to xn. In otherwords, an obvious necessary condition for the existence of an nth degree polyno-mial solution is that our equation be satisfied, for large x, by the maximum powerxn of this polynomial solution. So all we need to do is insert xn in (6.9) and requirethat the equation be satisfied for large x. The condition “for large x” really meansthat we must keep only the highest power of x when inserting xn in (6.9), since itis the highest power that dominates in this limit. Indeed, if we set H ∼ xn in (6.9)we obtain

n(n − 1)xn−2 − 2nxn + (2E − 1)xn = 0and if we now ignore the xn−2 term, as being negligible compared to xn, we find

(−2n + (2E − 1))xn = 0 ⇒ E = n + 12,

which is precisely what we wanted to show.The act of applying this necessary condition—namely, that the equation be sat-

isfied by the highest power of the polynomial for large x—is so simple that, witha little experience, we can obtain the desired eigenvalues by a simple inspectionof the equation. We just need to get used to identifying the terms of the equationthat give us the highest power of x, and replacing xn only in these terms. Forexample, it is clear in the present case that the highest power originates from theterms −2xH′ and (2E − 1)H ; substituting H ∼ xn in these terms yields the resultright away.


We can summarize our findings so far as follows: In the so-called natural sys-tem of units of the problem, where ℏ = m = 𝜔 = 1, the energy eigenvalues of theharmonic oscillator are given by the expression

En = n + 12, n = 0, 1, 2,…

The corresponding eigenfunctions are given by

𝜓n(x) = e−x2∕2Hn(x),

where Hn(x) are polynomials of degree n, which are even or odd, depending onn. According to this, the polynomials Hn(x) satisfy the equation

H′′n − 2xH′

n + 2nHn = 0,

which is known in the literature as the Hermite equation (its solutions are theso-called Hermite polynomials Hn).

Table 6.1 lists the first five normalized eigenfunctions of the harmonic oscilla-tor. We leave it to the readers to confirm the correctness of these expressions.STEP 4: Restoration of dimensions.We begin with the expression for the energy eigenvalues

En = n + 12,

which in ordinary units should be written as

En =(

n + 12

)

𝜖,

where 𝜖 is the natural energy unit of the problem—that is, the unique combina-tion of ℏ, m, and𝜔with dimensions of energy. That such a combination exists andis indeed unique is a direct consequence of the well-known (Section 1.4) “funda-mental theorem” of dimensional analysis:

Given three dimensionally independent physical quantities, we can alwaysconstruct uniquely any other physical quantity up to a dimensionlessmultiplicative constant.

Table 6.1 The first five eigenfunctions ofthe harmonic oscillator

𝜓0(x) =1

4√𝜋

e−x2∕2

𝜓1(x) =4

√4𝜋

xe−x2∕2

𝜓2(x) =1

√2 4√𝜋

(2x2 − 1)e−x2∕2

𝜓3(x) =1

√3 4√𝜋

(2x3 − 3x)e−x2∕2

𝜓4(x) =1

2√

6 4√𝜋

(4x4 − 12x2 + 3)e−x2∕2


In our case, the combination of ℏ, m, and 𝜔 with dimensions of energy (with anumerical multiplicative factor equal to 1) is

𝜖 = ℏ𝜔,

so that the energy eigenvalues in ordinary units are

En = (n + 1∕2)ℏ𝜔, n = 0, 1, 2, · · · .

What about the restoration of ordinary units in the wavefunctions? As we willexplain shortly, this task is achieved with the substitution

𝜓(x) → 1√

a𝜓

(xa

)

, (6.14)

where 𝜓(x) is the dimensionless form of the wavefunction and a is the charac-teristic length of the problem (i.e., the unique combination of ℏ, m, and 𝜔 withdimensions of length). Indeed, the substitution (6.14) expresses both the factthat a one-dimensional wavefunction has dimensions of L−1∕2—hence the factor1∕

√a—as well as the general requirement that the variables of all mathematical

functions—for example, sin, cos, exp, and so on—appearing in physical formulasbe dimensionless quantities. We thus need to replace x with x∕a.

As usual, we can construct a combination with the desired physicaldimensions—in our case, the length a—either by using the familiar sys-tematic approach2 or by combining known physical formulas that include ℏ, m,𝜔, and the desired length a. We follow the latter approach and equate quantitiesof the same physical dimension, as follows

ℏ𝜔 = energy andp2

2m=

(h∕𝜆)2

2m∼ ℏ2

ma2 = energy,

whence we obtain ℏ𝜔 = ℏ2∕ma2. Solving for a, we find for the characteristiclength of the problem the expression

a =√

ℏ

m𝜔.

As a simple example, let us restore the ordinary units for the ground-state wave-function, which we found earlier to be

𝜓0(x) =1

4√𝜋

e−x2∕2, (6.15)

so that, according to (6.14),

𝜓0(x) =1

√

a√𝜋

e−x2∕2a2,

or, equivalently,

𝜓0(x) = 4

√m𝜔𝜋ℏ

e−m𝜔x2∕2ℏ2. (6.16)

2 We can write a = ℏ𝜇m𝜈𝜔𝜆 and calculate the exponents 𝜇, 𝜈, 𝜆 from the set of three equationsobtained by requiring that the right-hand side has the desired dimension of length, L1M0T0.


In practice, however, restoring ordinary dimensions in wavefunctions is unneces-sary and even hampers calculations. It is much simpler to use the dimensionlessexpressions throughout and to only recover ordinary units after obtaining thefinal results. The following example describes a related case.

Example 6.1 Calculate the position and momentum uncertainties for theground state of the harmonic oscillator.

Solution: To calculate Δx and Δp, we invoke the dimensionless form (6.15),because it is clearly much simpler than (6.16). We thus find

Δx = 1√

2, Δp = 1

√2,

which in ordinary units becomes

Δx = 1√

2a, Δp = 1

√2

p0,

where a (=√ℏ∕m𝜔) is the length unit of the problem, while p0 is the correspond-

ing unit of momentum, which can be easily constructed as follows:

p = h𝜆⇒ p0 = ℏ

a= ℏ

√ℏ∕m𝜔

=√ℏm𝜔.

To sum up, we obtained the following results:

Δx = a√

2=√

ℏ

2m𝜔, Δp =

p0√

2=√ℏm𝜔

2.

The readers will surely appreciate the calculational advantages of the dimen-sionless version of the solutions, whose form in ordinary units becomes rapidlycumbersome as we move on to higher eigenfunctions.

We should now address a question that may linger from the discussion so far.When restoring dimensions in the eigenvalues and eigenfunctions via the substi-tutions

En → En𝜖, 𝜓n(x) →1√

a𝜓n

(xa

)

,

we used as energy and length units, respectively, the quantities

𝜖 = ℏ𝜔 and a =√

ℏ

m𝜔, (6.17)

which are the unique combinations of ℏ, m, and 𝜔 with the appropriate dimen-sions. But there is an extra step we implicitly took: We set the multiplicativeconstant equal to one. So the question arises: Why should we set this numericalfactor equal to one, of all possible values? For example, why did we not choose asenergy and length units the quantities

𝜖 = 2𝜋ℏ𝜔 = h𝜔 and a =√

2𝜋ℏm𝜔

=√

hm𝜔

(6.18)

6.3 Discussion of the Results 177

instead of (6.17)? The answer is quite simple. Had we used expressions (6.18)instead of (6.17), we would have obtained, for the energy eigenvalues, for example,the expression

En =(

n + 12

)

𝜖 =(

n + 12

)

h𝜔 =(

n + 12

)

2𝜋ℏ𝜔,

which is surely wrong, since for ℏ = m = 𝜔 = 1 it gives En = 2𝜋(n + 12), in

direct contradiction with what we found earlier by solving the Schrödingerequation for these “special” values of its parameters. The conclusion is com-pletely general: In the process of restoring ordinary physical dimensions, wemust choose expressions for the characteristic units of the problem that havenumerical coefficients equal to one. Our results thus reproduce those weobtained by solving the Schrödinger equation with its three parameters set equalto unity.

Problems

6.1 Find the third-degree polynomial solution of Hermite’s equation and verifythat the wavefunction 𝜓3(x) given in Table 6.1 is correct, including normal-ization.

6.2 The state of a particle at a given time, in the system of natural units of theharmonic oscillator (ℏ = m = 𝜔 = 1), is described by the wavefunction

𝜓(x) = Ne−𝜆x2.

Calculate, as a function of𝜆, the probability to measure the value E0 = 1∕2 ofits ground state. What is the probability that an energy measurement yieldsthe value E1 = 3∕2 of the first excited level?

6.3 Find the state of minimum energy for the potential of the figure on

V(x)

1

2k x2

x

the left, where the shading signifies that the poten-tial becomes infinite for x < 0. What is the completeset of eigenvalues and eigenstates?

6.3 Discussion of the Results

We will now attempt a systematic analysis of our results to highlight their physicalsignificance and some of their special features. We begin with the shape of theeigenfunctions.


6.3.1 Shape of Wavefunctions. Mirror Symmetry and the Node Theorem

We could have predicted that the wavefunctions ought to have the general shapeof Figure 6.3 without actually solving the problem. Specifically, we could haveguessed that they are alternately even and odd, and that their number of nodesincreases by one as we move from the wavefunction of the ground state (zeronodes) to those of higher states.

As we have stressed elsewhere, since the potential is independent of the signof x, the same must be true for physically measurable quantities like the positionprobability density P(x). Therefore, the eigenfunctions of such a problem shouldgive equal probabilities of finding the particle in positive versus negative x values.We must then have

P(−x) = P(x) ⇒ |𝜓(−x)|2 = |𝜓(x)|2 ⇒ 𝜓(−x) = ±𝜓(x),

which says that the wavefunctions must be even or odd. But apart from their sym-metry, the number of nodes of the wavefunctions also has a simple physical expla-nation. As we previously noted (Section 4.2.2), successive eigenfunctions of aparticle in a one-dimensional box have the characteristic shape of standing waveson a vibrating string. The first eigenfunction corresponds to fitting a half-wave inthe box, the second eigenfunction corresponds to fitting two half-waves, and like-wise for higher states. As can be seen in Figure 6.3, this general picture appliesalso for the eigenfunctions of a harmonic oscillator subject to the following plau-sible modifications: (i) The eigenfunctions do not terminate at the limits of theclassically allowed region (where the related energy level “crosses” the potentialcurve); instead, they have exponentially decaying “tails” that extend, in princi-ple, to infinity. (ii) The successive half-waves that “form” the wavefunction inside

V(x)

E0 =

E1 =

E2 =

E3 =

ℏω

ℏω

ℏω

ℏω

12

32

52

72

x

Figure 6.3 The first four eigenfunctions of the harmonic oscillator. The first eigenfunction iseven with zero nodes, the second eigenfunction odd with one node, the third again even withtwo nodes, and so on.


the classically allowed region no longer have the simple sinusoidal shape of theone-dimensional box, since the potential now depends on x, and so does thewavelength of the particle. Indeed, energy conservation implies that

p2

2m+ V (x) = E ⇒ p = p(x) =

√2m(E − V (x)),

and therefore

𝜆 = 𝜆(x) = hp(x)

= h√

2m(E − V (x)),

so the eigenfunctions inside a spatially varying potential V (x) have the expectedwavelike shape, albeit with a wavelength that depends on x.

6.3.2 Shape of Eigenfunctions for Large n: The Classical Limit

It follows from the discussion that, as we go away from the origin, consecutivehalf-waves of an energy eigenfunction tend to “open” up, since the local momen-tum decreases and the corresponding local wavelength increases. This trend isalready evident for the last two eigenfunctions of Figure 6.3 and becomes quiteobvious for large quantum numbers (Figure 6.4 depicts the n = 20 case).

It is also clear from the figure that consecutive peaks of the wavefunction 𝜓20increase as we approach the inversion points where the local speed vanishesin the classical case. Here is why. In the limit of large quantum numbers—andn = 20 is definitely not small—the quantum wavefunction should roughlyreproduce classical behavior, whereby the particle spends more time in regionsof low speed; hence the probability of locating it there is greater. So, it is quiteexpected that, as we approach the boundaries of the classical oscillation, theeigenfunctions for large n will increasingly have broader and higher peaks.

The relationship between the classical motion and the eigenfunctions for largen becomes even more apparent if we plot together the quantum probability

41 41

ψ20(x)

x

–

Figure 6.4 The harmonic oscillator eigenfunction with a quantum number n = 20. As wemove away from the origin, the local wavelength of successive half-waves increases, whiletheir corresponding “height” also increases. Thus, in the limit of large n, classical behavior isrestored: The probability of finding the particle increases in regions where the local speed issmall, that is, near the boundaries of the classical oscillation.


ψ20(x)

x

Pcl(x)

2

Figure 6.5 Comparison between the quantum and classical probability densities for theharmonic oscillator state with n = 20. The quantum distribution oscillates symmetrically aboutthe corresponding classical curve. In the limit of large n, where quantum oscillations cease tobe observable, the classical curve can be regarded as a kind of average of the quantumdistribution.

density Pn(x) = |𝜓n(x)|2 and the corresponding classical quantity

Pcl(x)dx = 2 dtT

=2(dx∕u(x))

T(6.19)

⇒ Pcl(x) =2

Tu(x). (6.20)

Even though classical motion is fully deterministic, Pcl(x) describes in a proba-bilistic way the degree of residence of the particle in any infinitesimal region ofthe oscillation interval. The degree of residence is equal to the fraction of time,relative to the full period, that the particle “spends” in this particular region. (Thefactor of 2 in (6.19) expresses the fact that the particle passes through the sameregion twice in one period.)

In the dimensionless system of units we are using, we have T = 2𝜋∕𝜔 = 2𝜋 and

u(x) =√

2(E − V (x))m

|||||m=1,V=x2∕2

=√

2E − x2,

so that, according to (6.20), we find

Pcl(x) =1

𝜋√

2E − x2(−

√2E < x <

√2E).

The comparison between Pcl and the quantum probability distribution for n = 20(⇒ 2E = 41) is shown in Figure 6.5.

6.3.3 The Extreme Anticlassical Limit of the Ground State

We have seen that quantum wavefunctions must approximate classical behaviorfor large n values, and by the same token we expect that deviations from thisbehavior will grow stronger as we go to smaller n (i.e., lower eigenstates). It is


Pcl(x) =

P0(x) =

1

1

1–1

1–x2

e–x2

π

π

x

Figure 6.6 Comparison between the quantum and classical probability densities for theground state of the harmonic oscillator. In contrast to the classical case, the quantum particleis much more likely to be found near the origin than near the endpoints of the classicaloscillation. Thus, when the particle is in the ground state, it behaves in the most extremeanticlassical manner.

therefore not surprising that the most anticlassical behavior is exhibited by theground state of the oscillator, as can be seen in Figure 6.6.

6.3.4 Penetration into Classically Forbidden Regions: What Fraction of Its“Lifetime” Does the Particle “Spend” in the Classically Forbidden Region?

Following the discussions of the previous chapters and this one, penetration intoclassically forbidden regions must be taken as an essential feature of quantummechanics, much like quantization of the energy spectrum (for bound states),or quantum resistance to confinement, which causes the rise of the ground-stateenergy above the bottom of the corresponding potential. In the case at hand, pen-etration is surely evident, since the wavefunction extends throughout the realaxis, while the classically allowed region is limited to the range [−a, a], wherea is the amplitude of the classical oscillation and is determined by the familiarcondition

V (a) = E ⇒12

a2 = E ⇒ a =√

2E =√

2n + 1.

Therefore, the classically allowed interval for the ground state (n = 0) of theharmonic oscillator is [−1, 1], and the probability of finding the particle in theclassically forbidden region |x| > 1 is given by the expression3

P[|x| > 1] = 1 − P[|x| < 1] = 1 − ∫1

−1|𝜓0(x)|2dx, (6.21)

3 We recall that the symbol P[range of x values] denotes the probability of finding the particle inthis particular interval.


Table 6.2 The probability of finding the particle in the classicallyforbidden region for several states of the harmonic oscillator

State Probability (%)

n = 0 15.7n = 1 11.2n = 2 9.5n = 3 8.5n = 4 7.9n = 5 7.4n = 6 7.0n = 7 6.7n = 8 6.4n = 9 6.2

which takes into account that P[|x| > 1] is complementary to the probability offinding the particle in the classically allowed interval |x| < 1.

In Problem 6.9, we suggest a simple way to approximately calculate the integralon the right-hand side of (6.21). We will thus obtain the numerical result

P[|x| > 1] ≈ 1 −√

1 − e−1 ≈ 0.205 ≡ 20.5%,

which is rather close to the exact value

P[|x| > 1] = 0.157 ≡ 15.7%

one finds using a computer algebra software such as Mathematica or Matlab.Therefore, we can say that in the ground state of a harmonic oscillator, the particlespends 15.7% of its life in the classically forbidden region! The same software canbe used to compute the corresponding fraction for any other oscillator state wechoose (see Table 6.2 for the first 10 states). As one would expect, the probabilityof locating the particle in the classically forbidden region decreases with increas-ing n, since it should tend to zero for large n, where classical physics applies andno penetration is allowed. But it is worth noting that this decrease is very slow, ascan be confirmed from further numerical calculations. For example, for n = 30the probability of penetration into the classically forbidden region decreases tojust 4.2%, while for n = 50 it is still 3.2%!

Note also that these probabilities, being dimensionless quantities, do notdepend on the oscillator parameters ℏ, m, and 𝜔. Their values are the same forany value of mass, classical frequency 𝜔, or even Planck’s constant! Can youexplain why this fact is remarkable and suggest an explanation?

6.3.5 A Quantum Oscillator Never Rests: Zero-Point Energy

The quantum oscillator is a prototype system for the phenomenon of quantumresistance to confinement, according to which the particle keeps moving even in


the state of “maximum rest”—that is, in the state of minimum total energy. As wefound earlier, this so-called zero-point energy is given by the formula

E0 = 12ℏ𝜔 = 1

2ℏ

√km.

Interestingly, this expression satisfies both the classical (E0 → 0 for ℏ→ 0 orm → ∞) and the strong quantum limit, which requires that the minimumallowed energy goes up as Planck’s constant increases or as the particle massdecreases (i.e., E0 → ∞ for ℏ → ∞ or m → 0). Actually, the origin ofzero-point energy is known to us. It stems from the uncertainty principle, whichprohibits the particle from being at rest at the point of minimum potentialenergy (i.e., at x = 0), since the kinetic energy becomes infinitely large then.The total energy thus becomes infinite, instead of being minimized. The three“scenarios” of Figure 6.7 show how the particle “searches for”—and eventually“finds”—the state of maximum possible rest!

The use of anthropocentric expressions in these descriptions—the particle“searching” for its state!—is intentional. It vividly describes the minimizationprocedure responsible for the formation of the ground-state wavefunction ofany quantum system. In this particular case, the procedure can be expressedquantitatively as follows: Let Δx ≈ a be the position uncertainty of the particleand Δp ≈ ℏ∕a its corresponding momentum uncertainty, according to theapproximate form of Heisenberg’s uncertainty principle, written as Δx ⋅ Δp ≈ ℏ.The mean energy of the particle is then given by the expression

V(x) V(x) V(x)

ψ(x)

ψ(x)

ψ(x)

x x x

Figure 6.7 How the state of minimum energy is reached for the harmonic oscillator. Scenario(a): In order for the particle to get closer to the bottom of the potential and thus minimize itspotential energy, it “forms” a strongly localized wavefunction. But in doing so, its kineticenergy increases so much that it cancels out the benefit of the lowering of the potentialenergy. Scenario (b): In order for the particle to avoid excessive localization—and theassociated increase of its kinetic energy—it “forms” a very broad wavefunction. Such awavefunction, however, raises the potential energy excessively, since it enhances theprobability of finding the particle away from the bottom of the potential. The total energy isnot minimized now either. Scenario (c): Finally, the minimization of the total energy is achievedby balancing the competing requirements of the potential and kinetic energy terms. Theactual wavefunction of the particle is neither too narrow nor too broad. It has the optimalextent.


E = ⟨H⟩ =⟨

p2

2m+ 1

2m𝜔2x2

⟩

= 12m

⟨p2⟩ + 12

m𝜔2⟨x2⟩ (6.22)

= 12m

(Δp)2 + 12

m𝜔2(Δx)2 (6.23)

= ℏ2

2ma2 + 12

m𝜔2a2, (6.24)

where, in going from (6.22) to (6.23), we used the fact that ⟨p⟩ = 0 and ⟨x⟩ = 0to write ⟨p2⟩ = (Δp)2 and ⟨x2⟩ = (Δx)2. The mean total energy of the particle as afunction of a can thus be written as

E = E(a) = ℏ2

2ma2 + 12

m𝜔2a2.

This expression is a sum of two terms—kinetic and potential—with a competingdependence on the confinement length a. As a decreases, the potential energydecreases, but the kinetic term then increases; and vice versa. The minimumenergy is attained when

dEda

= 0 ⇒ a = a0 =√

ℏ

m𝜔,

and the minimum value is

Emin = E(a0) = ℏ𝜔,

which differs from the exact value (by a factor of 2) because the given calculationis approximate, and can only serve as an order-of-magnitude estimate.

All these considerations are summarized graphically in Figure 6.8.

6.3.6 Equidistant Eigenvalues and Emission of Radiation from a QuantumHarmonic Oscillator

Looking at Figure 6.3 or the expression En =(

n + 12

)

ℏ𝜔, we can immediatelysee that the eigenvalues of a harmonic oscillator are equidistant, with the distancebetween them equal to ℏ𝜔. This is a very special feature of the parabolic potentialand, as we will show, it is the quantum analog of an equally distinct property ofthe corresponding classical problem: The oscillation period in a parabolic poten-tial is independent of the oscillation amplitude or the energy of the particle. Theperiod is always the same regardless of the oscillation amplitude (which is why apendulum can function as a clock). It therefore follows that, if the classical parti-cle is charged, it will emit electromagnetic radiation at the constant frequency ofits oscillation.

Let us now examine what happens in the quantum problem. Emission of radia-tion occurs in the form of indivisible light quanta (i.e., photons) when the charged


E(a)

E(a) ≈2ma2

mω2a2

2

1+

aa0

ℏ2

Figure 6.8 The mean total energy of a quantum oscillator as a function of the “size” a of itswavefunction. If the wavefunction is very localized (small a), the potential energy decreases,but then the kinetic energy increases excessively. Conversely, if the wavefunction is veryextended (large a), the kinetic energy goes down, but the potential energy grows greatly. Theminimum total energy is attained for an intermediate value of a (=

√ℏ∕m𝜔) and the

corresponding minimum energy is equal to ℏ𝜔.

particle performs quantum leaps4 from a higher to a lower energy level. As weknow, the emitted photon carries away the energy difference between the twolevels. Therefore, if the transition occurs between two successive levels, say, n andn − 1, then the energy of the emitted photon is equal to ℏ𝜔. Its frequency is then𝜔, which is the same as for the classical oscillator and the corresponding classi-cal radiation. So classical and quantum physics happen to agree on the frequencyof the emitted radiation in this case. Actually, this agreement is mandatory onlyin the limit of large quantum numbers, where quantum predictions must coin-cide with the corresponding classical ones, as we have repeatedly discussed. (Thisrequirement is often called the correspondence principle and was formulated byBohr, who made extensive use of it.) Such an agreement is guaranteed only if wedemand quantum transitions to satisfy the selection rule

Δn = 1,which means that

the only allowed transitions between the states of a quantum harmonicoscillator are those for which the quantum number n changes by one—thatis, transitions between successive levels.

The correspondence principle does not of course preclude transitions withΔn > 1 for small quantum numbers, where agreement between quantum and

4 The term “transitions” is also often used for the “quantum leaps”—or “quantum jumps”—ofparticles from one energy state to another.


xn = 0

n = 1

n = 2

n = 3

n = 4

n = 5

n = 6

Figure 6.9 Allowed and forbidden transitions in the harmonic oscillator. The only allowedtransitions are between neighboring levels (Δn = 1). All other transitions (Δn > 1) areforbidden; they are depicted here with “crossed-out” arrows and without any indication ofemitted photons. Since the oscillator levels are equidistant, the allowed transitions produceradiation of a single frequency, as in the corresponding classical problem.

classical physics is not mandatory. For example, in the nonclassical regime ofsmall n, we could have transitions with, say, Δn = 2—from n = 3 to n = 1. Forsuch transitions, the energy of the emitted photon will be 2ℏ𝜔 and its frequency2𝜔, which is not observed in the classical spectrum. Nevertheless, the exactquantum mechanical theory of radiation, which we discuss in Chapter 16, showsthat transitions with Δn > 1 do not take place even for small quantum numbers.The selection rule Δn = 1 holds for all n.

The fact that the harmonic oscillator has equidistant eigenvalues should bequite transparent by now, at least for large n: In that regime, there is no otherway to reconcile quantum theory with classical mechanics whereby a harmonicoscillator has only one frequency for all oscillator amplitudes.

These results on radiation emission from a harmonic oscillator are summarizedin Figure 6.9.

Problems

6.4 Use the parabolic approximation to calculate the first two or three energylevels of the potential V (x) = V0e𝜆x2 . What can you say about the higherenergy levels of this potential vis-a-vis those of the harmonic oscillator? Arethey equidistant? Explain.

6.5 Use the uncertainty principle to estimate the ground-state energy for thepotential V = gx4.


6.4 A Plausible Question: Can We Use the PolynomialMethod to Solve Potentials Other than the HarmonicOscillator?

Some inquiring readers may have the following question at this point. Canone also apply the polynomial method to other potentials to obtain an exactsolution—namely, the eigenvalues and eigenfunctions in closed analytical form?For example, can we apply the polynomial method to potentials like, say, gx4 orgx6, to find their eigenvalues and eigenfunctions?

This is the topic of the following chapter, where, starting from the abovemen-tioned question, we construct a systematic theory of the polynomial method thatallows us to decide whether a potential, like those mentioned, is exactly solvable;and if yes, to expeditiously calculate its solution without any prior knowledgeabout the differential equations involved. But to keep this special chapter as a“gift” for the mathematically inclined readers, we had to arrange things so it canalso be comfortably omitted by all others. All material following Chapter 7 is thusbased only on the simple techniques we developed so far. In other words, in solv-ing problems like the hydrogen atom, we will take it as given that polynomialsolutions of the corresponding equation exist, and apply the relevant necessarycondition to find the eigenvalues in just one line of algebra. We will also pro-ceed to construct at least the first two or three polynomials, both to make surethat polynomial solutions indeed exist and to expressly give the wavefunction ofthe most basic states of the quantum system at hand: its ground state and thefirst couple of its excited states. Using this simplified version of the polynomialmethod, the exact solution of the “classic” problems of quantum mechanics willprove much easier than in classical mechanics! And it will become easier stillfor those readers who venture into the more systematic techniques presented inChapter 7.

Further Problems

6.6 The shifting of a quantum system’s energy levels due to a homogeneouselectric field is called the Stark effect. (The corresponding shift is calledthe Stark shift.) The simplest case of the Stark effect relates to a chargedharmonic oscillator in a homogeneous electric field pointing along theoscillation axis. The potential for this problem has the form

V = 12

kx2 − qx, (1)

where q is the charge of the oscillating particle and is the intensity of theelectric field. Show that the presence of the additional term −qx in (1)causes a constant shift

ΔE = −q22

2kto all energy levels of the harmonic oscillator. By definition, ΔE is the Starkshift of the problem.


6.7 The state of a harmonic oscillator at a certain time is described by the infi-nite superposition of eigenstates

𝜓(x) = N∞∑

n=0

𝜆n∕2√

n!𝜓n(x),

where 𝜆 is a given, but arbitrary, real number.(a) Calculate the average energy ⟨E⟩ and the uncertainty ΔE of the oscil-

lator in this state.(b) Set 𝜆 to a specific value, for example, 𝜆 = 2. Find the most probable

energy value that can be measured and the probability for this mea-surement.

6.8 The state of a particle at a given time and in the natural unit system of aharmonic oscillator is described by the wavefunction

𝜓(x) = Nx2e−x2∕2. (1)

All of the following statements are wrong. Explain why, without perform-ing any calculation.(a) The mean energy of the particle in state (1) is (i) ⟨E⟩ = 3, (ii) ⟨E⟩ = 1∕4.(b) The mean kinetic energy of the particle in state (1) is ⟨K⟩ = 5∕2.(c) The mean momentum of the particle in state (1) is ⟨p⟩ = 2.

6.9 A particle is in the ground state of a harmonic oscillator. What is theprobability of finding it in the classically forbidden region of motion?Hint: You can find an approximate value for the relevant integral ∫ 1

−1 e−x2 dx,with a trick similar to what one uses for calculating ∫ ∞

−∞ e−x2 dx. That is,instead of integrating over the surface of a square, you integrate over asuitable circular disc.

6.10 A particle with mass m performs a three-dimensional motion under theinfluence of the potential

V = 12

kx2 + 12

ky2 + 12

kz2 = 12

m𝜔2(x2 + y2 + z2),

which is known as the three-dimensional harmonic oscillator. Find itseigenvalues and eigenfunctions and sketch the energy-level diagram forthe first five levels, denoting also their degeneracy. Work in the system ofunits where ℏ = m = 𝜔 = 1.

6.11 A particle with mass m is moving under the influence of the potential(anisotropic harmonic oscillator)

V (x, y, z) = 12

kx2 + 12

ky2 + 2kz2.

Draw the energy diagram for the first four levels and calculate the eigen-function of the ground state in the system of units where ℏ = m = k = 1.


6.12 A harmonic oscillator is in the superposition state

𝜓 = 1√

3𝜓0 +

√23𝜓1.

Somebody calculated the energy uncertainty ΔE for this state and claimsto have found that ΔE = 3. Can you argue why this claim is clearly wrong?

6.13 Successive energy measurements for the same physical state of a harmonicoscillator yielded only the two values E0 = 1∕2 and E1 = 3∕2, with proba-bilities P0 = 1∕3 and P1 = 2∕3, respectively.(a) Give the most general state of the oscillator consistent with the

measurement data.(b) Determine this state exactly if you are also given one of the following:

(i) ⟨x⟩ = 0,(ii) ⟨x⟩ = 1∕3,

(iii) ⟨p⟩ = 0,(iv) ⟨p⟩ =

√2∕3.

6.14 At a given moment in time, the state of a harmonic oscillator is a superpo-sition of its ground and first excited states. Determine this state exactly ifyou are also given one of the following pairs of data:(a) ⟨E⟩ = 1, ⟨x⟩ = 1∕2,(b) ⟨x⟩ = 1∕2, ⟨p⟩ = 1∕2,(c) ⟨E⟩ = 5∕4, ⟨p⟩ = 0.

6.15 Consider a harmonic oscillator whose state is described at a given momentin time by the wavefunction (in units where ℏ = m = 𝜔 = 1)

𝜓(x) = N(2x + i) e−x2∕2. (1)

Calculate the following:(a) ⟨x⟩,Δx,(b) ⟨p⟩,Δp,(c) ⟨E⟩,ΔE.Then, calculate the following time-evolved mean values:(d) ⟨x⟩t ,(e) ⟨p⟩t ,(f ) ⟨E⟩t .What is the most probable energy value that could be measured when thesystem is in the state described by (1)? What is the corresponding proba-bility of measuring this value?

191

7

The Polynomial Method1: Systematic Theory andApplications

7.1 Introduction: The Power-Series Method

Having solved the quantum oscillator problem—at least for the low-lyingstates—let us now look at the formal method of solution. Recall that the equationwe are dealing with, namely, the Hermite equation

H′′ − 2xH′ + (2E − 1)H = 0, (7.1)

is a linear and homogeneous differential equation with variable coefficients,whose only general method of solution is the power-series method. Here is howit works: We write the solution H(x) as an infinite power series of the form

H(x) =∞∑

k=0akxk , (7.2)

insert it into Eq. (7.1), and examine how to choose the unknown coefficients akso that the equation is satisfied. Given that

H′ =∑

kakkxk−1, H′′ =

∑

kakk(k − 1)xk−2,

the insertion of (7.2) into (7.1) gives∑

kk(k − 1)akxk−2 − 2x

∑

kkakxk−1 + (2E − 1)

∑

kakxk = 0. (7.3)

The original equation will be satisfied when the total coefficient—let us call itbk—of an arbitrary power xk on the left-hand side of (7.3) vanishes. This totalcoefficient is derived from the three partial series of (7.3) as follows: For the firstseries, the general power is xk−2, so to find the coefficient of xk we must shift kby two units in the corresponding coefficient k(k − 1)ak . The contribution of thisseries to the total coefficient of xk is then

k(k − 1)ak|k→k+2 = (k + 2)(k + 1)ak+2. (7.4)

1 This chapter can be safely omitted by readers interested in applications. But readers who aspire tounderstand the basic method of how to obtain exact solutions of the Schrödinger equation will findsome key answers here.


192 7 The Polynomial Method: Systematic Theory and Applications

For the second series in (7.3), multiplication by x restores xk as the general power,so the contribution of this series to the total coefficient will be

−2kak . (7.5)

Likewise, the third series contributes to the total coefficient bk the term

(2E − 1)ak . (7.6)

Adding (7.4)–(7.6), we obtain the following expression for bk :

bk = (k + 1)(k + 2)ak+2 − 2kak + (2E − 1)ak ,

which should now be set equal to zero to satisfy Eq. (7.3) and thus also Eq. (7.1).On doing so and solving for ak+2, we obtain the recurrence relation

ak+2 = 2k − (2E − 1)(k + 1)(k + 2)

ak . (7.7)

Formula (7.7) determines the solution completely, because it provides all the coef-ficients of the series (7.2) if we know the first one (which can always be set equalto unity, since the equation is homogeneous). We can also see from (7.7) that theseries has a step of two (it “jumps” from k to k + 2), and hence it will only containeither even or odd powers of x, depending on the “starting point.” If the seriesstarts with the zeroth power (i.e., with the constant term a0), it will contain onlyeven powers. If it starts with the first power of x (i.e., with the coefficient a1), itwill contain only odd powers. (The reason the starting power for Eq. (7.1) is x0 orx1 will become apparent later.) Consequently, the solutions will be either even orodd, as expected.

A closer look at the recurrence relation (7.7) also suggests that for

2E − 1 = 2n ⇒ E = En = n + 12, (7.8)

whence

ak+2 = 2(k − n)(k + 1)(k + 2)

ak , (7.9)

the series terminates at a polynomial of degree n. (For k = n the recurrence for-mula gives null and, therefore, all higher coefficients also vanish.) We thus arrivedat the result we already knew for the energy eigenvalues of the harmonic oscilla-tor, but with a systematic procedure that establishes its correctness beyond anydoubt.2 Furthermore, we have now obtained a recurrence relation that allows usto promptly calculate the coefficients of polynomial solutions for any value of n.For example, for n = 2 (and a0 = 1) the recurrence relation (7.9) gives

(

ak+2 = 2(k − 2)(k + 1)(k + 2)

ak

)|||||k=0

⇒ a2 = 2(0 − 2)1 ⋅ 2

⋅ 1 = −2,

2 It is useful to recall that what we showed in the previous chapter is simply the following: Ifpolynomial solutions of arbitrary degree n exist, then E must take values from the discrete sequence(7.8). However, as we stressed back then, this condition is necessary but not sufficient to guaranteethe existence of polynomial solutions. It is true that we had a sort of practical certainty thatpolynomial solutions do exist, since we actually constructed the first three of them. But practicalcertainty does not constitute a proof, as we know!

7.1 Introduction: The Power-Series Method 193

while all higher coefficients (a4, a6, a8, etc.) vanish. We thus have

H2 = a0 + a2x2 = 1 − 2x2 ∼ 2x2 − 1, (7.10)

which is the result we had already obtained in the previous chapter.Do not be surprised, however, if upon asking your symbolic calculator to

provide the Hermite polynomial H2(x), you obtain an expression different from(7.10) by an overall constant. The reason for this is simple. Given that theHermite polynomials are solutions of a homogeneous equation, they are definedup to a multiplicative constant, which may well depend on n. This constantis chosen in the literature on the basis of a “normalization condition” thatdefines the Hermite polynomials uniquely. So when we use the symbol Hn(x) wealways refer to a completely determined polynomial. What this “normalizationcondition” is and what properties these standard Hermite polynomials have, arethe kind of questions discussed in standard textbooks of mathematical physics.

We now come to a key question that may have occurred to many readers:Can we solve the Schrödinger equation for potentials “similar” to the harmonicoscillator using the same procedure, namely, the polynomial method? Forexample, can we use this method to solve for potentials like V = gx4 or V = gx6?At first glance, one may be tempted to answer yes. After all, the polynomialmethod, as presented earlier, seems to be entirely general. So, after performingthe necessary dimensional simplification (ℏ = m = g = 1), we just need to findthe asymptotic behavior at infinity—which would surely be an exponential of theform exp (−𝜆x𝜇) with suitable values for 𝜆 and 𝜇—and write the full solution inthe familiar form

𝜓(x) = 𝜓∞(x)F(x) = e−𝜆x𝜇F(x). (7.11)

Here, F(x) is the complementary function (analogous to H(x) for the harmonicoscillator), and is expected to be a polynomial of all possible degrees, for the phys-ical reasons already explained. As an illustration of where things can go wrongwith this approach, let us take V = 1

2gx6, where the factor 1

2is introduced for later

convenience. As you can see by a direct substitution in the Schrödinger equation,the asymptotic factor now has the form 𝜓∞(x) = exp (−x4∕4), and hence (7.11)is written as

𝜓(x) = e−x4∕4F(x),

which, upon insertion in the Schrödinger equation

𝜓 ′′ + (2E − x6)𝜓 = 0 (ℏ = m = g = 1),

yields the following new equation for F :

F ′′ − 2x3F ′ + (2E − 3x2)F = 0. (7.12)

However, you can easily verify—for example, by checking whether the zeroth-and first-degree polynomials F0 = 1 and F1 = x satisfy (7.12)—that the givenequation does not have polynomial solutions!

Therefore, we cannot always apply the polynomial method, as one would beinclined to assume from the deceptive generality of its pronouncement. Far frombeing self-evident, the existence of polynomial solutions is actually a rare event


that occurs under very special conditions. These conditions are investigated inthe following section.

7.2 Sufficient Conditions for the Existenceof Polynomial Solutions: Bidimensional Equations

The question before us is the following. Why does the polynomial methodwork for the potential V = kx2∕2, but not for V = gx6∕2? To put it differently:Why does the Hermite equation (7.1) have polynomial solutions, while thecorresponding equation for gx6∕2—that is, Eq. (7.12)—does not? Let us see why.First, recall that when we applied the power-series method to (7.1) we arrived ata recurrence relation with two terms, which has the general form

ak+𝓁 = f (k)ak , (7.13)

where f (k) is some function of k—in fact, a rational function—and 𝓁 is the step ofthe series. When𝓁 > 1, the series contains only those powers of x that result fromthe starting power by increments of 𝓁. For Eq. (7.7) we had 𝓁 = 2, so the serieswas indeed “advancing” with step 2, since it contained only even or odd terms.But a recurrence relation of the form (7.13) guarantees polynomial solutions pro-vided the factor f (k), which connects one coefficient with the next, vanishes forsome value of k, in which case all subsequent coefficients also vanish and theseries terminates at some polynomial. In contrast, when instead of (7.13) we havea recurrence relation with three terms, which has the general form

ak+2 = f (k)ak+1 + g(k)ak ,

then even if both functions f (k) and g(k) vanish for some k, this only ensuresthat ak+2 vanishes, but not necessarily all subsequent coefficients. For example,ak+3 is generally nonzero, since it depends not only on its immediately precedingterm ak+2, which vanishes, but also on ak+1, which does not. And this is thecrucial difference between Eqs (7.1) and (7.12). The former equation leads to arecurrence relation with two terms, while the latter leads to a correspondingrelation with three terms, as you can see by applying again the power-seriesmethod.

We thus arrive at the following question: Can we tell a priori (by mere inspec-tion, if possible) whether an equation leads to a recurrence relation with two,three, or more terms? To be clear, we are dealing with linear and homogeneousequations with polynomial coefficients, like (7.1) or (7.12), and our basic reason-ing is as follows. The number of terms in the recurrence relation must be equal tothe number of distinct displacements d1, d2,… , dn effected on the general termxk of the power-series solution, y(x) =

∑akxk , when it is inserted in the equation.

Assuming, as we did, that our equation has polynomial coefficients, it will onlycontain terms of the general form

xny(m) ≡ xn dmydxm . (7.14)

7.2 Sufficient Conditions for the Existence of Polynomial Solutions: Bidimensional Equations 195

Each one of these terms causes a displacement d on the exponent of the arbitrarypower xk , where

d = n − m, (7.15)

which is also the physical dimension of the term (7.14) if we treat x as havingdimensions of length and y as a dimensionless quantity. The integer number d isalso the length dimension of the corresponding operator

L = xn dm

dxm ,

whose action on the arbitrary power xk yields

Lxk = A(k)xk+d (d = n − m,A(k) = k(k − 1)… (k − (m − 1))),

that is, it shifts its exponent by d (as should be expected on dimensional groundsalone). It is clear from this discussion that the application of the power-seriesmethod to equations with polynomial coefficients leads to a recurrence relationwith n terms, where n is the number of distinct dimensions the equation con-tains. At this point, it is necessary to introduce some new terminology: We call adifferential equation of this type unidimensional, bidimensional, tridimensional,and so on. The most interesting special case of n-dimensional equations are thebidimensional ones. According to the discussion, they lead to recurrence rela-tions with two terms and can, therefore, in principle, have (the highly desirable)polynomial solutions. The trivial class of unidimensional equations has no recur-rence relation at all, and the solution reduces to a single power. Such equationsare known in the literature as Euler equations. In the light of this discussion, thedifference between Eqs (7.1) and (7.12) is now clear. Equation (7.1) is a bidimen-sional equation with dimensions

d1 = −2 (term H′′), d2 = 0 (terms − 2xH ′, (2E − 1)H),

while (7.12) is a tridimensional equation with dimensions

d1 = −2 (term F ′′), d2 = 0 (term 2EF), d3 = +2 (terms − 2x3F ′,−3x2F).

We can now say right away that Eq. (7.1), being bidimensional, can have polyno-mial solutions, while (7.12), being tridimensional, cannot. We can also say that thedifference between the dimensions d1 and d2 of a bidimensional equation givesthe “distance” between the distinct power terms xk+d1 and xk+d2 that arise whenwe substitute the series in the equation, and is thus equal to the step, 𝓁, of theresulting recurrence relation. That is,

𝓁 = d2 − d1, (7.16)

where, by convention, d2 denotes the highest dimension and d1 the lowest one.Indeed, for the Hermite equation (where d1 = −2 and d2 = 0), Eq. (7.16) yields𝓁 = 2, as it should (see Eq. (7.7)). For another well-known equation of mathe-matical physics, y′′ − xy = 0, namely, the Airy equation, (7.16) tells us right awaythat its power-series solution will advance with step equal to 𝓁 = 1 − (−2) = 3.


It follows from the definition of a bidimensional equation, Ly = 0, that itsdifferential operator, L, can always be written as

L = L1 + L2, (7.17)

where L1 is the (unidimensional) operator of the terms with the low dimensiond1, and L2 the corresponding operator with the high dimension d2. For example,for the Hermite equation (7.1) we have

L1 = d2

dx2 (d1 = 2) and L2 = −2x ddx

+ (2E − 1) (d2 = 0).

The significance of writing L in the form (7.17) (i.e., separating the lowest-dimension terms of the equation from the highest-dimension ones) can now beseen from the simple fact that the unidimensional equations

L1xs = 0, L2x𝜈 = 0 (7.18)

give the power behaviors of the solution y(x) for small and large x, respectively.In other words, the former equation in (7.18) yields the starting-power term xs

of the series solution

y(x) = xs∑

kakxk , (7.19)

while the x𝜈 term is the terminating power of the series if such a terminationindeed occurs. If not, then this power represents just one of the asymptoticbehaviors for large x, which are not always of exponential form. For example,the former equation in (7.18) arises readily if we recall that for small x, the series(7.19) is dominated by the lowest power xs and that, as the operator L1 + L2 actson xs, the lowest power that emerges (and which is the only surviving power forsmall x) is xs+d1 , which results from the action of L1 on xs. So, for the equationto be satisfied for small x, the power xs alone must satisfy the equation L1xs = 0.We can apply similar reasoning for large x. We thus arrive at the following basictheorem for the existence of terminating (i.e., polynomial) solutions.

Theorem 7.1 A sufficient condition for a linear and homogeneous equation withpolynomial coefficients to have terminating solutions is that: (i) It is bidimensionaland (ii) the terminating power 𝜈 differs from the starting power s by an integernumber of steps. That is,

𝜈 = s + n𝓁, n = 0, 1, 2,… , (7.20)

where s and 𝜈 are determined from the conditions (7.18), and 𝓁 is given by (7.16).A very simple result, indeed.

Problems

7.1 You are given the equations

(1 + x2)y′′ + 4xy′ + 𝜆y = 0 (𝜆 = −n(n + 3)) (1)

7.3 The Polynomial Method in Action: Exact Solution of the Kratzer and Morse Potentials 197

y′′ − 3xy′ + 𝜆y = 0 (𝜆 = 3n) (2)y′′′ − xy′ + 𝜆y = 0 𝜆 = ({3n, 3n + 1, 3n + 2}) (3)

Show that each equation has polynomial solutions for the listed parametervalues. Note that n is a positive integer or zero.

7.3 The Polynomial Method in Action: Exact Solutionof the Kratzer and Morse Potentials

We will now apply the abovementioned techniques to the Schrödinger equationfor two potentials that allow for a more accurate description of molecular oscil-lations compared to the harmonic oscillator of Chapter 6. In effect, we are goingto solve the following two examples.

Example 7.1 The Kratzer potential V (x) = Gx2 −

gx(G, g > 0).

V(x)

x0 = 2G / g x

E < 0

Vmin = –g2 / 4G

Mathematical formulation of the problem

Schrödinger

equation

Boundary conditions

for bound states

ψ″

ψ (0) = ψ (∞) = 0

ψ = 0+2m

E – +Gx2ħ2

gx

Solution: First, we choose a system of units where ℏ = m = g = 1, in which casethe Schrödinger equation is written as

𝜓 ′′ + 2(

E − Gx2 + 1

x

)

𝜓 = 0, (1)

where the only free parameter is G, with dimensions [G] = ℏ2∕m. This meansthat G is dimensionally dependent on ℏ and m, so we could not have chosen Ginstead of g to form the basic triad of dimensionally independent parameters forthe problem. Our next step is to calculate the asymptotic behavior of the solutionfor large x by examining the corresponding limiting form of (1), as follows:

𝜓 ′′∞ + 2E𝜓∞ = 0, 2E = −𝛾2 ⇒ 𝜓∞(x) = e±𝛾x, (2)

where, as usual, we write 𝜓(x) as𝜓(x) = e−𝛾xF(x). (3)

Upon inserting this in (1), we obtain

F ′′ − 2𝛾F ′ +(2

x− G

x2

)

F = 0, (4)


which is a bidimensional equation with equal-dimension terms grouped asfollows:

F ′′, − Gx2 F (d1 = −2), −2𝛾F ′,

2x

F (d2 = −1).

In contrast, the initial Eq. (1) was tridimensional with dimensions d1 = −2(terms 𝜓 ′′, −2G𝜓∕x2), d2 = −1 (term 2𝜓∕x), and d3 = 0 (term 2E𝜓). Wethus realize that, from a purely mathematical perspective, the removal ofthe asymptotic factor at infinity turned our equation from tridimensional tobidimensional. Now, Eq. (4) has terminating solutions when the terminationcondition 𝜈 = s + n𝓁 is satisfied, with 𝓁 = d2 − d1 = −1 − (−2) = 1. Moreover, sand 𝜈 are readily calculated from the conditions (7.18), which give

L1xs = 0 ⇒ (xs)′′ − 2Gx2 xs = 0 ⇒ (s(s − 1) − 2G)xs−2 = 0

⇒ s2 − s − 2G = 0 ⇒ s =1 ±

√1 + 8G2

(5)

and

L2x𝜈 = 0 ⇒ −2𝛾(x𝜈)′ + 2x(x𝜈) = (−2𝛾𝜈 + 2)x𝜈−1 = 0

⇒ −2𝛾𝜈 + 2 = 0 ⇒ 𝛾 = 1𝜈

(𝜈 = s + n), (6)

where L1 and L2 are the (unidimensional) operators corresponding tothe lowest-dimension terms (F ′′,−2GF∕x2) and highest-dimension terms(−2𝛾F ′, 2F∕x) of (4), respectively. By selecting s as the positive root in(5)—otherwise the wavefunction 𝜓(x) would not only fail to vanish at x = 0but would actually be infinite there—the termination condition 𝜈 = s + n(n = 0, 1, 2,…), in conjunction with (6) and 2E = −𝛾2, yields the followingexpression for the energy eigenvalues

En = −12

1(s + n)2 , n = 0, 1, 2,… (7)

To convert this result to familiar units we need to multiply expression (7) with theenergy unit 𝜖 of the problem—that is, the unique combination of the parametersℏ,m, and g with dimensions of energy. Now, 𝜖 can be readily found by equatingthe quantities ℏ2∕ma2 and g∕a—where a is length—since they both have dimen-sions of energy. (Why?) We thus obtain

ℏ2

ma2 =ga⇒ a = ℏ2

mg⇒ 𝜖 =

ga||||a=ℏ2∕mg

=mg2

ℏ2 . (8)

But to fully restore dimensions, expression (5) for s (with the positive sign)should stay dimensionless, so the quantity G inside the square root should bereplaced by G∕G0, where G0 (= ℏ2∕m) is the unique combination of ℏ,m, and g


with dimensions of G. Our final result for the eigenvalues is thus written as

En = −mg2

2ℏ2(s + n)2 , n = 0, 1, 2,… , (9)

where

s =1 +

√1 + (8mG∕ℏ2)

2. (10)

For the ground state of the system in particular (n = 0), we find

E0 = −mg2

2ℏ2s2 = −2mg2

ℏ2(1 +√

1 + (8mG∕ℏ2) )2(11)

and, therefore, in the classical limit ℏ → 0 or m → ∞, we have

E0(ℏ→ 0 or m → ∞) = −g2

4G= Vmin, (12)

so the particle indeed falls to the “bottom” of the potential in the classical limit.Note also that for G = 0, whereby the Kratzer potential reduces to the

one-dimensional Coulomb potential (with g = e2), we have s = 1, and theeigenvalue formula (9) is then written as

En = − me4

2ℏ2(n + 1)2 , n = 0, 1,… . (13)

This expression is identical to the Bohr result, and also to the purely quantummechanical one we will encounter later, with the obvious difference of having(n + 1) in the denominator, since the starting value of the quantum number nhere is 0 instead of 1.

Even though the Kratzer potential has the general form of a molecular poten-tial—that is, the potential between atoms in a diatomic molecule (see Chapter 13for details)—it does not decay exponentially for large x (i.e., large r) as expectedfor such potentials. For this reason, another potential has been proposed andproved very useful: the so-called Morse potential

V (x) = V0(e−2𝜆x − 2e−𝜆x),

which results from the more general expression

V (x) = Ae−2𝜆x − Be−𝜆x

by a suitable displacement of x so that the potential minimum (i.e., theequilibrium position of the molecule) occurs for x = 0. Therefore, our secondexample is to apply what we learned so far to find the exact solution of thispotential. Note that the potential is no longer a rational or polynomial function,so we must first employ a suitable change of the independent variable inthe Schrödinger equation to transform it into an equation with polynomialcoefficients. Let us begin.


Example 7.2 The Morse potential: V (x) = V0(e−2𝜆x − 2e−𝜆x)

V(x) Parabolic

approximation

E < 0

–V0

x

Mathematical formulation of the problem

Schrödinger

equation

Boundary conditions

for bound states

ψ″

ψ (–∞) = ψ (∞) = 0

ψ = 0+2m

E – V0(e–2λx – 2e–λx )ħ2

Solution: To simplify dimensions, we first select as our basic triad the parametersℏ,m, and 𝜆 (i.e., we set ℏ = m = 𝜆 = 1), so the Schrödinger equation takes theform

𝜓 ′′ + 2(E − V0e−2x + 2V0e−x)𝜓 = 0. (1)

The equation is transformed into one with polynomial coefficients upon thefollowing change of independent variable

z = e−x ∶ x = −∞ ⇒ z = ∞, x = +∞ ⇒ z = 0 (2)

for which we obtain

𝜓 ′′ =(

dzdx

)2

�� + d2zdx2 �� = z2�� + z�� ,

whence (1) is transformed to

z2�� + z�� + 2(E − V0z2 + 2V0z)𝜓 = 0, (3)

where the dots denote differentiation with respect to the new variable z.Note now that (3) is a tridimensional equation with dimensions zero (terms

z2�� , z�� , 2E𝜓), one (term 4V0z𝜓), and two (term −2V0z2𝜓). We hope, however,that the removal of the asymptotic behavior at infinity (z → ∞) will transformit into a bidimensional equation. This asymptotic behavior will surely have anexponential form because the highest-dimension operator of (3), L2 = −2V0z, isof zeroth order (it contains no derivatives), so the condition L2x𝜈 = 0 has no solu-tions. Therefore, both behaviors of (3) for large z will have an exponential form.So let us try an exponential of the form 𝜓∞ ∼ exp(𝜇z)—we leave it to the readersto check the more general form𝜓∞ ∼ exp(𝜇z𝜌)—for which one can easily confirmthat it satisfies (3) for large z, provided that 𝜇 = ±𝛾 = ±

√2V0. We thus have

𝜓∞(z) ∼ e±𝛾z (𝛾 =√

2V0), (4)

and, therefore, the substitution we should make in (3) is

𝜓(z) = e−𝛾zF(z), (5)


so the equation now takes the form

z2F + (z − 2𝛾z2)F + (2E + (2𝛾2 − 𝛾)z)F = 0, (6)

which is indeed bidimensional as hoped. The lowest-dimension terms are z2F ,zF , and 2EF , with dimension d1 = 0, while the highest-dimension terms are−2𝛾z2F and (2𝛾2 − 𝛾)zF , with dimension d2 = 1. The step of the equation is𝓁 = 1.

The rest of the algebra is rather easy. We have (note that the dots outside theparentheses denote differentiation with respect to z)

• L1zs = 0 ⇒ z2(zs)•• + z(zs)• + 2E(zs) = (s(s − 1) + s + 2E)zs = 0⇒ s2 + 2E = 0 (7)

• L2z𝜈 = 0 ⇒ −2𝛾z2(z𝜈)• + (2𝛾2 − 𝛾)z ⋅ z𝜈 = (−2𝛾𝜈 + (2𝛾2 − 𝛾))z𝜈+1 = 0

⇒ 𝜈 = 𝛾 − 12

(8)

• 𝜈 = s + n𝓁|𝓁=1 = s + n ⇒ s = 𝜈 − n = 𝛾 − 12− n (9)

and because of (7) and (9), we finally obtain

En = −12

s2 = −12

(

𝛾 − n − 12

)2(𝛾 =

√2V0), (10)

which gives us the allowed energies of the problem in units whereℏ = m = 𝜆 = 1.Note, however, that here (in contrast to the potential in the previous example) nhas an upper bound given by the inequality

n ≤ 𝛾 − 12=√

2V0 −12, (11)

since s in (9) must be positive. (Why?) We thus have a finite number of boundstates, in contrast to the Kratzer potential, where their number was infinite.Finally, we leave it to the readers to show that, in familiar units, we have

En = −ℏ2𝜆2

2m

(√2mV0

ℏ𝜆− n − 1

2

)2

. (12)

This expression can be equivalently written as

En = −V0 + ℏ𝜆√

2V0

m

(

n + 12

)

− ℏ2𝜆2

2m

(

n + 12

)2, (13)

which is the same as the empirical expression

En = −V0 + ℏ𝜔0

(

n + 12

)

− 𝜖

2

(

n + 12

)2(14)

used by spectroscopists to describe the vibration spectrum of many diatomicmolecules. The first term in (14) is the minimum potential energy of the molecule,the second term represents the quantized energies of small oscillations about theequilibrium position (where the actual molecular potential is adequately approx-imated by a harmonic oscillator), while the third term describes deviations from


the harmonic oscillator spectrum and from its very special feature, namely, thatlevels are equidistant. Therefore, the parameter 𝜖, due also to the square depen-dence of the corresponding term on the quantum number n, describes the con-stant step by which the distance between successive vibrational levels is shortenedas we go to higher frequencies. This effect (i.e., the increasing close packing ofthe levels) should be expected, since the right branch (i.e., for large r) of a realmolecular potential does not increase as sharply as a harmonic oscillator—fromsome point onward the increase is curbed and the potential becomes almosthorizontal for large r. We leave it for the readers to show that the coefficientof the second term in (13)—without ℏ—is simply the frequency of small oscil-lations about the minimum of the Morse potential, and also to discuss why ourtheoretical prediction 𝜖 = ℏ2𝜆2∕m, has this particular dependence on 𝜆 and m. Atthis point, the interested readers are invited to show also the following: All poten-tials of the form V = gx𝜇 with 𝜇 > 2, which thus increase at infinity faster thanthe harmonic oscillator, have a spectrum that becomes less dense as the energyincreases; while potentials that rise at infinity at a slower pace than the harmonicoscillator (𝜇 < 2), have a spectrum that gradually becomes more dense. The har-monic oscillator itself—with equidistant levels—lies at the borderline betweenthese behaviors.

7.4 Mathematical Afterword

There is a purely mathematical background to the previous discussion. We areinterested not only in equations with polynomial solutions but more generallyin equations (linear with variable coefficients) that are exactly solvable. Weare not going to discuss this topic here, but the readers may begin to dwell onthis question by taking a look at the following list of some known equations ofmathematical physics.

(A) (1 − x2)y′′ − 2xy′ + 𝜆y = 0 (Legendre)(B) xy′′ + (1 − x)y′ + 𝜆y = 0 (Laguerre)(C) (1 − x2)y′′ − xy′ + 𝜆y = 0 (Chebyshev)(D) x2y′′ + xy′ + (x2 − 𝜈2)y = 0 (Bessel)(E) x(1 − x)y′′ + (𝛼x + 𝛽)y′ + 𝜆y = 0 (Hypergeometric)(F) xy′′ + (𝛼x + 𝛽)y′ + 𝜆y = 0 (Confluent hypergeometric)

We observe that all these equations are bidimensional. So these famousequations are important, not because they are named after a great mathe-matician who worked on them but because they are bidimensional. They areimportant on purely mathematical grounds. Their solution by the power-seriesmethod leads to a recurrence relation with two terms, which allows for thegeneral coefficient of the series to be determined in closed form. This, of course,is the most exact representation of a function one can ask for, aside frompolynomials or rational functions. A pertinent example to clarify this point is


the power series f (x) =∑

xk∕k!, which represents the exponential function ex.Note, however, that even if we knew nothing about the function representedby this series, we would still be able to compute its numerical values with anydesired accuracy, and also show that it has all the symbolic properties we usuallyassociate with the exponential function f (x) = ex. Namely, that

f (x) f (y) = f (x + y), f ′(x) = f (x).

A power series∑

akxk with explicitly known general coefficients ak thereforeprovides complete knowledge (both numerical and symbolic) of the function itrepresents. The significance of bidimensional equations is now plain to see: Theyare exactly solvable.

The outline of a more systematic presentation of bidimensional equations alongthese lines is given in Problems 7.3–7.6.

Further Problems

7.2 Solve the Schrödinger equation for the potential

V(x)

Vmin =

x

x0 =4 2g / k

2gk 0 < x < ∞V (x) =g

+x2

1

2kx2 ⇒

and show that the allowed energies (for ℏ = m = k = 1) are given by theformula

En =√

1 + 8g2

+ 2n + 1, n = 0, 1, 2,…

Confirm that in the classical limit the particle falls to the bottom of thepotential.

7.3 From the examples of exactly solvable potentials you encountered so far(e.g., V = kx2∕2, V = (g∕x2) + kx2∕2, V = (G∕x2) − (g∕x)), and also forthose that cannot be solved exactly (e.g., V (x) ∼ x6), can you infer somepractical rule to distinguish exactly solvable potentials from others? Notethat in all these cases, the Schrödinger equation is tridimensional, but hasalso an additional feature shared by exactly solvable potentials only. Whatis this feature?


7.4 As we shall see in Chapter 9, the solution of the Schrödinger equation in acentral potential leads to the so-called associated Legendre equation

(1 − 𝜉2)Θ′′(𝜉) − 2𝜉Θ′ +(

𝜆 − m2

1 − 𝜉2

)

Θ(𝜉) = 0, (1)

where 𝜉 = cos 𝜃, and Θ(𝜉) gives the dependence of the wavefunction on theangle 𝜃 in spherical coordinates r, 𝜃, 𝜑. The full wavefunction 𝜓(r, 𝜃, 𝜑) hasthe form

𝜓(r, 𝜃, 𝜑) = R(r) Θ(𝜉) eim𝜑 (m = 0,±1,±2,…), (2)

where 0 ≤ 𝜃 ≤ 𝜋 ⇒ −1 ≤ 𝜉 ≤ 1.What you are asked to do here is find the solutions of (1) that are finite inthe range −1 ≤ 𝜉 ≤ 1 and especially at the endpoints.

As a first step in solving (1), observe—by multiplying both sides by1 − 𝜉2 so that its coefficients become polynomials—that we are dealingwith a tridimensional equation, which can be made bidimensional ifwe remove the asymptotic behavior at the endpoints ±1 of the naturalrange −1 ≤ 𝜉 ≤ 1, which is the analog of the range −∞ < x < +∞ of theSchrödinger equation for the harmonic oscillator.(a) Use the substitution z = 1 − 𝜉 to shift the point 𝜉 = 1 to the origin,

and show that near this point, the solutions are powers of z (i.e., ofthe form zs) with exponents s = ±|m|∕2. For symmetry reasons, thesame behavior is expected at the point 𝜉 = −1. By retaining the positiveexponent (since we are interested in solutions that are finite at 𝜉 = ±1),we can write the sought asymptotic behavior at the endpoints of therange −1 ≤ 𝜉 ≤ 1 as

(1 − 𝜉)|m|∕2(1 + 𝜉)|m|∕2 = (1 − 𝜉2)|m|∕2,

so that (in analogy to what we did so far) we must now make thefollowing substitution in (1):

Θ(𝜉) = (1 − 𝜉2)|m|∕2F(𝜉). (3)

(b) Use the substitution (3) to write Eq. (1) as

(1 − 𝜉2)F ′′ − 2(|m| + 1)𝜉F ′ + (𝜆 − |m|(|m| + 1))F = 0, (4)

which is indeed a bidimensional equation. Because the points 𝜉 = ±1are singular points, this equation has finite solutions only in polynomialform. Apply what you learned in this chapter to show that suchsolutions exist only when 𝜆 in (4) takes the discrete sequence of values:

𝜆 = 𝓁(𝓁 + 1), 𝓁 = n + |m| (n = 0, 1,…),

where n is the degree of the corresponding polynomial.(c) Let us now view this discussion in the context of the Schrödinger

equation for the solvable potentials we encountered in Problem 7.3.It becomes clear that there is a class of tridimensional equations thatreduce to bidimensional ones in a systematic manner. We simplyremove the asymptotic behavior of the solution at the endpoints of the


natural range of the problem, which are also the singular points of theequation. Tridimensional equations for which this can be done will becalled here associated tridimensional equations, in obvious referenceto the associated Legendre equation, which is a representative case.We leave it to the readers to verify that in all known cases the follow-ing statement is true: All associated tridimensional equations haveequidistant dimensions.

7.5 According to our discussion in the main text, the power behavior for thesolutions of a bidimensional equation at the origin and at infinity emergewhen we solve the unidimensional equations L1xs = 0 and L2x𝜈 = 0, whereL1 and L2 are the respective lowest- and highest-dimension componentsof the operator L in the equation Ly = 0. But while the operator L1 alwayshas the full order of the equation (in which case the condition L1xs = 0yields as many starting powers as the number of independent solutions tothe equation), the operator L2 may or may not have the full order of theequation, so that (for second-order equations) we have the following cases:

(i) L2 is of second order,(ii) L2 is of first order,

(iii) L2 is of zeroth order.In the first case, the condition L2x𝜈 = 0 has two roots, so both behaviors atinfinity are powers of x. In the second case, one behavior is a power of x andthe other an exponential. In the third case (where the condition L2x𝜈 = 0 hasno solution), both behaviors are exponentials of the general form exp(𝜇x𝜌),where 𝜇 and 𝜌 are calculated by direct substitution into the equation.

Given the significance of asymptotic behavior at infinity, as this emergedfrom the exercises we solved up to now, it is plausible to further distin-guish bidimensional equations into three kinds, depending on whether theirgroup of highest-order terms is of second order (first kind), first order (sec-ond kind), or zeroth order (third kind).(a) Accordingly, characterize all equations (A), (B), (C), (D), (E), and (F) of

Section 7.4 as of the first, second, or third kind.(b) For the Bessel equation, in particular, find its asymptotic behaviors for

small and large x. Then show that the Bessel equation is the most gen-eral bidimensional equation of the third kind with step 2, whose startingpowers are symmetrically arranged about zero.

(c) Show also that the hypergeometric equation is the most generalbidimensional equation of the first kind with step 1, with one of itsstarting powers at zero and its finite singular point at the standardposition x = 1. What is the corresponding statement for the confluenthypergeometric equation?

7.6 Use a simple argument to convince yourselves (without any calculations)that the following substitutions

y = x𝜇Y , z = xm (1)


for the dependent and independent variables, respectively, retain the bidi-mensional character and the kind of an equation. The former substitutionsimply shifts by 𝜇 the exponents of the power behaviors at zero and infinityof the initial equation. That is,

s′ = s − 𝜇, 𝜈′ = 𝜈 − 𝜇, (2)

while the latter changes the step from 𝓁 to

𝓁′ = 𝓁m. (3)

In view of this, are you surprised that all bidimensional equations of thefirst kind reduce to the hypergeometric equation, all of the second kindto the confluent hypergeometric equation, and all of the third kind tothe Bessel equation? Can you explain why the general solution y(x) of anarbitrary bidimensional equation of the third kind and with step 𝓁 canalways be written in terms of the Bessel functions Z𝜈(x)3 as

y(x) = x𝜇Z𝜈(𝜆x𝓁∕2)? (4)

As for the parameters 𝜇 and 𝜈, they are obtained by comparing theasymptotic behaviors of the two sides of (4) for small x, and the result youare asked to show is

𝜇 =s1 + s2

2, 𝜈 =

s1 − s2

𝓁, (5)

where s1 and s2 are the starting powers of the given equation. How wouldyou calculate the remaining parameter 𝜆? Apply the discussion to the Airyequation y′′ − xy = 0. Note that 𝜈 in (4) and (5) is the index of the Besselfunction (or its order, as it is also called) and is unrelated to the exponentof the power behaviors at infinity that appears in (2).

3 It is customary to denote as Z𝜈(x) the general solution c1J

𝜈(x) + c2Y

𝜈(x) of the Bessel equation of

order 𝜈, where J𝜈(x) is the well-behaving solution and Y

𝜈(x) the diverging solution of the equation

for x → 0.

207

8

The Hydrogen Atom. I: Spherically Symmetric Solutions

8.1 Introduction

The hydrogen atom is the “poster child” of quantum physics—and for goodreason. Apart from its historic importance in the discovery of the laws ofquantum mechanics and the development of quantum theory, it is also the onlyatom for which the Schrödinger equation can be solved exactly (at least, if weignore all but electrostatic interactions between the electron and the proton).The hydrogen atom is thus an ideal platform for the study of quantum theoryand the thorough testing of its implications.

From a mathematical perspective, our task now is to solve the three-dimensional Schrödinger equation

∇2𝜓 + 2mℏ2 (E − V )𝜓 = 0, (8.1)

where

V =q1q2

r= (e)(−e)

r= −e2

r(8.2)

is the potential energy—or, simply, the potential—of the electrostatic interactionbetween the electron and the proton in the cgs system of units. Note also that,due to its much bigger mass, we regard the proton as practically idle at the origin.Of course, in a rigorous treatment within classical mechanics, we would placethe origin at the center of mass of the electron–proton system and substitute theelectron mass me in (8.1) with the so-called reduced mass m of the system, whichis defined as

1m

= 1me

+ 1mp

⇒ m =memp

me + mp. (8.3)

But since we can always substitute the electron mass with the reduced mass atthe end of the calculation, we prefer—for simplicity—to treat the proton as aninfinitely heavy particle at the origin, in which case the mass m in (8.1) is theelectron mass. Given now that the potential (8.2) is a simple function of the dis-tance r from the origin, the appropriate coordinates for the solution of (8.1) are


208 8 The Hydrogen Atom. I: Spherically Symmetric Solutions

e

x = r sin θ cos ϕy = r sin θ sin ϕz = r cos θ

r

z

p y

x

θ

ϕ

r

Figure 8.1 Cartesian and spherical coordinates, and the relation between them. The proton(p) is considered fixed at the origin, while the electron (e) is at some point r with sphericalcoordinates r, 𝜃, 𝜙.

spherical coordinates, not Cartesian ones. In spherical coordinates, the Laplacian∇2 takes the form

∇2 = 1r𝜕2

𝜕r2 r + 1r2

(1

sin 𝜃𝜕

𝜕𝜃sin 𝜃 𝜕

𝜕𝜃+ 1

sin2 𝜃

𝜕2

𝜕𝜙2

)

, (8.4)

and the wavefunction 𝜓 is now a function of the spherical coordinates r, 𝜃, 𝜙(Figure 8.1). Since Eq. (8.1) involves more than one variable, it is a partialdifferential equation and, as such, its solution is not expected to be simplerthan the solution of the one-dimensional Schrödinger equation. Nevertheless,we can improve things drastically if we limit our search, to begin with, tothe subset of the solutions of (8.1) that depend only on r—that is, 𝜓 = 𝜓(r).For these so-called spherically symmetric solutions, the wavefunction has noangular dependence and, therefore, the probability distribution of the electronis spherically symmetric about the nucleus. The significance of sphericallysymmetric solutions (other than their mathematical simplicity) lies in the factthat they include—as first within their ranks—the most important state ofthe hydrogen atom: its ground state.1 Thus, by concentrating on sphericallysymmetric solutions, we can readily obtain the quantum mechanical pre-diction for the main features—for example, size and ionization energy—ofthe most elementary atom. Nevertheless, the search for solutions withangular dependence is equally necessary and will be dealt with in the nextchapter.

1 In effect, we are invoking here a well-known (and physically plausible) general theorem that theground state of any quantum system (with a finite number of particles) has the full symmetry of thesystem.

8.2 Solving the Schrödinger Equation for the Spherically Symmetric Eigenfunctions 209

8.2 Solving the Schrödinger Equation for theSpherically Symmetric Eigenfunctions

For 𝜓 = 𝜓(r), the action of the angular part of the Laplacian (8.4) on the wave-function 𝜓 yields clearly zero. We thus have

∇2𝜓 =(

1r𝜕2

𝜕r2 r)

𝜓 ≡ 1r(r𝜓)′′

and the Schrödigner equation for the hydrogen atom takes now the much simplerform

1r(r𝜓)′′ + 2m

ℏ2

(

E + e2

r

)

𝜓 = 0, (8.5)

which is an ordinary differential equation for the unknown function𝜓(r). We canfurther simplify (8.5) by multiplying both sides by r, so that the term r𝜓 appearson both terms

(r𝜓)′′ + 2mℏ2

(

E + e2

r

)

(r𝜓) = 0. (8.6)

By changing the dependent variable through the relation

y = r𝜓, (8.7)

Eq. (8.6) becomes

y′′ + 2mℏ2

(

E + e2

r

)

y = 0, (8.8)

which has exactly the form

y′′ + 2mℏ2 (E − V (r))y = 0

of a one-dimensional Schrödinger equation, where the variable is now r insteadof x, the wavefunction is y(r), and the potential is the Coulomb potentialV (r) = −e2∕r. This situation differs from the usual one-dimensional Schrödingerequation in that the variable r now takes values in the semi-infinite interval

0 < r <∞

and the boundary conditions to be satisfied at both ends are

y(0) = 0, y(∞) = 0.

The first boundary condition follows directly from (8.7), while the second oneexpresses the fact that we are seeking the bound states of the electron, that is,states for which the wavefunction must vanish at infinity. The radial wavefunc-tion y will also then vanish at infinity, since 𝜓 decays exponentially fast and anexponential decay dominates over the linear growth of the factor r in (8.7).

From here on we can proceed to solve the Schrödinger equation by followingthe standard four steps we also took for the harmonic oscillator. But we now add afifth step, namely, we will provide experimentally testable numerical predictionsstemming from the solution. In Section 8.3 we discuss the results in detail.


STEP 1: Dimensional simplification.As in the case of the harmonic oscillator, the Schrödinger equation for this prob-lem contains only three parameters—ℏ,m, and e—which we can always set tounity by a suitable choice of the three basic units L,M, and T . If, for example, wetake as our unit of mass the electron mass, then this parameter will no longer beequal to 0.91 × 10−27 g, as in the cgs system of units, but equal to unity. We canthus always set

ℏ = m = e = 1,

and write the radial Schrödinger equation in the dimensionless form

y′′ +(

2E + 2r

)

y = 0. (8.9)

STEP 2: Removing the asymptotic behavior at infinity.To ensure that the solution vanishes at +∞, we study (8.9) in that limit (r → ∞),where it takes the form

y′′∞ + 2Ey∞ = 0. (8.10)

Here, the term 2E must be negative, since we are interested in bound states whoseenergies are always lower than the value of the potential at infinity (Figure 8.2).

The fact that the energy must be negative for bound states to exist is also directlyevident from Eq. (8.10), since it is only then that the solutions take an exponentialform and thus may decay at ∞. Otherwise (i.e., for 2E = k2 > 0), the solutions aresinusoidal—sin kr or cos kr—and extend throughout space without decay. If wenow set

2E = −𝛾2,

Eq. (8.10) takes the form

y′′∞ − 𝛾2y∞ = 0

and has the two solutions

y∞ = e±𝛾r.

V

E < 0

V = –re2

r

Figure 8.2 The Coulomb potential.Bound states exist only when thetotal energy of the electron liesbelow the limiting value of theCoulomb potential at infinity. Inother words, for bound states, thetotal energy must be negative.


Of these two solutions, we retain only the decaying one and write the full radialwavefunction2 y(r) as

y(r) = e−𝛾rF(r), (8.11)

hoping that the new unknown function F(r) has a polynomial form for the rea-sons we presented in Chapter 6. Upon substituting (8.11) into (8.9), and recallingthat 2E = −𝛾2, we obtain the new equation for F ,

F ′′ − 2𝛾F ′ + 2r

F = 0, (8.12)

which we will now examine for polynomial solutions.

STEP 3: Search for polynomial solutions.According to the relevant necessary condition we discussed in Chapter 6, if (8.12)is satisfied by a polynomial of degree n, then, for large r, it must surely be satisfiedby the maximum power rn of that polynomial, because the rn term dominates inthat limit. We thus introduce the term rn into (8.12) and require that the equationbe satisfied for large r. We then find

��n(n − 1)rn−2 − 2𝛾nrn−1 + 2r

rn = 0

⇒ (−2𝛾n + 2)rn−1 = 0 ⇒ 𝛾 = 1n, (8.13)

where we dropped the power rn−2, since, for large r, it is negligible compared torn−1, which appears in the other two terms.

Equation (8.13), combined with 2E = −𝛾2, yields

En = − 12n2 , n = 1, 2,… , (8.14)

which are the allowed energy values for the spherically symmetric eigenstates ofthe hydrogen atom.

This is a fundamental physical result. Before we examine it in detail, we willcomplete the mathematical study of the polynomial solutions. To begin with, wenote that, as is evident from Eqs (8.13) and (8.14), the quantum number n cannotbe zero. So its starting value is n = 1, which means that only polynomial solutionsof degree equal to or greater than one are allowed.

Actually, the same conclusion follows independently from (8.11) in combi-nation with the boundary condition y(0) = 0, which also implies that F(0) = 0.Therefore, the polynomial F(r) must have the form

F(r) = a1r + a2r2 + · · · + anrn,

so its first term is not the constant a0, but a linear term in r. It follows, then, thatthe maximum power of that polynomial ought to be equal to or greater than 1.

2 We use the term “radial wavefunction” for the auxiliary function y = r𝜓 , since, as we will shortlysee, it has all the properties of a genuine one-dimensional wavefunction in the semi-infinite range0 < r < ∞: It vanishes at both ends and its square gives the probability per unit of radial distance tofind the electron at “position” r. Conversely, 𝜓(r) is the actual three-dimensional wavefunction ofthe electron; its square gives the probability per unit volume to find the electron near a point at adistance r from the origin.


To confirm the existence of polynomial solutions, we will now constructexplicitly the first two of these, and use them to obtain the correspondingeigenfunctions 𝜓1 and 𝜓2.(a) n = 1: Ground state

In this case, the polynomial Fn = F1 has a degree equal to 1—recall that n is thedegree of the polynomial—and begins with the first power of r. That is,

F1(r) = r

and since y(r) = e−𝛾rF(r), y(r) = r𝜓(r) and 𝛾 = 1∕n, we successively obtain

y1(r) = re−r

and

𝜓1(r) = Ne−r.

In the last line we included the normalization factor N because our equationsare homogeneous and, therefore, their solutions can always be multiplied by anynumber.(b) n = 2: First excited state

The polynomial F2 has a degree equal to 2 now. Since its first term is the linearone, it takes the form

F2(r) = r + ar2, (8.15)

where we set the coefficient of r equal to unity for the well-known reasonswe explained before. If we now insert (8.15) into (8.12)—with 𝛾 = (1∕n)|n=2 =1∕2—we obtain

2a − (1 + 2ar) + 2r(r + ar2) = 0

⇒ (2a − 1 + 2) + (−2a + 2a)r = 0 ⇒ a = −1∕2

⇒ F2(r) = r − 12

r2.

Consequently,

y2(r) = e−r∕2(

r − 12

r2)

and

𝜓2(r) = N(

1 − r2

)

e−r∕2.

Note here that the negative value of the coefficient a we found earlier (a = −1∕2)is no accident: Had a been positive, the polynomial F2(r) would have no zeroesin the physically acceptable range of r (r > 0) and thus the corresponding wave-function 𝜓2 would not have the one node required by the relevant theorem andthe physical arguments associated with it. (What are these arguments?)

Moreover, the normalization coefficients in solutions 𝜓1 and 𝜓2 are easilycalculated once we realize that, for a spherically symmetric wavefunction𝜓 = 𝜓(r), the volume integral of the probability density |𝜓(r)|2 can be calculated


using spherical shells as volume elements—that is, dV = 4𝜋r2dr (= the volumeof a spherical shell extending from r to r + dr)—so that

∫ |𝜓(r)|2dV = 4𝜋 ∫∞

0|𝜓(r)|2r2dr.

The integrals that arise in these expressions are all of the form ∫ ∞0 rne−𝜆rdr and

are readily calculated by performing n successive differentiations, with respect to𝜆, of the “source” integral

∫∞

0e−𝜆r dr = 1

𝜆,

which yields

∫∞

0rne−𝜆rdr = n!

𝜆n+1 .

Upon performing the integration, the normalization coefficients for 𝜓1 and 𝜓2turn out to be 1∕

√𝜋 and 1∕

√8𝜋, respectively. The normalized eigenfunctions

are then

𝜓1(r) =1

√𝜋

e−r

and

𝜓2(r) =1

2√

2𝜋

(

1 − r2

)

e−r∕2.

In Table 8.1, we list the first three normalized eigenfunctions with sphericalsymmetry.

STEP 4: Restoring dimensions.As in the harmonic oscillator, we can restore the usual units for the energy eigen-values via the substitution

En → 𝜖En, (8.16)

where 𝜖 is the characteristic energy unit for the problem, that is, the unique com-bination of ℏ,m, and e—the parameters we set equal to unity—with dimensionsof energy.

Table 8.1 The first three normalized, spherically symmetriceigenfunctions of the hydrogen atom (in a.u.).

𝜓1(r) =1

√𝜋

e−r

...............................................................................................................𝜓2(r) =

14√

2𝜋(2 − r) e−r∕2

...............................................................................................................𝜓3(r) =

181

√3𝜋

(27 − 18r + 2r2) e−r∕3


Restoring the usual units in eigenfunctions will be slightly different than in thesimilar case for the harmonic oscillator: Since we now have a three-dimensionalproblem, the wavefunctions must have a physical dimension L−3∕2, so that theirsquare gives the probability per unit volume. Therefore, if a is the characteristiclength of the problem, we can restore the dimensions in the wavefunctions viathe substitution

𝜓n(r) −−−−→ a−3∕2𝜓n(r∕a). (8.17)

To calculate a, we equate the two quantities having the same dimensions, namely,e2∕a (electrostatic energy) and ℏ2∕ma2 (kinetic energy due to the uncertaintyprinciple), and obtain

e2

a= ℏ2

ma2 ⇒ a = a0 = ℏ2

me2 , (8.18)

which is identical to the Bohr radius of the old quantum theory, as it should, sincethis is the sole combination of ℏ,m, and e with dimensions of length. By insertinga of (8.18) into either e2∕a or ℏ2∕ma2(since both have dimensions of energy), weobtain the characteristic energy of the problem

𝜖 = me4

ℏ2

so that the substitution (8.16) in the energy eigenvalues (8.14) yields

En = −me4

2ℏ21n2 , n = 1, 2,… ,

which is the exact same expression of Bohr’s theory for the allowed energy levelsof the hydrogen atom! This is more of a remarkable coincidence than an antici-pated one: Since Bohr’s theory is regarded today as a semiclassical quantizationcondition, we expect it to be accurate only in the region of relatively large quan-tum numbers. Schrödinger’s equation, on the other hand, is expected to alwaysprovide accurate predictions, if quantum mechanics is indeed the correct theoryof nature. Therefore, the fact that the quantum mechanical prediction for theallowed energies of the hydrogen atom coincides with the prediction of Bohr’stheory is more of a coincidence than an expected physical fact. Besides, as wewill see shortly, there are other predictions of Bohr’s theory for the hydrogen atomthat are erroneous and differ from their quantum mechanical counterparts.

To restore the usual units for the wavefunctions, the substitution (8.17) for theground state of the atom yields (see Table 8.1)

𝜓1(r) =1

a0√𝜋a0

e−r∕a0 , (8.19)

where a0 = ℏ2∕me2 is the Bohr radius.

STEP 5: Numerical predictions.Given the values of ℏ,m, and e,

ℏ = 1.056 × 10−27 erg s, m = me = 0.91 × 10−27 g, e = 4.802 × 10−10 esu,


all we have to do is insert them in the expressions for the characteristic units oflength (ℏ2∕me2) and energy (me4∕ℏ2) and see what we get. Actually, this calcu-lation is unnecessary, since these are the same expressions appearing in Bohr’stheory and whose numerical values are

a0 = ℏ2

me2 = 0.529 × 10−8 cm ≈ 0.5 Å

and

𝜖 = me4

ℏ2 = 0.436 × 10−10 erg = 27.2 eV.

With regard to the allowed energies of the spherically symmetric states of theatom, the exact quantum mechanical prediction is thus,

En = −13.6n2 eV (n = 1,… ,∞),

from which we can deduce a key property of the atom, namely, the ionizationenergy, to be

WI = −E1 = 13.6 eV,

in excellent agreement with experiment.As for the atomic size, we first ought to clarify the exact meaning of this con-

cept. In Bohr’s theory its meaning is unambiguous: The atomic radius is the radiusof the first Bohr orbit along which the electron is moving in the ground stateof the atom. In quantum mechanics, the “motion” of the electron in the groundstate is described by the wavefunction (8.19), which corresponds to a “probabilitycloud” that decays rapidly as we go away from the nucleus, but which, never-theless, extends to infinity, in principle. How are we to define the atomic size inthis case? Clearly, the definition is a matter of convention. For example, we coulddefine as atomic radius the distance (from the nucleus) within which the proba-bility of finding the electron is, say, 90%. Instead of this definition, it is customaryto define the atomic radius in the same manner we define the characteristic lengthor the characteristic time for any phenomenon governed by a law of exponentialdecay. A representative example is the law of discharge of a capacitor through aresistor, which is described by the relation

q(t) = q0e−t∕RC = q0e−t∕𝜏 (𝜏 = RC),

where 𝜏 = RC is the characteristic time of discharge, that is, the time we need towait until the charge in the capacitor has been reduced to 1∕e (= 1∕2.7) of itsinitial value. In practice, we often use the concept of characteristic time to referto the time we need to wait until the phenomenon “disappears,” even though weactually need to wait two or three times as long until the quantity we are inter-ested in can be considered to be practically zero.

In the same spirit, we define the atomic size to be the characteristic lengthof the exponential law (8.19), that is, the distance from the nucleus at whichthe electron’s wavefunction has dropped to 1∕e of its “initial” value. With this


definition, the size of the hydrogen atom in quantum mechanics—the so-calledatomic radius—is equal to

a = a0 = ℏ2

me2 = 0.529 ≈ 0.5 Å,

which is the Bohr radius of the old quantum theory. But we stress once againthat the notion of the “size of the atom” is very different in the two theories. And,needless to say, elaborate measurements of the electronic cloud about the nucleusfully confirm the quantum mechanical prediction.

8.2.1 A Final Comment: The System of Atomic Units

We shall close this section with a few remarks on the broader use of the system ofunits (ℏ = m = e = 1) we adopted in this chapter, which is widely used in atomicphysics and is abbreviated as

a.u. (≡ atomic units).

In this system, the unit of measurement of any physical quantity is the uniquecombination of ℏ,m, and e with dimensions of that physical quantity (and with anumerical coefficient of unity). Accordingly, the quantities

a0 = ℏ2

me2 = 0.5 Å and 𝜖 = me4

ℏ2 = 27.2 eV

are just the atomic units of length and energy. As for notation, the customaryconvention is to accompany our results with the letters “a.u.,” so as to clarify thatwe are talking about their numerical value in the atomic system, not about plainnumbers. For example, when we calculate the mean distance of the electron fromthe nucleus—using the “dimensionless” wavefunction 𝜓1(r) = e−r∕

√𝜋—and

obtain the result ⟨r⟩ = 3∕2, it is more appropriate to write

⟨r⟩ = 32

a.u.

or

⟨r⟩ = 32

a.u. ≡ 1.5 a0,

since we know that the atomic unit for length is the Bohr radius a0. In a similarfashion, the result E1 = −1∕2 for the energy of the ground state of the atom shouldbe actually written as

E1 = −12

a.u. ≡ −12⋅ 27.2 eV = −13.6 eV,

where we utilized the fact that the atomic unit of energy is 27.2 eV, also called aHartree.3 In other words,

1 a.u. of energy ≡ 1 hartree = 27.2 eV,

so we could say, for example, that the ionization energy of the hydrogen atom ishalf a Hartree.

3 Douglas Hartree is well known for his various contributions in the development of atomicphysics, one of which is the computational method that bears his name.


Problems

8.1 Find the third-degree polynomial solution of Eq. (8.12) and verify that thewavefunction 𝜓3(r) given in Table 8.1 is correct (including normalization).

8.2 Calculate (in eV) the mean values of the potential and kinetic energies ofthe electron in the ground state of the hydrogen atom.

8.3 A hydrogen atom collides with another atom. Immediately after thecollision, its state is given by the wavefunction

𝜓(r) = Ne−𝜆r (a.u.). (1)

Calculate, as a function of 𝜆, the probability that a measurement of theenergy in state (1) yields the value E1 = −1∕2 of the ground state of the atom.What would the electronic wavefunction be after such a measurement?

8.4 Calculate the mean distance of the electron from the nucleus in the secondexcited state of the hydrogen atom described by the wavefunction 𝜓2(r) ofTable 8.1.

8.3 Discussion of the Results

8.3.1 Checking the Classical Limit ℏ→ 0 or m → ∞ for the Ground Stateof the Hydrogen Atom

To begin analyzing our results, let us check whether the formulas

a0 = ℏ2

me2 = size of atom in its ground state

and

E1 = −me4

2ℏ2 = ground state energy

behave correctly in the classical limit. This limit requires that the electron fallsat the bottom of the potential—in this case, it falls onto the nucleus (r = 0)—andstays there. So, for ℏ → 0 or m → ∞, we must have

a0 → 0, E1 → −∞

because when the electron falls onto the nucleus, the size of the atom will actuallyvanish and its energy will be minus infinity. Clearly, these formulas for a0 and E1satisfy this requirement.

8.3.2 Energy Quantization and Atomic Stability

Having identified from the outset the stability of atoms—especially with regard tocollisions—as the central mystery of the atomic world, let us see whether we canadd anything to that discussion now that the solution of the hydrogen atom—orat least, a part of it—is known to us. In fact, we do not have much to add to


our discussion in Chapter 1 regarding the fundamental role quantization playsin guaranteeing atomic stability. Since the energy of the electron is quantized,the atom is necessarily stable in its ground state, because the electron can moveneither “down” to a state of lower energy (there is none), nor “up” to higherenergy, due to the large energy gap between the ground and the first excited state.Let us recall here that the energies exchanged during atomic collisions at roomtemperature (T ≈ 300 K) are on the order of

kT|T≈300 K ≈ 140

eV,

whereas the energy difference between the ground state and the first excitedstate—the so-called minimum excitation energy4—is

ΔE = E2 − E1 = (−3.4 eV) − (−13.6 eV) = 10.2 eV,

which is much greater. Therefore, thermal collisions are not able to excite elec-trons and the atoms emerge completely unchanged by them.

We should add a disclaimer to the preceding discussion. Given that the allowedenergies we found so far (En = −13.6 eV∕n2) correspond only to spherically sym-metric solutions, we are not in a position yet to know for sure whether some solu-tions with angular dependence have energy eigenvalues closer to the ground-stateenergy than the state with n = 2.5 If there are such eigenvalues, then the minimumexcitation energy will be smaller than what we previously calculated. Actually,as we show in the following chapter, there is no such possibility. The formulaEn = −13.6 eV∕n2 gives the complete set of allowed energies of the atom. Thesolutions with angular dependence are simply added to the other solutions, sothat from n = 2 onward there are more than one eigenstates with the same energy.Now that the energy spectrum for the hydrogen atom is fully known, we candepict it in Figure 8.3.

8.3.3 The Size of the Atom and the Uncertainty Principle: The Mysteryof Atomic Stability from Another Perspective

Following a related discussion in Chapter 3 (and also in Chapter 1, based on thewave–particle duality), we now recognize the mechanism responsible for the sizeof the atom. When the atom is in its ground state, it “chooses” its size so as to min-imize its total energy. The way this works is shown in Figure 8.4, where shadedareas depict regions of space about the nucleus where the wavefunction has suf-ficiently large values and where there is a high probability of finding the electron.A more accurate depiction would, of course, require gradual shading that weak-ens as we go away from the origin up to two or three atomic radii. In this way, wecould represent graphically the fact that the probability of locating the electron

4 It is also called the first excitation energy, since there are also excitations from the ground state toeven higher states (n = 3, 4,…).5 We take it as granted here that none of the solutions with angular dependence can have energylower than E1 = −13.6 eV and be thus the ground state of the atom. One reason for this is that theground state always has the full symmetry of the problem, so, in this case, it must be sphericallysymmetric. Another reason is that the solution 𝜓1 we have obtained has no nodes and, therefore, issurely the state with the lowest possible energy.


n = 4n = 3

E4 = – 0.85 eVE3 = – 1.5 eV

E2 = – 3.4 eV

E1 = – 13.6 eV

n = 2

n = 1

Min

imum

exc

itatio

n en

ergy

Ioni

zatio

n en

ergy

ΔE

= 1

0.2

eV

WI =

13.

6 eV

Figure 8.3 Energy-level diagram for the hydrogen atom. The diagram shows the energy levelsof the hydrogen atom and the two basic quantities derived from it: the minimum excitationenergy ΔE and the ionization energy WI . The atom in its ground state is stable against thermalcollisions, since ΔE ≫ (kT)room: It can move neither down in energy (there is no available state)nor up (due to the large energy gap separating the ground state from the first excited state).

(a) (b) (c)

Figure 8.4 How the electron chooses its ground state in the hydrogen atom. Scenario (a): To“capitalize” on the attraction from the nucleus—that is, to minimize its potential energy—theelectron “forms” a very localized wavefunction, which, however, results in a large increase of itskinetic energy due to the uncertainty principle. Scenario (b): To avoid being highly localized,the electron “blows up” its wavefunction and thus lowers its kinetic energy. But it thenincreases its potential energy by being far away from the nucleus. Scenario (c): The atom finallyachieves the state of minimum total energy by balancing the competing demands of thekinetic and potential energy terms. In this balanced state, its wavefunction is neither too“tight” nor too “loose”: it has the optimum size. As we see, in choosing its ground state, theelectron obeys a kind of “Goldilocks principle”: Its wavefunction is at first too localized, thentoo extended, and finally, about right.


extends throughout all space, albeit in a decaying manner. But for convenience,the shaded areas in Figure 8.4 extend only up to the distance from the nucleusthat we consider to be practically the atomic radius.

The mechanism of searching for the minimum energy, shown qualitativelyin Figure 8.4, can be quantified as follows: If a is the atomic size—that is, theradius of the shaded region that graphically describes its wavefunction—then theelectronic kinetic energy due to the uncertainty principle will be approximatelyequal to

K ≈ ℏ2

2ma2 ,

while its potential energy will be

V ≈ −e2

a. (8.20)

Here, we took into account the fact that the electron’s wavefunction extends alsoto distances greater than a, so we consider (8.20) as a reasonable estimate of itspotential energy. The total energy of the electron is then

E(a) ≈ ℏ2

2ma2 − e2

a

and its minimum value is achieved when

dEda

= 0 ⇒ − ℏ2

ma3 + e2

a2 = 0 ⇒ a = a0 = ℏ2

me2 ,

that is, when the size of the atom is equal to the Bohr radius. We convey all thisinformation in Figure 8.5.

E(a)

E(a) ≈ ℏ2

2ma2 a

aa0

e2

Figure 8.5 Total energy of the hydrogen atom as a function of its possible size. Becomingeither too small or too large is energetically costly for the atom. When it gets too small, itlowers its potential energy, but the kinetic energy increases dramatically; when it gets toolarge, it lowers its kinetic energy, but its potential energy increases rapidly (it becomes lessnegative). The total energy is minimized when the atomic radius becomes equal to the Bohrradius.


8.3.4 Atomic Incompressibility and the Uncertainty Principle

There is another basic property of atoms that is also directly related to theirstability: their incompressibility. It is the most evident feature of macroscopicmatter in solid or liquid form. Although it consists of atoms that are literallyhollow—recall that the nuclear radius is barely 10−13cm, while the atomic radiusis 10−8cm!—macroscopic matter has a tremendous resilience to external pres-sures. It behaves as if its basic ingredients—atoms or molecules—are completelycompact and incompressible.

It turns out that the uncertainty principle readily explains atomic incompressi-bility as well. If we attempt to “compress” the atom, its electron will be “squeezed”into less space and its kinetic energy will increase due to the uncertainty principle.The readers can easily show (see Problem 8.5) that to compress the atom to halfits size one should exert pressure on the order of a million atmospheres!

8.3.5 More on the Ground State of the Atom. Mean and Most ProbableDistance of the Electron from the Nucleus

Given the wavefunction of the ground state of the electron, we can readily findits average distance from the nucleus, by using the mean-value formula

⟨r⟩ = ∫ 𝜓∗(r𝜓) dV = ∫∞

0r|𝜓(r)|2 ⋅ 4𝜋r2 dr

= ∫∞

0r

(

1√𝜋

e−r

)2

4𝜋r2 dr

= 4∫∞

0r3e−2r dr = 4 ⋅

3!24 = 3

2a.u. ≡ 3

2a0.

We have thus found that the mean distance of the electron from the nucleus inthe hydrogen atom is 1.5 Bohr radii.

But the most probable distance is different, and can be obtained from the radialprobability distribution

(r) = |𝜓(r)|24𝜋r2 = 4r2e−2r, (8.21)

which gives the probability, per unit radial length, to locate the electron at a dis-tance between r and r + dr from the nucleus (i.e., inside a spherical shell rangingfrom r and r + dr). In (8.21), the term |𝜓(r)|2 describes as usual the probabilityper unit volume, while the term 4𝜋r2 results from having divided the volume ofthe spherical shell 4𝜋r2dr by the infinitesimal radial length dr in order to obtainthe probability per unit length.

As is evident from the plot of (8.21) shown in Figure 8.6, there is only onemaximum for (r). It can be readily found by requiring that

ddr

= 0 ⇒ 8re−2r − 8r2e−2r = 8r(1 − r)e−2r = 0

⇒ r = 1 ≡ 1 a.u. ≡ a0.

Thus, the most probable distance from the nucleus is one Bohr radius. This isexactly the location where the electron of the Bohr theory was supposed to be!


(r) = 4r2 e – 2r

(r)

r = 1 a.u. ≡ a0

r

Figure 8.6 Radial probabilitydensity in the ground state ofthe hydrogen atom. The mostprobable distance of theelectron from the nucleus isr = 1 ≡ a0, which is equal tothe radius of the first Bohr orbitin the old quantum theory.

8.3.6 Revisiting the Notion of “Atomic Radius”: How Probable Is It to Findthe Electron Within the “Volume” that the Atom Supposedly Occupies?

We saw earlier how to define the “atomic radius” when the wavefunction decaysexponentially. In atomic units, this radius is r = 1; in standard units, it is oneBohr radius. In this context, we can view the atom as a sphere of radius r = 1 (ina.u.) inside which we expect the electron to lie most of the time. Let us calculatethe corresponding probability, assuming for now that the atomic radius is a freevariable R. Then

P(R) ≡ P[r ≤ R] = ∫R

0|𝜓(r)|24𝜋r2 dr = 4∫

R

0r2e−2r dr

⇒ P(R) = 1 − (1 + 2R + 2R2)e−2R, (8.22)

where we performed two integrations by parts.6 Note that the result (8.22) satis-fies the basic test P(∞) = 1, while for R = 1 it yields

P(1) = 1 − 5e2 ≈ 32.3%,

which is an extraordinary result. It tells us that inside the supposed volume of theatom—a sphere equal to the Bohr radius—the electron spends a mere 32.3% of itstime! All the rest of the time—that is, 67.7%—the electron spends it outside theatom! The conclusion one can draw here is clear. In reality, the atom is much largerthan what the conventional definition of the atomic radius indicates. Table 8.2fully supports this conclusion, as it shows the values of the function P(R) for Rranging from 1 to 10. It is evident from Table 8.2 that we ought to “increase” thesize of the atom up to three Bohr radii for the probability of finding the electron“inside” the atom to surpass 90%. (The probability is exactly 90%when the atomicradius is 2.66 a.u. = 2.66 a0.)

In hindsight, we should not be surprised by how low the probability P(1) is.This has to do with the increased “weight” the three-dimensional volume elementdV = 4𝜋r2dr assigns to the greater values of r, thus weakening considerably

6 We can also calculate an indefinite integral of the general form ∫ rne−𝜆rdr by successivedifferentiations, with respect to 𝜆, of the “source” integral ∫ e−𝜆rdr = − 1

𝜆e−𝜆r , just as we did for the

corresponding definite integral.


Table 8.2 Probability P(R) of finding the electron inside a sphere of radius R aboutthe nucleus, when it occupies the ground state of the hydrogen atom.

Radius (R) Probability (%) Radius (R) Probability (%)

1 32.3 6 99.942 76.2 7 99.993 93.8 8 99.9984 98.6 9 99.99975 99.7 10 ∼ 100

the exponential decay of the wavefunction. It is the same basic mechanism thatexplains why the maximum of the radial probability density lies far from theorigin. But this is a topic deserving some further attention.

8.3.7 An Apparent Paradox: After All, Where Is It Most Likely to Findthe Electron? Near the Nucleus or One Bohr Radius Away from It?

First, what is the paradox? For the wavefunction 𝜓1(r) = e−r∕√𝜋, the probability

density of finding the electron somewhere is

P = |𝜓1(r)|2 = 1𝜋

e−2r, (8.23)

which decays exponentially as we go away from the origin. Clearly, then, themost probable place to find the electron is in the immediate vicinity of thenucleus. But this statement “obviously” contradicts our earlier assertion, thatthe most probable distance of the electron from the nucleus is one Bohr radius.The contradiction is only an apparent one and is due to the different meaningof the expressions (8.21) and (8.23): the former has its maximum at r = 1, thelatter at r = 0. Expression (8.23)—the square of the wavefunction—gives us theprobability per unit volume; while (8.21) is the probability per unit radial length(≡ radial probability density) and, as such, it accounts for all “sightings” of theelectron inside the spherical shell ranging from r to r + dr. Since the volume4𝜋r2dr of the shell increases as r2 as we go away from the origin, this increasecompensates for the decrease of the probability per unit volume and shifts themaximum value away from the origin (Figure 8.7).

8.3.8 What Fraction of Its Time Does the Electron Spend in the ClassicallyForbidden Region of the Atom?

In classical mechanics, the electron in the hydrogen atom can venture away fromthe nucleus up to a distance rc (“c” for “classical”) given by the well-known relation

V (r) = E ⇒ −1r= − 1

2n2 ⇒ r = rc = 2n2 a.u.

For example, for the ground state of the atom (n = 1) we have rc = 2 a.u. Butin quantum mechanics, the electron can be found outside that radius, and the


dr

dV = 4πr2 dr

Figure 8.7 Resolution of a paradox. Even though the probability per unit volume to find theelectron is greater near the nucleus (the “dots” get much denser there), the most probabledistance is actually away from the origin (at r = 1) because the radial probability is beingmaximized there. All the dots that lie inside the volume dV = 4𝜋r2dr of a spherical shellextending from r to r + dr contribute to the value of the radial probability. Since the volume ofthis shell increases with r, it overcompensates, up to some value r, the loss of probability perunit volume.

Table 8.3 Probability of finding the electron in the classicallyforbidden region for various eigenstates of the hydrogen atom.

State Probability (%) State Probability (%)

n = 1 23.8 n = 6 12.4n = 2 18.6 n = 7 11.7n = 3 16.0 n = 8 11.2n = 4 14.4 n = 9 10.7n = 5 13.2 n = 10 10.3

probability for this to happen—that is, for the electron to be in the classicallyforbidden region—is (see Table 8.2) equal to

P[r ≥ 2] = 1 − P[r ≤ 2] = 1 − P(2)= 100% − 76.2% = 23.8%.

Thus, in the ground state of the atom, the electron spends 23.8% of its time in aregion where it is classically forbidden to be. For the excited states of the atom,the respective probabilities are given in Table 8.3.

In complete analogy to what we found for the harmonic oscillator (Table 6.2),the probability for the electron to lie in the classically forbidden region decreaseswith n (as it should), but this decrease occurs at a much slower pace thanexpected. Apparently, this intrusion into classically forbidden regions “dies hard”as we move toward the classical limit (large n) and the classical order is graduallyrestored.


8.3.9 Is the Bohr Theory for the Hydrogen Atom Really Wrong? Comparisonwith Quantum Mechanics

As we noted earlier, both the theory of Bohr and modern quantum mechanicspredict exactly the same energy eigenvalues—the same energy spectrum—for thehydrogen atom. This means that we could not distinguish the two theories byspectroscopic measurements alone. How can we then tell which theory is cor-rect? One could think that in one theory the electron moves along a specific orbitin space—a circle of radius a0—whereas in the other, the position of the elec-tron is described by a statistical distribution with a considerable chance to find it“outside” the immediate vicinity of that circular orbit.

Therefore, any observation that would detect the electron outside an “orbitaltube” about the Bohr radius would automatically disprove Bohr’s theory. In prac-tice, it is much easier to measure not the position of a particle but rather itsenergy or its angular momentum, largely because these are conserved quantities.Let us see, then, whether the angular momentum of the electron can offer usthe experimental basis to prove one or the other theory wrong. We recall thatin Bohr’s theory the angular momentum of the ground state is 𝓁 = nℏ|n=1 = ℏ,whereas in modern quantum mechanics the corresponding prediction is 𝓁 = 0,as we will show right away. We are going to prove that the ground state of thehydrogen atom—and more generally, every spherically symmetric wavefunction𝜓(r)—is an eigenstate of the angular momentum operators with an eigenvalueequal to zero. This implies that the following equations hold,

𝓁x𝜓(r) = 0, 𝓁y𝜓(r) = 0, 𝓁z𝜓(r) = 0.

Proof : Due to symmetry, it suffices to prove any one of these relations, say, thefirst one. We have

𝓁x 𝜓(r) = −iℏ(

y 𝜕𝜕z

− z 𝜕𝜕y

)

𝜓(r) = −iℏ(

y𝜕𝜓𝜕z

− z𝜕𝜓𝜕y

)

, (8.24)

which for 𝜓 = 𝜓(r) becomes

𝜕𝜓

𝜕z= d𝜓

dr𝜕r𝜕z

= 𝜓 ′(r) zr,

𝜕𝜓

𝜕y= d𝜓

dr𝜕r𝜕y

= 𝜓 ′(r)yr,

so that Eq. (8.24) may be written as

𝓁x𝜓(r) = −iℏ(

y𝜓 ′(r)zr− z𝜓 ′(r)

yr

)

= −iℏ 𝜓 ′(r)r

(yz − zy) = 0,

which is what we wanted to show. Since the ground state of the hydrogenatom is spherically symmetric, it is a state where the electron has zero angularmomentum. □

The disparity between modern quantum mechanics and Bohr’s theory is nowclear and unambiguous and can therefore be tested experimentally. A pertinentexperimental test is discussed in Chapter 10, where it is also made absolutely clearwhy Bohr’s theory does not correspond to reality. It is a wrong theory.


Problems

8.5 As we mentioned before, the uncertainty principle makes atoms so incom-pressible that it would take an external pressure on the order of millions ofatmospheres to reduce by one-half the radius of an atom such as hydrogen.Prove this assertion with a simple order-of-magnitude calculation.Hint: Recall the relation p dV = −d𝑈 and use a rough estimate for thevolume decrease dV of the atom and the corresponding increase d𝑈 in theelectron’s energy. What would you say for a choice |dV | ∼ Å3 and |d𝑈 | ≈ afew eV?

8.6 Find the most probable distance (or distances) from the nucleus of anelectron in the n = 2 spherically symmetric state of the hydrogen atom.

8.4 What Is the Electron Doing in the Hydrogen Atomafter All? A First Discussion on the Basic Questions ofQuantum Mechanics7

We conclude this first study of the hydrogen atom by posing two questions thatmay have occurred to some readers. We begin with the first question:

Question 1: What is the electron actually doing in the ground state of the hydrogenatom? Or, to put it differently: What does the hydrogen atom look like in its groundstate?

Perhaps this question becomes clearer if we pose it first for a classical planetarysystem, say, our own. If we were asked by a friend “How does our solar systemlook like?” we would probably answer something like this: It consists of a sun(at its center), then at such a distance from the sun lies a planet (Mercury) thatmoves along an elliptical orbit and which right now is at such a position, but afterthis amount of time it would be at that position, and so on, for each planet inthe solar system. In the same manner we would answer the question “What is acertain planet doing?” We would say that it moves along an elliptical orbit thathas these features, and that right now it is to be found at this point in its orbitmoving at that speed, but after this amount of time it would be at that positionmoving at that speed, and so on. And if our friend asked us how she could checkour answers, we would reply “Just look through a telescope!” These are the sortof “automatic” answers we would provide to each question like “what does a cer-tain physical system look like?” or “what is a component of that physical systemdoing?” We would describe the current positions and velocities of the “particles”comprising the system, and their trajectories, as these are obtained from solv-ing the classical equations of motion. The promptness of our response “just lookthrough a telescope!” reflects the most self-evident assumption across all classical

7 The “practically minded” readers could skip reading this section. Nonetheless, a casual readingmight be useful.

8.4 What Is the Electron Doing in the Hydrogen Atom after All? 227

physics: That a physical system exists independently of the observer and the act ofobserving it simply confirms this independent existence. That is to say, the systemcan be described independently of the process of observing it.

Let us now revisit Question 1, in view of the preceding discussion. The firstthing to realize is that the answer, now, is by no means obvious. Since the notionof the orbit in the atom is no longer meaningful—it is not allowed by the uncer-tainty principle—what is there to say about “how the electron behaves” inside theatom? How would we describe to someone what the hydrogen atom “looks like”in its ground state? Let us first try to answer the question for ourselves. Whatdo we think the electron is actually “doing” in the hydrogen atom when nobodyis looking? It is almost certain that if we were to answer spontaneously—that is,without a self-imposed censorship—then the picture in our mind will be roughlythis: The electron swirling around the nucleus on a tangled orbit and at such a“crazy” pace that all we can see is how often it passes through each region ofspace. And this is more or less how we would “explain” the probability distribu-tion that emerges from the wavefunction of the ground state of the atom: Thisdistribution is the statistical outcome of an extremely complicated motion thatwe do not know right now—nor can we observe it experimentally—but which atsome point in the future will become known to us; and at that moment, we will nolonger require all these probabilities to describe the atom. It is clear, however, thatthis “private picture” for the atom—a picture that we all, perhaps, entertain—isirreconcilable with the uncertainty principle. Why? Because the very notion ofthe orbital motion of the electron is experimentally unattainable in the contextof the uncertainty principle. Nobody will ever be able to observe such an orbit,unless, of course, it is proved at some point that the uncertainty principle is notquite correct. This would also directly imply that quantum mechanics is no longera fundamental theory, but, rather, the approximate description of some deeperreality.

But as long as quantum mechanics—and along with it, the uncertaintyprinciple—remains experimentally unshakable, then a “private picture” likethe one described (the electron swirling about the nucleus in a “jumbled-up”orbit) has no theoretical justification whatsoever. It belongs to the realm ofpure metaphysics; to that realm of ideas for which it is in principle impossibleto come up with experimental evidence, for or against. But if we insist thatphysics is an experimental science—as we should!—then the use of pictures ordescriptions that have absolutely no chance of being experimentally verified, hasno meaning at all. Even further: It is not only the pictures or the descriptionsthat have no meaning, but even the very questions that lead us, inescapably, tosuch pictures and descriptions. In physics—as in any experimental science—weare only allowed to ask those questions that can be experimentally answered.Any questions that cannot be tested by experiment should not be acceptedas scientific questions; they ought to be rejected as such. And what does thisdiscussion imply for the issue at hand? (What is the electron doing, etc.)Quite simply, that the very question ought to be rejected! It is not a physicalquestion.

For those readers who are unpleasantly surprised by this answer, let us stressright away—increasing their discomfort even further!—that many conceptual


questions in quantum mechanics are actually answered this way. We “answer”the questions by rejecting them! The rationale for such rejections (assuming it isvalid) is that the experimental verification of any possible answer is theoretically(i.e., in principle) impossible.

The conclusion to our discussion is indeed unpleasant. The question “whatis the electron doing inside the hydrogen atom?”—which is really the mostself-evident question we can pose for the atom—ought to be rejected asunanswerable within the context of quantum mechanics. It is a metaphysicalquestion!

But is there anything at all we can ask about the hydrogen atom that would notbe rejected as pure metaphysics? What is a quantum mechanically viable descrip-tion of the ground state of this atom? The answer lies in the basic principle ofquantum mechanics, namely, that the wavefunction of a physical system providesthe most complete description we can ever have about it. But the wavefunctionis not a description of the system per se; it does not tell us “what the systemitself is doing.” The wavefunction only describes the possible types of behaviorif we were to observe the system one way or another. Let us take, for example,the ground state of the hydrogen atom, which has the well-known wavefunction𝜓1(r) = e−r∕a0∕

√

𝜋a30. All we can say about this system is that, if we try to measure

the electron’s position around the nucleus by a series of measurements on identi-cally prepared atoms, we will obtain a different result in each measurement, butall these results together will have the distribution described by the wavefunc-tion of the electron. Thus, we would obtain a distribution of points (dots) as inFigure 8.7, or a radial distribution as in Figure 8.6.

A general conclusion emerges from this discussion: In quantum mechanics,it is completely nonsensical to speak of a physical system in itself. Instead, wecan only speak about the relation of the system to a device used to observeit. Thus, in complete contrast to classical physics, in quantum mechanicsthe observation process is an integral part of the description of the physicalsystem.We hope that the discussion has shed some light as to how we should approachquestions such as Question 1. Let us now extend our discussion to the keyfeature of quantum mechanics—its probabilistic interpretation—from which allits conceptual “singularities” stem. The question, now, is the following:

Question 2: Given that the hydrogen atom is the simplest possible physicalsystem—one in which all factors influencing its behavior are known to us—whatis the reason for our inability to accurately predict the motion of its electron?What is it that we do not know? And if we actually knew that, could we make adescription of the atom without statistical indeterminacy?

As before, we will try to make better sense of the question by placing it in aclassical context first. Let us note at the outset that to speak in probabilistic termsabout a physical problem is not unusual in classical physics. A typical exampleis the flip of a coin. Our inability to predict the outcome—heads or tails—isknown. All we can say (provided the coin is not biased) is that the two possibleoutcomes are equally probable: 50% heads and 50% tails. But here, our inability

8.4 What Is the Electron Doing in the Hydrogen Atom after All? 229

to predict the outcome has a well-known cause: the imperfect knowledge of thephysical conditions influencing the coin’s motion—for example, the exact initialconditions of the flipping, the variations in the density and pressure of the air,any possible unevenness of the ground, and so on. If all these parameters wereknown to us, we would be able, at least in principle, to write down (and solvewith a supercomputer) the classical equations of motion of the coin, and thus topredict the outcome of a certain flip, given the specific conditions at the time.Our conclusion from this example is general: In classical physics the emergenceof probabilities—that is, uncertainties—is always the result of incompleteknowledge.

Returning now to Question 2, we see at once how paradoxical the emergenceof probabilities is in such an elementary physical system as the hydrogen atom.What is that incomplete knowledge causing statistical behavior in this case?What is it that we do not know about the hydrogen atom and, as a consequenceof this ignorance, we are unable to accurately predict the position of theelectron?

There are two possible answers to this question with regard to the origins ofquantum mechanical probabilities. The first answer—the so-called Copenhageninterpretation—was put forward by Bohr, Born, and Heisenberg, and is the gener-ally accepted interpretation of quantum mechanics. Although it is already knownto us, let us repeat it here.

Answer 1: (Copenhagen interpretation) The quantum mechanical probabilitiesare not the result of incomplete knowledge; they are fundamental. The probabilisticbehavior is an intrinsic feature of nature at the microscopic level.

The second answer represents the “opposition” to the Copenhagen interpretationand is associated with Einstein, de Broglie, and, later on, David Bohm (whoactually was its most systematic proponent). In broad terms, this point of viewcan be phrased as follows:

Answer 2: (The “opposition”) Quantum mechanical probabilities are not funda-mental. Instead, they result from the existence of some hidden variables, which,had they been known, would have uniquely specified the behavior of a quantumsystem. At a deeper level, nature is once again deterministic.

Simply put in Einstein’s own words, “God does not play dice [with the world].”But despite the passionate support of this alternative point of view by its instiga-tors, it has never been possible to formulate it as a complete alternative theorythat could stand up to the “official” quantum mechanics. Nevertheless, the debatekept ongoing until 1964, when John Bell succeeded in formulating a condition, thefamous Bell’s inequalities (Bell, J., 1964. Physics 1, 195), based on which it becamepossible to put the disagreement between the two interpretations to the experi-mental test. The question to be tested was: Are there hidden variables or not? Apertinent experiment was devised and performed in 1982 by Aspect, Dalibard,and Roger,8 and its results—confirmed repeatedly by later experiments—wereunequivocal: There are no hidden variables. Quantum mechanical probability

8 Aspect, A., Dalibard, J., and Roger, G. (1982) Phys. Rev. Lett. 49, 1804.


is not the result of incomplete knowledge. It is a fundamental feature of theworld we live in.

Further Problems

8.7 The electron and its antiparticle, the positron (a particle of equal mass andopposite charge to the electron), form a bound system known as positron-ium. Calculate the following:(a) The dissociation energy of the positronium in its ground state.(b) The mean distance between the electron and the positron in the same

state.

8.8 As we know from elementary electrostatics, when a charged particleapproaches a metallic wall, it is attracted to it due to the image chargeon the other side of the wall. Assuming that the charged particle is anelectron, calculate (in eV) the work needed to remove it from the metallicsurface when the electron is in the state of maximum “attachment” to thesurface. What is the mean distance of the electron from the surface in thisstate?

8.9 A particle of mass m is free to move inside a sphere of radius a, but cannotexit the sphere. It is thus under the influence of the (central) potential

V (r) =

{0, r < a∞, r > a.

Calculate the energy of its ground state.

8.10 A particle of mass m is under the influence of the central potential

V (r) =

{0, r < aV0, r > a,

(1)

which is a “spherical” potential well of finite depth. Calculate—with agraphical solution if necessary—the energy eigenvalues of all the s states(𝓁 = 0) of the particle. Use the solution you obtained to show that thepotential (1) can have bound states only if its parameters V0 and a satisfythe inequality

a√

2mV0

ℏ2 >𝜋

2.

231

9

The Hydrogen Atom. II: Solutions with AngularDependence

9.1 Introduction

In this chapter, we conclude the quantum mechanical study of the hydrogen atom,by searching for the complete set of solutions to the problem. This set includes,as a special case, the spherically symmetric solutions we discussed in the pre-vious chapter. As we will shortly see, a large part of the solution process (andthe solution itself ) is independent of the specific form of the atomic potential,V (r) = −e2∕r, and depends only on the fact that it is a central potential. Sincesuch a potential is only a function of r, it describes a force field that is alwaysdirected toward (or away from) the center of attraction (or repulsion) at r = 0; itis thus a central force field.

The special importance of central potentials, in both classical and quantummechanics, stems from the fact that when a particle moves in such a potential, itsangular momentum is always conserved. Indeed, starting from the definition ofangular momentum1

𝓵 = r × p

and taking the derivative of both sides, we obtain consecutively

d𝓵dt

= drdt

× p + r ×dpdt

= 𝒗 × p + r × F = 0 + 0 = 0,

where, in the last expression, we took into account that 𝒗 || p ⇒ 𝒗 × p = 0 andF || r ⇒ r × F = 0, since the force in a central potential has radial direction.

In case it is not clear to the readers, let us note that the connection betweencentral potentials and central force fields is a direct consequence of the factthat forces are always normal to the equipotential surfaces of the correspondingpotential. For a central potential, the equipotential surfaces are spheres, sincethe relation V (r) = constant implies r = constant. And the lines normal to asphere’s surface always go through the sphere’s center, which, in this case, is theorigin (Figure 9.1).

1 Throughout this chapter we will use the term angular momentum as a synonym to orbital angularmomentum, since the distinction from spin—another form of angular momentum—is not necessaryat this stage.


232 9 The Hydrogen Atom. II: Solutions with Angular Dependence

V (r) = constant

F

Figure 9.1 Equipotential surfaces andforces in a central potential. Theequipotential surfaces (V(r) =constant) are spheres centered at theorigin (r = 0). Since the forces arenormal to the equipotential surfaces,the force lines always pass throughthe center. In other words, these arecentral forces.

Because angular momentum is conserved, it plays a key role in the classicalproblem of motion in a central force field, a celebrated example of which isthe motion of planets under the sun’s gravitational attraction. The readers mayrecall that some features of planetary motion—for example, the law of equalareas (Kepler’s second law)—result directly from the conservation of angularmomentum and not from the specific form of the force law. These features mustthus be present for the whole class of central forces, not for gravity alone.

Angular momentum plays a similar role in the corresponding quantummechanical problems. As we will soon see, some key features of the solutions—forexample, their angular dependence—do not depend on the specific form of thepotential V (r), but only on the fact that it is a central potential, a property thatensures angular momentum conservation.

Therefore, in order to clarify which features of the solution apply to all cen-tral potentials, and not just to the Coulomb potential, it makes sense to retainthe general form V (r) of the potential function for as long as possible, and onlyemploy its specific form (i.e., V = −e2∕r) at the last stage of the solution.

With this in mind, this chapter is naturally divided in two sections. Section 9.2pertains to the part of the solution process that is common to all central poten-tials, while in Section 9.3 we focus on the Coulomb potential and, in particular,the hydrogen atom.

Those readers on a more introductory course in quantum physics could skimthrough Sections 9.2.1–9.2.3, and focus on Section 9.2.4, where we summarizethe basic results from solving the Schrödinger equation in an arbitrary centralpotential. From there on, they will be able to follow the full solution of thehydrogen atom and the physical interpretation of its results, an essential step forunderstanding fundamental features of our world, such as the periodic table ofelements, which we discuss in Chapter 12.

9.2 The Schrödinger Equation in an Arbitrary CentralPotential: Separation of Variables

9.2.1 Separation of Radial from Angular Variables

The equation we have to solve now is∇2𝜓 + (𝜖 − 𝑈 (r))𝜓 = 0, (9.1)

9.2 The Schrödinger Equation in an Arbitrary Central Potential: Separation of Variables 233

where

𝜖 = 2mEℏ2 , 𝑈 (r) = 2mV (r)

ℏ2 ,

while the Laplacian in spherical coordinates is written as

∇2 = 1r𝜕2

𝜕r2 r + 1r2

(1

sin 𝜃𝜕


𝜕𝜃+ 1

sin2 𝜃

𝜕2

𝜕𝜙2

)

. (9.2)

Here, the unknown function 𝜓 = 𝜓(r, 𝜃, 𝜙) no longer depends solely on r, as inthe previous chapter, but on all three spherical coordinates r, 𝜃, 𝜙. Equation (9.1)is thus a partial differential equation for which the only general method of solu-tion is the separation of variables. As we saw in Section 2.5.2, the basic idea of themethod is the following. We write the solution as a product of three functions,each of which depends only on one variable; then, we substitute this expression inthe equation, hoping it will “separate” into three ordinary differential equations,one for each of the three functions in the product. In the case at hand, this meanswriting the unknown function 𝜓(r, 𝜃, 𝜙) in the separable form

𝜓(r, 𝜃, 𝜙) = R(r)Θ(𝜃)Φ(𝜙)

where, for notational clarity, we denoted each of the three separate functions bythe capital letter corresponding to its respective variable. It is actually more prac-tical to make the separation of variables in two steps. First, we separate only theradial variable r from the angular variables 𝜃 and 𝜙, by writing the solution as

𝜓(r, 𝜃, 𝜙) = R(r)Y (𝜃, 𝜙) (9.3)

and we then carry out the second separation, by setting

Y (𝜃, 𝜙) = Θ(𝜃)Φ(𝜙).

With ∇2 given by (9.2), Eq. (9.1) is now written as2

(1r𝜕2

𝜕r2 r + 1r2

(1

sin 𝜃𝜕


𝜕𝜃+ 1

sin2 𝜃

𝜕2

𝜕𝜙2

)

+ (𝜖 − 𝑈 (r)))

𝜓 = 0

or, equivalently, as(

1r𝜕2

𝜕r2 r + (𝜖 − 𝑈 (r)) + 1r2

(1

sin 𝜃𝜕


𝜕𝜃+ 1

sin2 𝜃

𝜕2

𝜕𝜙2

))

𝜓 = 0,

and in more compact form as(

L + 1r2 Λ

)

𝜓 = 0 or L𝜓 + 1r2 Λ𝜓 = 0, (9.4)

2 It is common practice in the theory of linear (and homogeneous) differential equations to use thecompact symbolic form Ly = 0, where L is the differential operator of the equation. For example, thesymbolic form of the equation xy′′ − 2xy′ + y = 0 is

(

x d2

dx2 − 2x ddx

+ 1)

y = 0,

where the differential operator is

L = x d2

dx2 − 2x ddx

+ 1.


where L and Λ are the operators

L = 1r𝜕2

𝜕r2 r + (𝜖 − 𝑈 (r)) (9.5)

and

Λ = 1sin 𝜃

𝜕


𝜕𝜃+ 1

sin2 𝜃

𝜕2

𝜕𝜙2 . (9.6)

As we will see shortly, the compact form (9.4) simplifies things significantly,since we no longer have to specify the detailed form of the operators L andΛ; the only thing that matters is that L acts on the radial variable r, while Λacts on the angular variables 𝜃 and 𝜙. Therefore, we can plug (9.3) in (9.4)to obtain

(LR)Y + 1r2 R(ΛY ) = 0. (9.7)

The key idea now is to write (9.7) so that its variables are separated: A functionthat depends only on r is on one side, and a function that depends only on 𝜃 and𝜙 is on the other. This is easily achieved if we multiply (9.7) by r2 and then divideby the product 𝜓 = RY . Thus, we obtain

r2(LR)YRY

+ R(ΛY )RY

= 0 ⇒r2(LR)

R= −ΛY

Y. (9.8)

The last equation in (9.8) requires two functions of different independent vari-ables to be equal. But the only way for this to happen is if both functions are equalto a common constant (known as the separation constant). Therefore, Eq. (9.8)can only be true if

r2(LR)R

= −ΛYY

= 𝜆, (9.9)

where 𝜆 is the separation constant for this problem. Equation (9.9) is actuallyequivalent to the two equations

r2(LR) = 𝜆R (9.10)

and

ΛY = −𝜆Y , (9.11)

both of which have a simple general form. They are eigenvalue equations for theoperators = r2L and Λ, with opposite eigenvalues.

If we now divide each side of (9.10) by r2 and recall the explicit form of theoperator L from Eq. (9.5), we obtain

1r(rR)′′ + (𝜖 − 𝑈 (r))R = 𝜆

r2 R.

Multiplying each term by r so as to form everywhere the product rR, we get

(rR)′′ + (𝜖 − 𝑈 )(rR) = 𝜆

r2 (rR), (9.12)


whence, by a change of dependent variable

y = rR, (9.13)

and upon carrying all terms on the left-hand side, we can write Eq. (9.12) as

y′′ +(

𝜖 − 𝑈 (r) − 𝜆

r2

)

y = 0.

If we now substitute 𝜖 = 2mE∕ℏ2, 𝑈 = 2mV∕ℏ2, we obtain

y′′ + 2mℏ2

(

E − V (r) − ℏ2𝜆

2mr2

)

y = 0, (9.14)

which is a one-dimensional Schrödinger equation with respect to the radial vari-able r. The potential

V (r) = V (r) + ℏ2𝜆

2mr2

is known as the effective potential, for reasons we discuss in the following section.It is important to stress at this point that, because the angular Eq. (9.11) is

independent of the potential V (r), the angular function Y (𝜃, 𝜙) is the same forall central potentials. Hence, if we solve Eq. (9.11) with Λ given by (9.6), wewill find at once the angular dependence of the wavefunction for every centralpotential. Then the only equation that remains to be solved is the radial equation(9.14) for the potential at hand, V (r), with the eigenvalues 𝜆—obtained fromEq. (9.11)—being the same for all problems. The significance of this importantresult will become apparent as we continue our study of central potentials.

9.2.2 The Radial Schrödinger Equation: Physical Interpretation of theCentrifugal Term and Connection to the Angular Equation

First, let us note that the result (9.14) includes, as a special case, the correspondingradial Schrödinger equation of the previous chapter (Eq. (8.8) for V (r) = −e2∕r).Indeed, if we wanted to use the current method to retrieve all solutions with noangular dependence, we would set Y (𝜃, 𝜙) = constant, so that ΛY = 0 and 𝜆 = 0,due to (9.11). But for 𝜆 = 0, Eq. (9.14) becomes identical to the correspondingradial Schrödinger equation of the previous chapter, where 𝜓(r) = R(r), sinceY = constant. The radial wavefunction (9.13) satisfies the boundary conditions

y(0) = 0, y(∞) = 0,

which are appropriate for bound motion in the interval 0 < r < ∞with an impen-etrable wall at r = 0. As noted in the previous chapter, the function y(r) will haveall the features of a genuine one-dimensional wavefunction in the above range.

But the greater significance of the radial Schrödinger equation studied here,compared to the one in the previous chapter, stems from the new term

Vnew(r) =ℏ2𝜆

2mr2 , (9.15)

which is added to the initial potential V (r) to form the effective potential. Thephysical significance of the term (9.15) is better understood if we recall the corre-sponding problem in classical mechanics. This so-called centrifugal term always


arises when we attempt to separate the radial from the angular motion, whichtakes us to a reference frame that follows the instantaneous rotation of the par-ticle. In this case, we add to the potential a new energy term that corresponds tothe centrifugal force and is equal to the rotational energy

𝓵2

2I= 𝓵2

2mr2 , (9.16)

where I = mr2 is the moment of inertia of the particle with respect to the origin(i.e., the center of rotation) and𝓵 is the constant vector of its angular momentum.

We remind the readers that the expression 𝓵2∕2I for the rotational energyresults directly from the corresponding expression of the translational energyp2∕2m, with the standard substitutions

p −−−−→ 𝓵, m −−−−→ I(momentum −−−−→ angular momentum) (mass −−−−→ moment of inertia)

A comparison of (9.15) with (9.16) suggests the relation

𝓵2 = ℏ2𝜆,

which, however, we must reinterpret quantum mechanically, since in quantumphysics, 𝓵2 is no longer a numerical quantity but a differential operator. Thenatural interpretation here is that ℏ2𝜆 represents the eigenvalues of this oper-ator. If so, then the physical meaning of the angular Eq. (9.11) becomes moretransparent once we multiply it by −ℏ2, so that

(−ℏ2Λ)Y = ℏ2𝜆Y ,

where ℏ2𝜆 appears now naturally as the eigenvalue of the operator −ℏ2Λ. We arethus led to hypothesize that

𝓵2 = −ℏ2Λ = −ℏ2(

1sin 𝜃

𝜕


𝜕𝜃+ 1

sin2𝜃

𝜕2

𝜕𝜙2

)

, (9.17)

so that the angular equation ΛY = −𝜆Y we obtained from the separation ofvariables is, essentially, the quantum mechanical eigenvalue equation of thephysical quantity 𝓵2. The proof of (9.17) is actually rather cumbersome. Itrequires starting from the known expressions for the quantum mechanicaloperators 𝓁x,𝓁y,𝓁z in Cartesian coordinates

𝓁x = −iℏ(

y 𝜕𝜕z

− z 𝜕𝜕y

)

, 𝓁y = −iℏ(

z 𝜕𝜕x

− x 𝜕𝜕z

)

,

𝓁z = −iℏ(

x 𝜕𝜕y

− y 𝜕𝜕x

)

converting the differentiations with respect to x, y, and z to differentiations withrespect to r, 𝜃, and 𝜙, and forming the operator expression 𝓵2 = 𝓁2

x + 𝓁2y + 𝓁2

z toverify that it is indeed identical to the right-hand side of Eq. (9.17), as we assumed.

We leave it to the readers to complete the detailed steps for this proof. Here, wepresent only the expressions of the operators 𝓁x,𝓁y,𝓁z in spherical coordinates,


since we will need them later on:

𝓁x = iℏ(

sin𝜙 𝜕

𝜕𝜃+ cos𝜙

tan 𝜃𝜕

𝜕𝜙

)

, 𝓁y = iℏ(

− cos𝜙 𝜕

𝜕𝜃+ sin𝜙

tan 𝜃𝜕

𝜕𝜙

)

,

(9.18)

𝓁z = −iℏ 𝜕

𝜕𝜙. (9.19)

Note that these expressions contain differentiations with respect to 𝜃 and 𝜙,but not r. This is not surprising. The quantum mechanical operators of quantitiesthat describe motion, such as momentum and angular momentum, should con-tain derivatives of only those variables that truly vary during the type of motionassociated with each physical quantity. For example, the operator px = −iℏ 𝜕∕𝜕xcontains a derivative with respect to x, since this is the only coordinate that variesin a motion along the x-axis. Likewise, since the angular momentum 𝓁z cor-responds to rotation around the z-axis, which causes only the angle 𝜙 to vary,the associated operator (9.19) contains only the derivative with respect to 𝜙. Forthe same reason, the expressions (9.18) are not expected to contain a derivativewith respect to r, since the rotational motion, which is associated with angularmomentum, causes only 𝜃 and 𝜙 to vary, but not r.

Based on this discussion and the relationp2

2m= − ℏ2

2m∇2,

it seems plausible that the radial part in the expression (9.2) of the Laplacian issimply the “component” of the kinetic energy due to the radial motion of the par-ticle, while the angular part is the “component” due to rotational motion (i.e., theparticle’s angular momentum).

9.2.3 Solution of the Angular Equation: Eigenvalues and Eigenfunctionsof Angular Momentum

Now that we know the physical significance of the angular Eq. (9.11), we proceedto solve it. We can rewrite the equation explicitly as

(1

sin 𝜃𝜕


𝜕𝜃+ 1

sin2 𝜃

𝜕2

𝜕𝜙2 + 𝜆)

Y (𝜃, 𝜙) = 0, (9.20)

which is a partial differential equation that can be solved with the method ofseparation of variables, just like the original Schrödinger equation. For the samereasons as before, we write Eq. (9.20) in the more compact form

(

L + 1sin2 𝜃

𝜕2

𝜕𝜙2

)

Y = 0, (9.21)

where L is the operator

L = 1sin 𝜃

𝜕


𝜕𝜃+ 𝜆

that acts only on the variable 𝜃.


By setting now

Y (𝜃, 𝜙) = Θ(𝜃)Φ(𝜙)

in Eq. (9.21), we obtain

(LΘ)Φ + 1sin2 𝜃

ΘΦ′′ = 0

from where we see that the variables 𝜃 and 𝜙 can indeed be separated if wemultiply first by sin2 𝜃 and then divide by Y = Θ ⋅Φ. Thus, we have

sin2 𝜃(LΘ)Θ

+ Φ′′

Φ= 0 ⇒

sin2 𝜃(LΘ)Θ

= −Φ′′

Φ, (9.22)

where the last equation requires the equality of two functions of different inde-pendent variables: a function of 𝜃 (left-hand side) and a function of𝜙 (right-handside). As we know by now, this can only be true if both functions are equal to thesame constant. It thus follows from Eq. (9.22) that

sin2 𝜃(LΘ)Θ

= −Φ′′

Φ= 𝜇,

whence two equations emerge, namely,

Φ′′ + 𝜇Φ = 0 (9.23)

and

sin2 𝜃(LΘ) = 𝜇Θ ⇒ LΘ = 𝜇

sin2 𝜃Θ

⇒

(1

sin 𝜃d

d𝜃sin 𝜃 d

d𝜃+ 𝜆 − 𝜇

sin2 𝜃

)

Θ = 0. (9.24)

9.2.3.1 Solving the Equation for 𝚽We will now show that Eq. (9.23) has physically acceptable solutions only if theseparation constant 𝜇 takes the discrete values 𝜇 = m2, where m is a positive ornegative integer number (m = 0,±1,±2,…). The physical requirement leading tothese m values is that the wavefunction

𝜓(r, 𝜃, 𝜙) = R(r)Θ(𝜃)Φ(𝜙)

be a single-valued function of space, which means that it must have a single valueat any one point of space. Accordingly,

Φ(𝜙 + 2𝜋) = Φ(𝜙),

since the rotation by 2𝜋 brings us back to the same point in space (Figure 9.2).In other words, Φ(𝜙) must be a periodic function with period 2𝜋. This has two

implications: (i) The constant 𝜇 in (9.23) cannot be negative, since the solutionswould then take the form of real exponentials, and exponential functions of thistype are not periodic. (ii) Since 𝜇 is positive, we may set 𝜇 = m2, so the solutionstake the form

cos m𝜙, sin m𝜙.


Figure 9.2 Why wavefunctions should beperiodic in 𝜙. When 𝜙 changes by 2𝜋 we areback to the starting point. Therefore, for thewavefunction to be a single-valued functionof space, it has to remain the same under thechange 𝜙 → 𝜙 + 2𝜋.

x

y

r

z

2π

θ

φ

Now, the constraint that solutions have a period of 2𝜋 leads directly to the con-dition

cos m𝜙 = cos m(𝜙 + 2𝜋) = cos(m𝜙 + 2𝜋m),

which can only be satisfied if m is an integer number, since the cosine (and thesine) is a periodic function with period equal to any integer multiple of 2𝜋. Forreasons that will immediately become clear, it is preferable to choose as solutionsof (9.23), not the sine or the cosine, but the complex exponential eim𝜙, whichclearly satisfies the eigenvalue equation of the operator 𝓁z = −iℏ 𝜕∕𝜕𝜙, witheigenvalues ℏm for any sign of m; that is,

𝓁z eim𝜙 = ℏm eim𝜙.

Unwittingly, we have just solved the eigenvalue problem for the projection ofangular momentum onto an axis, and found that it takes the quantized values

ℏm ∶ eigenvalues of 𝓁z,

where m is an integer number (including zero), known as the quantum numberof the angular momentum projection, or—more simply—the magnetic quantumnumber, for reasons we will explain later on.

9.2.3.2 Solving the Equation for 𝚯We still need to solve Eq. (9.24). Our first step is to turn it into an equation withrational coefficients3 using a suitable change of variables.

It is not hard to see that the correct choice is

𝜉 = cos 𝜃, (9.25)

for which we havedΘd𝜃

= dΘd𝜉

d𝜉d𝜃

= − sin 𝜃 dΘd𝜉

⇒1

sin 𝜃d

d𝜃= − d

d𝜉.

3 Since Eq. (9.24) is a second-order linear differential equation with variable coefficients, we cansolve it only if its coefficients are rational functions, whereby we can apply the power-series methodto arrive at an exact solution (e.g., in polynomial form).


The differential operator of (9.24) is then written as1

sin 𝜃d

d𝜃sin 𝜃 d

d𝜃≡ 1

sin 𝜃d

d𝜃⏟⏞⏟⏞⏟

−d∕d𝜉

sin2𝜃⏟⏟⏟

1−𝜉2

1sin 𝜃

dd𝜃

⏟⏞⏟⏞⏟

−d∕d𝜉

≡ dd𝜉

(1 − 𝜉2) dd𝜉

and thus the equation with respect to 𝜃 takes the much simpler form4

(d

d𝜉(1 − 𝜉2) d

d𝜉+ 𝜆 − m2

1 − 𝜉2

)

Θ(𝜉) = 0,

or, equivalently, the form of the so-called associated Legendre equation

(1 − 𝜉2)Θ′′ − 2𝜉 Θ′ +(

𝜆 − m2

1 − 𝜉2

)

Θ = 0, (9.26)

which includes, as a special case for m = 0, the Legendre equation

(1 − 𝜉2)P′′ − 2𝜉 P′ + 𝜆 P = 0. (9.27)

We shall first search for the physically acceptable solutions of (9.27), from whichthe allowed values of the parameter 𝜆will emerge. We will then see that the phys-ically acceptable solutions of Eq. (9.26) are produced from the solutions of (9.27)via the formula5

Θ(𝜉) = (1 − 𝜉2)|m|∕2 d|m|P(𝜉)d𝜉|m|

, (9.28)

that is, by differentiating |m| times the solutions of the Legendre equation andsubsequently multiplying them with the factor (1 − 𝜉2)|m|∕2.

As we search for the physically acceptable solutions of Eq. (9.27), let us firstnote that the change of variables (9.25) maps the range of 𝜃 (0 ≤ 𝜃 ≤ 𝜋) onto therange

−1 ≤ 𝜉 ≤ 1, (9.29)

whose endpoints ±1 are physically acceptable points, since they correspond tothe positive and negative z-semiaxis respectively. So, we ought to look for thosesolutions of Eq. (9.27) that remain finite in the full range (9.29) and, especially,at its endpoints. Our emphasis on the endpoints 𝜉 = ±1 is no accident; they arethe singular points of Eq. (9.27). We remind the readers that for a second-orderlinear equation, written in the standard form

y′′ + p(x)y′ + q(x)y = 0,

4 We remind the readers that the differentiations in the expression of a differential operator mustbe carried out only when the operator is “called” to act on a certain function. For example, in ourcase, the action of the differential operator in the given parentheses, on the function Θ(𝜉), leads to

(d

d𝜉(1 − 𝜉2) d

d𝜉

)

Θ = ((1 − 𝜉2)Θ′)′ = (1 − 𝜉2)Θ′′ − 2𝜉Θ′.

5 We urge those readers who do not like to use ready-made formulas—an attitude stronglyencouraged in this book—to look at Problem 7.4 for an alternative approach that makes no use ofthis formula and allows them to obtain the solution with no outside help.


the singular points are those where at least one of the coefficients p(x) and q(x)diverges. Their significance lies in the fact that they are the only points wherethe solution may diverge. At all other points—the so-called ordinary points ofthe equation—the solution is finite and has derivatives of all orders, provided itscoefficients are functions with similar properties in the vicinity of these points.

If we now write Eq. (9.27) in the standard form

P′′ − 2𝜉1 − 𝜉2 P′ + 𝜆

1 − 𝜉2 P = 0

we see at once that 𝜉 = ±1 are indeed singular points. The solution will thusdiverge there in general, unless we do something to prevent this. Now, an obviousset of solutions of Eq. (9.27) that do not diverge at the points ±1, are the polyno-mials. So let us examine whether Eq. (9.27) has polynomial solutions for some“allowed values” of the parameter 𝜆. As a first step, we check for a polynomialsolution of zeroth degree, that is, a constant, which we can always set equal tounity for the known reasons (the equation is homogeneous, etc.). Setting

P(𝜉) = P0(𝜉) = 1in Eq. (9.27) yields

(1 − 𝜉2) ⋅ 0 − 2𝜉 ⋅ 0 + 𝜆 ⋅ 1 = 0 ⇒ 𝜆 = 0,which means that this solution does exist, provided 𝜆 is zero.

We should also note that (9.27) is invariant under the change 𝜉 → −𝜉6 (ithas mirror symmetry, as we say), so its solutions are necessarily even or odd.Therefore, if it has polynomial solutions, these should be even or odd, dependingon their degree. Polynomials of even degree are even and will thus only haveeven powers, while polynomials of odd degree are odd and contain only oddpowers. If, therefore, Eq. (9.27) is solved by a first-degree polynomial, this wouldhave to be of the form

P1(𝜉) = 𝜉

and Eq. (9.27) will then be satisfied if(1 − 𝜉2) ⋅ 0 − 2𝜉 ⋅ 1 + 𝜆𝜉 = 0 ⇒ 𝜆 = 2,

which means that a first-degree polynomial solution also exists, if 𝜆 is equal to 2.For a polynomial P𝓁(𝜉) of arbitrary degree 𝓁7 all we need to do is apply the

necessary condition of Chapter 6.

If an equation has a polynomial solution of degree n, then for large x, theequation must be satisfied by the maximum power xn of the polynomial solution.(Necessary condition)

6 This is true for every linear and homogeneous equation where the derivatives of even (odd) orderhave even (odd) coefficients. The reason is that the change 𝜉 → −𝜉 causes a sign flip on thederivatives of odd order, while it leaves intact the derivatives of even order. For example, inEq. (9.27), the sign flip in the first derivative will be cancelled out by the similar change in thecoefficient −2𝜉.7 For the degree of the polynomial we use the symbol 𝓁, instead of the usual n, because, as we willsee shortly, 𝓁 plays the role of the quantum number of angular momentum 𝓵 (more specifically, itsmagnitude).


The only new element in this case is that the variable 𝜉 is limited in the range−1 ≤ 𝜉 ≤ 1, so it cannot reach the “large-𝜉” limit of the above-stated condition.But the constraint −1 ≤ 𝜉 ≤ 1 results from the physical range of 𝜃 (0 ≤ 𝜃 ≤ 𝜋)through the change 𝜉 = cos 𝜃. From a mathematical perspective, however, it isnot possible for a polynomial solution to satisfy an equation only inside a finiterange but not outside it. The polynomial will necessarily be a solution in the fullinterval −∞ < 𝜉 <∞.

Based on this, we insert in (9.27) the maximum power 𝜉𝓁 of the candidate poly-nomial solution and require that it must satisfy the equation for large enough 𝜉.We then have

(1 − 𝜉2)𝓁(𝓁 − 1)𝜉𝓁−2 − 2𝜉 𝓁𝜉𝓁−1 + 𝜆𝜉𝓁 = 0

and, if we retain only the highest power term (in this case, 𝜉𝓁), we obtain

(−𝓁(𝓁 − 1) − 2𝓁 + 𝜆)𝜉𝓁 = (−𝓁(𝓁 + 1) + 𝜆)𝜉𝓁 = 0⇒ 𝜆 = 𝓁(𝓁 + 1),

where

𝓁 = 0, 1, 2,… .

From a physical perspective, this result is truly remarkable. Given that the eigen-values of the squared magnitude 𝓵2 of the angular momentum are ℏ2𝜆, the result𝜆 = 𝓁(𝓁 + 1) tells us that𝓵2 is a quantized physical quantity whose allowed valuesare

ℏ2𝓁(𝓁 + 1) ∶ eigenvalues of 𝓵2,

where the quantum number 𝓁—known as the angular momentum quantumnumber (more precisely, the quantum number of the magnitude of angularmomentum)—takes all positive integer values from zero to infinity.

To rigorously prove this, we have, of course, to show that the necessary condi-tion we employed is also sufficient for the existence of polynomial solutions forevery 𝓁 (not just 𝓁 = 0 and 𝓁 = 1), and that polynomial solutions are indeed theonly ones satisfying the physical constraints of the problem. We will not attemptto fill those mathematical gaps here (for those interested, see Chapter 7), but wewill proceed to present the full solution. We trust that the physical completenessof the end result will leave no doubt in the readers’ mind about its mathematicalrobustness.

If we now take expression (9.28)—which connects the solutions of (9.26) withthose of (9.27)—as given, then the solution of the angular equation ΛY = −𝜆Y isreadily obtained. The solutions Θ(𝜉) of (9.26) that are well behaved at the “tricky”points 𝜉 = ±1 are given by the formula

Θ𝓁m(𝜉) ≡ P𝓁m(𝜉) = (1 − 𝜉)|m|∕2 d|m|P𝓁(𝜉)d𝜉|m|

, (9.30)

where P𝓁m(𝜉) is the standard notation for the solutions, which are known asassociated Legendre polynomials, even though they are, in fact, polynomials,only when m is an even number.


The reason for the appearance of the absolute value of m in the right-hand sideof Eq. (9.30) is that the associated Legendre equation contains only the squareof m and, as a result, the solutions for positive and negative m are necessarilythe same.

Apart from the mathematical details, expression (9.30) embodies an importantphysical conclusion. Since a polynomial cannot be differentiated more times thanits degree (beyond that, differentiations yield zero), expression (9.30) requiresthat

|m| ≤ 𝓁, (9.31)

or, equivalently,

−𝓁 ≤ m ≤ 𝓁,

which tells us that for a given 𝓁, m assumes all integer values from −𝓁 to +𝓁.For example, for 𝓁 = 0 we necessarily have m = 0, for 𝓁 = 1 we have the threevalues

m = −1, 0, 1 (𝓁 = 1),

for 𝓁 = 2 we obtain the five values

m = −2,−1, 0, 1, 2 (𝓁 = 2),

while for an arbitrary 𝓁 the number of m values is

d𝓁 = 2𝓁 + 1. (9.32)

From a physical perspective, inequality (9.31) expresses the fact that the eigen-values of the projection of a vector cannot be greater than the vector’s length.And yet, the relation (𝓁z)max = ℏmmax = ℏ𝓁, combined with |𝓵| = ℏ

√𝓁(𝓁 + 1),

reveals that the maximum projection of the vector 𝓵 onto the z-axis is alwayssmaller than its length! We will come back to this point once we present the fullsolution.

To recap, we have found that the angular dependence of the wavefunctions inany central potential has the form

Y (𝜃, 𝜙) = Y𝓁m(𝜃, 𝜙) = P𝓁m(cos 𝜃) eim𝜙.

The functions Y𝓁m are commonly known as spherical harmonics, since theypertain to vibrations of a spherical surface. They describe the distortions of thesurface as it vibrates with a single frequency (normal modes of vibration). Justlike sinusoidal functions describe the normal modes of an oscillating string,spherical harmonics play the same role for a vibrating spherical surface. Theyare the harmonics of the sphere.

9.2.4 Summary of Results for an Arbitrary Central Potential

At this point, a brief summary of the results is needed for two reasons:

First. To show the readers—especially those who found the solution processtedious—that the obtained results can be codified in just a few basic formulaswith a simple physical meaning.


Second. To provide those readers on a more introductory course in quantumphysics, who may have skimmed over the preceding sections, with a referencepoint for the remainder of the book.

We now summarize our findings as follows.In any central potential V (r), the solution 𝜓(r, 𝜃, 𝜙) of the Schrödinger

equation has the form

𝜓(r, 𝜃, 𝜙) = R(r)Y𝓁m(𝜃, 𝜙), (9.33)

where the angular functions Y𝓁m(𝜃, 𝜙), known as spherical harmonics, arecommon for all central potentials. They are given by the relation

Y𝓁m ∼ P𝓁m(cos 𝜃)eim𝜙, (9.34)

where P𝓁m(𝜉) (𝜉 = cos 𝜃) are the so-called associated Legendre polynomials,which are related to the common Legendre polynomials P𝓁(𝜉) via the formula

P𝓁m(𝜉) ∼ (1 − 𝜉2)|m|∕2 d|m|P𝓁(𝜉)d𝜉|m|

. (9.35)

The polynomials P𝓁(𝜉) are even or odd solutions (depending on their degree 𝓁)of the Legendre equation

(1 − 𝜉2)P′′𝓁 − 2𝜉 P′

𝓁 + 𝓁(𝓁 + 1)P𝓁 = 0. (9.36)

From a physical perspective, the spherical harmonics Y𝓁m(𝜃, 𝜙) are the commoneigenfunctions of the quantum mechanical operators

𝓁z = −iℏ 𝜕

𝜕𝜙(9.37)

and

𝓵2 = −ℏ2(

1sin 𝜃

𝜕


𝜕𝜃+ 1

sin2 𝜃

𝜕2

𝜕𝜙2

)

(9.38)

with corresponding eigenvalues ℏm and ℏ2𝓁(𝓁 + 1). That is,

𝓁zY𝓁m = ℏmY𝓁m (9.39)

and

𝓵2Y𝓁m = ℏ2𝓁(𝓁 + 1)Y𝓁m. (9.40)

The integer numbers 𝓁 and m are known as the quantum number of angularmomentum (or, more precisely, the “quantum number of the magnitude ofangular momentum”) and the quantum number of the projection of the angularmomentum onto the z-axis, respectively.

Since the quantum number 𝓁 is the degree of a polynomial—the Legendrepolynomial P𝓁(𝜉)—it can only take the values

𝓁 = 0, 1, 2,… ,∞.[allowed values of 𝓁 ] (9.41)

On the other hand, because 𝜓 has to be a single-valued function in space, theangular function exp (im𝜙) in (9.34) ought to be periodic with a period 2𝜋


(remember that 𝜙 is the angle of rotation around the z-axis). This leads to thecondition

eim𝜙 = eim(𝜙+2𝜋) ⇒ e2im𝜋 = 1,

which can only be satisfied ifm = 0,±1,±2,…

[allowed values of m] (9.42)

Moreover, for a given 𝓁, the relation P𝓁m ∼ (1 − 𝜉2)|m|∕2d|m|P𝓁∕d𝜉|m| implies that

|m| ≤ 𝓁 ⇒ −𝓁 ≤ m ≤ 𝓁, (9.43)

which further limits m to integer values between −𝓁 and +𝓁 in unit steps, sincem is integer. We thus obtain

m = −𝓁,… ,+𝓁[allowed values of m for a given 𝓁 ] (9.44)

Note also that, given the form of the operator 𝓁z (𝓁z = −iℏ 𝜕∕𝜕𝜙), the depen-dence of the spherical harmonic Y𝓁m on the angle𝜙 (i.e., the complex exponentialeim𝜙) is to be expected. Actually, it is directly analogous to the expression eikx ofthe momentum eigenfunctions.

Having obtained the angular dependence of the solution, which is common forall central potentials, we only need to find the radial function R(r) for the specificform of the potential V (r). We will determine R(r) through the auxiliary function

y(r) = rR(r), (9.45)

which is known as the radial wavefunction, since it satisfies the radial Schrödingerequation

y′′ + 2mℏ2 (E − V (r))y = 0, (9.46)

where

V (r) = V (r) + ℏ2𝓁(𝓁 + 1)2mr2 (9.47)

is the so-called effective potential we obtain by adding to the original potentialthe centrifugal term

𝓵2

2I≡ ℏ2𝓁(𝓁 + 1)

2mr2

associated with the rotational energy of the particle at a distance r from the origin.(Here, I = mr2 is the particle’s moment of inertia at that distance.) The relationy(r) = rR(r) implies that the radial wavefunction y(r) for the bound states satisfiesthe boundary conditions

y(0) = 0, y(∞) = 0, (9.48)

which are conditions for a truly one-dimensional problem defined on the interval0 < r < ∞ with an impenetrable wall at r = 0.

From a practical perspective, the preceding discussion and, especially, thesolution of the angular equation, provide us with a set of general results we will


henceforth use as a starting point for every central potential we wish to solve.In each such case (e.g., the Coulomb potential), we will proceed directly to solvethe radial equation for the given potential V (r).

Problems

9.1 Find the second-degree polynomial solution of Eq. (9.27) and then use (9.28)to construct the associated Legendre polynomials P20, P2,±1, and P2,±2. Dothe same for the 𝓁 = 3 case.

9.2 Use the results you found, and those in the text for 𝓁 = 1, to construct thespherical harmonics up to 𝓁 = 3.

9.3 Show that the spherical harmonic Y𝓁𝓁—the one that corresponds tomaximum projection onto the z-axis—has the form

Y𝓁𝓁 = N sin𝓁𝜃 ⋅ ei𝓁𝜙.

9.4 Starting from the expression of 𝓁z in Cartesian coordinates, prove that itsform in spherical coordinates is given by (9.19).

9.3 The Hydrogen Atom

9.3.1 Solution of the Radial Equation for the Coulomb Potential

Based on our discussion so far, the only task remaining to complete the quan-tum mechanical picture of the hydrogen atom is to solve the radial Schrödingerequation, with V (r) being the potential due to the Coulomb interaction betweenthe electron and the nucleus. The equation we need to solve is thus

y′′ + 2mℏ2 (E − V (r))y = 0,

where

V (r) = −e2

r+ ℏ2𝓁(𝓁 + 1)

2mr2 , (9.49)

so that, in explicit form, the equation becomes

y′′ + 2mℏ2

(

E + e2

r− ℏ2𝓁(𝓁 + 1)

2mr2

)

y = 0, (9.50)

while the boundary conditions to be satisfied arey(0) = 0, y(∞) = 0.

We can now solve (9.50) by applying the procedure we developed in Chapter 6.

STEP 1: Simplifying physical dimensions.Let us use the atomic system of units, namely,

ℏ = m = e = 1,

9.3 The Hydrogen Atom 247

V (r)

E < 0

r

V (r) = − e2

r+

2 ( + 1)2mr2 ,𝓁 𝓁ħ

Figure 9.3 Effective potential for the hydrogen atom. The region of bound states lies belowthe value of the potential at infinity, that is, at negative energies.

so that Eq. (9.50) takes the simpler form

y′′ +(

2E + 2r− 𝓁(𝓁 + 1)

r2

)

y = 0. (9.51)

STEP 2: Removing the asymptotic factor.Since we are interested in bound states, we have E < 0 (see Figure 9.3), so wecan set

2E = −𝛾2 (9.52)

and thus rewrite Eq. (9.51) as

y′′ +(

−𝛾2 + 2r− 𝓁(𝓁 + 1)

r2

)

y = 0. (9.53)

The asymptotic form of Eq. (9.53), for large r, is

y′′∞ − 𝛾2y∞ = 0,

whose solutions are the two exponentials

y∞ = e±𝛾r,

of which we retain only the decaying one. We thus write the full solution y(r) as

y(r) = e−𝛾rF(r) (9.54)

in the hope that the complementary function F(r) has a polynomial form, forreasons well known by now. By inserting now (9.54) into (9.53), we obtain theequation for F

F ′′ − 2𝛾F ′ +(

2r− 𝓁(𝓁 + 1)

r2

)

F = 0 (9.55)

and proceed to find its polynomial solutions.

STEP 3: Searching for polynomial solutions.According to the familiar necessary condition, Eq. (9.55) will have a polynomialsolution of degree n only if it is satisfied for large r by the maximum power


rn of such a polynomial. By setting F ∼ rn in Eq. (9.55) and retaining only thehighest-power term, we obtain

(−2𝛾n + 2) rn−1 = 0 ⇒ 𝛾 = 1∕n,

so that, given Eq. (9.52), the allowed energies of the atom are given by

En = − 12n2 .

This is the same formula as the one we found in the previous chapter for thespherically symmetric solutions. In the present case, however, for a given n (i.e.,for the same energy eigenvalue), there are more than one eigenfunctions thatdiffer in the two other quantum numbers (𝓁 and m) needed to fully determinethe solution. To see how this happens we have to revisit Eq. (9.55)—an equationthat depends on 𝓁—to seek the starting power of its polynomial solutions. Letus assume that rs is this starting power, so the corresponding polynomial is

F(r) = rs + · · · + anrn. (9.56)

To find s, we apply to the starting power rs the same condition applied to theterminating power rn, but with the obvious difference that we are now talkingabout small values of r, not large ones. In other words, we realize that for smallr, where the low powers of r prevail, the starting power rs will dominate thepolynomial (9.56) and satisfy (9.55) on its own. We thus set F ∼ rs in Eq. (9.55),retain only the lowest power that appears, and require that its coefficient vanishso that the equation is satisfied. We therefore obtain

s(s − 1)rs−2 − 2𝛾srs−1 + 2rs−1 − 𝓁(𝓁 + 1)rs−2 = 0⇒ (s(s − 1) − 𝓁(𝓁 + 1)) rs−2 = 0 ⇒ s(s − 1) = 𝓁(𝓁 + 1)

⇒ s1 = 𝓁 + 1, s2 = −𝓁.

From these two values of s we reject the latter, since it leads to a divergingsolution at r = 0. Hence, we find

s = 𝓁 + 1,

which means that the polynomial solutions of (9.55) begin with the power r𝓁+1

and terminate at rn. That is,

Fn𝓁(r) = r𝓁+1 + · · · + anrn.

Given now that the starting power of a polynomial cannot be greater than itsterminating power, the inequality

𝓁 + 1 ≤ n

necessarily holds.Thus, for a given n, the allowed values for 𝓁 are

𝓁 = 0, 1,… , n − 1,

which correspond to different states of the atom, all with the same energy.The radial wavefunctions y(r) = e−𝛾rF(r) thus depend on both quantum

numbers n and 𝓁—that is, y(r) = yn𝓁(r)—and have the form

yn𝓁(r) = e−r∕nFn𝓁(r),


where Fn𝓁 are the polynomial solutions of (9.55) that begin with the power r𝓁+1

and terminate with rn. We can then obtain the radial function R(r) from y(r) viathe expression y = rR, so that

Rn𝓁(r) =1r

yn𝓁(r) = e−r∕n 1r

Fn𝓁(r). (9.57)

The eigenfunctions of the hydrogen atom are therefore written as

𝜓n𝓁m(r, 𝜃, 𝜙) = Rn𝓁(r)Y𝓁m(𝜃, 𝜙)

and thus depend on three quantum numbers: First, the so-called principalquantum number n, which determines the electron’s energy through the relation

En = − 12n2 a.u.;

second, the quantum number 𝓁, which determines the magnitude of the angularmomentum vector via the expression

|𝓵| = ℏ√𝓁(𝓁 + 1); (9.58)

and third, the quantum number m, which determines the projection of theangular momentum onto the z-axis through the formula8

𝓁z = ℏm. (9.59)

The quantum numbers n,𝓁, and m are subject to the constraints we mentionedbefore: For any n (n = 1, 2,…), the number 𝓁 takes all integer values from zeroto n − 1, while for any 𝓁, the number m takes the 2𝓁 + 1 integer values from−𝓁 to +𝓁.

STEP 4: Restoring ordinary physical dimensions.In view of the related discussion of the previous chapter, we can now readilyrestore the physical dimensions using the familiar substitutions

En → 𝜖En

and

𝜓(r, 𝜃, 𝜙) → 1a3∕2

0

𝜓

(r

a0, 𝜃, 𝜙

)

,

where 𝜖 = me4∕ℏ2 = 27.2 eV is the atomic unit of energy, and a0 = ℏ2∕me2 =0.529 ≈ 0.5 Å is the atomic unit of length, namely, one Bohr radius. Note herethat, since the angles 𝜃 and 𝜙 are dimensionless quantities, they are not affectedas we restore dimensions in the wavefunctions; only the radial variable r and thecorresponding radial part R(r) of the wavefunction are properly rescaled, becausethey carry a physical dimension of length and length−3∕2, respectively.

9.3.2 Explicit Construction of the First Few Eigenfunctions

Let us now apply the abovementioned general procedure to construct the firsttwo eigenfunctions (n = 1 and n = 2), which are basically the only ones we willneed later on. We begin with the case n = 1.

8 It is implied here that the symbols |𝓵| and 𝓁z in formulas (9.58) and (9.59) refer to thecorresponding quantities, not the quantum mechanical operators.


9.3.2.1 n = 1 ∶ The Ground StateAs we have seen, for n = 1, the quantum number 𝓁 can only take the value 𝓁 = 0,since 𝓁 ≤ n − 1 ⇒ 𝓁 ≤ 0. But for 𝓁 = 0 we also have m = 0, so for n = 1 there is aunique solution (𝜓100), which describes the ground state of the atom. Therefore,just as in the previous chapter, we have

𝜓100 = 1√𝜋

e−r . (a.u.)

We remind the readers that, for 𝓁 = 0 (⇒ m = 0), the corresponding sphericalharmonic Y00 is a constant. So, the respective wavefunctions

𝜓n00 = Rn0(r)Y00 ≡ 𝜓n(r)

have no angular dependence and are thus identical to the spherically symmetricsolutions 𝜓n(r) of the previous chapter. By the way, the fact that Y00 = constantresults also from the formulas (9.34) and (9.35), if we note that for 𝓁 = 0 (hencem = 0) we have P0(𝜉) = 1 ⇒ P00 = constant ⇒ Y00 = constant. The general con-clusion from this discussion is important and worth stating explicitly:

All states of zero angular momentum, including the ground state of the atom,are spherically symmetric.

9.3.2.2 n = 2 ∶ The First Excited StatesThe first excited states form the quadruplet

𝜓200, 𝜓21,−1, 𝜓210, 𝜓211

since for n = 2, the quantum numbers 𝓁 and m can take the values

𝓁 = 0 ⇒ m = 0 or 𝓁 = 1 ⇒ m = −1, 0, 1.

This feature, namely, the existence of more than one eigenfunction with the sameenergy, is known as degeneracy. We discuss it in detail later on. As for the form ofthe eigenfunctions, which is our main concern here, we have from the previouschapter

𝜓200 ≡ 𝜓2(r) =1

2√

2𝜋

(

1 − r2

)

e−r∕2 (a.u.).

For the remaining three states with 𝓁 = 1 (i.e., 𝜓210 and 𝜓21±1) we need thespherical harmonics Y10 and Y1±1. From expressions (9.34) and (9.35) we find

Y10 = P10, P10 = P1(𝜉) = 𝜉 ⇒ Y10 = cos 𝜃, (9.60)

while for Y11 we have successively

Y11 = P11(𝜉)ei𝜙

and

P11(𝜉) = (1 − 𝜉2)1∕2 dP1(𝜉)d𝜉

= (1 − 𝜉2)1∕2|||𝜉=cos 𝜃

= sin 𝜃

⇒ Y11 = sin 𝜃 ei𝜙, ⇒ Y1,−1 = sin 𝜃 e−i𝜙. (9.61)

To completely specify the eigenfunctions 𝜓210 and 𝜓21±1, we also need the radialfunction Rn𝓁 for our case of n = 2 and 𝓁 = 1. Note that Rn𝓁 is independent of the


quantum number m, so it is the same for all eigenfunctions sharing the same nand 𝓁. Now, given that the starting and terminating powers of the polynomial Fn𝓁are r𝓁+1 and rn, respectively, it is clear that for n = 2 and 𝓁 = 1, these two powersare identical. Therefore, the corresponding polynomial F21 has the form

F21 = r2,

whence, from formula (9.57), we get

R21(r) = re−r∕2,

so that, because of (9.60) and (9.61), we obtain

𝜓210 = Nre−r∕2 cos 𝜃, 𝜓21±1 = Nre−r∕2 sin 𝜃e±i𝜙.

The normalization factor N , which is not necessarily the same in both cases men-tioned, can be easily calculated if we recall that, in spherical coordinates, thevolume element is

dV = r2 sin 𝜃 dr d𝜃 d𝜙

so that, for the wavefunction 𝜓210, say, we have

∫ |𝜓210|2dV =∫ N2r2e−rcos2𝜃 ⋅ r2 sin 𝜃 dr d𝜃 d𝜙

= N2(

∫∞

0r4e−rdr

)

⋅(

∫𝜋

0cos2𝜃 sin 𝜃 d𝜃

)

⋅(

∫2𝜋

0d𝜙

)

= N2 ⋅ 4! ⋅(

∫1

−1𝜉2 d𝜉

)

⋅ 2𝜋 = N2 ⋅ 32𝜋 = 1 (𝜉 = cos 𝜃)

⇒ N = 14√

2𝜋⇒ 𝜓210 = 1

4√

2𝜋re−r∕2 cos 𝜃

and, similarly, we find

𝜓21±1 = 18√𝜋

re−r∕2 sin 𝜃e±i𝜙.

In Table 9.1, we collect our results for the eigenfunctions with n = 1 and n = 2.To construct the eigenfunctions for states with larger n (n = 3, 4,…), we pro-

ceed in a similar manner as above, except we have more calculations to perform.

Table 9.1 Eigenfunctions for the ground and first excited state of thehydrogen atom (in a.u.).

𝓵 = 0 𝓵 = 1

n = 1 𝜓100 = 1√𝜋

e−r —

n = 2 𝜓200 = 12√

2𝜋

(

1 − r2

)

e−r∕2 𝜓210 = 14√

2𝜋re−r∕2 cos 𝜃,

𝜓21±1 = 18√𝜋

re−r∕2 sin 𝜃e±i𝜙


If we do not wish to make use of tables available in the literature, we will havefirst to construct the polynomials Fn𝓁(r) and P𝓁(𝜉)—by direct substitution to thecorresponding Eqs (9.55) and (9.27)—and then apply formulas (9.57), (9.34), and(9.35) as before. One state for which we can easily construct its eigenfunction isthe state of maximum angular momentum for a given n. For example, for n = 3,this state is 𝜓322, whereby both quantum numbers 𝓁 and m take their maximumvalues. In this case, the polynomial F32 takes the form F32 = r3, since its highestand lowest powers are the same (rn|n=3 = r𝓁+1|𝓁=2 = r3). The associated Legendrepolynomial P22(𝜉) becomes then

P22(𝜉) = (1 − 𝜉2)d2P2(𝜉)

d𝜉2 ∼ 1 − 𝜉2 = sin2 𝜃

because, whatever the expression for the Legendre polynomial P2(𝜉) is, it is apolynomial of second degree and, therefore, its second derivative is a constant.We thus obtain

Y22 = P22e2i𝜙 = sin2 𝜃 e2i𝜙

and

R32(r) =1r

e−r∕3 F32(r) = r2e−r∕3

so that

𝜓322 = R32(r)Y22 ∼ r2e−r∕3sin2 𝜃 e2i𝜙.

We hope that this discussion has familiarized the readers with the solutions of thehydrogen atom and the steps one follows to construct them. Once this comfortlevel is attained, one can use tables such as Table 9.1 to retrieve the results needed.In this spirit, we list also in Table 9.2 all the spherical harmonics up to 𝓁 = 2, andin Table 9.3 all the radial functions Rn𝓁 up to n = 3, normalized in the standardfashion found in the literature.

Table 9.2 Spherical harmonics for 𝓁 = 0, 1, 2.

𝓵 = 0 𝓵 = 1 𝓵 = 2

m = 2 — — Y22 =√

1532𝜋

sin2 𝜃e2i𝜙

m = 1 — Y11 =√

38𝜋

sin 𝜃ei𝜙 Y21 =√

158𝜋

cos 𝜃 sin 𝜃ei𝜙

m = 0 Y00 = 1√

4𝜋Y10 =

√3

4𝜋cos 𝜃 Y20 =

√5

16𝜋(3cos2 𝜃 − 1)

m = −1 — Y1,−1 = −√

38𝜋

sin 𝜃e−i𝜙 Y2,−1 = −√

158𝜋

cos 𝜃 sin 𝜃e−i𝜙

m = −2 — — Y2,−2 =√

1532𝜋

sin2 𝜃e−2i𝜙


Table 9.3 Radial functions Rn𝓁 for n = 1, 2, 3 (in a.u.).

𝓵 = 0 𝓵 = 1 𝓵 = 2

n = 1 R10 = 2e−r — —

n = 2 R20 = 1√

2

(

1 − r2

)

e−r∕2 R21 = r2√

6e−r∕2 —

n = 3 R30 = 23√

3

(

1 − 2r3

+ 2r2

27

)

e−r∕3 R31 = 8r27

√6

(

1 − r6

)

e−r∕3 R32 = 4r2

81√

30e−r∕3

As for spherical harmonics, their standard normalization condition is

∫ |Y𝓁m|2 dΩ = 1 (9.62)

where dΩ = sin 𝜃 d𝜃 d𝜙 is the angular part of the three-dimensional volumeelement dV = r2 sin 𝜃 dr d𝜃 d𝜙. As one can easily verify (it follows directly fromthe definition dΩ = dS∕r2, in conjunction with dV = dS dr), the differential dΩis simply the infinitesimal solid angle formed on the surface of a sphere whenthe angular coordinates 𝜃 and 𝜙 of an arbitrary point on the sphere vary by d𝜃and d𝜙, respectively. The expression (9.62), combined with the normalizationcondition ∫ |𝜓|2dV = 1 for the three-dimensional wavefunction 𝜓 = Rn𝓁Y𝓁m,leads directly to the condition

∫∞

0r2R2

n𝓁 dr = ∫∞

0y2

n𝓁 dr = 1,

which has the expected form for the radial wavefunction yn𝓁 = rRn𝓁 . As wehave often stated, the radial wavefunction has all the characteristics of aone-dimensional wavefunction in the range 0 < r < ∞ and, therefore, satisfies acorresponding one-dimensional normalization condition.

An observant reader may have noticed the difference in sign (−1)|m| betweenthe spherical harmonics Y𝓁m and Y𝓁,−m, as shown in Table 9.2. This differencebears no physical significance; it is merely a convention used in the literature tosimplify some relations between spherical harmonics.

We complete our study with the graphical representations of the first few radialfunctions Rn𝓁 for 𝓁 ≠ 0, shown in Figure 9.4.

Problems

9.5 In the (2,1,0) state of the hydrogen atom, the wavefunction of the electronhas the form

𝜓210 = Nre−r∕2 cos 𝜃 (a.u.).

Calculate the following:(a) The mean distance of the electron from the nucleus.


0.15 R21 (r)

R31 (r)

R32 (r)

r (a.u.)

0.10

0.05

–0.25

5 10 15 20 25

Figure 9.4 The first few radial functions Rn𝓁(r) with 𝓁 ≠ 0.

(b) The mean potential energy and the mean kinetic energy of the electronin units of electron volts (eV).

(c) The probability to find the electron inside a double cone of 60∘ angleabout the z-axis.

9.6 The state of the electron in a hydrogen atom is described, at a certainmoment, by the wavefunction

𝜓 = N(𝜓100 + 2𝜓211 + 𝜓32,−1), (1)

where𝜓100, 𝜓211, and𝜓32,−1 are normalized eigenfunctions. Working in a.u.,calculate:(a) The normalization coefficient N such that (1) is also a normalized state.(b) The mean values ⟨𝓵2⟩, ⟨𝓁z⟩, and ⟨E⟩, and the uncertainty Δ𝓁z.

9.7 Find the third-degree polynomial solutions of Eq. (9.55), use them toconstruct the three radial functions R30, R31, and R32 of Table 9.3, and makesure they are properly normalized.

9.8 Verify that the spherical harmonics with 𝓁 = 2 given in Table 9.2 areproperly normalized.


9.3.3.1 The Energy-Level DiagramWe begin our discussion with the energy-level diagram of Figure 9.5, where thevarious energy levels corresponding to different 𝓁 values are shown in separatecolumns, to underline the fact that each column represents the energy-level


E

– 0.85 eV

– 1.5 eV

– 3.4 eV

– 13.6 eV n = 1

n = 2

n = 3

n = 4

ℓ = 0 ℓ = 1 ℓ = 2 ℓ = 3

Figure 9.5 Energy-level diagram for the hydrogen atom. Each column in the diagramrepresents the energy levels for a given 𝓁 and can be regarded as the separate energy diagramfor the corresponding effective potential V𝓁(r).

diagram of the corresponding effective potential

V𝓁(r) = −e2

r+ ℏ2𝓁(𝓁 + 1)

2mr2 .

Based on this depiction, the lowest state of each column should be the groundstate of the potential V𝓁(r), and should therefore lie above the potential’s bottom,that is, above the minimum of the corresponding curve V𝓁(r) (see Figure 9.3). Weleave it to the readers to verify this assertion as an exercise (we suggest the use ofatomic units, for simplicity).

9.3.3.2 Degeneracy of the Energy Spectrum for a Coulomb Potential:Rotational and Accidental DegeneracyOne of the most striking features of the energy spectrum of the hydrogen atomis its rich degeneracy. Apart from the ground state, which is unique, all otherstates are degenerate, so there exist more than one eigenfunction with thesame energy.

For example, for the n = 2 level, the order (or degree) of the degeneracy (i.e., thenumber of distinct eigenfunctions with the same n) is equal to 4, since we have


one state for 𝓁 = 0 and three states for 𝓁 = 1. Similarly, the order of the degen-eracy for the n = 3 level is equal to 9, since we should now add the five statescorresponding to 𝓁 = 2. For arbitrary n, the order of degeneracy is equal to

dn =n−1∑

𝓁=0(2𝓁 + 1), (9.63)

since for each n, the number 𝓁 takes the values 𝓁 = 0,… , n − 1, and for each 𝓁there exist 2𝓁 + 1 states corresponding to distinct values of the quantum numberm (remember m = −𝓁,… ,𝓁). Now, since the right-hand side of Eq. (9.63) is thesum of an arithmetic progression, it is equal to the half-sum of the first and lastterms, multiplied by the number of terms. The result is

dn = n2. (9.64)

What is the reason for this degeneracy? In general, degeneracy is related to acertain symmetry (or symmetries) of the problem. The higher the symmetry, thegreater the degeneracy of the energy spectrum. In this case, the problem has anobvious spherical symmetry,9 since the Coulomb potential is a central potentialfor which all spatial directions are equivalent. It is thus reasonable to expect allstates 𝜓n𝓁m that differ only in the value of the quantum number m to have thesame energy, since the various m values correspond to different orientations ofthe angular momentum vector in space, which clearly cannot affect the electron’senergy. There is thus a general conclusion we can draw here: Due to the rotationalsymmetry of any central potential, its energy eigenvalues depend on the quantumnumbers n and 𝓁, but not on m. In other words, the order of the degeneracy is2𝓁 + 1, which is equal to the number of eigenfunctions 𝜓n𝓁m corresponding todifferent values of m for a given 𝓁. Because it stems from the spherical symmetryof the system, this rotational degeneracy, as it is called, will be present in all centralpotentials, as we explained.

But the energy eigenvalues of the hydrogen atom are also independent of 𝓁 (inaddition to m). Therefore, there are now states of different𝓁 with the same energy!This additional degeneracy, known as accidental or hydrogenic degeneracy, is aspecial feature of the Coulomb potential and constitutes the quantum analog ofanother special property of this potential in classical mechanics: It has closednoncircular orbits, much like the elliptic orbits of planets in the sun’s gravitationalfield. The existence of such orbits is due to the fact that in a Coulomb poten-tial, V = −g∕r, there is yet another quantity that is conserved—besides angularmomentum, whose conservation is due to the spherical symmetry of the problem.Specifically, the so-called Lentz vector

M = 𝒗 × 𝓵 + g rr

(9.65)

is conserved, and a direct consequence of its conservation is that the major axisof a possible elliptical orbit stays constant in space, and, therefore, the orbitrepeats itself (i.e., it is closed). When there exist conserved quantities that arenot related to a geometric symmetry of the problem (e.g., rotational symmetry)

9 The terms spherical and rotational symmetry are equivalent.


but result from the specific form of the potential, we say there is a dynamicalsymmetry, and we call the corresponding degeneracy dynamical. The accidentaldegeneracy in the hydrogen atom is thus just a special case of a dynamicaldegeneracy.

9.3.3.3 Removal of Rotational and Hydrogenic DegeneracyThe previous discussion has some important practical implications. Since weconcluded that the degeneracy with respect to m is due to rotational symme-try, and the degeneracy with respect to 𝓁 is due to the coulombic character ofthe atomic potential, we are also led to conclude that when these conditions nolonger apply, the corresponding degeneracies are removed. For example, we canbe confident that when the hydrogen atom is placed in a magnetic field, whichclearly destroys the rotational symmetry of the problem, the rotational degen-eracy will be removed and the states of different m will have different energy.Indeed, this is the well-known Zeeman effect, which we will study in detail in thefollowing chapter. In Chapter 12 we will also see that in many-electron atoms,each electron can be regarded as if it moves in an effective potential that incorpo-rates the attraction from the nucleus and the repulsions from all other electrons.Since this effective potential is no longer purely Coulombic, we should expectthe removal of hydrogenic degeneracy and the creation of n different energy lev-els for any given n. Thus, in many-electron atoms, the energy eigenvalues dependon both n and 𝓁. We conclude that there exists a direct connection between thesymmetry of the problem and the degeneracy of its energy spectrum. Removingthis symmetry results in the removal of the corresponding degeneracy. We thushave the following cause-and-effect relationship:

Symmetry −−−−→ degeneracyRemoval of symmetry −−−−→ removal of degeneracy.

9.3.3.4 The Ground State is Always Nondegenerate and Has the FullSymmetry of the ProblemThe above statement is actually a general theorem, applicable to all quantumsystems with a finite number of particles. Clearly, the theorem holds for thehydrogen atom, whose ground state is indeed nondegenerate and has fullrotational symmetry, since it does not depend on the angles 𝜃 and 𝜙, but onlyon r. From a physical perspective, the absence of degeneracy in the groundstate has a profound implication for our world. Had the ground state of itsatom been degenerate, hydrogen would no longer have a definite chemicalbehavior. It would participate in various chemical compounds with a multi-tude of forms, whose number would actually be infinite, since every linearcombination of degenerate states is also a viable physical state with the sameenergy.

The fact that the ground state always has the same symmetry as the potentialis also plausible from a physical perspective. Because nonsymmetrical wavefunc-tions have more pronounced spatial variations, their kinetic energy is greater thanthat of wavefunctions with higher symmetry. Clearly, therefore, the state of min-imum energy should also be a state of maximum symmetry.


Table 9.4 Spectroscopic notation for various atomic states

𝓵 = 0 𝓵 = 1 𝓵 = 2 𝓵 = 3 𝓵 = 4 𝓵 = 5

s p d f g h

9.3.3.5 Spectroscopic Notation for Atomic StatesFor the remainder of our discussion, it would be useful to introduce an alternativenotation for atomic states. This widely used convention employs letters of thealphabet to denote states of definite quantum number 𝓁, as shown in Table 9.4.

The letters s, p, d, f , and so on, derive from standard spectroscopy terminology,where s stands for “sharp,” p for “principal,” d for “diffuse,” f for “fundamental,”and so on. For states of given n and 𝓁, the established convention is to write thequantum number n, followed by the spectroscopic symbol for the quantum num-ber 𝓁. For example, we denote the ground state of the atom (n = 1,𝓁 = 0) as 1s;the state with n = 2 and 𝓁 = 0 as 2s; the state n = 2, 𝓁 = 1 as 2p; and so on.

9.3.3.6 The “Concept” of the Orbital: s and p OrbitalsLet us clarify right away that the orbital is not a new concept (which is why weused quotes). It is merely another word to describe the quantum state of an atom,or, equivalently, the wavefunction that describes this state. Actually, the term isinteresting from a psychological perspective! On one hand, it is reassuringly rem-iniscent of the familiar concept of the classical orbit. At the same time, the ending“-al” reminds us that we are not dealing with a classical orbit here, but with itsquantum analog. In practice, we typically use the term orbital to refer to a stan-dard two-dimensional sketch, which pictures, with appropriate shading, a regionin space where the wavefunction has non-negligible values and where the particleis thus most likely to be found.

The fact that we speak of s orbitals, p orbitals, and so on, implies that the quan-tum number 𝓁 is the key factor responsible for an orbital’s shape. Indeed, for eachs-type orbital (1s, 2s, 3s, etc.), the wavefunction of the electron depends only on r,so the corresponding orbital has a spherically symmetric shape. Conversely, forall orbitals with 𝓁 ≠ 0 (i.e., p, d, f , etc.), the corresponding wavefunctions haveangular dependence and therefore their shapes in space are no longer spherical,but show a strong preference toward certain directions. Note also that the stan-dard sketches of atomic orbitals depict the probability amplitude—that is, thewavefunction itself—and not the corresponding probability density. We wouldtherefore normally expect these orbitals to show positive and negative signs, inorder to denote those regions where the probability amplitude is positive or neg-ative; but this is usually omitted in practice. The reason for choosing to sketchthe probability amplitude, and not the probability itself, is to be able to use thesesketches as probability waves that can be superimposed and exhibit interferenceeffects, just like classical waves. We will discuss such superpositions when weexamine the quantum theory of chemical bonds, especially carbon bonds. As wewill see then, the phenomenon of hybridization is a direct result of the superpo-sition of atomic orbitals.


= = 1 –

1s orbital

e– r e– r/211s

1 r22 2ππ

2s orbital

2sψψ

Figure 9.6 1s and 2s orbitals. Both orbitals are spherically symmetric, but the 2s orbital has a“rarefied probability” in the vicinity of the node, r = 2, of the corresponding wavefunction, andit changes sign across this node. But such differences among ns orbitals (n = 1, 2, 3,…) in howrarefied they locally are, are secondary compared to their common feature of sphericalsymmetry. It is thus customary to graphically depict all ns orbitals using only the sketch of theirfirst member: the 1s orbital.

We proceed to sketch some typical orbitals, starting with the 1s and 2s orbitalsshown in Figure 9.6.

Let us now look into p orbitals. We begin with 𝜓210, whose wavefunction

𝜓210 = 14√

2𝜋re−r∕2 cos 𝜃 (9.66)

is real and can be sketched as described earlier. If we ignore momentarily theangular factor cos 𝜃 of (9.66), the radial part re−r∕2 would look like the 1s orbitalbut with a “hole” in the middle, due to the factor r that produces a probabilityrarefaction near the origin. If we now allow the factor cos 𝜃 to “act” on the sketch,it will produce an angular modulation of its density, enhancing the regions nearthe z-axis and weakening those away from it. Thus, only two probability lobes10

around the z-axis survive in the end. These lobes have opposite sign, since in thepositive semiaxis, cos 𝜃 is equal to +1, while in the negative semiaxis, it is equalto −1 (Figure 9.7).

Notice now that the wavefunction 𝜓210 ≡ 𝜓2pz= Nre−r∕2 cos 𝜃 can also be

written as

𝜓2pz= Nze−r∕2. (9.67)

Since there is nothing special about the z-axis, it is clear that if we substitute zwith x or y, the wavefunctions we obtain

𝜓2px= Nxe−r∕2, 𝜓2py

= Nye−r∕2 (9.68)

will also be solutions of the Schrödinger equation, with the same energy as (9.67)and the same magnitude of angular momentum, since both quantities, H and𝓵2, are invariant under rotation and thus insensitive to our choice of axes. Theonly difference between states (9.67) and (9.68) pertains to the projection of their

10 It should be clear that the term probability is to be interpreted here as probability amplitude.More accurate expressions such as “rarefaction of probability amplitude,” “probability amplitudelobes,” and so on, are avoided, as they are rather cumbersome.


ψ2pz = Nr*e(–r/2) × cos θ =

z z z

cos θ×

+

–

= ≡

Figure 9.7 Construction of the 2pz orbital from its wavefunction. Due to the cos 𝜃 factor, onlytwo lobes (of opposite sign) survive around the z-axis. Thus, the orbital has strongdirectionality along the z-axis, a fact denoted by the index z in 2pz . All other states with 𝓁 = 1and m = 0 (but with greater n) have the same general form, but with more lobes on each sideof the z-axis. Again, since the dominant feature is the strong directionality around the z-axis(the factor cos 𝜃 is the same in all cases), it is customary to graphically represent all suchorbitals with the 2pz orbital, and use the notation pz without reference to the quantumnumber n.

angular momentum, where the physical meaning stays the same, but refers to adifferent axis. In other words, (9.67) represents an electronic state with 𝓁z = 0,while (9.68) represent states with 𝓁x = 0 and 𝓁y = 0, respectively.

Let us now examine how the states (9.68) relate to the states

𝜓21±1 = Nre−r∕2 sin 𝜃e±i𝜙,

which correspond to a specified projection 𝓁z = ±ℏ on the z-axis. First, considerthat

x = r sin 𝜃 cos𝜙 = r sin 𝜃 ei𝜙 + e−i𝜙

2from where we obtain

𝜓2px= 1

√2

(𝜓211 + 𝜓21,−1

)

and similarly,

𝜓2py= 1

i√

2

(𝜓211 − 𝜓21,−1

)

where the factor 1∕√

2 was introduced for normalization,11 and the complex ito make 𝜓2py

real. The states (9.68) are thus just suitable linear combinations of𝜓21±1, with the added feature of being real functions, which, together with (9.67),form a triplet of states with the same physical meaning with respect to the axesx, y, z.

Therefore, for 𝓁 = 1, we have three orbitals (px, py, pz), which have the sameshape as pz, but point along the respective x, y, and z axes (Figure 9.8).

Finally, let us note here another difference between s orbitals and all others,besides the lack of directionality. For s orbitals the wavefunction never vanishesat the origin, while for all other orbitals (p, d, f , etc.) it vanishes with an r𝓁 behav-ior, which becomes more intense as the electronic angular momentum grows.

11 We remind the readers that the normalization factor N for any linear combination oforthonormal eigenfunctions 𝜓 = N(c1𝜓1 + · · · + cn𝜓n) is equal to N = (|c1|

2 + · · · + |cn|2)−1∕2.


Figure 9.8 The triplet of px , py , pz orbitals.

x

y

zpz

py

px

From a physical perspective, this feature makes sense: The greater the angularmomentum of an electron, the greater its mean distance from the nucleus and,therefore, the lower the probability to locate the electron near the origin.

9.3.3.7 Quantum Angular Momentum: A Rather Strange VectorPerhaps the most impressive result from the solution of the hydrogen atom (orany central potential for that matter) is the quantization of the electron’s angularmomentum. Both the magnitude of this vector and its projection onto a certainaxis are quantized quantities with allowed values

|𝓵| = ℏ√𝓁(𝓁 + 1) , 𝓁 = 0, 1, 2,…

and

𝓁z = ℏm, m = −𝓁,… ,+𝓁

where |𝓵| and 𝓁z are used again as classical symbols of the correspondingquantities and not as quantum operators. It is instructive to compare (linear)momentum and angular momentum from the perspective of quantization. Bothare vector quantities, but momentum (both its magnitude and its components)takes continuous values, while angular momentum is quantized. Why thisdifference? The most basic reason is that for momentum, the range of valuesavailable to the pertinent variables (x, y, or z) is unlimited (spanning the wholerange from −∞ to +∞), while for angular momentum, the relevant variablesare angles that vary within a finite range. It is this constraint that leads toquantization, as the readers can verify by studying the eigenvalue problems fortwo representative quantities, for example, pz and 𝓁z.

At a more mathematical level, the cause of this fundamental difference liesin the fact that, while the components of the momentum vector commutewith each other, those of the angular momentum do not (see the discussionin the last section of Chapter 3). In fact, angular momentum quantization canemerge directly from the commutation relations of its components, for example,[𝓁x,𝓁y] = iℏ 𝓁z, and so on, without making any reference to the specific form ofthe operators 𝓁x,𝓁y,𝓁z.

Even though the use of classical pictures for quantum results is a rather pre-carious practice, it can be quite useful when our aim is not to present a quantumeffect as if it were classical, but to bring out the fundamental differences between


z

0

==

z

ℏ

2 ℏ6 ℏ

–2ℏ

2ℏ

–ℏ

0

ℏ

–ℏ

ℓ ℓ

ℓ = 1 ℓ = 2

Figure 9.9 The quantized vector of angular momentum for 𝓁 = 1 and 𝓁 = 2. Due to thequantization of its projection onto a particular axis, the angular momentum “vector” (to theextent that depicting it as a vector has a meaning) can only take discrete orientations in space.We call this effect “quantization of orientation” or “space quantization.” Note that the vector 𝓵is never aligned with the z-axis, which is a rather thought-provoking feature. Even when theprojection of the vector onto the z-axis is maximized, it forms a nonzero angle with it, since wealways have (𝓁z)max = ℏ𝓁 < |𝓵| = ℏ

√𝓁(𝓁 + 1).

the two. In Figure 9.9 we present such a classical picture for the quantized vectorof angular momentum.

As the caption of Figure 9.9 indicates, apart from the quantization of orienta-tion, the most nonclassical feature of the quantum angular momentum vector isthat it is absolutely impossible to align it on an axis, since the greatest value ofits projection onto the axis (𝓁z)max = ℏ𝓁 is always smaller than the magnitude|𝓵| = ℏ

√𝓁(𝓁 + 1) of the vector 𝓵. As the readers may suspect, this impossibility

stems from the fact that the components of angular momentum (in contrast tothose of momentum) do not commute (see Section 3.5.2), and cannot thereforebe simultaneously measured with absolute precision. So, even though the com-ponents 𝓁x and 𝓁y, have vanishing mean values in a given state 𝜓n𝓁m, they alsohave strong quantum fluctuations—that is, ⟨𝓁2

x⟩ = ⟨𝓁2y ⟩ ≠ 0—that prevent them

from being strictly zero, which makes full alignment with an axis impossible.Let us also add that, in the limit of large quantum numbers 𝓁, the quantum

peculiarities of the vector 𝓵 subside and are gradually replaced by the classi-cal behavior. For example, since for large 𝓁 we have |𝓵| = ℏ

√𝓁(𝓁 + 1) → ℏ𝓁 ≈

(𝓁z)max, the vector 𝓵 will indeed be aligned with the z-axis in the state of maxi-mum projection onto this axis. Furthermore, the relative discontinuities in𝓁z (i.e.,the difference between successive values of 𝓁z, divided by the magnitude of 𝓵)

Δ𝓁z

|𝓵|= ℏ

ℏ√𝓁(𝓁 + 1)

−−−→𝓁→∞

0


become gradually smaller and vanish completely in the classical limit 𝓁 → ∞.However strange the quantum world is (the behavior of the angular momentumvector being among its most extreme peculiarities), it eventually finds a way torecover classical normality in the appropriate limit.

9.3.3.8 Allowed and Forbidden Transitions in the Hydrogen Atom:Conservation of Angular Momentum and Selection RulesDespite the rather large energy gap between the ground and excited statesof the hydrogen atom, it is, of course, possible to excite the atom, either viahigh-temperature thermal collisions, or through irradiation with photons ofsufficiently high energy. What does the electron do when it reaches an excitedstate? This may seem an odd question to ask, since the answer is well known:Surely the electron will move to lower-energy states, or directly to the groundstate, by shedding its extra energy in the form of a photon. This is Bohr’s answer,on the basis of which he was able to explain remarkably well the spectrum of thehydrogen atom. And yet this general answer is not only insufficient, it can alsobe wrong, unless it is accompanied by some additional rules that distinguishbetween those quantum transitions that may be realized (allowed transitions)and those that may not (forbidden transitions), at least within the limits ofsome approximation. Just like in the harmonic oscillator, these selection rules,as they are known, are given in the form of allowed changes in the quantumnumbers involved in the quantum mechanical description of the physical systemof interest. For the hydrogen atom, the selection rules that define the allowedtransitions are

Δ𝓁 = ±1, Δm = 0,±1, Δn = arbitrary, (9.69)

where the second rule is a rather plausible consequence of the first one. There-fore, the allowed de-excitations of the electron are those for which the quantumnumber of the magnitude of the angular momentum changes by one, while theprincipal quantum number n is under no constraint.

In Figure 9.10 we apply the abovementioned rule, showing the allowed transi-tions with solid lines, and the forbidden ones with dashed lines.

Figure 9.10 Allowed and forbiddentransitions in the hydrogen atom.Allowed transitions satisfy the selectionrule Δ𝓁 = 1 and are shown with solidlines, while forbidden transitions do notobey the above rule and are shown withdashed lines. The figure shows onlytransitions originating from the 2s, 2p,3p, and 3d levels.

3s

2s

1s

3p 3d

2p

Δℓ

= 0

Δℓ

= 1

Δℓ = 1

Δℓ = 1Δℓ = 2

Δℓ

= 1

Δℓ =

2


Although this is a more complex topic (which we will examine from first prin-ciples in Chapter 16), there is actually a simple physical consideration of theselection rules (9.69) that renders them rather plausible. The key idea is angu-lar momentum conservation, combined with the fact that the photon is a carrierof intrinsic angular momentum (the so-called spin) with quantum number s = 1.Since the photon emitted from a de-excitation carries with it a unit of angularmomentum (its spin), the angular momentum of the atom is expected to changeby the same amount, so that Δ𝓁 = 1.

In view of this interpretation, it becomes clear, for example, that the transition2s→ 1s cannot be realized, as it would surely violate angular momentum conser-vation. The atom has zero angular momentum both in its initial (2s) and final (1s)state, and, as a result, the emission of a photon with angular momentum (≡spin)equal to unity cannot happen. But then the question arises: What will an electrondo once it occupies the 2s state? Will it stay there forever, since its de-excitationis deemed impossible? If this really were the answer, its implications for chem-istry would be spectacular. Because we would then have two hydrogen atoms: thesmall atom and the large one. That is, the atom in the 1s state, and the atom withan electron “trapped” in the 2s state that is unable to climb down from there! Onecan only imagine the great variety of chemical compounds that would be possi-ble if each hydrogen atom of the existing compounds came in either a “small” ora “large” variety!12

Of course, nothing of this sort happens, since hydrogen does not stay in the2s state forever. It has a mean life of 0.12 s in this state, after which it emits twophotons of opposite spin, as shown in Figure 9.11.

This discussion contains a bigger message. What we regard as “forbidden tran-sitions” are not actually completely forbidden. They are simply much harder totake place because they require mechanisms with a much smaller probability tooccur than the mechanism that corresponds to allowed transitions. We revisitthis important topic in Chapter 16.

In closing, let us stress again the importance of the hydrogen atom as a quantumsystem that is both very simple and representative of the quantum world. It is noexaggeration to say that to understand the hydrogen atom well is to make a bigstep toward the complete understanding of quantum laws.

2s

1s

– ℏ + ℏ

Figure 9.11 The “forbidden” transition 2s→ 1s. Thetransition is actually possible by emission of two photons ofopposite spin (±ℏ), so as not to violate angular momentumconservation. But since the emission of two photons is a“difficult” process, the lifetime of the electron in the 2s stateis huge by atomic measures: approximately 0.12 s!

12 Plausible objections to these arguments (e.g., de-excitation through collisions or via thedegenerate 2p state) need not be discussed here.


Problems

9.9 Verify that the lowest-energy level for a given 𝓁 is always higher that theminimum of the respective effective potential V𝓁(r).

9.10 Prove that the Lentz vector given in (9.65) is indeed a conserved quantityfor a particle moving in a Coulomb potential of the form V = −g∕r.

9.11 What is the “angle”—if there is a meaning for it in quantum mechanics—between the angular momentum vector and the z-axis in the state ofmaximum m?

Further Problems

9.12 A particle of mass M is constrained to move on a circle of radius a.Calculate the allowed values of its energy and sketch the correspondingenergy-level diagram, showing also the degeneracy of each level.

9.13 Solve the same problem as above, but with the particle moving freely onthe surface of a sphere of radius a.

9.14 The state of a particle of mass M that moves on the surface of a sphere withradius a is described, at a certain moment, by the wavefunction

𝜓(𝜃, 𝜙) = N(1 + cos 𝜃). (1)

(a) What is the probability to find the particle in the “northern hemi-sphere?”

(b) What are the allowed energy values of the particle, and their corre-sponding probabilities of being measured?

(c) What is the form of the wavefunction (1) after the lapse of time t?

9.15 Construct, for an arbitrary value of the quantum number n, the hydro-gen eigenfunction corresponding to the maximum possible value of thequantum numbers 𝓁 and m. Use your result to explain how the quantum“probability cloud” transforms into an “orbital tube” in the limit of large nvalues. How does this tube relate to the quantized orbits in Bohr’s theory?

9.16 Calculate the uncertaintiesΔx,Δy, andΔz in the ground state of the hydro-gen atom. You can easily perform this calculation if you use an appropriatesymmetry argument.

9.17 Calculate the mean values ⟨𝓁2x⟩ and ⟨𝓁2

y ⟩ in an arbitrary state of the hydro-gen atom. (There is a clever way of quickly performing both calculations.)Based on your result, what do you conclude regarding the inability of theangular momentum vector 𝓵 to align with the z-axis?


9.18 Find the eigenvalues and eigenfunctions of the following operators:(a) 𝓁2

x + 𝓁2y ,

(b) 𝓁2x + 𝓁2

y − 𝓁4z .

9.19 Consider a quantum system that has only rotational degrees of freedom(e.g., a rigid rotor). Calculate the allowed energy levels of such a system ifits Hamiltonian is given by one of the following expressions:(a) H = 𝛼𝓵2 + 𝛽𝓁z,(b) H = 𝛼𝓵2 + 𝛽(𝓁x + 𝓁y + 𝓁z).

9.20 A particle of mass m is subject to a central force F = −kr, where r is itsdistance from the attractive center located at the origin. Calculateits allowed energy values and give expressions for the eigenfunctionsof its ground and first excited states. The same potential, knownas the three-dimensional harmonic oscillator, can also be solved inCartesian coordinates, with a trivial reduction to the correspondingone-dimensional problem. Solve it also with this method, and comparethe results from both methods.

267

10

Atoms in a Magnetic Field and the Emergence of Spin

10.1 Introduction

Atoms do not “live alone” in space: They are constantly under the influence ofexternal fields, namely, electromagnetic force fields, since all other fundamentalforces (strong, weak, and gravitational) are overwhelmingly weaker on atomicscale. Among the electromagnetic forces that an atom experiences are thosedue to collisions with other atoms, which are mainly electric forces between thecharged constituents (electrons and nuclei) of the colliding atoms.

Thus, to completely understand the atom as a fundamental quantum system,we must also study how it is affected by external electromagnetic fields. Suchfields can be static (i.e., constant in time) or time dependent. The latter categoryincludes electromagnetic waves (microwaves, visible light, ultraviolet, x-rays,etc.), whose interaction with atoms typically results in transitions between theground state and an excited state. Whether these transitions can occur, and withwhat likelihood, is the subject of the theory of quantum transitions, which isactually a theory for the interaction between light and matter. In Chapter 16, wewill introduce the theory of quantum transitions, together with an approximatemethod for pertinent calculations (the so-called time-dependent perturbationtheory).

The much simpler category of static fields involves static electric fields or staticmagnetic fields. Actually, for our purposes, these fields can also be treated ashomogeneous in space, that is, independent of any spatial variables. The reasonis that any spatial variations of an external field (electric or magnetic) can hardlybe sensed by an atomic electron confined in the tiny space of an atomic volume.So, on an atomic scale, all external static fields can also be regarded as homoge-neous. Therefore, to study how static fields interact with atoms, we need to studythe following two problems in particular.

PROBLEM 1: The atom in a static and homogeneous electric field (Stark effect).

PROBLEM 2: The atom in a static and homogeneous magnetic field(Zeeman effect).

As we may expect, in both these problems the external field shifts the atomicenergy levels, so it must also shift its emission or absorption lines. Such shifts


268 10 Atoms in a Magnetic Field and the Emergence of Spin

were first observed, using spectroscopic techniques, by J. Stark (1913) andP. Zeeman (1895), respectively; hence, the corresponding effects are namedafter them.

In view of the present chapter’s title, readers may wonder why we focus hereon the second problem (Zeeman) instead of the first one (Stark). Actually,the Zeeman problem is easier to solve than the Stark problem. As we willshortly see, when we place a hydrogen atom in a (static and homogeneous)magnetic field, its wavefunctions remain unchanged, while its energy eigenvaluesare shifted by certain amounts that are readily calculated and are known asZeeman shifts.

In contrast, there is no simple solution for the electric field problem, wherethe calculation of the corresponding Stark shifts can only be done by employingapproximate methods (e.g., perturbation theory), which will be presented later inthe book and in the online supplement.

But there are more substantive reasons why the problem of an atom in a mag-netic field (the Zeeman problem) is much more interesting than the Stark prob-lem. The main reason is that an atomic electron, being a moving charged particle,behaves as a microscopic magnet. It has a magnetic moment 𝝁 that can be shownto be proportional to its angular momentum 𝓵 (that is, 𝝁 = 𝛾𝓵) with a pro-portionality coefficient 𝛾 = −e∕2mc. (Or, 𝛾 = −e∕2m in SI units.) Thus, when,say, a hydrogen atom is placed in a magnetic field B, its energy changes by thewell-known term1

𝑈 = −𝝁 ⋅ B = −𝛾𝓵 ⋅ B, (10.1)

which tells us something important: The magnetic field interacts directly withthe electronic angular momentum, and is thus a prime “tool” for measuring it(in the sense of being sensitive to it). Here is the measurement “procedure.” Tak-ing 𝓵 in expression (10.1) to be the electron’s angular momentum, we calculatehow this term affects the atomic energy levels (i.e., we calculate the Zeemanshifts) and compare with the shifts extracted from experimental spectroscopicdata. We can thus verify experimentally the validity of the quantum theory ofangular momentum as an integral part of the quantum mechanical picture of theatom.

Actually, the Zeeman shifts caused by (10.1) can be calculated in just one line.Indeed, if we assume the magnetic field to point along the z-axis, we can rewrite(10.1) as

𝑈 = −𝛾B𝓁z. (10.2)

Now, since the eigenfunctions 𝜓n𝓁m for an arbitrary central potential are alsoeigenfunctions of 𝓁z with eigenvalues ℏm, the term (10.2) simply adds to theatom’s initial eigenvalues En𝓁 the quantities −𝛾Bℏm, where m is the quantumnumber of the z component of the angular momentum. The new eigenvaluesbecome thus

E′n𝓁m = En𝓁 − 𝛾 ℏBm, (10.3)

1 This term is the magnetic analog of 𝑈 = −d ⋅ , which describes the interaction energy betweenan electric dipole with dipole moment d and an electric field .

10.1 Introduction 269

where we took into account the fact that the initial eigenvalues En𝓁 depend also onthe quantum number 𝓁, as is the norm for an arbitrary central potential. Note,by the way, that the naming of m as a magnetic quantum number is now fullyjustified. The significance of this quantum number emerges when we place theatom in a magnetic field.

The result (10.3) tells us that under the influence of a magnetic field, the rota-tional degeneracy is lifted, and each level of a given 𝓁 is replaced by 2𝓁 + 1 dif-ferent levels, whose energies depend on the quantum number m, as shown inFigure 10.1.

Thus, when we place atoms in a magnetic field, the single spectral line associ-ated with the transition of the electron from an 𝓁 = 1 level to an 𝓁 = 0 level isreplaced by three lines, since the 𝓁 = 1 level itself splits into three levels (Zeemansplitting), while the 𝓁 = 0 level remains unchanged. In other words, the quanti-zation of angular momentum is directly reflected in the Zeeman effect.

This discussion has another important implication. If, in addition to its orbitalangular momentum, the electron were to also have an intrinsic angular momen-tum—which is, actually, the electon’s spin—then this would readily register in theZeeman spectrum of the atom. So, if we wanted to discover spin on our own, thenthe first experiment that comes to mind is to study the spectrum of an atom in amagnetic field.

The outline of this chapter begins to take shape. Our first step is to support thenotion of the atomic electron as a magnet, and prove the basic relation 𝝁 = 𝛾𝓵by calculating the value of the gyromagnetic ratio 𝛾 . Next, we will elaborate onthe Zeeman effect (formulas (10.1), (10.2), (10.3), and Figure 10.1) and comparethe theoretical prediction with the experimental data. Even though this compar-ison may seem disappointing at first, it will lead to a fundamental discovery: theelectron spin. Then, we will reaffirm this discovery with an experiment specifi-cally designed for this purpose (utilizing a magnetic field again). Finally, we willpresent an introduction to the quantum theory of spin, which we will developfurther in the online supplement of this chapter.

E0

B = 0 B ≠ 0

m = –1

m = 0𝓁 = 1

m = 1

(γ > 0)

E0 + h γ B

E0 – h γ B

E0

Figure 10.1 Lifting of the rotational degeneracy when an atom is placed in a magnetic field.A single 𝓁 = 1 level gives rise to a triplet of levels that are symmetrically arranged with respectto the initial level (Zeeman splitting).


10.2 Atomic Electrons as Microscopic Magnets:Magnetic Moment and Angular Momentum

As we saw earlier, atomic electrons behave as microscopic magnets, since, beingcharged particles in motion, they produce a magnetic field. And as we know,the basic quantity we need in order to describe the magnetic field of an arbi-trary, localized current distribution is its magnetic moment. If we know the mag-netic moment of the current distribution, we can calculate its magnetic field atdistances large enough compared to the spatial extent of the distribution. Thisso-called “far field” resembles closely the field of an ideal magnetic dipole, whichin turn looks the same as the electric field from an ideal electric dipole. Tra-ditionally, it used to be customary to depict magnetic dipoles as bar magnets,while in the modern theory of magnetism it is more common to depict them ascurrent-carrying loops, as shown in Figure 10.2.

If we wanted to follow closely the analogy with the electric dipole, we wouldpicture a bar magnet as a dipole with negative and positive “magnetic charges”concentrated in its south and north pole, respectively. We would also define themagnetic moment of this dipole as a vector that points from the south to thenorth pole (i.e., from the negative to the positive “charge”) and has a magni-tude equal to the product of the “charge” times the distance between the poles(i.e., the length of the magnet). But in the modern picture of the current-carryingloop, we define the magnetic moment as a vector perpendicular to the plane of

(a) (b)

μ μμ = qm a

N

Sa

μ = i A

i

c

Figure 10.2 The magnetic dipole in its traditional and modern depiction (bar magnet andcurrent-carrying loop, respectively). (a) If magnetic charges existed, we would define themagnetic moment as a vector that points from the south to the north pole (inside the magnet)and has a magnitude 𝜇 = qma, where ±qm are the magnetic charges at the edges of the dipoleand a is the dipole length. (b) The magnetic moment is perpendicular to the plane of thecurrent-carrying loop and has a magnitude 𝜇 = iA∕c in cgs units—𝜇 = iA in SI—where i is thecurrent and A is the surface area of the loop.

10.2 Atomic Electrons as Microscopic Magnets: Magnetic Moment and Angular Momentum 271

the loop, with a magnitude equal to

𝜇 = iA (SI) (10.4)

or

𝜇 = iAc

(cgs), (10.5)

where i is the current and A the surface area of the loop.We make a note here that the factor 1∕c arises when we recast formula (10.4)

from SI into cgs units, since in both systems the expression −𝝁 ⋅B describesenergy, but the fields B in the two systems are related via a factor of c. In otherwords, we have Bcgs = cBSI, as we can deduce by comparing the correspondingexpressions for the Lorentz force in the two systems

F = q𝒗 × B (SI), F = q𝒗c× B (cgs).

Now, since we must have

𝜇cgs ⋅ Bcgs = 𝜇SI ⋅ BSI ,

it follows that

𝜇cgs =1c𝜇SI .

Based on the above discussion, we shall now prove that the magnetic moment 𝝁produced when a charged particle of mass m and charge q moves along a closedorbit is related to its angular momentum, 𝓵, via the expression

𝝁 =q

2mc𝓵. (10.6)

The proof is straightforward for the case of the circular orbit shown in Figure 10.3.Given that the vectors 𝝁 and 𝓵 are clearly parallel in this case, we only need to

prove relation (10.6) with regard to their magnitudes. The following successiveequalities hold

𝜇 = iAc

=qT

⋅𝜋r2

c=

q(2𝜋r∕𝑣)

⋅𝜋r2

c=

q2c𝑣r

=q

2mc(m𝑣r) =

q2mc

𝓁, (10.7)

Figure 10.3 Relation between magnetic moment 𝝁 andangular momentum 𝓵 for a charged particle. The vectors𝝁 and 𝓵 are parallel and 𝝁 = (q∕2mc)𝓵.

μ

υ

r

q, mq

2mc=

𝓁

μ 𝓁


where we used the fact that the motion of a particle of charge q along a circu-lar orbit corresponds to a current i = q∕T , where T = 2𝜋r∕𝑣 is the period of itsmotion, r the radius of the orbit, and 𝑣 the speed. Since 𝝁 and 𝓵 are collinear,(10.7) clearly proves (10.6).

In more advanced textbooks, (10.6) is proven directly at a quantum mechani-cal level, without relying on the shape or even the notion of the particle’s orbit,which is meaningless in quantum mechanics anyway. In light of this proof, (10.6)describes a proportionality relation between the dynamic quantities of magneticmoment and angular momentum. The proportionality coefficient is known as thegyromagnetic ratio, where the term gyro (Greek for “round”) relates to angularmomentum and the term magnetic to the accompanying magnetic moment. Forthe electron in particular, its gyromagnetic ratio 𝛾 = q∕2mc (q = −e) is negativeand (10.6) is thus written as

𝝁 = − e2mec

𝓵, (10.8)

where me denotes the electron mass, to avoid confusion with the magnetic quan-tum number m.

One direct consequence of (10.6) (or (10.8) for that matter) is that the magneticmoment of (charged) particles is quantized, similarly to their angular momentum.For example, it follows from (10.8) that the projection of the electronic magneticmoment onto the z-axis is

𝜇z = − e2mec

𝓁z = − e2mec

ℏm = −𝜇Bm, (10.9)

where 𝜇B is the quantity

𝜇B = eℏ2mec

, (10.10)

which is known as the Bohr magneton, and is the established unit for the mea-surement of atomic magnetic moments. According to (10.9), a component of themagnetic moment of the electron can only take values that are integer multiplesof the Bohr magneton. In other words, 𝜇B is the quantum of magnetic moment, inthe same manner that Planck’s constant ℏ is the quantum of angular momentum.No measurement can ever yield an electronic magnetic moment smaller than 𝜇B,just as no measurement can yield a nonzero orbital angular momentum smallerthan ℏ.

The magnitude of the vector𝝁 is also quantized, as can be seen from the relation

|𝝁| = 𝜇B

√𝓁(𝓁 + 1),

which is the exact analog of

|𝓵| = ℏ√𝓁(𝓁 + 1),

with the Bohr magneton in the place of Planck’s constant. However, we shouldstress that while Planck’s constant (the quantum of angular momentum) is thesame for all particles, the Bohr magneton (the quantum of magnetic moment) isdifferent for different particles, since it depends on their mass, according to the

10.2 Atomic Electrons as Microscopic Magnets: Magnetic Moment and Angular Momentum 273

relation 𝜇B = eℏ∕2mc. In this spirit, it is customary to use the term “Bohr magne-ton” only for the electron, while for the proton we call the corresponding quantitynuclear magneton, and denote it as 𝜇N = eℏ∕2mpc.

The numerical value of the Bohr magneton is found from formula (10.10) to be

𝜇B = 0.93 × 10−20 erg/G (cgs)= 0.93 × 10−23 J/T (SI)= 5.8 × 10−9 eV/G, (practical units)

where we listed the (rounded) value of𝜇B in cgs and SI units, and in the often-usedpractical units of eV/G. To convert between different units we recall the relations

1 erg = 10−7 J, 1 eV = 1.6 × 10−12 erg, 1 G = 10−4 T.

The existence of a basic unit of atomic magnetism (this is the meaning of theBohr magneton, after all) has direct implications for the intensities of macro-scopic magnetic fields we can achieve in practice. A strong magnetic field is pro-duced mainly by the alignment of atomic magnets at the interior of a magneticmaterial when the latter is placed inside the coil of an electromagnet. So, themagnetic field B is maximized when the magnetization M of the material (i.e.,the magnetic moment per unit volume) takes its maximum value

M ≈ 𝜇B ⋅ n, (10.11)

which is realized when all atomic magnets inside a cubic centimeter of the mate-rial (let us call their number n) are aligned with the coil’s magnetic field H . Incgs units, the quantities B (total field), H (field due to external currents), and M(magnetization) are related via the expression

B = H + 4𝜋M (cgs), (10.12)

whereas we also have

B = 𝜇H,

where 𝜇, the magnetic permeability of the material, is a number much greaterthan unity. We thus have H ≪ B, so (10.12) can be approximated, using also(10.11), as

B ≈ 4𝜋M ≈ 4𝜋𝜇Bn. (10.13)

The maximum realizable value of the magnetic field is obtained from (10.13) uponsetting

𝜇B ≈ 10−20 erg/G, n ≈ 1024 atoms∕cm3

⇒ Bmax ≈ 105 G = 10 T,

where the value we used for n (≈ 1024 cm−3) follows directly from the size ofthe atoms (a ≈ a0 ≈ 1 Å = 10−8 cm), which, in turn, is a direct consequence ofquantum laws, and, in particular, the value of Planck’s constant.

Thus, the magnetic fields we can realize in a laboratory by using magnetic mate-rials can reach up to a few hundred thousand gauss (i.e., a few tens of tesla). Thislimit directly reflects the quantum nature of atomic magnetism.


10.3 The Zeeman Effect and the Evidencefor the Existence of Spin

Based on our discussion in the introduction, if we place an atom in a magneticfield B, its energy changes by the quantity

𝑈 = −𝝁 ⋅B, (10.14)

so the new Hamiltonian of the atom becomes

H′ = H − 𝝁 ⋅B, (10.15)

where H is the Hamiltonian before we applied the field and 𝝁 = (−e∕2mec) 𝓵 isthe electron’s magnetic moment. Assuming that the magnetic field points alongthe z-axis (this is a matter of convention, after all), we can write (10.15) as

H′ = H + eB2mec

𝓁z, (10.16)

whence it is evident that the eigenfunctions 𝜓n𝓁m of the original Hamiltonian Hare also eigenfunctions of H′. The corresponding eigenenergies are derived from(10.16) by substituting the operator 𝓁z with its eigenvalues ℏm. Such a substitu-tion is allowed, since 𝜓n𝓁m are eigenfunctions, not only of the initial HamiltonianH but also of the operators 𝓵 2 and 𝓁z, with eigenvalues ℏ2𝓁(𝓁 + 1) and ℏm,respectively. If we thus let H′ act on the eigenfunctions 𝜓n𝓁m, we obtain

H′𝜓n𝓁m =(

H + eB2mec

𝓁z

)

𝜓n𝓁m = H𝜓n𝓁m + eB2mec

(𝓁z 𝜓n𝓁m)

= En𝓁 𝜓n𝓁m + eB2mec

ℏm 𝜓n𝓁m

=(

En𝓁 +eB

2mecℏm

)

𝜓n𝓁m,

which confirms the assertion that the initial eigenfunctions are also eigenfunc-tions of the new Hamiltonian, with the new eigenvalues being

E′n𝓁m = En𝓁 + 𝜇BBm, (10.17)

where we used the Bohr magneton expression 𝜇B = eℏ∕2mec.Note that instead of the hydrogen atom with the Coulomb potential we consid-

ered here the more general case of an atom in an arbitrary central potential. Asa result, the corresponding energy eigenvalues E = En𝓁 depend on the quantumnumber 𝓁. Our theory can then be applied to atoms, such as the alkali atoms,which are similar to hydrogen as they have a single electron in their outer shell.The outer shell electron in such atoms experiences a central potential that isattractive but does not have the Coulomb form due to the cloud of electrons inthe inner shells that occupies the space between the nucleus and the outermostelectron.

If we now set

𝜖 = 𝜇BB,

10.3 The Zeeman Effect and the Evidence for the Existence of Spin 275

which is the so-called Zeeman energy, we can rewrite (10.17) asE′

n𝓁m = En𝓁 + 𝜖 m, m = −𝓁, · · · + 𝓁 (10.18)to obtain something we already knew: When we place an atom in a magnetic field,the rotational degeneracy is lifted, and each level of a given 𝓁 splits into a bunchof 2𝓁 + 1 levels that are symmetrically arranged with respect to the original level.Each new level is equidistant from its adjacent levels, with an energy separationequal to the Zeeman energy. One example of this Zeeman splitting for 𝓁 = 1 (butwith 𝛾 > 0) is shown in Figure 10.1, while for 𝓁 = 2 (and 𝛾 < 0) the emergingbunch of levels is shown in Figure 10.4, where the reference level E0 is taken tobe zero.

The experimental test of the above prediction is readily done using spectro-scopic measurements, as follows. The sample atoms (in this case, hydrogenatoms) are placed in a discharge tube that lies between the poles of a magnet.Then, an electrical discharge excites the atoms, causing them to emit radiationas they return to their initial state. The observation of this radiation yields thespectrum in the form of bright lines on a dark background. So, the questionfor the experimentalist is how these spectral lines are modified when we obtainthe spectrum in the presence of a magnetic field. The answer is provided inFigure 10.5, where we sketched two allowed transitions (𝓁 = 1 → 𝓁 = 0 and𝓁 = 2 → 𝓁 = 1) before and after we switch on the magnetic field.

As is evident from the figure, the splitting of the initial spectral line into a tripletis a consequence of the selection rule Δm = 0,±1. Transitions with Δm = 0 cor-respond to a vanishing frequency shift, that is, to the middle component of theline; transitions with Δm = 1 correspond to the left component; and transitionswith Δm = −1 correspond to the right component. This holds for all allowedtransitions 𝓁 → 𝓁 ± 1 (Δ𝓁 = ±1) for arbitrary 𝓁. Note also that the quantity

𝜔L = 𝜖

ℏ= eB

2mec,

B = 0 B ≠ 0

𝓁 = 2

ϵ = μBB

2ϵ

0

ϵ

– ϵ

– 2ϵ

m = 1

m = 2

m = 0

m = –1

m = –2

Figure 10.4 Zeeman splitting for a level with 𝓁 = 2. When we place the atom in a magneticfield, the rotational degeneracy of a level with quantum number 𝓁 is lifted and 2𝓁 + 1 newlevels appear, symmetrically arranged with respect to the original level. The distance betweenadjacent levels is 𝜖 = 𝜇BB.


B = 0

ω0 – ωL ω0 – ωL ω0 + ωL

ω0ω0 + ωL

m = 0

Δm

= 1

Δm

= 1 Δm

= 0

Δm

= –

1

Δm

= 0

Δm

= –

1

m = 1

m = 2

m = 1

m = 0m = –1

m = –2

m = 0

m = –1

m = 1m = 0

m = –1

ωL =ω0

B ≠ 0 B = 0 B ≠ 0

𝓁 = 1 𝓁 = 2

𝓁 = 1𝓁 = 0

ϵh

Figure 10.5 The Zeeman effect. Owing to the selection rules Δ𝓁 = ±1, Δm = 0,±1, eachspectral line becomes a triplet, with the middle line positioned at the original one and theother two lines placed symmetrically on either side. The distance between the “Zeemancomponents” of each spectral line is proportional to the magnetic field.

known as the Larmor frequency, corresponds to the classical frequency of pre-cession of the angular momentum vector for a charged particle in a magneticfield B. (The vector 𝓵 rotates with frequency 𝜔L = |𝛾|B = |q|B∕2mc on a cone ofconstant angle around the magnetic field direction.)

But is the splitting of spectral lines large enough to be experimentally observ-able? This depends on the magnitude of the Zeeman energy, 𝜖 = 𝜇BB, relative tothe energy difference between the transition levels, which is typically on the orderof a few eV. With 𝜇B = 5.8 × 10−9 eV/G, the Zeeman energy, for a typically strongmagnetic field B = 104 G, is equal to

𝜖 = 5.8 × 10−5 eV ≈ 10−4 eV,

which means that the relative shift in wavelength (i.e., Δ𝜆∕𝜆) is of the order of10−4 to 10−5 (equal to the ratio of the Zeeman energy to the energy differencebetween the two initial levels) and can be easily observed even with the technicalcapabilities of the nineteenth century! If such a high precision seems surprisingto some readers, let us simply mention that close-lying wavelengths are resolvedusing interference measurements, which are very precise indeed.

So, after all, what is the result of the “Zeeman experiment” for the hydrogenatom? The answer is simple: A complete disaster for our theoretical approach! Theexperimental spectrum of the hydrogen atom in a typical magnetic field with an

10.3 The Zeeman Effect and the Evidence for the Existence of Spin 277

intensity up to 104 G bears no resemblance whatsoever to our theoretical predic-tion. The characteristic “Zeeman triplet” is nowhere to be seen in the spectrum.Instead, spectral lines split into many more components, usually of even num-ber: 4, 6, 8, 10, and so on. For example, the first line of the Lyman series, whichcorresponds to the transition 2p → 1s, splits into 10 components, instead of theexpected 3 components that are unambiguously predicted by theory. The oddthing is that the characteristic “Zeeman triplet” appears indeed in some of thelines of the Zeeman spectrum of many-electron atoms, while the remaining spec-tral lines of those atoms split into many more components, as in the hydrogenatom (Figure 10.6). To describe this situation, experimentalists use the terms nor-mal Zeeman effect (for lines split into triplets) and anomalous Zeeman effect (forlines split into more components).

The above discussion leads inescapably to a quandary. There are two possi-ble answers for the failure of quantum theory to correctly describe the Zeemanspectrum.

Answer 1: The quantum mechanical theory of angular momentum is wrong.Answer 2: The quantum mechanical theory of angular momentum is correct, butwe have not taken into account some other factor that apparently influences theZeeman spectrum in a way similar to angular momentum.

Readers may have already guessed that we will go with the second answer. Afterall, one does not abandon a fundamental theory that has scored many successfulpredictions at the first sign of difficulty, especially when the problem that causedthe crisis (the anomalous Zeeman effect) offers clear hints toward its own resolu-tion. In other words, it is clear that the deviations from the normal Zeeman effectcan have no other origin than an additional angular momentum that couples tothe magnetic field in the same way as our familiar (orbital) angular momentum,and which has thus a similar effect in the observed spectrum. If we bring to mindthe classical planetary model again, where it is only natural for planets to haveadditional rotational motion around their own axis, then the notion of electronicspin sounds almost self-evident. Moreover, it emerges directly from the experi-mental data for the hydrogen spectrum in a magnetic field.

It is thus probable that the electron also has, apart from its orbital angularmomentum, an intrinsic angular momentum, or spin, which we could imagineclassically as the result of a rotation about its axis. In the following section, weshall describe an experiment that allows us to test this assumption unequivocally.

Figure 10.6 Three groups of spectral lines of a many-electron atom in the presence of amagnetic field. The left “line” is related to the normal Zeeman spectrum, while the other twopertain to the “anomalous” Zeeman spectrum. The distances between the original lines, as wellas the distances between the Zeeman components of each line, are shown out of scale.


10.4 The Stern–Gerlach Experiment: UnequivocalExperimental Confirmation of the Existence of Spin

As the previous discussion showed, the Zeeman experiment provides onlyindirect support for the existence of spin, and is thus in principle unsuitablefor its direct observation. Actually, the Zeeman effect is not even the idealexperiment for measuring the orbital angular momentum of the electron and itsquantization. Here is why. In the Zeeman experiment, we only measure energydifferences between quantum levels, and, in fact, only between those levels thatcan be “coupled” via allowed transitions, based on the selection rules Δ𝓁 = ±1,Δm = 0,±1. We are thus unable to directly observe the group of 2𝓁 + 1 Zeemanlevels corresponding to each 𝓁; had we been able to observe these levels, thentheir sheer number would yield at once the value of the quantum number 𝓁. Inthe Zeeman spectrum there is no such possibility, since, for all values of 𝓁, thespectral line associated with the transition 𝓁 → 𝓁 ± 1 always shows the samecharacteristic triplet.

Therefore, to directly observe spin, we need a “smarter” experiment, whichwe will present in the following section. But to better understand the basic ideabehind this experiment, it makes sense to look first for a plausible theoreticaldescription of spin. With such a description at hand, we can proceed to designan experiment that can verify or falsify the existence of spin.

10.4.1 Preliminary Investigation: A Plausible Theoretical Description of Spin

Starting from the basic conclusion of the previous section, namely, that spin is aform of angular momentum, it is plausible to assume that it can also be describedby a vector that has properties similar to the orbital angular momentum 𝓵. Letus denote this vector by s. Then, in analogy to the known quantization rules of 𝓵,namely,

𝓵 2 = ℏ2𝓁(𝓁 + 1), 𝓁 = 0, 1, 2,…

and

𝓁z = ℏm𝓁 , m𝓁 = −𝓁,… ,+𝓁,

we can assume that the following also hold:

s2 = ℏ2s(s + 1)

and

sz = ℏms, ms = −s, · · · ,+s,

where s is the quantum number of the magnitude of spin (or, more simply, thespin quantum number) and m𝓁 , ms are the respective quantum numbers for theprojections of 𝓵 and s onto the z-axis. (We used the subscripts 𝓁 and s for nota-tional clarity.) The quantum number ms, similar to its orbital counterpart m𝓁 ,can assume all values from −s to s, in unit steps. This implies that the interval

10.4 The Stern-Gerlach Experiment 279

from −s to s, which equals 2s, is an integer number, and we must have

2s = integer number

⇒ s =integer

2= 0, 1

2, 1, 3

2, 2,…

Therefore, the quantum number s of spin (in contrast to the quantum number 𝓁)can take, not only the integer values 0, 1, 2,…, but also the half-integer values

s = 12,

32,

52, · · ·

Note that the half-integer values, which emerge naturally if we assume that thequantization rules of angular momentum are valid in general, are not possible fororbital angular momentum, because the requirement that the wavefunction besingle valued stipulates that m𝓁 and, concomitantly, 𝓁 be integers. Since no suchconstraint holds for spin, the existence of half-integer values cannot be excludedand remains to be decided experimentally. The smallest half-integer value of thespin quantum number s is 1∕2, in which case ms takes the two values

ms = −12, ms =

12

(s = 1∕2),

with corresponding values for the projection sz

sz = −ℏ2, sz =

ℏ

2,

while for the magnitude of the vector s we have

|s| = ℏ√

s(s + 1)||||s=1∕2

= ℏ

√3

2.

Therefore, any particles in nature with spin s = 1∕2 will have only two spin orien-tations, sz = ℏ∕2 and sz = −ℏ∕2, known as spin up and spin down, respectively.

If the electron turns out by experiment to have spin s = 1∕2, then, since it is acharged particle, it will also have a magnetic moment 𝝁s due to its spin, for thesame reason that it has a magnetic moment 𝝁𝓁 due to its orbital angular momen-tum. And just as 𝝁𝓁 is given by

𝝁𝓁 = 𝛾𝓁𝓵(

𝛾𝓁 = − e2mec

)

we can also expect the following relation to hold for spin

𝝁s = 𝛾ss,

where the gyromagnetic ratio of spin 𝛾s may or may not be equal to the gyromag-netic ratio 𝛾𝓁 = −e∕2mec of the orbital angular momentum. This matter is to besettled by experiment also.

How do we design such an experiment? To avoid the problems encounteredin the Zeeman experiment, we need to find a way to directly measure the energyeigenstates, not merely the transitions between them. For example, in the groundstate 1s of the hydrogen atom, the electron has zero orbital angular momen-tum (𝓁 = 0). Therefore, if spin did not exist, the atom would have zero magnetic


moment and would thus not interact with an external magnetic field at all. But ifspin does exist, the atom in its ground state will have a magnetic moment𝝁s = 𝛾s sand will thus behave as a microscopic magnet that can interact with and be “de-tected” by an external magnetic field. So the pertinent experimental questionis this: Can we verify whether a hydrogen atom in its ground state is magneticor not? The answer is given by the Stern–Gerlach experiment, which is basedon the following simple idea: If a beam of “tiny magnets” (i.e., hydrogen atoms)passes through an inhomogeneous magnetic field, then the “magnets” that areoriented parallel to the field will acquire a negative energy (𝑈 = −𝝁 ⋅B < 0) andmove to the direction of even more negative energy, that is, in the direction that Bincreases. In contrast, those “tiny magnets” that happen to be oriented antiparal-lel to the magnetic field will move to the direction of decreasing field strength, inorder to lower their positive energies. In both cases, we made use of the generalprinciple that motion always occurs in the direction of decreasing energy.

10.4.2 The Experiment and Its Results

In view of the above discussion, the setup of the experiment we seek begins toemerge. A beam of hydrogen atoms2 in their ground state is to be sent throughan inhomogeneous magnetic field. An example of such a field is formed betweenthe poles of a horseshoe magnet, where one pole is wedge shaped and the otheris flat (see Figure 10.7). If the atomic beam passes through the magnetic fieldwithout splitting, then surely the atoms have no magnetic moment and thereforeno spin. But if a splitting of the beam occurs, then spin must indisputably exist.In this case, the spin quantum number s would be given by the condition

2s + 1 = number of beam components,

Figure 10.7 The Stern–Gerlach experiment for the hydrogen atom. Because the field isinhomogeneous, atoms in the beam experience a force in the vertical (z) directionFz = −g𝜇Bms(dB∕dz) that is proportional to the quantum number ms. Consequently, the beamsplits into 2s + 1 components, where s is the quantum number of the electronic spin. In theactual experiment, the beam splits in two components. Therefore, the electron carries anintrinsic angular momentum (spin) with quantum number s = 1∕2.

2 The original experiment by O. Stern and W. Gerlach was performed in 1922 using silver atoms.The experiment with hydrogen was conducted by T. E. Phipps and J. B. Taylor (1927).

10.4 The Stern-Gerlach Experiment 281

since the splitting of the beam into a discrete number of components results fromthe quantization of the spin orientation into 2s + 1 directions, equal to the pos-sible values of the quantum number ms.

To proceed further in our analysis, we need an expression for the interactionenergy between spin and the magnetic field. We assume that the field points alongthe z-axis, that is,

B = z B(z),

but that its magnitude varies with z. If we further assume that the magnet has theshape shown in Figure 10.7, then we have dB∕dz > 0, since the field grows as weapproach the wedged pole (i.e., as z increases).

As for the magnetic moment of spin, 𝝁s = 𝛾s s, we mentioned before that wecould have 𝛾s = 𝛾𝓁 = −e∕2mec (this is the obvious assumption to make). However,this assumption may not hold if spin turns out to be something different fromorbital angular momentum. To allow for this possibility, we set

𝝁s = −g e2mec

s,

where g is a numerical factor (known as Lande factor) to be determined by exper-iment. With B and 𝝁s as above, the interaction energy of the magnetic field in theStern–Gerlach apparatus with a hydrogen atom is written as

𝑈 = −𝝁s ⋅B = g e2mec

B(z)sz = g e2mec

B(z) ℏms

= g eℏ2mec

B(z)ms = g𝜇BB(z)ms,

where we used the relation sz = ℏms and the definition of the Bohr magneton(𝜇B = eℏ∕2mec). But since the energy of the atoms inside the magnetic fielddepends on z according to

𝑈 (z) = g𝜇BB(z)ms,

the atoms will experience a force

Fz = −d𝑈dz

= −g𝜇BmsdBdz, (10.19)

whose magnitude and sign depend on the value of the quantum number ms. Now,if the atomic beam is not polarized, it will contain atoms with all possible spinorientations (i.e., all possible values of ms), and will thus split into as many com-ponents as the possible ms values, since every subset of atoms with the samems will experience a different force. Therefore, by the time the beam exits theStern–Gerlach device, it will have split into 2s + 1 components, which will forman equal number of spots on the screen at the rear end of the device.

Figure 10.7 shows the outcome of the experiment for a beam of hydrogen atomsin their ground state. The number of spots that form on the screen is clearly equalto two; therefore, the quantum number s of the electronic spin has the value

s = 1∕2.


Some technical details of the experiment are worth clarifying. Since hydrogenatoms are electrically neutral (so they cannot be accelerated by an electric field),the production of an atomic beam can only be accomplished with a collimationdevice (not shown in Figure 10.7). The atoms exit from a small aperture of an“oven” with a wide-angle distribution of directions and a typical mean thermalspeed corresponding to the oven’s temperature. They then go through a numberof narrow slits that allow the passage of only those atoms that move largely parallelto the horizontal axis. Thus a collimated beam is ultimately formed, albeit witha much lessened intensity, since a large fraction of the initial atoms is lost in allother directions. Note also that the shape of the beam’s trace on the screen uponexiting the magnet will not look as in Figure 10.7, but more like Figure 10.8a,actually. This shape of the trace results from the fact that the cross section of thebeam is the same as that of the slit. So, atoms on the beam’s periphery will deviateless, since they are exposed to a smaller field gradient than atoms at the center,which travel right below the apex of the magnet’s wedge. We should also notethat, had there been no quantization at all, the trace of the beam would look asin Figure 10.8b, since all intermediate deviations would then be allowed.

As for the maximum value of the deviation d that occurs at the center of thebeam, it can be calculated as follows. Let us assume that g is known, and that thegradient of the magnetic field takes a reasonable value of, say, dB∕dz ≈ 104 G/cm,so that the force Fz (formula (10.19) with ms = 1∕2) is also known. This force willcause a transverse motion of the atoms, so that their total displacement uponexiting from the magnetic field is equal to d = at2∕2, where a = Fz∕mH is theacceleration of the transverse motion (mH ≈ mp is the mass of the hydrogen atom)and t = L∕𝑣 is the time the atoms spend inside a magnet of length L as they movewith a typical thermal speed 𝑣.

Using the typical values

dB∕dz ≈ 104 G/cm, L = 30 cm, 𝑣 ≈ 3 × 105 cm/s,

and for g = 1 (a “reasonable” value for g), it turns out that

d ≈ 0.15 cm,

which implies, to begin with, that the Stern–Gerlach experiment is easily realiz-able. As the two beams exit the magnet, they will be separated from each other byabout three-tenths of a centimeter. (This separation can be increased if we place

d

(a) (b)

Figure 10.8 The actual trace of the beam in the Stern–Gerlach experiment: (a) in quantumphysics, (b) in classical physics.

Problems 283

the screen a little further away from the magnet’s exit, provided our apparatuscan maintain vacuum conditions throughout the path of the atoms.)

Alas, this prediction is wrong! It turns out that, no matter how we performthe experiment, the observed distance between the two traces is always twice thetheoretical prediction for g = 1. This implies that the actual value of g is not g = 1,but g = 2! We thus have

𝝁s = −g e2mec

s||||g=2

= − emec

s,

whereas for the orbital angular momentum we had

𝝁𝓁 = − e2mec

𝓵.

We infer from the experiment that the gyromagnetic ratio of spin is twice thatof the orbital angular momentum. In simpler terms, spin is twice more “mag-netic,” so to speak, than orbital angular momentum. For a set “amount” of angularmomentum, spin produces twice as strong a magnetic field. This effect is knownas the spin magnetic anomaly and is one of the “mysteries” quantum mechanics iscalled in to solve. (These “mysteries” are indeed solved, but only in the frameworkof relativistic quantum mechanics, as expressed in the celebrated Dirac equation.)

Before we conclude this section, let us consider a question that may haveoccurred to some readers. As we noted earlier, from the perspective of anatomic electron, all external magnetic fields can be treated as homogeneous,since any field inhomogeneities will be completely unnoticeable within thetiny space of a cubic angstrom wherein an atomic electron is moving. And yet,in the experiment we have just described, we made use of precisely such aninhomogeneous field. Is this not an obvious inconsistency? The answer is no. Inthe Stern–Gerlach experiment, the inhomogeneities in question are not what anelectron experiences as it moves in the confined space of an atom but what anatom (i.e., a tiny “magnet”) experiences as it moves in a region of macroscopicdimensions that extends from the flat pole of the magnet to the wedged pole.

Evidently, in experiments with atomic beams, the inhomogeneity of the field isexperienced by the moving atoms in the beam. In fact, the inhomogeneity is thedecisive factor in such experiments.

Problems

10.1 Sketch the energy levels that arise due to the Zeeman effect from twostates with 𝓁 = 3 and 𝓁 = 2. Next, show that, because of the selection rulesΔ𝓁 = 1, Δm = ±1, 0, the initial spectral line associated with the transition𝓁 = 3 → 𝓁 = 2 appears now as a triplet. Does this also happen for other 𝓁values for the initial and final states?

10.2 In a Stern–Gerlach experiment with hydrogen atoms in their ground state,the gradient of the field is |dB∕dz| = 5 × 103 G/cm, the speed of the atomsin the beam is 2 × 105 cm/s, the length of the magnet is 20 cm, and the


distance the atoms traverse once they exit the magnetic field (i.e., until theyhit the screen) is another 20 cm. Calculate the maximum distance betweenthe two traces that appear on the screen.

10.5 What is Spin?

Since it is now evident that spin exists and we have to deal with it from now on, itis important to examine what spin really is. We will address this question in threestages (or levels), with the corresponding section headings as described below:

1. Spin is no self-rotation.2. How is spin described quantum mechanically?3. What spin really is.

Interestingly, the first level of our understanding of spin is a kind of negativeresult, namely, the realization that spin cannot be what at first we might thinkit is: a rotation of the electron around its axis. The next level of understand-ing is to provide a quantum mechanical description of this physical quantity: toexplain, for example, what kind of operators the spin components sx, sy, sz are andwhere exactly they act. We know that the components 𝓁x,𝓁y,𝓁z of orbital angularmomentum are differential operators that act on three-dimensional wavefunc-tions 𝜓(x, y, z). What is the corresponding quantum mechanical description ofspin?

But the mathematical description of a physical quantity does not automati-cally provide its physical meaning as well. Thus we proceed to a third level ofunderstanding, where we address the question of the “nature of spin.” To besure, the complete understanding of spin is a topic for more advanced textbooksthan this one. But for now, the interested readers may study the pertinent onlinesupplement of this chapter.

10.5.1 Spin is No Self-Rotation

We will now show that, even though the usual picture for spin (the electron spin-ning around its own axis) seems plausible, it is also wrong. Here are the two basicarguments.

Argument 1: If spin were due to the rotation of the electron about its axis, thespeed of any point on the equator of the electron would be greater—in fact muchgreater—than the speed of light.Argument 2: If we were to adopt the classical picture of spin as a self-rotatingsphere, then we would expect that, as electrons collide with each other orwith different particles, higher states of self-rotation would arise, and we wouldobserve different values of the quantum number s that go beyond s = 1∕2. But thisnever happens! The quantum number of spin is “frozen” at its one and only value(e.g., s = 1∕2) and is as permanent a feature of the electron as its mass or charge.

Let us examine Argument 1 a little closer. The angular momentum of a rigidbody rotating about an axis is I𝜔, where I is the moment of inertia with respect

10.5 What is Spin? 285

to this axis, and 𝜔 is the angular frequency of rotation. For the electron we thusobtain

I𝜔 = sz =ℏ

2, (10.20)

where I = 2ma2∕5 is the moment of inertia of the electron, if the latter is treatedas a sphere of radius a and total mass m that is uniformly distributed throughoutits volume. But whatever the mass distribution, we would always have I = kma2,where k is a number slightly less than unity. So, for an order-of-magnitude calcu-lation, we can set I ≈ ma2 and sz ≈ ℏ, so that (10.20) becomes

ma2𝜔 ≈ ℏ ⇒ ma𝑣 ≈ ℏ, (10.21)

where 𝑣 = a𝜔 is the speed of rotation at the equator of the electronic sphere. Butwhat is the value of a? According to results from high-energy experiments withelectrons, if there is a radius for an electron (contemporary local quantum fieldtheories actually treat it as a point particle) then it must surely be smaller than10−17 cm! With m ≈ 10−27 g, ℏ ≈ 10−27 erg s and a ≈ 10−17 cm, formula (10.21)yields

𝑣 = ℏ

ma= 1017 cm∕s ≈ 3 × 106c,

which tells us that in order for the electron to have an intrinsic angular momen-tum of order ℏ, the speed of rotation at its equator must exceed the speed of lightby three million times!

Argument 2 is equally damning for the picture of spin as a self-rotation. If theelectrons were more or less similar to “billiard balls,” we would not be able toexplain how their rotational state stays invariant: Even though they collide witheach other all the time and with immense speeds, their spin goes neither up (noteven with quantum leaps) nor down. These peculiar billiard balls appear to bedestined to forever rotate with the same angular velocity!

The conclusion from the above discussion is unequivocal: Spin is not therotation of an electron about its axis. The real understanding of spin begins themoment we reject the classical picture.

10.5.2 How is Spin Described Quantum Mechanically?

We can easily arrive at the correct quantum mechanical description of spin if wefollow “literally” the analogy with the quantum mechanical formalism we alreadyknow. Consider, for example, the meaning and use of a one-dimensional wave-function 𝜓(x). Here the quantum mechanical quantity of interest (the positionof the particle) takes a continuous set of values, in the range −∞ < x < +∞. Thevalues of 𝜓(x) at various points give the probability amplitudes (more accurately:the probability density amplitudes) to find the particle in the immediate neigh-borhood of any one of the possible values of x.

In the case of spin, the quantity measured is its projection onto the z-axis,which has just two possible values: sz = ℏ∕2 and sz = −ℏ∕2, namely, spin upand spin down, respectively. Thus, in this case, instead of the infinite numberof probability amplitudes 𝜓(x), we need only two probability amplitudes: One


for the possibility of measuring the projection of spin to be sz = ℏ∕2 (let us callthis probability amplitude a) and another (let us call it b) for the possibility ofmeasuring sz = −ℏ∕2. The amplitudes a and b are in general complex numbers,similar to the values of 𝜓(x). The corresponding probabilities

P+ = probability of finding the spin up

and

P− = probability of finding the spin down

are given by the formulas

P+ = |a|2, P− = |b|2

and, since there are only two possible outcomes (spin up or spin down), thenormalization condition becomes

P+ + P− = |a|2 + |b|2 = 1.

Now, because we only need two (complex) numbers to describe spin in quantummechanics, it is natural3 to place these two numbers in a column vector

X =(

ab

)

(10.22)

and summarize the above discussion by saying that the spin state of a particlewith s = 1∕2 is described by a complex column vector with two componentsthat also satisfies the normalization condition

X†X = (a∗, b∗)(

ab

)

= a∗a + b∗b = |a|2 + |b|2 = 1,

where

X† = (a∗, b∗)

is the hermitian conjugate vector of X. The latter is formed by taking the complexconjugates of the elements of the initial vector and converting the column intoa line.4

Two interesting special cases of the so-called state vector (10.22) are the vectors

X+ =(

10

)

, X− =(

01

)

,

whose physical meaning is the following. The first vector describes the electron’sspin-up state (sz = ℏ∕2), while the second vector describes its spin-down state

3 Actually—as we shall see in detail later in Section 10.7—it is also natural (and in some casesadvantageous) to regard the wavefunction of spin as a function X(𝜇) of the discrete variable𝜇 = ±1∕2 (=possible values of sz for ℏ = 1). The values of X(𝜇) are the probability amplitudes to findthe particle with spin up or down, respectively, that is, X(1∕2) = a, X(−1∕2) = b.4 This is done in much the same way that we form the hermitian conjugate—or adjoint— of anarbitrary n × m matrix A. That is, we form the complex conjugates of its elements (matrix A∗) andthen convert the lines into columns. If the matrix is square, the conversion of lines into columns isequivalent to transposing the matrix with respect to the main diagonal.


(sz = −ℏ∕2). Clearly, any arbitrary vector X can always be written as a linear com-bination of X+ and X−, that is, as

X = a(

10

)

+ b(

01

)

=(

ab

)

,

since the act of multiplying a column vector by a number is equivalent to mul-tiplying all the vector’s elements by that number, while adding two columns isequivalent to adding their respective elements.

Now that we settled the question of how to describe the spin quantum mechan-ically via the state vector X, the next question is, what is the form of the quantummechanical operators sx, sy, sz? This is easy to answer. Since sx, sy, sz act on thestate vectors X that are column vectors with two components, they can only besquare matrices of dimensions 2 × 2. Their action on the X vectors is equivalentto the usual multiplication of a matrix by a column.

Let us remind the readers here that the multiplication of an n × n square matrixA by a column vector X with n components is done as follows:

i-th line

=

which means that the ith element of the column that results from the multiplica-tion is equal to the product of the ith line of the matrix A times the given columnX. As usual, the product of a line times a column is the sum of the products oftheir respective elements. Here is an example:

(1 22 −1

) (1

−1

)

=(

1 ⋅ 1 + 2(−1)2 ⋅ 1 + (−1) ⋅ (−1)

)

=(−1

3

)

.

Aside from requiring that the matrices sx, sy, sz have dimensions 2 × 2, it is alsoreasonable to impose the following requirements on them:

(a) The matrices sx, sy, sz must be hermitian5 so that their eigenvalues, which aremeasurable quantities, are real numbers.

(b) All three matrices sx, sy, sz must have eigenvalues ±ℏ∕2, since all axes areequivalent (even though we choose one of them as the “axis of quantization,”since it is impossible to simultaneously measure all three spin components).

5 We remind the readers that a matrix is called hermitian if it is equal to its hermitian conjugatematrix, that is, if A† = A. This definition implies that a matrix is hermitian if its diagonal elementsare real numbers, while the elements that are symmetric with respect to the diagonal are complexconjugate numbers. For a matrix with real elements to be hermitian, it suffices that it is symmetricwith respect to the main diagonal. We also note the basic property of hermitian matrices, namely,that they have real eigenvalues and orthogonal eigenvectors. The significance of hermitian matricesin quantum mechanics stems from the fact (see the proof in Section 2.6.4) that all quantummechanical operators that describe physical quantities (and which are hermitian operators, as weknow) can be represented by suitably chosen hermitian matrices.


(c) The matrices sx, sy, sz must satisfy the commutation relations

[sx, sy] = i ℏsz, and cyclic permutations, (10.23)

which are the characteristic feature of angular momentum as a quantummechanical quantity, as we have discussed in Section 3.5.2.

Since all three spin matrices have eigenvalues ±ℏ∕2, we can write them in theform

sx =ℏ

2𝜎x, sy =

ℏ

2𝜎y, sz =

ℏ

2𝜎z, (10.24)

where 𝜎x,y,z are the so-called Pauli matrices, which are also hermitian with eigen-values ±1, and dimensionless, since Planck’s constant accounts for the physicaldimension of sx,y,z. Moreover, if we substitute expressions (10.24) in the commu-tation relations (10.23), we see at once that the Pauli matrices satisfy the relations

[𝜎x, 𝜎y] = 2i𝜎z, and cyclic permutations. (10.25)

Since we chose the z-axis as our “quantization axis,” we expect that the matrix𝜎z has the simplest possible form (i.e., it is diagonal), so that

𝜎z =(

1 00 −1

)

, (10.26)

where the diagonal elements are the matrix eigenvalues. If this matrix form iscorrect, then the following eigenvalue equations must hold

sz X± = ±ℏ2

X± ⇒ 𝜎z X± = ±X±,

because the vectors X± represent eigenstates of sz with eigenvalues±ℏ∕2. Clearly,for 𝜎z as in (10.26), the equations 𝜎z X± = ±X± are satisfied.

A heuristic way to arrive at the matrices 𝜎x and 𝜎y (and then check their validity)is to argue as follows. Given that the axes x, y, and z are orthogonal, the corre-sponding matrices ought to be as “orthogonal” to each other as possible, that is,as “different” as possible. So, if 𝜎z is diagonal, the matrices 𝜎x,y should have thehighest possible non-diagonal form. That is, they should have vanishing diago-nal elements to ensure zero “overlap” with 𝜎z. Furthermore, their off-diagonalelements should be real numbers in the first case (matrix 𝜎x) and complex conju-gate numbers in the second (matrix 𝜎y). In this way, the two matrices, 𝜎x and 𝜎y,are not only hermitian, but also maximally different: One of them is purely real,the other purely imaginary. If we now add the requirement that the matrices 𝜎x,yought to have eigenvalues ±1—whereby their determinant, which is always givenby the product of the eigenvalues, is equal to −1—we arrive at the matrix forms

𝜎x =(

0 ±1±1 0

)

, 𝜎y =(

0 ±i∓ i 0

)

,

which are the well-known Pauli matrices, provided we select the positive sign forthe first matrix and the negative sign for the second one, namely,

𝜎x =(

0 11 0

)

, 𝜎y =(

0 −ii 0

)

.


To fully verify the above matrix forms, we need to check whether the commu-tation relations (10.23) or (10.25) are satisfied; after all, this is the fundamentalcriterion for a vector quantity to be called angular momentum. Let us look at thefirst relation of (10.25) in particular. We have

𝜎x𝜎y =(

0 11 0

) (0 −ii 0

)

=(

i 00 −i

)

𝜎y𝜎x =(

0 −ii 0

) (0 11 0

)

=(−i 0

0 i

)

so that for the commutator [𝜎x, 𝜎y] = 𝜎x𝜎y − 𝜎y𝜎x, we obtain

[𝜎x, 𝜎y] = 𝜎x𝜎y − 𝜎y𝜎x =(

i 00 −i

)

−(−i 0

0 i

)

=(

2i 00 −2i

)

= 2i(

1 00 −1

)

= 2i 𝜎z ,

which is precisely the commutation relation (10.25).For easy reference, we collect our results in Table 10.1, while in Table 10.2 we

present some useful properties of the Pauli matrices and encourage readers toverify them. (We recall that the trace of a matrix A, denoted Tr A, is the sumof its diagonal elements, which is also equal to the sum of its eigenvalues if thematrix is diagonalizable.)

Table 10.1 Spin matrices and Paulimatrices.

Spin matrices Pauli matrices

sx =ℏ

2

(0 11 0

)

𝜎x =

(1 00 −1

)

sy =ℏ

2

(0 − ii 0

)

𝜎y =

(0 −ii 0

)

sz =ℏ

2

(1 00 − 1

)

𝜎z =

(1 00 −1

)

Table 10.2 Useful properties of Pauli matrices.

1 det 𝜎x = det 𝜎y = det 𝜎z = −12 Tr 𝜎x = Tr 𝜎y = Tr 𝜎z = 03 𝜎2

x = 𝜎2y = 𝜎2

z = 14 𝜎x𝜎y = −𝜎y𝜎x, and cyclic permutations5 𝜎x𝜎y = i𝜎z, and cyclic permutations


We should stress again that it is the spin matrices that have physical meaning(they represent the operators of a physical quantity), while the Pauli matrices aresimply used as a substitute to save us the trouble of carrying around Planck’sconstant and the factor 1∕2.

As far as formalism is concerned, spin matrices are used in much the sameway as all other operators of physical quantities. Thus the key function of suchoperators, which is to allow the calculation of mean values via the formula

⟨A⟩ = ∫ 𝜓∗(A𝜓) dx,

leads to the following expression in the case of spin:

⟨si⟩ = X†siX, i ≡ x, y, z,

where X =(

ab

)

is the given state vector and X† = (a∗, b∗) is its hermitian con-

jugate. The following example should help clarify how these ideas are applied inpractice.

Example 10.1 Calculate the mean values of the spin components sx, sy, sz for astate described by the vector

X = 1√

5

(12

)

. (1)

Solution: First, we note that the vector (1) is normalized: The sum of the squaresof its components equals unity, as it should, since this sum represents the totalprobability of finding the particle with spin up or spin down. For the mean valueof sz we have

⟨sz⟩ =1√

5(1, 2)ℏ

2

(1 00 −1

)1√

5

(12

)

= ℏ

10(1, 2)

(1 00 −1

) (12

)

= ℏ

10(1, 2)

(1

−2

)

= ℏ

10(−3) = − 3

10ℏ,

while for ⟨sx⟩ we find

⟨sx⟩ =ℏ

10(1, 2)

(0 11 0

) (12

)

= ℏ

10(1, 2)

(21

)

= ℏ

104 = 2

5ℏ.

Finally, for ⟨sy⟩ we obtain

⟨sy⟩ =ℏ

10(1, 2)

(0 −ii 0

) (12

)

= ℏ

10(1, 2)

(−2i

i

)

= ℏ

10⋅ 0 = 0.

The fact that ⟨sy⟩ vanishes should not surprise us. Since the matrix sy is purelyimaginary and the vector X is real, if the mean value were nonzero then it wouldactually have to be purely imaginary, which is nonsensical. (Readers may recallthat we used a similar argument about the mean value of momentum in caseswhere the wavefunction was real.)


The negative sign of ⟨sz⟩ is also expected, since state (1) gives a higher probabil-ity of measuring the spin down rather than up. For the z component in particular,the mean value of spin can be also calculated using the statistical formula

⟨sz⟩ = P+

(ℏ

2

)

+ P−

(

−ℏ2

)

= 15ℏ

2+(4

5

) (

−ℏ2

)

= − 310ℏ,

which agrees with the above result.

10.5.3 What Spin Really Is

Previously, we explained that spin is not a self-rotation, and discussed how todescribe it quantum mechanically. But if we were to ask ourselves, “what is spin?,”it is quite likely that we would find ourselves muttering some serious-soundinggeneralities with buzzwords such as spin operators, Pauli matrices, state vectors,etc., in a manner more revealing of our own (current) ignorance than the truenature of spin.

Now, it would be tempting to attribute to spin itself our own difficulty in pro-viding a decent answer to the above question. To say that it is the peculiar natureof this new property of the electron that does not fit the classical picture of aself-rotating spherical particle.

But if spin itself is to “blame,” then surely we ought to be able to answer thisquestion: “What is the orbital angular momentum of the electron in quantummechanics?” And yet, we experience the same kind of difficulty to provide adecent answer to this question just as in the case of spin. What could we say aboutorbital angular momentum? Perhaps that “it is the operator we obtain from theclassical expression 𝓵 = r × p when we replace r and p with the correspondingquantum mechanical operators?” Surely this cannot be considered an acceptableanswer.

We thus realize that the problem lies not in the peculiar nature of spin, assome may believe, but in the very essence of the quantum mechanical descrip-tion of the physical world. It is the abstract mathematical nature of quantummechanics that does not allow a palpable representation of physical quantitiesin the manner familiar to us from classical physics. Come to think of it, the for-malism of quantum mechanics—especially the probabilistic interpretation of thewavefunction—makes no mention of the objective nature of quantum systemsand quantum quantities, but only of what we find when we observe them: whatare the possible outcomes of our measurements and their corresponding prob-abilities. We thus arrive at the heart of the matter, namely, the central role ofmeasurement in quantum mechanics. This role implies that the measurementprocess should be regarded as a constituent element of the description of thequantum mechanical system—a constituent of the very understanding of whata quantity or a physical state actually is. In effect, all concepts and expressionsused in quantum mechanics must have an experimental meaning. They are to bethought of in the context of an actual or thought experiment that attributes anempirical meaning to them.

We summarize the above discussion as follows:In quantum mechanics, physical quantities are defined by the way they aremeasured.


Well, can we now say what spin really is? In general, this question has noexperimental meaning and is thus not a valid question. But if we ask “what is aparticle with spin s?” or, more precisely, “what is the meaning of the statementthat a particle has spin s?,” then the answer is simple: A particle has spin s whenthe passage of a beam of such particles through a Stern–Gerlach apparatus yields2s + 1 traces. Thus, we say that a particle has spin zero when the Stern–Gerlachexperiment for the corresponding beam yields one trace; spin 1∕2 when weobtain two traces; spin s = 1 for three traces, and so on. Viewed in this manner,spin becomes completely analogous to orbital angular momentum. Thereagain, the sensible question to ask is not about the nature of orbital angularmomentum in general but about the meaning of specific statements, such as,“A quantum system is in a state with orbital angular momentum 𝓁.” Again, theanswer is the following: We say that a quantum system is in a state with orbitalangular momentum 𝓁 when the passage of a beam of such systems through aStern–Gerlach apparatus yields 2𝓁 + 1 traces. (There is an inherent assumptionhere that the system does not carry both spin and angular momentum, or elsethe Stern–Gerlach apparatus measures the total angular momentum of thesystem, which we will discuss shortly.)

In this spirit, the statement that “a quantum system is in a p state” is equivalentto the statement that “the Stern–Gerlach experiment for this system yielded threetraces.”

Here is another statement made frequently: “An electron with spin 1∕2 is inthe spin-up state.” What does it mean? It simply means that when a beam of suchelectrons (all with spin up) passes through a Stern–Gerlach apparatus, we onlyobserve one trace, which is actually closer to the side of the flat pole of the magnet.(Why?)

A general conclusion emerges from this discussion. If, in quantum mechanics,physical quantities “are defined only through measurements,” then there is abso-lutely no conceptual difference between orbital angular momentum and spin.Intrinsic angular momentum (i.e., spin) has the same phenomenological featuresand the same experimental signature as orbital angular momentum, the only dif-ference being the possibility of half-integer values for the quantum number s.

10.6 Time Evolution of Spin in a Magnetic Field

Let us assume that at a given moment in time, an electron with spin up along thez-axis is placed in a magnetic field B that points along another axis, say, the x-axis.What will happen to the spin of the electron? Will it remain oriented along thez-axis, or will it begin to “move?” And if it is the latter, then what orientation willthe spin have after some time t? Let us put our question in quantum mechanicalterminology:What is the probability, after time t, of finding the electron with spin up along theoriginal axis?

Since our question pertains to evolution in time, we need to solve thetime-dependent Schrödinger equation

iℏ dXdt

= HX, (10.27)

10.6 Time Evolution of Spin in a Magnetic Field 293

where, since we are now only interested in the evolution of spin, the unknownwavefunction is simply the spin wavefunction

X(t) =(

a(t)b(t)

)

.

The Hamiltonian H now takes the form of a 2 × 2 matrix, since

H = −𝝁s ⋅B = −(

− emec

s)

⋅B||||| B = x B

s = (ℏ∕2)𝝈

= eℏ2mec

⋅ B ⋅ 𝜎x

= 𝜇B ⋅ B ⋅ 𝜎x = 𝜖𝝈x = 𝜖

(0 11 0

)

(𝜖 = 𝜇BB).

Thus, we can write the Schrödinger equation as follows,

iℏX = iℏ(

ab

)

= 𝜖

(0 11 0

) (ab

)

= 𝜖

(ba

)

or, equivalently, if we set 𝜖∕ℏ = 𝜔,

ia = 𝜔b, ib = 𝜔a, (10.28)

which is a system of first-order linear differential equations, with the componentsa(t) and b(t) of the state vector X(t) as the unknowns. The system is to be solvedsubject to the initial condition

X(0) =(

10

)

⇒ a(0) = 1, b(0) = 0,

which expresses the fact that for t = 0 the electron is in the spin up state alongthe z-axis. We can easily solve the system (10.28) by substituting b from thefirst equation (b = ia∕𝜔) into the second equation. This yields an equation withrespect to a,

i(ia∕𝜔)⋅ = 𝜔a ⇒ a + 𝜔2a = 0 (10.29)

subject to the initial conditions

a(0) = 1, a(0) = 0, (10.30)

of which the latter condition follows from b(0) = 0, in conjunction with the firstequation of the system (10.28). The general solution of (10.29) is

a(t) = c1 cos 𝜔t + c2 sin 𝜔t,

which, upon application of the initial conditions (10.30), yields c1 = 1 and c2 = 0,so that

a(t) = cos 𝜔t, (10.31)

while from the relation b = ia∕𝜔, we obtain for the function b(t),

b(t) = −i sin 𝜔t. (10.32)


1P+(t)

T

t

P–(t)

= ωπ

Figure 10.9 Time evolution of spin in a magnetic field. P+(t): The probability to find anelectron having again spin up along the z-axis as it evolves in a magnetic field B that pointsalong the x-axis. P−(t): The probability of a “spin flip” after the lapse of time t. Bothprobabilities oscillate periodically with frequency 𝜔′ = 2𝜔 and period T = 2𝜋∕𝜔′ = 𝜋∕𝜔.

As we know, the physical meaning of the solution is that the squares of the abso-lute values of a(t) and b(t) are the probabilities to find the particle with spin upor down along the z-axis after the lapse of time t. That is,

P+(t) = |a(t)|2 = cos2 𝜔t (10.33)

and

P−(t) = |b(t)|2 = sin2 𝜔t, (10.34)

where the sum of these probabilities at any time t equals unity, as expected.Let us now return to our original question, namely, the probability of finding

the electron after time t having again spin up along the z-axis, which is given by(10.33). In Figure 10.9 we sketch this probability, together with the complemen-tary probability (10.34) of a “spin flip” after time t.

As we shall see in the online supplement of Chapter 16, the motion of spin invarious kinds of magnetic fields is one of the most fascinating topics in quantumphysics, and has significant practical applications also, such as nuclear magneticresonance spectroscopy (NMR), magnetic resonance imaging and so on.

Readers may be pleased to know that our discussion so far, even though incom-plete from a mathematical perspective, suffices for the handling of all pertinentapplications of spin.

Problems

10.3 The spin state of a particle at a certain moment in time is given by thecolumn vector

X = 1√

2

(1i

)

.

Calculate the mean values and the corresponding uncertainties of the spincomponents, and comment on your results.

10.7 Total Angular Momentum of Atoms: Addition of Angular Momenta 295

10.4 Somebody claims that the (unnormalized) spin vectors

X ∼(

1i

)

and Y ∼(

1−i

)

describe states with a definite spin projection onto a certain axis. Howcan you test this claim and how can you infer which axis it refers to?Does your answer have anything to do with the result of the previousproblem?

10.5 The spin state of a particle with s = 1∕2 is given by the (unnormalized)vector

X ∼(

1 + 2i2

)

.

Calculate the probabilities of finding the particle with spin up or spin downalong the(a) z-axis,(b) x-axis,(c) y-axis.

10.6 Somebody claims to have calculated the quantities ⟨sx⟩ and Δsx for thestate of the previous problem and found ⟨sx⟩ = 3ℏ∕5 and Δsx = 3ℏ∕2.Explain why these results are clearly wrong, without repeating thecalculation. What would you have said if the claimed results had been⟨sx⟩ = ℏ∕5 and Δsx = ℏ?

10.7 Use the solution we found in the main text for the evolution of spin in amagnetic field

X(t) =(

cos 𝜔t−i sin 𝜔t

)

to calculate the mean spin vector

⟨s⟩ =(⟨sx⟩, ⟨sy⟩, ⟨sz⟩

)

at a time t after positioning the spin in parallel to the z-axis. Use a simplepicture to describe the result of your calculation.

10.7 Total Angular Momentum of Atoms: Additionof Angular Momenta

10.7.1 The Eigenvalues

Given that the electron has both orbital and spin angular momentum, the ques-tion arises as to what would be the total angular momentum of the atom in anarbitrary state that has nonzero orbital angular momentum. Let’s rephrase this


question in quantum mechanical terminology: Given the quantum numbers 𝓁and s that determine the magnitude of the vectors 𝓵 and s via the relations

𝓵 2 = ℏ2𝓁(𝓁 + 1), s2 = ℏ2s(s + 1),what are the allowed values of the quantum number j that defines the magnitudeof the total angular momentum

j = 𝓵 + sthrough the relation

j 2 = ℏ2j(j + 1)?The answer is simple and can be given in the form of a theorem, as follows.6

Theorem 10.1 The quantum number j of the total angular momentum can takeall values from jmin = |𝓁 − s| to jmax = 𝓁 + s, in unit steps.

In other words,j = |𝓁 − s| ,… ,

⏟⏟⏟

unit steps

𝓁 + s. (10.35)

The validity of the above theorem goes beyond the orbital angular momentumand the electron spin; it extends to all kinds of angular momenta and arbitraryvalues of the quantum numbers that determine their magnitudes. In other words,if we have

j = j1 + j2,

then the possible values of j span the range from |j1 − j2| to j1 + j2 in unit steps,while the vectors j1 and j2 can represent orbital or spin angular momenta witharbitrary quantum numbers j1 and j2.

In the classical limit of very large 𝓁 and s, the law of addition of angularmomenta (10.35) is rather obvious. In this limit (whereby j is generally also large)the magnitudes of the vectors j,𝓵, and s are given by the relations

|j| = ℏj, |𝓵| = ℏ𝓁, |s| = ℏs (classical limit).Therefore, if the vectors 𝓵 and s are parallel (𝓵 ↑↑ s), we obtain for their sum

⇒ = + s = z + sz = ( + s)z ≡ jz⇒ j

j= jmax = + s,

sz

l l

l

` ` ` `l

while if the vectors are antiparallel (𝓵 ↑↓ s), we obtain

⇒ = + s = z − s z = ( − s)z ≡ jz⇒ j

j= jmin = − s

sz

s( ).l l

ll

` ` ` `l

6 See Problems 10.12 and 10.13 for a sketch of its proof beyond the explanation in the classical limitgiven here.


The “proof” is complete (in the classical limit) if we observe that, because theprojections of 𝓵 and s onto the z-axis vary in integer steps, the same is expectedto hold for the possible values of the quantum number j, which will thus rangefrom its minimum to its maximum value, in unit steps.

Actually, we can verify the validity of the law for the addition of angularmomenta (10.35) even for arbitrary (nonclassical, i.e., small) values of 𝓁 and s,as follows: Since for given 𝓁 and s, the vectors 𝓵 and s have 2𝓁 + 1 and 2s + 1orientations in space, respectively, the number of initial states is (2𝓁 + 1)(2s + 1),which is equal to the number of ways the different orientations for the twovectors can be combined. After the angular momenta are added, we have (2j + 1)possible orientations for each value of j given by (10.35). Thus the number ofpossible states after the addition of the angular momenta 𝓵 and s is given by thesum

𝓁+s∑

j=𝓁−s(2j + 1) (𝓁 > s),

which is easy to calculate.7 The result is (2𝓁 + 1)(2s + 1) as we should expect: Thenumber of states before and after the addition ought to be the same. The followingexample provides a simple application of the above ideas.

Example 10.2 Calculate the possible values for the quantum number of thetotal angular momentum in the following two cases:(a) 𝓁 = 1, s = 1∕2,(b) 𝓁 = 3, s = 2.

Verify that the number of states before and after addition is the same.

Solution: In case (a) (i.e., 𝓁 = 1, s = 1∕2), we have

jmin = |𝓁 − s| = 12, jmax = 𝓁 + s = 3

2,

which are actually the only possible values of j, since they differ by one. In case(b) (i.e., 𝓁 = 3, s = 2), we find

jmin = |𝓁 − s| = 1, jmax = 𝓁 + s = 5and the possible values of j are now

j = 1, 2, 3, 4, 5.

7 It is the sum of the terms of an arithmetic sequence, which is equal to the half-sum of the first andlast terms, times the number of terms. That is,

jmax∑

jmin

(2j + 1) =(2jmin + 1) + (2jmax + 1)

2⋅ (jmax − jmin + 1)

|||||jmin=𝓁−sjmax=𝓁+s

(𝓁 > s)

= (2(𝓁 − s) + 1) + (2(𝓁 + s) + 1)2

⋅ (𝓁 + s − (𝓁 − s) + 1)

= (2𝓁 + 1) ⋅ (2s + 1)


Let us check that the number of states before and after addition in case (b) staysthe same.

• Number of states before addition

(2𝓁 + 1) ⋅ (2s + 1) = (2 ⋅ 3 + 1)(2 ⋅ 2 + 1) = 35.

• Number of states after additionPossible values of j Number of states: (2j + 1)

j = 1 3j = 2 5j = 3 7j = 4 9j = 5 11

Total 35

The equality of the number of states (before and after addition) confirms thecorrectness of the calculation.

The allowed j values in case (a) (j = 1∕2 and j = 3∕2) correspond to the total angu-lar momentum of the electron of the hydrogen atom in a p state (say, the 2p state),where the orbital angular momentum and the spin are both nonzero, so we mustadd them to obtain the total angular momentum of the atom. According to theabove discussion, there exist two different 2p states in the hydrogen atom. Thesestates are denoted as 2p1∕2 and 2p3∕2, where the subscript specifies the value of j,the quantum number of the total angular momentum.

Does the manner in which two angular momenta are added affect the energyof the electron? If yes, which one of the two states (2p1∕2 or 2p3∕2) has the lowestenergy? The answer emerges readily, once we recall that the electron, owing to itsorbital “motion” and spin, carries two magnetic moments,𝝁𝓁 and𝝁s, respectively,which are given by the formulas

𝝁𝓁 = − e2mc

𝓵, 𝝁s = − emc

s. (10.36)

The orbital magnetic moment 𝝁𝓁 in (10.36) is considered as anchored to thecenter of the atom (as if it resulted from a classical circular current around thenucleus), while 𝝁s is considered anchored to the electron, for obvious reasons. Ifwe now recall the classical expression V = 𝝁1 ⋅𝝁2∕r3 for the interaction energybetween two magnetic moments at a distance r, then the interaction of the twomagnetic moments of the electron will add to the atom an energy term equal to

VLS =𝝁𝓁 ⋅𝝁s

r3 = e2

2m2c2𝓵 ⋅ sr3 , (10.37)

which is actually known as spin–orbit interaction (or coupling). For anorder-of-magnitude estimate, in atomic units, we set in (10.37) r ≈ a0 = 1, |𝓵| ≈|s| ≈ ℏ = 1, e = 1, m = me = 1 and also c = 𝛼−1 = 137, since 𝛼 = e2∕ℏc = 1∕137.We thus obtain

VLS ≈ 12𝛼2 a.u. ≈ 1

2

( 1137

)2⋅ 27.2 eV ≈ 10−4 eV, (10.38)


where we have ignored the sign of the inner product 𝓵 ⋅ s, since we are onlyinterested in the order of magnitude of the result. As we might have expected,the magnetic interactions in the hydrogen atom are insignificant compared tothe electrostatic interaction, which is thus the dominant force in atomic scales(although this ceases to be true for heavy atoms). Despite being small, the mag-netic spin–orbit interaction causes shifts in the energy levels of the atom and,concomitantly, in its emission and absorption spectra. These shifts had alreadybeen measured by the end of the nineteenth century and became known as thefine structure of the spectrum (hence the naming of the constant 𝛼 = e2∕ℏc asthe fine structure constant). As can be seen from (10.38), the value of 𝛼 is indeedassociated with tiny shifts of energy levels, and hence of spectral lines, of the atom.

We are now ready to address our original question, namely, which one of thetwo states (2p1∕2 or 2p3∕2) has lower energy. Clearly, it is the 2p1∕2 state, sinceits total angular momentum has the smallest of the two possible values (j = 1∕2as opposed to j = 3∕2), so the vectors 𝓵 and s are added at an obtuse angle(say, antiparallel), which makes their inner product negative, and their interac-tion energy (10.37) also negative. If the readers harbor any lingering doubt aboutthis fact, they are encouraged to go through the following example.

Example 10.3 For given values of the quantum numbers 𝓁 and s, calculate (i)the allowed values of the inner product𝓵 ⋅ s and (ii) the angle between the vectors𝓵 and s in the 2p1∕2 and 2p3∕2 states of the hydrogen atom.

Solution: To answer question (i), we take the square of relation j = 𝓵 + s andobtain

j 2 = 𝓵 2 + s 2 + 2𝓵 ⋅ s, (1)

where we took into account the fact that the vectors 𝓵 and s commute, sincethey act on different variables, so we can apply to them all the usual algebraicoperations and identities. It is clear from (1) that, because the inner product 𝓵 ⋅ sis expressed in terms of 𝓵 2, s 2, and j 2 (specifically, 𝓵 ⋅ s = (j 2 − 𝓵 2 − s 2)∕2),its values are determined once the magnitudes of these vectors are known. Thusthe allowed values of the inner product are given by substituting in the aboveexpression the allowed values (i.e., eigenvalues) of 𝓵 2, s 2, and j 2. We thus obtain

𝓵 ⋅ s = 12( j 2 − 𝓵 2 − s 2) = ℏ2

2(j(j + 1) − 𝓁(𝓁 + 1) − s(s + 1)). (2)

And now for question (ii). The angle 𝜃 between the vectors 𝓵 and s is definedthrough the classical relation

𝓵 ⋅ s = |𝓵| ⋅ |s| cos 𝜃,

whence, in conjunction with (2), we have

cos 𝜃 = 𝓵 ⋅ s|𝓵| ⋅ |s|

=j(j + 1) − 𝓁(𝓁 + 1) − s(s + 1)

2√𝓁(𝓁 + 1)

√s(s + 1)

,


so that, for the two states of interest to us, we obtain

2p3∕2(𝓁 = 1, s = 1∕2, j = 3∕2) ⇒ cos 𝜃 = 1√

6⇒ 𝜃 = 65.9∘

2p1∕2(𝓁 = 1, s = 1∕2, j = 1∕2) ⇒ cos 𝜃 = −√

23

⇒ 𝜃 = 144.74∘.

We thus confirm that, for j = 1∕2, the inner product 𝓵 ⋅ s is negative and thecorresponding angle is obtuse. Note that there is never a complete alignment(parallel or antiparallel) of the added vectors. For the same reason that anangular momentum cannot fully align with an axis (see Figure 9.9), any twoangular momenta can never be completely aligned either. Even in the state ofmaximum alignment (i.e., the state of maximum total angular momentum),the two angular momenta that are being added form an angle with each otherthat is never 0∘.

10.7.2 The Eigenfunctions

One last issue warrants further discussion. We ought to say a little more aboutthe wavefunctions that describe a state with a given total angular momentum,when two angular momenta are added. First, we note that the spin wavefunctions(described earlier as column vectors) can be viewed equivalently as functionsX(𝜇) of the discrete variable 𝜇 = ±1∕2 (which accounts for the two possible val-ues of sz). These functions take as values the probability amplitudes of finding theparticle with spin up or down, respectively. We thus have

X(1∕2) = a, X(−1∕2) = b

and the correspondence with the wavefunction 𝜓(x) becomes now clear. Theposition variable x is replaced by the discrete variable 𝜇 (since the quantity ofinterest has now a discrete spectrum) and, instead of 𝜓(x), we now use the func-tion X(𝜇), which can also be represented as a column vector, as we know. As forthe spin wavefunctions of systems with two particles, these are functions of thetwo discrete variables 𝜇1 and 𝜇2; that is, they have the form

X = X(𝜇1, 𝜇2)

similarly to the corresponding position wavefunctions 𝜓 = 𝜓(x1, x2).We now proceed to construct wavefunctions with definite total angular

momentum in the simplest possible case: a system of two particles withs1 = s2 = 1∕2.

Example 10.4 Construct the eigenfunctions of definite total spin for a systemof two particles with spin 1∕2.

Solution: The possible values of the quantum number S of the total spin S =s1 + s2 are given by the rule S = |s1 − s2|,… , s1 + s2. In our case (s1 = s2 = 1∕2),we obtain

S = 1 and S = 0,


where the former value pertains to a state often described as that of “parallelspins” (pictorially denoted as ↑↑), and the latter value to a state of “antiparallelspins” (denoted as ↑↓). Again, we stress that in quantum mechanics it is impos-sible to have spins completely parallel or antiparallel to each other.

The corresponding spin wavefunctions will, clearly, be three for the caseS = 1 (as many as the possible values of projection onto the z-axis for an angularmomentum with quantum number equal to 1), and one for S = 0 (since thespin vector and its projection vanish in this case). If we denote these states inthe same manner as the spherical harmonics Y𝓁m𝓁

(i.e., XS,mS), then the four

wavefunctions we seek are

X11,X1,−1,X10⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟

S=1

and X00⏟⏟⏟

S=0

.

The first two of these (X11 and X1,−1) are readily constructed as follows:

X11(𝜇1, 𝜇2) = X+(𝜇1)X+(𝜇2) (1)

and

X1,−1(𝜇1, 𝜇2) = X−(𝜇1)X−(𝜇2), (2)

that is, as products of the spin wavefunctions of the two particles, since the rel-evant events (i.e., spin orientation for each particle) are independent, and hencethe corresponding probability amplitudes are multiplied. For the cases (1) and(2) in particular, it is clear that there is only one combination of individual spinswith the desired projection onto the z-axis: To obtain the state with projection+1, both particles must be in the state with “spin up,” while for the −1 projection,both spins must be in the state with “spin down.” But the other two wavefunctions(X10 and X00, which both correspond to zero total projection) can be realized intwo ways: Either particle #1 has spin up and particle #2 has spin down, or, con-versely, particle #1 has spin down and particle #2 has spin up. The correspondingwavefunctions for these two cases are

X+(𝜇1)X−(𝜇2), X−(𝜇1)X+(𝜇2),

while any linear combination of them will also be a wavefunction with projectionof total spin equal to zero. The desired wavefunctions X10 and X00 will thus havethe form

𝛼X+(𝜇1)X−(𝜇2) + 𝛽X−(𝜇1)X+(𝜇2), (3)

where 𝛼2 + 𝛽2 = 1 for normalization. Now, since the wavefunction X10 has thesame magnitude of total spin as the above states (1) and (2), and can be obtainedfrom them with a suitable rotation in space, it must have the same symmetry asthe states (1) and (2).8 It should be symmetric to the interchange of variables of the

8 This is the same rotation that changes a vector from being parallel to the z-axis (projection +1) tobeing vertical to the axis (projection 0). How such a rotation “acts” on a spin wavefunction is beyondthe scope of our discussion (this is a topic for more advanced textbooks), but it is clear that this kindof transformation cannot alter the symmetry of the wavefunction.


two particles. This property implies that from the full set of linear combinations(3) we ought to select the symmetric one for the X10 state

X10(𝜇1, 𝜇2) =1√

2(X+(𝜇1)X−(𝜇2) + X−(𝜇1)X+(𝜇2)), (4)

and the antisymmetric one for X00,

X00(𝜇1, 𝜇2) =1√

2(X+(𝜇1)X−(𝜇2) − X−(𝜇1)X+(𝜇2)), (5)

in order to make it orthogonal to (4), as it should be. (Why?)We conclude with a few more mathematical observations that seem necessary,

with regard to the meaning of products such as (1) and (2). Note first that, hadwe insisted on representing the spin wavefunctions as column vectors, then aproduct such as X(1)

+ X(2)+ (where the upper indices characterize different particles)

would make no sense, mathematically: What is the meaning of the product of twocolumns? But if we returned to the original meaning of the spin wavefunctionsX(𝜇) as functions of the discrete variable 𝜇, then the interpretation of productssuch as the one above poses no challenge. It is, simply, the familiar product of twofunctions. Note also that in the notation Xs,ms

(𝜇) (or XS,mS(𝜇1, 𝜇2), for two parti-

cles) the indices refer to the specific spin state described by the wavefunction (sothey take definite values in each case), while 𝜇 is the variable of the function andalways assumes the two values 𝜇 = ±1∕2 (or 𝜇 = −s,… ,+s for arbitrary spin).For example, the two basic spin wavefunctions

X+(𝜇) ≡ X1∕2,1∕2(𝜇) and X−(𝜇) ≡ X1∕2,−1∕2(𝜇)are two specific functions of the variable 𝜇; in particular,

X+(𝜇) =

{1, 𝜇 = 1∕20, 𝜇 = −1∕2

, X−(𝜇) =

{0, 𝜇 = 1∕21, 𝜇 = −1∕2

.

In the same manner, you can see which specific functions of the two discretevariables𝜇1 and𝜇2 are (1) and (2), or (4) and (5). For instance, you may write downwhich values the function X10(𝜇1, 𝜇2) assumes for all possible combinations of itsvariables. This useful exercise could help you make sure that you have indeedunderstood the meaning of the expressions we wrote above. If you are interestedin a more abstract mathematical description of the pertinent products (the keyterm here is tensor product) then you are advised to consult a more advancedtextbook on quantum mechanics.

Problems

10.8 Show that, for arbitrary values of the quantum numbers 𝓁 and s (orbitalangular momentum and spin), the magnitude of the total magneticmoment of the electron is given by

|𝝁| = 𝜇B(2j(j + 1) + 2s(s + 1) − 𝓁(𝓁 + 1))1∕2,

where 𝜇B (= eℏ∕2mec) is the Bohr magneton. Apply this result to thestates 2p1∕2 and 2p3∕2 of the hydrogen atom.


10.9 Two neighboring—and practically stationary—nuclei are character-ized by spin vectors s1 and s2, and spin quantum numbers s1 = 1 ands2 = 7∕2, respectively. The magnetic interaction between the two nucleiis described by the following Hamiltonian:

H = Aℏ2 s1 ⋅ s2,

where A = 4 meV. Calculate the allowed energy levels of the system, aswell as the corresponding value of the total-spin quantum number, S, ineach case.

10.10 When spin–orbit coupling is taken into account, it may be shown that theprojections of the spin and angular momentum vectors onto the z-axisare no longer conserved quantities. Since the total angular momentumvector, j = 𝓵 + s, remains a conserved quantity, the state of the atom willbe characterized by the quantum numbers of its conserved quantities,namely, the quantum number n (due to energy conservation), the quan-tum numbers 𝓁 and s (due to conservation of the magnitude of the orbitalangular momentum and spin vectors), and the quantum numbers j andmj (due to conservation of both the magnitude and z-projection of thetotal angular momentum vector).

Given that the spin quantum number s is “frozen” to the value s = 1∕2,the wavefunction describing an arbitrary state of the atom can be writ-ten as 𝜓 = 𝜓n𝓁jmj

, which includes both the spatial wavefunction and thewavefunction describing the electron’s spin state. Based on the above, youare asked to do the following:(a) Write down the wavefunctions 𝜓21,3∕2,3∕2 and 𝜓21,3∕2,−3∕2.(b) As you will quickly realize, determining the wavefunction 𝜓21,3∕2,1∕2

is a more difficult task, as the state it describes has mj = 1∕2, which,given that mj = m𝓁 + ms (since jz = 𝓁z + sz), can be realized withmore than one combination of m𝓁 and ms values. If you identifythese combinations, it might not surprise you to see an expressionsuch as

𝜓21,3∕2,1∕2 =√

23𝜓210X+ + 1

√3𝜓211X−, (1)

or

𝜓21,1∕2,1∕2 = 1√

3𝜓210X+ −

√23𝜓211X−. (2)

Can you show that you can go from (1) to (2) by performing a simplecalculation?

Further Problems

10.11 Show that the components of the total angular momentum j = 𝓵 + ssatisfy the basic commutation relations

[jx, jy] = iℏjz, and cyclic permutations.


Therefore, the corresponding vector can indeed be called angularmomentum.

10.12 For a two-electron system (say, a He atom) the total orbital angularmomentum 𝓵 is equal to the sum of the orbital angular momenta 𝓵1and 𝓵2 of the two electrons. That is, 𝓵 = 𝓵1 + 𝓵2. Show that the product𝜓n1𝓁1m1

(r1) ⋅ 𝜓n2𝓁2m2(r2) of the single-particle eigenfunctions 𝜓n1𝓁1m1

and 𝜓n2𝓁2m2is also an eigenfunction of the projection 𝓁z = 𝓁1z + 𝓁2z

with eigenvalue ℏm, where m = m1 + m2. Explain why this result isqualitatively evident.

10.13 We can use the previous result (m = m1 + m2) to prove the law of addi-tion of angular momenta (𝓁 = |𝓁1 − 𝓁2|,… ,𝓁1 + 𝓁2), based on the fol-lowing simple idea. Given two quantum numbers 𝓁1 and 𝓁2 (and there-fore, given also two sets of values of the quantum numbers m1 and m2)we can write down the full set of possible values of m (= m1 + m2) andthen examine which values of 𝓁 are needed to produce this set of m val-ues. One systematic way to do this is to write down the values of m inan array, with m1 and m2 denoting the line and column respectively, andplace their sum in the element (m1,m2) of the array. Construct such anarray for the case 𝓁1 = 3, 𝓁2 = 2, and explain what you see. In particular,describe what values of m you encounter as you traverse the first line ofthe array from left to right, then the last column from top to bottom, andso on, for the rest of the array. What do you conclude?

10.14 Apart from spin–orbit coupling, a more accurate description of theenergy spectrum of hydrogen requires another type of magnetic inter-action, the so-called hyperfine interaction, to be taken into account.The term hyperfine interaction pertains to the coupling between themagnetic moments of the proton and the electron. Even though thedetailed form of the hyperfine interaction potential, VHF, can be derivedfrom first principles, for our purposes we will adopt the followingphenomenological model:

VHF = Aℏ2 S ⋅ I, (1)

where S and I are the spin vectors of the electron and the proton,respectively. The system’s total spin vector is commonly denoted byF (F = S + I). The parameter A appearing in (1), which has units ofenergy, is a measure of the hyperfine-interaction strength.(a) What is the order of magnitude of the parameter A? How does this

compare to the typical order of magnitude of spin–orbit interactions?Express your answer in units of eV in both cases.

(b) Is there any relation between the results you obtained in (a) and thewell-known 21 cm hydrogen line that is ubiquitous in both interstel-lar and intergalactic space? How large must A be to explain the wave-length value of this spectral line?

305

11

Identical Particles and the Pauli Principle

11.1 Introduction

We will now discuss the second foundational premise—the first one was theuncertainty principle—of quantum theory: the Pauli principle. This will allow usto pursue, in the following chapters, our ultimate goal, which is to understandthe structure of matter from first principles. To understand the structure ofatoms—and how their basic properties are mapped in a periodic table—andproceed from there to construct the quantum theory of the chemical bond andextend it to crystalline solids. Let us begin.

11.2 The Principle of Indistinguishability of IdenticalParticles in Quantum Mechanics

The concept of identical particles is surely the same in both classical and quantummechanics. We call identical all those particles that share the exact same physicalproperties: mass, charge, spin, baryon or lepton number, and any other quantumnumber required for their complete identification. Put in a different way, all par-ticles of the same species are identical: all electrons, all protons, all photons, andso on.

But there is one fundamental difference between classical and quantummechanics when it comes to distinguishing identical particles of a physicalsystem. In classical mechanics, we can always tell one identical particle fromanother because of the uniqueness of their orbits that allows us to know at anymoment which one is particle #1, which one is particle #2, and so on. In contrast,in quantum mechanics, it is impossible to distinguish between particles of thesame physical system (e.g., electrons in an atom), since these are describedby overlapping wavefunctions that allow the particles to be found at the samepoint in space, which renders their identification impossible. This fundamen-tal difference between classical and quantum mechanics is demonstrated inFigure 11.1.


306 11 Identical Particles and the Pauli Principle

11

2 2

(a) (b)

33

Figure 11.1 The fundamental difference between classical and quantum mechanics withrespect to the distinguishability of identical particles. The electrons of a classical atom in(a) can always be distinguished owing to the uniqueness of their trajectory. In contrast, it isimpossible to distinguish them in the corresponding quantum atom (b), since the electronsare now described by overlapping wavefunctions, and can thus be found at the same point inspace.

Therefore, in quantum mechanics, it is impossible to distinguish between iden-tical particles of the same physical system. This fundamental feature is known asthe principle of indistinguishability of identical particles.

A direct consequence of this principle is that identical particles of the samephysical system, which therefore have mutually overlapping probability clouds,must be described in such a way that it is fundamentally impossible to say whichone is #1, #2, and so on. While this requirement may sound trivial, it has dramaticconsequences on the quantum mechanical description of a system of identicalparticles, as we will see shortly.

11.3 Indistinguishability of Identical Particlesand the Pauli Principle

Let us examine, for simplicity, a system of two identical particles, say, the twoelectrons of a helium atom. If we ignore the electron spin, we can describe thissystem by a wavefunction 𝜓(r1, r2) whose amplitude squared,

P(r1, r2) = |𝜓(r1, r2)|2, (11.1)

yields the probability density P(r1, r2) to find particle #1 in the vicinity of point r1and particle #2 in the vicinity of point r2.

But since the particles are assumed to be identical and, therefore, indistinguish-able in a quantum mechanical context, the probability density (11.1) should notdepend on their enumeration, namely, which one we call particle #1 or #2. There-fore, the result (11.1) should be invariant to the interchange of labels of the twoparticles, that is, the change r1 → r2 and r2 → r1, or, more succinctly, r1 ↔ r2.Imposing this requirement on (11.1), we obtain

P(r2, r1) = P(r1, r2) ⇒ |𝜓(r2, r1)|2 = |𝜓(r1, r2)|2,

which is certainly satisfied when

𝜓(r2, r1) = ±𝜓(r1, r2), (11.2)

11.4 The Role of Spin: Complete Formulation of the Pauli Principle 307

that is, when the wavefunction of the two particles is either symmetric (+ sign)or antisymmetric (− sign) with respect to the interchange of its variables.

The result (11.2), which emerged as a sufficient but not necessary condition,can be proved in a much more general way, as follows. Since particles #1 and #2are indistinguishable, the wavefunction 𝜓(r2, r1) we obtain by exchanging labels1 and 2 should be physically equivalent to 𝜓(r1, r2). The two wavefunctions𝜓(r2, r1) and 𝜓(r1, r2) must be therefore related via the expression

𝜓(r2, r1) = ei𝛼𝜓(r1, r2), (11.3)

since, as we know, two wavefunctions that describe the same physical state (i.e.,they are physically equivalent) can only differ by a constant phase factor. If wefurther request that (11.3) be valid when we interchange once again the labels ofthe two particles (1 ↔ 2), we obtain

𝜓(r1, r2) = ei𝛼𝜓(r2, r1),

which, in conjunction with (11.3), leads to

e2i𝛼 = 1 ⇒ ei𝛼 = ±1,

so that (11.3) takes the final form

𝜓(r2, r1) = ±𝜓(r1, r2),

which says once again that: The wavefunction of a system of two identical particlesmust be either symmetric or antisymmetric with respect to the interchange of itsvariables.

11.4 The Role of Spin: Complete Formulationof the Pauli Principle

The extension of the above analysis to particles with spin—since we have elec-trons in mind—requires the expressions of spin wavefunctions with a given totalspin, which we gave in the previous chapter (Section 10.7.2). Actually, all we needto know about those expressions is their symmetry type (symmetric or antisym-metric), which we can readily deduce using simple, intuitive arguments.

We recall that the spin state of a particle with s = 1∕2 can be described by acolumn vector X =

(ab

)

, which can also be viewed as a function X(𝜇) of the dis-crete variable 𝜇 (with 𝜇 = ±1∕2), where the values of X(𝜇) give the probabilityamplitudes to find the particle with spin up or down, respectively. That is,

X(1∕2) = a, X(−1∕2) = b.

For a system of two particles with s = 1∕2, the spin wavefunction (just like thecorresponding spatial wavefunction) is a function X(𝜇1, 𝜇2) of the two discretevariables 𝜇1 and 𝜇2. The total wavefunction of the system can then be written as

𝜓(r1, 𝜇1; r2, 𝜇2) = 𝜓(r1, r2)X(𝜇1, 𝜇2), (11.4)


that is, as a product of the spatial and spin wavefunctions, since the two “mo-tions” (the motion of the particle in space and its spin orientation) are normallyindependent from each other1 and the corresponding probabilities should thusbe multiplied.

If we now require that the wavefunction (11.4) satisfy the principle of indistin-guishability of identical particles, we arrive, as before, at the condition

𝜓(r2, 𝜇2; r1, 𝜇1) = ±𝜓(r1, 𝜇1; r2, 𝜇2), (11.5)namely, that the wavefunction ought to be either symmetric or antisymmetricwith respect to the interchange of spatial and spin variables of the two particles.

A question arises naturally at this point. Which one of the two signs (plus orminus) should we choose in (11.5)? The answer is provided by the followingstatement, known as the Pauli principle, or, more precisely, the generalized Pauliprinciple (to disambiguate from the Pauli exclusion principle, which actuallyfollows from it, as we will see shortly):

PAULI PRINCIPLE The wavefunction of a system of identical particles withinteger spin (s = 0, 1, 2,…)—the so-called bosons2—is symmetric with respectto interchange of their variables. Conversely, a system of identical particleswith half-integer spin (s = 1∕2, 3∕2,…)—the so-called fermions3—is describedby wavefunctions that are antisymmetric with respect to interchange of theirvariables.

According to the above principle, electrons are described by antisymmetricwavefunctions, since their spin of 1∕2 means they are fermions. But to appre-ciate the dramatic consequences of this antisymmetry, it is useful to first obtainthe symmetry type of the spin wavefunctions X(𝜇1, 𝜇2) in the following cases:

Case I: The two electrons have parallel spins (total spin S = 1).Case II: The two electrons have antiparallel spins (total spin S = 0).4

If we denote as X↑↑(𝜇1, 𝜇2) and X↑↓(𝜇1, 𝜇2) the spin wavefunctions for thesetwo cases, the following equations must hold (we derived them in the previouschapter, where we used the standard notation X1,±1,X10 for X↑↑, and X00 for X↑↓)

X↑↑(𝜇2, 𝜇1) = X↑↑(𝜇1, 𝜇2)

1 The spin orientation of atomic electrons depends only on their mutual magnetic interactions,which are significantly weaker than their electrostatic interactions and can thus be ignored to firstapproximation (Section 10.7). As a result, the spatial motion of electrons in the atom does not affecttheir spin at all.2 Named after the Indian physicist Satyendra Nath Bose, who first suggested the correctstatistics—known today as Bose–Einstein statistics—for these particles, especially photons.3 Named after the Italian–American physicist Enrico Fermi (Nobel Prize, 1938), who proposedanother statistics for electrons—the so-called Fermi statistics. Fermi also laid the foundations for thetheory of weak interactions and is credited for the creation of the first nuclear reactor.4 As we saw in the previous chapter, the quantum number S for the total spin S = s1 + s2 of twoparticles takes the values

S = s1 + s2,… , |s1 − s2|,

where s1 and s2 are the spin quantum numbers of each particle. So, for s1 = s2 = 1∕2, we have thetwo cases S = 1 or, S = 0 respectively known as parallel and antiparallel spins.

11.4 The Role of Spin: Complete Formulation of the Pauli Principle 309

Table 11.1 Symmetry type of wavefunctions in a system of two electrons.

Total spin Spin wavefunction Spatial wavefunction Total wavefunction

S = 1 (parallel spins) Symmetric Antisymmetric AntisymmetricS = 0 (antiparallel spins) Antisymmetric Symmetric Antisymmetric

and

X↑↓(𝜇2, 𝜇1) = −X↑↓(𝜇1, 𝜇2).

In words, the “wavefunction” for parallel spins (actually, the triplet of the corre-sponding wavefunctions) is symmetric with respect to their interchange, whilethe wavefunction for antiparallel spins is antisymmetric. Actually, we can arriveintuitively at this conclusion. Indeed, the arrangement of two parallel vectors (↑↑)remains invariant when they are interchanged, whereas the antiparallel arrange-ment (↑↓) changes sign, since the upward vector becomes downward and viceversa:

↑↓interchange−−−−−−−−→ ↓↑= − ↑↓ .

It is straightforward to apply the above considerations to a two-electron system;the results are summarized in Table 11.1.

The take-home message of Table 11.1 is the following. Since the total wavefunc-tion Ψ = 𝜓(r1, r2)X(𝜇1, 𝜇2) must always be antisymmetric—remember, electronsare fermions—the symmetry type of the spatial wavefunction is always oppositeto that of the spin wavefunction, and is therefore determined by the relative orien-tation of the two electron spins. Thus, if the electrons have parallel (antiparallel)spins, then their spatial wavefunction is necessarily antisymmetric (symmetric).

But the symmetry of the spatial wavefunction 𝜓(r1, r2)—namely, whetherit is symmetric or antisymmetric—has direct consequences for the motion ofelectrons in space. If 𝜓 is antisymmetric, then

𝜓(r2, r1) = −𝜓(r1, r2),

so that, for r1 = r2 = r, we find

𝜓(r, r) = −𝜓(r, r) ⇒ 𝜓(r, r) = 0,

which means that the probability amplitude to find both electrons at the samepoint in space is identically zero! It is not difficult to recognize that this propertyis a manifestation of the Pauli exclusion principle. Indeed, since the two particleshave the same spin, they cannot be found at the same point in space, because ifthey were, they would have had the same “quantum numbers,” that is, the exactsame physical features: Immediately after a measurement found them at the samespot and with equal spin, the two particles would be described by the same quan-tum state, which is certainly forbidden by the exclusion principle, as we knowfrom basic physics.

Furthermore, note that the vanishing of a function such as 𝜓(r1, r2) for r1 = r2implies that 𝜓(r1, r2) also takes very small values when the distance |r1 − r2|


between the two points r1 and r2 is small. In other words, when the two par-ticles (say, two electrons) have parallel spins, they stay away from each other!Conversely, when the particles have antiparallel spins, they show no such ten-dency to mutually avoid each other. The spatial wavefunction 𝜓(r1, r2) is thensymmetric—that is, 𝜓(r2, r1) = 𝜓(r1, r2)—and need not vanish when r1 = r2, ortake small values when the particles are close together. In fact, symmetric wave-functions generally tend to take large values when the particles are in proximity(i.e., for small |r1 − r2|) and thus favor their coexistence in the same region ofspace. The conclusion is simple, yet astounding: Parallel spins avoid each other,while antiparallel spins prefer to be near each other!

A direct consequence of the above discussion in the case of electrons isthat the parallel orientation of their spins is energetically favorable, since theelectrons then stay apart and reduce their electrostatic repulsions. The existenceof permanent magnets—the familiar ferromagnets—is a spectacular manifes-tation of this mechanism. To decrease their electrostatic repulsions (and thusminimize their energy), the conduction electrons of a ferromagnet align theirspins—those microscopic magnets—creating a powerful magnet of macroscopicscale. It may seem counterintuitive, but macroscopic magnetism results fromthe synergy of electrical forces and the Pauli principle. The atomic magnetsare not aligned because of their own magnetic interactions but because of theelectrostatic repulsion of electrons, which is minimized when their spins becomeparallel, since the exclusion principle forces the electrons to stay away from eachother.

11.5 The Pauli Exclusion Principle

We will now show that the well-known Pauli exclusion principle—namely, thattwo electrons in the same atom can never have the same quantum numbers—isa direct consequence of the (generalized) Pauli principle we introduced above,applied to a system of identical independent fermions.

First, let us clarify that we call the particles of a system “independent” if theymove under the influence of the same external potential but have no mutual inter-actions.

If V (r) is the common external potential, the Hamiltonian of this system iswritten as

H =p2

1

2m+ V (r1)

⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟H1

+p2

2

2m+ V (r2)

⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟H2

+ · · · +p2

N

2m+ V (rN )

⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟HN

= H1 + H2 + · · · + HN , (11.6)that is, as the sum of N independent Hamiltonians that are essentially the same,except that each one has the variable of a different particle. Because of this inde-pendence, it is not hard to see that the eigenvalues E and the eigenfunctions𝜓(r1,… , rN ) of the total Hamiltonian H are respectively given by the expressions

E = E1 + E2 + · · · + EN (11.7)

11.5 The Pauli Exclusion Principle 311

and

𝜓(r1,… , rN ) = 𝜓1(r1) · · ·𝜓N (rN ), (11.8)

which tell us the following:The total energy of a system of independent particles is the sum of the energies ofits constituent particles, while the total wavefunction is the product of the wave-functions of the individual particles.

Proof : We have to show that H𝜓 = E𝜓 with H,E, and 𝜓 given by (11.6), (11.7),and (11.8), respectively. We have

H𝜓 = (H1 + · · · + HN )𝜓1 · · ·𝜓N = H1(𝜓1 · · ·𝜓N ) + · · · + HN (𝜓1 · · ·𝜓N )= (H1𝜓1) · · ·𝜓N + · · · + 𝜓1 · · · (HN𝜓N )= E1𝜓1 · · ·𝜓N + · · · + EN𝜓1 · · ·𝜓N

= (E1 + · · · + EN )𝜓1 · · ·𝜓N = E𝜓. □

The key point in the above sequence of calculations5 is the fact that eachsingle-particle Hamiltonian Hn(n = 1,… ,N) acts only on the correspondingsingle-particle eigenfunction 𝜓n(rn), which satisfies by definition the eigenvalueequation Hn𝜓n = En𝜓n.

The practical implication of the above theorem is that, to find the eigenval-ues and eigenfunctions for a system of identical independent particles, we simplyhave to solve the single-particle Schrödinger equation

(

− ℏ2

2m∇2 + V (r)

)

𝜓(r) = E 𝜓(r)

for the particular external potential V (r), and then “populate” the resulting eigen-states with the available particles of the system. The total eigenfunction is thenthe product of the individual eigenfunctions and the total eigenvalue is the sumof the individual eigenvalues.

From a physical perspective the above implication is rather evident. When theparticles of a system do not interact with each other, they do not affect the motionof one another. Therefore, the probabilities to find the particles somewhere inspace are independent and are thus to be multiplied, as it always happens forprobabilities of independent events. For the same reason, we should also multi-ply the corresponding probability amplitudes, that is, the individual eigenfunc-tions. When there is no interaction energy term, it is also clear that the system’stotal energy is simply the sum of the energies of its constituent particles, as inEq. (11.7).

Let us now examine the case of atomic electrons. These are definitely notindependent particles, since they interact through electrostatic forces. Theirinteraction energy is equal to

V (r1, r2) =e2

|r1 − r2|,

5 For those readers who have found it difficult to follow the proof for arbitrary N , we suggest theyrepeat it for N = 2, so that H = H1 + H2, E = E1 + E2, and 𝜓 = 𝜓1𝜓2.


where r1, r2 are the position vectors for a particular pair of electrons. Butif this interaction is ignored, or if it is taken into account in a way we willdiscuss in the next chapter—the crucial concept here being that of an effectivepotential that comprises both the attraction from the nucleus and the electronicrepulsions—then the atomic electrons can indeed be regarded as independentparticles, and the above analysis applies in full.

In particular, let us consider the simplest many-electron atom (the heliumatom) and assume that one of its two electrons is in the 𝜓𝛼 state, while the otheris in the 𝜓𝛽 state. Here, 𝛼 and 𝛽 are “collective indices” representing the triplets

𝛼 ≡ n1𝓁1m1, 𝛽 ≡ n2𝓁2m2

of the quantum numbers n𝓁m that are required to describe the eigenfunctions ofa hydrogen-like system.6 According to the above discussion, the wavefunction ofthe two-electron system has the form

𝜓𝛼𝛽(r1, r2) = 𝜓𝛼(r1)𝜓𝛽(r2), (11.9)which is the product of the individual wavefunctions. Given now that the elec-trons are identical particles, we cannot tell #1 from #2, which implies that anequally acceptable wavefunction to (11.9) is the following:

𝜓𝛽𝛼(r1, r2) = 𝜓𝛽(r1)𝜓𝛼(r2) ≡ 𝜓𝛼(r2)𝜓𝛽(r1), (11.10)with the roles of the two electrons interchanged. Specifically, in (11.9) electron#1 is placed in the state 𝛼 and electron #2 is in state 𝛽, while in (11.10) the reversehappens: #1 is placed in the 𝛽 state and #2 is in the 𝛼 state.

But since (11.9) and (11.10) are independent solutions of the Schrödingerequation with the same eigenvalue E = E𝛼 + E𝛽 , any linear combination of themwill also be a solution7

𝜓(r1, r2) = c1𝜓𝛼𝛽 + c2𝜓𝛽𝛼 = c1𝜓𝛼(r1)𝜓𝛽(r2) + c2𝜓𝛽(r1)𝜓𝛼(r2), (11.11)where the constants c1 and c2 must now be selected so that the spatial wave-function 𝜓(r1, r2) has the symmetry required by the Pauli principle, as shown inTable 11.1. Therefore, from the linear combinations (11.11) we need only thosecases that give a symmetric wavefunction 𝜓S,

𝜓S = 1√

2(𝜓𝛼𝛽 + 𝜓𝛽𝛼) =

1√

2(𝜓𝛼(r1)𝜓𝛽(r2) + 𝜓𝛽(r1)𝜓𝛼(r2)), (11.12)

or an antisymmetric wavefunction 𝜓A,

𝜓A = 1√

2(𝜓𝛼𝛽 − 𝜓𝛽𝛼) =

1√

2(𝜓𝛼(r1)𝜓𝛽(r2) − 𝜓𝛽(r1)𝜓𝛼(r2)), (11.13)

where the factor 1∕√

2 is introduced for normalization. We recall that a linearcombination of normalized wavefunctions must be divided by the square root of

6 We remind the readers that a hydrogen-like system is an atom with only one electron, but withnuclear charge Ze.7 Since the Schrödinger equation H𝜓 = E𝜓 is linear and homogeneous, any linear combination ofits solutions—with the same eigenvalue—is also a solution.

Problems 313

the sum of the squares of the coefficients, to produce again a normalized wave-function. Here, the wavefunctions 𝜓𝛼𝛽 and 𝜓𝛽𝛼 are assumed to be normalized, sothe normalization factor is equal to 1∕

√2. Let us also note that the states 𝜓𝛼𝛽

and 𝜓𝛽𝛼 derive from one another via the interchange r1 ↔ r2, so their symmet-ric and antisymmetric combination will be given by their sum and difference,respectively.

Based on the above, the wavefunctions for the two cases of total spin (S = 1 andS = 0) are given by the expressions

𝜓↑↑ = 𝜓A(r1, r2)X↑↑(𝜇1, 𝜇2) (11.14)

and

𝜓↑↓ = 𝜓S(r1, r2)X↑↓(𝜇1, 𝜇2). (11.15)

Note now that when 𝛼 ≡ 𝛽, that is, when the two spatial wavefunctions areidentical, their antisymmetric combination (11.13) is identically zero. In this case,the only possible state of the system is (11.15), and the particles have antiparallelspins. This is simply an example of the Pauli exclusion principle, which states that“the coexistence of two electrons in the same quantum state is impossible. Thetwo electrons must differ by at least one quantum number.”

This conclusion is so significant that it is worthwhile repeating the reasoningbehind it. The idea is straightforward. When the two particles have the same spa-tial quantum numbers (𝛼 = 𝛽), their spatial wavefunctions are identical and wecan form only a symmetric combination from them—the product 𝜓𝛼(r1)𝜓𝛼(r2).The antisymmetry of the total wavefunction requires then that the spin wave-function be antisymmetric, so the spins are antiparallel.

We thus see that the Pauli exclusion principle is merely a consequence of thegeneralized Pauli principle for the special case of a system of two independentfermions. The generalization to more than two independent particles requiresthe use of group theory, and we will not present it here. Nevertheless, it is evidentthat the basic rule “one electron per state” will continue to apply for a system withan arbitrary number of particles. (Readers are encouraged to argue why this is thecase.)

Problems

11.1 Ten identical noninteracting particles coexist in a one-dimensional poten-tial box whose ground state energy is equal to E1 = 1eV. The lowest pos-sible total energy of the particles is equal to 22 eV. Find the spin of eachparticle and the minimum excitation energy of the system.

11.2 Eight electrons—whose mutual repulsions we consider to be very small,but not exactly zero—move on a circle of radius 2 Å. Calculate, in eV, thelowest total energy, the total spin, and the minimum excitation energy ofthe system.


11.3 Twelve weakly repelling, identical particles with spin 1∕2 move on the sur-face of a sphere. What is the total spin in the lowest energy state of thesystem?

11.4 Consider a system of two identical, spin-1/2 particles moving in onedimension. Which of the following wavefunctions can describe such asystem?(a) 𝜓↑↑(x1, x2) = N(x2

1 + x22)e

−𝜆(x21+x2

2)∕2,(b) 𝜓↑↓(x1, x2) = N sin[k(x1 − x2)] e−𝜆|x1−x2|,(c) 𝜓↑↓(x1, x2) = N cos[k(x1 − x2)] e−𝜆|x1−x2|,(d) 𝜓↑↑(x1, x2) = N(x1 − x2) e−𝜆(x1−x2)2∕2,(e) 𝜓↑↓(x1, x2) = N

x1 − x2

cosh[𝜆(x1 − x2)].

In the above expressions, the up/down arrows denote whether the spinsof the particles are parallel or antiparallel, while x1 and x2 are the positionvariables of the particles.

11.5 Two identical, noninteracting spin-1/2 particles occupy the ground andfirst excited states of an infinite potential well, respectively. Write downthe wavefunction describing the two-particle system if the total spin is(a) S = 1, (b) S = 0. Which of the two total-spin states will have the lowestenergy if the two particles are weakly attracting each other?

11.6 Which Particles Are Fermions and Which AreBosons

Let us now examine the implications of the above discussion for the varioustypes of identical particles encountered in nature. First of all, electrons havespin s = 1∕2 and are thus fermions, as we could have expected. For if electronswere bosons, then atomic states could be populated to no limit and the groundstate of a many-electron atom would have all the electrons at the lowest energylevel. This in turn would result in ever-shrinking atomic radii, their size beinginversely proportional to the atomic number Z.8 The atomic volumes wouldthen vary as 1∕Z3, which means that the densities of macroscopic matter wouldspan six orders of magnitude, from g/cm3 to tons/cm3! Much more importantly,the periodic table would not exist and nobody would be around to talk aboutit. Therefore, electrons must surely be fermions. The very fact that we existconfirms that this is indeed the case.

Another important class of particles are photons, the particles of light. Whatkind of spin—integer or half-integer—do you think they should have? Clearly,the photon spin has to be an integer; otherwise, photons would not be allowed tomove collectively and produce a detectable macroscopic wave. In other words,

8 For a hydrogen-like system (which is the model for an arbitrary atom if we ignoreelectron–electron repulsions) the “radius” of the 1s orbital is given by the relationa0(Z) = ℏ2∕me2Z = a0∕Z, where Z is the atomic number of the nucleus.

11.6 Which Particles Are Fermions and Which Are Bosons 315

photons, as well as all other carrier particles of fundamental fields, must bebosons, so as to allow their limitless co-occupation of the same quantum state,which is necessary for the creation of a macroscopic classical field. In particular,the spin of a photon must be equal to one (s = 1). Why?

As for the basic nuclear particles (proton and neutron), they also have tobe fermions. Had the opposite been true, all nuclear matter of the universewould eventually “collapse” to a “nuclear droplet” of unimaginable density,since nothing would prevent the unlimited aggregation of nucleons at theground state formed by their mutual nuclear attractions. Thus there wouldalways be an energy motive for nucleons to aggregate in ever greater num-bers, up to the point where all nuclear matter around would be trappedtogether! In contrast, if nuclear particles are fermions—and indeed theyare—then only finite-sized nuclei can be formed, since the Pauli principle forcesthem to populate ever higher energy levels, causing nuclear radii to increasemonotonically—or almost monotonically—up to the point where short-rangenuclear forces no longer suffice to keep nucleons together and the nucleusdissociates.

As we see, the existence of two types of identical particles—fermions andbosons—is absolutely critical for the structure of our world. In particular, allparticles that are building blocks of matter ought to be fermions, and all particlesthat are carriers of fields ought to be bosons. Otherwise, nobody would bearound to study them!

The following example should help readers appreciate what we have discussedso far.

Example 11.1 Two identical noninteracting fermions with spin 1∕2 “coexist”in a one-dimensional potential box of length L. Write down the spatial wave-function of the system in the following cases: (a) The particles occupy the state oflowest total energy for the system. (b) The particles are in the first excited stateof the system and have a total spin equal to (i) S = 1, (ii) S = 0. Answer the samequestions if the particles of the system are bosons with zero spin.

Solution: We recall first that the eigenvalues and the (normalized) eigenfunctionsof the one-dimensional box are given by the expressions

En = ℏ2𝜋2

2mL2 n2 = n2E1

(

E1 = ℏ2𝜋2

2mL2

)

and

𝜓n =√

2L

sin n𝜋xL,

where L is the length of the box and n = 1, 2,… ,∞. It is clear that in thestate of minimum total energy, the two particles occupy the ground state ofthe box (n = 1) with opposite spins, so their spatial wavefunction is given by theproduct

𝜓(x1, x2) = 𝜓1(x1)𝜓1(x2) =2L

sin𝜋x1

Lsin

𝜋x2

L,

which is clearly a symmetric function under the interchange x1 ↔ x2. So, whenwe combine the spatial wavefunction with the spin wavefunction X↑↓(𝜇1, 𝜇2) of


opposite spins (S = 0), we obtain an antisymmetric total wavefunction, in accor-dance with the Pauli principle for fermions. Now, the first excited state of thesystem corresponds to placing one particle at the ground level n = 1 and the otherat the n = 2 level. In this case the spatial wavefunction of the system can have oneof the two forms

𝜓S(x1, x2) =1√

2(𝜓1(x1)𝜓2(x2) + 𝜓1(x2)𝜓2(x1))

= 1√

22L

(

sin𝜋x1

Lsin

2𝜋x2

L+ sin

𝜋x2

Lsin

2𝜋x1

L

)

(symmetric)

or

𝜓A(x1, x2) =1√

2(𝜓1(x1)𝜓2(x2) − 𝜓1(x2)𝜓2(x1))

= 1√

22L

(

sin𝜋x1

Lsin

2𝜋x2L

− sin𝜋x2

Lsin

2𝜋x1L

)

(antisymmetric).

The choice of one or the other form depends of course on the relative orienta-tion of the spins of the two particles: If the spins are parallel (so that S = 1), thespin wavefunction is symmetric and the spatial wavefunction must thus have theantisymmetric form 𝜓A(x1, x2), and vice versa for antiparallel spins (S = 0).

0

(a) (b)

2S = 1

PA(x1, x2) = ψA2(x1, x2) Ps(x1, x2) = ψs

2(x1, x2)

1

2S = 0

1

1 1

01 1

x2 x2

x1 x1

Second diagonal First diagonal

Figure 11.2 Density plots for the probability distributions PA and PS corresponding to paralleland antiparallel spin arrangement for two particles that populate the first two levels of aninfinite potential well. In the parallel arrangement, “hills” are found along the second diagonal(where the distance |x1 − x2| is large), while in the antiparallel arrangement, the probabilitydensity is maximized when the interparticle distance is small, that is, along the first diagonal.Therefore, parallel spins (a) like to stay apart, while antiparallel spins (b) prefer to get together.Note that the horizontal and vertical axes represent x1 and x2, respectively.

11.7 Exchange Degeneracy: The Problem and Its Solution 317

If, now, the particles are bosons with zero spin, then their spatial wavefunctionin the state of lowest total energy is the same as before, while for the first excitedstate of the system, the spatial wavefunction—which is the full wavefunction now,that is, there is no spin part—will necessarily take the symmetric form 𝜓S(x1, x2).

This example is also suitable for a graphical demonstration of a property wementioned earlier, namely, that for particles with s = 1∕2, parallel spins “are keptapart,” while antiparallel spins “tend to approach each other!” To this end, weneed to plot the two-variable functions

PS(x1, x2) = 𝜓2S (x1, x2), PA(x1, x2) = 𝜓2

A(x1, x2)as a so-called density plot on a computer. Such plots show the values of a functionF(x, y) on the x–y plane using gradual shading and with the following convention:The bright white regions represent the “hills” (where the function takes large val-ues), while the very dark regions correspond to “valleys” (where the value of thefunction tends to zero). Figure 11.2 shows a density plot for L = 1 using Mathe-matica (the particular command is DensityPlot).

11.7 Exchange Degeneracy: The Problemand Its Solution

There is one more thing worthy of a comment. As we saw earlier—formula(11.11)—we may know all quantum numbers that determine the state of eachof the two particles separately, but unless we invoke the Pauli principle, weare still unable to fully determine the state of the system. All superpositionstates (11.11) are equally plausible and no further measurement can distinguishany one of them as the correct quantum state that corresponds to the full setof measurements performed. In other words, there is a residual degeneracy,which—in contrast to all kinds of degeneracy we encountered so far—cannotbe lifted by further measurements, simply because no such measurementsexist. Had they existed, we would have been able to distinguish particle #1from particle #2, in contrast to the assumption that these are identical—andhence indistinguishable—particles. This is the well-known exchange degeneracy,and it reveals a serious gap in the theory with regard to the description ofidentical particles. Therefore, to describe identical particles in the context ofquantum theory we need an additional postulate that would allow us to uniquelydetermine the system’s wavefunction from all degenerate states (11.11). And aswe know, this additional postulate is the Pauli principle.

Nonrelativistic quantum mechanics, which is the subject of this book, identifiesthe problem and makes some progress toward its resolution—namely, that ourwavefunctions have to be symmetric or antisymmetric—but fails to suggest howto select one symmetry type over another. This issue is definitively settled in thecontext of relativistic quantum mechanics (Dirac equation), which provides alsoa resolution to the problems of spin and its singular magnetic behavior (spin’smagnetic anomaly). But in the context of nonrelativistic quantum mechanics, thePauli principle is a statement that cannot be proved, and thus needs to be addedto the three postulates of Section 2.7 as the fourth postulate.


Postulate 4: The Pauli principle. All identical particles with integer spin aredescribed by wavefunctions that are symmetric with respect to exchange (i.e.,interchange) of their variables, and all identical particles with half-integer spinhave wavefunctions that are antisymmetric.

And there we have it: The complete and definitive formulation of quantummechanics in four postulates. Can this theory succeed in describing the real worldin all its richness and complexity? This is the question to be answered in thenext—and last—part of the book.

Further Problems

11.6 2N identical particles with spin s = 1∕2 and mass m are trapped in aone-dimensional box (tubule) of length L.(a) Show that the lowest total energy of the system is given by the

expression

Etot = 2E1

(13

N3 + 12

N2 + 16

N)

,

where E1 is the ground state energy of the box.(b) Calculate now the average energy per particle (𝜖 = Etot∕N) in the limit

of large N . Show that the result, if we ignore numerical coefficients onthe order of one, is the same as what we would find assuming the par-ticles “split” the available space equally among themselves, and occupyin pairs little “tubules” of length a = L∕(N∕2). The conclusion of thisproblem has a general validity:For order-of-magnitude estimates, and for a large number of indepen-dent fermions, the Pauli principle is equivalent to the assumption thatthe particles in the system share the available space equally amongthemselves.In Chapter 15 we will have the opportunity to discuss some remarkableconsequences of this fundamental inference.

11.7 A billion noninteracting electrons are enclosed in a cubic box with sidesof 1000 Å. Apply the final conclusion of the previous problem to estimate,in eV, the average kinetic energy of each electron at the temperature ofabsolute zero. What is the broader conclusion here? Can we compress afermionic system as much as we wish, even if we ignore completely anyrepulsion between the particles? What can you infer for the case of deadstars? Can gravity cause the “collapse” of a star when the latter runs out ofnuclear fuel and there is no longer any thermal pressure to sustain it?

11.8 The first excited state of the He atom corresponds to the state whereone of its two electrons goes from the 1s to the 2s state (while the other


electron remains in the 1s state). In fact, there exist two excited statesof He corresponding to this electron configuration, depending on therelative orientation of their spin vectors. These states are referred to asorthohelium when S = 1 (parallel spins), and parahelium when S = 0(antiparallel spins).

Consider the following questions:(a) Is there an energy difference between these two states? If yes, which

one has a lower energy?(b) What would you reply if you were told that the above energy differ-

ence is on the order of 10−4 eV (which is the typical magnitude of themagnetic interaction energy in atoms)?

(c) What would be the wavefunction describing the electrons in eachcase if electron–electron repulsions were ignored? Use atomic unitsto express your answer.

321

Part III

Quantum Mechanics in Action: The Structure of Matter

323

12

Atoms: The Periodic Table of the Elements

By convention sweet is sweet, bitter is bitter, hot is hot, cold is cold, andcolor is color. But in reality there exist only atoms and the void.1

Democritus

12.1 Introduction

We now enter the most important part of the book. Here, quantum theoryis called upon to justify the great expectations that marked its ascent and todemonstrate its explanatory power in the realm where classical physics failsirrevocably: the structure of matter. In this part we will not deal, of course, with allmaterial structures—which range from quarks to the whole universe—but onlywith structures that, while being simplest in form, are most interesting in termsof physics or technological applications: atoms, molecules, and crystalline solids.

It is only natural to begin our study of the structure of matter with atoms. Thisis the realm where classical physics suffered its crashing defeat and where crucialexperimental data and ideas emerged, guiding the search for the new theory.

Quantum mechanics originated from the atom (this microscopic planetarysystem) in the same way that classical mechanics emerged from the observationof our macroscopic solar system. Recall that the collapse of the Aristotelian lawof motion, which postulated that a constant force is needed to sustain uniformmotion, was caused by the irrefutable observation that we live on a moving Earth(without losing the ground under our feet when we jump!)—an observationthe (then) New Mechanics successfully explained (via the principle of inertia)together with the detailed motion of planets in the solar system.

For a single-electron atom (i.e., hydrogen or a hydrogen-like ion) a quantummechanical explanation has already been given. It addressed the fundamentalquestions regarding atomic stability and provided detailed predictions for theexcited states of hydrogen-like atoms, which are resoundingly confirmed byexperiment.

Our main interest in this chapter will thus be on many-electron atoms.

1 Durant, W. (1939) The Story of Civilization: Part II – The Life of Greece, New York: Simon andSchuster.


324 12 Atoms: The Periodic Table of the Elements

12.2 Arrangement of Energy Levels in Many-ElectronAtoms: The Screening Effect

A many-electron atom is a very complex physical system. Its electrons sense notonly the attraction from the nucleus but also their mutual electrostatic repul-sions, and it is the latter that make it impossible to solve the Schrödinger equationexactly. When electrons interact with each other, the electronic wavefunction canno longer take the simple form of a product, as in (11.8), but becomes instead acomplicated function of all 3N variables r1,… , rN , which satisfies a Schrödingerequation with the same number of variables. For example, for a light atom suchas oxygen, we would have 24 variables, rendering impossible the search for anexact solution in this case. Nevertheless, we can embark on an approximatedescription of the atom and achieve our main goal, namely, to obtain a qualitativeunderstanding of the basic features of the periodic table of elements. The mainidea is the following. From the point of view of an arbitrary electron, the presenceof all other electrons can be approximated by a spherically symmetric “cloud”of negative charge that envelops the positively charged nucleus. We can thusassume that the electrons of the atom move as if they were independent—thatis, noninteracting—particles under the influence of a common effective potentialthat combines both the electrostatic attraction from the nucleus and the mutualrepulsions of the electrons. What does this effective central potential look like? Atvery short distances from the nucleus, our electron penetrates through the nega-tive cloud of the other electrons and “sees” the full charge+Ze of the bare nucleus.But as the electron moves away from the center, the nucleus is gradually screenedby the remaining Z − 1 electrons. At very large distances (r → ∞) these otherelectrons cancel out the charge of the nucleus, except for one electronic unit.

We can thus write the effective potential V (r) in the form of a screenedCoulomb potential

V (r) = −eQ(r)r

, (12.1)

where Q(r) is the effective charge each electron sees at a distance r from thenucleus. Now, the function Q(r) is not known in advance. Clearly, however, itmust be a decreasing function of r, with the following limits at zero and infinity:

Q(r) −−→r→0

Ze, Q(r) −−→r→∞

e.

Had we ignored completely the repulsions among electrons, the effectivepotential V (r) would have been a bare Coulomb potential

VC(r) = −Ze2

r, (12.2)

which is depicted in Figure 12.1, together with the screened potential (12.1), tohighlight their differences.

How does screening (i.e., the electron–electron repulsions) affect thearrangement of the energy levels? This is the key question for our understandingof the periodic table of the elements. For if we know the arrangement of theenergy levels, then all we have to do is to successively populate them with

12.2 Arrangement of Energy Levels in Many-Electron Atoms: The Screening Effect 325

Figure 12.1 Screened and bareCoulomb potentials. The twopotentials coincide near theorigin, while at greater distances,the screened potential is alwayshigher than the bare potential,since it corresponds to adecreased electrostatic attractionfrom the nucleus. Bare Coulomb potential

V(r)

Screened Coulomb potential

r

Figure 12.2 Hydrogen-likeenergy levels (bare Coulombpotential). The main feature ofthis diagram is thehydrogen-like degeneracy, thatis, the coincidence of energylevels with 𝓁 ranging from zeroto n − 1 for a particular n.

3s

2s

1s

3p 3d

2p

electrons, taking into account the limitations imposed by the Pauli exclusionprinciple.

We can readily infer the arrangement of the energy levels, by qualitativelycomparing the screened Coulomb potential with the bare potential, for whichwe know the exact solution. The first few hydrogen-like levels (i.e., thosecorresponding to the bare Coulomb potential) are shown in Figure 12.2.

A direct consequence of screening is the overall rise of hydrogen-like levels.We can understand this intuitively once we realize that the screened Coulombpotential is everywhere higher than the bare potential, so it “pushes” up allenergy levels associated with it. Another way to see this is to realize thatelectron–electron repulsions contribute positively to the energy of the atom, sothey must cause a rise of the energy levels when taken into account.

However, this rise is not the same for all levels. Let us examine, for example,the 2s and 2p levels, which are degenerate in the case of a bare Coulomb poten-tial. For the 2s state the angular momentum is 𝓁 = 0, while for the 2p state it is𝓁 = 1. As we explained in Section 9.3.3.6, when the quantum number of angu-lar momentum increases, the electron is found further away from the immediatevicinity of the nucleus. We remind the readers that the hydrogen eigenfunctions𝜓n𝓁m go to zero at the origin as r𝓁 , that is, they vanish faster for large 𝓁. For 𝓁 = 0


4s4p 4d 4f

3d3p2p3s

2s

1s

Figure 12.3 Energy-leveldiagram for many-electronatoms. Because of screening,the hydrogen-like degeneracyis lifted and the levels for aparticular n are ordered inincreasing values of thequantum number 𝓁.

the wavefunction does not vanish at the origin and there is a finite probabilityto find the electron “on” the nucleus. Therefore, an electron in the 2s state hashigher probability to be near the nucleus than an electron in the 2p state. This,in turn, implies that the 2s electron largely penetrates the negative cloud of theother electrons and is thus exposed to almost the full attraction of the unscreenednuclear charge. In contrast, the 2p electron spends most of its time away from theimmediate vicinity of the nucleus, so it is subjected to a weakened nuclear attrac-tion. We are naturally led to conclude that the 2s level lies lower than the 2p level.

We thus see that electron–electron repulsions lift the hydrogen-likedegeneracy—in a rather predictable way—in addition to causing a generalrise of the hydrogen-like levels. Specifically, the new levels that emerge from thesplitting for a given n are arranged in increasing values of the quantum number𝓁. But the rotational degeneracy remains, since the screened Coulomb potentialis still central, that is, rotationally symmetric. The main conclusion is presentedin Figure 12.3.

As for the eigenfunctions of the atom, these will still have the general form

𝜓n𝓁m(r, 𝜃, 𝜙) = Rn𝓁(r)Y𝓁m(𝜃, 𝜙)

because the screened Coulomb potential is a central potential and thereforeour discussion from the first part of Chapter 9 applies. That is, the angularpart of the eigenfunctions remains the same—and is described by the sphericalharmonics Y𝓁m—while the particular form of the central potential V (r) affectsonly the radial functions R(r) = Rn𝓁(r), which are determined by solving theradial Schrödinger equation

y′′ + 2mℏ2 (E − V (r))y = 0,

where

V (r) = V (r) + ℏ2𝓁(𝓁 + 1)2mr2

and

y(r) = rR(r).

Given now that the general shape of atomic orbitals does not depend on thedetails of the radial function R(r), these orbitals should look the same as those

12.3 Quantum Mechanical Explanation of the Periodic Table: The “Small Periodic Table” 327

for the hydrogen atom and reflect the key feature of each orbital, namely, thetype of angular dependence. Thus, all s orbitals (i.e., 1s, 2s, 3s, etc.) are sphericallysymmetric and correspond to zero angular momentum (𝓁 = 0) for the electron.In contrast, spherical symmetry is absent for p orbitals (𝓁 = 1, i.e., 2p, 3p, 4p,etc.) and also those with even higher quantum number 𝓁.

Moreover, p orbitals constitute a degenerate triplet of states (px, py, pz) withthe same general shape and exactly the same physical content with respect to thethree axes. Each of these p orbitals has zero projection of angular momentumonto the corresponding axis. Thus, the px orbital corresponds to a state withangular momentum magnitude equal to |𝓵| = ℏ

√𝓁(𝓁 + 1)||

|𝓁=1= ℏ

√2, and with

vanishing projection on the x-axis (𝓁x = 0), and likewise for the py and pz orbitals.The corresponding plots of these orbitals are the same as those given in Figure 9.8.

12.3 Quantum Mechanical Explanation of the PeriodicTable: The “Small Periodic Table”

The periodic appearance, as the atomic number increases, of elements with sim-ilar chemical properties, is one of the most startling characteristics of the atomicworld. It is for this reason that the formulation of the periodic table, by the Russianchemist Dmitri Mendeleev in 1869, gave a great boost to the atomic hypothesis.From that point in time, the atomic concept gradually became mainstream. Andyet, for many decades thereafter, the periodic table remained a purely empiri-cal ordering of the elements, without any theoretical justification; just anothermystery of the “wonderland” of the atomic world.

In this section, we will see that quantum theory elucidates the periodic table insuch a spectacular way that is perhaps unprecedented in the history of science.For pedagogical reasons though, we will not examine here the whole periodictable but only its first 18 atoms, which include some of the most importantelements found in nature, such as hydrogen, oxygen, nitrogen, and carbon. This“small periodic table” is shown in Table 12.1, where for each element we havelisted its chemical symbol, name, and ionization energy in electron volts.

We recall that the ionization energy is the energy required for the removal ofone electron from the outer shell of a neutral atom. The magnitude of this quantityis a first “index” of the chemical behavior of the atom, as will become clear in thenext chapter, when we present a detailed discussion on the formation of chemical

Table 12.1 The small periodic table.

Z = 1 (H)Hydrogen

13.6

Z = 2 (He)Helium

24.6Z = 3 (Li)Lithium

5.4

Z = 4 (Be)Beryllium

9.3

Z = 5 (B)Boron

8.3

Z = 6 (C)Carbon

11.3

Z = 7 (N)Nitrogen

14.5

Z = 8 (O)Oxygen

13.6

Z = 9 (F)Fluorine

17.4

Z = 10 (Ne)Neon21.6

Z = 11 (Na)Sodium

5.1

Z = 12 (Mg)Magnesium

7.6

Z = 13 (Al)Aluminum

6.0

Z = 14 (Si)Silicon

8.1

Z = 15 (P)Phosphorus

10.5

Z = 16 (S)Sulfur10.4

Z = 17 (Cl)Chlorine

13.0

Z = 18 (Ar)Argon15.8


H He

1s (Z = 1)

1s (Z = 2)

Figure 12.4 The modern quantum mechanical picture for the first two atoms of the periodictable. In hydrogen, the 1s orbital is occupied by one electron. In helium, the 1s orbital isoccupied by two electrons of opposite spins, which thus form a closed shell. The two orbitalshave the same shape but differ in size, since the corresponding Bohr radius—assuming we canignore electron–electron repulsions—depends on Z through the formulaa0(Z) = (ℏ2∕me2)|e2→Ze2 = a0(1)∕Z. Therefore, the 1s orbital of helium is about two timessmaller than the corresponding orbital of hydrogen.

bonds. For now, we will simply explain qualitatively the chemical significance ofthe ionization energy and how it tends to vary across the periodic table.

12.3.1 Populating the Energy Levels: The Shell Model

Having in mind the arrangement of energy levels in a many-electron atom(Figure 12.3), the “small periodic table” emerges as a mere consequence of thePauli principle. The first row of this table contains only two elements (hydrogenand helium) because this is the number of electrons that can be accommodatedin the first energy shell (i.e., the 1s state). Thus, for hydrogen we have oneelectron in this state, while for helium we have two electrons with opposite spins(Figure 12.4).

We remind the readers that the state of an atomic electron is fully determinedby the quadruplet of quantum numbers (n,𝓁,m𝓁 ,ms). The first three of thesedetermine the spatial wavefunction 𝜓n𝓁m (m ≡ m𝓁), while the fourth numberms determines the spin state and, in particular, the spin projection onto thez-axis. The value ms = 1∕2 corresponds to the state with sz = ℏ∕2 (spin up),while ms = −1∕2 corresponds to the state with sz = −ℏ∕2 (spin down). Since the1s state corresponds to a definite triplet of quantum numbers (n,𝓁,m) = (1, 0, 0),the two electrons in the ground state of the helium atom must differ in the fourthnumber ms: one electron has ms = 1∕2 (spin up), the other ms = −1∕2 (spindown).

Once we fill the 1s shell, we necessarily move on to n = 2. Actually, this is not asingle shell, since for n = 2, the number 𝓁 can take two values, 𝓁 = 0 and 𝓁 = 1.We thus obtain the two subshells 2s and 2p, of which the 2s subshell will getfilled first, since the 2p state is slightly higher in energy2 for reasons we alreadyexplained (see Figure 12.3). Therefore, for the elements with Z = 3 (lithium)and Z = 4 (beryllium) the electronic configuration (i.e., the way electrons are

2 We stress here the word slightly, since, if the 2p level were much higher than the 2s, then wewould have to treat the 2s as a separate shell, much like 1s.


arranged in different levels) is the following:

lithium (Z = 3) ∶ 1s2 2s1

beryllium (Z = 4) ∶ 1s2 2s2

}

electronic configuration

where we have adopted the spectroscopic notation for the occupied states andused superscripts to denote the number of electrons in each one. Thus theexpression 1s2 2s1 implies that the 1s state is populated with two electrons andthe 2s state with one electron. Similarly, the expression 1s2 2s2 2p3 describes theconfiguration for an atom with seven electrons. Two of these electrons are inthe 1s level,3 two in the 2s level, and the remaining three in the 2p level, whosecapacity is actually equal to 6, since it comprises three degenerate states 2px, 2py,2pz, each of which can hold two electrons of opposite spins. The total capacity ofthe n = 2 shell is thus equal to 8, because the 2s subshell can take two electronsand the 2p another six.

Using the above notation, the electronic configuration for the remainingelements of the second row (Z = 5 to Z = 10) can be written as follows:

Z = 5 (boron) ∶ 1s2 2s2 2p1

Z = 6 (carbon) ∶ 1s2 2s2 2p2

Z = 7 (nitrogen) ∶ 1s2 2s2 2p3

Z = 8 (oxygen) ∶ 1s2 2s2 2p4

Z = 9 (fluorine) ∶ 1s2 2s2 2p5

Z = 10 (neon) ∶ 1s2 2s2 2p6

Given now that a closed shell cannot accept any more electrons, one caneasily infer that atoms with closed shells are not particularly “keen” to formchemical compounds, since such chemical bonding requires the sharing ofelectrons between partnering atoms. It is therefore reasonable to expect thatthe elements with Z = 2 and Z = 10—which correspond to complete n = 1 andn = 2 shells—are chemically inert and are expected to be noble gases. And thefact that they are—element Z = 2 is helium and Z = 10 is neon, both of whichare known noble gases—is an irrefutable success of the quantum theory.

12.3.2 An Interesting “Detail”: The Pauli Principle and Atomic Magnetism

Let us now examine in greater detail how the 2p subshell gets filled, that is,how the degenerate levels 2px, 2py, 2pz are gradually filled. For example, shouldwe fill up one of these levels first—say, the 2px level—and then move on to thenext one? Or should we allocate one electron to each level in the first turn, thenanother electron to each level in the second turn, and so on? And if we are todo the latter—distribute one electron to each level at a time—then what wouldbe the relative orientation of the electron spins? For instance, if we have twoavailable electrons in the 2p level, as in the C atom, and place one electron in each

3 The terms level, shell (or subshell) are practically equivalent and will be used in turn to avoidrepeating the same words. The term shell is a reference to an “onion-like” picture for the atom, withconcentric “shells” corresponding to the successive energy states that have to be filled from theinside out, according to the Pauli principle.


of the 2px and 2py levels, how would their spins align, parallel or antiparallel?And does all this have any practical implication?

The answer, and the rationale behind it, is shown in Figure 12.5, where wesketch the occupied energy-level diagrams of the pertinent elements in thesecond row of the periodic table. These diagrams also provide a prediction forthe total spin of each atom, namely, S = 1∕2 for boron, S = 1 for carbon, S = 3∕2for nitrogen, and S = 1 for oxygen. Chemical data confirm this prediction, andin doing so they also confirm the Pauli principle, which “forces” electrons toalign their spins in parallel so that—due to the antisymmetric character of theirspatial wavefunctions—they are kept apart, thus minimizing their electrostaticrepulsions.

Z = 5 : Boron

Z = 7 : Nitrogen Z = 8 : Oxygen

Stot = 1/2

Stot = 3/2 Stot = 1

Stot = 1

2s 2s

1s

2s

1s 1s

2s

1s

2px

2px

2px2py 2py2pz

2py 2pz

2pz

2px 2py 2pz

Z = 6 : Carbon

Figure 12.5 Energy-level diagram for the ground state of the elements with atomic numbersfrom Z = 5 to Z = 8. Because of the Pauli principle, the placement of electrons in thedegenerate 2px , 2py , 2pz states has to be done as shown in the figure. We thus successivelyplace one electron on each level with spins in parallel, and when we are done with parallelarrangement we continue with antiparallel spins. This way of filling the levels—knownempirically as “Hund’s first rule”—is imposed by the minimization of the atomic energy, sinceparallel spins keep electrons further apart and thus decrease their electrostatic repulsions.


We thus establish empirically that the Pauli principle affects how electrons arepositioned not only in the same orbital, where it forces them to have oppositespins, but also in different orbitals, where it provides them with an energy incen-tive to align their spins and thus become microscopic magnets. This alignmentof spins implies the alignment of electronic magnetic moments and hence thecreation of an atomic magnetic moment, that is, paramagnetic atoms.4 Atomicmagnetism—and the magnetism of matter in general—is a true consequence ofthe Pauli principle.

12.3.3 Quantum Mechanical Explanation of Valence and Directionalityof Chemical Bonds

Let us now revisit the chemical properties of elements and examine the mostimportant such property: valence.

It follows from the previous discussion that valence is determined by thenumber of electrons in the outer shell of an element—the so-called valenceshell—since this number determines how many electrons the atom can give ortake as it forms a “chemical partnership” with other atoms. If the shell is less thanhalf filled (as in, e.g., Li, Be, or B), then the valence of the element is expectedto be equal to the number of outer electrons, while if the shell is more thanhalf filled, the valence is given by the number of empty states in the shell.5 Forexample, Li has valence 1, Be has 2, B has 3, C has 4, N has 3 (three empty states),O has 2 (two empty states), F has 1 (one empty state), and Ne has 0 (filled shell).And because the valence of hydrogen is (obviously) 1, we can safely predict theexistence of compounds such as

H2O, NH3, CH4, and so on.

Indeed, the existence of these compounds in nature corroborates our analysis.Actually, we can do better than guessing the valence of the elements, or, at

least, its typical value; we can also predict the directionality of their chemicalbonds, which, in turn, determines the configuration of atoms in space, that is,the geometric shape of molecules. A classic example is the water molecule. As wecan see from the corresponding energy-level diagram for oxygen (Figure 12.5),the valence orbitals of the oxygen atom are the 2px and 2py, since they havean empty slot and can thus receive the electrons of another atom. Therefore,it is reasonable to expect that the bonding of an oxygen atom with two atomsof a monovalent element (say, hydrogen) will occur along the two orthogonaldirections of the 2px and 2py orbitals, as shown in Figure 12.6.

Let us also note that the “bonding” between the valence orbitals of two atomsin a molecule produces a single molecular orbital on which the two bond

4 Atoms with nonzero magnetic moment are called paramagnetic, while those with vanishingmagnetic moment are called diamagnetic. Whether the paramagnetic atoms of a material will aligntheir magnetic moments to produce a macroscopic magnet depends critically on the nature of thematerial and its temperature, of course.5 Let us warn the readers here that this is a very superficial analysis of valence, since most atomshave more than one valence state depending on their chemical partners each time.


H

O

H

H2O

H

O

H

Figure 12.6 Quantum mechanical explanation of the geometric shape of the water molecule.Given that the valence orbitals of oxygen are 2px and 2py , oxygen is chemically reactive in twoorthogonal directions, along which it can form bonds with the 1s valence orbitals of the twohydrogen atoms. According to this scenario, the H2O (i.e., water) molecule has the shape of anisosceles right triangle with an oxygen atom on its apex and two hydrogen atoms at the endsof its base. Actually—for reasons we will explain later—the apex angle is not 90∘ butapproximately 105∘.

electrons—the valence electrons—move. As they do so, they cause a decreasein the energy of the molecule—compared to the two free atoms—because inthis way the valence electrons can move around both atoms and exploit thesimultaneous attraction of both nuclei.

12.3.4 Quantum Mechanical Explanation of Chemical Periodicity: The ThirdRow of the Periodic Table

We now move on to the third row of the (small) periodic table and examinewhether its elements have similar chemical properties with the elements directlyabove them, that is, in the same column of the table. But first, we ought toaddress an obvious question: Why does the third row not contain all elementscorresponding to the completion of all three subshells 3s, 3p, and 3d of the n = 3shell? This is because, even though the 3s and 3p levels are close enough to eachother to be regarded as belonging to the same shell, the 3d level lies much higher.By the time we populate it with electrons, the screening effect is so strong—thelarge value of 𝓁 keeps the 3d electrons far away from the nucleus—that the 3dlevel is raised much higher than the 3p and even higher than the 4s level of thenext shell. Actually, this last reversal in the order of the shells is responsible formost of the peculiarities of the periodic table beyond the element with Z = 18.

Based on the above, we conclude that, due to the large energy differencebetween the 3p and 3d states—compared to the much smaller differencebetween the 3s and 3p states—the third atomic shell includes only the 3s and3p subshells and not the 3d subshell. Thus the third row of the small periodictable is basically the exact analog of the second row, and elements of the samecolumn have similar physicochemical properties. For example, let us examinesulfur (Z = 16), just below oxygen. Its electrons are placed in successive atomiclevels as follows:

[S] = [Ne]3s2 3p↑x 3p↑

y 3p↑↓z (12.3)

where the brackets denote the electron configuration of the corresponding ele-ment, and we have also used the configuration of neon—[Ne] ≡ 1s2 2s2 2p6—for


brevity. Note that in (12.3) the electrons in the half-filled 3p state are denotedwith arrows, to describe not only their number, but also the orientation of theirspins.

A comparison with the electronic configuration of oxygen

[O] = [He] 2s2 2p↑x 2p↑

y 2p↑↓z (12.4)

shows that elements O and S are identical from a chemical perspective, since theyhave the same electronic configuration in their outer shell (the valence shell).They differ only in the quantum number n (n = 2 for O and n = 3 for S), which,however, does not affect the chemistry of an atom, since the general form of thevalence orbitals is the same. Both elements have a valence of 2, because they havetwo empty slots in their outer shells. Moreover, since the valence orbitals areessentially the same—2px, 2py in O and 3px, 3py in S—the geometry of chemicalbonds is the same for both elements. We can thus safely predict that sulfurforms with hydrogen a molecule—H2S, hydrogen sulfide—that has the sametriangular shape, and therefore very similar properties, as the H2O molecule.This is actually true. H2S does exist and has many of the remarkable propertiesof water, which we will present in detail in the next chapter. As is evident fromtheir electronic configurations—“formulas” (12.3) and (12.4)—sulfur and oxygenmust also have the same magnetic properties. Because of the Pauli principle, thetwo electrons in the half-filled px and py states have parallel spins, so both atomsare paramagnetic—that is, they have a nonzero magnetic moment and the sametotal spin value (Stot = 1).

The chemical periodicity between the second and third rows corresponds to aperiod of 8, which is the “length” of each row on the small periodic table. Thus,the chemical “twin” of an element in the second row can be found by shiftingits atomic number by 8. For example, the chemical analog of C (Z = 6) is theelement with atomic number Z′ = Z + 8 = 6 + 8 = 14, which is Si (silicon), andlies immediately below C in the periodic table. The electronic configurations ofthe two elements are

[C] = [He] 2s2 2p↑x 2p↑

y 2pz (Z = 6)[Si] = [Ne] 3s2 3p↑

x 3p↑y 3pz (Z = 14),

where, again, we denoted the occupation numbers of the outer subshell (2p or3p) with arrows, to specify the value of the total spin for the ground state of theatom. We also chose to include the unoccupied orbitals 2pz and 3pz, because theytoo belong to the valence shell and participate in the chemical life of the atom. Si,just like C, has a (typical) valence of 4 and forms, among other compounds, SiO2(silicon dioxide), the chemical analog of CO2 (carbon dioxide). Moreover, Si is aparamagnetic atom with total spin S = 1, just like C. And yet, for reasons that willtranspire in Chapter 14, C has a clear advantage over Si in terms of chemical prop-erties, which is precisely why it is the basis for the chemistry of life. This “chemicaldisadvantage” of Si becomes nevertheless an advantage when we look at itssemiconducting properties. Thus, Si wins over C as the basic element of moderntechnology and, perhaps, of a future artificial life that could be based on it!


12.3.5 Ionization Energy and Its Role in Chemical Behavior

We can add more substance to the preceding analysis of chemical behavior bydiscussing the role of a basic atomic parameter: the ionization energy of the atom.The main idea is the following. The smaller the ionization energy, the easier itis to extract from the atom an outer-shell electron, which can then participatein a chemical bond with another atom. Therefore, we expect atoms with smallionization energies to be “donors” of electrons in a chemical compound. Incontrast, atoms with large ionization energies are “unwilling” to give electronsto another atom, but are very “receptive” in accommodating electrons in theempty slots of their outer shells. We thus expect atoms with large ionizationenergies—excluding noble gases, of course—to act as acceptors of electrons inchemical compounds.

In view of the above discussion, it is important to understand the generalmechanism behind the value of the ionization energy and how this varies as wemove from one element to another in the periodic table. In particular, we wishto understand the general trend of ionization energies to increase, as we movefrom left to right in a particular row of the table. We also wish to explain whythis trend is interrupted at specific positions, which are actually the same for thesecond and third rows; for example, from Be to B, or from N to O—where theionization energy decreases slightly—and similarly in the third row. What is thequantum mechanical explanation for these changes?

Broadly speaking, there is a simple explanation for the general trend. As wemove along a particular row (i.e., within a given shell), n remains constant, whilethe effective nuclear charge (Zeff) the electrons experience continues to increase,since the atomic number increases.

As we continue filling the shell, we expect the ionization energy to increase,since the outer electrons of the atom become more tightly bound to its nucleusowing to the increase in the effective nuclear charge. To put it differently, becausewe keep n constant while Z—actually, Zeff—increases, the atomic levels generallysubside and ionization energies rise.

We can explain the deviations from this general trend just as easily. Forexample, the drop in the ionization energy from Be (9.3 eV) to B (8.3 eV) isdue to the fact that in Be the 2s subshell is filled, so the next electron—inthe B atom—ought to go to the 2p subshell, which is slightly higher than 2s.Therefore, despite the increase in nuclear charge, it is reasonable to expect thatthe outer-shell electron in the B atom is less firmly bound than in the Be atom,and therefore, less work is required for its extraction. However, the continuingincrease in nuclear charge becomes dominant in the subsequent atoms and thegeneral—increasing—trend for the ionization energy is restored. The next smalldrop is encountered as we move from N to O, for the following reason. Up untilN we had the option to place one electron on each of the 2px, 2py, 2pz orbitals,with their spins parallel, so the Pauli principle kept these electrons apart, thusminimizing their electrostatic repulsions. By the time we come to the O atom,we run out of this option, because we now have four outer electrons, two ofwhich must necessarily be placed in the same orbital with antiparallel spins. Asa result, these two electrons tend to remain in close proximity, thus increasing


25He

Li

Be

B

Na

Mg

Al

Si

P

S

Cl

Ar

C

N

O

F

Ne

WI(Z)

20

15

10

5

1

First row Second row Third row

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Z

eV

Figure 12.7 Ionization energy as a function of atomic number for the elements of the smallperiodic table.

the electron–electron repulsion. Therefore, the 2p level is slightly elevated andthe ionization energy decreases.

The same general trends are also observed in the third row of the periodictable; only here the ionization energies are slightly lower compared to those ofthe second row. All the above observations are summarized in Figure 12.7.

For a more quantitative treatment of the subject, we will need the expressionfor hydrogen-like atomic energies

En = −13.6n2 Z2

eff (eV),

where we have simply replaced Z with the effective charge Zeff6 of the nucleus as

experienced by the outer shell electrons. If n is the quantum number of the outershell, the ionization energy is given by

WI = −En = 13.6n2 Z2

eff (eV). (12.5)

Hence, it is clear that WI decreases dramatically when n increases—for example,from n = 2 to n = 3—even though this decrease is partially compensated by

6 Or, more accurately, the effective atomic number. The term effective charge is, however, moretransparent and becomes strictly correct if we use the electron charge as the unit of charge, whichimplies that the effective charge of the nucleus is Zeffe. In a similar manner, the atomic number Z ofan element can be called its bare nuclear charge, or, simply, its nuclear charge.


the expected increase of Zeff as Z becomes larger. (The screening effect cannotcompletely cancel the increase of the bare nuclear charge Z, and therefore, Zeff isin general a monotonically increasing function of Z.)

The elements of the first column of the periodic table—the so-calledalkalis7—form an ideal testbed of the above discussion, precisely because theyare similar to hydrogen in that they have one electron in their outer shell.Lithium, for example, has an ionization energy of 5.4 eV. We can provide a crudeexplanation of this number if we assume that the single valence electron of thisatom—the 2s electron—sees the nucleus as completely shielded by the two 1selectrons of its inner shell. Hence, the effective nuclear charge is equal to oneelectron unit (Zeff = 1), and we can apply formula (12.5) with n = 2 to obtain

WI(Li) = 3.4 eV,

which is a rather satisfactory result, especially if we consider the crudeness ofthe calculation. Now, looking back at our assumption, it is not difficult to alsosee why our result is smaller than the experimental value of 5.4 eV. The reason isthat the true effective charge Zeff should be greater than 1, because the screeningby the inner shell cannot be complete as the outer 2s electron can still reach thenucleus and briefly “see” the whole nuclear charge Z = 3. The correct value forZeff can be found if we insert the experimental value for the ionization energy in(12.5) and solve with respect to Zeff. We thus obtain

Zeff (Li) =√

WI (Li) n2

13.6

||||||n=2

= 1.26,

which is a plausible result, indeed. For the next alkali element (Na) withWI = 5.1 eV and n = 3, a similar calculation gives

Zeff (Na) = 1.83,

which is also a reasonable result, because the bare nuclear charge is now muchgreater—Z = 11 instead of Z = 3—and thus raises the effective charge experi-enced by the valence electron of the atom. According to the above discussion,and owing to their small ionization energies, all alkalis behave as electron donorsin their chemical compounds, and are actually monovalent elements. The exactopposite behavior is displayed by halogens (F, Cl, I, etc.), which lie on the otherside of the periodic table, one column to the left of the noble gases. Halogensare also monovalent elements—since they have one empty state in their outershell—but act as electron acceptors, since their ionization energy is very large,making it energetically favorable for them not to give but to receive electrons.It is therefore clear that when an alkali atom (e.g., Na) binds with a halogenatom (e.g., Cl), a strongly polar molecule is formed8 (NaCl), since the loosely

7 Actually, hydrogen does not belong to this group, owing to its highly peculiar properties, whichare caused by the very small size of its ion—it is merely a “bare” proton—and its very largeionization energy (13.6 eV). These properties render hydrogen a not-so-good donor of electrons; attimes, it can even be an acceptor.8 We know that the molecule of NaCl is only encountered when the compound is in very dilutegaseous form. In its solid form, NaCl—our familiar salt—forms an ionic crystal.


bound electron of Na “prefers” to hop to the empty state of Cl at much lowerenergy. We thus see that the motion of valence electrons in a chemical bond isnot necessarily distributed equally between the two bonding atoms, but shows apreference to one of them. Valence electrons prefer the “more attractive” of thetwo atoms and a first measure of this attractiveness is the ionization energy of theatom. The greater the ionization energy, the more willingly the correspondingatom receives an electron. Actually, this receptiveness of the atom dependsnot only on its ionization energy but also on its ability to hold on to the extraelectron. This ability—the so-called electron affinity—is measured by the workrequired to extract this electron. For hydrogen, for example, the addition of oneextra electron produces the negative ion H− and the work needed to extractit is 0.75eV, the tiny value of which explains why H is a nontypical element: Itcan be regarded neither as a good electron acceptor—due to its tiny electronaffinity—nor as a good donor—due to its large ionization energy.

To describe the relative ability of some atoms to attract electrons more stronglythan others, chemists use the semiempirical concept of electronegativity. Themore electronegative an atom is, the stronger it pulls electrons to its side ina chemical bond. As we discussed above, a first measure of electronegativityis the ionization energy of the atom. We can improve on this by also takinginto account the atom’s ability to hold on to an extra electron. We can thusregard electronegativity as a kind of average between the ionization energy ofthe neutral atom and the work needed to remove the extra electron from thenegative ion. In any case, it is evident that electronegativity shows the sametrends along the periodic table as ionization energy. As we move from left toright in a row of the periodic table, electronegativity increases.

To quantify the electronegativity of various elements, chemists have devisedan empirical electronegativity scale in which, for example, H has the value 2.1,F—the most electronegative of all atoms—has the value 4.0, while noble gasesare excluded for obvious reasons. In the case of our small periodic table theconventional electronegativity values are shown in Table 12.2. Using this table,we can easily predict the polarity of the chemical bond between two elements.The molecular orbital that hosts the two valence electrons of the bond will beenhanced in the vicinity of the most electronegative atom. Such polar bondsare called by chemists polar covalent, in contrast to nonpolar covalent bondsbetween atoms of the same kind, for which the motion of valence electrons issymmetrically distributed between the atoms. The ionic bond is the extremeexample of a polar covalent bond and corresponds to an almost complete

Table 12.2 Conventional electronegativity values forelements of the small periodic table.

H2.1

He

Li1.0

Be1.5

B2.0

C2.5

N3.0

O3.5

F4.0

Ne

Na0.9

Mg1.2

Al1.5

Si1.8

P2.1

S2.5

Cl3.0

Ar


transfer of valence electrons to the most electronegative atom. NaCl is a classicexample of an ionic bond.

Table 12.2 is to be used in the following way. If the difference between theelectronegativities of two elements is very small—say, less than half a unit—thenwe can practically regard their bond as nonpolar covalent. If the difference isgreater than one-half, but less than 2, we classify the bond as polar covalent,while for differences greater than 2, we are clearly in the realm of an ionic bond.

But let us not be misled by these descriptive categorizations of chemical bondsand lose sight of their fundamental unifying property, which stems from the factthat in all cases, valence electrons move so as to attain the state of lowest energy.This requirement forces electrons to distribute their motion equally between thetwo atoms (when these are the same or very similar), or to preferentially lie closerto the most attractive atom, that is, the most electronegative one. As we will seein greater detail in the following chapter, quantum mechanics offers a simpleand comprehensive explanation of the chemical bond, which renders obsoletethe phenomenology of the “old” chemistry, except for the purpose of easy com-munication. From a fundamental physical point of view, all chemical bonds arequalitatively the same.

12.3.6 Examples

We will now work through a few simple examples to help readers check theirunderstanding of the above discussion.

Example 12.1 Construct the occupied energy-level diagram (i.e., the elec-tronic configuration) of the element with Z = 7, and find (i) its total spin, and(ii) the valence and the directionality of its chemical bonds. Answer the samequestions for the element with Z = 16. Note: We intentionally avoided providingthe chemical symbols and names of these elements—even though it is clear thatwe are talking about nitrogen, N, and sulfur, S, respectively—to make it clear thatour prediction relies only on first principles and does not require any a prioriknowledge about these atoms.

Solution: Figure 12.8 shows the energy-level diagram and the electronicconfiguration for Z = 7, whence the following conclusions can be drawn. Tobegin with, the directionality of the chemical bonds of N is determined by thedirections of its three valence orbitals 2px, 2py, 2pz. Therefore, the molecule of acompound such as NH3—the familiar ammonia—has the shape of a triangularpyramid with N at its apex and the three hydrogen atoms at the corners ofits base. The N–H bonds point along three mutually orthogonal directions(Figure 12.9).

For the Z = 16 element, we can readily obtain its electronic configuration bywriting the sequence of successive quantum states 1s 2s 2p 3s 3p…, and fillingthem in this order with the maximum allowed number of electrons for eachstate: 2 for the s states and 6 for the p states. Thus, with 16 electrons at ourdisposal, we find

[Z = 16] = 1s2 2s2 2p6 3s2 3p4,


Figure 12.8 Occupied energy-level diagramfor the element with Z = 7 (nitrogen). ThePauli principle imposes parallel arrangementof electronic spins. As a result, the total spinof the atom is 3∕2 and its valence is equal to3, since there are three empty states in the2p valence shell.

2s2px 2py

Z = 7

2pz

1s

z

y

H

N

H

(a) (b)

HH H

N

H

x

Figure 12.9 The NH3 molecule according to quantum mechanics. Since the nitrogen valenceorbitals are 2px , 2py , 2pz , the bonding of nitrogen with the three hydrogen atoms takes placealong three orthogonal directions, as shown in (a). As a result, the molecule has thepyramid-like shape of (b).

whence we easily see that the valence shell is 3p and, since it has two emptystates, the valence of the element is equal to 2. Moreover, we can explicitly writethe orbitals of this shell and their spins as

3p4 ≡ 3p1x 3p1

y 3p2z ≡ 3p↑

x 3p↑y 3p↑↓

z .

We thus see that the element with Z = 16 has a total spin of 1. Clearly, we aredealing with sulfur (S), the homologous element to oxygen.

Example 12.2 In a hypothetical world the electron spin is equal to 3∕2. Whatare the atomic numbers of the first three noble gases in this world? Answer thesame question if the electron spin were equal to 2 (s = 2).

Solution: For an arbitrary value s of the spin quantum number, there are 2s + 1possible orientations for the vector s in space—that is, possible values forthe quantum number ms. Therefore, for s = 3∕2, we can place in the 1s shell2(3∕2) + 1 = 4 electrons, and another four electrons in the 2s shell and in eachof the 2px, 2py, 2pz shells. That is, each p state can take up to 12 electrons. Thusthe first atomic shell (1s) will get filled with 4 electrons, and the second shell willrequire 4 electrons for the subshell 2s and another 12 for 2p; and likewise for the


3s and 3p subshells of the third row. Consequently, the atomic numbers of thefirst three noble gases will be

Z = 4↑1s

, Z = 4↑1s

+ 4↑2s

+ 12↑

2p

= 20, Z = 20 + 4↑3s

+ 12↑

3p

= 36.

If we had to answer the same question for the electron spin being s = 2, then thecommon mistake would be to repeat the above procedure and simply change2s + 1 from 4 to 5, while there is a much more fundamental issue here. Inthis case the question itself is not valid, since for s = 2, electrons are bosonsand therefore the whole concept of filled shells—and of noble gases, let alonethe periodic table—has no meaning, since the capacity of each quantum statebecomes infinite.

Example 12.3 We know from experiments that the ionization energy ofthe helium atom is 24.6 eV. (i) Based on this fact, calculate the ground stateenergy of the atom. (ii) Give your theoretical prediction for this quantity ifelectron–electron repulsions can be ignored. (iii) If your prediction in (ii) is notsatisfactory, what would be the simplest thing to do to improve it?

Solution: Without its negative sign, the energy of the ground state of an atomis the work required to remove all its electrons. In the case of He, the removal ofone of its two electrons requires 24.6 eV (this is the ionization energy). For thesecond electron we need

13.6n2 Z2 eV

||||n=1,Z=2

= 54.4 eV

since we now have a hydrogen-like system with Z = 2. Therefore, the removalof both electrons requires in total 24.6 + 54.4 = 79 eV and so the energy of theground state of the He atom is

E1(He) = −79.0 eV (1)

which is essentially an experimental result because it was derived from theexperimental value for the ionization energy. If we ignore electron–electronrepulsions, our theoretical prediction for the quantity E1(He) becomes

E1(He)theory = − 13.6n2 Z2 eV

||||n=1

Z=2

⋅ 2 = −108.8 eV (2)

since we now have two electrons in the ground state of a hydrogen-like systemwith Z = 2. Compared to the experimental value (1), the theoretical prediction (2)is hardly a cause for celebration. The reason for failure is, evidently, the omissionof electronic repulsions. The latter can be approximately accounted for, if weassume that the two electrons of He move in the 1s orbital (Figure 12.4) and stayas far as possible from each other; that is, the mean electron–electron distance is

d = 2a0(Z)|Z=2 = 2a0

2= a0

12.4 Approximate Calculations in Atoms: Perturbation Theory and the Variational Method 341

so that the energy of their electrostatic repulsion is approximately equal toe2

d= e2

a0= 27.2 eV.

If we add this energy to the initial theoretical prediction (2), we obtain a morerealistic value for the energy of the ground state of the atom. We thus find

E1(He) = −108.8 + 27.2 = −81.6 eV,

which is much closer to the experimental value (1).

Problems

12.1 In a hypothetical world, electrons have spin 3∕2. Find the electronicconfiguration of the element with Z = 11. What is the total spin of theelement?

12.2 Consider a helium-like system (i.e., a He atom, or a heavier ion that hasbeen stripped of all but two electrons) of arbitrary atomic number Z.(a) Calculate the system’s ground state energy and eigenfunction (in a.u.),

assuming that electron–electron repulsions are negligible. Howdoes your prediction for the ground state energy compare with thefollowing experimental values?

E(He) = −79.0 eV, E(Li+) = −198.1 eV,E(Be++) = −371.6 eV, E(B+++) = −599.6 eV

(b) A reasonable estimate for the electron–electron interaction energy ofthe system can be obtained by assuming that the two electrons aremoving in diametrically opposite positions on the same Bohr orbit,so as to minimize their repulsion. Based on this simple model, whatis your new prediction for the ground state energy of an arbitraryhelium-like system? Find an analytic expression for this energy as afunction of Z, and compare your result with the experimental valuesmentioned in (a).

12.3 Potassium (K; Z = 19) is the next alkali metal after sodium (Na). Giventhat the ionization energy of potassium is equal to 4.34 eV, calculate theeffective nuclear charge, Zeff, “seen” by the electron in the outermost shellof this atom.

12.4 Approximate Calculations in Atoms: PerturbationTheory and the Variational Method

Our discussion of atomic structure so far suffers from a serious omission. Wehave not exposed the readers at all to the idea of approximate calculations.Actually, the readers may not even suspect that it would be impossible to


apply quantum mechanics to realistic physical systems—atoms, molecules,solids, and so on—if we confined ourselves strictly to exact solutions of theSchrödinger equation. The equation can be solved exactly for a many-electronsystem only if we ignore electron–electron repulsions. Only then do electronsmove as independent particles in the Coulomb potential of the nucleus andthe problem reduces (see Section 12.2) to the motion of one electron in thispotential. Unfortunately, the omission of electron repulsions leads almost alwaysto dubious results. Recall, for example, the previous problem for the He atom,where our theoretical prediction for the ionization energy was 54.4 eV, while theexperimental value is a mere 24.6 eV!

Evidently, unless we develop suitable approximate methods, we will not beable to apply quantum mechanics to the real world. But since a systematicpresentation of approximate methods is beyond the scope of this introductorytextbook, our aim in this section is to only provide a brief introduction to thistopic. More details can be found in the online supplement of this chapter.

As professed by the title of this section, there are basically two approximatemethods in quantum mechanics—the so-called perturbation theory and thevariational method. We will describe the main ideas behind these two methodsand see how we can apply them to a realistic problem: the helium atom orhelium-like systems.

12.4.1 Perturbation Theory

The principal idea of the method is simple. Suppose that for a certain HamiltonianH we can solve exactly the eigenvalue equation

H𝜓 = E𝜓

to find the eigenvalues and eigenfunctions of the problem. We now add to theHamiltonian H a new term 𝛿H that can be regarded as small compared to theoriginal Hamiltonian—hence the term perturbation—which is also why we usethe symbol 𝛿H to imply an “infinitesimal” change in the energy of the system. Butif the perturbation is very small, then the changes 𝛿𝜓 and 𝛿E to the wavefunction𝜓 and the energy E of the system should also be small. In other words, we cansay that the wavefunction for the new Hamiltonian H + 𝛿H will be 𝜓 + 𝛿𝜓with a corresponding new eigenvalue E + 𝛿E. We thus have the new eigenvalueequation

(H + 𝛿H)(𝜓 + 𝛿𝜓) = (E + 𝛿E )(𝜓 + 𝛿𝜓),

which we can expand as

H𝜓 + H(𝛿𝜓) + (𝛿H)𝜓 + (𝛿H)(𝛿𝜓) = E𝜓 + E(𝛿𝜓) + (𝛿E)𝜓 + (𝛿E)(𝛿𝜓).

The equation above can be drastically simplified if we recall that H𝜓 = E𝜓 ,and ignore to first approximation the underlined terms (𝛿H)(𝛿𝜓) and (𝛿E)(𝛿𝜓),since they are “differentials” of second order. We thus arrive at the much simplerequation

H(𝛿𝜓) + (𝛿H)𝜓 = E(𝛿𝜓) + (𝛿E)𝜓, (12.6)


which allows us to readily calculate the sought correction 𝛿E in the energy of thesystem, as follows. We take the inner product9 of both sides of (12.6) with thewavefunction 𝜓 , to obtain

(𝜓,H(𝛿𝜓)) + (𝜓, (𝛿H)𝜓) = E(𝜓, 𝛿𝜓) + 𝛿E(𝜓,𝜓)= E(𝜓, 𝛿𝜓) + 𝛿E, (12.7)

where in the last step we used the fact that 𝜓 is normalized, that is, (𝜓,𝜓) = 1.Now, because of the hermiticity of the operator H , the expression (𝜓,H(𝛿𝜓)) ofthe left-hand side is equivalently written as

(𝜓,H(𝛿𝜓)) = (H𝜓, 𝛿𝜓) = (E𝜓, 𝛿𝜓) = E(𝜓, 𝛿𝜓)

and cancels the first term of the right-hand side of (12.7). We thus obtain thefollowing simple expression for the sought correction—or, more precisely, thefirst correction, or the first-order correction10 —to the initial energy of the system

𝛿E = (𝜓, (𝛿H)𝜓) = ⟨𝛿H⟩, (12.8)

which tells us something simple:The first correction 𝛿E to the energy of a quantum system caused by a small

perturbation 𝛿H in its Hamiltonian is equal to the mean value of the perturbationwith respect to the unperturbed wavefunction.

Let us now see how we can apply this result for an approximate calculation ofthe ground-state energy of a helium atom. Actually, our calculation will be validalso for the so-called helium-like systems, such as the Li+ ion—singly ionizedlithium—the Be++ ion—doubly ionized beryllium—and so on, which are akin toHe, since they have two electrons around the nucleus. In all these systems, theunperturbed Hamiltonian H includes the kinetic energies and the Coulomb inter-action of the two electrons with the nucleus (of arbitrary atomic number Z), while

9 We recall (see Section 2.6.3) that the inner product of two wavefunctions 𝜓 and 𝜙 is defined as

(𝜓, 𝜙)def= ∫ 𝜓∗𝜙 dx,

so that, for a normalized wavefunction, we find

(𝜓,𝜓) = ∫ 𝜓∗𝜓 dx = ∫ |𝜓|2 dx = 1.

We also recall that, using the notation for the inner product, a hermitian operator is defined as

(𝜓,A𝜙)def= (A𝜓, 𝜙) ∶ hermitian operator,

that is, as an operator whose action can be transferred from one wavefunction of an inner productto the other.10 The usage of terms first correction or first-order correction implies that there are also correctionsof second order, third order, and so on, in a continuing sequence of successive corrections that cangive us the sought eigenvalue with any accuracy desired. Indeed, the systematic use of perturbationtheory in the online supplement of this chapter allows us, in principle, to obtain as many successivecorrections as needed, in order to calculate the energy of a quantum system with the accuracyrequired by experiments.


the electron–electron repulsion can be treated as a perturbation. We thus have

H =p2

1

2m+

p22

2m− Ze2

r1− Ze2

r2

= − ℏ2

2m∇2

1 −ℏ2

2m∇2

2 −Ze2

r1− Ze2

r2(12.9)

and

𝛿H = V = e2

|r1 − r2|, (12.10)

where r1, r2 are the position vectors of the two electrons with respect to thenucleus, which is fixed at the origin.

Given now that the Hamiltonian H is exactly solvable—since it describestwo independent electrons in the Coulomb potential of the nucleus—the aboveseparation into H and 𝛿H is in line with the general framework of perturbationtheory, except for one thing: The electrostatic repulsion between electrons—inthe He atom—can hardly be regarded as a small perturbation to the initial system,since it is of the same order of magnitude as the attraction of the electrons fromthe nucleus. The situation changes considerably as we move to larger helium-likesystems, where the nuclear attraction increases proportionally to the atomicnumber Z, while the electron–electron repulsions remain essentially unchanged(but not completely unchanged, because the electrons also come closer whenZ increases). We thus expect, in principle, that the application of perturbationtheory will give acceptable results for helium-like ions—especially the heavierones—but we cannot be sure about its accuracy for the He atom itself.

To apply formula (12.8), we need the unperturbed wavefunction𝜓 . Because thetwo electrons are independent, 𝜓 has the familiar product form

𝜓(r1, r2) = 𝜓1s(r1)𝜓1s(r2),

where 𝜓1s(r) is the hydrogen-like wavefunction of the 1s state for a nucleus witharbitrary atomic number Z. Given that for Z = 1 the wavefunction is

𝜓1s(r) =1

√𝜋 a3∕2

0

e−r∕a0 (Z = 1)

and that for arbitrary Z we have

a0(Z) =ℏ2

me2

||||e2→Ze2

= ℏ2

Zme2 =a0

Z,

the hydrogen-like wavefunction for arbitrary Z is equal to

𝜓1s(r) =1

√𝜋 (a0(Z))3∕2

e−r∕a0(Z) = Z3∕2√𝜋 a3∕2

0

e−Zr∕a0 ,

which becomes, in the atomic system of units (where a0 = 1),

𝜓1s(r) =Z3∕2√𝜋

e−Zr (a.u.). (12.11)


Thus the unperturbed wavefunction of an arbitrary helium-like system can bewritten—in atomic units—as

𝜓(r1, r2) =Z3∕2√𝜋

e−Zr1 ⋅Z3∕2√𝜋

e−Zr2 = Z3

𝜋e−Z(r1+r2) (a.u.). (12.12)

For the corresponding unperturbed energy, we have

E = 2E1s(Z) = 2(

−me4

2ℏ2

)|||||e2→Ze2

= −me4

ℏ2 Z2 = −Z2 (a.u.), (12.13)

where we multiplied with a factor of 2 because there are two electrons in the 1sstate around a nucleus with atomic number Z. All we have to do now to calculatethe perturbative correction 𝛿E is use (12.8) with 𝛿H as in (12.10)—in atomic units,that is, with e = 1—and 𝜓 as in (12.12). That is, we need to calculate the meanvalue

𝛿E =⟨

1|r1 − r2|

⟩

= ∫|𝜓(r1, r2)|2

|r1 − r2|dV1 dV2. (12.14)

This is a six-dimensional integral whose calculation may seem daunting to mostreaders but can be done nevertheless by clever use of an electrostatic analog sug-gested in the online supplement of this chapter. The result is

𝛿E = 58

Z.

Thus, the corrected value E′ = E + 𝛿E for the ground-state energy of ahelium-like system is given by the expression

E′ ≡ Epert = E + 𝛿E = −Z2 + 58

Z (a.u.),

which, in practical units of eV, is equal to

Epert =(

−Z2 + 58

Z)

⋅ 27.2 eV, (12.15)

where the index “pert” denotes the use of perturbation theory in obtaining theresult.

Table 12.3 compares the theoretical result (12.15) with experimental data foratomic numbers up to Z = 4.

The success of the calculation surpasses even our most optimistic expectation.Even for the He atom—which, as we said, is the worst case—the error is a mere5.3%, while for doubly ionized beryllium the error drops almost to 1%! This

Table 12.3 Comparison of perturbation theory calculations andexperimental data for helium-like systems up to Z = 4.

Z Eexp (eV) Epert =(−Z2 + 5

8Z)⋅ 27.2 eV Percent

error (%)

2 (He) −79.0 −74.8 5.33 (Li+) −198.1 −193.8 2.24 (Be++) −371.6 −367.2 1.2


resounding success confirms perturbation theory as a highly reliable method,provided the additional term 𝛿H is truly a small perturbation to the originalsystem.

12.4.2 Variational Method

As we saw earlier, if we ignore the repulsion between the two electrons in anarbitrary helium-like system, we can write the wavefunction of its ground statein the form

𝜓(r1, r2) = 𝜓1s(r1)𝜓1s(r2) =Z3∕2√𝜋

e−Zr1 ⋅Z3∕2√𝜋

e−Zr2 (a.u.), (12.16)

where Z is the atomic number of the nucleus. What would be a plausiblemodification of this wavefunction, if we took electron–electron repulsions intoaccount? If we think in terms of the screening concept we employed earlierto qualitatively predict the energy-level diagram in many-electron atoms, theanswer is straightforward. Between each electron in a helium-like atom and thenucleus lies a cloud of negative charge that is produced by the other electron ofthe system. As a result, each electron sees not the full charge Ze of the nucleusbut only a fraction of it. If we denote this effective charge as e, where issmaller than Z, then it is clear that the wavefunction of the system should alsobe modified. Instead of (12.16) it will take the form

𝜓(r1, r2) =3∕2√𝜋

e−r13∕2√𝜋

e−r2 (a.u.), (12.17)

which corresponds to a hypothetical nucleus with an effective atomic number instead of Z. What is the value of ? Since we seek the ground state of thesystem—that is, the lowest energy state—the best possible value for is the onethat minimizes the total energy of the system.

It is now clear how to proceed. Using the wavefunction (12.17) and the fullHamiltonian of an arbitrary helium-like system in atomic units

H = −12∇2

1 −12∇2

2 −Zr1

− Zr2

+ 1|r1 − r2|

we can calculate the average energy of the system

E = E() = ⟨H⟩

and find the value of —let us call it 0—that minimizes this expression. Thesought approximate value for the ground-state energy will then be

Evar = E(0) ≡ Emin,

where the index “var” stands for “variational,” from the term “variationalmethod,” which is how this approach is widely known. Let us recapitulate how itworks.

First, we choose a wavefunction that has as many of the features expected forthe ground state of the system as possible. In addition, this trial wavefunctioncontains one or more variational parameters that allow it to adapt in the best


possible way and become a state of minimum energy. We then calculate the aver-age energy of the system for the chosen wavefunction and find the parametervalues that minimize it. The minimum value of the average energy is the soughtapproximate value for the ground state energy of the system.

Let us now go back to helium-like systems and complete the calculation. Wecan simplify the algebra by writing the Hamiltonian H in the equivalent form

H = −12∇2

1 −12∇2

2 −r1

− r2

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

H0()

+ 1|r1 − r2|

+ ( − Z)r1

+ ( − Z)r2

.

Here, the Hamiltonian H0() corresponds to nuclear “charge” and its eigen-function is thus (12.17), while its eigenvalue−2 is obtained as before, albeit with instead of Z. It follows from our previous discussion that

E() = ⟨H0()⟩ +⟨

1|r1 − r2|

⟩

+ ( − Z)(⟨

1r1

⟩

+⟨

1r2

⟩)

= −2 + 58 + 2( − Z)

⟨1r

⟩

, (12.18)

where the average value ⟨|r1 − r2|−1⟩ is equal to 5∕8 as before—with instead

of Z. For the average values ⟨1∕r1⟩ and ⟨1∕r2⟩ we have⟨

1r1

⟩

=⟨

1r2

⟩

=⟨1

r

⟩

,

since the two electrons are completely equivalent. Therefore, we need to onlycalculate the average value of 1∕r using the wavefunction

𝜓1s(r) =3∕2√𝜋

e−r

for either electron. We thus obtain⟨1

r

⟩

= ∫∞

0|𝜓1s(r)|2

1r

4𝜋r2 dr = 3

𝜋4𝜋 ∫

∞

0re−2r dr

= 3

𝜋4𝜋 1!

(2)2 = ,which allows us to write (12.18) as

E() = −2 + 58 + 2( − Z) = 2 +

(58− 2Z

).The minimization condition gives then

dEd = 2 +

(58− 2Z

)

= 0 ⇒ = 0 = Z − 516

with a corresponding minimum energy

Emin = E(0) = −20 = −

(

Z − 516

)2,


Table 12.4 Comparison of variational calculations andexperimental data for helium-like systems up to Z = 4.

Z Eexp (eV) Evar = −(

Z − 516

)2⋅ 27.2 eV Percent

error (%)

2 (He) −79.0 −77.5 1.93 (Li+) −198.1 −196.5 0.84 (Be++) −371.6 −369.9 0.5

which is the approximate value for the ground-state energy using the variationalmethod. We thus obtain

Evar = −(

Z − 516

)2a.u. = −

(

Z − 516

)2⋅ 27.2 eV.

Table 12.4 compares the theoretical values with experimental data for atomicnumbers Z equal to 2, 3, and 4.

The accuracy of the variational method is impressive. In fact, there is nolimit to how accurate the method can be, as long as we choose variationalwavefunctions with the appropriate physical features and allow for severalvariational parameters instead of just one. With the help of a computer, themethod can provide results in astounding agreement with experiment. Thus,at least for the ground states of atoms, quantum mechanics has no “excuse” ifit ever fails to agree with experimental values, no matter how high the accuracyof the measurements is. Even a deviation in the 10th decimal digit—assumingexperiments of such high accuracy—suffices to discredit quantum theory—or, atleast, put it in serious doubt. Having survived such scrutiny for nearly 100 years,it is clear that quantum mechanics is not going away anytime soon. Quantummechanics lives on and invites us to new, even better calculations!

Problems

12.4 Use perturbation theory to calculate the correction to the ground-stateenergy of the harmonic oscillator, when we add to the parabolic potentialan extra term of the form gx4, where g can be regarded as sufficiently small.What does this mean? “Small” compared to what? You are advised to workin the natural system of units for the harmonic oscillator (ℏ = m = 𝜔 = 1)and restore ordinary units only in the final result. In a specific problem wehave m = mp = 1.6 × 10−24 g, 𝜔 = 1.3 × 1013 Hz, and g = 3 × 1020 cgs. Isperturbation theory applicable to this problem? Apply the same questionfor g = 5 × 1016 cgs.

12.5 Use the variational method to calculate—approximately—the ground-stateenergy of the harmonic oscillator. Choose a wavefunction that has thebasic features imposed by the form of the potential and depends on


a suitable variational parameter, which is to be determined by therequirement that the mean energy of the problem be minimized.

12.6 Apply the variational method to calculate the ground-state energy fora potential that cannot be solved exactly. Specifically, for the potentialV = gx4 of Problem 12.4—without the parabolic term kx2∕2, of course.Work in the system of units ℏ = m = g = 1 and compare your result tothe value E0 = 0.668 obtained from a numerical calculation.

12.7 In the first excited state 1s12s1 of the helium atom, the two electrons canhave parallel (S = 1) or antiparallel (S = 0) spins. The former state (1s↑2s↑)is known as orthohelium and the latter (1s↑2s↓) as parahelium.(a) Explain which one of these two states will have lower energy.(b) Apply perturbation theory to provide an expression—in the form of

an integral—for the energy difference between these two states.(c) What would you reply if you were told that the above energy

difference is of the order of 10−4 eV (which is the typical magnitudeof the magnetic interaction energy in atoms)?

Further Problems

12.8 Starting from the outermost occupied shell, the energy required tosuccessively extract the first, second, third, and so on, electron from anatom is called the first, second, third, and so on, ionization energy. Theenergy required to extract the next-to-last electron for the elements withZ ranging from Z = 5 (boron) to Z = 8 (oxygen) is listed below in unitsof eV:

B: 259.37, C: 392.09, N: 552.07, O: 739.29.

Use the general formula we derived in the text for helium-like systems topredict the above quantities and compare your results to the aforemen-tioned experimental values. What is the percent error in each case?

12.9 Use the variational method to determine the approximate form of thewavefunction describing the ground state of the helium atom.

12.10 The idea behind the variational method is based on the followingtheorem:

THEOREM: For any given (square integrable) wavefunction 𝜓 , the meanenergy E ≡ ⟨H⟩ = (𝜓,H𝜓) is always greater than or equal to the groundstate energy E0 of the respective quantum system.

Prove the above theorem, and then explain why its validity is as evident asthe following statement: The mean income in a country is always greaterthan or equal to the income of its poorest citizen.

351

13

Molecules. I: Elementary Theory of the Chemical Bond

… and as atoms move in vacuum, they collide and cluster, some jump back,others join and stay together depending on their shape, size, and kind. Thisis the way complex objects are formed.

Democritus

13.1 Introduction

In the previous chapter we studied atoms. As we discussed the periodic table ofthe elements, which classifies atoms according to their chemical behavior, wetook a first look at the nature of the chemical bond, the subject of this chapter.As you may recall, the general idea behind the quantum mechanical analysis ofchemical bonds is the following. When two atoms approach each other, theirouter electrons—the so-called valence electrons—cease to be localized in their“own” atom and “prefer” to move around both atoms so they can “enjoy” thesimultaneous attraction of both nuclei. As we will see in detail shortly, this“transfer” of valence electrons from one atom to the other is only possible inquantum mechanics, since it entails—at least during the molecule formationprocess—movement across the classically forbidden region between the twoatoms. Therefore, just like the very existence of atoms, their chemical behavioris also impossible in a purely classical context. In Chapter 12, we attempted ageneral description of the wavefunction—the so-called molecular orbital—forthe motion of valence electrons along a chemical bond. The basic idea is thatthe molecular orbital hosting the valence electrons is simply a superposition(i.e., a linear combination) of the atomic orbitals of the electrons participatingin the bond. Why is that so? In the vicinity of a particular atom, valenceelectrons feel mostly the attraction of the corresponding nucleus. Therefore,their wavefunction in that region must be identical to the corresponding atomicorbital, since the latter satisfies the Schrödinger equation near the atom. Themolecular wavefunction—that is, the solution of the Schrödinger equationfor the combined attractive potential generated by both nuclei—will thus be(approximately) a linear combination of the local solutions in the vicinity of


352 13 Molecules. I: Elementary Theory of the Chemical Bond

the two atoms, that is, a combination of the corresponding atomic orbitals.Specifically,

𝜓 = c1𝜓1 + c2𝜓2, (13.1)

where𝜓1 and𝜓2 are the local wavefunctions (the indices 1 and 2 refer to the atomsof the chemical bond), c1 and c2 are appropriate numerical coefficients, and 𝜓 isthe molecular wavefunction. This is the renowned method of linear combinationof atomic orbitals—best known by its acronym, LCAO—and will be our main toolfor constructing molecular wavefunctions in this chapter and the next.

The physical meaning of the coefficients c1 and c2 is that they describe thedegree of participation of the corresponding atomic orbitals 𝜓1 and 𝜓2 in theformation of the molecular orbital 𝜓 = c1𝜓1 + c2𝜓2. In particular, the squaresof these coefficients—P1 = |c1|

2, P2 = |c2|2—give the probabilities of finding the

valence electrons around each atom. In view of this interpretation, we expectthat1

|c1|2 + |c2|

2 = 1, (13.2)

which is simply a normalization condition for the molecular wavefunction.Clearly, the relative magnitude of c1 and c2 depends on the electronegativity

of the atoms in the bond. According to our discussion in the previous chapter,valence electrons prefer the more electronegative atom—the atom that attractsthem more strongly—so the corresponding atomic orbital will participate withgreater weight in the linear combination. The LCAO method allows thus a unifiedtreatment of the various types of chemical bonds, such as nonpolar covalent andionic bonds, to mention the most extreme cases. The differences between thesetypes of bonds lie simply in the relative magnitude of the coefficients c1 and c2,which determines the polarity of each bond, that is, the degree of asymmetry inthe valence electrons’ motion between the two atoms.

Our development of the theory of the chemical bond naturally splits into twoparts. First, in this chapter, we discuss the so-called elementary theory of thechemical bond. Then, in Chapter 14, we will analyze the two main deviations fromthat theory: hybridization—needed to understand the chemical behavior of car-bon and other elements—and delocalization—which enables the understandingof a broad class of organic molecules, such as benzene, but also most crystallinesolids, which can be viewed as giant molecules.

13.2 The Double-Well Model of Chemical Bonding

13.2.1 The Symmetric Double Well

As we noted earlier, a stable molecule is formed because the motion of valenceelectrons around both nuclei decreases the system’s energy compared to the two

1 In reality, as you can easily check, the relation holds provided the so-called overlap integral∫ 𝜓1𝜓2dx is practically negligible. This is only approximately true, as we will see shortly.

13.2 The Double-Well Model of Chemical Bonding 353

isolated atoms. Therefore, the key to understanding chemical bonding is the studyof how a particle moves in the field of two attractive centers, that is, a double well.If we assume, for simplicity, that each of the two wells is square, then our dou-ble well has one of the two shapes shown in Figures 13.1a,b. The former casecorresponds to homopolar molecules, where the two attractive centers are iden-tical. The latter case of wells with unequal depths pertains to polar molecules,where one atom is more attractive (i.e., more electronegative) than the other. Inthe same figure, and below each system of wells, we sketch a plausible shape forthe corresponding ground state wavefunction.

Apart from the ground state of the double well, we are also interested in itsfirst excited state, since the energy difference between these two states is thecrucial factor for the stability of the molecule. For the symmetric double well ofFigure 13.1a, it is easy to see that, because of mirror symmetry, the wavefunctionof the first excited state has the shape shown in Figure 13.2. It is an odd combina-tion of the eigenfunctions of the single wells, while their even combination is theground state.

Having found these plausible approximate shapes for the wavefunctions of thefirst two eigenstates of the symmetric double well, it is easy to calculate—again,approximately—the corresponding eigenenergies. For that matter, we use themean-energy formula

E = ⟨H⟩ = ∫ 𝜓∗(H𝜓) dx,

1

E0 E0

ψ1 ψ2

E2

ψ2

ψ1

E1

2

(a) (b)

2

1(ψ1 + ψ1)ψ =

ψ = c1ψ1 + c2ψ2

2

(c2 > c1)

1x

x x

x

Figure 13.1 The double-well problem as a simple model for the quantum mechanicaldescription of chemical bonds: (a) The symmetric double well. (b) The asymmetric double well.In both cases, energy minimization forces the particle’s wavefunction to have “hills” (i.e.,higher values) on the wells—to take advantage of low potential energy regions—and “valleys”(i.e., lower values) between the wells, where the potential energy is high. Owing to symmetry,the molecular wavefunction in the first case is an equal-weight superposition of thewavefunctions for single wells. In the asymmetric double well, the wavefunction of the deeperwell dominates (i.e., c2 > c1) because it is energetically favorable for the particle to spend more“time” in this region.


Region of well 1Region of well 2

1ψ–

(ψ1 – ψ2)=2

x

Figure 13.2 A plausible approximate form for the wavefunction of the first excited state for asymmetric double well. Since the wavefunction must resemble the local solution 𝜓1 or 𝜓2 inthe vicinity of the corresponding well, the first excited state ought to be the odd superpositionof 𝜓1 and 𝜓2, in order to have the expected symmetry and the required one node.

which, when𝜓 approximates an eigenfunction, yields the corresponding approx-imate eigenvalue. For 𝜓± = (𝜓1 ± 𝜓2)∕

√2 we find2

E = ∫ 𝜓±(H𝜓±) dx = 12 ∫ (𝜓1 ± 𝜓2)(H(𝜓1 ± 𝜓2)) dx

= 12 ∫ 𝜓1(H𝜓1) dx + 1

2 ∫ 𝜓2(H𝜓2) dx ± 12 ∫ 𝜓1(H𝜓2) dx

± 12 ∫ 𝜓2(H𝜓1) dx = 1

2(H11 + H22 ± H12 ± H21), (13.3)

where the integrals

Hij = ∫ 𝜓i(H𝜓j) dx, i, j = 1, 2

are what we called elsewhere (Section 2.6.4) the matrix elements of the opera-tor H . Note that the plus and minus signs in (13.3) correspond to even and oddeigenfunctions, 𝜓±, respectively.

Let us now assume that the two wells lie close enough for their eigenfunctions𝜓1 and 𝜓2 to overlap, but not so close that they influence each other strongly. Wecan then show the following approximate expressions for the matrix elements,

H11 = H22 ≈ E0, H12 = H21 ≈ −A,

where E0 is the eigenenergy of the single well and A is a positive number that ismuch smaller than E0.

Proof : If V1(x) and V2(x) are the potentials of the single wells, then the totalpotential V (x) and the corresponding Hamiltonian H of the double well are,respectively,

V (x) = V1(x) + V2(x), H =p2

2m+ V (x) =

p2

2m+ V1(x) + V2(x).

The following expressions must obviously hold:

H = H1 + V2 = H2 + V1 , H1𝜓1 = E0𝜓1 , H2𝜓2 = E0𝜓2 ,

2 Given that in any one-dimensional problem, all eigenfunctions of bound states are real, we do notneed to use the symbol for complex conjugation in the mean value formula.


where H1 and H2 are the Hamiltonians of the single wells

H1 =p2

2m+ V1(x), H2 =

p2

2m+ V2(x).

Based on the above, we find

H11 = ∫ 𝜓1(H𝜓1) dx = ∫ 𝜓1(H1 + V2)𝜓1dx

= ∫ 𝜓1(H1𝜓1) dx + ∫ 𝜓21 V2 dx = E0 ∫ 𝜓2

1 dx + ∫ 𝜓21 V2(x) dx

= E0 + ∫ 𝜓21 V2(x) dx ≈ E0.

In the last step we took into account the fact that the term ∫ 𝜓21 V2(x) dx is negli-

gible compared to E0, since𝜓21 is localized in the vicinity of well #1, whereas V2(x)

is localized in the vicinity of well #2.We thus have H11 ≈ E0 and, for the same reason, H22 ≈ E0. Moreover, it is easy

to see that the nondiagonal matrix elements H12 and H21 are equal and that

H12 = ∫ 𝜓1(H𝜓2) dx = ∫ 𝜓1(H2 + V1)𝜓2dx

= ∫ 𝜓1(E0 + V1)𝜓2dx < 0. (13.4)

since 𝜓1𝜓2 > 0 and E0 + V1 < 0, given that E0 < 0 and V1(x) ≦ 0. □

Based on the above, we can set

H11 = H22 ≈ E0 , H12 = H21 = −A (A > 0), (13.5)

where A is a positive quantity with dimensions of energy, which can be deter-mined from the details of the problem at hand (i.e., the eigenfunctions and poten-tials). Under the conditions we stated before, we have A ≪ E0, since the integrandin (13.4) takes very small values, as it is the product of two functions localized indifferent regions of the x-axis. Note, however, that this condition is often not quitemet in real molecules mainly because of the long range character of the Coulombpotential.

If we now plug relations (13.5) into (13.3), we obtain the approximate expres-sions for the eigenvalues:

E− = E0 − A, E+ = E0 + A,

where the lower value, E−, corresponds to the even superposition of the localwavefunctions, and the higher value, E+, to the odd one. In other words, E− and E+correspond to the ground state and first excited state of the double well, respec-tively.

We summarize the above results in Figure 13.3, where on either side we showthe levels of single wells (which in our case have the same energy), and, in themiddle, the levels of the corresponding double well. We also show the pertinent“atomic” and “molecular” wavefunctions.


E0

1

2

E– = E0 – A

E0

E+ = E0 + A

=

=

ψ1

ψ–

ψ+

(ψ1 – ψ2)

1

2(ψ1 + ψ2)

ψ2

Figure 13.3 Approximate solution (eigenvalues and eigenfunctions) for the symmetricdouble well, with the LCAO method. The main conclusion applies also to higher states and canbe stated as follows. When we bring close enough two identical wells (i.e., two identicalattractive centers) then each state of the single wells gives rise to two states for the doublewell, symmetrically positioned with respect to the original state. The correspondingeigenfunctions are given by the even and odd superpositions of the single-welleigenfunctions, respectively.

13.2.2 The Asymmetric Double Well

Let us now examine the case of two wells with unequal depths. Because mirrorsymmetry is now absent, we can no longer guess the form of the eigenfunctionsby symmetry arguments alone. Instead, we have to use the LCAO method infull force, which consists of the following steps. First, we require that the soughtapproximate solution 𝜓 = c1𝜓1 + c2𝜓2 satisfies (approximately) the Schrödingerequation

H𝜓 = E𝜓, (13.6)

where E are the eigenvalues of the double well. We then multiply both sides of(13.6) with 𝜓1 and subsequently with 𝜓2, and integrate over x from −∞ to +∞.We thus obtain

H11c1 + H12c2 = Ec1

H21c1 + H22c2 = Ec2

}

. (13.7)

Here, we used the definition Hij = ∫ 𝜓i(H𝜓j)dx, and also the relations

∫ 𝜓21 dx = ∫ 𝜓2

2 dx = 1, ∫ 𝜓1𝜓2 dx ≈ 0,

of which the latter reflects the fact that the eigenfunctions 𝜓1 and 𝜓2 have avery small overlap with each other, and thus the integral of their product—theso-called overlap integral—can be ignored to first approximation.3 Equations(13.7) form a linear system of two equations with two unknowns (c1 and c2),which can be written in matrix form as

HC = EC (13.8)

3 Actually, because the overlap integral S = (𝜓1, 𝜓2) = ∫ 𝜓1𝜓2dx is comparable to the matrixelement H12 = −A, a more consistent calculation would keep S as a (nonzero) parameter in theLCAO method. Nevertheless, we set S ≈ 0 in order to simplify the theory and make it morephysically transparent (as is usually done in introductory-level textbooks). In the online supplementof Chapter 14 we study the case S ≠ 0.


with

H =

(H11 H12

H21 H22

)

, C =

(c1

c2

)

. (13.9)

Equation (13.8) is an eigenvalue equation for the Hamiltonian matrix H . Solvingit yields the eigenenergies E and the corresponding eigenvectors C, that is, thecoefficients in the linear combination 𝜓 = c1𝜓1 + c2𝜓2.

As for the matrix elements Hij, the following approximate relations will hold(just as in the symmetric double well)

H11 = E1 , H22 = E2 , H12 = H21 = −A (A > 0),

the only difference here being that we have two distinct eigenvalues E1 and E2(Figure 13.1b), instead of a doubly degenerate eigenvalue E0. We can thus rewritethe eigenvalue Eq. (13.8) as

(E1 −A−A E2

)(c1

c2

)

= E

(c1

c2

)

⇒

(E1 − E −A−A E2 − E

)(c1

c2

)

= 0, (13.10)

where the last form is a homogeneous system of equations, which has a nontrivialsolution—that is, (c1, c2) ≠ (0, 0)—only if the determinant of the 2 × 2 matrix ofits coefficients vanishes:

det

(E1 − E −A−A E2 − E

)

= 0 ⇒ (E1 − E)(E2 − E) − A2 = 0

⇒ E ≡ E± =E1 + E2

2±

√(E1 − E2

2

)2

+ A2,

which can be rewritten as

E± = E0 ±√Δ2 + A2, (13.11)

where the quantities E0 and Δ are defined as

E0 =E1 + E2

2, Δ =

E1 − E2

2, (13.12)

that is, they are the half-sum and half-difference of the original eigenvalues,respectively. (We assume that well #2 is deeper than well #1, as in Figure 13.1,so that E2 < E1.) Note that for two identical wells—where E1 = E2 = E0 andΔ = 0—formula (13.11) reduces to our previous result, that is, E± = E0 ± A.Figure 13.4 shows the results (13.11) and (13.12) for the more general case of anasymmetric double well (with E2 < E1).

As for the ratio c2∕c1, which determines the degree of participation for eachatomic eigenfunction 𝜓1 and 𝜓2 in the molecular eigenfunction 𝜓 = c1𝜓1 + c2𝜓2,it is readily found from the system of Eq. (13.10) to be:

c2

c1=

E1 − EA

.


Energy level

of well 1

Energy level

of well 2

2

2

Energy levels

of the double well

++=

E1

E+

=E0

E1 + E2

E1 – E2

E0

Δ

Δ2 A2

+–=E–

E2

E0 Δ2

Δ

Δ =

A2

Figure 13.4 Energy levels of an asymmetric double well in the LCAO approximation. Bymoving in both wells at the same time, the particle lowers its energy even more than if itremained localized in the deeper well. We thus have E− < E2.

Therefore, for the two eigenstates of the double well—the ground state withE = E−, and the first excited state with E = E+—we obtain

(c2

c1

)

g=

Δ +√Δ2 + A2

A,

(c2

c1

)

exc=

Δ −√Δ2 + A2

A, (13.13)

where “g” and “exc” stand for “ground” and “excited” states, respectively. In thespecial case of the symmetric double well (where Δ = 0) we obtain (c2∕c1)g = 1and (c2∕c1)exc = −1. Combining these relations with the normalization conditionc2

1 + c22 = 1, we obtain c1 = 1∕

√2, c2 = ±1∕

√2, which are the familiar results

𝜓+ = 1√

2(𝜓1 + 𝜓2), 𝜓− = 1

√2(𝜓1 − 𝜓2)

we had obtained earlier using only symmetry arguments. But for the general caseof an arbitrary double well we need the actual calculations. For the ground statein particular, we need the formula

(c2

c1

)

g=

Δ +√Δ2 + A2

A= Δ

A+√

(ΔA

)2+ 1,

from where it follows that for Δ > 0 (i.e., E1 > E2), we always have (c2∕c1) > 1,which means that the eigenfunction 𝜓2 of the deeper well contributes more than𝜓1 to the molecular orbital 𝜓 = c1𝜓1 + c2𝜓2. This unevenness increases with theenergy difference between the levels of the two wells, as measured by the quantityΔ = (E1 − E2)∕2. Thus, we confirm quantitatively what we deemed qualitativelyevident from the beginning, namely, that the particle spends most of its time inthe vicinity of the deeper well (Figure 13.1b).

We conclude the section with some comments on terminology and a few lastremarks. First, looking at Figure 13.3—with the eigenvalues and eigenfunctionsof the symmetric double well—we note again that the lowest energy is realizedvia the even superposition of atomic orbitals, whereas their odd superpositionproduces a molecular level that lies higher than the corresponding atomic levels

Problems 359

(of the isolated wells). A bound state—that is, a stable molecule—is thus onlyformed for the molecular orbital 𝜓+ = (𝜓1 + 𝜓2)∕

√2, which is aptly called

bonding orbital. Conversely, the orbital 𝜓− = (𝜓1 − 𝜓2)∕√

2 is called antibondingbecause it does not lead to a stable chemical bond between the two atoms,and the corresponding state is thus an unstable excited state of the molecule.A similar observation can be made for the asymmetric double well, where thelower molecular level (Figure 13.4) lies below both atomic levels and leads tomolecule formation (bonding state), while the higher molecular level lies higherin energy than either atomic level and is thus an antibonding excited state.

Coming back to Figure 13.1—say, Figure 13.1a—we observe that, since theatomic level E0 has to lie lower than the edges of the well, the particle can movefrom one well to the other only by passing through the classically forbiddenregion between the two wells—that is, by quantum mechanical tunneling. Thistunneling process can only take place when the two wells get close enough toallow the wavefunction of one well to penetrate inside the other. This is how aparticle “feels” the presence of another well nearby, and utilizes it—by appro-priately spreading its wavefunction—to further lower its energy. Alternatively,we can say that the process of forming a molecule begins when the atomsapproach each other close enough to allow a noticeable overlap between theirwavefunctions. But even in the final form of the molecule, only a partial overlapwould be expected, since the mutual repulsion of the nuclei prevents them fromcoming too close. We can thus say that in the equilibrium state of the molecule,the atomic valence orbitals (i.e., the orbitals that participate in the chemicalbond) have a noticeable, but not very large, overlap.

Finally, let us stress again that in the preceding discussion, neither the particularshape of the wells—whether they are square, and so on—nor their dimensionalityis of particular significance. The conclusions drawn from this discussion can thusbe applied to actual molecules, which we will study next.

Problems

13.1 The “atomic levels” E1 and E2 of a system comprising two wells of unequaldepths (an approximate model for a diatomic molecule) have the valuesE1 = −4 eV and E2 = −10 eV. The Hamiltonian matrix element H12between the “atomic wavefunctions” 𝜓1 and 𝜓2 has been calculated to be−4 eV in the equilibrium configuration of the molecule.

(a) Calculate the eigenenergies for the ground state and the first excitedstate of the molecule. How much energy is gained upon formation of themolecule from two free atoms? Plot the energy diagram of the system.

(b) Construct the wavefunctions for the ground and the first excited stateof the molecule.

13.2 The distance between the two atoms in the equilibrium state of themolecule in the previous problem is equal to 1.2 Å. Calculate its dipolemoment in debye (recall that 1 D ≡ 1 D = 0.2e ⋅ Å).


13.3 At time t = 0, the state of a particle that moves in a system of two identicalwells is described by the wavefunction𝜓(0) = 𝜓1, which means that at thismoment the particle is localized in well #1. Find, using the LCAO method,the time-evolved form 𝜓(t) of the wavefunction, and use it to calculate theprobabilities P1(t) and P2(t) of finding the particle in well #1 and well #2,respectively. Plot P1(t) and P2(t) in a common diagram as functions of t.(As usual, you may assume the matrix element H12 = −A to be known).

13.3 Examples of Simple Molecules

13.3.1 The Hydrogen Molecule H2

According to the above discussion, the H2 molecule should look as in Figure 13.5.The picture has the following two direct implications.

(a) The bond length—that is, the distance between the protons—is slightly lessthan 1 Å, since the pairing 1s orbitals have a radius of a0 = 0.5 Å and theformation of a bond requires their noticeable (but not too large) overlap. So,a reasonable estimate for the bond length would be d ≈ (0.7–0.8) Å. Actually,the experimental value is 0.74 Å.

(b) The molecule is nonpolar, since the valence electrons move symmetricallyabout the two atoms, which means that the center of mass of the negativecharge will be at the center of the molecule (as will the center of mass for thepositive charge). The electric dipole moment of the molecule vanishes.

As we will see next, the polarity of a molecule affects directly the macroscopicproperties of the corresponding substance—for instance, whether the substanceis a solid, liquid, or gas, at normal temperatures. So let us examine this issuehere in greater detail. In the most general sense, the problem pertains to thenature of forces between molecules, the so-called van der Waals interactions.The topic of these interactions is not particularly simple and requires nontriv-ial quantum mechanical treatment. But a purely descriptive introduction sufficesfor our needs in this book. So, let us state outright that these interactions arepurely electrical in nature, not some kind of a novel fundamental force. The van

1s 1s

p

(a) (b)

p p p

Figure 13.5 Quantum mechanical picture of the H2 molecule. (a) The overlapping atomicorbitals participating in the bond. (b) A simplified sketch of the molecular orbital produced bycombining (“conjoining”) the atomic orbitals, where the two valence electrons occupy themolecular orbital with opposite spins. The H2 molecule is a classic example of covalent bonding.

13.3 Examples of Simple Molecules 361

der Waals forces arise due to the fact that, even though molecules are overallneutral, they consist of charged particles—namely, electrons and nuclei—whichaffect each other when the molecules are close enough. The interaction mech-anism is then determined by the molecules’ dipole moment. Two scenarios arepossible:

1. The molecules are polar, that is, they have a permanent (electric) dipolemoment.

2. The molecules are nonpolar, that is, they have no permanent (electric) dipolemoment.

In the first case, the nature of the van der Waals interaction is clearly the elec-trical force between the two dipoles.

In the second case, the interaction mechanism is similar; only now the molec-ular dipole moments are induced, not permanent. They are produced by the“displacement” of molecular orbitals caused by the interaction of neighboringmolecules. For example, when two H2 molecules come in close proximity, theyinduce a dipole moment on each other, and the interaction between theseinduced dipole moments is the cause of the attractive van der Waals forcebetween the two molecules. We should also mention, for completeness, the“mixed” case whereby one molecule has a permanent dipole moment andinduces a dipole moment on an otherwise nonpolar molecule nearby.

In terms of strength, it is clear that van der Waals interactions are strongestbetween polar molecules, and weakest between nonpolar molecules, assumingthat the molecules are of similar size in both cases. But when the interactingnonpolar molecules are very large, the induced dipole moments can also becomevery large and result in enhanced van der Waals forces. Roughly speaking, wecan say that the van der Waals interaction between nonpolar molecules of thesame species, being proportional to the inner product of their induced dipolemoments—which, in turn, are proportional to the linear size of the molecules—isproportional to the square of the linear size of the molecule and hence to its sur-face area (to the extent that the latter has a clear meaning for a complex molecule).

Based on the above, it is clear that intermolecular interactions for hydrogenought to be very weak, since H2 molecules are small and nonpolar. We thus expectthat hydrogen molecules remain in gas phase at room temperature (and for nor-mal atmospheric pressure). We would have to go down to very low temperaturesfor the weak van der Waals attraction to become dominant over thermal motionand turn the gas, first into a liquid, and then into a solid. Indeed, H2 condensesat −253 ∘C and solidifies at −259 ∘C.

Figure 13.6 shows hydrogen’s energy-level diagram, where the levels of freeatoms are shown on either side and the molecular levels in the middle.

We recall that the quantity A (or, more accurately,−A) is the matrix element H12of the molecular Hamiltonian between the atomic orbitals that form the bond. Soit is a quantity we can calculate and compare with experiment. For example, theexperimental value of 4.5 eV for the dissociation energy of the molecule shouldbe compared to the theoretical prediction 2A. However, we do not intend to putthe theory to a quantitative test in this book. First, because the calculation of Ais quite laborious—it can only be done numerically—and, second, because the


Energy level

of atom 1Energy level

of atom 2

Molecular energy levels

E0 + A

E0 – A

E0E0

Figure 13.6 Energy-level diagram of the H2 molecule. On either side we show the (groundstate) levels of the free atoms, while the first two molecular states are shown in the middle. Theenergy gain due to molecule formation is equal to 2E0 − 2(E0 − A) = 2A.

LCAO method, as described here, is not sophisticated enough to entertain suchexpectations. Note, for instance, that in all our discussions so far, we ignored theCoulomb repulsion between valence electrons and treated them as independentparticles that move in the attractive potential of the two nuclei. In an improvedversion of the method, we could, as we did in the case of atoms, continue to con-sider the valence electrons as independent particles, but have them move in ascreened potential that incorporates electron–electron repulsions, instead of theinitial bare potential of the nuclei. But such advanced calculations are beyond thescope of an undergraduate textbook. Here, it is sufficient to develop a qualitativeunderstanding of the basic idea. Our conclusion from this discussion is then thefollowing.

The LCAO method, as presented in this chapter, is suitable not for accuratequantitative predictions but for a qualitative description and understand-ing of the basic mechanisms of the chemical bond.

In this spirit, we will treat A as an input parameter of the problem with a spe-cific physical content: 2A is the energy gained by the valence electrons (i.e., bythe molecule) due to their delocalized motion between the two atoms. To put itdifferently, −2A is the energy 𝜖(R) of valence electrons measured with respect totheir energy in the corresponding free atoms (R is the distance between the pro-tons). The total energy E(R) of the molecule includes also the energy of repulsione2∕R between its nuclei, so we have

E(R) = e2

R+ 𝜖(R) = e2

R− 2A(R). (13.14)

Given now that A(R) increases as R decreases—because, then, the overlap ofatomic orbitals and the associated matrix element |H12| = A increases—the totalenergy E(R) of the molecule has the characteristic shape in Figure 13.7.

In reality, the above analysis for the total energy of the molecule—expression(13.14)—is correct only up to interatomic distances slightly greater than the equi-librium configuration of the molecule. Beyond this point, the molecule breaksup into its constituent atoms, which then interact through van der Waals forces.


Figure 13.7 The total energy E(R) of the hydrogenmolecule. E(R) is a sum of the term e2∕R for repulsionbetween nuclei, and the term 𝜖(R) = −2A(R) forattraction of the valence electrons by the protons.The minimum of the E(R) curve determines thelength of the molecule in its equilibriumconfiguration.

E(R)

e2/R

R0 R

(R) = –2 A(R)

E0

E0 + A

E0 – A

E0

Figure 13.8 Energy-level diagram of a hypothetical He molecule. The molecule cannot existbecause the energy gained by placing two electrons in the bonding state E0 − A is cancelled bythe necessary placement of the other two electrons in the antibonding state E0 + A.

As a result, the “tail” of the curve in Figure 13.7 follows the van der Waals expres-sion E(R) ∼ −R−6 for nonpolar molecules (or atoms).

13.3.2 The Helium “Molecule” He2

In this section, we aim to understand why it is impossible to form molecules fromatoms of noble (or inert) gases, which have filled outer shells.

The reason for this chemical “inertness” is depicted in Figure 13.8 for the sim-plest noble gas, helium.

Even if two He atoms come temporarily close together, they will again sepa-rate after their first collision with other atoms or molecules, since there is reallyno energy required to split them up. Helium will thus be found in nature only ingaseous atomic form, a behavior that persists down to temperatures near abso-lute zero, because its atoms—the smallest in nature—attract each other by theweakest possible van der Waals forces. Actually, helium cannot solidify even atabsolute zero, because the kinetic energy associated with the quantum motionof its atoms—the so-called zero point energy4 —is especially strong due to theirtiny mass.

4 This is the least possible “motion,” that is, the minimum kinetic energy that a quantum particlemust have due to the uncertainty principle, when it is confined in a finite region. The term is usedmostly for the minimum possible energy of a particle moving in the potential of a harmonicoscillator or near the bottom of an arbitrary potential.


13.3.3 The Lithium Molecule Li2

For Z = 3, the electronic configuration of the atom is

[Li] = 1s2 2s1

whence it is evident that the valence orbital is 2s and the valence of a lithium atomis 1. To confirm that only the outer shell plays a role in the chemical behavior ofthe element, let us examine the molecule’s energy diagram in Figure 13.9.

We can draw a general conclusion here. Chemical bonding is exclusively causedby the outer electrons of an atom, because only these electrons can produce anenergy gain for the formation of a molecule. All other electrons in the fully occu-pied inner levels make no energy contribution to a chemical bond, and can thusbe regarded as completely localized in their atom and unaffected by the presenceof neighboring atoms.

Here is a question for the readers. In Figure 13.9 the two molecular levels thatoriginate from the 1s atomic level are shown closer together—and are in factmuch closer—than those originating from the 2s level. Can you explain why?

13.3.4 The Oxygen Molecule O2

With eight electrons available (Z = 8), the successive filling of 1s, 2s, and 2p levelsleads to the following electronic configuration

[O] = 1s2 2s2 2p1x 2p1

y 2p2z .

Hence, the valence orbitals of oxygen are 2px and 2py, a fact that poses a problemwe have not encountered so far. How can we combine atomic orbitals now thatwe have two pairs of valence orbitals instead of one? In other words, how do wedeal with a double bond instead of a single one? Two possibilities are plausible, asshown in Figure 13.10.

In the standard terminology of chemistry, the strong bond resulting from thehead-on combination of two −s or p− orbitals is called a 𝜎 bond, while the bondresulting from the lateral combination of p orbitals is called a 𝜋 bond. Likewise,the corresponding molecular orbitals are called 𝜎 and 𝜋 orbitals, respectively.

2s 2s

1s 1s

Figure 13.9 Energy-level diagram for the lithium molecule. Because inner atomic levels arefully occupied, any energy gain can only result from the outer level 2s that is half-filled.Therefore the inner, filled levels do not affect the chemical behavior of the atom—since theydo not contribute energywise. Only the outer, half-filled 2s level (the so-called valence orbital)is important in this respect.


(a) (b)

Figure 13.10 Two possible ways to combine atomic orbitals and produce a double bond in anoxygen molecule. (a) The orbitals of one pair bind head-on (i.e., axially) with a strong overlap,thus forming a very strong bond, while the orbitals of the other pair bind sideways (i.e.,laterally) with very weak overlap and form a second, much weaker bond. Note that we haveused a series of vertical lines to denote the weak lateral overlap of orbitals in this case. (b)Another way to combine orbitals, where both bonds have intermediate strength. Generalenergy considerations show that the bonding configuration (a) is energetically favorable, sincein configuration (b), the produced molecular orbitals form an angle—that is, thecorresponding molecular wavefunction has abrupt spatial variations—and cause an excessiveincrease of the kinetic energy of valence electrons. Simple symmetry arguments lead to thesame conclusion. For instance, that the ground state always has the full symmetry of theproblem (in our case, rotational symmetry about the molecular axis) and no nodes, while thefirst excited state has one node—or one nodal surface in a three–dimensional problem. Thebonding configuration (a) satisfies these general requirements, while (b) does not.)

Note that the Greek letters 𝜎 and 𝜋 for molecular orbitals correspond to the let-ters s and p, respectively, for atomic orbitals. This is no accident. The 𝜎 and 𝜋orbitals are the molecular analogs of the atomic orbitals s and p. The 𝜎 orbitalsare symmetric under rotations around the molecular axis—just as s orbitals havethe full rotational symmetry of the atom—while 𝜋 orbitals comprise two lobes5

of opposite sign lying on either side of a plane (just like p orbitals in atoms).As for the nature of the various bonds in a chemical compound, it is clear that

all single bonds are strong 𝜎 bonds, since they form by the head-on overlap ofatomic orbitals. But in multiple bonds, only one is a 𝜎 bond, while the others areweak 𝜋 bonds, because the corresponding orbitals have no other option but tocombine sideways with a weak overlap.

Coming back to the O2 molecule, we can now say that it has the chemical rep-resentation

Oπσ O

where the thick and thin lines denote the strong 𝜎 and the weak 𝜋 bond,respectively.

With regard to physical properties, it is evident that O2 is a gas at room tem-perature, with a very low condensation point, because its molecules are nonpolarand interact—owing to their small size also—through very weak van der Waalsforces. Indeed, O2 condenses at −183 ∘C and solidifies at −218 ∘C.

5 We remind the readers that, because the plots of atomic or molecular orbitals represent theprobability amplitude (i.e., the wavefunction values) they bear signs in their various regions, eventhough this effect is not shown in the figures.


13.3.5 The Nitrogen Molecule N2

For Z = 7, the electronic configuration is

[N] = 1s2 2s2 2p1x 2p1

y 2p1z

and therefore nitrogen is a trivalent atom with valence orbitals 2px, 2py, and 2pz,each of which carries one electron. The nitrogen molecule will thus have the shapedepicted in Figure 13.11. The head-on overlap of one pair of valence orbitals formsa strong 𝜎 bond, while the sideways overlap of the other two orbital pairs formstwo weak 𝜋 bonds.

Since N2 is a small nonpolar molecule, it will have similar properties tomolecules such as H2 and O2, as confirmed by the data in Table 13.1.

The increase of the bond strength as we move from H2 to N2 is a direct con-sequence of the fact that the bond is single in H2, double in O2, and triple in N2.But there is no straightforward explanation for the fact that the increase of thebond strength from H2 to O2 is tiny compared to the increase from O2 to N2.To account for this fact, we have to take into consideration “details” such as thesize of pairing atoms, which affects electronic repulsions and other quantitiesinvolved in the pertinent calculations. As for bond lengths, the large differencebetween H2 and the other two molecules is due to the small size of the hydrogenatom, while the difference between O2 and N2 is largely due to the stronger triplebond of N2 that pulls atoms closer together. Finally, with regard to the meltingand boiling points, it is evident from Table 13.1 that they roughly increase withthe size of the corresponding molecules, as represented by their bond lengths.Indeed, being the smallest molecule of the three, H2 has the lowest condensation(or boiling) point, followed successively by N2 and O2. Finally, the reversal of theorder between O2 and N2 in the melting point is not a major concern, because

z

x x

z

yN2 = N N

π

σ

π

Figure 13.11 The formation of molecular orbitals in the N2 molecule.

Table 13.1 Comparison of basic properties for three nonpolar diatomic molecules.

Melting point (∘C) Boiling point (∘C) Bond length (Å) Bond strengtha) (eV)

H2 −259 −253 0.74 4.5O2 −218 −183 1.21 5.2N2 −210 −196 1.10 9.8

a) Bond strength is an alternative name for the dissociation energy of the molecule.


the size difference between the two atoms is so small that the “details” of theintermolecular interactions can no longer be ignored.

13.3.6 The Water Molecule H2O

Given that oxygen is a divalent atom (with valence orbitals 2px and 2py) andhydrogen is monovalent (with valence orbital 1s), they form a chemical com-pound of the type OH2 (one oxygen, two hydrogen atoms), with bonds along twoorthogonal directions, as in Figure 13.12.

But the accurate prediction of the shape of the molecule requires knowledge ofthe polarity of the oxygen–hydrogen bonds, which determines the overall polarityof the molecule, and is a crucial factor for the physical and chemical proper-ties of water. Indeed, the O—H bond is clearly polar. Given that oxygen is moreelectronegative than hydrogen (since it lies to the right of hydrogen in the peri-odic table; see also Table 12.2), the valence electrons in an O—H bond will movetoward the oxygen, leaving the hydrogen nucleus partially exposed. The O—Hbond will thus be a polar bond with negative charge around the O atom and pos-itive charge around the H atom. As a result, the shape of the H2O molecule willbe modified to look more as in Figure 13.13.

It is evident from the above discussion that H2O is a polar molecule, sincethe dipole moments of the two O—H bonds add up to a nonzero total dipolemoment. We note that the dipole moment of an electric dipole ( )

–q +q is definedby physicists as the vector D, with direction from the negative to the positivecharge of the dipole and magnitude D = q ⋅ a, where a is the distance betweenthe dipole charges. In contrast, chemists usually define the dipole moment in theopposite direction, that is, along the direction of motion for the valence electronsof the bond, which points from the electropositive to the electronegative atom.The dipole moment vector is denoted by a “ticked” arrow adjacent to the corre-sponding bond (with the tick mark closer to the positive charge), while anotherticked arrow on the side denotes the total dipole moment of the molecule. SeeFigure 13.14 for the case of an H2O molecule.

H

(a) (b)

H

H

O

O

90°H

Figure 13.12 The quantum mechanical explanation for the shape of the water molecule.Owing to the directionality of the valence orbitals in the O atom, its binding with two H atomscan take place along two perpendicular directions, as in (a). As a result, the water moleculelooks (to first approximation) like an isosceles orthogonal triangle, with the O atom at its apexand the two H species at the vertices of its base, as in (b).


H

(a) (b)

H

104.5°

H H

O

90°+ + + +

O

Figure 13.13 Actual shape of the water molecule. Because oxygen is more electronegativethan hydrogen, valence electrons move toward the O atom and produce an excess of negativecharge there, and a deficit of negative charge—that is, positive charge—around the H atom.The repulsion between the positively charged H atoms widens the angle of the molecule to itsfinal value of 104.5∘. For even greater angles, the energy gained by further distancing the Hatoms from each other is overcompensated by the energy penalty associated with thedeformation of the 2px and 2py valence orbitals of the O atom.

H

O

H

D1

D = D1 + D2

D2

+ +

Figure 13.14 Chemical notation for the dipole moments of bonds and the total dipolemoment of molecules (here, for H2O). The dipole moments point in the direction of motionof the valence electrons along the bond, which is opposite to the direction used byphysicists.

A plausible unit for molecular dipole moments in atomic units would be thequantity e ⋅ a0 (a0 is the Bohr radius), or perhaps the quantity e ⋅ Å, which is twicethe former value, since 1Å ≈ 2a0. Chemists have adopted as the unit of dipolemoment the debye,6 which is defined as

1 D ≡ 1 D = 10−18 cgs.

Since 1e = 4.8 × 10−10 esu, we have

1e ⋅ Å = 4.8 × 10−10 esu × 10−8 cm = 4.8 × 10−18 cgs⇒ 1e ⋅ Å = 4.8 D ⇒ 1 D ≈ 0.2e ⋅ Å ≈ 0.4 a.u.

The dipole moment of H2O is 1.85 D, a rather large value compared to thedipole moments of several (much bigger) organic molecules. But there are alsoother small molecules with dipole moments comparable to water, as shown inTable 13.2.

6 Named after P. Debye (1884–1966). Dutch physicist, 1936 Nobel prize in chemistry. He was oneof the founders of modern molecular and solid state physics, renowned both for his theoretical andexperimental skills.


Table 13.2 Dipole moments of representativepolar molecules.

Molecule Dipolemoment

(D)

H2O 1.85NH3 1.42H2S 0.97HCl 1.05

The properties of water, which are responsible for its key role as the liquidmedium of living cells, are now readily explained based on its molecular structureand, in particular, the fact that water is a small molecule with high polarity.

Two immediate examples of water’s amazing properties are its high specificheat (1 cal∕g K)7 and its unusually high latent heat of evaporation (≈ 540 cal∕g).Because of these properties water is an ideal thermostat, that is, it can absorbor emit large amounts of heat while its temperature barely changes. It is for thisreason that seaside regions have a mild climate.

How does this actually happen? It is all because of the high polarity of the watermolecule. When an amount of heat is absorbed by water, a large fraction of itis used to “detach” its polar molecules from each other, since they attract oneanother strongly. Only a small fraction of the heat is then left to contribute to therandom thermal motion of these molecules and cause the temperature increase.

The strong polarity of its molecule is also responsible for another remarkableproperty of water: The fact that it is an ideal solvent of other polar compounds.A typical example is the solvation of NaCl—the common salt. Let us recall thatthe sodium–chlorine bond is an extreme example of an ionic bond, because Naand Cl lie at the opposite ends of the third line of the periodic table, so theirdifference in electronegativity is maximal (= 2.1 units, according to Table 12.2).We may thus assume that in the NaCl molecule, the pair of valence electrons hascompletely moved to the Cl side, producing a negative Cl ion and leaving behinda positive Na ion.

Actually, due to the extreme ionic character of the Na—Cl bond, NaCl in itssolid form does not consist of molecules but forms a single crystal lattice of Naand Cl ions in alternating positions. Such an ionic solid dissolves readily in water,because the positive sodium ions (Na+) are attracted by the negatively chargedregion of a water molecule—near the oxygen atom—while the negative chlorineions (Cl−) are attracted by the positively charged region near the hydrogen atoms.The end result (also determined by entropy considerations) is the dissociation ofthe NaCl crystal into sodium and chlorine ions.

7 The fact that the specific heat of water is 1 cal∕g K is actually a consequence of the definition ofthe calorie (cal). One calorie is defined as the amount of heat required for the temperature of 1 g ofwater to increase by 1 ∘K.


At a more macroscopic level, the high polarity of the water molecule is mani-fested by its huge dielectric constant (𝜖 ≈ 80), which causes the Coulomb forcebetween two electric charges in water to drop by a factor of 80 compared to vac-uum; it becomes 80 times weaker!

Being an ideal solvent that remains in liquid state over a large range of temper-atures (from 0 ∘C to 100 ∘C) far away from absolute zero, water is best suited forbeing the liquid medium of life. It is the cell-filling liquid inside which the variouscellular substances are dissolved in sufficiently high concentrations, to allow allvital biological processes to take place, such as photosynthesis in plant cells, orprotein synthesis in our own bodies. Indeed, water is the liquid medium in ourcells, and in the cells of all living organisms. Quantum mechanics allows us tounderstand a little the vital role of the water molecule.

Water is also renowned for another unusual—though not unique towater—property with biological significance: the well-known “anomalousexpansion,” whereby its solid form—ice—is less dense than the liquid phase.Because of this property, ice does not sink in water, sparing most of marine lifefrom death during freezing conditions in winter.

This property has a simple explanation. When water is in solid form, itsmolecules arrange themselves in space in an orderly manner to minimize thetotal energy. Specifically, in order to minimize the total electrostatic energy ofthe system, H2O molecules are arranged in a way that brings their oppositecharged regions in close proximity. Thus a hydrogen atom of one moleculeapproaches the oxygen atom of another molecule, and so forth throughout thewhole crystal. As one can verify by constructing a simplified two-dimensionalanalog of an ice crystal using this principle, the structure that emerges containslarge voids between the molecules. Consequently, ice is less dense than water.Actually, this structure is also preserved “locally” in the liquid form of water andexplains some of its peculiar properties.

13.3.7 Hydrogen Bonds: From the Water Molecule to Biomolecules8

Actually, the above theoretical analysis of the structure and properties ofwater—in both its liquid and solid forms—is not complete without the consider-ation of another quantum mechanical mechanism, known as hydrogen bonding.Figure 13.15 can help us understand what this mechanism is about.

But there is more to the hydrogen bond than what can be inferred fromFigure 13.15. To begin with, it is a stronger bond than one might expect. Onereason for this is the very small size of the H ion, H+, which is actually aproton, and has thus essentially nuclear size, compared to the atomic-sizedions of other atoms. Owing to its small size, H+ can come quite close to thenegative oxygen ion of another atom, which causes a significant electrostaticattraction between the two ions. Another reason for the remarkable strength ofthe hydrogen bond is purely quantum mechanical, and its basic idea is shown inFigure 13.16.

8 This section can be omitted in a first reading.


H

O

H

H

H

O H O

H

O H

H

HH

O

+

+

+

+

+

+

+ +

+

+

–

–

–

–

–

–

–

–

––

Figure 13.15 A spatial ordering of H2O molecules that brings their oppositely charged regionsto proximity. Solid lines denote the standard polar covalent O—H bonds, while dashed linesshow the weaker coupling between an oxygen atom from one molecule and a hydrogen atomfrom another, caused by the electrostatic attraction of their opposite charges. Through thisdual action, each hydrogen atom acts as a kind of bridge between two oxygen atoms, hencethe term hydrogen bond.

H

H O

O

I

II

O

O

(a) (b)

Figure 13.16 The quantum mechanical mechanism of hydrogen bonding. (a) Under thesimultaneous attraction of two centers (the two oxygen atoms), the hydrogen atom performsa delocalized motion between the corresponding wells, interchanging continuously the typeof its “binding” to each oxygen ion (continuous line). Thus a hydrogen bond is actually aquantum superposition of forms I and II, as shown in (b). A more realistic description shouldalso take into account the lack of symmetry between the positions of the two oxygen atoms.The simplified picture above is thus only meant to explain the basic idea of hydrogen bonding.

The significant strength of the observed hydrogen bond (in the order of a fewkcal/mol ≈ a few tenths of an eV per bond) is a direct consequence of the abovequantum mechanical mechanism. By simultaneously moving between two attrac-tive centers, the hydrogen ion gains extra energy, in the same manner that valenceelectrons gain energy in a conventional chemical bond. Viewed from this perspec-tive, the hydrogen bond can be regarded simply as another type of chemical bond,with the particle exchanged between the partnering atoms being a proton, insteadof the valence electrons. We can now understand the unique role of hydrogenin facilitating this bond. Being the lightest of all atoms, hydrogen—actually, itsnucleus—has immensely higher probability than any other atom to cross the clas-sically forbidden region between the two wells of Figure 13.16 and enhance thebinding between the two oxygen atoms.


It follows from the above discussion that hydrogen bonds must exist, not onlybetween oxygen atoms in water but also between other species of atoms in dif-ferent compounds. For example, we have hydrogen bonds of the type N H N(simply denoted NHN) or even NHO, to name a couple of important cases.

But the culmination of the role of hydrogen bonding lies not so much in under-standing the properties of water, as in explaining biomolecules, such as proteinsand DNA. A quick browsing of a biochemistry book suffices to convince the read-ers of the prominent role of hydrogen bonds in producing the structural featuresof biomolecules that determine their biological functionality. For example, hydro-gen bonds are responsible for the helical structure of proteins and the pairing of“conjugate bases” in the double helix of DNA, two features of immense impor-tance in biology. See the pertinent Figure 13.17.

In the chemical formulas of Figure 13.17, we follow the standard conventionof not showing C or H atoms, except those that bind to atoms other than C. So,each node corresponds to a C atom, and if fewer than four lines (i.e., bonds) stemfrom that node, then a fourth line (bond) is also implied there, binding a C atomwith a H atom, since C is tetravalent. To clarify these conventions, we sketch inFigure 13.18 the full chemical formula for adenine.

Following the above analysis, we conclude that the backbone of biomoleculesis very stable because it is formed by strong chemical bonds (covalent or polarcovalent). But the winding of biomolecules in space—which controls theirspecialized chemical behavior—is determined by hydrogen bonds. These bondsare strong enough for the biomolecule to retain its structure under normalconditions in cells, but not too strong altogether, so they can be broken to allow

H

Adenine (A) Thymine (T) Guanine (G) Cytocine (C)

N

N N

H

HO CH3

O

H

H H

H

HH

H

H

H

N

NN

N N

N O

O

NN

N

NN

N

Figure 13.17 Hydrogen bonds between bases in the double helix of DNA. Adenine binds tothymine (A–T) and guanine binds to cytosine (G–C). Dashed lines denote hydrogen bonds.

HC

C

C

C C

N

N

N

N

N H

H

H

H

Figure 13.18 Conventions for the structural formulas oflarge organic molecules. The complete structural formula ofadenine (to be compared with Figure 13.17).


the molecule to change its shape when necessary. The whole “game” of life islargely based on this “plasticity” of hydrogen bonds.

13.3.8 The Ammonia Molecule NH3

Given its electronic configuration [N] = 1s2 2s2 2p1x 2p1

y 2p1z —from where we can

see that the valence orbitals are 2px, 2py, and 2pz—a nitrogen atom can formbonds with three hydrogen atoms along the x-, y-, and z-axes. The ammonia(NH3) molecule will thus have the geometry of Figure 13.19, as we already knowfrom a similar discussion in Chapter 12.

Being a small and strongly polar molecule—its dipole moment is equal to 1.42D, compared to 1.85 D for H2O—ammonia’s properties are expected to be similarto those of water, only a bit less pronounced. Even though ammonia is a gas atroom temperature, it condenses at a temperature rather close to 0 ∘C (−33 ∘C,to be precise), which is surely much higher than the corresponding condensationtemperatures of nonpolar molecules (see Table 13.1). Because of the polarity ofthe NH3 molecule, ammonia is also a very good solvent of polar substances, henceits use as a cleaning liquid.

We will now examine a particularly interesting property of the NH3 moleculeknown as nitrogen inversion (see Figure 13.20). The curve depicts the potentialexperienced by the N atom as it tries to move from one equilibrium position tothe other, across the potential barrier. During the first half of this motion, the Natom feels an uphill potential due to the strong deformation of the N—H bondsalong the way, while in the second half, the potential becomes downhill, since thebond deformation gradually relaxes and the molecule returns to an equilibriumconfiguration, which is actually a mirror image of the original one. As a result,the ground state of the ammonia molecule (in gas form) is neither the left nor the

z

y

xNH3

N

H

+ +

––

–

+

H

(b)(a)

H

Figure 13.19 Quantum mechanical analysis of the NH3 molecule. (a) Pairing between the 2px ,2py and 2pz valence orbitals of N and the 1s valence orbitals of the three H atoms. (b) The finalgeometry of the NH3 molecule. The molecule has the shape of a pyramid with the N atom atthe apex and the three H atoms at the vertices of the base. Since nitrogen is clearly moreelectronegative than hydrogen, N–H bonds are polar and NH3 is a polar molecule. Owing torepulsions between hydrogen atoms, the HNH angles exceed 90∘ (the experimental value is107∘).


H

E0 – AE0

E0 + A

N

H

H N

Figure 13.20 Nitrogen inversion in the ammoniamolecule. Owing to the finite energy barrier thatseparates its two mirror configurations, the ammoniamolecule’s true ground state is the even superpositionof these configurations, while the odd superpositionrepresents an excited state of the molecule. As a result,the NH3 vibrational levels split in two, a typical featurefor particle motion in a double well. (The matrixelement of the Hamiltonian H between the twolocalized vibrational states is −A.)

right configuration, but their symmetric (even) superposition

𝜓+ = 1√

2(𝜓L + 𝜓R). (13.15)

On the other hand, the antisymmetric (i.e., odd) superposition also exists

𝜓− = 1√

2(𝜓L − 𝜓R) (13.16)

and represents the excited state that results from this oscillation of the moleculebetween the left and right configuration. The situation here is the same as in ourearlier discussion of the chemical bond, where the even superposition of atomicorbitals led to the ground state of the molecule (the so-called bonding state),while the odd superposition represented its excited counterpart (antibondingstate).

If we now denote as E0 the energy of each localized solution 𝜓L and 𝜓R—thatis, the minimum energy the molecule needs to have as it “oscillates”9 aroundeach equilibrium configuration—then the states 𝜓+ and 𝜓− will have energiesE0 − A and E0 + A, respectively (see Figure 13.20). The energy difference of 2Abetween these states is roughly equal to 10−4 eV and corresponds to a wavelengthequal to

𝜆(Å) = 12 000𝜖(eV)

= 12 00010−4 = 1.2 × 108 Å = 1.2 cm,

which falls within the microwave region. The above analysis led in 1958 tothe operation of the first microwave amplifier using stimulated emission ofradiation—the renowned maser10—which paved the way, a few years later, forthe invention of the laser and the avalanche of technological and scientificdevelopments that ensued.

9 In the following section, we will study molecular oscillations around equilibrium configurations,which produce the so-called vibrational spectrum. We will also look at molecular rotations, whichgive rise to the so-called rotational spectrum of molecules.10 MASER≡microwave amplification by stimulated emission of radiation.


But the description of the NH3 molecule so far raises an obvious question.Given that in its true ground state—namely, (13.15)—NH3 is a nonpolarmolecule—since both the left and right configurations appear in (13.15) withequal weight—how are we to explain those properties of NH3 that are attributedto the strong polarity of its molecule?

The answer is simple. The double-sided nonpolar form of the molecule—(13.15)—is found only in the gas phase of NH3, where intermolecular interactions arenegligible and each molecule is free to find its lowest energy state. But in theliquid phase—and even more so in the solid phase—it is energetically favorablefor molecules to return to their one-sided, polar form, because they then interactwith each other through strong van der Waals (dipole–dipole) forces. Theenergy contribution of these forces is of the order of 10−1 to 10−2 eV per pair ofinteracting molecules,11 compared to a mere 10−4 eV due to the superpositionof its two polar forms.

Finally, we should note that as the molecule oscillates, the actual mass of theoscillating body is neither the mass of the N atom nor the mass of the three Hatoms, but the reduced mass of the N–3H system,

𝜇 =M1M2

M1 + M2=

mN ⋅ 3mH

mN + 3mH,

which is much closer to the sum of the three H masses, given that the N atom isquite heavier (mN ≈ 14mH).

Actually, the NH3 molecule is an ideal physical system to study the time evolu-tion of a quantum state, when the initial state is not an eigenstate of the system.See the following example.

Example 13.1 An NH3 molecule is initially in its one-sided polar form 𝜓L, thatis, 𝜓(0) = 𝜓L. What will its state be after the lapse of time t? What is the prob-ability of finding the molecule, after time t, again in state 𝜓L, or in its mirrorstate 𝜓R?

Solution: The true energy eigenfunctions of the molecule are not 𝜓L and 𝜓R but𝜓+ and 𝜓−. Therefore, to find the time-evolved form of 𝜓L, we have to write it asa linear combination of 𝜓+ and 𝜓−, which is easily achieved by solving the set ofEqs (13.15) and (13.16) with respect to 𝜓L. We thus obtain

𝜓L = 1√

2(𝜓+ + 𝜓−). (1)

11 This result is readily obtained from the expression for the interaction energy between twodipoles at distance r

𝑈int ≈d1 ⋅ d2

r3 ∼ d2

r3 ,

so that, for d ≈ (1 − 2) D ≈ 1 a.u. and r ≈ (3 − 5) Å ≈ 10 a.u., we find

𝑈int ≈12

103 a.u. = 27.2 eV103 ≈ 3 × 10−2 eV.


The time-dependent form of (1) is found by “time evolving” 𝜓+ and 𝜓− usingthe familiar exponential factor e−iEt∕ℏ for each one, where E = E0 − A for 𝜓+ andE = E0 + A for 𝜓−. We thus obtain

𝜓(t) = 1√

2(𝜓+e−i(E0−A)t∕ℏ + 𝜓−𝜖

−i(E0+A)t∕ℏ)

= e−iE0t∕ℏ√

2(𝜓+eiAt∕ℏ + 𝜓−e−iAt∕ℏ).

Since the common phase factor e−iE0t∕ℏ has no physical significance it can bedropped, and we write

𝜓(t) = 1√

2(𝜓+eiAt∕ℏ + 𝜓−e−iAt∕ℏ),

so that, if we now substitute 𝜓+ and 𝜓− from (13.15) and (13.16), we obtain

𝜓(t) = cos (At∕ℏ)𝜓L + i sin (At∕ℏ)𝜓R.

From the above expression, we conclude that the probabilities PL(t) and PR(t)of finding the molecule in the left or the right polar form, respectively, areequal to

PL(t) = cos2(Atℏ

)

, PR(t) = sin2(Atℏ

)

.

These probabilities are plotted as a function of time in Figure 13.21.Finally, note that the period of inversion—that is, the time needed for the

molecule to return to its initial state—is equal to

T = 2𝜋𝜔

= 2𝜋(2A∕ℏ)

= ℏ𝜋

A,

PL(t) PR(t)

T

t

A= π

Figure 13.21 Time evolution of the state of an ammonia molecule that was originally in its leftpolar form. PL(t) is the probability of finding the molecule in the same state L after time t, whilePR(t) is the probability for the molecule to undergo inversion and end up in its right polar state.(See also Figure 13.20.)

13.4 Molecular Spectra 377

and can thus be derived from the Bohr frequency𝜔 = 2A∕ℏ of the transition fromthe excited state to the ground state for the two levels shown in Figure 13.20. Whyshould we have anticipated this result?

Problems

13.4 Chemists use the symbol 𝛿 to denote excess (or deficit) of charge at theends of a polar chemical bond. So, if a is the bond length, then the dipolemoment is 𝛿 ⋅ a. Calculate the excess of charge 𝛿 for the O—H bond, giventhat the H—O—H angle in the H2O molecule is equal to 104.5∘, its dipolemoment is 1.85 D, and the length of the O—H bond is a = 0.96 Å.

13.5 In analogy to the compounds OH2 (≡ H2O) and NH3, we could have alsostudied the methane molecule CH4, where there are four H atoms, sincecarbon (Z = 6) is tetravalent. Explain why the valence of carbon has thisvalue, and employ the elementary theory of the chemical bond, as dis-cussed in the text, to predict the shape of the methane molecule. If youranalysis is correct, you will find that CH4 is a polar molecule and shouldthus be expected to have similar properties to water or ammonia. And yetmethane—the main constituent of natural gas—behaves as a typical non-polar molecule, as can be seen, for example, by its very low condensationpoint of−162 ∘C. What do you think is the origin of this spectacular failureof the elementary theory of the chemical bond?

13.4 Molecular Spectra

Let us begin with a simple question that may have occurred to some readers.How do we know the geometrical shape of various molecules and even theirquantitative features? For example, how do we know that the hydrogen moleculeis 0.74 Å long, or that the water molecule has a triangular shape with an angle104.5∘ between its bonds, an O–H bond length of 0.96 Å, and a dipole momentof 1.85 D? In the next chapter, we will discuss some cases of organic moleculeswhose structure can be deduced from purely chemical data. But our informationon the structure of molecules comes mostly from molecular spectra. (Nowadays,molecular spectra include also the so-called NMR spectrum, which we discussbriefly in the online supplement of Chapter 16.) For this reason, in this chapterwe will concern ourselves with the typical spectra associated with the main typesof molecular motion, namely, vibration and rotation. The corresponding spec-tra are (unsurprisingly) called vibrational and rotational, respectively, and, sincetheir frequencies lie in the infrared region, they are also known as infrared spec-tra. Previously, we encountered the electronic spectrum of atoms or molecules,which falls in the ultraviolet region as it relates to transitions of atomic or molecu-lar electrons between various electronic energy levels. But we will not discuss theelectronic spectrum here, whose basic features are already familiar to us. Actually,


M1

M1 + M2

r1

r1

r2

M2

M2

C.M.

=

a

a

M1 + M2

r2

M1= a

Figure 13.22 Mechanical model for adiatomic molecule. A weightless bar oflength a (equal to the bond length)with atomic masses M1 and M2 isattached to its ends. The system isknown as a rigid rotor. The center ofmass (CM) divides the line segmentjoining the atoms into intervals thatare inversely proportional to theatomic masses at its respective ends.

the molecular structure is more closely connected to the vibrational and rota-tional spectra, which we will study in the following sections, starting with thesimplest case of the rotational spectrum.

13.4.1 Rotational Spectrum

We will limit our study of molecular rotations and vibrations to diatomicmolecules. The case of larger molecules is beyond the scope of this introductorybook, even though the basic idea remains the same.

From a classical physics viewpoint, a diatomic molecule is modeled as a weight-less bar of a particular length—the bond length—at the ends of which the massesM1 and M2 of its two atoms are attached (Figure 13.22). In classical mechanics,the total kinetic energy of this bar—known as a rigid rotor—can be expressed asthe sum of two terms: the translational energy of the system as a whole, and therotational energy with respect to its center of mass. We thus have

Etot =p2

tot

2Mtot+ 𝓵 2

2I,

where ptot (= p1 + p2) is the total momentum of the rotor, Mtot (= M1 + M2) is itstotal mass, 𝓵 is the angular momentum of the relative motion of the two masses,and I is the moment of inertia for the system with respect to its center of mass.Since we are not really interested in the purely translational part of the totalenergy,12 we will focus only on the rotational energy of the molecule. We thuswrite

E = 𝓵 2

2I,

which is the rotational analog of the expression p2∕2m for the translationalenergy, with the standard substitutions

p −−−−→ 𝓵, m −−−−→ I(momentum −−−−→ angular momentum) (mass −−−−→ moment of inertia)

12 Translational energy, much like momentum, has a continuous spectrum in quantum mechanicsand behaves as in classical mechanics. For example, the translational energies (or speeds) of atomsor molecules in a gas follow the Maxwell–Boltzmann distribution at a given temperature. Incontrast, the quantity that differs markedly from its classical counterpart is rotational energy,because it is associated with the angular momentum vector, whose classical and quantum behaviorsare completely different.


The transition from classical to quantum mechanics is now straightforward. TheHamiltonian of the rotational motion of the molecule is given by the expression

H = 𝓵 2

2I, (13.17)

which means that the rotational energy is quantized, with allowed values

E𝓁 = ℏ2𝓁 (𝓁 + 1)2I

(𝓁 = 0, 1, 2,…), (13.18)

which are equal to the eigenvalues of the square of the angular momentumdivided by 2I. The eigenfunctions are clearly those of 𝓵 2, namely, the sphericalharmonics Y𝓁m.

We now have to calculate the moment of inertia with respect to the centerof mass of the molecule. From its definition, we know that the center of masslies on the line connecting the two atoms and its distance from each atom isinversely proportional to the corresponding atomic mass (Figure 13.22). We thushave for I,

I = M1r21 + M2r2

2 = M1

( M2

M1 + M2a)2

+ M2

( M1

M1 + M2a)2

=( M1M2

2

(M1 + M2)2 +M2M2

1

(M1 + M2)2

)

a2

=M1M2(M1 + M2)

(M1 + M2)2 a2 =M1M2

M1 + M2a2 = 𝜇a2,

where the quantity

𝜇 =M1M2

M1 + M2or 1

𝜇= 1

M1+ 1

M2

is our familiar reduced mass of the system, which arises whenever we “decouple”the total translational motion of a system of two interacting particles to studytheir relative motion.

The allowed values for the rotational energy of a diatomic molecule with lengtha and reduced mass 𝜇 are therefore given by the expression

E𝓁 = ℏ2

2𝜇a2 𝓁(𝓁 + 1) = 12𝜖 𝓁(𝓁 + 1)

(

𝜖 = ℏ2

𝜇a2 , 𝓁 = 0, 1, 2,…)

and the corresponding energy-level diagram is shown in Figure 13.23.As we know from previous discussions (e.g., Section 9.3.3.8), quantum transi-

tions between the above energy states obey the selection rule Δ𝓁 = 1.13 There-fore, for an arbitrary transition 𝓁 → 𝓁 − 1 we have

ΔE𝓁 = E𝓁 − E𝓁−1 = 12𝜖 𝓁(𝓁 + 1) − 1

2𝜖(𝓁 − 1)𝓁 = 𝜖𝓁,

13 The rule expresses in fact—under certain conditions that will be discussed in Chapter 16—theconservation of angular momentum. Since the emitted photon, which has spin one, “carries away”one unit of angular momentum, this amount ought to be subtracted from the initial angularmomentum of the atom or the molecule.


10 4

3

2

μa2

2

1

0

4

3

2

E

6

3

0

Figure 13.23 The quantized rotational levels of a diatomic molecule and the allowedtransitions between them, according to the selection rule Δ𝓁 = 1. The energies of the emittedphotons are integer multiples of 𝜖, and thus the rotational spectrum consists of a fundamentalfrequency f1 = 𝜖∕h and its integer multiples.

which means that the frequencies f𝓁 = ΔE𝓁∕h (𝓁 = 1, 2,…) of the emitted pho-tons are integer multiples of the first of them, f1 = 𝜖∕h. That is,

f𝓁 = 𝜖

h𝓁 = f1𝓁 ⇒ 𝜆𝓁 = c

f𝓁=

c∕f1

𝓁=𝜆1

𝓁(𝓁 = 1, 2,…).

The rotational spectrum of the molecule will thus comprise integer multiples of abasic frequency f1, or, equivalently, integer submultiples of a basic wavelength 𝜆1.

The position of rotational frequencies on the electromagnetic spectrum is read-ily obtained once we know the order of magnitude for rotational energies. The lat-ter is determined by the quantity 𝜖 = ℏ2∕I = ℏ2∕𝜇a2, where 𝜇 is a typical nuclearmass (at least 2000 times the electron mass) and a is typically of the order of 1 Å.We thus find

𝜖 = ℏ2

𝜇a2 ≡ ℏ2

mea20

(a0

a

)2 me

𝜇≈ 27.2 eV ⋅

(12

)2 12000

≈ 10−3 eV.

The basic wavelength 𝜆1 of the emitted photon is then

𝜆1(Å) = 12 000𝜖 (eV)

≈ 104

10−3 = 107 Å = 10−1 cm = 1 mm,

so the rotational spectrum lies in the millimeter range, which is often called farinfrared, in contrast to the so-called near infrared, which lies immediately belowthe visible range and “hosts” the vibrational spectrum of molecules, as we shallshortly see. Now, when the reduced mass of a molecule is much greater thanthe value we used above—where we set 𝜇 ≈ mp ≈ 2000me—then the basic wave-length 𝜆1, and the whole rotational spectrum, moves gradually toward the cen-timeter range, known as the microwave range.


As for the temperatures required to thermally excite molecular rotations, theyare roughly determined by the condition

kT ≈ 𝜖 ≈ 10−3 eV,

which, in conjunction with the familiar expression

(kT)T=12 000 K≈104 K ≈ 1 eV,

yields

T ≈ 104

103 ≈ 10 K,

which means, among other things, that at room temperature, the rotationalmotion of molecules is certainly “switched on.”

There is one more important point to be made on rotational spectra. For a rotat-ing molecule to emit radiation, it has to be polar, in which case it behaves as arotating dipole and thus radiates, according to electromagnetic theory, albeit withquantum leaps. Therefore, only polar molecules (such as HI, HF, HCl, CO) canreadily produce rotational (emission or absorption) molecular spectra. To obtainsuch spectra for nonpolar species (such as H2, N2, O2) requires special techniques(such as Raman spectroscopy) that are beyond the scope of this book.

Let us now apply what we learned above in the following example.

Example 13.2 The observed absorption spectrum for gaseous HCl in the deepinfrared contains the wavelength 𝜆1 = 0.48 mm and its first few integer submul-tiples. Explain the origin of this spectrum, and then use the above experimentalvalue for 𝜆1 to calculate the length of the molecule.

Solution: Since the observed absorption lines fall in the millimeter range andare integer submultiples of a basic wavelength, we can unequivocally concludethat we are dealing with the rotational absorption spectrum of the molecule. Thecalculation of the molecule’s length is thus easily done using the relations

ℏ2

𝜇a2 = 𝜖 = 12 000𝜆1(Å)

eV,

where 𝜇 ≈ mp, since Cl is 35 times heavier than hydrogen, and, therefore, thereduced mass of HCl

𝜇HCl =mH ⋅ mCl

mH + mCl=

mp ⋅ 35mp

mp + 35mp= 35

36mp ≈ mp

is very close to the hydrogen mass, that is, the proton mass. If you do the cal-culation within two decimal digits using the more exact formula for 𝜖 (i.e., 𝜖 =12 400∕𝜆1), you will find a = 1.27 Å, in excellent agreement with experimen-tal data.

The following example demonstrates how we can build on the above experi-mental result to obtain another useful parameter for the HCl molecule.


Example 13.3 According to Table 13.2, the dipole moment of a HCl moleculeis equal to 1.05 D. Combine this value with the result of the previous exampleto calculate the excess (or deficit) of charge, 𝛿+ or 𝛿−, in the two atoms of themolecule.

Solution: By definition, the dipole moment d of a molecule is d = 𝛿 ⋅ a and isequal to 1.05 D = 1.05 ⋅ 0.2 e ⋅ Å, since 1 D ≈ 0.2 e ⋅ Å, as we saw in Section13.3.6. We thus obtain14

𝛿 = da= 1.05 ⋅ 0.2 e ⋅ Å

1.27 Å= 0.17 e,

which tells us that the chemical bond in the HCl molecule is only 17% ionic.This small value is surprising at first, since hydrogen and chlorine lie on diamet-rically opposite columns in the periodic table, so the difference between theirelectronegativity—and hence their polarity—ought to be very high.

Although the above reasoning is in general correct, it provides an overestimatein the case of hydrogen, because this atom—in sharp contrast to all otherelements in the same column of the periodic table—has a high ionizationenergy and is thus not so willing to “lend” its electron to another—moreelectronegative—atom. In other words, hydrogen bonds with atoms of highelectronegativity, such as Cl, do not have the high degree of polarity one wouldexpect based solely on their positions in the periodic table. Which is whyhydrogen’s electronegativity—see Table 12.2—is disproportionately high, givenits position in the table. But if, on the other hand, we replaced H with anyother atom of the same column (say, Na), then the polarity of the bond withCl will increase dramatically. Indeed, in its molecular form—that is, in gaseousphase15—NaCl has a dipole moment 9 D (nine times greater than that of HCl!).So, with a corresponding bond length of 2.36 Å, this leads to an excess ofnegative charge toward Cl equal to

𝛿 = da= 9 D

a= 9 ⋅ 0.2 e ⋅ Å

2.36 Å= 0.76 e.

We see that the Na—Cl bond is 76% ionic.

13.4.2 Vibrational Spectrum

To be sure, no diatomic molecule is a simple “rigid rotor” as we previouslyassumed. Even though there is a configuration of stable equilibrium with thetwo atoms at a certain distance from each other, the atoms cannot come to restthere. Indeed, the Heisenberg uncertainty principle compels them to alwaysjiggle around their equilibrium positions. This vibrational motion of a diatomicmolecule—being a confined motion along the line connecting its atoms—isquantized, and the corresponding spectrum has the shape shown in Figure 13.24,where we have also sketched the molecular potential for the interaction between

14 Let us note again that the symbol 𝛿+ (𝛿−) is used by chemists to denote the excess (deficit) ofcharge in the atoms of a chemical bond.15 NaCl (which is solid at room temperature) melts at 801 ∘C and evaporates at 1413 ∘C.


Figure 13.24 The molecular potentialand the corresponding vibrationalspectrum for a diatomic molecule (in thiscase, HCl). Since the molecular potentialcoincides with that of a harmonicoscillator (dashed line) in the vicinity ofits minimum, the first few vibrationalenergy levels are almost equidistant.Only as we go to higher states do theenergy levels become gradually denser.

Cl

V(r)

H

r0

r

the two atoms. For simplicity, we assumed in the figure that the molecule is HCl,so we can regard Cl—which is much heavier than H—as fixed at the origin, inwhich case only the H atom vibrates.

As explained in the caption of Figure 13.24, the first few vibrational energy lev-els are given by the familiar formula of the harmonic oscillator

E𝑣 =(

𝑣 + 12

)

ℏ𝜔0, (13.19)

where we adopted the convention used in chemistry and denoted the vibra-tional quantum number with the letter 𝑣 (𝑣 ≡ vibration), instead of n, whichis reserved for the principal quantum number of electronic energy levels inatoms and molecules. Now, if the harmonic oscillator approximation were validthroughout—so that all levels in Figure 13.24 would be equidistant—then thevibrational spectrum would contain a single frequency𝜔0, equal to the frequencyof the corresponding classical oscillation of the molecule. This is because, as wediscussed in Chapter 6, transitions between the levels of a harmonic oscillatorobey the selection rule

Δ𝑣 = 1.

But since in reality the higher vibrational states are not equidistant—in fact, theybecome increasingly denser16 as we move upward in energy—the full vibrationalspectrum will contain more frequencies, smaller or even greater than𝜔0, becausethe selection rule Δ𝑣 = 1 is no longer strictly valid and transitions with Δ𝑣 > 1can also occur.

The energy range of the vibrational spectrum is readily obtained from thequantity

E𝑣 = ℏ𝜔0,

16 The spacing of energy levels for a particular potential satisfies the following rule. If the potentialrises more steeply than the potential of the harmonic oscillator, then its levels become gradually lessdense. Conversely, if the potential is less steep than that of the harmonic oscillator (and especially ifit tapers off and stops increasing altogether at some point), then its levels become gradually denser.See a related discussion in Chapter 7, where we also presented the exact solution of two realisticmolecular potentials.


which is the energy of the emitted photon and also the typical value for thevibrational energies of the molecule (e.g., ℏ𝜔0∕2 is the “zero-point energy” ofthe molecule). Given now that 𝜔0 =

√k∕𝜇, where 𝜇 is the reduced mass of the

molecule and k = V ′′(r0), what we need is an estimate for k. This can be easilyobtained from the relation

12

ka2 ≈ ka2 ≈ D = dissociation energy of the molecule ≈ Ee,

where we assumed that if we increase the distance of the two atoms by an amountequal to the length a of the chemical bond, the molecule will break up. Therefore,the work ka2∕2 needed for this displacement—provided again that the harmonicoscillator approximation is valid throughout; after all, we are only interestedin a rough estimate—must equal the dissociation energy D of the molecule,which in turn has a typical electronic value Ee, of the order of a few eV. We thusfind

k ≈Ee

a2 ⇒ 𝜔0 =√

k𝜇

≈

√

Ee

𝜇a2

⇒ E𝑣 = ℏ𝜔0 ≈ ℏ

√

Ee

𝜇a2 =√

ℏ2

𝜇a2

√Ee ≡

√me

𝜇

√

ℏ2

mea2

√Ee ≈

√me

𝜇Ee,

where, in the last step, we also took into account the fact that the bond lengtha is of the order of 1 Å, and hence the quantity ℏ2∕mea2, which represents thekinetic energy of an electron trapped in a region of size a, is of the same order ofmagnitude as Ee (i.e., a few eV). We thus obtain

E𝑣 ≈√

me

𝜇Ee.

Now, for Ee ≈ (3–4) eV and for one of the smallest possible values of 𝜇, say,𝜇 ≈ mp—note that for the H2 molecule 𝜇 is even smaller, that is, 𝜇 ≈ mp∕2—weget

E𝑣 ≈ 10−1 eV,

while for an arbitrary diatomic molecule we have in general

E𝑣 ≈ (10−2 − 10−1) eV.

We may thus conclude that the vibrational spectrum of diatomic molecules liesin the near-infrared region, that is, immediately below the visible spectrum.Figure 13.25 shows the relative position of the various spectral components—rotational, vibrational, or electronic—for a typical diatomic molecule.

For the HCl molecule, which we took as a reference point for our discussionso far, the basic vibrational frequency (corresponding to the transition 𝑣 = 1 →𝑣 = 0) has the experimental value

f = 8.66 × 1013 Hz (HCl),


Rotational spectrum

Far infrared

(∼ 10–3 eV)

Near infrared

(∼ 10–1 eV)UV

eV

Vibrational

spectrum

VisibleElectronic

spectrum

1.6 3

Figure 13.25 The three types of molecular spectra and their position in the electromagneticspectrum measured in eV. (The spectra are shown out of scale.)

which implies that the energy hf of the emitted photon (i.e., the typical vibrationalenergy E𝑣) is equal to

E𝑣 = hf = 0.36 eV,

a result that agrees with our order-of-magnitude estimate (E𝑣 ≈ 10−1 eV).Based on these estimates for vibrational energies, we can now explain why the

rotation of the molecule, for small 𝓁, does not cause a stretching of the “spring”connecting its atoms, and hence an increase of the molecular length. For small 𝓁,rotational energies are much smaller than the energy difference between vibra-tional levels. Accordingly, rotations do not stretch the molecule because theycannot excite the higher vibrational states. Owing to the quantization of its vibra-tional energies, the molecule remains frozen at the lowest vibrational state as itrotates, unless rotation becomes sufficiently strong (large 𝓁). Actually, becausethe vibrational energies of molecules with large reduced mass can be as low as10−2 eV, while rotational energies are typically a few meV, excitation of highervibrational states—that is, molecule stretching—can arise even for relatively low𝓁 values such as 𝓁 = 4 or 𝓁 = 5. This effect shows up in the rotational spectrum,where, even for low 𝓁 values, the observed frequencies cease to be exact inte-ger multiples of a basic frequency—as the rigid rotor model predicted—and thedifference between them starts to decrease, very slowly at first, but much morenoticeably when 𝓁 ≈ 10. We will present pertinent experimental data—for theHCl molecule—shortly, once we examine the combined vibrational and rota-tional spectrum.

13.4.3 The Vibrational–Rotational Spectrum

In the previous discussion, we examined separately the vibrational and rota-tional spectra. But the experimental observation of the vibrational emission orabsorption spectrum17 cannot be done in the absence of the rotational spectrum,because the photon emitted during a vibrational transition (Δ𝑣 = ±1) carriesangular momentum—its spin—and so the rotational motion of the moleculeshould also change (Δ𝓁 = ±1) to conserve the system’s angular momentum.

17 We remind the readers that absorption spectra are obtained in a certain frequency region bytransmitting an electromagnetic wave with a broad frequency spectrum through a molecular gassample, and observing the dark lines—that is, the missing frequencies—in the spectrum of thetransmitted wave.


Note also that at temperatures where the excitation of molecular vibrationalstates is possible, rotational states are also largely excited. Thus the total energyE𝑣,𝓁 of the molecule is the sum of two terms

E𝑣 =(

𝑣 + 12

)

ℏ𝜔0, E𝓁 = ℏ2𝓁(𝓁 + 1)2I

= 12𝜖𝓁(𝓁 + 1), (13.20)

so that

E𝑣,𝓁 =(

𝑣 + 12

)

ℏ𝜔0 +ℏ2𝓁(𝓁 + 1)

2I(𝑣,𝓁 = 0, 1, 2,…). (13.21)

Figure 13.26 shows the corresponding energy-level diagram, along with theallowed transitions according to the selection rules

Δ𝑣 = ±1, Δ𝓁 = ±1.

Here is how we construct the diagram. For a given 𝑣—that is, for a given vibra-tional state—the molecule can rotate with various values of 𝓁. Therefore, to everyvibrational level we add the corresponding values of the rotational energies. Forexample, when the molecule is in its ground vibrational state (𝑣 = 0), it can eitherhave no rotation at all (𝓁 = 0) or it can rotate with 𝓁 = 1, 2,…. Thus a group ofrotational states appear above the ground vibrational level (Figure 13.26).

Each of the resulting two groups of spectral lines—say, the right group—is iden-tical to the purely rotational group of lines, except that it is shifted to the right(i.e., to higher frequencies) by a quantity equal to the basic frequency 𝜔0 of thepurely vibrational spectrum, if this were possible to observe separately.18 Thusthe experimental observation of the vibrational–rotational spectrum allows usto concurrently observe both the rotational and the purely vibrational spectrumof the molecule, since the frequencies 𝜔0 of the latter are readily obtained as thecenter of the group of lines in Figure 13.26. Note also that the purely vibrationalspectrum consists not of a single line (located at the frequency 𝜔0) but of a num-ber of lines, for reasons we described earlier: deviations of the true molecularpotential from that of a harmonic oscillator, but also the appearance of transi-tions with Δ𝑣 > 1, since the corresponding selection rule Δ𝑣 = 1 is strictly validfor the parabolic potential only. Thus the complete vibrational–rotational spec-trum consists of a fine structure of rotational lines centered on each vibrationalline 𝜔1, 𝜔2,…, as shown in Figure 13.26. Now, if the resolution of our spectrom-eter is low, this fine structure will not show up in the spectrum. Instead, we willobserve only some thick lines (like bands) that depict, in a way, the purely vibra-tional levels of the molecule.

Another point worth commenting on here is the intensity of the observed spec-tral lines. Why are some absorption (or emission) lines so strong, while others aremuch weaker? Although the quantum mechanical “details” pertaining to the pre-cise shape of the molecular wavefunctions play a role here, the decisive factoris the relative population of the various states—that is, the fraction of the total

18 Note, incidentally, that the theory of quantum transitions, which we present in Chapter 16, canbe used to prove not only all these selection rules but also the fact that the purely rotationaltransitions can occur only for a polar molecule. But the physical explanations given here aresufficient for our purposes.


E1

4

4

3

2

1

0

3

2

1

0

E0

ω0

υ = 1

υ = 0

Figure 13.26 Absorption lines of the vibrational–rotational spectrum for a typical diatomicmolecule. Transitions with Δ𝓁 = +1 and Δ𝓁 = −1 produce two groups of spectral lines,positioned symmetrically on either side of a hypothetical spectral line 𝜔0, which correspondsto a forbidden transition between the purely vibrational levels E0 and E1. This transition is notobserved in the spectrum because it violates the selection rule Δ𝓁 = ±1, which has to besatisfied simultaneously with the rule Δ𝑣 = 1, for reasons we explained before.

number of molecules in each state, which is determined by the temperature ofthe system through the well-known Boltzmann factor exp (−En∕kT).

So, by observing the position but also the relative intensity of spectral lines,we can both identify the molecules of a gas and determine its temperature.It is for this reason that in astrophysics—where direct measurements areimpossible—molecular spectroscopy is the key tool for the determination of thechemical composition and temperature of interstellar and intergalactic gases.

Molecular spectra—and, in particular, vibrational–rotational spectra—are alsowidely used to detect isotopes of elements in a sample, because both the rota-tional and vibrational energy levels of a molecule depend directly on its reducedmass, that is, on the masses of its constituent atoms. Therefore, if an isotope ispresent in a sample, we can observe shifted lines in the absorption spectrum,the intensity of which can inform us of the isotope composition of our sample.


65

54

43

32

2

2

23 4 5

65

43

1

1

1 10

8.00 8.20 8.40 8.60

f0

8.80 9.00 9.20 ×1013 Hz

0

Figure 13.27 The vibrational–rotational absorption spectrum of the HCl molecule. Theappearance of a second peak for each line is due to the Cl37 isotope that constitutes 24.5% ofthe Cl in the HCl sample. We can clearly see two groups of lines (R-branch and L-branch)corresponding, respectively, to transitions Δ𝓁 = +1 and Δ𝓁 = −1. In the middle of the curvewe show the “central frequency” f0 = 𝜔0∕2𝜋 that corresponds to the forbidden purelyvibrational transition 𝑣 = 0, 𝓁 = 𝓁0 → 𝑣 = 1, 𝓁 = 𝓁0.

A good example to demonstrate these effects—especially the isotope shift—is theHCl molecule. Indeed, since 24.5% of the element Cl found in nature is in the Cl37

isotopic form, the spectral lines of HCl (in the vibrational–rotational spectrum)show up as double lines, as can be seen in the experimental curve of Figure 13.27.

Actually, the spectrum of HCl in Figure 13.27 calls for some further remarks.First, the (approximately) equal distancesΔf between neighboring peaks confirmthe formula f = 𝜖𝓁∕h for rotational frequencies, which indeed predicts a constantdistance between them, equal to (note that Δ𝜔 = Ω = 𝜖∕ℏ = ℏ∕𝜇a2)

Δf = Δ𝜔2𝜋

= Ω2𝜋

= ℏ

2𝜋𝜇a2 . (13.22)

Now, a closer inspection of the spectrum reveals that Δf is not strictly constant.For example, in the right branch (R-branch) of the spectrum, Δf appears todecrease as we move to the right, that is, toward rotational transitions withgreater 𝓁. This trend reflects the stretching of the molecule, that is, the increaseof the bond length a in (13.22), caused by the increasing centrifugal force due tofaster rotation. The opposite effect is observed in the left branch (L-branch) ofthe spectrum, where the distance between neighboring peaks appears to increaseas we move leftward. Can you think of a plausible explanation for this distinctbehavior of the two branches? Does it matter that the rotational transitions inthe L-branch are from higher to lower 𝓁 values (that is, 𝓁 → 𝓁 − 1), and viceversa for the R-branch (𝓁 → 𝓁 + 1)?

Let us apply the ideas discussed above in the following examples.

Example 13.4 Use the experimental data of Figure 13.27 to calculate the forceconstant k of the HCl molecule. (k is the proportionality constant in the expres-sion for the force F = −k(r − r0), valid for small displacements of atoms from


their equilibrium distance r0 in the molecule.) Express the result in the SI unit ofN/m for this quantity.

Solution: Using the relations

𝜔0 =√

k𝜇

= 2𝜋f0 = 2𝜋(8.66 × 1013 Hz)

and19

𝜇 =mH ⋅ mCl

mH + mCl≈

mp ⋅ 35mp

mp + 35mp= 35

36mp = 1.6260 × 10−27 kg

we obtain

k = 𝜇𝜔20 = 𝜇(2𝜋f0)2 = 1.626 × 10−27(2𝜋)2(8.66 × 1013)2

⇒ k = 480.9 N/m ≈ 481 N/m.

Note that k is a very useful parameter for a diatomic molecule because it is a goodmeasure of the force that keeps its atoms together.

And a question for the reader: If someone claimed that this value of k seemstoo large for a tiny system such as a molecule, how would you convince him heis wrong? Can you come up with a simple order-of-magnitude calculation basedon the simplest possible data?

Example 13.5 Chlorine can be found in nature in its two stable isotopes, Cl35

and Cl37, with concentrations ≈ 76% and 24%, respectively. A typical sample ofHCl gas that is used to obtain the infrared spectrum of Figure 13.27 will thuscontain the two Cl isotopes with these proportions. (a) Explain qualitatively whythe lower peaks in the spectrum lie to the left of the higher peaks. (b) Give a roughestimate for the distance between low and high peaks of the same spectral lineand compare with what you see in the figure.

Solution: According to our earlier discussion, the angular frequencies 𝜔𝓁 of thevibrational–rotational spectrum (say, its right branch) are given by the expression

𝜔𝓁 = 𝜔0 + 𝓁Ω, 𝓁 = 1, 2, (1)

where 𝓁 is the quantum number of the final rotational state in the transition𝓁 − 1 → 𝓁, and

𝜔0 =√

k𝜇, Ω = 𝜖

ℏ= ℏ

𝜇a2 . (2)

Both the force constant k and the distance a between atoms in the molecule areindependent of the reduced mass𝜇 of the molecule, because the molecular poten-tial is determined solely by the motion of valence electrons in the electrostaticfield of the nuclei.

19 We assumed for simplicity that the mass of Cl35 is 35mp, which is not strictly correct. Thecorrect calculation should involve the masses of H1 and Cl35 in atomic mass units u, which aremH = mp = 1.0078u and mCl35 = 34.9688u, respectively.


Given now that the reduced mass 𝜇′ of HCl37 is greater than that of HCl35,20 itis clear that the corresponding frequencies 𝜔0 and Ω for HCl37

𝜔′0 =

√k𝜇′ , Ω′ = ℏ

𝜇′a2

will be smaller than the values (2) for HCl35. As a result, the correspondingabsorption lines for HCl37 will shift to the left compared to the absorptionlines of HCl35. And they will have lower peaks because there are fewer HCl37

molecules in the sample than HCl35 molecules, so they will absorb less radiation.How pronounced is this “isotope shift?” It is clear from expressions (2) that the

relative change Δ𝜔∕𝜔 is of the same order of magnitude for both the vibrationaland rotational frequencies, 𝜔 and Ω, respectively. So we need only calculate thisshift for one of the two frequencies, say, the vibrational frequency 𝜔0 =

√k∕𝜇.

We have thenΔ𝜔0

𝜔0=𝜔0 − 𝜔′

0

𝜔0=

√k∕𝜇 −

√k∕𝜇′

√k∕𝜇

= 1 −√

𝜇

𝜇′

= 1 −

√

35∕3637∕38

= 0.0008 ≈ 10−3,

which tells us that the isotope shift of absorption lines is expected to be on theorder of one part per thousand. This prediction is confirmed by the spectrum ofFigure 13.27. Indeed, the distanceΔf between the low and high peaks in the figureis approximately equal to 0.01 × 1013 Hz—note that if the high peak is positionedat 8.60 × 1013 Hz, then the low peak is roughly at 8.58 × 1013 Hz or 8.59 × 1013

Hz—so that the relative shift Δf ∕f is on the order of 10−3, as predicted.

Problems

13.6 Choose the correct relations among the following. Here, the symbol𝜆𝓁→𝓁−1 denotes the wavelength of the photon emitted during the𝓁 → 𝓁 − 1 rotational transition.(a) 𝜆1→0 = 3𝜆3→2, (b) 𝜆2→1 = 3

2𝜆3→2,

(c) 𝜆5→4 = 54𝜆4→3, (d) 𝜆4→3 = 3

2𝜆6→5.

13.7 The vibrational frequency f of the diatomic molecules

HF, HBr, CO, NO

is, respectively, equal to

8.72, 7.68, 6.42, 5.63 (× 1013 Hz).

20 From the definition 𝜇−1 = M−11 + M−1

2 , it is clear that the reduced mass 𝜇 is always smaller thanthe smallest of the two masses, M1 and M2. So, if the smallest mass is M1, say, then 𝜇 approaches M1from lower values, as M2 increases.


Calculate the zero-point energies and the spring constants (in N/m) ofthese molecules.

13.8 In the H2 molecule, the equilibrium distance between its atoms is equal to0.74 Å. Calculate (in eV) the rotational energy of the molecule (around itscenter of mass) when the rotational motion of the molecule is describedby the quantum number 𝓁 = 1. Then, calculate the wavelength of theemitted photon when the system makes a transition from 𝓁 = 1 to theground state.

13.9 Which one of the O2 and N2 vibrational spectra do you expect to lie inhigher frequencies? Elaborate on your answer.

13.10 The vibrational frequency of the HCl molecule is equal to 8.66 × 1013Hz.Use the H−Cl bond length we found in Example 13.2 (a = 1.27 Å) to cal-culate the amplitude of its oscillation in the vibrational ground state, andthus determine how much the length of the molecule fluctuates as a frac-tion of the equilibrium value a.

Further Problems

13.11 Consider a system of three potential wells, as depicted below.

1 2 3x

The ground state energy of each well, E0, and the Hamiltonian matrix ele-ment between two neighboring wells, H12 = H23 = −A, are consideredto be known quantities. Use the LCAO approximation to calculate theenergy eigenvalues and the corresponding eigenfunctions in terms of E0and A. In which of the three wells is it more likely to find the particle if thesystem is in its ground state? Is your answer reasonable from a physicalpoint of view?

Note that the triple-well problem—as well as its generalization to N >

3 wells—will prove useful in the next chapter, where we will discuss theso-called delocalized chemical bonds, which constitute a fundamentaldeviation from the elementary theory of the chemical bond we presentedin this chapter.

13.12 (a) Provide a quantum mechanical analysis for the formation of theammonia (NH3) molecule and use it to predict its shape. (b) Basedon your analysis, would you believe someone who claims that NH3condenses at −209 ∘C? If not, what is the condensation temperatureyou would expect? (c) What can you say about the specific heat ofammonia? If someone claimed that the supply of the same amount


of heat causes one gram of ammonia to be heated more (i.e., reach ahigher temperature) than one gram of H2 or N2, would you believe heror him? (d) It is well known that ammonia is an excellent solvent. Canyou explain this property based on your earlier analysis of the structureof its molecule?

13.13 As we stated in the text, when two identical wells approach each other,then each level in the single well splits to two levels in the double well,which are symmetrically arranged with respect to the initial level. Thecorresponding wavefunctions are given, respectively, by the even andodd linear combinations of the wavefunctions of the single wells. Applythis recipe to sketch the complete energy diagram for a double-well sys-tem, when each well has two bound states. Follow the usual conven-tion and sketch also the corresponding wavefunctions on the energy lev-els. What should you be mindful of with regard to the energy differencebetween levels of the same pair?

13.14 A hydrogen molecule (H2) is in its rotational ground state. What is itsmost probable spatial orientation? Consider the same question if themolecule is in the rotational state where 𝓁 = 1 and m = 0.

13.15 Two hydrogen atoms (H) can combine to form a hydrogen molecule(H2 ≡ HH). It is also possible for atomic hydrogen to combine with thesecond most stable isotope of hydrogen, deuterium (D), whose nucleuscontains one proton and one neutron, to form a molecule known ashydrogen deuteride (HD). Similarly, two deuterium atoms can combineto form a deuterium molecule (D2 ≡ DD). If 𝜆1(H2) is the wavelengthof the photon emitted during the rotational transition 𝓁 = 1 → 𝓁 = 0 ofthe H2 molecule, calculate 𝜆1(HD) and 𝜆1(D2) in terms of 𝜆1(H2).

13.16 Use the solution of the asymmetric double-well problem to show thatit is impossible for two different noble gases (e.g., helium and neon) tochemically bond and form a molecule.

393

14

Molecules. II: The Chemistry of Carbon

14.1 Introduction

Without carbon and its compounds, chemistry would be a rather narrowdiscipline—not to mention that chemists would not exist!

Therefore, the real test of the elementary theory of the chemical bond,presented in the previous chapter, is to apply it to carbon compounds and seewhether it explains the fascinating chemistry of this unique element. As wementioned briefly in Chapter 13, the application of the elementary theory ofthe chemical bond to carbon compounds led to the realization that there aretwo basic deviations from it. These deviations, known as hybridization anddelocalization, are the subject of this chapter.

14.2 Hybridization: The First Basic Deviation fromthe Elementary Theory of the Chemical Bond

14.2.1 The CH4 Molecule According to the Elementary Theory:An Erroneous Prediction

We will now show that the application of the elementary theory of the chem-ical bond to carbon compounds leads to flawed conclusions. This failure callsfor a re-examination of the basic assumptions of the theory and a correspond-ing modification. For this purpose, it will suffice to examine the simplest of allorganic compounds, the well-known molecule of methane, CH4. We begin withthe electronic configuration of carbon (Z = 6),

[C] = 1s2 2s2 2p1x 2p1

y 2pz,

which readily tells us that carbon is a tetravalent element whose valence orbitalsare 2px, 2py (both half-filled), and 2pz (unoccupied). The chemical formula fora compound made of C and H atoms is thus CH4, while its geometric shape isshown in Figure 14.1.


394 14 Molecules. II: The Chemistry of Carbon

H

H

C H

HH

z y

x

(b)

H

H

H(a)

y

z

x

Figure 14.1 The CH4 molecule according to the elementary theory of the chemical bond.(a): Coupling between atomic orbitals of C and H. Since the 2pz orbital of C contains noelectrons, it couples to the 1s orbitals of two H atoms, which provide the two electrons thatoccupy the molecular orbital that is formed. (b): The geometric shape of CH4. Because thedipole moments of the two bonds on the x–y plane do not cancel each other out, the CH4molecule ought to be polar, with all the associated physical properties. And yet, CH4 isnonpolar. The elementary theory of the chemical bond fails spectacularly here.

The basic conclusion from the above discussion is that CH4 ought to be a polarmolecule—like H2O or NH3—with all the associated properties. For example,high melting and boiling points and high latent heats. This prediction is mani-festly wrong, if we recall, for example, that methane, which is the primary con-stituent of natural gas, is carried over via pipelines across areas such as Siberiaor Alaska, where temperatures fall to −70 ∘C, and yet it does not liquify! In fact,the boiling point of CH4 is −162 ∘C, compared to −33 ∘C for NH3 and 100 ∘Cfor water! Clearly, CH4 is unlike H2O or NH3, and behaves instead as a typicalnonpolar molecule. There is no doubt that the elementary theory of the chemi-cal bond fails badly for this molecule, whose actual shape does not resemble ourprediction (Figure 14.1b), but is depicted in Figure 14.2.

The tetrahedral structure of CH4 can be deduced from spectral measurements(including NMR spectra) and purely chemical data. For example, the existence ofa unique CH3X compound in nature—where X is a monovalent element such asCl—implies that the four possible positions of the X atom in the compound mustbe equivalent. The tetrahedral structure of CH4 is thus an indisputable experi-mental fact, which cannot be explained by the elementary theory of the chemicalbond. Clearly, we must modify the theory.

H

H

H

H

C

Figure 14.2 Actual structure of the methane molecule. Thecarbon atom sits at the center of a tetrahedron whose fourvertices are occupied by the H atoms. Owing to the symmetricarrangement of the H atoms, the dipole moments of the fourC–H bonds cancel each other out, and the CH4 molecule isnonpolar.

14.2 Hybridization: The First Basic Deviation 395

14.2.2 Hybridized Orbitals and the CH4 Molecule

Which assumption led us to the incorrect shape of Figure 14.1? It was the assump-tion that only the 2p (i.e., 2px, 2py, 2pz) states of C participate in the formation ofthe molecular orbitals of methane. To begin with, this is not an unreasonableassumption to make. Even though the n = 2 shell includes the 2s state, we cannormally ignore it, since it lies lower than the 2p state, as we know. But sup-pose the 2s level lies very close to 2p. Can we still ignore it? The answer arisesreadily if we consider the limiting case of degeneracy, namely, when the 2s and2p levels have the same energy. How would we analyze the chemical behavior ofcarbon then? The first thing to note in that case is that the valence orbitals can-not be chosen uniquely, since, due to degeneracy, every linear combination of the2s, 2px, 2py, and 2pz eigenstates is also an eigenstate with the same eigenvalue.1At this point the readers may wonder why we did not raise a similar issue whendealing with the degeneracy of the 2p (i.e., 2px, 2py, 2pz) states earlier. Actually,that issue was resolved on its own, so to speak. For the 𝓁 = 1 degeneracy, the bestchoice of valence orbitals are the px, py, pz (compared to any other linear combi-nation of them): Because their directions are orthogonal, these orbitals are as farapart from each other as possible, thus minimizing the electrostatic repulsionsbetween electrons in the corresponding bonds. So, in the 2p case, the correctvalence orbitals have already been found to be the 2px, 2py, and 2pz orbitals, wherex, y, and z represent three arbitrary orthogonal directions. What is needed nowis to examine what happens when we include the 2s state to the degenerate set.The new possibility that arises is to form linear combinations of both types ofstates (2s and 2p) in a way that may be advantageous in terms of chemical bond-ing. To explore this possibility, let us form the combination 𝜓 ∼ 𝜓2s + 𝜓2pz

andconsider the shape of the corresponding mixed orbital. Figure 14.3 depicts thisconstruction, where in the left-hand side we show the two superimposed orbitals(in the usual schematic way), and in the right-hand side we show a simplifiedpicture of their superposition. As you can see, the s2 orbital interferes construc-tively with the upper pz lobe—they have the same sign—and destructively withthe lower lobe. The resulting hybridized orbital thus comprises a large positivelobe on one side and a tiny negative lobe on the other. We customarily depictthis on paper as a strongly asymmetric “8” digit, as in Figure 14.3. In any case,our main conclusion is that the mixing (hybridization) of an s and a p orbitalproduces a strongly unidirectional orbital, which has an obvious advantage interms of chemical bonding. Indeed, recall that a chemical bond gets stronger asthe overlap of orbitals from the partnering atoms increases (for a certain sep-aration distance). Thus a hybridized orbital is clearly advantageous, since it iselongated—actually, its amplitude is increased—along one direction, while a purep orbital is equally distributed on both sides. Figure 14.4 highlights this differencebetween the hybridized and unhybridized orbitals, when a bond is formed with ans orbital, as in the case of interest here (CH4 molecule). It is clear that the mixingof the s and p orbitals produces orbitals of superior “breed” (hence the biological

1 Because all those states are solutions of the same linear and homogeneous equation(the Schrödinger equation) with the same energy.2 For simplicity, we omit the principal quantum number here.


2s 2pz = hybrid

Figure 14.3 The concept of hybridization. The superposition (i.e., hybridization) of the s and porbitals produces a strongly unidirectional orbital that is much more suitable for chemicalinteractions with other atoms. Of course, this sketch reflects a convention used in the literaturerather than an accurate depiction, which would require two equal-sized lobes but with muchmore intense shading on the upper lobe than the lower one.

(a) (b)

Figure 14.4 A “pure” p orbital (a) and a hybridized orbital(b) couple with the 1s orbital of another atom. The couplingis much stronger in case (b) because of the much higherintensity of the probability cloud in the upper lobe of thehybridized orbital. Even though this is not very clearlyshown in the figure, we could say that in case (b) we haveenhanced overlap between the participating orbitals, sincein the overlap integral the intensity of the probabilityclouds also plays a role, not only the spatial extent of theiroverlap. The chemical bond formed from hybridizedorbitals as depicted in (b) is thus much stronger.

term hybridization), which may well be better suited for the explanation of thepeculiar chemical behavior of carbon we encountered in the CH4 molecule. In thereal CH4 molecule, of course, the 2s and 2p states are not exactly degenerate. Buttheir energy difference is small enough that it is overcompensated by the energygain from the strengthening of the chemical bonds due to hybridization. So, itpays for a carbon atom to “lift” one of its 2s electrons up to the empty 2pz level(Figure 14.5) because the mixing of the four half-filled orbitals 2s, 2px,y,z produceschemical bonds of higher quality.

Our next step is to construct the linear combinations of 2s and 2p states thatmake up the valence orbitals of carbon in the methane molecule. Since we alreadyknow the shape of the molecule, it is not hard to realize that, here, carbon has fourhybridized orbitals along the directions of its tetrahedral structure. Therefore, ourfirst task is to construct hybridized orbitals in an arbitrary direction n. Towardthis end, we recall that the angular dependence of the 2px, 2py, 2pz states is givenby the spherical harmonics

Yx ∼xr, Yy ∼

yr, Yz ∼

zr, (14.1)

which describe, as we know from Chapter 9, states with 𝓁 = 1 and zero pro-jection of angular momentum along the corresponding axis. We can readily seefrom expressions (14.1) that the spherical harmonic with 𝓁 = 1 and a vanishing


2s2px 2py 2pz

1s

Figure 14.5 Occupied energy-level diagram in a hybridized state. A carbon atom can achievehybridization between its s and p orbitals by “lifting” one of the two 2s electrons up to theempty 2pz level. Thus—assuming we ignore the energy difference between the 2s and 2pstates—all states 2s, 2px , 2py , and 2pz , and all their linear combinations, are equally availablefor chemical bonding.

projection along an arbitrary direction n should have the form

Yn ∼ n ⋅ rr

= nxxr+ ny

yr+ nz

zr= nxYx + nyYy + nzYz = n ⋅Y ,

where n is the unit vector along that arbitrary direction, and Y is the vectorformed by the triplet Yx,Yy,Yz. That is, Y = (Yx,Yy,Yz). A wavefunction withquantum numbers n = 2, 𝓁 = 1, and mn = 0 is thus given by

𝜓2pn∼ R2p(r)Yn ≡ nx𝜓2px

+ ny𝜓2py+ nz𝜓2pz

.

At this point, let us simplify the notation and denote the states𝜓2s, 𝜓2px, 𝜓2py

, 𝜓2pz

as s, px, py, pz, respectively, with no reference to the principal quantum numbern, which is constant throughout (n = 2). We thus write

𝜓2s ≡ s, 𝜓2px≡ px, 𝜓2py

≡ py, 𝜓2pz≡ pz,

while for the pn ≡ 𝜓2pnstate, we have

pn = px ⋅ nx + py ⋅ ny + pz ⋅ nz = p ⋅ n,

where p = (px, py, pz). Clearly, the pn orbital has the same shape as the px, py, pzorbitals, but points in the direction of the unit vector n. So, to construct ahybridized orbital in the n direction, we simply need to form the combination

𝜓n = N(s + 𝜆pn), (14.2)

where N is the normalization constant and 𝜆 is a mixing parameter that deter-mines the degree to which the s and pn orbitals participate in the hybridized state(14.2). Apart from very small or very large values of 𝜆, for which (14.2) becomeseither a pure s or a pure pn state, for all other 𝜆 values the (14.2) orbital has thedistinctive shape of Figure 14.3, but is directed along n. In particular, for positive𝜆, the hybrid points along the positive n axis,3 while for negative 𝜆 it points inthe opposite direction.

3 The directionality of a hybridized orbital is determined by the orientation of its mostprominent lobe.


H

H

H

H

ψb

ψa

ψd

ψc

Figure 14.6 The methane molecule according tohybridization theory. The four hybridized orbitals of Cpoint from the center of a tetrahedron—shown indashed lines—toward its four vertices, where theymeet with the 1s orbitals of the four hydrogen atoms.

Returning to the methane molecule, we can now write the four valence orbitalsof carbon as

𝜓a = N(s + 𝜆pa), 𝜓b = N(s + 𝜆pb),𝜓c = N(s + 𝜆pc), 𝜓d = N(s + 𝜆pd),

(14.3)

where

pa = p ⋅ a, pb = p ⋅ b, pc = p ⋅ c, pd = p ⋅ d

and a, b, c, and d are four unit vectors forming equal angles with each other.(This defines uniquely the directions from the center of a tetrahedron to its ver-tices.) Actually, the fact that all hybridized orbitals in (14.3) share the same mixingparameter 𝜆 is a manifestation of the equivalence of the four methane bonds, aproperty depicted clearly in Figure 14.6.

We will now show that both the value of the parameter 𝜆 and the tetrahedralcharacter of the a, b, c, and d directions follow from a single mathematicalrequirement, whose physical meaning will be discussed below. The requirementstipulates that the four states 𝜓a, 𝜓b, 𝜓c, 𝜓d form an orthonormal basis in thesubspace defined by the four initial orthonormal states 𝜓2s ≡ s, 𝜓2px

≡ px,

𝜓2py≡ py, 𝜓2pz

≡ pz. As for normalization, it is clear that N in (14.3) must takethe value N = (1 + 𝜆2)−1∕2. To explore the orthogonality between the states(14.3), we invoke the familiar relations4

(s, s) = 1, (s, pi) = 0, (pi, pj) = 𝛿5ij i, j = 1, 2, 3 ≡ x, y, z,

so that

(pa, pb) = (p ⋅ a,p ⋅ b) =

(∑

ipiai,

∑

jpjbj

)

=∑

i,j(pi, pj)aibj =

∑

i,j𝛿ijaibj = a ⋅ b,

4 We recall that (see also Section 2.6.3) the simplified notation for the expression ∫ 𝜓∗𝜙 dx, knownas the inner product of the wavefunctions (or states) 𝜓 and 𝜙, is ∫ 𝜓∗𝜙 dx = (𝜓, 𝜙).5 The symbol 𝛿ij, known as Kronecker delta, is equal to zero for i ≠ j, and one for i = j. Its usefulnessis that it allows us to express in a compact way both the orthogonality of a set of vectors and the factthat they are normalized to unit length (i.e., they form an orthonormal set).


Figure 14.7 Tetrahedral geometry. Four unit vectors form aregular tetrahedron when they have a common origin and theirmutual angles are all equal.

dD

B

A

O

a

C

c

b

and, furthermore,

(𝜓a, 𝜓b) = N2(s + 𝜆pa, s + 𝜆pb)= N2 ((s, s) + 𝜆(s, pb) + 𝜆(pa, s) + 𝜆2(pa, pb)

)

= N2 ((s, s) + 0 + 0 + 𝜆2(pa, pb))

= N2 (1 + 𝜆2a ⋅ b).

Hence, in order that (𝜓a, 𝜓b) = 0, we must have

1 + 𝜆2a ⋅ b = 0 ⇒ a ⋅ b = −𝜆−2 (14.4)

and similarly for the other inner products of the 𝜓a, 𝜓b, 𝜓c, 𝜓d set. Thus, accord-ing to (14.4), all inner products of pairs of the unit vectors a, b, c, d are equal.This means that all their mutual angles are equal and, if we plot the vectors inspace with a common origin, as in Figure 14.7, their endpoints A,B,C, and Dform a regular tetrahedron. (If this does not seem evident to you, simply considerthat, since the angles between the unit vectors a, b, c, and d are equal, the trian-gles OAB,OBC,OCD, and ODA are also equal, which implies that the tetrahedraledges AB,BC,CD,CA,DB, and DA must be equal as well.)

We have shown that the orthogonality condition alone forces the fourhybridized orbitals 𝜓a, 𝜓b, 𝜓c, 𝜓d to arrange into a tetrahedral structure. Whatabout the mixing parameter 𝜆? This is readily calculated from (14.4), since it isrelated to the angle between the vectors a and b, which is uniquely determinedby the tetrahedral geometry. Indeed, by a clever choice of a quadruplet of vectorsa, b, c, d, that satisfy the condition a ⋅ b = b ⋅ c = · · · c ⋅ d—see, for example, thequadruplet given below—it can be easily shown that

cos (a, b) = −13= − 1

𝜆2 ⇒ 𝜆 =√

3,

where we selected the positive value of 𝜆 to ensure that the (14.3) orbitals pointin the positive direction of the vectors a, b, c, d of Figure 14.7. The value 𝜆 =

√3

tells us that the hybridized state

𝜓a =s + 𝜆pa√

1 + 𝜆2

||||||𝜆=

√3

= 12

(

s +√

3 pa

)

,

is one part s state and three parts p state. Such proportions imply that the s, px, py,and pz states participate equally in the formation of the hybrid, which makes


sense: Since we have three p states, it is reasonable that their collective partic-ipation in the hybrid is 3∕4, while the remaining 1∕4 is left for the s state.

If we now wish to describe the (14.3) states more explicitly, all we have to do ischoose any four vectors a, b, c, and d, which form equal angles with each other,and calculate the inner products pn = p ⋅ n, where n = a, b, c, d. The standardchoice in the literature is

a = 1√

3(1, 1, 1), b = 1

√3(1,−1,−1),

c = 1√

3(−1, 1,−1), d = 1

√3(−1,−1, 1).

We can derive the above b, c, d vectors from a, through mirror reflections withrespect to all but one axes at a time: y and z first, then x and z, and finally, x andy. We thus arrive at the four hybridized states

𝜓a = 12(s + px + py + pz), 𝜓b = 1

2(s + px − py − pz),

𝜓c =12(s − px + py − pz), 𝜓d = 1

2(s − px − py + pz).

(14.5)

Another suitable choice of vectors, corresponding to the geometry ofFigure 14.7, is

a = (0, 0, 1), b =

(

−√

23,−

√23,−1

3

)

,

c =

(2√

23

, 0,−13

)

, d =

(

−√

23,

√23,−1

3

)

.

The associated hybridized orbitals are then

𝜓a = 12

(

s +√

3 pz

)

, 𝜓b = 12

(

s −√

23

px −√

2 py −1√

3pz

)

,

𝜓c =12

(

s + 2√

23

px −1√

3pz

)

, 𝜓d = 12

(

s −√

23

px +√

2 py −1√

3pz

)

.

(14.6)

Let us now briefly discuss the physical meaning of the orthogonality conditionfor the hybridized orbitals of carbon. Basically, the orthogonality of any twowavefunctions expresses the fact that the corresponding states have nothing incommon. For example, if a particle is found in one of these two states, it haszero probability to be found in the other. In the present case, the orthogonalityof hybridized orbitals tells us that these chemical bonds are independent fromeach other. Electrons do not hop from one bond to another, and each bondcan thus be examined independently. This independence of the chemical bondsof carbon—among other elements—is a well-known empirical fact: The total


chemical energy of a molecule is the sum of the energies of its bonds. So, theorthogonality of the valence orbitals has direct physical implications,6 as we willsee in our study of hybridization below.

14.2.3 Total and Partial Hybridization

We saw earlier that we can treat the valence orbitals of carbon in the CH4molecule as another orthonormal basis in the four-dimensional space definedby the original, unhybridized states s, px, py, and pz. A distinctive feature ofthis basis is its complete hybridization. In other words, all members of theoriginal basis participate in the mixing. But we can just as well construct otherorthonormal bases, where one or two p orbitals do not participate—i.e., theyremain unhybridized—and examine whether these bases can be useful in thereal world.

In this spirit, let us examine the case whereby only one p orbital—say,pz—remains pure, while the other two orbitals—i.e., px, py—hybridize with s.The new basis is then

𝜓a = Na(s + 𝜆apa), 𝜓b = Nb(s + 𝜆bpb), 𝜓c = Nc(s + 𝜆cpc), 𝜓 = pz,

where pa = p ⋅ a, pb = p ⋅ b, pc = p ⋅ c, and a, b, c are three arbitrary unit vectorsin the x-y plane. The mixing parameters 𝜆a, 𝜆b, 𝜆c can also be arbitrary, unless wewant the three hybridized states𝜓a, 𝜓b, 𝜓c to be equivalent, in which case we musthave 𝜆a = 𝜆b = 𝜆c = 𝜆. In this last scenario—which has actually greater practicalinterest—the four basis states are

𝜓a = N(s + 𝜆pa), 𝜓b = N(s + 𝜆pb), 𝜓c = N(s + 𝜆pc), 𝜓 = pz,

where 𝜆 is determined by the orthogonality condition between these states. First,note that the unhybridized state 𝜓 = pz is orthogonal to all other states, sincethese contain only s, px, and py orbitals, which are all orthogonal to pz. Further-more, given that

(𝜓a, 𝜓b) = N2(1 + 𝜆2a ⋅ b),

the orthogonality of the hybridized triplet 𝜓a, 𝜓b, 𝜓c requires that

a ⋅ b = b ⋅ c = a ⋅ c = −𝜆−2,

which implies that all angles between the (coplanar) vectors a, b, and c are equalto each other, and therefore equal to 120∘. We then have

a ⋅ b = cos (a, b) = cos 120∘ = −12= − 1

𝜆2 ⇒ 𝜆 =√

2,

6 Actually, this argument is oversimplified. The lack of mixing between electrons of different bondsis due to the very small value of the matrix element of the molecular Hamiltonian pertaining to themolecular orbitals of two different bonds. But it is qualitatively plausible that molecular orbitalswith small mutual overlap (and thus orthogonal to each other) have correspondingly smallHamiltonian matrix elements.


where the value√

2 makes sense, again, for the same reason as before: The sorbital “partners up” with only two p orbitals, so the participation ratios mustbe 1∕3 and 2∕3 for the s and p orbitals, respectively. This situation is reflected inthe expression for the corresponding hybridized orbital


1 + 𝜆2

||||||𝜆=

√2

= 1√

3

(

s +√

2 pa

)

,

which differs from the analogous expression of the previous section because thevector a is now on the x–y plane and, therefore, only two p orbitals participate inthe hybrid with combined probability 2/3; of course, we have 1/3 for the s state.If we now choose the vector a on the x axis, and the b and c vectors so that theyform angles with a equal to 120∘ and 240∘, respectively, we obtain the states

𝜓a = 1√

3

(

s +√

2 px

)

, 𝜓b = 1√

3

(

s − 1√

2px +

√32

py

)

,

𝜓c =1√

3

(

s − 1√

2px −

√32

py

)

,

(14.7)

which, together with pz, make up an orthonormal quadruplet. This type ofhybridization, whereby only two p orbitals mix with the s orbital, is called sp2

hybridization, in contrast to the case we examined earlier, which is denoted assp3. Of course, there is also the sp1 hybridization, where mixing involves onlyone p orbital, say, px. In this case, the orthonormal quadruplet comprises thestates

𝜓1 = N1(s + 𝜆1px), 𝜓2 = N2(s + 𝜆2px), 𝜓3 = py, 𝜓4 = pz.

If we now request that the first two states be equivalent—i.e., they have the sameshape but different directionality—we are led to the constraint

|𝜆1| = |𝜆2| ⇒ 𝜆1 = 𝜆, 𝜆2 = −𝜆 (𝜆 > 0),

which implies that

𝜓1 = 𝜓+ = N(s + 𝜆px), 𝜓2 = 𝜓− = N(s − 𝜆px).

The orthogonality condition then gives

(𝜓+, 𝜓−) = N2(1 − 𝜆2) = 0 ⇒ 𝜆 = 1,

and the hybridized states become

𝜓+ = 1√

2(s + px), 𝜓− = 1

√2(s − px), (14.8)

where the+ and− indices denote the (positive or negative) direction of the corre-sponding hybrid along the x axis. The shapes of the sp2 and sp1 hybridized orbitalsare depicted in Figure 14.8.

Before considering further applications of hybridization theory, we willwork through two simple examples to familiarize the readers with the relevantmathematics.


(a): sp2 hybridization (b): sp1 hybridization

Figure 14.8 The sp2 and sp1 hybridized orbitals.

Example 14.1 Show that the two quadruplets of sp3 hybridized orbitals givenby (14.5) and (14.6) are orthonormal, and likewise for the triplet (14.7) of the sp2

hybridization.

Solution: Let us first recall that a set of vectors forms an orthonormal basis ifthe vectors are normalized and orthogonal to each other. The states of the (14.5)quadruplet are clearly normalized, since the sum of the squares of their coeffi-cients is equal to unity. The same is true for the (14.6) quadruplet. For the 𝜓bstate, for example, we have

14

⎛⎜⎜⎝

12 +

(

−√

23

)2

+ (−√

2)2 +

(

− 1√

3

)2⎞⎟⎟⎠

= 14

(

1 + 23+ 2 + 1

3

)

= 1

and similarly for the other states. With regard to orthogonality, let us notefirst—see also Section 2.6.3—that two superposition states

𝜓 = c1𝜓1 + · · · + cn𝜓n, 𝜙 = d1𝜓1 + · · · + dn𝜓n

are orthogonal to each other if the vectors (c1,… , cn) and (d1,… , dn) of theircoefficients are orthogonal. That is, if

c∗1d1 + c∗2d2 + · · · + c∗ndn = 0.

The above expression differs from the familiar inner product in that the vec-tors can now have complex components, which makes the conjugation symbolnecessary, as we discussed elsewhere.

Therefore, in order to show the orthogonality of, say, the (14.6) states, it sufficesto demonstrate the pairwise orthogonality of the four (four-dimensional) vectors

(

1, 0, 0,√

3)

,

(

1,−√

23,−

√2,− 1

√3

)

,

(

1, 2√

23, 0,− 1

√3

)

,

(

1,−√

23,√

2,− 1√

3

)

,


where we dropped the common factor 1∕2, since it obviously plays no role inorthogonality.

It is now easy to show—by multiplying the corresponding components andsumming up—that the above vectors are indeed orthogonal, as they should be.The same is true for the four states of (14.5), and the three states of (14.7).

Example 14.2 A valence electron of the carbon atom “resides” in an sp3

hybridized orbital

𝜓 = 12

(

s + 2√

23

px −1√

3pz

)

. (1)

(a) Calculate the probabilities of finding the electron in each of the s, px, py, andpz states. (b) What is the mean energy of the electron in the above state, if theenergies of the s and p states of C are 𝜖s and 𝜖p, respectively? (c) Which directionin space does the hybrid (1) point to?

Solution: It follows from (1) that the probabilities are as follows.(1

2

)2= 1

4∶ (s state),

(

12⋅ 2

√23

)2

= 23∶ (px state),

(

12⋅

(

− 1√

3

))2

= 112

∶ (pz state),

and zero for the py state, since it is absent from the superposition (1). The totalprobability for the electron to occupy any one of the p type states is equal to(2∕3) + (1∕12) = 3∕4, as we would have expected from an earlier discussion. (Ansp3 hybrid is one part s state, three parts p state. Can you recall why?) As a result,the mean energy of the electron is

⟨E⟩ = 14𝜖s +

34𝜖p.

Finally, the direction of the hybrid (1) is given by the three-dimensional vectorformed by the coefficients of px, py, pz in the superposition (1). This vector is

(√23, 0,− 1

2√

3

)

∼(√

2, 0,−12

)

, (2)

where, in the latter expression, we dropped the common factor 1∕√

3, as it doesnot affect the direction of the vector. It follows from (2) that the hybrid (1) lies inthe quadrant of positive x and negative z values on the x–z plane, and forms anangle of about 19.5∘ with the x axis.

14.2.4 The Need for Partial Hybridization: The Molecules C2H4, C2H2,and C2H6

For reasons that will become apparent shortly, partial hybridization is fullyutilized in compounds with double or triple bonds, for example, unsaturated


H

H

H(a) (b) H

H

H

CC

H

H

Figure 14.9 The need for partial hybridization in multiple bonds. The sp3 hybridization doesnot result in “good quality” bonds for a molecule such as ethene (C2H4) owing to the doublebond between the carbon atoms. Partial hybridization is necessary in this case.

hydrocarbons. The simplest molecule of this category is ethene (C2H4), whosestructural formula is shown in Figure 14.9a. It is not hard to see that the tetrahe-dral configuration for the valence states of carbon is completely unsuitable forthe formation of double bonds. One such possibility is shown in Figure 14.9b,where the carbon valence orbitals below the level of the page are shown indashed lines, while those above the page are shown in solid lines. The obviousdrawback of this arrangement is that it leads to “bent” molecular orbitals withvery low bonding energy,7 for reasons we discussed elsewhere (Section 13.3.4).At the same time, the possibility of combining head-on and sideways bonding(𝜎 and 𝜋 bonds, respectively) is of little use here, because the large angle (≈ 109∘)between the orbitals of a tetrahedral structure renders the sideways overlap oforbitals essentially impossible.

The way out of this impasse is to look for other possible valence states of car-bon (besides the tetrahedral state), corresponding to alternative hybridizationschemes. In the present case, the most suitable hybridization state for the forma-tion of a double bond is the sp2 state. Here, carbon atoms have three hybridizedvalence orbitals at an angle of 120∘ with each other, while a fourth valence stateis perpendicular to the plane of the other three and is associated with the unhy-bridized pz orbital. The formation of bonds in the ethene molecule is now clear.Two out of the three hybridized orbitals of each carbon atom bind with hydrogenatoms, while the third orbital binds head on (𝜎 bond) with its counterpart fromthe other carbon atom. The remaining two unhybridized pz orbitals bind sidewaysto produce a weak 𝜋 bond between the carbon atoms. The resulting molecularorbital complex is shown in Figure 14.10, where light shading denotes the weakoverlap of pz orbitals in a 𝜋 bond. Actually, as we saw in a related discussion forthe double bond of the O2 molecule in the previous chapter, the requirements forsymmetry and the right number of nodal surfaces lead us readily to the correctarrangement of orbitals in double or multiple bonds. With the pictorial repre-sentation of Figure 14.10 in mind, it is not hard to realize that to maximize thesideways overlap between unhybridized pz orbitals of C atoms the molecule oughtto be planar. And the fact that it is so is an excellent confirmation of our analysis.

7 Even though these “bent” bonds have low energy, they are inevitable in some organic molecules,due to their geometry. A typical example is cyclopropane, C3H6, whereby three C atoms occupy thevertices of an equilateral triangle.


HH

H H

x

y

zz

Figure 14.10 Hybridization in the C2H4 molecule. The two C atoms hybridize in an sp2 state,which results in a planar molecule.

Let us now examine what kind of orbitals are utilized toward bond formationin the acetylene molecule (C2H2), where carbon atoms bind together via a triplebond (Figure 14.11).

A basic conclusion can be drawn from the preceding discussion: A multiplebond can contain only one 𝜎 bond, with the other bonds being necessarily 𝜋bonds that form by sideways overlap of p orbitals. Indeed, there is no reasonfor hybridized orbitals to participate in 𝜋 bonds, since the sideways overlapleaves the key advantage of these orbitals (i.e., enhanced overlap) unutilized.It is only in head-on 𝜎 bonds that hybridized orbitals are fully exploited. So,the valence state of carbon in the acetylene molecule will look as in Figure 14.11.The hybridized orbitals 𝜓± = (s ± px)∕

√2 line up along the axis of the

molecule—the x axis here—to form 𝜎 bonds (C–C and C–H), while the unhy-bridized orbitals py and pz along the y and z axes will participate in two 𝜋 bondsbetween the carbon atoms. A direct consequence of this arrangement is thatthe acetylene molecule ought to be linear, as confirmed by experiments. It isalso evident that acetylene, together with other similar compounds, ought tobe a gas at ambient temperatures, since all dipole moments of the polar C–Hbonds cancel each other out owing to the symmetrical shape of the molecule.Indeed, the boiling points of C2H4 and C2H2 are well below zero, at −104 ∘C and

H H

z z

y y

xC C

Figure 14.11 Hybridization in the acetylene molecule. The hybridization state of C atoms issp1, since a triple bond can contain only one 𝜎 bond, with the other two bonds beingnecessarily 𝜋-type. As a result of this hybridization, the C2H2 molecule is linear.


H

H

H H(a) (b)

H

H

H

H

H

H

H

H

Figure 14.12 In the C2H6 molecule, the two hydrogen “tripods” are in a crosswise (staggered)arrangement in order to minimize their electrostatic repulsions. Note, by the way, that thesmall lobes of the hybrid orbital are not shown anymore, for simplicity.

−84 ∘C, respectively. Note also that, because of the way 𝜋 bonds are formed, theparts of the molecule on either side of the double or triple bond cannot rotaterelative to each other and around the molecular axis. Such rotational freedomexists only for a single 𝜎 bond, which does not prevent one part of the moleculefrom rotating around the bond axis—and even then, rotation is not entirelyfree. This partial hindrance of molecular rotation has interesting consequencesthat warrant a brief discussion. Let us consider the ethane molecule (C2H6),where—as in all saturated hydrocarbons—the valence state of carbon is sp3. As aresult, the molecule’s shape will be that of Figure 14.12.

Now, if rotation around the single carbon–carbon bond were uninhibited, thethree-dimensional shape of the molecule would not be uniquely determined,because the relative position of the two “tripods”—that are formed by thebonds of each carbon with hydrogen—would be arbitrary. Actually, owing tothe polarity of the C–H bond, the two hydrogen triplets must be in a “cross-wise” arrangement—not facing each other—in order to minimize their mutualrepulsions. In chemistry, this arrangement is called staggered conformation—asopposed to eclipsed conformation, when the two tripods face each other—andis customarily depicted as in Figure 14.12b, where the molecule is projectedonto a plane perpendicular to its axis, and the two triplets of C–H bonds areshown in gray and white tone, respectively. Therefore, the equilibrium positionof the molecule is not arbitrary, but is uniquely determined by the condition thatelectrostatic interactions between polar bonds be minimized. Actually, this is notquite correct. There is not one but three equilibrium positions, since there arethree distinct ways of arranging the two C–H bond triplets so that they dissecteach other’s angles. Which of these three equivalent equilibrium positions willthe molecule take? The readers may have guessed the answer, following ourearlier related discussion on ammonia. Since these three equilibrium positionsare separated by a finite potential barrier, the molecule does not remain indefi-nitely in any one of them, but spreads itself in all three positions simultaneously,to further lower its energy. Specifically, the potential sensed by one hydrogen“tripod” as it attempts to rotate with respect to the other tripod would look asin Figure 14.13, where the maxima correspond to the tripods facing each other


Staggered conformation

Eclipsed conformation

2π

V(ϕ)

ϕ

Figure 14.13 The triple potential wellexperienced by each hydrogen “tripod” of theethane molecule, as it rotates with respect tothe other tripod. Because of this potential, themolecular spectrum for torsional oscillationsaround the molecule’s axis consists of tripletsof adjacent levels.

and the minima to a “crosswise” configuration. Later in this chapter, when westudy in detail the motion in a multiple potential well, we will be able to betterunderstand the implications of the above analysis for the vibrational spectrum ofethane. As will become clear then, the triple potential well causes the vibrationalspectrum of the molecule to consist of triplets of adjacent levels. These levelscoalesce around energy values corresponding to the molecule remaining inany one of the three equilibrium configurations, and performing only small,localized, torsional oscillations around it. The experimental observation of thisfine structure in the vibrational spectrum of the ethane molecule is a directconfirmation of these subtle implications of quantum theory.

14.2.5 Application of Hybridization Theory to Conjugated Hydrocarbons

We now take a first look at a special class of carbon compounds, the so-calledconjugated hydrocarbons, whose distinct feature is the alternation of single anddouble bonds along the carbon chain, which can be open or closed. If the carbonchain is open, the conjugated hydrocarbon is called linear; if the chain is closed(i.e., it is a ring) it is called cyclic.

Two typical compounds of this class are the molecules of hexatriene and ben-zene, both with six carbon atoms. We examine first the benzene (C6H6) molecule,which has a closed chain and a simpler arrangement of valence orbitals than thehexatriene (C6H8) molecule, which is linear.

The structural formula of benzene, first proposed by August Kekulé in 1865, hasthe characteristic shape of a hexagon with alternating single and double bonds,as depicted in Figure 14.14.

Because of the double bonds, it is clear that carbon hybridizes into an sp2

state in the benzene molecule. Indeed, each carbon atom has three coplanarhybridized orbitals, forming an angle of 120∘ with each other, whereas thefourth, unhybridized orbital is perpendicular to the plane of the other three.The three hybridized orbitals are utilized in strong 𝜎 bonds C–C and C–H,while the unhybridized pz orbitals couple sideways and in pairs to form weak 𝜋bonds between C atoms.

The planar character of the benzene molecule is a direct consequence of thisanalysis. In order for the unhybridized pz orbitals to couple sideways, they musthave the same orientation, so the corresponding “shamrock-shaped” triplets ofhybridized orbitals should lie on the same x–y plane.

As for the hexatriene molecule, its structural formula and the configuration ofits valence orbitals are depicted in Figure 14.15.

There is one point in the discussion above that has a broader significance: Thedetermination of the hybridization state of carbon atoms. Given that only one of


C

CC

C C

C

H(a) (b)H

H

HH

H

σ σ

σ

σ

σ

σ

σ

σσ

σ

σ

σ ππ

π

Figure 14.14 Structural formula (a) and configuration of valence orbitals (b) in the benzenemolecule. Owing to the double bonds, carbon hybridizes into an sp2 state, while the need forsideways overlap of the unhybridized pz orbitals makes the molecule planar.

C C C C CC

H

H HH

H

(a)

(b)

H H

H σ

σσ

σσ

σ

σσ

σ

σσ

σσπ π π

Figure 14.15 The structural formula (a) and the configuration of valence orbitals (b) in thehexatriene molecule. The hybridization state of carbon is sp2, for the same reason as in thebenzene molecule. While this does not guarantee the planarity of the molecule—since it is stillpossible for parts of the molecule to rotate around the single 𝜎 bonds—the molecule isactually planar, for reasons we will explain later. Moreover, owing to the 120∘ angle betweenthe strong 𝜎 bonds, the actual spatial arrangement of the carbon atoms—the “spine” of themolecule, so to speak—does not follow a straight line (as implied by the simplified sketch ofthe structural formula) but “zigzags” at 120∘ angles.

the multiple bonds can be a strong 𝜎 bond while the others are weak 𝜋 bonds,we mark on the structural formula the type of each bond, and then simply addup the 𝜎 bonds of each C to deduce its hybridization state. So, for four 𝜎 bonds,the hybridization state is sp3; for three 𝜎 bonds, sp2; and for two 𝜎 bonds, sp1. Itis really that simple!

14.2.6 Energy Balance of Hybridization, and Application to InorganicMolecules

How does hybridization lead to energy gain? Let us take a look at the CH4molecule to find out. We have already presented the basic idea. In order forthe 2s orbital of carbon to join the game of chemical bonding, so to speak, oneof the two 2s electrons is pushed up to the empty 2pz level (Figure 14.5). If we


ignore, for a moment, the energy difference between these two levels, we realizethat carbon can now form linear combinations (i.e., hybrids) of the “degenerate”s and p orbitals, which it can then utilize as valence orbitals in bonding withother atoms. But since the 2s and 2p levels are not quite degenerate, there is aninitial cost of hybridization, equal to the energy required to raise the electronfrom the 2s to the 2p level. We thus have

Energy cost of hybridization∶ 𝜖p − 𝜖s,

where 𝜖s and 𝜖p are the energy eigenvalues of the 2s and 2p states. As for the energygain of hybridization, we expect it to result from the improved quality of carbon’schemical bonds, which is due to increased overlap between hybridized orbitalsand valence orbitals of other atoms participating in the bond (see Figure 14.4).As we already know, the energy gain of a chemical bond is equal to 2A, where Ais the absolute value of the matrix element of the Hamiltonian between the twoorbitals of the bond. So, the energy advantage of hybridization is determined byhow much the matrix element A increases owing to the utilization of hybridizedorbitals in a chemical bond, compared to the initial “down payment” 𝜖p − 𝜖s forhybridization.

The following example elucidates the energy logistics of hybridization.

Example 14.3 Calculate the mean energy of an electron in any hybridizationstate spn (n = 1, 2, 3). Use this result to confirm that the energy cost for carbon toutilize these hybridization states is equal to 𝜖p − 𝜖s.

Solution: For a hybridization state spn (n = 1, 2, 3) whose direction (i.e., orienta-tion in space) is a, the wavefunction is

𝜓a =s +

√n pa

√n + 1

(pa = p ⋅ a),

which means that the electron has a probability 1∕(n + 1) to be in the s state (withenergy 𝜖s) and a probability n∕(n + 1) to be in the p state (with energy 𝜖p). So, itsmean energy ⟨E⟩ = 𝜖spn is equal to

𝜖spn = 1n + 1

𝜖s +n

n + 1𝜖p =

𝜖s + n𝜖p

n + 1.

We thus have

𝜖sp3 =𝜖s + 3𝜖p

4, 𝜖sp2 =

𝜖s + 2𝜖p

3, 𝜖sp1 =

𝜖s + 𝜖p

2.

The total energy of the carbon atom in each of these hybridization states is

Etot(sp3) = 4𝜖sp3 + 0 ⋅ 𝜖p = 𝜖s + 3𝜖p

Etot(sp2) = 3𝜖sp2 + 𝜖p = 𝜖s + 3𝜖p

Etot(sp1) = 2𝜖sp1 + 2𝜖p = 𝜖s + 3𝜖p,

where we accounted for the fact that in the sp3 hybridization state, all four elec-trons of carbon occupy hybridized levels—and no electron lies in an unhybridized


p orbital—whereas in the other two hybridization states (sp2 and sp1), one and twoelectrons occupy unhybridized p orbitals, respectively.

As you see, the total energy of carbon is the same for all three hybridizationstates, and is equal to 𝜖s + 3𝜖p. So, the energy cost for their formation is also thesame, and is equal to

ΔE = Etot (after hybridization) − Etot (before hybridization)= (𝜖s + 3𝜖p) − (2𝜖s + 2𝜖p) = 𝜖p − 𝜖s,

which is precisely what we expected.

At this point, we need to address an important question: Is hybridization a priv-ilege exclusive to the carbon atom, and if so, why? If not, then how can we applyhybridization theory to inorganic molecules, such as water or ammonia?

To answer this question, we make use of the sp3 hybridization state to describethese molecules from a quantum mechanical perspective. The valence orbitalsare shown in Figure 14.16.

With regard to utilizing sp3 hybridization, at least, carbon has a clear advantageover oxygen or nitrogen, since it has as many electrons in the n = 2 shell as thereare sp3 orbitals, namely, four. So by having one electron populate each suchorbital, carbon can utilize all four of them as valence orbitals in bonding withmonovalent elements, such as H. In contrast, sp3 hybridization is only partiallyexploited in the N and O atoms, since some of their hybridized orbitals are filledand are thus unable to participate in energetically favorable chemical bonding.It is as if we paid to produce these hybridized orbitals and are now unable touse them! We conclude that, even though it is not the only atom whose orbitalshybridize, carbon capitalizes on hybridization by having four electrons in its outer

H

N

H

HH (b)(a)

O

H

Figure 14.16 Application of hybridization theory to the molecules H2O and NH3. (a) Using sp3

hybridization for the H2O molecule. Four of the six electrons in the n = 2 shell of O ([O] =[He]2s22p4) fill two of the four sp3 orbitals. The other two electrons half-fill the remaining twoorbitals, which therefore become the valence orbitals of O and bond with the 1s orbitals of H toform the H2O molecule. (b) An analogous picture for the NH3 molecule. Since there are fiveavailable electrons here—remember that [N]=[He] 2s2 2p3—only one sp3 orbital is filled; theother three hybrid orbitals remain half-filled, and bind with the 1s orbitals of H to form the NH3molecule.


shell. Other atoms, such as N and O, may still profit energetically from hybridiza-tion, albeit to a lesser degree than carbon. A direct confirmation of this effect isprovided by the measured angles between bonds in the H2O and NH3 molecules(Figure 14.16), which are very close to the values predicted by sp3 hybridization.In contrast, the elementary theory of unhybridized p orbitals predicts theseangles always at 90∘.8 For the H2O molecule, in particular, sp3 hybridizationtheory predicts an angle of 109∘, while the experimental value is 104.5∘.

Similarly, the theoretical prediction for the angle between two N–H bonds inNH3 is again 109∘ (this is the viewing angle from the center of a tetrahedrontoward any one of its edges), while the experimental value is 107∘. The closeagreement between the theoretical and experimental values for both moleculesprovides strong empirical evidence that the sp3 hybridization scheme is, indeed,the appropriate quantum mechanical description of these molecules.

There is one point in the preceding analysis that warrants clarification. Howdo we justify theoretically the use of sp3 hybridization—instead of sp2 or sp1—inthe NH3 and H2O molecules? Take, for example, the molecule of ammonia. Whyshould we employ sp3 hybridization, when only three of its four orbitals are tobe used in chemical bonding? Why not try sp2 hybridization instead, which, afterall, has as many hybridized orbitals as the ones to be used? The answer is unex-pectedly simple. In the sp2 hybridization state of N, the total energy of the atom’svalence electrons is greater than in the sp3 hybridization state. In other words,the sp2 hybridization state of N costs more than the sp3 state. We can easily showthis. In the sp3 state, where the mean energy per electron is 𝜖sp3 = (𝜖s + 3𝜖p)∕4,the total energy is

Etot(sp3) = 5𝜖sp3 = 54(𝜖s + 3𝜖p) =

54𝜖s +

154𝜖p. (14.9)

On the other hand, in the sp2 state, where 𝜖sp2 = (𝜖s + 2𝜖p)∕3, we get

Etot(sp2) = 3𝜖sp2 + 2𝜖p = (𝜖s + 2𝜖p) + 2𝜖p = 𝜖s + 4𝜖p, (14.10)

since two of the five valence electrons of N must occupy the unhybridized pzorbital, so that the other three electrons reside unpaired in the three sp2 orbitals,which are thus the valence orbitals of the atom. Note that if sp2 were the actualvalence state of N, the NH3 molecule would be planar, with N located at the centerof an equilateral triangle, and the three H atoms at the triangle’s vertices.

The results (14.9) and (14.10), and, especially, the fact that

Etot(sp2) − Etot(sp3) = 14(𝜖p − 𝜖s) ⇒ Etot(sp2) > Etot(sp3)

tell us that the sp2 valence state of N costs more than the sp3 state, while theenergy gain from chemical bonding is the same in both cases. We have thus shownthat the energetically favorable hybridization state of N in the NH3 moleculeis sp3, and in doing so, we obtained a complete theoretical justification for thegeometrical shape of the molecule, as depicted in Figure 14.16b. Actually, theinequality Etot(sp3) < Etot(sp2), which establishes the energetic advantage of sp3

8 Which forces us, after the fact, to invoke the repulsion between polar bonds in the molecule toaccount for angles greater than 90∘!

Problems 413

over sp2, is qualitatively evident. Given that the mean energy of an electron in ahybridized state is lower than in a pure p state, the sp2 hybridization of N is ener-getically costlier, since it requires that two electrons occupy the highest possiblestate, the p level.

It follows from the above discussion that in the H2O molecule also, sp3 is theenergetically favored hybridization state of O. As a result, the molecule’s shapemust surely be as described earlier (Figure 14.16a).

There is thus no doubt that the capability of C (Z = 6) to fully exploithybridization—a property that clearly distinguishes it from all its neighboringelements—renders C a clear winner in the “contest” for the most suitable elementfor the chemistry of life. The only other possible contender would be silicon,which lies exactly below carbon in the periodic table, has the same valence shell(albeit with n = 3), and is thus expected to have very similar chemical behaviorto carbon.

But silicon has a substantially greater energy difference between the s and pshells, so the cost of hybridization is much greater than for carbon, while theparameter A that measures the energy gain from chemical bonding, is not propor-tionally greater. The net gain from the use of hybridized orbitals in the chemicalbonds of silicon is thus much smaller.

We have arrived at an amazing, yet indisputable, conclusion: The fact that weare “carbonaceous” beings—that is, living beings based on the chemistry of C—isno accident. It is actually inevitable. For an element to fully exploit hybridization,it must have exactly four electrons in the n = 2 shell, plus two more in n = 1, thatis, a total of six. Therefore, the element in question has Z = 6!

Problems

14.1 Somebody claims that the hybridized orbital𝜓 = N(5s + 3px + 4py)

is an sp2 orbital, since it consists only of px and py orbitals in the x–y plane.Do you agree? If not, then how would you characterize this orbital: sp2, sp3,or sp1? Apply the same reasoning to characterize the following hybridizedorbitalsA∶ 𝜓 = N

(

s −√

2 px

)

,

B∶ 𝜓 = N(

s +√

2 px + py

)

,

C∶ 𝜓 = N(

2s + px −√

2 py + pz

)

.Give the direction of each hybrid and find the mean energy of an electrondescribed by the hybrid.

14.2 Construct the following: (a) An sp3 hybrid that points along the x axis. (b)An sp2 hybrid that points in the (1, 2,−2) direction. (c) An sp1 hybrid thatpoints in the (1,−1, 1) direction.

14.3 Use hybridization theory to predict the geometric shape of the molecules

CO2, CH2O, C2O2, C2H2O.


In particular, state which molecules are linear (i.e., straight), planar, orthree dimensional. Examine the possibility for some molecules to havemore than one isomer; in other words, look for molecules with the samechemical, but different structural formulas.

14.4 You are given the following empirical data for the strength and length ofcarbon-carbon bonds in three different organic molecules:

Molecule Strength (kcal/mol) Length (Å)Ethane (C2H6) 90 1.54Ethylene (C2H4) 146 1.33Acetylene (C2H2) 200 1.20

What can you comment about these data? Convert kcal/mol toeV/molecule to confirm that the above values are reasonable.

14.3 Delocalization: The Second Basic Deviation fromthe Elementary Theory of the Chemical Bond

14.3.1 A Closer Look at the Benzene Molecule

We saw earlier—in Section 14.2.5 and Figure 14.14—that hybridization canexplain, in principle, the main structural features of the benzene molecule,as these appear in Kekulé’s formula. But in spite of explaining these featurescorrectly—including the planar character of the molecule—hybridization theoryalso leads to a conclusion that is in sharp contrast with experimental data:Given that double bonds are stronger than single bonds, the correspondingsides of the hexagonal molecule ought to be shorter than the others, so themolecule surely cannot be a regular hexagon. And yet, it is! We actually knowthis both from modern experimental techniques—such as the NMR spectrumof the benzene molecule—and from chemical data, namely, the number ofdistinct isomers of dibromobenzene C6H4Br2, which is the compound formedwhen two hydrogen atoms of benzene are replaced by two bromine atoms.As can be clearly seen in Figure 14.17, there ought to be four different iso-mers with the above formula, since the two bromine atoms can be placed inneighboring hexagonal positions (1,2-dibromobenzene), next-nearest neigh-boring positions (1,3-dibromobenzene), or diametrically opposite positions(1,4-dibromobenzene). But for 1,2-dibromobenzene, there would be two dif-ferent isomers, since the two bromine atoms can be placed across either asingle or a double bond (Figure 14.17). And yet there is only one such isomer. Weare forced to conclude that, contrary to our theoretical prediction, the benzenemolecule has the shape of a regular hexagon.

So where did we go wrong in our reasoning? One clue is provided by empiricalfacts, namely, that the single and double bonds on the hexagonal chain are indis-tinguishable, which means the double bonds have sort of “spread out” over theentire molecule. Here lies the failure of another basic assumption in the elemen-tary theory of the chemical bond, namely, that the motion of valence electrons

14.3 Delocalization: The Second Basic Deviation 415

Br Br Br Br

Br

Br

Br

BrHH

H H H H

H H

H H

HH

HHH

H

(a)

1,2-Dibromobenzene 1,3-Dibromobenzene 1,4-Dibromobenzene

(b)

Figure 14.17 The four possible isomers of dibromobenzene. Since we can detect only onedibromobenzene of the 1,2- type in the laboratory, we are led to conclude that edges withsingle and double bonds are equivalent, as are therefore all the hexagon edges. The benzenemolecule is thus a regular hexagon.

is confined between neighboring atoms, or, equivalently, that chemical bondsare always localized. (This erroneous assumption is reflected in how we have sofar depicted a bond, namely, a small line segment linking the two “partnering”atoms.) But looking back at the basic quantum mechanical model of the chem-ical bond—a particle moving in a double potential well—we realize at once thatthere is nothing “sacred” about the idea of chemical bonds being localized. If cer-tain molecules can allow some valence electrons to move across more than twoatoms—by “hopping” from one atom to another—then this is exactly what theywill do to further lower their energy. After all, is this not what electrons do ina typical chemical bond? Do they not move from one well to another in tryingto lower their energy compared to localized atomic states? Evidently, a particlemoving in a multiple potential well will not restrict itself within just two of thewells, but will spread its wavefunction across all wells, to exploit their simulta-neous attraction and attain the lowest possible energy. The equivalence betweensingle and double bonds in benzene implies, therefore, that double bonds havesomehow spread throughout the hexagonal chain, and the same spreading effectapplies to the corresponding electrons whose wavefunctions must also extendthroughout the entire chain. In hindsight, this arrangement can be understoodeasily if we envisage removing the six electrons from the weak 𝜋 bonds9 and let-ting them “resettle” in the molecule in a way that minimizes their energy. Clearly,the electrons will “see” a cyclic system of six attractive centers—located at the car-bon atoms—and will spread their wavefunctions along the entire sextet of wells,to profit from their attraction, rather than confine themselves pairwise betweenneighboring wells. We can arrive at the same conclusion by noticing that the elec-trons of unhybridized pz orbitals (Figure 14.14b) can just as well hop onto eitherof their neighboring orbitals, and from then on to their next neighbors and so on,until they traverse the entire hexagonal chain. Clearly, electrons cannot affordto “hang out” in pairs between adjacent wells, nor can we force them to do thatanyway. For even if we tried, sooner or later they would hop onto the “wrong”orbital—away from the orbital pair of their “own” bond—and continue hopping

9 Chemists call these 𝜋 electrons, and we will use the same term hereafter.


H(a) (b)H

H

H H

C+

+

+

+

+ +

C

C C

CC

H

Figure 14.18 The delocalization mechanism in the benzene molecule. (a) Imagine strippingthe molecule momentarily from its six 𝜋 electrons, leaving behind an equal number of carbonions; this is how the molecule would look then. Subsequently, the six ions act as attractivecenters—i.e., potential wells—that draw the π electrons to perform delocalized motion alongthe entire hexagonal chain, to achieve the lowest possible energy. (b) The six pz orbitals of themolecule form a hexagonal array, along which delocalized motion is bound to occur, as the πelectrons can “hop” from one orbital to another with the same probability.

along, until they perform a completely delocalized motion throughout the entirehexagonal chain. Figure 14.18 summarizes the above discussion.

At this point, readers may wonder why delocalization occurs only for electronsof weak 𝜋 bonds and not for electrons of strong 𝜎 bonds between C atoms, orbetween C and H. The answer is simple and is given in Figure 14.19.

The take-home message from the above discussion is that the second basicdeviation from the elementary theory of the chemical bond—the delocalizationeffect—is most vividly manifested in the array of 𝜋 orbitals and the corresponding

H

C C

C

C C

C

: Transitions of high probability

: Transitions of low probability

Figure 14.19 Why delocalization is impossible for 𝜎 electrons in the benzene molecule. Thereis considerable overlap between atomic orbitals forming a 𝜎 bond, but negligible overlapbetween atomic orbitals of different bonds. Consequently, the probability for 𝜎 electrons tohop onto a neighboring bond (dotted arrows) is far smaller than the probability of themstaying on the same bond and hopping from one of its orbitals onto the other (solid arrows).So the 𝜎 bonds are localized and thus independent from one another, precisely as predictedby traditional chemistry.


𝜋 electrons that occupy them. And the ideal physical system to study delocaliza-tion is the benzene molecule.

14.3.2 An Elementary Theory of Delocalization: The Free-Electron Model

Based on our discussion so far, the quantum mechanical study of delocaliza-tion in benzene—or in similar molecules—would require that we solve (usingthe LCAO method) the problem of a particle moving in multiple wells, arrangedcyclically or linearly, to cover all possible cases that arise in practice. But it isuseful and instructive to employ first a much simpler model that focuses on thedominant feature of the problem: the motion of 𝜋 electrons in a cyclic (i.e., closed)chain, which can be roughly approximated by a circle. The idea is to ignore theindividual centers of attraction (i.e., the individual wells on the perimeter of themolecule) and assume that the 𝜋 electrons move freely on a circular tubule whoseradius is roughly equal to the circle circumscribed onto the regular hexagon ofthe molecule.

The quantum mechanical treatment of such a problem is rather straightfor-ward. First, let us consider a classical particle of mass 𝜇 that moves on a circle ofradius a, and has thus a rotational energy

E =𝓁2

z

2I=

𝓁2z

2𝜇a2 , (14.11)

where I = 𝜇a2 is the particle’s moment of inertia with respect to the center of thecircle. Note that formula (14.11) contains only the 𝓁z component of the particle’sangular momentum, since the circular motion takes place on the x–y plane, andthus the angular momentum points along the z axis.

Moving on to quantum mechanics, the angular momentum 𝓁z is now a quan-tized quantity with eigenvalues ℏm, and the particle’s allowed rotational energiesare given by the formula

E = Em = ℏ2

2𝜇a2 m2 = 𝜖m2, m = 0,±1,±2,… , (14.12)

where 𝜖 = ℏ2∕2𝜇a2. Let us now apply all this to the benzene molecule. All weneed to do is populate the (14.12) levels with the six available electrons, and seeif we can correctly predict an essential property of the molecule, such as the firstexcitation energy, whose experimental value is 4.8 eV. Using (14.12), we constructthe energy-level diagram of the molecule in Figure 14.20.

Given now the experimental value for the radius of the circumscribed circle onthe hexagon—which is equal to the hexagon’s side, 1.39Å—our numerical predic-tion for the energy of the first molecular excitation becomes

ΔE(a.u.) = 3ℏ2

2mea2

|||| ℏ=1,me=1

a=1.39∕0.52

= 0.218 a.u. = 0.218 ⋅ 27.2 eV = 5.93 eV,

which is in satisfactory agreement with the experimental value of 4.8 eV.


m = ±3

m = ±2

m = ±1

m = 0

9ϵ

4ϵ

ΔE = 3ϵ

ϵ

0

Figure 14.20 The occupiedenergy-level diagram of 𝜋electrons in the benzenemolecule according to thefree-electron model. The firstexcitation energy is equal toΔE = 3𝜖 = 3ℏ2∕2𝜇a2, where𝜇 = me is the electron mass.

But the model of free electrons moving in a circle can also explain easily anotherspecial feature of the aromatic hydrocarbons, which is known as the aromaticityrule, or Hückel’s rule. The rule states that a closed chain of conjugated hydrocar-bons is especially stable when the number N of its 𝜋 electrons is equal to

N = 4n + 2, n = 1, 2,…In other words, stability is enhanced when the number of sides—that is, the num-ber of carbon atoms—is equal to 6, 10, 14, and so on. For a physical explanationof this rule we turn to the energy-level diagram of Figure 14.20, which helps usrealize that the greatest stability occurs for rings with filled energy shells. Suchrings contain 4n + 2 electrons, since the ground level can hold two electrons, andevery other level can accommodate four electrons, since it is doubly degenerate.So it takes (4n + 2) 𝜋 electrons to completely fill the ground state and n excitedstates. This is equal to the number of sides in the polygonal chain—hence thenumber of carbon atoms—since the weak 𝜋 bonds cover half the polygon edges,and each bond contains two electrons.

14.3.3 LCAO Theory for Conjugated Hydrocarbons. I: Cyclic Chains

We will now see how the LCAO theory, which we developed in the previouschapter, can be extended to treat delocalized motion in more than two attractivecenters that are arranged linearly or circularly. We begin with the latter case,which can be readily applied to the benzene molecule. As we argued earlier,the wavefunction describing such a delocalized motion is given by the linearcombination

𝜓 =N∑

n=1cn𝜓n, (14.13)

where 𝜓n are the eigenfunctions of the individual wells—the atomic orbitals ifyou will—and cn the unknown coefficients of the superposition. As always, thewavefunction (14.13) must satisfy the Schrödinger equation

H𝜓 = E𝜓 ⇒ H

(∑

ncn𝜓n

)

= E

(∑

ncn𝜓n

)

. (14.14)


By taking now the inner product of both sides with an arbitrary local eigenfunc-tion 𝜓m, we arrive10 at the equation

∑

mHnmcm = Ecn. (14.15)

This equation can be written equivalently as

HC = EC, (14.16)

which is an eigenvalue equation for matrices, with

H =

⎛⎜⎜⎜⎜⎝

H11 · · ·H1N

· · · · · · · · ·

HN1 · · ·HNN

⎞⎟⎟⎟⎟⎠

and C =

⎛⎜⎜⎜⎜⎝

c1

⋮

cN

⎞⎟⎟⎟⎟⎠

, (14.17)

where Hnm are our familiar “matrix elements”

Hnm = (𝜓n,H𝜓m) = ∫ 𝜓∗n (H𝜓m) dx,

a term that becomes now fully justified since these are, after all, the elements of amatrix.

We further note that—for the same reasons as in Section 13.2, Eq. (13.5)—wecan set

Hnn ≈ E0, Hn,n±1 = −A, Hnm = 0 (m ≠ n, n ± 1), (14.18)

where we assumed that all attractive centers—that is, wells—are identical, as isthe case in benzene.

For reasons that will soon become apparent, the explicit form (14.15) ispreferred over the matrix form (14.16). So we substitute the (14.18) relations in(14.15) to obtain

E0cn − Acn+1 − Acn−1 = Ecn,

which can also be written as

(E − E0)cn + A(cn+1 + cn−1) = 0. (14.19)

Equation (14.19) is a difference equation, the discrete analog of a differentialequation. The unknown function cn ≡ c(n) depends now on an integer, not acontinuous variable.

10 In taking this step, we assumed again that—even though there can be no chemical bondingwithout it—the overlap between eigenfunctions of neighboring atoms is so small that we are allowedto set

(𝜓n, 𝜓m) = ∫ 𝜓∗n𝜓m dx = 0 (n = n ± m).

We are careful, however, not to set to zero the matrix elements Hnm = ∫ 𝜓∗n (H𝜓m) dx, as we would

then have neither delocalization nor even a chemical bond.


But (14.19) is also a linear difference equation with constant coefficients,11 sowe can look for solutions of exponential form

cn = ein𝜃, (14.20)

which is the discrete analog of the exponential substitution y(x) = e𝜆x (obviously,x ↔ n) we typically use for a linear differential equation with constant coeffi-cients. As for the complex form of (14.20), this reflects the physical equivalence ofall vertices of a polygonal chain, such as benzene, which implies that all probabil-ities Pn = |cn|

2 (n = 1,… ,N) must be equal. In any case, in using the complexform we did not preempt the final result, since 𝜃 might as well be a complexnumber.

If we now insert (14.20) in (14.19), we obtain the equation

E = E0 − 2A cos 𝜃, (14.21)

from which we can calculate the energy eigenvalues if we know the allowed valuesfor the parameter 𝜃. This is easily done once we realize that for a closed periodicchain of N carbon atoms—for benzene, N = 6—the coefficients cn must satisfythe periodicity condition

cn+N = cn, (14.22)

since, if we move by N sites along the chain, we return to where we started. Byapplying now the condition (14.22) on (14.20) we obtain

ei(n+N)𝜃 = ein𝜃 ⇒ eiN𝜃 = 1,

which is satisfied only if

N𝜃 = 2k𝜋 ⇒ 𝜃 = 𝜃k = 2k𝜋N

(k = 0, 1,… , (N − 1)). (14.23)

Here, the restriction to the first N integer values of k follows from the fact that forevery k ≥ N , the parameter 𝜃 differs by 2𝜋 from one of its previous values, so itdoes not contribute a physically distinct solution cn = ein𝜃 . (Besides, for a systemof N wells, we expect to find N independent solutions, which is the number ofallowed values for 𝜃.)

We can now insert (14.23) in (14.21) to obtain the eigenvalues

Ek = 0 − 2A cos 2k𝜋N

(k = 0, 1,… ,N − 1), (14.24)

where we denoted the initial atomic level as 0—instead of E0 as before—to avoidconfusion with the eigenvalue Ek for k = 0. We can provide a simple graphicalconstruction of the N eigenvalues of (14.24) as follows. First, we draw a circleof radius 2A, with its center located at a height 0 (along a vertical energy axis).We then mark on the circle the points that correspond to angles 𝜃k = 2k𝜋∕N(k = 0, 1,… ,N − 1), as measured from the circle’s lowest point. Finally, the pro-jections of these points—which form a regular N-gon, that is, a polygon with Nangles—onto the vertical energy axis give us the allowed energies of the problem.

11 It is a linear equation because it does not contain nonlinear terms such as c2n, cncn−1, c3

n, and soon; and it has constant coefficients because all coefficients of the unknown function cn (or cn±1) areindependent of the variable n.


E

(a) (b)

Ɛ0 + 2A Ɛ0 + 2A

Ɛ0 + A

Ɛ0 – A

Ɛ0 – 2A

2A

θ

Ɛ0 + A

Ɛ0

Ɛ0 – A

Ɛ0 – 2A

Figure 14.21 (a) Graphical construction of the energy eigenvalues for delocalized motion inthe benzene molecule. (b) The occupied energy-level diagram of the molecule.

In the case of benzene (N = 6), the “angular step” Δ𝜃 in the aboveconstruction—that is, the angular distance between two successive markedpoints on the circle—is equal to 2𝜋∕6 = 60∘. We thus obtain the pictorialrepresentation of eigenvalues shown in Figure 14.21, where we also show theenergy-level diagram of the molecule, populated by its six 𝜋 electrons.

The main feature of the energy-level diagram of Figure 14.21b is the doubledegeneracy of all states except those with the minimum and maximum eigenval-ues. As one can easily see, this feature applies for all even values of N , while forodd values, the nondegenerate state at the top of the diagram is missing. (Theenergy-level diagram for the N = 5 case is a good example. Construct it.) Thedouble degeneracy of the excited states (with the exception of the top state foreven N) was also a distinctive feature of the energy-level diagram of Figure 14.20for the free-electron model, which gave us the “4n + 2 rule” for the stability ofcyclic chains. So, we now realize that this rule is predicted also by the LCAOmodel, the application of which to conjugated hydrocarbons is actually known tochemists as “Hückel’s theory.”12

The energy-level diagram of Figure 14.21b can also help us calculate animportant physical quantity known in chemistry as stabilization energy, whichdescribes the additional energy gain for benzene due to the delocalization of 𝜋electrons. To calculate this quantity we need to find the total energy of the six 𝜋electrons before and after delocalization, and take their difference. For the stateafter delocalization, the energy-level diagram 14.21.b tells us that

Eafter = 2(0 − 2A) + 4(0 − A) = 60 − 8A. (14.25)

12 We remind the readers that the defining feature of conjugated hydrocarbons is the alternation ofsingle and double bonds on a carbon chain, which may be open or closed. Conjugated hydrocarbonswith a closed chain that satisfy the 4n + 2 rule are also known as aromatic hydrocarbons. All carbonchains with alternating single and double bonds are called conjugated chains.


For the state before delocalization, we get

Ebefore = 3 ⋅ 2(0 − A) = 60 − 6A (14.26)

since, in this case, every electron pair of a localized 𝜋 bond fills the bottom level0 − A of a double potential well. Therefore each pair has energy 2(0 − A), andthe total energy for all three pairs is 6(0 − A).

We subtract (14.25) from (14.26) to find the stabilization energy of benzene

ΔE = 2A, (14.27)

which is equal to the energy gained by two electrons that pair up in a localized 𝜋bond. We thus arrive at a very interesting conclusion:

Owing to delocalization, the benzene molecule appears to gain one extra 𝜋 bond.This additional stability of benzene—and other similar compounds—had been

known to organic chemists well before the advent of quantum mechanics, andcould not be explained by traditional chemistry. One manifestation of this sta-bility is the unusual reluctance of benzene to participate in reactions that affectthe conjugated structure of its hexagonal ring. In contrast to other unsaturatedhydrocarbons13 —which react readily with H to replace weak 𝜋 bonds betweenC atoms with strong 𝜎 bonds between C and H—benzene resists strongly theseso-called addition reactions, and prefers participating in substitution reactions,where one or more of the peripheral hydrogen atoms is replaced by anothermonovalent element. A typical example of such a reaction is

C6H6 + Br2FeBr3−−−−→C6H5Br + HBr,

which occurs readily in the presence of iron as a catalyst. In contrast, the followingaddition reaction

C6H6 + Br2FeBr3−−−−→C6H6Br2

never happens, because, as we now realize, its final product would be themolecule whereby the hexagonal sequence of pz orbitals is disrupted and theability of electrons to delocalize is thus lost.

HH

H H

H H

C

C C

C

CC

BrBr

13 A hydrocarbon is called unsaturated when it has double or triple bonds, whose carbon atomshave thus not exhausted their potential for bonding with the maximum possible number ofhydrogen atoms.


Using the above results and additional data, chemists have calculated thestabilization energy of benzene—that is, the extra energy gain associated withdelocalization—and found it equal to14

ΔE = 1.6 eV,

so that, from (14.27), we must have

2A = 1.6 eV. (14.28)

But if we go back to the energy-level diagram of Figure 14.21b, we see immedi-ately that the first excitation energy is also given by 2A and is thus equal to 1.6eV,according to (14.28). This number is in complete disagreement with the exper-imental value of 4.8eV we quoted earlier. The origin of this (rather spectacular)failure is a feature we mentioned before. The LCAO model (in its present ele-mentary form) ignores electronic repulsions completely and is thus unsuitablefor more accurate quantitative calculations; it remains useful, instead, for a qual-itative (and, at best, semi-quantitative) understanding of the basic mechanismsof the chemical bond.

We now turn our attention to the eigenfunctions Ψk for the delocalized motionof 𝜋 electrons on the benzene ring. These eigenfunctions are given (for eacheigenvalue Ek , k = 0,… , 5) by the formula

Ψk = 1√

6

6∑

n=1ein k𝜋

3 𝜓n.

In particular, the eigenfunction of the ground state (k = 0) is given by

Ψ0 = 1√

6(𝜓1 + 𝜓2 + · · · + 𝜓6),

where, as expected, all six atomic pz orbitals are weighted equally in the linearcombination. Given the shape of pz orbitals, the wavefunction Ψ0 of the firstmolecular orbital of benzene should look as in Figure 14.22.

We leave it for the readers to consider what the higher molecular orbitals ofbenzene look like—for example, whether there are sign changes in their superpo-sition of pz orbitals; why they have complex coefficients and whether we can treattheir real and imaginary parts separately; what physical advantage is retainedby our using their complex form; and so on. We conclude this section with abrief mention of resonance, a concept used widely in chemistry to qualitativelydescribe delocalization. A rough depiction of the idea for the case of benzeneis given in Figure 14.23, whence we realize that the resonance concept does not

14 Actually, chemists measure energies not in units of eV/molecule but in kcal/mol. The conversionrelation is

kcal∕mol = NA ⋅ eV∕molecule, (1)

where NA(= 6.02 × 1023) is Avogadro’s number. Using the definitions of cal = 4.18 J andeV = 1.6 × 10−19 J we obtain from (1) the equivalence

1 eV∕molecule = 23 kcal∕mol.


H

(a) (b)

HH

H

H

CC

CC

CC

H

Figure 14.22 The molecular orbital for the ground state of the delocalized motion of 𝜋electrons in a benzene molecule. (a) The probability amplitude of the pair of π electronsoccupying this orbital resembles two tori that lie above and below the molecule’s hexagon.The torus above (below) the hexagon has positive (negative) sign. (b) The modern chemicalsymbol for delocalized chemical bonds in the benzene molecule.

ΨI

(ΨI + ΨII)Ψ =

I

1

2

ΨII

II

Figure 14.23 The concept of resonance. When a conjugated molecule has more than onepossible locations for its 𝜋 bonds—that is, more than one possible structures, or forms—thenits actual state is a suitable quantum superposition of these forms. In the case of benzene,where there are two, entirely symmetrical, possible forms—also known as Kekuléstructures—the actual state of the molecule is described by their symmetric superposition.Moreover, because none of these forms is an eigenstate of the molecular Hamiltonian, if themolecule is found momentarily in either of these, it will start to oscillate periodically betweenthe two forms—hence the term resonance to describe this motion. After spending some timein this oscillatory “motion,” the molecule will emit the extra energy of the initial form—via a

photon—and fall to its ground state Ψ = (1∕√

2)(ΨI + ΨII) whereby no periodic oscillationtakes place. While in the ground state, the molecule has an equal probability to be in form I orform II at any given moment.

really contribute new knowledge, and is certainly not a calculational tool such asthe LCAO theory. But it is useful for readers to be familiar with the concept toavoid confusion when they encounter it as a descriptive tool in chemistry books.

14.3.4 LCAO Theory for Conjugated Hydrocarbons. II: Linear Chains

We can easily extend the above results to linear chains.15 First, we recall the form(14.20) of the general solution of the difference equation (14.19), which can also

15 Perhaps the term “open chains” would be more suitable to avoid the impression that we aretalking about straight chains, which is not the case. What we have in mind typically is “zig-zagchains” with equidistant vertices that accommodate the pz orbitals, allowing electrons to hop fromone vertex to another with the same probability throughout the chain.


be written as a linear combination of sine and cosine terms,

cn = 𝛼 sin n𝜃 + 𝛽 cos n𝜃, (14.29)

since both the real and imaginary parts of (14.20) are also solutions of (14.19).Moreover, the form (14.29) is clearly more suitable for a linear chain, in whichcase the function cn must satisfy the boundary conditions

c0 = 0, cN+1 = 0, (14.30)

which imply the termination of the chain on the left of well #1 and on the rightof well #N . (The particle cannot move further to the left of the first well, hencec0 = 0, or further to the right of the last well, hence cN+1 = 0.) If we now apply thefirst of the (14.30) conditions to (14.29) we get 𝛽 = 0, while if we apply the secondcondition, we obtain

cn|||n=N+1

= 𝛼 sin(N + 1)𝜃 = 0 ⇒ (N + 1)𝜃 = k𝜋

⇒ 𝜃 = 𝜃k = k𝜋N + 1

(k = 1,… ,N), (14.31)

where we restricted k to the first N integer values for the same reason as before.Note that k starts from k = 1, because for k = 0 the cn solution is identicallyzero, and must thus be rejected. Based on (14.31), expression (14.21) gives forthe energy eigenvalues of the chain the formula

Ek = E0 − 2A cos k𝜋N + 1

(k = 1,… ,N), (14.32)

where we have retained the symbol E0 for the energy level of the single well, sincethere is no risk now of notational confusion.

The coefficients cn ≡ c(k)n corresponding to the Ek eigenvalue are

c(k)n = 𝛼 sin nk𝜋N + 1

, (14.33)

while the corresponding wavefunctions Ψk16 are given by the formula

Ψk =√

2N + 1

N∑

n=1sin nk𝜋

N + 1𝜓n (k = 1,… ,N), (14.34)

where we have also calculated the normalization coefficient 𝛼 of (14.33) to beequal to

√2∕(N + 1). The readers are encouraged to verify this normalization

and also check the correctness of the above general result, by an indepen-dent calculation—for example, by direct diagonalization of the Hamiltonianmatrix—in the special case of N = 3.

We will now apply the general theory of a linear chain of wells to the simplestconjugated hydrocarbon, the butadiene molecule. We present our analysis in theform of an example to prepare readers for similar problems later on.

16 Note that here, as in the case of benzene earlier, we write Ψk for the eigenfunctions of themultiple well and 𝜓n for the eigenfunctions of the individual wells.


Example 14.4 You are given the structural formulaH

HH

C C C C

H

HH

σ

σσ σ

σ

σσ

σ

σπ π

(1)

of the butadiene molecule. Apply what you have learned thus far from the theoryof the chemical bond to predict the exact shape of the molecule. If you concludethat delocalization of 𝜋 electrons takes place, then how much additional energydoes the molecule gain as a fraction of the energy of a 𝜋 bond?

Solution: The first step for the theoretical study of any organic molecule is tofind the hybridization state of each carbon atom in it. We can do this easily byrecalling that a multiple C bond contains only one strong 𝜎 bond—formed byappropriate hybridized orbitals—while the other bonds are weak 𝜋 bonds thatare formed though the sideways overlap of p orbitals. For the butadiene molecule,the characterization of bonds as 𝜎 and 𝜋 is shown in the structural formulaabove.

Thus, each carbon atom forms three strong 𝜎 bonds and one weak 𝜋 bond,so its hybridization state is sp2 (hybridization on a plane with three hybridizedorbitals). As a result, the shape of the molecule is

H H

HC

C

C

C

H

H H Free rotationaround the bond

(2)

However, representation (2) of the molecule does not determine uniquely themolecular geometry, because the freedom of rotation around the simple bondbetween the second and third carbon atoms implies that the “second half” of themolecule does not have to lie on the same plane with the first half. But if themolecule assumed a planar form, the unhybridized orbitals of all four C atomswould be aligned in parallel, as shown in Figure 14.24. In that case, delocalizedmotion of 𝜋 electrons throughout the chain would be possible, thus reducingfurther the energy of the molecule.

So, without proceeding to any calculation, we can infer that—owing todelocalization—butadiene is a planar molecule. Indeed, the planar shape is inagreement with experimental data.

To determine the additional stabilization energy of the molecule due to delo-calization, we need to calculate the relevant energy levels using formula (14.32)for N = 4. We find

E1 = E0 − 1.62A, E2 = E0 − 0.62A, E3 = E0 + 0.62A, E4 = E0 + 1.62A,

so the occupied energy-level diagram of 𝜋 electrons is that of Figure 14.25.


Figure 14.24 p orbitals of the butadiene molecule in its planar form. In this arrangement, all porbitals are parallel to each other and thus have the same sideways overlap. This means thatdelocalized motion throughout the chain is not only possible but also inevitable. As for thezigzag shape of the chain, this clearly does not affect the probability of an electron hoppingfrom one p orbital to another.

E0

E0 + 1.62A

E0 + 0.62A

E0 – 0.62A

E0 – 1.62A

Figure 14.25 The occupied energy-level diagram for the delocalized motion of 𝜋 electrons inbutadiene.

We encourage the readers to show, by using the diagram, that the additionalenergy gain for the molecule due to delocalization is roughly equal to a quarter ofthe strength of a 𝜋 bond between carbon atoms. The gain is not negligible—whichis why it affects both the shape and the chemical properties of butadiene—but itis markedly smaller than in benzene. Can you tell why this should be expected?In any case, you can find the answer in the following section.

14.3.5 Delocalization on Carbon Chains: General Remarks

We conclude this discussion with some general remarks on the consequences andprerequisites of delocalization along carbon chains and on the role of the corre-sponding molecules in nature. Our main conclusion is that delocalization lowersthe energy of the electrons—and the distance between their energy levels—byallowing them to move across regions far wider than the range of typical localizedbonds, whose length never exceeds 1–2 Å. As a result of this energy downscaling,the electronic absorption spectrum shifts gradually—as the carbon chain growslarger—from the ultraviolet toward the visible range, reaching eventually the vis-ible limit for molecules with 10 to 20 C atoms in the chain. So, we should not besurprised that all biological molecules that function either as natural photocol-lectors (e.g., chlorophyll) or as pigments in various fruits, vegetables, or flowers(e.g., 𝛽-carotene for carrots) have a common morphological feature, namely, aconjugated chain, open or closed. The same is true for all organic pigments thatare used widely in industry and in everyday life. Actually, besides shifting the


absorption range toward the visible region, delocalization increases significantlythe absorption efficacy of the molecule, since it provides its electrons a muchgreater range to move in, rendering them much more susceptible to the incidentelectromagnetic field. Thus, the conjugated molecules—molecules with alternat-ing simple and double bonds on a carbon chain—are most effective collectors ofvisible light.

It is worth mentioning that—as shown in the last example— the “beneficial”effects of delocalization are much more pronounced in closed chains (i.e.,rings) than in open chains. We can see why this happens if we compare theenergy-level diagram for motion on a circular tubule (Figure 14.20) with thecorresponding diagram for a straight tubule of the same length. For the circulartubule the ground state lies at zero, while for a straight tubule—that is, aone-dimensional potential box—the energy is always positive because of theuncertainty principle.17 In addition, the excited states of the circular tubule aredoubly degenerate and can thus accommodate twice as many electrons asthe straight tubule, for which there is no degeneracy. Because of these crucialdifferences—present also in the LCAO model—𝜋 electrons have lower energylevels in closed systems, which become thus energetically favored over opensystems. Cyclic chains are far more stable than linear chains.

There is a self-evident assumption in all the above discussion: The crucial fea-ture for delocalization is the conjugated structure of the carbon chain, not the side“species” of the chain, which we have assumed here to be hydrogen atoms, but canoften be larger complexes or even organic groups. As an example, Figure 14.26depicts the light-collecting molecule of our eye (it is a variant of vitamin A).

For the sake of completeness, let us now examine whether the alternation ofsingle and double bonds (what we called conjugated structure) is necessary for theexistence of delocalization in carbon chains. The answer is negative. A relevant

H

H H H H

H C

C

C

C

CCCC

H2

CH3 CH3

CH3

CH3 CH3

CH2

CH2H

CCCCO C

Figure 14.26 The light-collecting molecule for vision (structural formula). It is a typicalconjugated system with eleven carbon atoms on the main chain. The molecule is originallyattached on an enzymatic catalyst from which it detaches upon absorption of a photon. Thedetachment activates the enzyme and triggers a series of chemical reactions, whose finalproduct is an electrical signal that is transmitted to the brain via the optical nerve.

17 Since a circular tubule has no boundary, the particle moving inside it “feels” no spatial constraintand is thus not required by the uncertainty principle to have a nonzero minimum kinetic energy.


H

H

C C C C

H

H

(a) (b)

H

C C C C

H

Hy

yy zzzyy zzz z

H

Figure 14.27 The two possible conformations of the C4H4 molecule (structural formula(14.35)). Conformation (a): 4–2 delocalization system. The molecule is planar. Conformation (b):3–3 delocalization system. The end C–H bonds form planes that are perpendicular to eachother.

example is the molecule shown below,

H

H

C

sp2 sp2sp1 sp1

C C C

H

H (14.35)

whose carbon chain has successive double bonds. Evidently, the two outer C atomsof the chain have sp2 hybridization, while the two central atoms are in an sp1 state.Therefore the chain is a straight line, but the planes of hydrogen atoms at the twoends can have either one of the conformations shown in Figure 14.27.

As the above figures demonstrate, delocalization definitely takes place, but itcan do so over two independent rows of p orbitals: a row of pz orbitals and arow of py orbitals. Thus, two possibilities arise. Delocalization can occur eitheron the 4–2 system, with four orbitals along the z axis and two along the y axis,or on the 3–3 system, with three orbitals along the z axis and three more alongthe y axis. We can calculate the energies of these two arrangements using thecorresponding energy-level diagram—the readers are encouraged to do this—toidentify the system with the lowest energy. The result is this: The lowest energyis achieved for the 4–2 system, so the molecule must be planar. Indeed, it is!

We conclude that delocalization is not limited to conjugated chains, but takesplace whenever there are rows of parallel p orbitals along which the electronscan move by “hopping” from one orbital to another. This hopping movementhas impressive implications. Even crude features of a large number of organicmolecules—such as their shape or their ability to absorb light—are ultimatelydetermined by the delocalized motion of electrons along suitable carbon chains.Delocalization—just like hybridization—is thus such a fundamental deviationfrom the elementary theory of the chemical bond that we are hardly justified incalling it a “deviation.”

14.3.6 Delocalization in Two-dimensional Arrays of p Orbitals: Grapheneand Fullerenes

In all the above applications, delocalization took place on a one-dimensionalarray of p orbitals, which may be open or closed. But the chemistry of carbonoffers a much wider range of possibilities, many of which have only recently been


discovered, while others possibly still await their discovery. One such possibilityis the delocalization on two-dimensional periodic arrays of p orbitals that appearin pure carbon compounds such as graphene or fullerenes. First, let us remind thereaders that the classic (and precious) form of pure carbon is diamond—from theGreek adamas, which means “untamable, invincible”—where C atoms with sp3

hybridization form a three-dimensional lattice of very strong 𝜎 covalent bonds.The great strength of these bonds explains both the unique hardness of diamondand its absolute indifference to forming chemical partnerships with other atomsor molecules. Another common form of pure carbon is graphite, whose atomshave sp2 hybridization, as shown in Figure 14.28. Here, carbon forms essentiallygigantic two-dimensional molecules—molecular layers that cover an infinitearea in principle—which stick together by van der Waals forces to form thecorresponding solid (graphite). As a result of its “construction,” graphite hasstrongly anisotropic properties. For example, it is very hard inside each molecularsheet—harder than diamond, in fact—but also very soft in allowing the relativeshifting of the sheets—or even their detachment, as demonstrated by the easewith which a pencil (that has a graphite core) writes on paper. Indeed, the term“graphite” draws its etymological origins from the Greek word graphein (“towrite”) because of its use as pencil “lead.” Perhaps more important than graphiteitself is graphene, which is a single sheet of graphite. As Figure 14.28 suggests,the electrons of pz orbitals delocalize across the entire plane of graphene, sincethey can thus hop from one orbital to another in more than one direction.

The ease with which electrons can move on the x–y plane is directly respon-sible for the high conductivity of graphene. The comparison with diamond isimpressive. For graphene, the value of the in-plane resistivity is of the order of10−6 Ω cm, while for diamond, it is of the order of 1020 Ω cm! We thus realize that

Figure 14.28 Graphene (a single sheet of graphite). Carbon atoms are sp2 hybridized, and,accordingly, form a two-dimensional hexagonal “honeycomb” lattice. Perpendicular to theplane of the honeycomb lattice lie the unhybridized pz orbitals, forming also atwo-dimensional network on which the 𝜋 electrons can become completely delocalized.Graphene was isolated and characterized for the first time in 2004 by Andre Geim andKonstantin Novoselov (Nobel Prize, 2010), and its spectacular properties are currently one ofthe most active fields of research.


Figure 14.29 The C60 fullerene. The carbonatoms lie on the vertices of a polyhedron thatconsists of 12 regular pentagons and 20regular hexagons. Carbon is in an (almost) sp2

hybridization state and forms three strong 𝜎bonds with neighboring C atoms. The (almost)unhybridized p orbitals are perpendicular tothe molecule’s circumscribing sphere. Thus a“forest” of p orbitals is formed, on whichcomplete delocalization of electrons takesplace.

the motion of delocalized electrons on large scales—which is also the essenceof the so-called metallic bond in the solid state—can be used to explain theconductivity in solids.

We now proceed to examine whether delocalization can exist not on an infiniteplane, but on a finite two-dimensional surface, such as the surface of a sphere.This possibility arises in a finite form of pure carbon—actually, a molecule of purecarbon—that was synthesized for the first time in 1985 by R. F. Curl, H. W. Kroto,and R. E. Smalley, a feat that earned them the 1996 Nobel prize in chemistry.This molecule has the chemical formula C60, and is actually the first member of awhole family of similar molecules that are collectively known as fullerenes, namedafter the American architect R. Buckminster Fuller, who designed domes with asimilar geometrical structure.18 The most popular fullerene is carbon-60 (C60),shown in Figure 14.29.

Given that an LCAO calculation for an orbital “forest” as in C60 is not easy—itrequires, among other things, a nontrivial use of group theory to exploit the highsymmetry of the molecule—it is useful to employ the very simple free-electronmodel for the description of delocalization, an approach we followed earlier forthe benzene molecule. In the fullerene case, we assume that the sixty valence elec-trons originating from an equal number of carbon p orbitals move effectively asfree particles on the surface of the molecule’s circumscribing sphere. The allowedenergies for this type of motion are calculated from the classical expression ofrotational energy

E = 𝓵2

2I= 𝓵2

2mea2 .

Here, we only need to replace 𝓵2 with the quantum expression ℏ2𝓁(𝓁 + 1) to get

E𝓁 = ℏ2

2mea2 𝓁(𝓁 + 1) = 12𝜖𝓁(𝓁 + 1) (𝓁 = 0, 1, 2,…),

18 Actually, the geometric structure itself was first discovered—many, many years ago(!)—byArchimedes. Fullerene C60 is merely one of the 13 semi-regular convex polyhedra of Archimedes,which are also known as Archimedean solids.


where 𝜖 = ℏ2∕mea2, and a is the radius of the molecular sphere that has an experi-mental value of 3.5 Å. The degeneracy of a level with quantum number 𝓁 is 2𝓁 + 1and, therefore, the level can accommodate—taking spin into account—a total of2(2𝓁 + 1) electrons. So, the electronic capacities of successive energy states onthe sphere areState: 𝓁 = 0 𝓁 = 1 𝓁 = 2 𝓁 = 3 𝓁 = 4 𝓁 = 5Capacity: 2 6 10 14 18—

↓ 22—↓

Partial Sum → 50 72With a total of sixty available electrons, all states up to 𝓁 = 4 will be filled (by 50electrons), while the remaining ten electrons will go to the 𝓁 = 5 state, which willthus not be fully occupied, since it can accommodate a maximum of 22 electrons.

The above considerations are valid in the free-electron model, where the prob-lem has full spherical symmetry. But the actual C60 molecule has clearly lowersymmetry than a sphere, a fact that ought to reduce the degeneracy of its energyspectrum,19 and especially of the 𝓁 = 5 state, which is the most important statefor the electronic properties of the molecule. And this is what actually happens.According to theoretical calculations based on group theory, the degeneracy ofthe𝓁 = 5 state is partially lifted, and the 22 initial states, are split into two bunchesof 10 and 12 states respectively, with a gap of about 1.5 eV between them. Whichmeans that the remaining 10 𝜋 electrons suffice to fill the lower energy subshellof the 10 degenerate states that resulted from the partial lifting of the originaldegeneracy.

It is now straightforward to explain the high structural and chemical stabilityof the C60 molecule. It results, on one hand, from the large energy gain due tothe extended delocalization, and on the other, from the filling of the last occu-pied shell, which renders the molecule almost as chemically inert as noble gasatoms. For instance, the C60 molecule has no energy incentive to use its pro-tracted p orbitals toward the formation of chemical bonds with other atoms—say,hydrogen—because this would ruin delocalization along with all its associatedadvantages.

One point that warrants further elaboration is the actual hybridization stateof carbon atoms. This cannot be a pure sp2 state, because the three strong 𝜎bonds that “spread out” from each carbon atom are not completely coplanar, butfollow the directions of the edges of a convex polyhedron. In chemistry this sit-uation is often described by saying that these are not pure sp2 bonds, but havealso a little bit of an sp3 character. From a fundamental point of view, there isnothing peculiar about hybridized bonds having nontypical directions. If we goback to the related discussion of Section 14.2.3 we will recall that these typi-cal directions resulted from the assumption that all hybridized bonds of each

19 Recall the related discussion of Sections 9.3.3.2 and 9.3.3.3. Degeneracy is always linked to thesymmetry of the problem. The greater the symmetry, the richer the degeneracy of an energyspectrum. When this symmetry is lowered, we expect a corresponding decrease—but notnecessarily a complete lifting—of degeneracy. More often than not, we do not need to fully solve theSchrödinger equation to predict the initial and residual degeneracy. Instead, we can use grouptheory (a mathematical tool used mainly to extract the implications of symmetry, not the detailedcharacteristics of the problem).


Figure 14.30 Why pentagons are needed. Fivehexagons around a pentagon leave gaps amongthem. When this planar structure is folded tobecome part of a convex polyhedron, thehexagons can touch each other without beingdistorted. In contrast, structural distortions willarise if we try the same folding with a lattice ofregular hexagons that fill the plane.

type are equivalent. This allowed us to equate all the corresponding coefficients𝜆a, 𝜆b, 𝜆c—for hybridization in two dimensions—and also 𝜆d for three dimen-sions. But if we remove this assumption—as we should, when carbon forms 𝜎bonds that are not equivalent—then we can very well construct two-dimensionalhybrids with angles different from 120∘, or even three-dimensional hybrids with-out tetrahedral directionality. In the present case of the C60 molecule, the hybridswill be roughly as in sp2 hybridization, but will also have a small component ofthe pz orbital, to allow for a slight inward bent of the hybridization plane. Onceagain, we realize how critical it is not to memorize rules blindly—as if they werefundamental laws—but to keep in mind the assumptions and caveats in settingthem up.

Let us now discuss briefly the geometry of the fullerene C60. Why does it havethis particular geometry and not another? Why are there 12 pentagonal and 20hexagonal faces? And, more fundamentally, why are pentagons needed in the firstplace? Can we not construct the fullerene by simply “folding” a graphene sheet,as smoothly as possible, around the surface of a sphere? The answer is hinted atin the last of the above questions, and shown pictorially in Figure 14.30.

So the pentagons are needed to allow the formation of a polyhedron thatconsists mostly of hexagons—recall that we seek to stay as close as possible tothe graphene structure. The question now becomes: How many pentagons areneeded? The answer can be deduced from the following theorem:

Theorem 14.1 The formation of any convex polyhedron that consists solely of(regular) pentagons and hexagons requires always 12 pentagons.

Proof : We begin with the celebrated Euler’s theorem, which applies to an arbi-trary convex polyhedron, and states that

F + V − E = 2,

where F is the number of faces, V is the number of vertices, and E is the numberof edges. Let us now assume that the polyhedron consists of p pentagons and hhexagons. We then have

F = p + h, (a)


while the number of vertices is

V = 13(5p + 6h) (b)

since each pentagon has 5 vertices, each hexagon 6 (hence a sum of 5p + 6h), buteach vertex belongs to 3 of these polygons—hence the division by 3 in order toobtain the actual number V of vertices of the polyhedron. In complete analogy,the number E of edges is given by

E = 12(5p + 6h) (c)

since each pentagon has 5 and each hexagon 6 edges, respectively, while everyedge is shared by 2 polygons, hence the factor 1∕2 in (c). We can now insert (a),(b), and (c) in Euler’s theorem to obtain

p + h + 13(5p + 6h) − 1

2(5p + 6h) = 2

⇒ p + h − 16(5p + 6h) = 2 ⇒

16

p = 2

⇒ p = 12,which means that the Euler condition requires always 12 pentagons and an arbi-trary number of hexagons. For p = 12 and V = 60—we are mainly interested infullerene C60—formula (b) yields readily h = 20. So the fullerene C60 indeed con-sists of 12 regular pentagons and 20 regular hexagons, and this is its structure.It is one of Archimedes’ 13 semi-regular polyhedra! Ancient mathematics meetsmodern physics beautifully in this remarkable molecule. □

Let us also stress the usefulness of the idea that the closed structures ofpure carbon—that is, fullerenes—are basically “smoothened” foldings ofgraphene around a sphere (the pentagons merely facilitate the “smoothen-ing”). Here, we started with graphene because it is planar—as a result of sp2

hybridization—and thus forces all unhybridized p orbitals to be parallel, whichguarantees the best possible conditions for delocalization and for the associatedstabilization of the molecular structure. So, the concept of “folded graphene”allows us to retain the fundamental features of carbon’s planar form also forfinite, closed structures. The validity of this approach is confirmed by oursuccessful prediction of the main features of C60, and even by the practical wayof producing fullerenes: One bombards graphite with the appropriate laser light!

We can explore the concept of “folded graphene” further, and imagine foldingaround simpler surfaces, say, a cylinder. Here—due to the greater geometricalsimilarity between a cylinder and a plane—the process could allow graphenefolding without the distortions that forced us to use pentagons, as in the case offolding around a sphere. Indeed, folding graphene around a cylinder is possibleand produces a new family of carbonaceous structures, of infinite length in prin-ciple, which are known as carbon nanotubes. Folding into a nanotube can occurin various ways—all of which are very interesting mathematically—that allow thesmooth linking of the ends of graphene without deformation of the hexagonalstructure. Actually, the particular way of constructing a nanotube—a kind of heli-cal rotation with a controlled step—predetermines the electrical properties of thenanotube, that is, whether it is a conductor or a semiconductor.

Problems 435

It may be that carbon, having won the battle against silicon for the element oflife, is about to beat silicon again, winning back the title as the basic element oftechnology! Perhaps the “tiny robots” of the future will not be based on silicon,as we hastened to speculate earlier, but will be carbonaceous, like their creators!

Problems

14.5 Calculate the energy eigenvalues and sketch the correspondingenergy-level diagrams for all linear and cyclic systems of N wells,with N ranging from 3 to 6. Confirm, based on these diagrams, the validityof the following general rules:(a) The algebraic sum of all “shifts” from the initial level of the single well

(E0 or 0) is equal to zero.(b) In a linear system of wells, all levels are nondegenerate and arranged

symmetrically with respect to the initial level E0. This fact implies thatthe middle level always lies at E0 for odd N .

(c) In the cyclic system of wells, all levels above the ground state are doublydegenerate, except the highest state, which is nondegenerate for evenN . Moreover, in the cyclic system of wells with even N , the levels arealways arranged symmetrically with respect to the initial level (0). Canyou prove the above for arbitrary N?

14.6 We can certainly apply the free-electron model on “linear” conju-gated systems, if we assume that they can be roughly approximated asone-dimensional potential boxes—straight tubules—whose length L isequal to the length of the particular chain. Apply this simple model onthe hexatriene molecule to show that your theoretical prediction for themaximum wavelength absorbed by the molecule is given by the expression𝜆 = 8L2∕7𝜆C, where 𝜆C is the Compton wavelength of the electron and L(= 7.25Å) is the length of the molecule. Compare your prediction withthe experimental value 𝜆 = 2580Å.

14.7 Examine the possibilities for delocalized motion of 𝜋 electrons on each ofthe molecules

H2C = C = C = · · · = C⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟

n C atoms

H2

for n = 3, 4, 5, and 6. Which of these molecules are planar? In each case,what is the extra energy gain due to delocalization as a fraction of theenergy of a localized 𝜋 bond? Can you guess the general rule for theexistence of planarity in this family of molecules?

14.8 Calculate (as a fraction of a 𝜋 bond) the extra energy gain due to delo-calization in the hexatriene molecule (Figure 14.15). How good is yourprediction, given that the experimental value for the wavelength of themolecule’s first excitation energy is 2580 Å?


14.9 Apply the free-electron model on C60 (a = 3.5Å) to calculate (a) the totalenergy of the system of 60 electrons, (b) the energy of the electrons atthe last-occupied level, (c) the energy difference between the last-occupiedand the first-unoccupied shell. How reliable do you deem your last resultin view of our discussion in the text about the remarkable stability of C60?

Further Problems

14.10 As we discussed in the text, if we remove the requirement that hybridsof the same type (sp1, sp2, or sp3) be equivalent—which led us to equatethe corresponding mixing coefficients 𝜆a, 𝜆b, 𝜆c,…—then the possibilityof nontypical hybridized orbitals arises. Such orbitals do not belong tothe established types. A pertinent example is given by the triplet

𝜓a =12

(

s +√

3 px

)

, 𝜓b = 12

(

s − 1√

3px + 2

√23

py

)

,

𝜓c =1√

2

(

s − 1√

3px −

√23

py

)

.

(a) For each of these orbitals apply the “diagnostic criterion” we used ear-lier to show that none of them belongs to the established hybrid types.

(b) Show that the above three orbitals—they all lie in the x–y plane—arenormalized and mutually orthogonal. Hence, they can be utilizedtoward the formation of planar 𝜎 bonds, which cannot be completelyequivalent, owing to differences among the participating atoms.What are the directions of these bonds?

(c) Can you think of some cases—e.g., molecules with polar bondsbetween C and other elements—where the modification of standardsp2 hybrids is inevitable? How would you describe the correspondingbonds?

14.11 Immediately after a sudden breakup—that is, dissociation—of a moleculethat includes C in sp3 hybridization, the wavefunction of a valenceelectron retains its previous form

𝜓 = N(s + px + py − pz) (1)

but describes now a purely atomic electron.(a) What is the mean energy of the electron in state (1)?(b) What is the probability that a measurement yields zero angular

momentum (𝓁 = 0) for this electron? What is the probability ofmeasuring an angular momentum of 𝓁 = 1 or 𝓁 = 2?

(c) How would the state (1) evolve with time if we ignored the process ofspontaneous de-excitation?

(d) Assume that there remain N atoms and corresponding electrons instate (1) after the dissociation. How many photons will be emittedwhen spontaneous de-excitation finally takes place?


14.12 As we stated in the text (and asked you to confirm in Problem 14.10),we can construct nontypical hybridized orbitals whose mutual angles aredifferent than in standard hybrid types. In order to study more systemat-ically the construction of such generalized hybrids—on the x–y plane, atfirst—we begin with the familiar expressions


1 + 𝜆2, 𝜓b =

s + 𝜇pb√

1 + 𝜇2, 𝜓c =

s + 𝜈pc√

1 + 𝜈2, (1)

where the mixing coefficients 𝜆, 𝜇, and 𝜈 are now different, and all unitvectors a, b and c lie in the x–y plane. Show the following:(a) The requirement for orthogonality of the hybrids 𝜓a, 𝜓b, and 𝜓c leads

to the conditions

a ⋅ b = − 1𝜆𝜇, b ⋅ c = − 1

𝜇𝜈, c ⋅ a = − 1

𝜈𝜆. (2)

Given that the sum of the angles 𝜃ab = 𝜃, 𝜃bc = 𝜙, and 𝜃ca = 𝜔 is equalto 2𝜋, the above conditions are mutually compatible only for 𝜆, 𝜇, and𝜈 such that

𝜆2𝜇2𝜈2 − 𝜆2 − 𝜇2 − 𝜈2 = 2. (3)

(b) If we retain 𝜆 and 𝜇 as independent parameters—we know from (3)that 𝜈2 = (𝜆2 + 𝜇2 + 2)∕(𝜆2𝜇2 − 1)—and we take the x axis as the adirection, then the three generalized hybrids sp2 are given by theexpressions

𝜓a =1

√1 + 𝜆2

(s + 𝜆px), 𝜓b = 1√

1 + 𝜇2

(

s − 1𝜆

px +√𝜆2𝜇2 − 1𝜆

py

)

,

𝜓c =

√

𝜆2𝜇2 − 1(𝜆2 + 1)(𝜇2 + 1)

(

s − 1𝜆

px −𝜆2 + 1

𝜆√𝜆2𝜇2 − 1

py

)

.

(c) Show that in the limit 𝜆→ ∞, the above generalized hybrid orbitalsreduce to

𝜓a → px, 𝜓b →s + 𝜇py√

1 + 𝜇2, 𝜓c →

s − (1∕𝜇)py√

1 + (1∕𝜇)2,

which represent an unhybridized px orbital and a generalized sp1

hybrid. Apropos, show that for two sp1 hybrids (𝜓1 ∼ s + 𝜆1px,𝜓2 ∼ s + 𝜆2px), the orthogonality condition (𝜓1, 𝜓2) = 0 yields𝜆1𝜆2 = 1 → 𝜆2 = −1∕𝜆1. In your opinion, what is the morphologicaldifference between a generalized and a standard pair of sp1 hybrids?

(d) Show that if we require that all angles between hybrids 𝜓a, 𝜓b, and 𝜓cbe equal, then we must have 𝜆 = 𝜇 = 𝜈 and, based on (3), 𝜆 =

√2. We

thus recover the standard sp2 hybridization.(e) Explore other limiting or special cases of the above general expres-

sions and summarize your conclusions. Can you generalize to threedimensions?


14.13 Consider the symmetric triple-well system, and assume that the particleis initially located in the “atomic” state of the first well, which is describedby the wavefunction 𝜓1. Show that the probability of finding the particlein well #1, #2, or #3 after time t is given by the following expressions:

P1(t) = cos4(𝜔t2

)

, P2(t) =12

sin2𝜔t, P3(t) = sin4(𝜔t2

)

,

where 𝜔 = A√

2∕ℏ.

439

15

Solids: Conductors, Semiconductors, Insulators

15.1 Introduction

We will now continue our discussion from the previous chapter and bringits main theme—the quantum theory of the chemical bond—to its naturalconclusion: the study of the most extreme form of delocalization, as observedin the electrons of most crystalline solids. The motion of these electrons canextend throughout the crystal, giving rise to gigantic valence orbitals and cor-respondingly large molecules. For metallic crystals, the macroscopic chemicalbond thus formed is known as the metallic bond. Now, what is really interestingabout crystalline matter is not so much the typical chemical questions (e.g., tocalculate from first principles the crystal structure and the cohesive energy ofa material) but the unusual physical properties due to the delocalized motionof electrons at a macroscopic scale. The crystalline solid is clearly a new formof matter—third, in terms of size, after the atom and the molecule—which,because of its periodicity, can actually be described in rather simple terms. Andthe central element in such a description is the so-called band structure of theenergy spectrum, which we will now discuss.

15.2 Periodicity and Band Structure

Our problem here is to predict the energy spectrum of an infinite chain ofidentical wells, which we use as the one-dimensional model of a crystallinesolid, with the atoms located at the centers of the wells. The new feature in thisproblem is that the potential extends throughout the entire x-axis (−∞ < x <∞)and is a periodic function of x, with a period equal to the distance a between twosuccessive wells.

That is,

V (x + a) = V (x), −∞ < x < ∞.

A direct consequence of periodicity, which can be proved for all periodicpotentials, is that the energy spectrum has the renowned band structure: Itconsists of bands of allowed energies separated by energy gaps wherein no stateexists.


440 15 Solids: Conductors, Semiconductors, Insulators

This fundamental characteristic of the spectrum can be easily understood if wetreat the infinite chain of wells as the limiting case of a finite chain, whose numberof wells is allowed to increase indefinitely. What happens then is qualitativelyevident. As we saw in the previous chapter, when we bring N wells together, eachenergy level E0 of the single well gives rise to N levels of the multiple-well system,all of which lie within the energy range

E0 − 2A < E < E0 + 2A, (15.1)

since they are given by the formula Ek = E0 − 2A cos(k𝜋∕(N + 1)) (k = 1,… ,N),and thus lie between the extreme eigenvalues1

E0 − 2A cos 𝜋

N + 1< E < E0 + 2A cos 𝜋

N + 1.

The above interval lies, in turn, inside the range (15.1), which it approachesasymptotically as N → ∞. But as N increases, so does the number of levelswithin the finite range (15.1), which means that the distance between levelscontinuously decreases, until it eventually vanishes in the limit of infinite N .Evidently, each level of a single well gives rise to a continuous band of energystates of the infinite chain of wells, whose width is given by relation (15.1).So, barring an overlap between bands originating from different levels (suchoverlaps often appear in realistic systems), we expect the energy spectrum of theone-dimensional crystal to have the band structure of Figure 15.1.

Our first remark about this spectrum is that the width of its bands increases aswe move up in energy. This is a direct consequence of (15.1), which says that thewidth 4A of each band is proportional to the matrix element A = |(𝜓n,H𝜓n+1)|,which increases for higher levels, as their wavefunctions are more extended andthus overlap much more with wavefunctions centered on neighboring wells.Therefore, successive bands become broader and the corresponding energy gapsbecome narrower as the energy increases. Our second remark pertains to whathappens in the range above the wells’ edges (where, by convention, E > 0). For

E3

E2

E1

Figure 15.1 Band structure of the energy spectrum in a periodic potential. Bands form aroundthe discrete energy levels of the single well. But they also form in the range of the (singlewell’s) continuous spectrum.

1 Because cos(N𝜋∕(N + 1)) = − cos(𝜋∕(N + 1)), we have Ek|k=N = E0 − 2A cos(N𝜋∕(N + 1)) =E0 + 2A cos(𝜋∕(N + 1)).

15.3 Band Structure and the “Mystery of Conductivity.” 441

a finite number of wells, the spectrum is continuous in this range. But in thelimit of infinite N , this spectrum also breaks up in bands, which are separatedby gradually decreasing energy gaps to restore the spectrum’s continuity at highenergies. The appearance of bands in the continuous part of the spectrum doesnot follow from our previous discussion, but results from a general mathemat-ical analysis, available in the online supplement of this chapter. This analysispredicts the existence of bands in the continuous spectrum for practically allone-dimensional periodic potentials (except for a few very special cases).

To summarize, periodicity produces a new type of energy spectrum—the bandspectrum—with dramatic consequences for the dynamics of electrons, as wewill see.

15.3 Band Structure and the “Mystery of Conductivity.”Conductors, Semiconductors, Insulators

15.3.1 Failure of the Classical Theory

We will now see that the band structure in the energy spectrum can help usexplain readily one of the greatest mysteries of solid-state matter: the enormousvariations of its electrical conductivity which spans thirty orders of magnitude!Compared to other solid-state properties, such a range of values is truly vast.But the essence of this mystery can only be grasped if we try for a moment tothink in terms of the classical model of electrical resistance, which attributes it tocollisions of electrons with the positive ions of the crystal lattice. In the contextof this classical mechanism, there is little room for large variations of resistivityvalues among solids. As for the “explanation” quoted frequently in elementarytextbooks, namely, that conductors have “free electrons” while insulators do not,this is merely a tautology. We can call a material a conductor or an insulator if itselectrons are “mobile” or “immobile,” respectively, but this does not bring us anycloser to understanding the gigantic variations in the “mobility” of electrons fromone material to another. Why are there certain materials (metals) where the elec-trons move through the lattice practically uninhibited, whereas in other materials(insulators) their motion is practically impossible?

Another problem for classical physics pertains to the dependence of the spe-cific resistance on temperature. In the classical picture, resistance must alwaysincrease with temperature, since electrons collide more frequently with the ionsof a hotter lattice. And yet, there exists a large class of poor conductors (e.g.,crystalline silicon) whose resistance clearly drops as the temperature increases!

The classical model fails miserably also in the case of good conductors. Forexample, the conductivity of copper at room temperature is 2

𝜎 ≈ 6 × 105 (Ω cm)−1 ≈ 5 × 1017 esu,

2 We remind the readers that the specific resistance 𝜌, as defined by the well-known relationR = 𝜌L∕S, has dimensions of Ω m (in the SI system of units), even though in practice we often usethe unit Ω cm for convenience. The corresponding unit of conductivity 𝜎 = 𝜌−1 is (Ω cm)−1. Toconvert the latter to cgs units—or Gauss–cgs, the electrostatic system of units (esu)—we write

(Ω cm)−1 = 0.898 × 1012 esu ≈ 1012 esu.


whereas at 4 K (liquid helium temperature) the conductivity increases by fiveorders of magnitude! Such a steep increase of 𝜎 cannot be accounted for in clas-sical physics, as we will now see. First, we will show that classical theory leads tothe so-called Drude formula, namely,

𝜎 = ne2𝜏

m, (15.2)

where m and e are the mass and charge of the electron, respectively, n is the num-ber of conduction electrons per unit volume, and 𝜏 is their mean free time, that is,the average time between two successive collisions of an electron with the lattice.To arrive at (15.2) we begin from the definitions

j = 𝜎 , j = neu ⇒ 𝜎 = neu , (15.3)

where j is the current density, ne is the charge density, and u is the average speedacquired by electrons as they are accelerated by an electric field in the timeinterval 𝜏 between two successive collisions with the lattice. (Here, we assumedthat after a collision, the electron loses all “memory” of its directed motion causedby the electric field in the previous interval.) We then have

u = acceleration ⋅ 𝜏 = Fm𝜏 = e

m𝜏,

whence (15.3) leads to the Drude formula, as promised.In order now to use (15.2), we need an estimate for 𝜏 . We have

𝜏 = 𝓁𝑣, (15.4)

where 𝓁 is the mean free path and 𝑣 is the mean thermal speed of the electrons.The latter can be calculated from the relation

12

m𝑣2 = 32

kT ⇒ 𝑣 =√

3kT∕m. (15.5)

Let us now make the plausible assumption that the mean free path 𝓁 is tem-perature independent and has a typical value in the order of a few times theinteratomic distance. We then obtain from (15.2), (15.4), and (15.5) the ratio ofconductivities at two different temperatures, namely,

𝜎(T1)𝜎(T2)

=

√T2

T1. (15.6)

For the two temperatures of interest to us (room temperature 300 K and liquidhelium temperature 4 K) we thus obtain

𝜎(4 K) ≈ 10𝜎(300 K), (classical theory)

whereas from experimental data we know that

𝜎(4 K) ≈ 105𝜎(300 K). (experiment)

Clearly, the classical theory cannot explain the observed dramatic increase ofconductivity at low temperatures. But in fairness, let us note that the classicalformula (15.2) is not so bad at “high” temperatures (T ≈ 300 K), where it gives


the correct order of magnitude for 𝜎, assuming that 𝜏 = 𝓁∕𝑣, 𝑣 =√

3kT∕m ≈107 cm∕s, and 𝓁 ≈ a few atomic diameters ≈ 10 Å. As for the electron density n,this is approximately equal to 1023 cm−3, if we consider that the radius of a cop-per atom is 1–2 Å and that each such atom contributes one conduction electron.The precise value of n is 8.5 × 1022 cm−3, which means that the average distanceof Cu atoms in the crystal is about 2 Å, which is roughly the same as their atomicradius. In such a close-packed lattice of ions, it is not surprising that an electroncan only travel a distance of 4–5 atomic diameters before it collides with an ion.Therefore, the estimate 𝓁 ∼ 10 Å is reasonable. Bearing all the above in mind andusing the known values of m and e, we see immediately that the Drude formulagives the correct order of magnitude for the electrical conductivity of copper atroom temperature, which is about 6 × 105 (Ω cm)−1. But at low temperatures,things are drastically different. To obtain an acceptable estimate for 𝜎 at liquidhelium temperature, we need to consider that the mean free path of electronshas increased by four orders of magnitude! That is,

𝓁(4 K) ≈ 104𝓁(300 K) ≈ 105 Å ≈ 50 000 atomic diameters,

while for ultrapure Cu samples, the measured conductivity implies a mean freepath on the order of a centimeter! We thus arrive at a conclusion for the “mysteryof conductivity,” which in somewhat dramatic terms can be put as follows: Atlow temperatures the motion of the electrons in a conductor is almost completelyunhindered. Electrons can traverse macroscopically large distances without evercolliding with the ions!

15.3.2 The Quantum Explanation

We will now see how easy it is to explain the mystery of conductivity in the contextof quantum theory. The basic elements for this explanation are the band structureof the spectrum and the Pauli exclusion principle. To get to the point, let us sayat once that the key issue is whether the highest occupied band of the crystal isfully occupied or not (Figure 15.2).

Suppose that we have the case (a), whereby the highest occupied band is com-pletely filled with available electrons. (The shading in the figure depicts the occu-pied part of the band.) Such a crystal is necessarily an insulator, since its electronscannot absorb energy from an external electric field (and thus accelerate), as theonly energy states available to them are at least 1 eV away, in the next band. Incontrast, in case (b), the electric field can easily accelerate electrons of the par-tially occupied band, because the electrons can absorb the work produced by the

Figure 15.2 Energy bands in an insulator (a) and aconductor (b). The highest occupied band in aninsulator is completely filled with electrons, but in aconductor, it is only partially occupied, thus enablingthe movement of electrons, that is, current flow.

(b)(a)


field and move to the next available (unoccupied) states. The material is now aconductor.

It is worth analyzing further the radical difference between the above explana-tion and the classical picture, which attributes the disparity between conductorsand insulators to the presence or absence of free electrons. The quantum expla-nation is essentially the exact opposite of the classical picture. If we regard asfree those electrons that do not belong to a particular atom but are shared bythe whole crystal, then according to quantum theory (for the one-dimensionalmodel we are discussing) there is no difference whatsoever between conductorsand insulators. In both cases, electrons have “broken free” from individual atomsand perform a delocalized motion throughout the crystal chain; their wavefunc-tions are not localized but extend unabated throughout the crystal. So the causeof the different electrical behavior between conductors and insulators lies notwhere classical theory puts it but in mechanisms completely alien to the classi-cal context. The insulating behavior is not the result of the presumed absence offree electrons but of the stringent constraints imposed by the Pauli principle ontheir motion. When an energy band is filled, none of its electrons can change itsstate of motion, since all other nearby states are “taken” and the exclusion princi-ple forbids double occupancy. Nature has found an impressively effective way tofreeze the motion of electrons inside an insulating material.

Assume now that the energy gap between the highest occupied band of aninsulator and the next unoccupied band is not 4–5 eV, as is often the case, but inthe range of 1 eV (Figure 15.3). Such an energy gap may be significantly greaterthan the mean thermal energy at room temperature (kT ≈ 1∕40 eV), but isstill narrow enough to allow a finite fraction of electrons to cross it. If N is thenumber of these excited electrons, then according to Boltzmann’s law we have

N = N0 e−ΔE∕kT ,

where ΔE = Eg is the width of the energy gap and N0 the number of electronsat the upper edge of the occupied band. Actually, it turns out that the expressionfor N has the form

N = N0 e−ΔE∕2kT , (15.7)which means that the energy gap is effectively halved. The reason for this is nottrivial. It has to do with the so-called recombination process, which describes themutual annihilation of an electron and a hole. The probability for such an eventis expected to be proportional to the product of the electron and hole densitiesne ⋅ nh, or simply to n2, since ne = nh in general. But the equilibrium populationwill be reached when the previous recombination rate becomes equal to thecreation rate of new electron–hole pairs, which is proportional to the Boltzmannfactor exp(−Eg∕kT). So we must have

n2 ∝ e−Eg∕kT ⇒ n ∝ e−Eg∕2kT .

For an order-of-magnitude estimate, we set in (15.7) ΔE = Eg ≈ 1 eV,kT ≈ (1∕40) eV, and assume that only a small fraction of the valence electrons(those in a very thin slice at the top of the valence band) are thermally excited. Fora realistic value of N0 (N0 ≈ 1018 cm−3), and using the formula ex ≈ 10x∕2.3, we get

N ≈ 1018 ⋅ e−20 ≈ 1018 × 10−9 = 109 electrons,


Conduction band

Valence band

(a) T = 0

Eg

(b) T ≠ 0

Figure 15.3 Energy bands in a semiconductor for T = 0 and T ≠ 0. When the energy gapbetween the valence and conduction bands is sufficiently small (Eg < 2 eV), a non-negligiblefraction of valence electrons is thermally excited to the conduction band, where the electronscan easily move, as can the holes they leave behind. The material behaves then as asemiconductor.

while for a gap of 4 eV the corresponding number is practically zero,N ≈ 10−18 electrons!

We conclude that, when the energy gap is sufficiently narrow (Eg ≈ 1–2 eV), asmall but non-negligible number of electrons from the highest occupied band ofthe insulator—the so-called valence band—can cross the forbidden energy gapand reach the so-called conduction band (Figure 15.3), where there are availablestates and the Pauli principle can no longer block their motion. In this case, thecrystal does not behave as a perfect insulator, but rather as an “imperfect con-ductor,” hence the term semiconductor for such a material. In fact the numberN ≈ 109 of electrons in the conduction band is infinitesimally small compared tothat of a conductor, where N ≈ 1022 cm−3. So it is necessary to note here thatthe conductivity of semiconducting materials is mainly due to quantum statesinside the gap created by “foreign” atoms in the crystal—the so-called impurities.Unfortunately, the beautiful physics and technology of doped semiconductors isoutside the scope of this book.

The difference between conductors and semiconductors is, in fact, qualitative,not quantitative. The dependence of conductivity on temperature illustrates thisdifference nicely. In conductors, the increase of temperature lowers conductivitydue to collisions with the more vigorously vibrating crystal ions. But in semi-conductors, an increase of temperature raises conductivity, as the number ofthermally excited electrons (i.e., available carriers of electric current) increases.

There is one more qualitative difference between conductors and semiconduc-tors. In conductors, only electrons carry current (and thus contribute to the con-ductivity), whereas in semiconductors, current flows also in the form of holes, theempty states left behind in the valence band when electrons are thermally excited.Other electrons of the valence band can now move to occupy the empty statesand carry current in the process. But as they do so, new empty states are created,which move in the opposite direction. This picture is reminiscent of what happenswhen a seat at the end of a line of occupied chairs becomes available, and people


move successively to occupy the empty seat next to them. Here, the whole processcan be described much more efficiently as the movement of the empty seat (thehole), instead of the combined movement of all people from one chair to another.

The analogy with the valence band of a semiconductor is evident: The motionof numerous electrons as they try to occupy a handful of empty states (holes) canbe equivalently described by the reverse motion of these holes. It is also clear thatthe holes are to be treated as positively charged particles, since they move in theopposite direction compared to electrons under the influence of an electric field.

We conclude that the conductivity of a semiconductor is due to two types ofcarriers: the negatively charged electrons of the conduction band and the posi-tively charged holes of the valence band. The presence of both types of carriers isirrefutably confirmed in a classic experiment of solid-state physics, the celebratedHall experiment. An elegant description of this experiment can be found in TheFeynman Lectures on Physics, vol. III, Chapter 14, Addison Wesley, 1971.

What remains now is to see how we can a priori decide the type of conductivebehavior (conductor, semiconductor, or insulator) of a given material when itsconstituent atoms are brought together to form a solid. In our one-dimensionalmodel, this is easy to answer. An energy band will get filled depending on whetherthe corresponding atomic level is filled. This conclusion is easily drawn if wereturn to the one-dimensional model of a finite chain of wells and take the infinitelimit.3 For example, if we have the one-dimensional analog of the helium atom,where each well has a doubly occupied (and hence filled) level, then when twosuch wells are brought together, we obtain a pair of levels with a total capacity of4, which fill up with the four available electrons. For three helium wells, we getthree levels and six available electrons, so the levels again fill up. In the generalcase of N wells, there are N levels in total, and they are completely filled with the2N available electrons. Clearly, the same happens in the limit of an infinite chainwe are interested in. We conclude that the energy band produced from a doublyoccupied state of a single well (atom) is completely filled.

But if the initial “atomic” level is half filled (i.e., it has only one electron) thenthe corresponding energy band will also be half filled. So, a completely filledatomic level leads to a one-dimensional crystal that is an insulator—or, of course,a semiconductor, if the gap is small—whereas a half-filled atomic level resultsin a conductor. The only exception to this rule is when the valence and conduc-tion bands become sufficiently broad so they overlap, in which case the insulatorbecomes again a conductor.

But the question remains: Why do the electrons of a half-filled band of aconductor have this extraordinary mobility, which allows them to traverse—especially at low temperatures—macroscopically large distances inside thecrystal without colliding with the ions? To elucidate this fundamental “mystery”of conductivity we will take a deeper look at the band structure and itseigenfunctions.

3 This procedure is clearly necessary, because an energy band has an infinite capacity and can thusaccommodate an infinite number of electrons. So we need to start with the finite case, becausewithout it, it is impossible to “compare” these two infinite numbers.

15.4 Crystal Momentum, Effective Mass, and Electron Mobility 447

15.4 Crystal Momentum, Effective Mass, and ElectronMobility

As the readers may have expected, our starting point will be the followingrelations from the previous chapter

𝜓 =∑

ncn𝜓n, E = E0 − 2A cos 𝜃, (15.8)

cn = 𝛼ein𝜃 + 𝛽e−in𝜃, (15.9)

which are valid both for a finite and an infinite chain of wells, with only theboundary conditions being different in the two cases. For a finite chain (openor closed), the boundary conditions imposed on the coefficients (15.9) lead toquantization of the parameter 𝜃 and hence of the energy eigenvalues (15.8). Foran infinite chain, all we can request is that the wavefunction (15.8) remain finitethroughout the range −∞ < x < +∞. This softer requirement is automaticallysatisfied by the general form of the coefficient (15.9) for any real number 𝜃. Theenergy spectrum within each band is thus continuous (as we predicted) andthe corresponding states can be characterized by the (continuous) parameter𝜃. In particular, for every 𝜃 we have one energy eigenvalue and two linearlyindependent eigenfunctions (our familiar double degeneracy of the continuousspectrum) that are given by the expressions

𝜓+ =+∞∑

n=−∞ein𝜃𝜓n , 𝜓− =

+∞∑

n=−∞e−in𝜃𝜓n , (15.10)

and which are, obviously, complex conjugates of each other (i.e., 𝜓+ = 𝜓∗−).

Actually, both wavefunctions (15.10) can be described by the expression

𝜓 = 𝜓𝜃 =∑

ein𝜃 𝜓n , (15.11)

where the angle 𝜃 takes values in the symmetric interval

−𝜋 ≤ 𝜃 ≤ 𝜋. (15.12)

Consequently, each energy eigenvalue,

E = E𝜃 = E0 − 2Acos 𝜃, (15.13)

is associated with two opposite values of 𝜃, which correspond to the two distincteigenfunctions given by expression (15.11) for 𝜃 and −𝜃. The graphical represen-tation of the function E = E(𝜃) is given in Figure 15.4. The same graph applies forthe other bands as well, with the obvious substitution of the parameters E0 andA by the values corresponding to the higher atomic levels. The shaded region onthe vertical axis shows the energy range of this particular band.

But what is the physical meaning of the parameter 𝜃? Let us first say that 𝜃 doesnot need to have a particular physical meaning! Since the energy spectrum is(piecewise, at least) continuous, there must be a continuous parameter to char-acterize or index its states, in the same manner that discrete energy states areparametrized by discrete quantum numbers. Nevertheless, 𝜃 does have a physicalmeaning, which, despite its approximate character, is central to all the qualitative


E(θ)

−π +π

E0

E0 + 2A

E0 – 2A

θ

Figure 15.4 Dependence ofthe electronic energy on thecontinuous parameter 𝜃 thatcharacterizes energy states ina one-dimensional crystal.Since 𝜃 is linked to the crystalmomentum k through therelation 𝜃 = ka, the figure alsoshows the relation E = E(k)between energy andmomentum in the crystal.

features of the motion in a periodic potential. Specifically, we will see that if,instead of 𝜃, we introduce a new parameter k via the relation4

𝜃 = ka ⇒ k = 𝜃

a, (15.14)

where a is the period of the lattice, then k can be regarded as a type of momen-tum of the electrons in the crystal. This interpretation emerges naturally from theobservation that the eigenfunctions of momentum in one dimension

𝜓k(x) = eipx∕ℏ = eikx (p = ℏk)

have the following property:

𝜓k(x + a) = eik(x+a) = eika ⋅ eikx = eika 𝜓k(x), (15.15)

which says that an arbitrary translation by a in the x-direction leaves the eigen-function intact, except for a phase factor.

What is particularly interesting is that the above property is valid also forthe eigenfunctions (15.11) of a crystal, but with the crucial difference that thedisplacement a must now be equal to the lattice period or an integer multiple ofit. The proof of this assertion is based on the relation

𝜓n(x + a) = 𝜓n−1(x), (15.16)

which expresses the simple fact that the local eigenfunctions 𝜓n(x) can bederived from one another through a mere translation by the lattice period a. So,if we perform the translation x → x + a in (15.11), we find

𝜓𝜃(x + a) =∑

ein𝜃 𝜓n(x + a) =∑

ein𝜃 𝜓n−1(x)

=∑

ei(n+1)𝜃 𝜓n(x) = ei𝜃∑

ein𝜃 𝜓n(x) = ei𝜃 𝜓𝜃(x),

where, in the second line, we used the fact that the sum extends from n = −∞ ton = +∞, so that every translation of n (including n → n + 1) is allowed. We havethus shown that the eigenfunctions (15.11) satisfy the relation

𝜓𝜃(x + a) = ei𝜃 𝜓𝜃(x), (15.17)

4 Since 𝜃 is dimensionless, k has dimensions of inverse length, that is, dimensions of a wavenumber.Hence our choice of the symbol k.


whence we obtain

𝜓𝜃(x + Na) = eiN𝜃 𝜓𝜃(x), (15.18)

where a is the lattice period and N an arbitrary integer (positive or negative).A direct comparison of (15.17) and (15.18) with (15.15) reveals that if, instead

of using 𝜃 in the parametrization of the eigenstates, we use the variable k = 𝜃∕a,then k will have similar properties to momentum. Indeed, by introducing k, rela-tions (15.17) and (15.18) can be rewritten as

𝜓k(x + a) = eika 𝜓k(x) and 𝜓k(x + Na) = eiNka 𝜓k(x), (15.19)

whence the similarity to (15.15) becomes evident.Nevertheless, there are obvious reasons why the characterization of k as

momentum cannot be strictly correct. Only plane waves eikx (where is aconstant) have definite momenta (they are momentum eigenstates), and thewavefunctions (15.11) are certainly not plane waves. Moreover, relation (15.15),which also uniquely defines the momentum eigenstates, does not hold for states(15.11) unless a is an integer multiple of the lattice constant. Such differenceswith the ordinary momentum have important consequences. For example, whilefor free electrons, k extends throughout the range −∞ < k < +∞, in the case ofa crystal it is restricted—because of (15.12)—inside the so-called Brillouin zone

−𝜋a≤ k ≤ 𝜋

a. (15.20)

Another difference with ordinary momentum lies in the energy–momentumrelation, which for free electrons has the familiar form E = ℏ2k2∕2m, while forelectrons moving inside the crystal it becomes

E = E0 − 2A cos ka. (15.21)

To see what kind of qualitative implications we get from the interpretation of kas momentum,5 let us examine the form of (15.21) in the limit of small k, that is,near the bottom of the corresponding band. We then have

E = E0 − 2A(

1 − (ka)2

2+ · · ·

)

≈ (E0 − 2A) + Aa2k2

and by omitting the constant term E0 − 2A we obtain E ≈ Aa2k2.This expressionhas the same form as the familiar energy–momentum relation E = ℏ2k2∕2m forfree electrons, and becomes identical to it if we set

Aa2k2 = ℏ2k2

2m∗ ⇒ Aa2 = ℏ2

2m∗ ⇒ m∗ = ℏ2

2Aa2 , (15.22)

where m∗ is the so-called effective mass of the electron as it moves inside the crys-tal. Therefore, if the above interpretation of k as momentum is qualitatively cor-rect, then, at least for small momenta, the effect of the crystal field on the electroncan be described through the substitution of its mass by the effective mass (15.22).In other words, to describe the motion of an electron inside the crystal under the

5 We remind the readers that k has the physical dimensions of a wavenumber and can thus beregarded as momentum only if we set ℏ = 1 (otherwise we should have p = ℏk).


influence of an external field, we can ignore the presence of the lattice and simplyassume that the particle has an effective mass instead of its physical mass.

The qualitative correctness of this interpretation is confirmed directly byrelation (15.22), according to which the effective mass is inversely proportionalto the matrix element A that determines how easily the electron hops from onewell to another. The formula (15.22) tells us therefore that as A increases, thatis, as an electron hops more easily from one well to another, its effective massdecreases or, equivalently, its mobility increases. Conversely, when A is small,the hopping probability is reduced and the effective mass increases. All thesearguments are plausible and support our treatment of k as a kind of crystalmomentum for the electron.

Numerically, the effective mass of the electron is, in general, of the same orderof magnitude as its physical mass. This can be readily seen from formula (15.22),whose form in atomic units (can you explain why?) is

m∗ = 1

2 A(eV)27.2

(aa0

)2 a.u. = 1

2 A(eV)27.2

(aa0

)2 me,

whence—for typical values of A (a few eV) and a (1–3 Å)—we easily see that m∗

is indeed about the same as the physical electron mass.But the best justification of k as momentum stems from the realization, which

we will prove shortly, that the eigenfunctions 𝜓k(x) of the crystal can always bewritten in the form

𝜓k(x) = eikx uk(x), (15.23)

where uk(x) is a periodic function of x with a period a. The above relation is noneother than the famous Bloch’s theorem for an arbitrary periodic potential:

BLOCH’S THEOREM The energy eigenfunctions𝜓(x) of a periodic potential canalways be written in the form

𝜓(x) = eikx u(x),

where u(x) is a periodic function with the same period as the potential.

Within the LCAO approximation, which we use here, this crucial property ofthe eigenfunctions in a one-dimensional crystal (namely, that uk(x) in (15.23) isperiodic) emerges readily if we require that (15.23) satisfy the relation𝜓k(x + a) =eika𝜓k(x). Indeed, if we insert (15.23) in (15.19) we obtain

eik(x+a) uk(x + a) = eika eikx uk(x) ⇒ uk(x + a) = uk(x).

From a physical perspective, the relation (15.23) tells us something very interest-ing: The wavefunctions that describe electronic motion inside a crystal are similarto those of free electrons with definite momentum

𝜓k(x) = eikx, (15.24)

with the difference that the amplitude is no longer a constant but a periodicfunction, with a period equal to that of the crystal. This periodic variation of


(= uk(x)) reflects the spatial amplitude modulation induced on the electronicwave by the crystal lattice. It is also important to note that, owing to theperiodicity of uk(x), the wavefunction (15.23) cannot contain the factor e−ikx,whose presence is inevitable for motion in a typical localized potential well, as itdescribes the possibility for the particle to be reflected and move backward.6 Therelation (15.23) tells us therefore that reflection—and scattering in general—isimpossible inside a periodic potential. The particle moves inside the crystal witha constant “momentum” and is otherwise oblivious to the crystal’s presenceexcept for a periodic modulation in the amplitude of the associated wave. Thismodulation causes a change in the so-called dispersion relation E = E(k), fromE = ℏ2k2∕2m for free motion to E = E0 − 2Acos ka for motion inside the crystal.Thus the movement of a particle inside a periodic potential faces no resistancewhatsoever! An external field can accelerate the particle until its momentumreaches the edge of the Brillouin zone, where the above analysis fails, and thetraveling Bloch waves become instead standing waves, giving rise to the so-calledBloch oscillations. But for small momenta, electrons move essentially unhinderedinside a crystal. As a result, the mean free path becomes infinite, as does theconductivity!

This remarkable conclusion can be analyzed from another perspective thatis better suited for 2D or 3D crystals. Basically, we are dealing with a generalproperty for wave propagation in a periodic medium. Owing to periodicity, thesecondary waves (recall the Huygens principle) emitted from the atoms of sucha medium reconstruct the original wavefront that triggered them. In this way,the wave propagates inside the periodic medium unimpeded, that is, withoutscattering (Figure 15.5). If we now replace the classical wave with the electronicprobability wave, we obtain a qualitative explanation for the remarkable mobilityof electrons in a periodic crystal. The only obstacles to this motion, causingscattering of the electrons in random directions, can result from deviationsfrom perfect periodicity. Thus the resistance of a real conductor is ultimately theresult of random impurities and thermal motion, both of which cause randomdeviations from the crystalline structure. Given the above, it no longer seemsodd that at low temperatures and for very pure samples, the resistance dropsrapidly. In a high-quality crystal the electronic quantum waves propagate almostas in free space!

Figure 15.5 How oblique scattering is avoided during wavepropagation in a periodic medium. The secondary quantum wavesemitted from the atoms reconstruct the original plane wave and theelectrons keep moving forward without oblique scattering.

6 An exception is the very special class of the so-called reflectionless potentials.


Problems

15.1 Show that, in cgs units, the electrical resistivity 𝜌 has dimensions of time.Use this fact to show that the atomic unit of electrical resistivity is 𝜌0 =ℏ3∕me4, while the corresponding atomic unit of electrical resistance R isR0 = ℏ∕e2.

15.2 Consider an infinite, periodic chain of identical potential wells, eachof which has only two bound states with energies E1 = −12 eV andE2 = −6 eV. The parameter A for those bound states (i.e., the matrixelement with the corresponding state of the neighboring well) is equal to

A(1) = 0.5 eV, A(2) = 2 eV.(a) Sketch the band structure diagram of the system, and state which of

the following energy values are allowed:E = −9,−12,−1,−10.5,−15 eV.

(b) If the one-dimensional “atom” represented by each well has two elec-trons, what type of material does this particular crystal correspondto? A conductor, an insulator, or a semiconductor? Consider the samequestion if the “atom” has three electrons.

(c) Calculate the effective mass of the conduction electrons if the afore-mentioned crystal is a semiconductor.

15.3 Use the Drude formula to estimate the conductivity 𝜎 of a conductor ata temperature of 2000 K, where the electron mean free path is equal to3 Å, and the electron density is equal to 5 × 1022 cm−3. What is the valueof 𝜎 if the same conductor is cooled to a temperature of 20 K? Note that1 esu-𝜎 = (1∕9) × 10−11 (Ω cm)−1 ≈ 10−12 (Ω cm)−1.

15.4 The conductivity of a good conductor is on the order of 1013 (Ω cm)−1 atliquid-helium temperature. Use the Drude formula to estimate the meanfree path of the conduction electrons in such a crystal. What conclusionscan you draw from your result?

15.5 Consider a one-dimensional crystal formed by potential wells that haveonly one bound state with energy E0 = −5 eV, and where the energy bandformed around this energy level has a width equal to 4 eV.(a) If the distance between the wells is equal to a, write down the disper-

sion relation for the electrons in this particular crystal.(b) Use the LCAO approximation to determine the wavefunction of the

electrons moving at the bottom, and in the middle of the above energyband.

(c) What is the energy of an electron with a crystal momentum k = 𝜋∕3amoving in the above crystal?

15.6 The dispersion relation of the conduction band in a one-dimensional semi-conductor is given by the expression

E(k) = −4 − 3 cos k,

15.5 Fermi Energy and Density of States 453

where E and k are measured in eV and Å−1, respectively. Calculate theeffective mass of a conduction electron in this semiconductor.

15.7 Consider the following wavefunctions:(a) 𝜓(x) = eikx sin 𝜋x

2a,

(b) 𝜓(x) = eikx sin 𝜋xa

,(c) 𝜓(x) = eikxsin2 𝜋x

a.

Which of the above satisfy Bloch’s theorem for a one-dimensional crystalwhose lattice constant is equal to a?

15.5 Fermi Energy and Density of States

15.5.1 Fermi Energy in the Free-Electron Model

We will now introduce two key concepts for the description of electronic motioninside a crystalline solid. We begin with the so-called Fermi energy, which isan elementary but very useful concept. Given that electrons are fermions, theirplacement in the available energy levels of a lattice must obey the Pauli exclusionprinciple. As a result, higher and higher energy levels become occupied until werun out of available electrons. The energy of the last level thus occupied is theFermi energy (or Fermi level). This is shown in Figure 15.6 for a set of levels thatlooks “continuous,” as happens in a typical crystalline solid wherein the distancebetween successive levels in a band is practically zero. Evidently, the above elec-tronic configuration is valid only at absolute zero, since at finite temperatures,those electrons in the vicinity of the Fermi level can be thermally excited to higher,unoccupied states. Thus the Fermi energy is the energy of the highest occupiedstate in a system of fermions at absolute zero.

At this point, readers may rightfully wonder how useful it is to introduce aconcept that is well-defined only at very low temperatures. The answer lies—aswe will shortly see—in the fact that for typical metals the Fermi energy is inthe range of a few eV, while the mean thermal energy at room temperature is amere 1∕40 of an eV. This means that even at relatively high temperatures, theelectronic behavior of the material is ultimately determined by a very thin energyinterval—of width ΔE ≈ 1∕40 eV below the Fermi level—whose electrons can be

EF

Unoccupied states

Occupied states

Fermi energy

Figure 15.6 Fermi energy of a crystalline solid at absolute zero temperature.


z

y

x

L

L

L

Figure 15.7 A cubic-shaped solid for the free-electron model.Electrons are assumed to move as free particles inside a cubicbox that represents the bulk of the solid body.

thermally excited and “jump” out of the “frozen sea” of occupied states, towardthe unoccupied energy states.

Let us now see how we can calculate the Fermi energy in the simplest possiblemodel of a crystalline solid, where electrons can be treated as free particles.This free-electron model is the exact analog of the model we used in the previouschapter to describe the delocalized motion of electrons in organic molecules.To apply the model to a three-dimensional crystal, we assume the crystal hasa cubical shape with edge length L. Our problem reduces then to finding thehighest occupied level for a system of N particles with spin 1∕2, which movefreely inside the cubic box shown in Figure 15.7. The first step is to calculatethe allowed energies for a particle trapped inside the box, in which case theboundary conditions are that the wavefunctions vanish at the walls. Actually,this problem can be solved readily once we realize that in each direction x, y,and z, the particle “feels” an infinite potential well, for which the eigenenergiesare given by the formula E = ℏ2𝜋2n2∕2mL2, as we know. Given now that themotions in these directions are independent from each other, the total energy ofthe particle is simply the sum of three similar terms and is given by

E = Ex + Ey + Ez =ℏ2𝜋2

2mL2 n2x +

ℏ2𝜋2

2mL2 n2y +

ℏ2𝜋2

2mL2 n2z

= ℏ2𝜋2

2mL2 (n2x + n2

y + n2z ), (15.25)

while for the eigenfunctions we will have the product form (𝜓 = 𝜓nx𝜓ny

𝜓nz)

𝜓 = 𝜓nxnynz=

(√2L

)3

sinnx𝜋x

Lsin

ny𝜋yL

sinnz𝜋z

L, (15.26)

where nx, ny, and nz are positive integer numbers. The results (15.25) and (15.26)can also be written in the physically more transparent form

E = ℏ2

2mk2 = ℏ2

2m(k2

x + k2y + k2

z )

𝜓 =(2

L

)3∕2sin kxx sin kyy sin kzz,

(15.27)

which tells us that the particle “moves” as if it were free inside the box—thisis why the expression of its energy is the same as in free motion—but its


“momentum” p = ℏk is quantized due to the confinement within the walls. Sothe components of the wavevector k take the quantized values

kx =nx𝜋

L, ky =

ny𝜋

L, kz =

nz𝜋

L. (15.28)

Now that we know the allowed energy levels, all we need to do is populate themwith electrons (two electrons per level) and see where the process terminates. Butthe application of this recipe is highly nontrivial in this case, since the expres-sion (15.25) for the energy eigenvalues has rich degeneracy7 for which no closedformula exists. Thus the exact calculation of the last occupied level is extremelyarduous even for a small number of particles, and becomes completely intractablefor a macroscopically large number of them.

It turns out that in this last case of a macroscopic system—which is ourmain interest, after all—the calculation of the Fermi energy can be done muchmore easily, due to the fact that for a macroscopically large box, the allowedvalues (15.28) of the wavevector k—which characterize the allowed states in thebox—are so close to each other that they practically form a continuum. Andsince the distance between two successive allowed values of each componentof k is constant and equal to 𝜋∕L, we can argue that in the three-dimensionalspace with axes kx, ky, kz, the set of allowed points has a constant density, and thevolume corresponding to each point is equal to

Ω0 = (Δkx) (Δky) (Δkz) =(𝜋

L

)3= 𝜋3

V, (15.29)

where Δkx = Δky = Δkz = 𝜋∕L is the constant distance between two successivepoints on the corresponding axes, while V = L3 is the box volume.

The above considerations are depicted in Figure 15.8, where, for clarity, we limitourselves to the two-dimensional problem. The little square box in the figureshows the size of the “cell” that contains one particular allowed point. Note thatall allowed points lie within the first quadrant of the plane, since all componentsof k are positive. Coming back to the three-dimensional case, we can now see thatthe number of allowed points inside a particular region of k space is determined

Figure 15.8 “Allowed points” for kx and ky in kspace, for a free particle in a two-dimensionalbox. The “volume” (actually, here an “area”) thatcorresponds to each point is equal toΩ0 = (𝜋∕L)2.

ky

kx

kF

kFLπ

Lπ

LΩ0 =2π

7 That is, there are many different triplets of quantum numbers nx, ny, and nz that give the sameenergy eigenvalue.


by how many times the volume Ω0 = 𝜋3∕V of the “unit cell” fits inside the totalvolume Ω of the said region.8 We thus have

Number of allowed points inside a volume Ω = ΩΩ0, (15.30)

which makes the calculation of the Fermi energy fairly easy. First of all, the setof points corresponding to occupied states will now be the positive one-eighth(octant) of a sphere, since the energy E = ℏ2k2∕2m depends only on the magni-tude |k| = k of the wavevector k. If we denote the radius of this sphere as kF, thenthe Fermi energy is given by the expression

EF = ℏ2

2mk2

F , (15.31)

where the quantity kF is called the Fermi wavenumber. (It is the magnitude of thewavevector of electrons in the highest occupied state.)

We can now calculate kF by making the self-evident observation that the totalnumber of states in the first octant of the Fermi sphere must equal the total num-ber N of available electrons. All we then need to do is apply formula (15.30), takingalso into account the fact that to each point in k space correspond two allowedstates of different spin. We thus obtain a relation for the Fermi wavenumber kF

N = 2Ω(kF)Ω0

, (15.32)

whereΩ0 = 𝜋3∕V andΩ(kF)(

= 18

43𝜋k3

F

)

is the volume of a spherical octant withradius kF. We thus have

N = 218

43𝜋k3

F

𝜋3∕V= V

3𝜋2 k3F ⇒ kF = (3𝜋2n)1∕3, (15.33)

where n = N∕V is the number of electrons per unit volume in our crystal.Finally, by inserting (15.33) in (15.31) we obtain the Fermi energy

EF = ℏ2

2m(3𝜋2n)2∕3. (15.34)

Before we proceed to a qualitative analysis of this result, let us undertake thecorresponding calculations for the two-dimensional and one-dimensional cases.These are relevant to the motion of electrons on thin films and thin wires, twosystems of great current interest and promise for a number of technologicalapplications.

Equation (15.32) is still the basic formula, except that now the expressions forthe “volumes” Ω(kF) and Ω0 are different in each case. In two dimensions we have

Ω0 = area of unit cell in the (two-dimensional) k space

= (Δkx) (Δky) =𝜋

L⋅𝜋

L= 𝜋2

S

8 In the same way that the number of sand grains in a big box of volume V is given by the ratio V∕𝑣,where 𝑣 is the volume of a typical sand grain. Clearly, it does not matter if some grains in theboundary of the box do not fit perfectly inside.


and

Ω(kF) = area of a quadrant for a circle of radius kF = 𝜋k2F∕4.

We thus find from (15.32)

N = 2𝜋k2

F∕4𝜋2∕S

= S2𝜋

k2F,

so for the Fermi energy we obtain

EF = ℏ2

2mk2

F = ℏ2

2m(2𝜋n), (15.35)

where n = N∕S is the two-dimensional electron density, that is, the number ofelectrons per unit area of our two-dimensional crystal.

For the one-dimensional case we have

Ω0 = length of unit cell in the (one-dimensional) k space = Δk = 𝜋∕L

and

Ω(kF) = length of positive half-interval of “radius” kF = kF

⇒ N = 2kF

𝜋∕L= 2L𝜋

kF ⇒ kF = 𝜋n2

⇒ EF = ℏ2

2mk2

F = ℏ2

2m

(𝜋n2

)2, (15.36)

where n = N∕L is the one-dimensional electron density, that is, the number ofelectrons per unit length of our crystalline wire.

15.5.2 Density of States in the Free-Electron Model

Besides the Fermi energy, another quantity of interest is the electronic density ofstates as a function of energy, because it informs us of the number of states (andhence electrons) that lie inside any small energy range below the Fermi energy.If (E) is the total number of states with energies ranging from 0 to E, then thenumber of states inside the interval from E to E + ΔE is equal to

Δ = (E + ΔE) − (E).

Therefore, the density of states (i.e., the number of states per unit energy interval)is given by the expression

𝜌(E) = limΔE→0

ΔΔE

= d (E)dE

(15.37)

and can thus be found from the function (E), which can be calculated eas-ily. Actually, this is the same calculation we made earlier for the Fermi energy,except that now the radius of the spherical octant in the formula (15.32) is nolonger kF but k, while the energy–momentum relation is E = ℏ2k2∕2m. We thushave

(E) = 2 Ω(k)Ω0

= 218

43𝜋k3

𝜋3∕V= V

3𝜋2 k3


(a) (b) (c)

EEE

ρ(E) ρ(E) ρ(E)3D 2D 1D

EF EF EF

Figure 15.9 The density of states as a function of energy in the free-electron model.

and if we express k in terms of E through the relation E = ℏ2k2∕2m, we get

(E) = V3𝜋2

(2mℏ2

)3∕2E3∕2, (15.38)

whence (15.37) yields for the density of states the expression

𝜌(E) = V2𝜋2

(2mℏ2

)3∕2E1∕2, (15.39)

the graphical representation of which is shown in Figure 15.9a. The figure alsoshows the corresponding results for the one- and two-dimensional cases, whichare given by the expressions (as the readers are encouraged to show for them-selves)

2D ∶ 𝜌(E) = mS𝜋ℏ2

( (E) = mS𝜋ℏ2 E

)

, (15.40)

1D ∶ 𝜌(E) = L𝜋

(2mℏ2

)1∕2E−1∕2

(

(E) = 2L𝜋

(2mℏ2

)1∕2E1∕2

)

, (15.41)

where 2D and 1D denote the two- and one-dimensional cases, respectively.Clearly, it follows from the above that

(EF) = N , (15.42)

where N is the total number of electrons.Another useful quantity is the total energy𝑈 of the electron gas, which is given

by the general expression9

𝑈 = ∫EF

0𝜌(E) EdE. (15.43)

When we apply (15.43) to the three-dimensional case (𝜌 = AE1∕2, = 23AE 3∕2,

and A = (V∕2𝜋2)(2m∕ℏ2)3∕2 for convenience), we find

𝑈 = A∫EF

0E 3∕2 dE = 2

5AE 5∕2

F = 35

(23

AE 3∕2F

)

EF

= 35 (EF)EF = 3

5NEF, (15.44)

9 This is analogous to the formula for the mean value ⟨A⟩ = ∫ P(a)a da of a statistical quantity Awhose probability distribution is P(a).


Figure 15.10 Typical experimentalcurve for the density of states in theconduction band of athree-dimensional conductor.

E

EFEmin Emax

ρ(E)

which means that the mean electron energy (E = 𝑈∕N) is equal to

E = 35

EF, (15.45)

which is a reasonable result, given the shape of the curve 𝜌(E) in Figure 15.9a,which gives more weight to the energy range near the Fermi level. Thecorresponding results for the two- and one-dimensional cases are, respectively,

2D ∶ 𝑈 = 12

NEF , E = 12

EF (15.46)

1D ∶ 𝑈 = 13

NEF , E = 13

EF. (15.47)

Let us now say a few words about how to apply the above ideas to a realcrystal, whose electrons are no longer free but subject to a periodic potential.Figure 15.10 depicts a typical curve for the density of states in the conductionband of a three-dimensional conductor, and the position of the Fermi energy. Aspecial feature of this curve that should be stressed here is that at the band edges,the density-of-states curve has the same shape as the corresponding curve forthe free-electron model near E = 0, which is the only band edge in that case.10

So, for the curve of Figure 15.10, we have𝜌(E) −−−−→

E→E+min

(E − Emin)1∕2, 𝜌(E) −−−−→E→E−

max

(Emax − E)1∕2,

while at intermediate energies inside the band, the function 𝜌(E) can have a com-plicated form depending on the particular crystal. Similar observations hold forthe one- and two-dimensional cases, where the density of states at the band edgeshas again a similar shape as that of the free-electron model. As for the position ofthe Fermi level in Figure 15.10, this can be calculated by taking into account onlythe conduction electrons that populate the corresponding band. We thus have

∫EF

Emin

𝜌(E) dE = N ,

where N is the number of conduction electrons. Let us also note that in mostreal solids the electronic density of states is derived from experimental measure-ments, since it directly relates to the emission or absorption of radiation fromthe crystal.

10 The reason for this is clear. At least in the lower edge of a band—that is, for small k—electronsbehave as free particles, as we have already shown.


EF

E

ρ(E)

Figure 15.11 Experimental determination of the density of states. Data from x-ray spectra ofatoms in a crystal allow us to determine the electronic density of states of the band where thetransitions originated from.

The simplest example of this process is depicted in Figure 15.11 and pertainsto emission of x-rays from a crystalline conductor. As in the case of individ-ual atoms, x-rays are emitted when an inner-shell electron is removed from theatom11 and a valence electron “rushes” to take its place, shedding the correspond-ing energy difference (which is of the order of a keV) in the form of a photon.The only difference in the case of a crystal is that the outer atomic shell (thevalence shell) has become a band, while for the inner shells (and especially theinnermost one) we can safely assume they have negligible broadening, and thusbehave as lines of vanishing thickness. It follows from the above that in a crys-tal, the x-ray spectrum—assuming we ignore its continuous part that is causedby the braking radiation, or Bremsstrahlung, of the impinging electrons—is notdiscrete, as in the case of isolated atoms. Instead, the spectrum consists of con-tinuous parts whose positions are determined by the inner shells involved in thetransition, and whose widths are determined by the width of the occupied regionof the conduction band. (The latter is equal to the Fermi energy of the prob-lem.) Moreover, the intensity of the emitted x-rays in the various parts of thespectrum depends directly on how densely populated by electrons is the regionof the conduction band where the electrons originated from. The experimen-tally measured variations of intensity with respect to energy can thus be usedto deduce the density of states, directly from experiment. Diagrams of the typeshown in Figure 15.11—where the density of states is drawn alongside the asso-ciated band—are thus very useful, since they provide information not only aboutthe position of the emission and absorption spectra but also about the spectralintensities in the various energy regions.

15.5.3 Discussion of the Results: Sharing of Available Space by the Particlesof a Fermi Gas

The results of the previous section call for further discussion, because this isthe first time we encounter quantitative consequences of the Pauli principle fora macroscopically large number of fermions. Let us begin with the expression

11 In x-ray-producing devices, this is achieved by bombarding the anode with high energy(∼10 keV) electrons.


EF = (ℏ2∕2m) (3𝜋2n)2∕3 for the Fermi energy in a three-dimensional box. Theessence of this result becomes clear if, instead of the particle density n = N∕V ,we use its inverse

1n= V

N= 𝑣,

which is the “specific volume” 𝑣, that is, the volume per particle inside the box. Ifwe think of 𝑣 as the volume of a tiny cubic cell of edge length a, we have 𝑣 = a3

and the Fermi energy formula becomes

EF = ℏ2

2ma2 (3𝜋2)2∕3. (15.48)

If we now ignore the numerical factor (3𝜋2)2∕3—which is approximately equalto 10—the result (15.48) expresses the simple fact that the Fermi energy is roughlyequal to the kinetic energy that each electron would have if it were confined in thevolume available to it. We see, therefore, that the exclusion principle has effec-tively caused the sharing of available space by the existing pairs of electrons withopposite spin, and the confinement of each pair in its “own” region in space. Thisconclusion, while by no means obvious, makes sense. Even though the Pauli prin-ciple is usually expressed as the impossibility of two fermions to coexist in thesame energy state, it is also known that, due to the antisymmetric character ofthe fermionic wavefunction, two electrons cannot share the same region in spaceeither. It is thus not at all surprising that the “distribution” of electrons in the vari-ous energy states has approximately the same effect on energy as their placementin separate regions of the available volume. In the case of a metal with one con-duction electron per atom, it is clear that the volume available to each electron isthe volume of an atom, in which case the parameter a in (15.48) is on the orderof an Å, and the Fermi energy falls necessarily in the range of a few eV. And sincethe mean electron energy is basically equal to the Fermi energy EF (E = 3EF∕5),we can infer that at absolute zero, the electrons of a metal will have a kineticenergy equivalent to a temperature of a few tens of thousands of degrees kelvin!12

If the positive ions were not there to stabilize the structure, the pressure exertedby these electrons on the walls of the box would be truly colossal, on the order ofhundreds of thousands of atmospheres! You can actually obtain this estimate eas-ily if you recall that the particle density of our electron gas is roughly equal to thatof dense matter,13 that is, about a thousand times greater than the density of airunder normal conditions, while the temperature corresponding to a mean kineticenergy of a few eV is two orders of magnitude greater than room temperature (300K). As a result, the pressure of the electron gas is greater than atmospheric pres-sure by a factor of 103 × 102 = 105, reaching the order of hundreds of thousandsof atmospheres, as we asserted.

Let us stress again that this colossal pressure is not exerted outward, becausethe electrons are held together by the very intense electrostatic attraction ofthe ionic lattice. But if, instead of electrons, we had a neutral Fermi gas of thesame density, then it would be practically impossible to confine it in the box. The“natural tendency” of fermions to avoid one another is so powerful that, despite

12 Recall that 1 eV of energy corresponds roughly to a temperature of 12 000 K.13 That is, on the order of one atom per cubic Å.


the availability of a large space, they manage to confine themselves in suchmicroscopic regions that the quantum resistance to confinement (uncertaintyprinciple) manifests itself with extraordinary ferociousness. This outwardpressure is known in the literature as degeneracy pressure, or Fermi pressure, andis a purely quantum mechanical phenomenon.

The above considerations are very useful also for quick estimates of the Fermienergy in any dimension. The main idea is to start with the formula

E = ℏ2

2ma2 (15.49)

for the kinetic energy (based on the uncertainty principle) of a particle confinedwithin a region of linear size a. We can then calculate a from the relation

ad ≈ volume per particle = VN

= 1N∕V

= 1n

⇒ a ≈ n−1∕d, (15.50)

where d = 1, 2, 3 is the dimensionality of the box that contains the fermions. If weinsert (15.50) in (15.49), we finally obtain the following approximate expressionfor the Fermi energy

E ≈ EF ≈ E ≈ ℏ2

2mn2∕d.

For d = 1, 2, 3, the above relation agrees with the expressions we found earlier,except for the accompanying numerical factor that can be ignored for rough,order-of-magnitude estimates. (Recall that we did the same in obtaining esti-mates of the kinetic energy, ℏ2∕2ma2, where we did not use the exact expressionℏ2𝜋2∕2mL2 in the one-dimensional box, and also ignored the extra factor of 3 inthe three-dimensional case. Indeed, carrying over such details is a pedantic actthat runs contrary to the whole spirit of order-of-magnitude estimates.)

Another point worth commenting on is the dependence of the density of states𝜌(E) on the volume of the box and, in particular, the fact that 𝜌 is proportional tovolume. Indeed, from the relation (15.39) of the previous section we see that

𝜌(E) = Vg(E),

where g(E) is the function

g(E) = 12𝜋2

(2mℏ2

)3∕2√E ,

which denotes the number of states per energy interval and per unit volume ofthe crystal. In other words, the number Δ of states inside an energy intervalΔE is equal to

Δ = 𝜌(E)ΔE = g(E)V ⋅ ΔE,

and is thus proportional to both the energy range ΔE and the volume of the box.From a physical perspective, the proportional increase of the number of stateswith volume is reasonable. It indicates, for example, that when the volume of thesample is doubled, the number of electrons within a particular energy range isdoubled as well. This feature ensures that certain properties of the material showthe expected proportional dependence on the volume of the sample.


In the online supplement of this chapter we are going to apply the ideas of thissection to some of the most exotic forms of dense matter, namely, white dwarfsand neutron stars, to explain their successful resistance against gravitational col-lapse, and why it fails when their mass exceeds a critical value.

15.5.4 A Classic Application: The “Anomaly” of the Electronic Specific Heatof Metals

We will now discuss a simple physical problem where the consequences ofthe Pauli principle for a macroscopic collection of fermions manifest in astriking way. The problem is the contribution of electrons to the specific heat ofconductors, for which the classical prediction at room temperature is 100 timesgreater than the experimental value(!) and becomes completely unreliable attemperatures close to absolute zero. In light of our discussion so far concerningthe behavior of a Fermi gas, it is not difficult to trace the source of this “anomaly.”To understand what is going on, let us first look at the classical analysis of theproblem. The main finding is that the electrons of a conductor move insidethe crystal essentially as if they were free. Consequently—according to theequipartition theorem—each electron has a mean thermal energy equal to 3kT∕2(kT∕2 per degree of freedom) and the total energy of the electron gas is thus

𝑈 = N 32

kT , (15.51)

where N is the number of electrons in a particular crystalline sample. It followsfrom (15.51) that the specific heat under constant volume is independent oftemperature and has the constant value

CV =(𝜕𝑈

𝜕T

)

V= 3

2Nk, (15.52)

which is in stark contrast with experimental data, as shown in Figure 15.12.The quantum explanation of this paradoxical behavior is simple. As we men-

tioned earlier, the Pauli principle causes those electrons lying deep inside theFermi sea to remain “frozen” even at relatively high temperatures, for the simplereason that the thermal energy kT does not suffice to lift them above the Fermilevel and place them in available energy states. Therefore, the only electrons thatcan “heat up” (i.e., absorb thermal energy) and contribute to the specific heat ofour electron gas are those within an energy distance ΔE ≈ kT immediately below

Figure 15.12 Electronic specificheat of metals. Comparison of theclassical prediction and a typicalexperimental curve for theelectronic specific heat of metals asa function of temperature.

Cv

T

Classical theory prediction

Typical experimental curve

Nk3

2


the Fermi level. If ΔN is their number at some temperature T , then the quantumprediction for the specific heat would be smaller than the classical one by the fac-torΔN∕N , which is the fraction of electrons that truly partake in thermal motion.We can get a rough estimate of this fraction if we assume the distribution 𝜌(E) inFigure 15.9a to be constant (i.e., independent of E), in which case the ratio ΔN∕Nis equal to the ratio of the areas of two rectangular regions with equal height andwidths ΔE ≈ kT and EF, respectively. We thus find

ΔNN

≈ kTEF.

At room temperature (T ≈ 300 K ⇒ kT ≈ 1∕40 eV) and for a typical Fermienergy EF ≈ 3 eV we get

CV (quant.)CV (class.)

≈ ΔNN

≈ kTEF

≈1∕40

3≈ 1

100, (15.53)

which is exactly the order of magnitude of the ratio between the experimen-tal value and the classical prediction. Moreover, relation (15.53) reveals a lineardependence of the specific heat on temperature over a very broad temperaturerange (from absolute zero and up to thousands of degrees). Specifically, we have

CV (quant.) ≈ CV (class.) ⋅ kTEF

= 32

Nk kTEF

∼ T ,

an unequivocal prediction that is beautifully confirmed by experiment. Note alsothat if we wanted to characterize—using an equivalent temperature—the inten-sity of electronic motion due to the Pauli principle, we could use the so-calledFermi temperature

kTF = EF (Fermi temperature),which is on the order of tens of thousands of degrees, since 1 eV correspondsto 12 000 K and the Fermi energy EF for most metals is greater than 3 eV! Theelectrons of a metal at room temperature are thus immensely “hotter” (due to thePauli principle) than any blacksmith could ever hope to heat them in a furnace!

We conclude this discussion with Figure 15.13, which depicts the distributionof electrons at various energies for a temperature above absolute zero. As shownin the figure, when the material is heated, a fraction of electrons that lie in anarrow zone below the Fermi level are elevated above it. But the bulk of thedistribution—that is, electrons that lie considerably deeper than the Fermilevel—remains unaffected.

T = 0

T ≠ 0

E

EF

ρ(E)Figure 15.13 The occupancyof the states at a temperatureabove absolute zero.

Problems 465

Finally, this is a good time to examine the role of the Fermi energy in the keyproblem of conductivity. The main thing to realize here is that the speed 𝑣 in theformula 𝜏 = 𝓁∕𝑣 connecting the mean free time to the mean free path is not themean thermal speed, as the classical theory has it, but the Fermi velocity, whichis 10 to 20 times greater, as we shall now see. Indeed, since EF is typically a feweV, while the mean thermal energy at room temperature is a mere 1∕40 eV, wefind

𝑣F

𝑣th=

√EF

Eth≈√

3 eV(1∕40) eV

≈ 10.

For the mean free path we thus have𝓁 = 𝑣F𝜏 (15.54)

instead of 𝓁 = 𝑣th𝜏 , as before. And given that 𝜏 does not change—since wedecided to have it determined from the experimental value for conductivitybased on the Drude formula— formula (15.54) shows that the classical estimate𝓁(Cu) ≈ 10 Å should be revised upward by a factor of 10. That is,

𝓁 ∼ 100 Å,which is a rather large value and shows that even at room temperature, themobility of electrons—and hence, the conductivity—inside a metallic crystal isremarkably high.

The general conclusion is unequivocal. The free-electron model, althoughsimplistic, successfully predicts some important solid-state properties(for metals, at least) because it incorporates the combined action of theuncertainty and Pauli principles.

Problems

15.8 The Fermi energy of a system of noninteracting spin-1/2 particles movingin a one-dimensional box is equal to EF.(a) What will the new Fermi energy be if the number of particles in the

system is doubled?(b) How much bigger must the box get for the Fermi energy to remain

equal to EF after the number of particles in the system has beendoubled?

(c) Consider the same questions as in (a) and (b) for a two- andthree-dimensional box.

15.9 As we explained in the text, the density of states 𝜌(E) is proportional to thevolume V of a material. Therefore, the density of states per unit volume(i.e., the quantity g(E) = 𝜌(E)∕V ) is independent of V . As a result, in thefree electron model, the function g(E)will depend only on the parametersℏ, m, and E.(a) Invoke the fundamental theorem of dimensional analysis to prove

that g(E) will necessarily have the formg(E) ∼ ℏ𝛼m𝛽E𝛾 , (1)


and then show that the constants 𝛼, 𝛽, and 𝛾 will—for purelydimensional reasons—have the following values:

𝛼 = d, 𝛽 = d2, 𝛾 = d

2− 1,

where d (= 1, 2, 3) is the dimensionality of the material.(b) Use Eq. (1) to show that the relation between the mean energy E and

the Fermi energy of the system is

E = dd + 2

EF. (2)

Does the validity of Eq. (2) depend on the numerical coefficient that isimplicit in Eq. (1) but cannot be determined on dimensional grounds?

(c) As a last step in employing dimensional analysis, calculate, as a func-tion of d, the exponents in the relation

EF = ℏ𝜇m𝜈n𝜆,where n = N∕V is the material’s electron density. Is there anythingnoteworthy about your results?

15.10 The Fermi energy of electrons in a conductor is EF = 2 eV.(a) Calculate the Fermi temperature of the conductor, and then obtain

a rough estimate for the percentage of thermally excited electrons ifthe conductor is at room temperature.

(b) Suppose that an enormous “cosmic force” (e.g., the gravity of a deadstar) compresses a sample of the above material to 1∕8 of its originalvolume. What would its Fermi energy be in that case?

15.11 An astronomically large number of fermions with spin 1∕2 and mass minteract only through gravity and make up a “dead” stellar object—forexample, a neutron star—with a total mass M. Estimate its radius,R0, assuming this “object” has reached equilibrium under the actionof its self-gravity and the Fermi pressure of its particles. Your answershould be expressed as a function of M, m, and the appropriate physicalconstants.

Further Problems

15.12 To familiarize yourselves with the conversion of electric units from thecgs (esu) to the SI system, you are asked to fill in the following:

1 esu-q = ………………… C (charge)1 esu-I = ………………… A (current)

1 esu-V = ………………… V (potential)1 esu-R = ………………… Ω (resistance)1 esu-𝜌 = ………………… Ω cm (resistivity)1 esu- = ………………… V∕m (electric field)


For the above conversions, you will only need the (rounded off) value ofCoulomb’s constant, kC = 1∕4𝜋𝜖0 = 9 × 109 N m2∕C2.

15.13 Consider a (one-dimensional) semiconductor whose conduction bandranges from −7 eV to −1 eV, and whose lattice constant is a = 3 Å.Write down the dispersion relation for this band, and calculate theeffective mass of a conduction electron. What is the momentum k (inunits of 𝜋∕a) of an electron located in the middle of the conductionband? What is the corresponding form of the wavefunction describingsuch an electron in the LCAO approximation?

15.14 Let us assume that, in the LCAO approximation, the electrons of theprevious semiconductor are described by the following wavefunction:

𝜓(x) =∑

n

(

1 + i√

2

)n

𝜓n(x).

What is the momentum k (in units of 𝜋∕a) and the energy E of theseelectrons?

15.15 Consider a conductor whose Fermi energy is equal to 2.5 eV. If theconductor has a temperature of 40 K, how does the specific heat of itselectrons compare to the corresponding classical prediction? At whattemperature do the two predictions become comparable in magnitude?

469

16

Matter and Light: The Interaction of Atoms withElectromagnetic Radiation

16.1 Introduction

When matter and light, the two fundamental constituents of nature, encountereach other, exciting things can happen. Thus our principal pledge in this book—toprovide a comprehensive introduction to quantum mechanics and the structureof matter—would remain incomplete without a chapter on the interaction ofatoms with electromagnetic radiation.

The term “radiation” suggests that we treat light as it really is: a flow ofphotons—quantum particles that have also a wavelike behavior described byMaxwell’s equations. Fortunately, this quantum nature of light can be ignoredfor a broad range of applications. Since photons—unlike electrons—are bosons,they can populate in unlimited numbers a quantum state, which is thus realizedas a classical electromagnetic (EM) wave. So, when the intensity of a light beamthat strikes an atom is sufficiently high (that is, when the beam comprises amacroscopically large number of photons), we can treat the beam classically andonly use quantum mechanics to describe the atom.

The mathematical description of light–matter interaction becomes particularlysimple when the wavelength of the incident light beam is much greater thanthe size of the atom. This is surely the case for visible light, whose wavelength(𝜆 ∼ 6000 Å) is roughly ten thousand times greater than atomic dimensions(a ≈ 1 Å).

Under these conditions, we can ignore the spatial variations of the inci-dent wave inside the atom and assume that atomic electrons are subject tohomogeneous (but time-dependent) electric and magnetic fields. Furthermore,since magnetic forces on electrons are smaller than electric forces by a factorof 𝑣∕c1—and 𝑣∕c = 𝛼 = 1∕137 for the hydrogen atom—we can justifiablyignore magnetic field effects. We can thus assume that the atom is subject to ahomogeneous electric field of the form

(t) = 0 cos𝜔t,

1 We can easily derive this relation from the fact that the intensities of the electric and magneticfield of a monochromatic EM wave are equal in the cgs system. We thus obtain Fel. = q andFmagn. = q 𝒗

c× B ⇒ Fmagn. = (𝑣∕c)Fel. for the corresponding forces on a charged particle.


470 16 Matter and Light: The Interaction of Atoms with Electromagnetic Radiation

where 𝜔 is the frequency of the incident beam. If, now, the electric field is polar-ized along the z axis, then the force on the electron due to a potential V (z, t) is

−𝜕V𝜕z

= q(t)|||q=−e

= −e(t),which means that

V (z, t) = e(t)z.Therefore, for (t) = 0 cos𝜔t, we find

V (z, t) ≡ V (t) = e0 z cos𝜔t. (16.1)

The total Hamiltonian of the atom (say, hydrogen) then takes the form

H ≡ H(t) = H0 + V (t) ≡(

p2

2m− e2

r

)

+ e0 z cos𝜔t,

where H0 (= p2∕2m − e2∕r) is the unperturbed Hamiltonian of the system andV (t) = e0 z cos𝜔t is the time-dependent perturbation due to the interactionbetween the incident electromagnetic field and the atom.

In view of the above discussion, the physical question that arises is thefollowing: For an atom initially in a state 𝜓i, what is the probability of finding theatom in a state 𝜓f at a time t after the incidence of a light beam on it? (Here, iand f stand for initial and final, respectively.)

This question pertains to what is known in quantum physics as the problem ofquantum transitions, which is the type of problem we encounter when an elec-tromagnetic beam impinges on an atom. We are looking for the atomic transitionprobabilities from an initial state—typically, the ground state—to an excited state.

In principle, the theoretical calculation of transition probabilities is notdifficult. First, we need to solve the time-dependent Schrödinger equation

iℏ𝜕𝜓𝜕t

= H(t)𝜓 (16.2)

in combination with the initial condition

𝜓(0) = 𝜓n, (16.3)

where, for simplicity, we omit the spatial variable of the wavefunction 𝜓(r, t),and write 𝜓(r, t) ≡ 𝜓(t). We also use, again for simplicity, just one integerquantum number n to characterize the initial state 𝜓n of the atom. Clearly,𝜓n is an eigenstate of the unperturbed Hamiltonian H0 with an eigenvalue En(i.e., H0𝜓n = En𝜓n).

Given that the eigenstates 𝜓m (m = 1,… ,∞) of H0 form a complete basis, thesolution 𝜓(t) of (16.2), subject to the initial condition (16.3), can be written as asuperposition (with infinite terms in general) of the 𝜓m eigenstates, namely,

𝜓(t) =∞∑

m=1cm(t)𝜓m.

The coefficients cm can be calculated from the familiar formula

cm(t) = (𝜓m, 𝜓(t)) ≡ ∫ 𝜓∗m(r)𝜓(r, t) dV

16.2 Resonance, Scattering, Ionization, and Spontaneous Emission 471

and will in general depend on time, since the expanded function 𝜓(t) dependson time also. The physical meaning of the coefficients cm(t) is known to us. Thesquares of their absolute values provide the probabilities for the atom to be, aftertime t, in an eigenstate of quantum number m. Given that the atom was initiallyin the eigenstate n, the quantities |cm(t)|2 are nothing but the sought transitionprobabilities Pn,m(t), also denoted as Pn→m(t). We thus have

Pn→m(t) = |cm(t)|2 = |(𝜓m, 𝜓(t))|2,

and the problem of quantum transitions is fully reduced to calculating thesolution 𝜓(t) for the time-dependent Schrödinger equation.

However, it is practically impossible to find an exact solution of (16.2).Since the Hamiltonian H(t) of this problem depends on time, the method ofseparation of variables—the only general method to exactly solve linear partialdifferential equations—is no longer applicable. Nevertheless, we can always findan approximate solution, using the fact that the electric force exerted by theEM wave on the electron is much smaller than the Coulomb force exerted onit by the nucleus. We can thus treat the additional potential (16.1) as a smallperturbation to the dynamics of the atom and employ an approximate methodsimilar to that of Chapter 12. This approximate method is known as the theoryof time-dependent perturbations and will be presented at the end of the chapter,for reasons that will become apparent shortly.

Without an approximate method of calculation at hand, how can we proceed?We will follow the engineering approach. That is, first, we will make a list of thephenomena of interest—in the present case, the main physical processes that cantake place when a light beam hits an atom. Then we will identify the physicalquantities we need for a quantitative description of these phenomena. Finally, wewill return to theory to investigate how to calculate these quantities.

16.2 The Four Fundamental Processes: Resonance,Scattering, Ionization, and Spontaneous Emission

Even though we are treating the incident beam as a classical EM wave, we shouldalso remember that it actually consists of individual photons, a fact we ought totake into account when necessary. In the context of such a semiclassical treatmentof the EM field, a monochromatic beam with frequency 𝜔 comprises photons ofenergy ℏ𝜔. When one of these photons is absorbed by an atomic electron, it addsℏ𝜔 to the electron’s energy and thus raises the electron to a level that lies higherthan the original state by the same amount of energy. As a result, we obtain thethree possibilities shown in Figure 16.1, where we assume, for simplicity, thatthe electron is originally in the ground state of a hydrogen atom. The figuredemonstrates what can happen. In cases I and II, the photon has the energyneeded to transfer the electron to a real (i.e., actual) atomic state of the discreteor continuous spectrum, respectively. In contrast, in case III, the electron istransferred to an energy “level” that does not correspond to any atomic state. Andsince it cannot remain in this virtual state, the electron emits back the absorbedphoton and returns to the ground state after a very short time (on the order


I.a.: Resonant absorption

ℏω = E2 – E1

E3

WI

E2

E1

III: Scattering

ℏω < E2 – E1

II: Ionization

ℏω > WI

ω

ω

Figure 16.1 The three fundamental processes that take place when a photon hits an atomthat is initially in its ground state.

of 10−14 s). Since the re-emission takes place in an arbitrary direction, this processis the well-known scattering of light by an atom or any other quantum system.

Symbolically, we can represent the above three processes as follows.

I ∶ 𝛾 + A → A∗ II ∶ 𝛾 + A → A+ + e− III ∶ 𝛾 + A → 𝛾 + A

where A∗ is the standard notation for an atom A in an excited state.What happens when the incident photon in case I finds the electron not in

the ground state but in an excited state with energy E2? In this case, the reverseprocess, known as stimulated emission, can also take place. The process is shownin Figure 16.2, together with the familiar spontaneous decay, or spontaneousemission, which occurs without an incident photon.

Given that stimulated emission is, simply, the reverse process of resonantabsorption, we can treat both processes as different manifestations of the samephysical process, which we call resonance. Thus, we arrive at four fundamentalprocesses, three of which—resonance, scattering, and ionization—are stimulated(i.e., they are triggered by the action of an external EM field), while the fourthprocess—spontaneous emission—can also occur in the absence of an externalfield.

I.b: Stimulated emission IV: Spontaneous emission

ω ω ωω

Figure 16.2 The twofundamental processes thattake place when the atom isalready in an excited state.

16.3 Transition Rate, Effective Cross Section, Mean Free Path 473

16.3 Quantitative Description of the FundamentalProcesses: Transition Rate, Effective Cross Section, MeanFree Path

16.3.1 Transition Rate: The Fundamental Concept

Let us consider a type I or II process as shown in Figure 16.1. These processes,as well as all others mentioned in the previous section, are governed by a simpleempirical law, known as the law of constant rates.

The law of constant rates: The fraction of atoms of an irradiated sample thatundergo a certain transition per unit time is constant and characteristic of theparticular process.

That is, we haveΔN∕NΔt

= −Γ, (16.4)

where N = N(t) is the population of the initial state at time t, ΔN is the pop-ulation change in time Δt due to transitions to the final state (hence ΔN < 0),and Γ is the constant rate of the particular process. If the specific conditions in acertain process do not permit any returns to the initial state, then the differentialequation form of Eq. (16.4) is N = −ΓN , whose solution is

N(t) = N0e−Γt, (16.5)

which expresses the fact that the initial population in this case undergoes anexponential decay with time.

An obvious case where returns to the initial state are impossible is spontaneousemission. Here, (16.5) is written as

N(t) = N0e−Γspt = N0e−t∕𝜏sp ,

where Γsp is the rate at which spontaneous emission occurs and 𝜏sp = Γ−1sp is the

characteristic de-excitation time. We can easily show that the latter is equal tothe mean lifetime of the atom in the particular excited level.

Another process for which returns to the initial state are effectively excludedis the photoelectric effect (i.e., ionization). In this case, electrons are excitedto a state of the continuous spectrum, so they are described by extendedwavefunctions (roughly similar to plane waves), which facilitate the electrons’rapid removal from the vicinity of the atom.

But since returns to the initial state are the norm for resonance processes,the time evolution of electron populations cannot be governed by a purelyexponential law, but must have a more complex form, as shown in the followingexample.

Example 16.1 For the two-level system shown in the figure, the processes thattake place are resonant absorption and stimulated emission, at the same rateΓ; and spontaneous emission, at a rate Γsp. If all atoms of the sample—let theirnumber be equal to N0—are initially in the ground state, find the populations of


the two levels after time t. In particular, calculate the equilibrium populations, ifan equilibrium is indeed reached.

Γ Γ Γsp

N0 – N(t)

N(t)

Solution: For the population N(t) of the lower level, the following rate equationholds:

N = −ΓN + (Γ + Γsp)(N0 − N), (1)

where the first term on the right-hand side represents the population decreasedue to departures from this level (hence the negative sign), while the second termdescribes arrivals from the upper level (hence the positive sign). Now, departuresfrom a level are always proportional to the population of that level (i.e., the levelfrom which the corresponding arrow in the diagram is drawn). In the presentcase, the number of arrivals to the lower level—second term on the right-handside of (1)—is proportional to the population N0 − N(t) of the upper level. Theirrate is equal to Γ + Γsp, since these arrivals can occur through either stimulatedor spontaneous emission.

If we rearrange terms in the standard form

N = −(2Γ + Γsp)N + (Γ + Γsp)N0 (2)

we realize that (2) is really just a first-order, linear, inhomogeneous differentialequation, which can be solved easily2 —together with the initial conditionN(0) = N0—to obtain

N(t) =ΓN0

2Γ + Γspe−(2Γ+Γsp)t +

(Γ + Γsp)N0

2Γ + Γsp,

which is graphically sketched in Figure 16.3.Note that the equilibrium population

N∞ = limt→∞

N(t) =(Γ + Γsp)N0

2Γ + Γsp

is greater than N0∕2, as it should. (Why?) Moreover, for Γsp = 0, we haveN∞ = N0∕2, which is also an expected result. Note, finally, that if we were mainlyinterested in the equilibrium population, we could obtain it readily by settingN = 0 in (1) or (2), and solving for N (≡ N∞).

2 We simply add to the general solution of the homogeneous equation any particular solution of thefull equation. The simplest such solution is a constant. What is its value?


N(t)

t

N0

N0/2

N∞

Figure 16.3 Time evolution of the ground-state population for a two-level system, whichundergoes resonant absorption and stimulated emission at rate Γ and spontaneous emissionat rate Γsp.

16.3.2 Effective Cross Section and Mean Free Path

We will now show how to use the rate Γ of a process to construct two newphysical quantities—the so-called effective cross section 𝜎 and the mean free path𝓁—that are more useful from an experimental perspective. First, the effectivecross section expresses, in the form of an appropriate area, the extent of thetarget atom as viewed by an incident photon, for the process of interest tooccur. A large effective cross section means that the process is very probable,and vice versa. As we will now see, in order to ascribe to the (effective) crosssection the above physical meaning—note we often omit the word “effective” forbrevity—we must define it through the relation

𝜎 = ΓF, (16.6)

where Γ is the rate of the process3 and F is the flux of photons in the incidentbeam. For a beam of photons with energy 𝜖, this flux is related to the beamintensity I—that is, the amount of power that crosses a unit area perpendicularto the direction of the beam—via the expression

F = I𝜖.

Clearly, F measures the number of photons per unit time that cross a unit areaperpendicular to the direction of the light beam. We can easily see that thequantity 𝜎 in (16.6) has, indeed, the physical dimension of an area, and providesa geometrical measure for the probability of a particular photoreaction. Thereasoning behind the definition (16.6) is explained in Figure 16.4.

If the photons of the beam propagate through a medium (e.g., a gas) with adensity of n atoms per cm3 and each atom has a cross section 𝜎, then—accordingto a similar discussion in Section 1.3.2—the mean free path of the photons (i.e.,

3 All processes under discussion can be described as reactions of the form

𝛾 + A → “something,”

where A represents an arbitrary atom or a molecule. The specific nature of the reaction “products”does not affect our discussion here.


I σ Figure 16.4 The cross section of a process. Theeffective area 𝜎 through which the atom collects thenecessary “reaction” energy of one photon in thecharacteristic time 𝜏 = Γ−1 of the particular process.This implies thatI ⋅ 𝜎 ⋅ 𝜏 = 𝜖 ⇒ 𝜎 = 𝜖∕(I ⋅ 𝜏) = 1∕((I∕𝜖) ⋅ Γ−1) = Γ∕F

σ

Δ

ΔI

I

I= −ΔS

S

S

= −σ · n · Δx · SS

= −σnΔx

⇒ I (x) = −σnI

⇒ I(x) = I0e−nσx def

= e−

⇒ =1

nσ

Figure 16.5 Proof of the exponential decay law I(x) = I0e−x∕𝓁 . By the time the light beam hastraversed a slice of thickness Δx and area S of the material, the fractional decrease ΔI∕I of itsintensity must be equal to the fractional area ΔS∕S “blocked” by the cross sections of atomscontained in this slice.

the mean distance they travel before they “hit” an atom and trigger the processof interest) is equal to

𝓁 = 1n𝜎. (16.7)

We can derive a more precise experimental meaning for 𝓁 by examining how theintensity I of a light beam decays as the beam enters a material of density n, where-upon it interacts with the atoms and loses some of its photons. As explained inFigure 16.5, the intensity I(x) of the beam decays according to the exponential law

I(x) = I0 e−x∕𝓁 ≡ I0 e−𝜇x (𝜇 = 𝓁−1),

whence we can see that the mean free path is simply the characteristic decaylength, or, equivalently, the inverse of the decay coefficient𝜇, which can be readilydeduced from experiment.

16.3.3 Scattering Cross Section: An Instructive Example

We know that visible-light photons have energies of the order of 2 eV—rangingfrom 1.6 to 3 eV, actually—and that the first excitation energy for the vastmajority of molecules in the atmosphere is greater than 3 eV. Therefore, assolar light propagates in the atmosphere, the only process that can take place isscattering. We will now show that a mere glance at the daytime cloudless sky,which shows a bright solar disk and a much weaker diffuse light throughout the


atmosphere, provides sufficient information for an estimate of the effective crosssection of light scattering by molecules in the sky. Namely, we will show that

𝜎 ≈ 10−26 cm2.

The idea is straightforward. From the picture of the sky we just described, itfollows that the mean free path 𝓁 of solar light in the atmosphere must beon the order of the atmosphere’s height H . If the mean free path were muchsmaller than H (𝓁 ≪ H), solar light would have diffused completely by the timeit reached the ground, and the sky would look entirely different: The wholeatmosphere would be “bathed” in diffuse light, while the solar disk would hardlybe visible. (Just to be clear, what we have in mind here is the portion of theatmosphere that contains most of its mass and can thus cause scattering—thatis, the troposphere, whose height is roughly 10 km.)

If, in contrast, the mean free path were much greater than the atmosphere’sheight (𝓁 ≫ H), then the scattering of solar light during its full journey to theground would be negligible. The sky would look similar to what the astronautssee from space: a blinding solar disk and almost sheer darkness everywhere else.

We thus infer from the observed view of the sky that the mean free path of solarlight in the atmosphere (i.e., troposphere) is on the order of its height, namely,

𝓁 ≈ H ≈ 10 km.Given now that 𝓁 = 1∕n𝜎 and n = natm ≈ 1020 cm−3, we have

𝜎 ≡ 𝜎sc =1

n𝓁= 1

1020 cm−3 × 106 cm≈ 10−26 cm2.

The result is spectacular. To solar light, the atoms or molecules of the atmosphereappear as if they were 100 000 times smaller than their actual size. Their effectiveradius is R ≈

√𝜎 ≈ 10−13 cm, which is on the order of a nuclear radius.

In retrospect, this huge difference between the geometrical and effective crosssections should come as no surprise. As we stressed earlier, the effective crosssection is a measure of the likelihood of a certain process, so it depends largelyon the nature of this process and how easily it can take place for a particularatom or molecule. In general, light scattering by an atom is an unlikely process,for the simple reason that the atom is electrically neutral and cannot thereforeinteract with the electromagnetic field of incident light—especially when thewavelength of light is thousands of times greater than the atomic radius, thusdiminishing the ability of light to “see” the internal structure of an atom and getscattered by its charged constituents. So we should expect a vanishing value forthe effective cross section to begin with. Then, on second thought, we wouldimprove on this expectation, because the wavelength of light, although large, isnot infinite, so the photons have a finite resolving power after all. If this analysisis correct, then scattering should be more pronounced for shorter wavelengths,and this is actually observed experimentally. The blue color of the sky is preciselydue to the fact that blue photons, having shorter wavelengths, undergo strongerscattering than red photons.

Evidently, the effective cross section describes the interaction of light withmatter in a very tangible way. So it is worth examining how to theoreticallycalculate it for one of the basic processes, namely, resonance.


Problems

16.1 A gas composed of atomic hydrogen is bombarded by a ray of photonswith wavelength 𝜆 = 800 Å. After time t = 10 ns, it is observed that halfthe atoms in the gas have been ionized. Calculate the rate Γ of this process,and then determine the photon mean free path if the particle density of thegas is n = 1010 cm−3.

16.2 An atom in its first excited state at t = 0 is subjected to an externalelectromagnetic field, which causes resonant transitions with rate Γbetween the ground state and the first excited state, while spontaneousemission with rate Γsp also occurs. What is the probability of finding theatom in the first excited state after time t?

16.3 An external EM field causes resonant absorption and stimulated emissionwith rate Γ between levels 1 and 2, as in the following figure. At the sametime, spontaneous emissions take place at the 2 → 3 and 3 → 1 transitionswith rates 𝛾1 and 𝛾2, respectively. (The spontaneous emission at the 2 → 1transition can be ignored.)2

3

1

Γ

γ1

γ2

Γ

(a) Write the set of differentialequations satisfied by the popu-lations N1(t), N2(t), and N3(t) ofthe three levels at time t. Doesyour set of equations conservethe total number of particles?

(b) Based on the above set of equations and the initial conditions

N1(0) = N0, N2(0) = N3(0) = 0

calculate the populations of the three levels after infinite time, i.e.,when equilibrium has been reached. Examine the special case whenall rates Γ, 𝛾1, and 𝛾2 are equal.

16.4 According to the law N(t) = N0 e−Γspt , the probability of finding theelectron at the first excited state after time t is proportional to theexponential e−Γspt . If we treat the lifetime as a (continuous) statisticalvariable, its distribution will thus have the form P(t) = A e−Γspt , where Ais an appropriate normalization factor. Find A, and calculate the averagelifetime 𝜏sp = ⟨t⟩ to show that it is equal to Γ−1

sp .

16.4 Matter and Light in Resonance. I: Theory

16.4.1 Calculation of the Effective Cross Section: Fermi’s Rule

As we saw in the preceding discussion, the most interesting phenomenologicalquantities for the description of quantum transitions are not the time-dependent

16.4 Matter and Light in Resonance. I: Theory 479

transition probabilities Pn→m(t) but the corresponding transition rates Γn→m,which are constant. We also realized that we can use the rate Γ of a certaintransition to define physical quantities—such as the scattering cross section orthe mean free path—that can be readily deduced from experimental data orempirical facts.

Clearly, therefore, the main objective of a theory of quantum transitions is thecalculation of the rates Γn→m for processes that take place when light interactswith atoms. As a typical example, we will examine the resonance process ofFigure 16.1. Our question is how to calculate from first principles (i.e., basedonly on the Schrödinger equation) the rate of a resonant transition, such as the1s → 2pz in the hydrogen atom. Once again, we will use the engineering approachto address this question. That is, we will take as granted a known formula for Γand try to understand it as much as necessary to solve the problem of interest.In our case, this formula is known as Fermi’s rule, or even Fermi’s golden rule,to underscore its wide applicability and remarkable practical importance for avariety of problems. The formula is the following:

Γ = 2𝜋ℏ|𝑈fi|

2𝜌(Ef ), (16.8)

where we set Γi→f ≡ Γ for simplicity. Our problem here is to calculate theeffective cross section for a transition such as 1s → 2pz in the hydrogen atom.First, let us explain the meaning of the various symbols on the right-hand sideof Eq. (16.8). Here, i and f denote the initial and final states of the transition,respectively. That is, 𝜓i = 𝜓1s and 𝜓f = 𝜓2pz

in our case. Upon recalling thenotation of the previous chapter—where 𝜌(E) denoted the density of states inthe vicinity of an energy E—we realize that 𝜌(Ef ) in (16.8) represents the densityof states in the vicinity of the final state. But what could the term “density ofstates” mean for a resonant transition where the destination level is unique? Theanswer is unexpectedly simple. The final state may indeed be single—𝜓2pz

inour case—but it extends over an energy range ΔE, which is equal to the naturallinewidth owing to the energy–time uncertainty relation for an excited level. Wethus have “one state within the energy interval ΔE,” so the density of states 𝜌(Ef )in (16.8) must be equal to

𝜌(Ef ) =1ΔE

. (16.9)

Actually, it is better to write 𝜌 ≈ 1∕ΔE, since it is customary not to preciselydefine ΔE but only as an estimate.

What is the meaning of the quantity 𝑈fi? Based on the symbols used in(16.8)—remember that 𝑈 usually denotes energy or potential—it is evidentthat 𝑈fi are the matrix elements (between the initial and final states) of theperturbation 𝑈 associated with the action of the external EM field on theatom. We saw earlier that—for an electric field polarized along the z axis—thisperturbation takes the form

V (z, t) ≡ V (t) = e0 z cos𝜔t. (16.10)

Clearly, the operator 𝑈 in (16.8) cannot be equal to V (t), because in that case,the matrix elements 𝑈fi ≡ Vfi(t) would depend on time, and the rate Γ would not


be a constant, as it should be. Could 𝑈 then be the time-independent part of theoperator (16.10), that is, e0z? Actually, 𝑈 is half this quantity, as we will explain.That is,

𝑈 = 12

e0 z. (16.11)

The qualitative explanation of the coefficient 1/2 is interesting and worth dis-cussing. Given that the operator of (16.10) “triggers” the transition, let us examinehow it acts on the (time-dependent) wavefunction

𝜓i(t) = e−iEit∕ℏ𝜓i

of the initial state. But first, we must write the operator in the complex form

V (t) = e0 z e−i𝜔t + ei𝜔t

2= 1

2e0 z e−i𝜔t + 1

2e0 z ei𝜔t (16.12)

as appropriate for time-dependent problems in quantum mechanics. We obtain

V (t)𝜓i(t) =12

e0 z 𝜓i e−i(Ei+ℏ𝜔)t∕ℏ + 12

e0 z 𝜓i e−i(Ei−ℏ𝜔)t∕ℏ.

The main feature of this expression is the appearance of two time exponentialswith “energies” Ei + ℏ𝜔 and Ei − ℏ𝜔, which correspond to the incident photonenergy ℏ𝜔 being “added” to or “subtracted” from the initial state, respectively.This “addition” is possible when the electron is initially in the ground state,whereby we have resonant absorption, while the “subtraction” may occur whenthe electron already resides in the excited state of the transition, whereby stimu-lated emission is induced. The decomposition of the interaction operator (16.12)in two terms—one with a positive and one with a negative frequency—hasthus a straightforward physical interpretation. One term is responsible forresonant absorption and the other term accounts for stimulated emission. Thesetwo reverse processes make up the phenomenon we called resonance. Surelyit is no coincidence that the corresponding operators (V+ = 1

2e 0zei𝜔t and

V− = 12e 0ze−i𝜔t) are hermitian conjugates of each other.

Now that the origin of the “one half” coefficient in (16.11) has been explained,the rest is merely algebra. For an arbitrary resonant transition 𝜓i → 𝜓f we have

𝑈fi = 12

e0 zfi ≡ 12

e0 (𝜓f , z𝜓i),

in which case Fermi’s rule yields

Γ = 2𝜋ℏ

||||

12

e0 zfi||||

2 1ΔE

= 𝜋

2e22

0

ℏΔE|zfi|

2 (16.13)

and the only remaining task is to calculate the matrix element zfi of the positionoperator z between the initial and final states of the transition.

But instead of Γ, we will focus on the resonance cross section 𝜎R = Γ∕F . Beingindependent of the intensity of the incident beam—and hence of the electric fieldintensity—𝜎R is an intrinsic characteristic of the atom undergoing the particular


transition. To calculate 𝜎 ≡ 𝜎R, we need, apart from Γ, the flux density F of theincident beam. We have

F =intensity of the beam

energy of photon= I𝜖=

c(20∕8𝜋)𝜖

=c2

0

8𝜋𝜖. (16.14)

Here, we have expressed the energy flux I—that is, the intensity of the beam—viathe general formula j = 𝜌𝑣, which yields the flux density—or simply the flux—ofany physical quantity distributed in space with a density 𝜌 and traveling with a(local) speed 𝑣. As an electromagnetic wave, light travels with speed c, while itsenergy density in space—in the cgs system, where = B for a monochromaticwave—is

u = 2(t)8𝜋

+ B2(t)8𝜋

= 2(t)4𝜋

=2

0

4𝜋cos2𝜔t.

The time-average of u over one period is

u =2

0

4𝜋cos2𝜔t =

20

8𝜋,

since cos2𝜔t = 1∕2. So, the energy flux (i.e., intensity) of the incident beam isu ⋅ c = c 2

0∕8𝜋, and hence the flux density of photons is indeed given by (16.14).If we now insert (16.13) and (16.14) into expression 𝜎 = Γ∕F , we obtain the

resonance cross section

𝜎R = 4𝜋2𝛼 |zfi|2 𝜖

ΔE, (16.15)

where 𝛼 = e2∕ℏc is the well-known fine structure constant.

16.4.2 Discussion of the Result: Order-of-Magnitude Estimates andSelection Rules

The result (16.15) is interesting for a number of reasons. First, it confirmsthe expected independence of the resonance cross section from the intensityof the incident beam, and hence from the intensity of its electric field. Suchan independence was expected, because both Γ and F—recall the definition𝜎 = Γ∕F—are proportional to 2

0 . Therefore, 𝜎 must be a geometric quantitythat depends only on those intrinsic characteristics of the atom that pertain toeach particular transition. The reason why 𝜎 was defined through the formula𝜎 = Γ∕F is now evident: to obtain a quantity that is independent from theintensity of the incident beam.

Actually, the most interesting aspect of (16.15) is its order-of-magnitudeprediction for 𝜎. This can be easily obtained if we realize that the matrix elementzfi—a quantity with dimensions of length—should have a value on the order of

zfi ≈ a0, (16.16)

since the Bohr radius a0 is the only “length” we can construct from the physicalconstants related to the atom. (We will provide more details for this estimateshortly; but for now, let us just hold on to the above purely dimensionalargument.)


Using (16.16), and setting 4𝜋2𝛼 ≈ 1∕3 ≈ 1 in (16.15), we obtain the roughestimate

𝜎R ≈ a20𝜖

ΔE, (16.17)

which says that the resonance cross section is greater than the atom’s geometriccross section a2

0 by a factor equal to the ratio of the energy of the incident photon𝜖 and the natural width ΔE of the excited level of the transition. And since weusually have 𝜖 (≡ E2 − E1) ≈ a few eV and ΔE ≈ 10−7 eV, (16.17) yields

𝜎R ≈ (106 − 107)a20,

which is a remarkable result. It stipulates that, in resonance, the effective targetarea presented by the atom to the incident photon is a few million times greaterthan its geometric cross section! It is as if the atomic radius expands a thousandtimes to “capture” the incident photon. In contrast, when the incident photonis not in resonance with any of the excited levels, the atomic radius “shrinks” upto a hundred thousand times to “avoid” the photon! As a result, the cross section𝜎(𝜖) must exhibit enormous fluctuations as a function of the energy 𝜖 of theincident photon. The effect is shown in Figure 16.6, where the vertical axis is inlogarithmic scale.

What remains is to explain in more detail our estimate for the matrix elementzfi (zfi ≈ a0) and the conditions under which it applies. For this purpose, we will

σ (E)

σ R ≈ 10–10 cm2

σ sc ≈ 10–26 cm2

σ p ≈ 10–16 cm2

E1 E2 E3 WI

E

Figure 16.6 Effective cross section of the atom–photon interaction as a function of incidentphoton energy. The effective cross section skyrockets to high values in the immediate vicinityof atomic bound states (resonances), while it almost vanishes away from them, in which casethe (very weak) process of scattering occurs (hence the index “sc”). Only near the ionizationthreshold the effective cross section coincides, roughly, with the geometric cross section ofthe atom. Beyond this threshold, there is the continuum of unbound states and the processthat takes place is the ionization or photoelectric effect, hence the index ‘p’ (photoelectric) inthe corresponding cross section. Note also that resonances become gradually weaker andbroader. Can you explain why?


make use of the relation

zfi = ⟨z⟩𝜓=(𝜓i+𝜓f )∕

√2, (16.18)

which tells us that the matrix element zfi is equal to the average value of z in thesuperposition state𝜓 = (𝜓i + 𝜓f )∕

√2, which results from an equal-weight linear

mixing of the two participating states. We can prove (16.18) very easily usingthe definition zfi = (𝜓f , z𝜓i) = ∫ 𝜓∗

f z𝜓i dV , once we consider that the atomiceigenfunctions are always odd or even, so the so-called diagonal matrix elementsof z, zii ≡ ⟨z⟩𝜓=𝜓i

= ∫ |𝜓i|2z dV and zff ≡ ⟨z⟩𝜓=𝜓f

= ∫ |𝜓f |2z dV vanish. (We

encourage the readers to complete the proof along these lines.)The qualitative conclusion of (16.18) is the following. While for the atomic

eigenfunctions𝜓i and 𝜓f , the mean value of z vanishes because of their reflectionsymmetry (i.e., r → −r), for a superposition of states with different symmetry(even and odd), the mean position of the electron is shifted off the origin.Naturally, this mean shift ⟨z⟩ cannot be much different than the size of thesuperimposed orbitals. In particular, for the first few eigenstates of the hydrogenatom—whose spatial extent is roughly 2–3 Bohr radii—the mean shift ofthe electronic cloud when their superposition is formed must also be of thesame order of magnitude (i.e., one Bohr radius). A typical example of such asuperposition is shown in Figure 16.7 for the 𝜓1s and 𝜓2pz

states in the 1s → 2pztransition of the hydrogen atom.

Even though the above estimate zfi ≈ a0 is plausible and has a simple physicalexplanation, we should not forget that it is valid only for the first few eigenstatesof the atom. We will return to this point later.

16.4.3 Selection Rules: Allowed and Forbidden Transitions

We saved for the end of this section the important case of a vanishing matrixelement zfi, whence the transition i → f does not even take place. It is a forbiddentransition, at least within the approximations we have employed. An obvious class

++ =

= +1

z α0≈

z

2

+ +

–

–

ψ1sψψ2pz

ψ1s ψ2pz

= 0 z = 0 z ≠ 0

Figure 16.7 Qualitative estimate of the matrix element zfi. The superposition of the two statesinvolved in the 1s → 2pz transition produces an asymmetric “hybrid” for which the meanposition of the electron is no longer at the origin, as was the case for each of the two states ofthe transition that had mirror symmetry. The mean displacement of the electronic cloud dueto superposition is of the same order as the spatial extent of the participating orbitals, andhence comparable to a Bohr radius

(⟨z⟩ ≈ a0

).


of forbidden transitions involves pairs of initial and final states with the sametype of symmetry (i.e., states that are both even or odd).4 The integrand in theexpression

zfi = ∫ 𝜓∗f z 𝜓i dV

is then odd, since the product 𝜓∗f 𝜓i is necessarily even, while z is odd. As a result,

the integral over all space vanishes. We can also show that for an arbitrary tran-sition n𝓁m → n′𝓁′m′, the matrix element zfi ≡ zn′𝓁′m′;n𝓁m is nonzero only if thefollowing selection rules apply:

Δ𝓁 = ±1, Δm = 0. (16.19)

The second part of (16.19) is easily shown if we recall that the dependence of thewavefunctions on angle 𝜙—the familiar exp (im𝜙) factor—is determined solelyby the quantum number m. Therefore, and due also to the separation of variablesin the eigenfunctions,5 the matrix element zfi

zfi = zn′𝓁′m′; n𝓁m = ∫ 𝜓∗n′𝓁′m′ z 𝜓n𝓁m dV

can be written as a product of three integrals (with respect to r, 𝜃, and𝜙), namely,

zfi = ∫(Rn′𝓁′ (r)P𝓁′m′ (cos 𝜃)eim′𝜙

)∗r cos 𝜃(Rn𝓁(r)P𝓁m(cos 𝜃)eim𝜙) r2 sin 𝜃drd𝜃d𝜙

=(

∫∞

0R∗

n′𝓁′Rn𝓁r3 dr)(

∫𝜋

0P∗𝓁′m′P𝓁m cos 𝜃 sin 𝜃 d𝜃

)

(

∫2𝜋

0(eim′𝜙)∗(eim𝜙)d𝜙

)

,

whence it is evident that the selection rule for m is solely determined by the inte-gral over 𝜙

∫2𝜋

0(eim′𝜙)∗eim𝜙 d𝜙 = ∫

2𝜋

0ei(m−m′)𝜙 d𝜙, (16.20)

which is nonzero only if m′ = m (⇒ Δm = 0), as the readers can confirm,either via a direct calculation, or by recalling the orthogonality theorem foreigenfunctions—here, of the operator 𝓁z—with different eigenvalues.

So far, we have assumed for simplicity that the polarization of the incidentEM wave—that is, the direction of its electric field—is along the z axis. But this

4 When a wavefunction is even or odd, we say it has definite parity. Specifically, it has parity +1 if itis even and −1 if it is odd. For any central potential, the eigenfunctions have a definite parity (sincethe potential remains invariant under the change of r to −r), which is solely determined by thequantum number 𝓁, and is given by (−1)𝓁 . States with even 𝓁 are even and states with odd 𝓁 areodd. (We encourage the readers to show this.)5 We recall that 𝜓n𝓁m(r, 𝜃, 𝜙) = Rn𝓁(r)P𝓁m(cos 𝜃)eim𝜙, where P𝓁m(cos 𝜃) are the associated Legendrepolynomials that depend only on 𝜃, not on 𝜙.


need not always be the case. For an arbitrary direction, the previous expressionV (z, t) = ez(t) for the interaction energy can be generalized to

V = e r ⋅ (t) ≡ −d ⋅ (t), (16.21)

where d = −er is the dipole moment of the atom in the standard convention,which has the vector pointing from the negative (electron) to the positive (pro-ton) charge, hence the negative sign. Actually, (16.21) is the familiar expression ofthe interaction energy between a dipole and an electric field. In the special caseof a monochromatic wave, we have (t) = 0 cos𝜔t, and (16.21) takes the form

V = −d ⋅ 0 cos𝜔t,

so the operator 𝑈 in Fermi’s rule becomes

𝑈 = −12

d ⋅ 0 = 12

e r ⋅ 0 .

If the electric field of the incident wave is polarized along the x axis, we have0 = x0, so that

𝑈 = 12

e 0x,

which is directly analogous to the expression 𝑈 = e 0z∕2 we had from equation(16.11). The only difference lies in replacing z with x or y, when the EM fieldis polarized along the x or y axis, respectively. And the only practical conse-quence of this replacement is that the selection rule for m includes now thecase Δm = ±1, as is evident from the expressions for the x and y operators inspherical coordinates

x = r sin 𝜃 cos𝜙 = r sin 𝜃 ei𝜙 + e−i𝜙

2,

y = r sin 𝜃 sin𝜙 = r sin 𝜃 ei𝜙 − e−i𝜙

2i,

which depend on 𝜙 through the exponential terms ei𝜙 and e−i𝜙 that correspondto m values ±1. The integral over𝜙 in the matrix element xfi ≡ xn′𝓁′m′;n𝓁m includesnow—in addition to the eigenfunctions eim′𝜙 and eim𝜙 of 𝓁z—the dependence ofx on 𝜙, so it takes the form

∫2𝜋

0(eim′𝜙)∗ ei𝜙 + e−i𝜙

2eim𝜙 d𝜙,

which, as one can easily show, is nonzero only for m′ = m ± 1, that is, Δm = ±1,as stated above. So, for an arbitrary polarization state of the incident EM wave,the only pairs of states involved in electronic transitions are defined by theselection rules for m, namely, Δm = 0,±1 so the rules (16.19) are now written as

Δ𝓁 = ±1, Δm = 0,±1, (16.22)

and are, of course, identical to those we listed earlier in this book for spontaneoustransitions. At that point, we had explained these selection rules as a consequenceof the conservation of angular momentum. The present analysis includes theproof of the m selection rules but also clarifies the conditions under which both


the 𝓁 and m rules apply: The wavelength of the incident light must be muchgreater than the dimensions of the atom. That is,

𝜆 ≫ a0. (16.23)When this condition is fulfilled, the atom is picked out by the EM wave as anelectric dipole and their mutual interaction has the so-called electric-dipole form(16.21), where the electric field (t) = 0 cos𝜔t of the wave is taken as constantwithin the atom, since—according to (16.23)—its spatial variations are impercep-tible there. The above description is known, for obvious reasons, as the electricdipole approximation, and the corresponding allowed transitions—that is, thosethat obey the selection rules (16.22)—are known as electric-dipole transitions.Transitions that do not satisfy these rules are not absolutely forbidden; they justhave a much smaller probability to occur compared to dipole transitions.

Note that a formal proof of the selection rule Δ𝓁 = ±1 in (16.22) has not beenprovided so far, for the following reasons:(a) The proof is not so easy, even though the general idea remains the same. To

furnish a proof, we have to show that the matrix elements xfi, yfi or zfi arenonzero only if the quantum numbers 𝓁i and 𝓁f of the initial and final statesof the transition differ by one; that is, Δ𝓁 = |𝓁f − 𝓁i| = 1.

(b) The selection rule Δ𝓁 = ±1 is, nevertheless, a plausible consequence ofΔm = 0,±1, once we realize that this triplet of m values is naturally asso-ciated with a quantum number equal to one for the magnitude of the Δ𝓵vector.

We conclude the present discussion with a pertinent example.

Example 16.2 Calculate the matrix element zfi for the 1s → 2pz transition ofthe hydrogen atom and confirm that the result is consistent with the order-of-magnitude estimate we found earlier for this quantity. The (normalized)wavefunctions of the states involved in the transition are

𝜓1s =1

√𝜋

e−r, 𝜓2pz≡ 𝜓210 = 1

4√

2𝜋re−r∕2 cos 𝜃 (a.u.).

Solution: We have

z2pz ,1s = ∫ 𝜓2pzz𝜓1sdV = ∫

14√

2𝜋re−r∕2 cos 𝜃 ⋅ r cos 𝜃 ⋅ 1

√𝜋

e−rr2 sin 𝜃dr d𝜃d𝜙⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟

dV

= 14𝜋

√2

(

∫∞

0r4e−3r∕2dr

)(

∫𝜋

0cos2𝜃 sin 𝜃d𝜃

)(

∫2𝜋

0d𝜙

)

= 14𝜋

√2

4!(3∕2)5

(

∫1

−1𝜉2 d𝜉

)|||||

⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟

2∕3

𝜉=cos 𝜃⋅ 2𝜋 =

27√

235 a.u. =

27√

235 a0.

For the numerical value we findz2pz ,1s = 0.745 a0,

which agrees with the order-of-magnitude estimate we gave earlier.

16.5 Matter and Light in Resonance. II: The Laser 487

As for the effective cross section 𝜎R of this particular process, you can calculateit based on the experimental value of 1.52 × 10−9 s for the lifetime 𝜏sp of the 2pzlevel.

16.5 Matter and Light in Resonance. II: The Laser

16.5.1 The Operation Principle: Population Inversion and the ThresholdCondition

The huge value of the resonance cross section is such a remarkable effect thatit ought to have some important application. Such an application indeed exists,but its significance is such that to call it merely an “application” is a grossunderstatement. It is a scientific and technological discovery that is comparableonly with the advent of the transistor. In retrospect, the main idea seems quitesimple. When a photon is “in resonance” with the atoms of some medium,the effective size of the atoms becomes so large that the photon interacts withthem almost instantaneously, before it transverses even a few atomic diameters.For example, even for a medium that is a thousand times less dense than air(i.e., n ≈ 1017 cm−3), the mean free path for a resonant photon is

𝓁 = 1n ⋅ 𝜎

≈ 11017 × 10−10 ≈ 10−7 cm = 10 Å,

where we considered as typical a resonance cross section that is 106 times greaterthan the geometric cross section a2

0 ≈ 10−16 cm2 of the atom. “Picking out” theatoms as a thousand times greater (in radius) than what they actually are, thephoton naturally “feels” as if it moves in a space-filling medium that is more akinto a liquid than a very dilute gas. Let us examine how we can exploit this peculiareffect.

To begin with, we already know what this resonant photon does when it hitsan atom (which is a very frequent event as noted above). If the photon finds theatom in its “ground” state6 —let E1 be its energy—it lifts it up to an excited levelwith energy, say, E2. If it finds the atom in state E2, it causes it to fall back to E1.The photon can thus induce the two reverse processes we termed previouslyresonant absorption and stimulated emission. In the former case, the initialphoton is absorbed and disappears, while in the latter case, the number ofphotons is doubled, since the stimulated emission produces a second photon inaddition to the initial one (Figure 16.8).

It is now evident that for amplification of the initial photon to occur—whichis, after all, the objective of a laser cavity—the number n2 of atoms in the excitedstate (per cm3) must be greater than the number n1 of atoms in the ground state.In other words, there must be what we call a population inversion, so that theinitial photon has a greater probability of encountering an excited atom and beamplified than finding an atom in the ground state and being absorbed. Now,population inversion contradicts the laws of thermodynamics—which stipulatethat n2∕n1 = e−(E2−E1)∕kT ⇒ n2 < n1—and cannot thus be sustained without a

6 As we will soon explain this state is not the true ground state of the system.


Fully reflectivemirror

Semi-reflectivemirror

Transparent walls

Pumping energy

Atoms in an excited state

Atoms in the “ground” state

Spontaneous emissionStimulated emission

Figure 16.8 The operation principle of a laser cavity. If there is population inversion in theactive medium of the cavity—that is, more atoms in the excited state than in the groundstate—a photon produced from a spontaneous emission can be multiplied exponentially,giving rise to a cascade of photons through stimulated emission. The process can go on onlyas long as population inversion is sustained via a continuous energy supply to the cavity,known as pumping.

continuous supply of energy. This process (i.e., the continuous supply of energyneeded to maintain inversion) is known as pumping, and is a key element for theoperation of a laser cavity, such as the one shown schematically in Figure 16.8.As for the initial resonant photons that trigger the amplification process, theyare produced by the spontaneous emission between those levels used in theoperation of the laser. Most of the photons produced in the cavity are emitted inrandom directions and get lost by escaping through the (transparent) side walls.But those (very few) photons that propagate along the axis have a considerableprobability to be significantly amplified by producing “copies” of themselves viastimulated emission from excited atoms in the cavity. And precisely in order toprolong the amplification process reflective mirrors are attached on both endsof it, one of which is a little transparent, to allow the produced beam to exit thecavity (Figure 16.8).

At this point of our discussion, it is expedient to have in mind a numericallyhandy example of a laser cavity. As such we choose a “tubule” of length L = 5 cmwith an exit mirror that has a reflectance R = 0.99 ≡ 99%. Since the transparencyof the mirror is then T = 1 − R = 0.01 ≡ 1%, a photon can escape from thecavity after impinging a hundred times on its walls, on average. So, the totaldistance the photon travels in the cavity is a hundred “back and forth” trips, thatis, 100 × 2L = 100 × 10 cm = 103 cm = 10 m. Henceforth, we will denote thiseffective length with the letter d.

We are now ready to formulate and prove the following necessary and sufficientcondition for the operation of a laser cavity.

Threshold condition: For the successful operation of a laser cavity, its inversiondensity Δn = n2 − n1 must not only be positive but should exceed a critical valueΔnc equal to

Δnc =1

d ⋅ 𝜎R, (16.24)


where 𝜎R is the effective resonance cross section for the pair of levels involved, andd is the total distance traveled by a photon in the cavity.

The proof of (16.24) is simple. The basic idea is that the propagation of a reso-nant photon inside a cavity with inversion density Δn is practically equivalent topropagation in a medium that contains only excited atoms with density Δn. Thereason for this equivalence is that atoms in the ground state simply “neutralize”an equal number of excited atoms—since all photons produced by the latter are“consumed” by the former—and what matters eventually is the excess densityΔn of excited atoms compared to atoms in the ground state. When a resonantphoton propagates in such a medium—with only excited atoms of densityΔn—its mean free path is equal to

𝓁 = 1Δn ⋅ 𝜎

, (16.25)

where we wrote 𝜎 instead of 𝜎R, for simplicity (and will do so henceforth). Wecan now derive the threshold condition in a straightforward manner. For thelaser to operate, we must have

𝓁 ≤ d, (16.26)

to allow for the initial photon to hit an excited atom and reproduce itself beforeit completes its trip of length d inside the cavity. With 𝓁 given from (16.25),(16.26) yields

Δn ≥ 1𝜎d,

which means that for the laser to function as an amplifier,7 the minimum allowedvalue of Δn is Δnc = 1∕𝜎d, as we promised to show. For our prototype cavitywith L = 5 cm, R = 0.99 ⇒ T = 0.01 ⇒ d = 1000 cm, and for 𝜎 ≈ 10−10 cm2,formula (16.24) gives a value of the critical density equal to

Δnc = 107 cm−3,

which is low enough to suggest that the pumping power needed for laseroperation is practically attainable. To estimate that power, let us recall that thecontinuous operation of the laser requires the renewal of population inversion inapproximately the time it takes the resonant photon to travel the whole distanced in the cavity. In fact, it is exactly our limited ability to maintain continuouspumping that hinders the operation of a laser with Δn values much greaterthan Δnc, and restricts it instead to inversion densities that are only marginallygreater than the critical value. For example, let us examine what happens whenΔn = 20Δnc. We then have 𝓁 = d∕20, so the initial photon impinges on theexcited atoms twenty times on average, each time producing an additionalphoton that can do the same thing. After 20 generations of self-replications, thefinal number of identical photons produced is 220. Actually, because 𝓁 is only a

7 The acronym (LASER ≡ Light Amplification by Stimulated Emission of Radiation) makes plainthat a laser is, by definition, a photon amplifier.


mean free path, the correct amplification factor G—G from “gain”—is not 220,or more generally 2d∕𝓁 , but

G = ed∕𝓁 . (16.27)

This result is readily derived from the formula I(x) = I(0)e−n𝜎x we showed earlier,with the substitution n → −Δn, which converts the exponential decay to anexponential amplification. We then find

I(x) = I(0)eΔn𝜎x = I(0)ex∕𝓁 (𝓁 = 1∕Δn ⋅ 𝜎)

⇒ G(x) = I(x)I(0)

= ex∕𝓁 ,

from which we can derive (16.27) as a special case for x = d. In our numericalexample—where Δn = 20Δnc = 2 × 108 cm−3—the final number of photons(per cm3) produced from a single initial photon is

n = e20 ≈ 1020∕2.3 ≈ 109 ≈ 5Δn.

This value implies that while the initial photon is inside the cavity, which lastst𝛾 = d∕c ≈ 3 × 10−8 s, the initial population inversion must be renewed at leastfive times, corresponding to a pumping power of n ⋅ 𝜖𝛾∕𝜏𝛾 ≈ 10 mW, the power ofa typical laser pointer. But if we sought an operating value of, say, Δn = 100Δnc,then the pumping power would skyrocket to values much greater than the poweremitted by the sun! Evidently, the operation of the laser is practically attainablefor inversion density values only a few times—about an order of magnitude,say—greater than the critical inversion density.

Note also that the threshold condition (16.24) applies only in the case of an exitmirror with reflectance very close to unity. (Why is this so?). In the more generalcase, where both mirrors are partially reflective with arbitrary reflectivities R1and R2, the laser operation condition can be written as

R1R2 e2Δn𝜎L ≥ 1, (16.28)

where the left-hand side is the total amplification factor of the beam during one“back-and-forth” trip inside the cavity. Specifically, R1 and R2 are the decay fac-tors of the beam due to partial reflection by the mirrors, and e2Δn𝜎L ≡ eΔn𝜎x|x=2Lis the amplification factor in the distance 2L due to population inversion. Thus(16.28) expresses the obvious requirement that the total amplification factorafter each “back-and-forth” trip be greater than or equal to unity. We can thenuse expression (16.28) to obtain the critical inversion density

Δnc =ln(1∕R1R2)

2L ⋅ 𝜎. (16.29)

For the special case we mentioned earlier (R1 = 1, R2 = R = 1 − T , and T ≪ 1),and given the approximate expressions for small T , ln(1 − T) ≈ −T , Eq. (16.29)leads to the simpler formula

Δnc =T

2L ⋅ 𝜎,

which agrees with our previously found result, considering that d = 2L∕T .


16.5.2 Main Properties of Laser Light

At this point, we will briefly examine whether the above principles of laseroperation can adequately explain the known properties of laser light. Let usanalyze these properties one by one.

16.5.2.1 Phase CoherenceWe already have a rudimentary understanding of this property. Contrary tocommon incandescent light bulbs, where atoms radiate spontaneously andwithout any phase relation to each other, all de-excitations in a laser cavityare stimulated and the photons produced are exact copies—in frequency andphase—of the initial photon. Consequently, all individual “wavelets” that rep-resent the emitted photons are added in phase, producing an extremely intenseand coherent EM wave.

16.5.2.2 DirectionalityIn sharp contrast to common light beams from parabolic mirrors, the direc-tionality of laser beams is spectacular. They can travel several kilometers andyet “open” up by only a few centimeters off the axis of the beam. This property,which is exploited in numerous applications, is a direct consequence of theway a laser cavity functions. In particular, a laser amplifies only those photonsthat are sufficiently aligned with the cavity axis to remain inside it, and whichtravel the whole distance d to attain the maximum possible amplification andproduce the maximum possible number of “descendants.” In contrast, thoseinitial photons that are not highly aligned with the axis diverge considerablyafter a few “back-and-forth” trips, hit the transparent side walls, and exit thecavity, leaving behind a negligible number of descendants, which will have thesame destiny. The laser cavity is thus a “ruthless” aligner of photons, constrainedonly by the diffraction limit set by the wave nature of the photon.

16.5.2.3 IntensityAside from lasers used in large industrial facilities, the beam intensity of atypical laser of continuous operation8 is unremarkable. Typically, it rangesfrom 10−3 W∕cm2 to 104 W∕cm2, and is thus comparable with the intensity of“standard” light sources, such as the sun, whose surface light intensity is of theorder of a few kilowatts per square centimeter. But the really impressive featureof a CW laser is that its intensity is emitted within an extremely narrow spectralrange, which is actually much narrower—as we will shortly see—than even thelinewidth Δf of the corresponding excited state. As a result, the energy emittedby a laser within its emission bandwidth can be up to 12 orders of magnitudegreater than the energy emitted by the sun, per square centimeter of its surface,in the same frequency range. Such huge intensities (per unit frequency interval)can occur in nature only under Big Bang conditions.

8 In the case of the so-called pulsed lasers, which do not radiate continuously but in pulses, theintensity of a particular pulse can be much greater than that of a continuous wave (CW) laser.


16.5.2.4 MonochromaticityMonochromaticity is the key feature of laser light. The spectral width of lightproduced by a laser cavity can be up to a million times smaller than thelinewidth9 of the excited state from which the laser transition occurs. There aretwo mechanisms responsible for this remarkable property. The first mechanismhas to do with the laser cavity itself, which can accommodate only those photonswhose frequencies fn = n(c∕2L) are consistent with the formation of standingEM waves inside the cavity.10 For the cavity of our example (L = 5 cm) theinterval Δf between two consecutive allowed frequencies is

𝛿f = fn+1 − fn = c2L

= 3 × 1010 cm s−1

10 cm= 3 × 109 Hz.

If we ignore Doppler broadening and assume that the excited level of the laserhas its natural width (Δf = ΔfN ≈ 107 Hz), then within this very narrow rangeof 107 Hz, there will lie at most one allowed frequency of the cavity. If noallowed cavity frequency exists in this linewidth range, then laser operation isimpossible to begin with. But if one allowed cavity frequency falls within thelinewidth range, the emitted light will be purely monochromatic—assumingwe ignore other sources of broadening (e.g., due to thermal fluctuations of thelength of the cavity). Of course, if we take Doppler broadening into account, thenatural linewidth is replaced by ΔfD ≈ 1000ΔfN ≈ 1010 Hz, and there is enoughroom in this widened spectral band for three allowed cavity frequencies. Butthe intensities of these additional frequencies are unequal, owing to the secondmechanism of laser spectral narrowing we will describe right now.

The second mechanism relates to a fact we have ignored thus far. Namely, thebroadening of the excited atomic levels due to the uncertainty principle—whichconverts them from mathematical (i.e., infinitely thin) lines to bands—isnot homogeneous throughout each “band.” Instead, it is described by a lineshape—or line profile—function 𝜌(E), a kind of fractional density of states, whichtells us what “fraction” of each single state “spreads out” over an infinitesimalenergy region around the center of the “line” of interest. The implications of thisgradual decay of the density of “states” are dramatic. Because of the dependenceof Γ on 𝜌(E)—recall that Γ(E) = (2𝜋∕ℏ)|𝑈fi|

2𝜌(E)—photons that are in resonancewith the center of the line (Figure 16.9) have a greater Γ and hence a larger 𝜎.As a result, they are amplified much more—in fact, exponentially more—as theypropagate inside the cavity, compared to those photons that are in resonancewith the tails of the line profile. Therefore, even if the mechanism of frequencyselection due to the cavity were not operative, the spectral width of laser lightwould still be spectacularly smaller than the width of the corresponding atomicstate from which the laser light originated. We summarize all the above inFigure 16.9.

Before closing this brief introduction to laser physics, let us tend to an issuethat is a common source of confusion. The readers may have erroneously

9 Actually, the linewidth ΔEN we mention here is not the natural linewidth due to the uncertaintyprinciple—for which a typical value is in the order of ΔfN ≈ 107 Hz—but the much larger width(by about a thousand times) caused by Doppler shifts due to the thermal motion of atoms at roomtemperature. The typical order of magnitude for this so-called Doppler broadening is ΔfD ≈ 1010 Hz.10 Just as in standing waves in a string, we must have L = n(𝜆∕2) = n(c∕2f ) ⇒ f = n(c∕2L).


E0

E1

E = 0

ρ(E)

ΔE

Figure 16.9 Line shape function and its role in the spectral “narrowing” of laser light. Photonsthat originate from denser regions of the line (i.e., from the line center) interact much morestrongly with the atoms of the active medium, so they are amplified much more than“peripheral photons.” For example, in the case shown here, the cross section of the “peripheralphoton” is roughly three times smaller compared to a central photon, and will thus leavebehind three fewer generations of descendants. Thus the laser tube produces light in anextremely narrow frequency band around the center of the initial line.

understood—in spite of our relevant warning earlier—that laser operationrequires only two levels: the ground state and an excited state. Actually, thisis not true. With only two levels available, population inversion would beunfeasible, because for every possible pumping mechanism, the transition rates1 → 2 and 2 → 1 are the same, and the final state of equilibrium would entailpopulation equality, not inversion. Laser operation requires the participationof at least three levels, while a setup of four energy levels is usually the mostsuitable. Figure 16.10 shows how such a laser works.

3

0

1

2Laser

transition

Pum

pin

g

Rapid

Rapid

Transition

Transition

Figure 16.10 A four-level laser. Population inversion in this setup is not only feasible but alsorelatively easy to achieve, because the levels involved in pumping (0 and 3) are different fromthe lasing levels (1 and 2). For example, a rapid “emptying”—through spontaneous emissionto the ground state—of the lower laser level facilitates the sustainment of inversion, while itdecreases considerably the pumping requirements also. Pumping is facilitated even furtherwhen “level” 3 is actually a “group of levels,” the majority of which may spontaneously decaytoward the upper lasing level.


But how do we know which spontaneous emissions are possible and which onesare not, and how slow or quick each one is? Knowing these answers we would beable to select the most appropriate material and energy levels, and “design” anefficient laser, or improve the one we have. Even though the following sectiondoes not aspire to transform readers to “laser engineers,” it will, in effect, helpthem understand the fundamental physical process of spontaneous emission.

16.6 Spontaneous Emission

We will again adopt the engineering approach, to study another basic process ofthe laser operation: spontaneous emission. Following our familiar recipe, we willstart from the “ready-to-use” formula for the rate of this process

Γsp = 4𝜔3

3ℏc3 d2fi, (16.30)

and then try to obtain a sufficient level of understanding to correctly apply it.Our experience to date should come in handy here. First, the frequency 𝜔 canonly be the Bohr frequency

𝜔 ≡ 𝜔if = (Ei − Ef )∕ℏ

of the i → f transition—from state 𝜓i to state 𝜓f —while dfi must be thematrix element of the dipole moment operator d = −er11 between these states.Specifically, we must have

dfi = −erfi = −e(xfi, yfi, zfi)⇒ d2

fi = e2(x2fi + y2

fi + z2fi),

where we assumed, for simplicity, that the wavefunctions of the initial and finalstates are real, which makes the matrix elements of the x, y, and z operatorsalso real. A straightforward application of the above discussion is given in thefollowing example.

Example 16.3 Calculate the rate of spontaneous emission for the 2pz state ofthe hydrogen atom, and the corresponding mean lifetime 𝜏sp, whose experimentalvalue is 1.597 ns.

Solution: Actually, the calculation is trivial. We already know from Example16.2 that

zfi ≡ z1s,2pz=

27√

235 a.u. =

27√

235 a0 .

The corresponding matrix elements for operators x and y vanish here, since bothstates involved in the transition have m = 0, and hence Δm = 0. (As we saw

11 The matrix element of a vector operator, for example, the position operator r, is itself a vectorwhose components are the matrix elements of the operator components. For example, we haverfi = (xfi, yfi, zfi).

16.6 Spontaneous Emission 495

earlier, the operators x and y have nonvanishing matrix elements only betweenstates with Δm = ±1.) We thus have

d2fi = e2(x2

fi + y2fi + z2

fi) = e2(

0 + 0 + 215

310 a20

)

= 215

310 e2a20

and expression (16.30) gives then

Γsp = 217

311

e2a20

ℏc3 𝜔3.

Since ℏ𝜔 = E2 − E1 = (3me4∕8ℏ2), we find

Γsp =(2

3

)8𝛼5 mc2

ℏ

and numerically

Γsp = 0.626 × 109 s−1.

The corresponding mean lifetime is

𝜏sp = Γ−1sp = 1.597 × 10−9 s,

in perfect agreement with the experimental value.

But the most important feature of formula (16.30)—the one we should keepin mind for a quick qualitative understanding of spontaneous emission—isthe third-power dependence of the decay rate on the Bohr frequency of thetransition, or, equivalently, on the energy difference of the corresponding energylevels. This feature, combined with the fact that at least for the first few excitedstates of most atoms the dfi matrix elements are all of the order of ea0, leads usto conclude that the key factor determining the order of magnitude for the rateof a spontaneous emission is the energy difference between the relevant levels.The transition rate is proportional to the third power of the energy differenceand the lifetime is inversely proportional to the third power of this difference.As a result, if an excited atomic electron has to choose between two (or more)allowed transitions, it will most likely choose the one with the highest energydifference, with a probability that is proportional to the third power of the ratioof the corresponding energy differences. Figure 16.11 illustrates all this with anexample.

We can also use the “cubic-power law” to quickly estimate the rates of sponta-neous emission or the corresponding lifetimes of excited states associated withnon-electronic degrees of freedom. A typical example is the excited rotationalstates of diatomic molecules, where the energy gaps between levels are—as wesaw in Chapter 13—at least a thousand times smaller than the correspondingelectronic energy gaps. For such energy differences, the spontaneous emissionrates and lifetimes are up to (1000)3 ≈ 109 times smaller and greater, respectively.Therefore, if the typical lifetimes of excited electronic states are on the orderof 10−8 s, the corresponding lifetimes of excited rotational states, for smalldiatomic molecules, are on the order of 10−8 s × 109 ≈ 10 s. Rotational states arelong lived precisely because they are so closely packed together.


1s

2s

3s

2p

3p 3d1.9 eV

12.1 eV

Figure 16.11 The “cubic-power law” of spontaneous emissions. Atomic electronsprefer—their preference obeying a cubic-power law—those quantum leaps that have thehighest energy difference. In the case shown here, only about four out of a thousand electronsprefer the small leap (3p → 2s) over the large one (3p → 1s).

The above arguments might create the misleading impression—if we forgetour basic assumption that |dfi| ≈ ea0 ≈ constant—that large leaps are alwayspreferred over smaller ones. Surely this is not the case for highly excited atomicstates—the so-called Rydberg states—where the requirement for the correctclassical limit (Bohr’s correspondence principle) mandates that transitions toneighboring energy levels are far more likely than to more distant ones. Forexample, the transition from level n = 50 to n = 49 is much more preferredover the transition to n = 1, so the emitted photon has the classically expectedfrequency. Now, the reason for this change of preference lies with the matrixelement dfi, which becomes negligible when one of the transition states lies veryhigh in the spectrum and the other very low, which in turn causes an almostvanishing overlap of the corresponding wavefunctions. For example, considerthat the wavefunction of a Rydberg state of very high n (and 𝓁 = n − 1) looks inessence like a huge orbital tube—something like a bicycle inner tube—aroundthe corresponding Bohr trajectory (rn = n2a0). In contrast, the wavefunction ofthe ground state looks like a tiny sphere at the center of the atom. Since these twowavefunctions have practically zero overlap, the corresponding matrix elementdfi is negligible and it is almost impossible for the transition to occur.

Having obtained some understanding for spontaneous transitions, can wego one step further and seek some kind of proof for formula (16.30)? We aretalking about “some kind of proof,” and not actually a proof , because, in reality,spontaneous emissions cannot even take place without the quantization of theEM field,12 which renders impossible their full quantitative description withoutthe prior development of a quantum theory for the EM field, also known asquantum electrodynamics.

12 One can understand this point from the time dependence 𝜓n(r, t) = 𝜓n(r)e−iEnt∕ℏ of the atomiceigenstates predicted by the usual form of the Schrödinger equation, where only the electrostaticinteraction between electrons and the nucleus has been taken into account. This form does notallow any time variation in the distribution of the electronic cloud, so it cannot allow for any kind ofradiation. Therefore 𝜏 = ∞ and ΔE = 0, in agreement with the fact that we are talking about energyeigenstates.

16.6 Spontaneous Emission 497

Nevertheless, we can arrive at “some kind of proof” of formula (16.30) if webegin with the classical relation

P =𝜔4d2

0

3c3 (16.31)

for the time average of the power emitted by an electric dipole that performsa harmonic oscillation at frequency 𝜔, that is, d(t) = d0 cos𝜔t. Assuming, fornow, that (16.31) also applies in the quantum case—by replacing the classicaldipole moment d0 with an appropriate quantum mean value—the mean time 𝜏spof spontaneous emission can be calculated from the expression

P ⋅ 𝜏sp = ℏ𝜔, (16.32)

which yields the time it takes for an atom to emit all the energy of a photon byradiating with a rate P. The rate Γsp = 𝜏−1

sp of spontaneous emission will then begiven by

Γsp = Pℏ𝜔

=𝜔3d2

0

3ℏc3 ,

a formula that takes its final quantum form with the substitution

d0 → 2⟨d⟩ = 2dfi . (16.33)

The substitution makes sense, except for the factor of “2,” which appears for thesame reasons that a similar numerical coefficient appeared in Fermi’s rule. Inany case—remember, we only promised “some kind of proof,” not a proof!—thefinal quantum formula for the spontaneous emission rate is

Γsp = 4𝜔3

3ℏc3 |dfi|2, (16.34)

where dfi is the vector formed by the matrix elements of the dipole momentoperator (d = −er) between the initial and final states. Clearly, dfi is the samequantity that appears in stimulated transitions, and which is equal, as we sawabove, to the mean dipole moment of the atom in an equal-weight superpositionof the states involved.

At this point, it is worth pausing for a moment to focus on the physicalmechanism of spontaneous emission, as described by relation (16.32). Thephysical picture that emerges is remarkable in its simplicity. During spontaneousemission, the atom behaves as a classical oscillating dipole—a microscopicdipole antenna—but, at the same time, it satisfies the quantum constraint thatEM radiation can be emitted only in the form of indivisible light quanta of energyℏ𝜔. And here is how the oscillating dipole “handles” this restriction: It retains theradiated energy until it accumulates the required amount of one light quantum,and only then does it release the photon into space! In the context of this picture,the time for spontaneous emission to occur is simply the waiting time for theaccumulation of the energy of the photon to be emitted. This “proof” helps usmake sense of how the atom behaves as an oscillating electric dipole. It tells usthat we can think of the spontaneous emission process as the result of somekind of “spontaneous mixing” of the two transition states. In this way, the atom


acquires a dipole moment that is, actually, oscillating, because the wavefunctionsvary in time. For example, in the 2pz → 1s transition of the hydrogen atom, theequal-weight superposition of the two relevant wavefunctions

𝜓 = 1√

2(𝜓1s + 𝜓2pz

)

produces—as we have already seen—a dipole along the z axis, which oscillateswith the Bohr frequency of the transition, since

𝜓(t) = 1√

2(𝜓1s e−iE1t∕ℏ + 𝜓2pz

e−iE2t∕ℏ) = e−��iE1t∕ℏ√

2(𝜓1s + e−i(E2−E1)t∕ℏ𝜓2pz

)

⇒ 𝜓(t) = 1√

2(𝜓1s + e−i𝜔t𝜓2pz

).

The phase factor e−i𝜔t gives rise to a continuous oscillation of the dipole, since,for example, for t = 0, the hybrid 𝜓(0) = (𝜓1s + 𝜓2pz

)∕√

2 is directed along thepositive z axis, while for t = 𝜋∕𝜔 = T∕2, it is aligned in the opposite direction,since 𝜓(T∕2) = (𝜓1s − 𝜓2pz

)∕√

2. Finally, according to a similar calculation inSection 2.5.6, the mean value of the dipole moment dz = −ez along the z axisafter time t is

⟨dz⟩ ≡ ⟨−ez⟩t = −e⟨z⟩t = −e⟨z⟩0 cos𝜔t,

or

⟨d⟩t = ⟨d⟩0 cos𝜔t,

where we have set dz ≡ d for simplicity, since this is the only component of thedipole moment with a nonvanishing mean value for this particular transition.

What can we conclude from this discussion? During the 2pz → 1s transition,the atom behaves indeed as a dipole oscillating along the z axis and has theexpected Bohr frequency 𝜔 = (E2 − E1)∕ℏ of the transition. But if this picture iscorrect then the angular distribution of the emitted radiation must also resemblethat of a classical dipole: It must vanish along the z axis and reach a maximumin the x–y plane. Indeed, this behavior is observed for a collection of radiatinghydrogen atoms if they are all in the 2pz state. The vast majority of photons areemitted perpendicular to the z axis, and no photon is emitted along this axis. Andsimilar is the case where the atoms’ initial state is 2px or 2py. In contrast, if the 2slevel happens to be the excited state of the atom, there can be clearly no radiation,since the superposition of two spherically symmetric wavefunctions—𝜓1s and𝜓2sin this case—always produces a spherically symmetric charge distribution, whichcannot radiate—even if it oscillates—according to classical electromagnetism.

We thus see that a semiclassical approach to the problem of spontaneous emis-sion provides a quick and easy method to arrive at important results—includingthe polarization state of the emitted photons—that would otherwise require thefull arsenal of quantum electrodynamics. Moreover, it offers a qualitative pictureof great insight and predictive power. In our opinion, the picture of the atom asa “quantum antenna,” which oscillates like a classical dipole but emits radiation

16.7 Theory of Time-dependent Perturbations: Fermi’s Rule 499

in the form of “quanta,” is one of the most powerful pictures we could hope toobtain for the atomic world.

Problems

16.5 Calculate the matrix element z2pz ,3s and use it to find the 3s → 2pz sponta-neous emission rate in the hydrogen atom. You can select the appropriatewavefunctions from the text. How does your answer compare to the resultfor the 2pz → 1s transition?

16.6 A hydrogen atom undergoes the 2pz → 1s spontaneous emission process.In which directions will most of the photons be emitted? What can you sayabout their polarization?

16.7 Consider a laser cavity of length L = 50 cm and diameter D = 1 cm.The cavity’s mirrors have reflectances R1 = 1 and R2 = 0.99, and theeffective resonance cross section 𝜎R between the relevant states is equalto 10−10 cm2.(a) Calculate the cavity’s critical inversion density.(b) If the cavity operates at an inversion density that is 30 times greater

than the critical value you just calculated, and assuming that popula-tion inversion is sustained through continuous pumping, calculate theamplification factor of the initial photon. How much power is requiredfor the necessary continuous pumping such that the laser cavity keepsworking?

(c) Calculate the angular opening of the ray exiting the laser cavity andmake sure that it does not violate the diffraction limit.

(d) Finally, calculate the number of frequencies the above laser cavity can“accomodate,” assuming that the higher laser level has a spectral widthΔf = 109 Hz, and the lower one Δf ≈ 0.

16.7 Theory of Time-dependent Perturbations:Fermi’s Rule

16.7.1 Approximate Calculation of Transition Probabilities Pn→ m(t) for anArbitrary “Transient” Perturbation V(t)

We will now address the pending open questions of this chapter with a briefintroduction to the theory of time-dependent perturbations. The main plan hasbeen outlined in the beginning of the chapter. We have to solve—with someapproximate method—the time-dependent Schrödinger equation

iℏ𝜕𝜓(t)𝜕t

= H(t) 𝜓(t) (16.35)

with the initial condition𝜓(0) = 𝜓n, (16.36)


where 𝜓(r, t) ≡ 𝜓(t) and

H(t) = H0 + V (t),

while𝜓n is the eigenfunction of the unperturbed Hamiltonian H0 with eigenvalueEn. If the solution 𝜓(t) of Eq. (16.35)—with the initial condition (16.36)—isknown, the required transition probabilities Pn→m(t) are given by

Pn→m(t) = |cm(t)|2 = |(𝜓m, 𝜓(t))|2,

where cm(t) are the coefficients of the expansion of 𝜓(t) in eigenfunctions of theunperturbed Hamiltonian H0

𝜓(t) =∑

mcm(t) 𝜓m, (16.37)

and can be readily calculated from the familiar inner product formula

cm(t) = (𝜓m, 𝜓(t)),

which projects the state vector 𝜓 onto the basis vector 𝜓m. Note that in all theabove, the symbols 𝜓n and 𝜓m denote the time-independent forms of the eigen-functions 𝜓n(r) and 𝜓m(r) of the unperturbed Hamiltonian H0. Note also thatthe coefficients cm(t) depend indirectly on the index n of the initial state (16.36)of the system. We can thus write cm(t) ≡ cnm(t), or even better, cm(t) ≡ cn→m(t),to express the physical significance of these coefficients as transition amplitudesfrom state n to state m.

Our goal is to calculate these amplitudes, at least when the time-dependentperturbation V (t) is much weaker than the potential in the time-independentHamiltonian H0 of the system. For this purpose, we use (16.35) to construct aset of equations for cm(t), by taking the inner product of both sides with theeigenfunctions 𝜓m, and writing also 𝜓(t) in the expanded form (16.37), with thesummation index m replaced with k. We thus find

iℏ cm(t) =∑

kHmk(t) ck(t), (16.38)

which is consistent with our discussion in Chapter 2 on the representationof quantum mechanical operators as matrices and wavefunctions as columnvectors, respectively. Since the components of these column vectors are thecoefficients of the wavefunction expansion in the basis set of eigenfunctions,(16.38) is really the set of equations

iℏ ddt

⎛⎜⎜⎜⎝

c1(t)⋮

cm(t)⋮

⎞⎟⎟⎟⎠

=⎛⎜⎜⎜⎝

H11 · · · H1k · · ·· · · · · · · · · · · · · · ·

Hm1 · · · Hmk · · ·· · · · · · · · · · · · · · · · · ·

⎞⎟⎟⎟⎠

⎛⎜⎜⎜⎝

c1(t)⋮

ck(t)⋮

⎞⎟⎟⎟⎠

(16.39)

or, in more compact form,

iℏ C(t) = H(t) C(t), (16.40)


where C(t) is the column vector for the wavefunction 𝜓(t)13 and H(t) the matrixrepresentation of the Hamiltonian of the problem, with elements

Hmk = (𝜓m,H𝜓k)

in the chosen basis 𝜓k . Consequently, Eq. (16.40) or (16.39)—and (16.38), ofcourse—is merely the transcription of the original Schrödinger equation inmatrix form.

Given now that H = H0 + V (t), and that the basis functions 𝜓k are eigenfunc-tions of the H0 operator, we have

Hmk = (H0)mk + Vmk = Em 𝛿mk + Vmk , (16.41)

where 𝛿mk is the familiar Kronencker delta, defined as

𝛿mk =

{0, m ≠ k1, m = k.

We can now insert (16.41) in (16.38) to obtain

iℏ cm(t) = Em cm(t) +∑

kVmk(t)ck(t), (16.42)

which is an exact equation—a set of equations, actually—and completelyequivalent to the original Schrödinger equation.

The next step is to bring (16.42) to a form that is more suitable for an approx-imate solution. The underlying idea is simple: If the perturbation V (t) wereabsent, the time evolution of cm(t) would be determined by the unperturbedHamiltonian of the system as follows,

cm(t) = ame−iEmt∕ℏ,

where am are constants. Therefore, when we add the external perturbation V (t),it is reasonable to employ a change of variables in (16.42) of the form

cm(t) = am(t)e−iEmt∕ℏ, (16.43)

where the coefficients am(t) are now functions of t and describe the “part” oftime evolution that results from the external perturbation V (t). We can theninsert (16.43) in (16.42) to obtain a new set of equations

iℏ am(t) =∑

kVmk(t)ei𝜔mk tak(t), (16.44)

where 𝜔mk = (Em − Ek)∕ℏ are the Bohr frequencies of the unperturbed system.The system of equations (16.44) can now be solved in conjunction with the initialconditions

am(0) = 𝛿mn, (16.45)

that is,

an(0) = 1, and am(0) = 0 for m ≠ n, (16.46)

13 Because (16.40) is fully equivalent—both in content and form—to the initial time-dependentSchrödinger equation (16.35), we often use the symbol 𝜓(t) instead of C(t).


which express the requirement that the solution we are seeking, namely,

𝜓(t) =∑

mcm(t) 𝜓m ≡ ∑

mam(t)e−iEmt∕ℏ𝜓m,

be identical to the eigenfunction 𝜓n for t = 0. Clearly, if the perturbation V wereabsent, the functions am(t) would be constant and equal to their initial values(16.45) or (16.46). That is, we would have a(0)

m (t) = 𝛿nm, where the upper index (0)

denotes the “zeroth-order approximation” to the problem, which would be thesolution in the absence of V .

To take a step further, we can set

am(t) = a(0)m + a(1)

m (t), (16.47)

where a(1)m (t) are the first-order corrections, which represent the (small) change to

the time-independent initial coefficients a(0)m caused by the perturbation V (t). The

final step is to insert (16.47) in (16.44) and equate those quantities of the two sidesthat are of the same order of magnitude. Since the perturbation V and the firstcorrection a(1)

m are “differentials of first order”—hence their product is a “differ-ential of second order”—while the coefficients a(0)

m are of zeroth order, we obtain

iℏ a(1)m =

∑

kVmk(t) ei𝜔mk t a(0)

k ≡ ∑

kVmk(t) ei𝜔mk t 𝛿kn

⇒ iℏ a(1)m = Vmn(t) ei𝜔mnt

⇒ a(1)m (t) = − i

ℏ ∫t

0Vmn(t) ei𝜔mnt dt. (16.48)

If we now omit the approximation index and use the familiar symbol for thetransition amplitude between two states, (16.48) can be written as

an→m(t) = − iℏ ∫

t

0Vmn(t) ei𝜔mnt dt. (16.49)

Using (16.49), we finally obtain the following first-order approximation for thetransition probabilities

Pn→m(t) =1ℏ2

|||||∫

t

0Vmn(t) ei𝜔mnt dt

|||||

2

. (16.50)

This formula is our first—but not final—result. To familiarize ourselves with it,we will pause for a moment to solve a simple problem.

Example 16.4 The ground state of a hydrogen atom is subject to an electricfield pulse that points along the z axis and varies in time as

(t) = 0e−𝜆t2.

Calculate the probability of finding the electron of the atom in the 2pz state afterthe pulse “goes off.” What are the corresponding probabilities for the 2px and 2pystates?

Solution: We need to apply formula (16.50) with

V (t) = V (z, t) = e(t)z.


The relevant time interval should cover the whole duration of the pulse, so it mustextend from−∞ to+∞. For simplicity, let us set𝜔 = 𝜔21 ≡ (E2 − E1)∕ℏ and write(16.50) as

P1s→2pz= 1ℏ2

|||||∫

+∞

−∞e(t)z2pz ,1s ei𝜔t dt

|||||

2

= e2

ℏ2 z22pz ,1s

|||||∫

+∞

−∞(t) ei𝜔t dt

|||||

2

or

P1s→2pz= e2

ℏ2 z22pz ,1s| (𝜔)|2,

where (𝜔) is the so-called Fourier transform of the (t) function, defined by theintegral

(𝜔) = ∫+∞

−∞(t) ei𝜔t dt.

When (t) = 0 e−𝜆t2 , we can use the generalized Gaussian integral

∫+∞

−∞e−𝛼x2+𝛽x dx =

√𝜋

ae𝛽2∕4𝛼

to find (here, we have 𝛼 = 𝜆, 𝛽 = i𝜔)

(𝜔) = 0

√𝜋

𝜆e−𝜔2∕4𝜆.

Using the known value of 27√

2 a0∕35 for the matrix element z2pz ,1s (≡ z1s,2pz), we

obtain the final result

P1s→2pz= 𝜋

215

310

a0 20

𝜆me−𝜔2∕2𝜆.

The corresponding probabilities for the transitions 1s → 2px and 1s → 2py arezero. Why?

16.7.2 The Atom Under the Influence of a Sinusoidal Perturbation:Fermi’s Rule for Resonance Transitions

Let us now return to our main goal, which was to prove Fermi’s rule. First, weneed to apply formula (16.50) to a general sinusoidal perturbation14

V (t) = V0 cos𝜔t, (16.51)

where V0 is a time-independent hermitian operator, and see what happens.The sinusoidal time dependence of the perturbation V (t) is, of course, what isexpected when a photon of frequency 𝜔—and energy ℏ𝜔—impinges on an atom,and its electric field interacts with the electrons. We should stress, however,that the proof of Fermi’s rule requires—for reasons that will become apparentshortly—more than just a “simple application” of perturbation theory for asinusoidal time dependence as in (16.51). It takes a little more elaborate analysis

14 Evidently, this form includes the special case of the electric dipole approximation for the EMfield, where V0 = −d ⋅ 0 = er ⋅ 0.


than that. The best way to proceed is to start with a careful examination of theformula we seek to prove

Γ = 2𝜋ℏ

|𝑈fi|2 𝜌(Ef ), (16.52)

and make a road map of how to get there. To begin with, the most peculiar featureof (16.52) is the presence of the density of states 𝜌(Ef ) at energies near the finallevel of the transition. The term density of states implies that we are talking hereabout transitions to the continuous spectrum, since this is the only energy rangefor which the term applies fully (as we discussed in Section 15.5.2). A broaderinterpretation of the term was also given already in Section 16.4.1, where wetreated a broadened—due to the uncertainty principle—discrete energy level asa kind of a continuous band (similar to the energy bands in a crystal) and usedformula (16.52) with 𝜌(Ef ) = 1∕ΔE. What we want to clarify now is that (16.52)applies, without qualifications, solely for transitions to the continuous spectrum.But our main task here is to prove that it can be also applied to discrete levels,subject to the condition

ΔE ≫ |𝑈fi|. (16.53)

Equation (16.53) expresses the reasonable requirement that the width of the des-tination level be much larger than the strength of the perturbation as measuredby the relevant matrix element 𝑈fi. In this case, the level can be treated as a kindof a continuous band. Figure 16.12 summarizes these arguments.

But there is another peculiar feature of formula (16.52) we need to analyzebefore attempting to prove it using perturbation theory: Unlike expressions(16.49) and (16.50), which will be our starting point below, (16.52) does not

Ef

ℏω

Ei

Figure 16.12 Fermi’s rule: The rule applies literally for transitions toward dense groups of finalstates, namely, transitions to the continuous spectrum. But it can also be used for transitions todiscrete states, provided the natural width ΔE of the destination level is much larger than the“strength” |𝑈fi| of the external perturbation, whereby the broadened line appears like acontinuous band compared to the perturbation.


refer to transition probabilities but to transition rates. Therefore, the followingquestion has to be answered first: How can we obtain a transition rate from atransition probability of the form Pi→f (𝜏), where 𝜏 is the interaction time of thesinusoidal perturbation?

The answer has been somewhat implied in what we said earlier in the chapterconcerning the experimental meaning of the parameter Γ. Based on the empiricallaw of constant rates, the population of the initial level will undergo an exponen-tial decay of the form Ni(t) = N0e−Γt , if the i → f transition is one way—that is,if the reverse transition is excluded. We then have for the destination level f

Nf (t) = N0 − Ni(t) = N0(1 − e−Γt)

and the transition probability Pi→f after time t = 𝜏 is equal to

Pi→f =(

proportion of atoms found after time 𝜏in level f , having started from level i

)

=Nf (𝜏)

N0

⇒ Pi→f (𝜏) = 1 − e−Γ𝜏 . (16.54)

We should recall at this point that we are trying to calculate Γ in the context offirst-order perturbation theory, so we only allow for linear terms with respect toΓ to be present. It is thus necessary to use the Taylor expansion of the exponentialfunction and write (16.54) in the form

Pi→f (𝜏) = Γ𝜏 + · · · , (16.55)

where the higher order terms—(Γ𝜏)2, (Γ𝜏)3, and so on.—can be ignored if thecondition

Γ𝜏 ≪ 1 (16.56)

is satisfied. The conclusion is unequivocal. To prove Fermi’s rule for resonancetransitions within the time-dependent perturbation theory (and for sinusoidalperturbations) we need to show that the transition probability Pi→f (𝜏) is pro-portional to the interaction time of the perturbation. Once this is shown, theproportionality coefficient is the rate Γ we are looking for.

But the above arguments also suggest a more accurate physical interpretationof the transition probability Pi→f (𝜏). Since we are dealing with transitions to thecontinuous spectrum of some sort, it is meaningless to talk about a transition toa particular final state, but only to a group of final states in the immediate vicinityof the “exact” destination level of the transition. The latter is determined by therelation

Ef = Ei + ℏ𝜔, (16.57)

which expresses energy conservation for a sinusoidal perturbation thatconsists—in its quantum version—of quanta of energy ℏ𝜔. Actually, eventhe energy of these quanta is not rigorously defined. As is evident from themathematical description of the interaction, and also in actual experiments, theperturbation occurs over a finite time interval 𝜏 , which necessarily introduces(through the energy-time uncertainty principle) an uncertainty to the energy 𝜖


of the incident photon—and hence to the final energy Ef of the electron—of theorder of Δ𝜖 ≈ ℏ∕𝜏 .

We thus realize that we are not dealing with a single transition, but with agroup of transitions to levels that lie near the “exact” final level (16.57). So, toobtain the physically correct answer, it is not sufficient to calculate the transitionprobability Pi→f (𝜏) to the “exact” final state, but to sum over all possible f statesand obtain a weighted average of the form

Pi→f (𝜏) =∑

fPi→f (𝜏) = ∫ Pi→f (𝜏) 𝜌(Ef ) dEf , (16.58)

where the summation is transformed into an integral—by using the (fractional)density of states 𝜌(Ef )—while the “final” energy Ef now becomes a free integrationvariable that is not constrained by the conservation relation (16.57). As we willshortly see, the expression (16.57) for Ef is readily obtained from the functionPi→f (𝜏) for large 𝜏 . As for the integration limits in (16.58), they can now extendfrom −∞ to +∞ to include all probable—and improbable!—final states. Theweight of all these final states in the determination of the final result Pi→f (𝜏) isdetermined by the functions Pi→f (𝜏) and 𝜌(E), both of which are sharply peakedaround the “destination level” of the transition. The line shape function 𝜌(E) hasthe sharply localized form shown in Figure 16.9, while the transition probabilityfunction Pi→f (𝜏) is also focused around that level (Ef = Ei + ℏ𝜔) and becomesincreasingly sharper as 𝜏 increases and the uncertainty Δ𝜖 ≈ ℏ∕𝜏 in the photon’senergy decreases.

We are almost ready to calculate the quantity we are after. We begin by writingformula (16.49) for the transition amplitude ai→f (𝜏) as

ai→f (𝜏) = − iℏ ∫

𝜏∕2

−𝜏∕2Vfi(t) ei𝜔fit dt, (16.59)

which we then apply to a sinusoidal perturbation of the form (16.51). As usual,this type of function can be expressed as the sum of two complex terms

V (t) = 𝑈e−i𝜔t + 𝑈ei𝜔t, (16.60)

where 𝑈 = V0∕2 is a time-independent operator. Using the above expression forV (t) we can rewrite (16.59) as

ai→f (𝜏) = − iℏ𝑈fi

(

∫𝜏∕2

−𝜏∕2ei(𝜔fi−𝜔)t dt + ∫

𝜏∕2

−𝜏∕2ei(𝜔fi+𝜔)t dt

)

. (16.61)

The key message in this expression will become apparent if we use the well-knownformula for the delta function (the proof is given below)

∫+∞

−∞ei𝜔t dt = 2𝜋𝛿(𝜔), (16.62)

and take, for a moment, the limit 𝜏 → ∞ to find

ai→f (∞) = − iℏ𝑈fi(2𝜋𝛿(𝜔fi − 𝜔) + 2𝜋𝛿(𝜔fi + 𝜔)). (16.63)


Equation (16.63) has the following consequences: (i) If Ef > Ei (⇒ 𝜔fi > 0), thesecond term in (16.63) is necessarily zero, since (𝜔fi + 𝜔) is positive and hence𝛿(𝜔fi + 𝜔) is zero then. (ii) If Ef < Ei (⇒ 𝜔fi < 0), the first term 𝛿(𝜔fi − 𝜔) vanishesand the second term 𝛿(𝜔fi + 𝜔) survives. In case (i) (i.e., for Ef > Ei), the finalexpression for the transition amplitude is

ai→f (∞) = −2𝜋iℏ𝑈fi 𝛿(𝜔fi − 𝜔), (16.64)

and the presence of the delta function 𝛿(𝜔fi − 𝜔) mandates that

𝜔fi − 𝜔 = 0 ⇒Ef − Ei

ℏ− 𝜔 = 0 ⇒ Ef = Ei + ℏ𝜔,

which is the result we expect in the infinite 𝜏 limit: The atom has absorbed theincident photon and has been raised to an excited atomic state. In contrast,the condition 𝜔fi + 𝜔 = 0 of case (ii) leads to the relation Ef = Ei − ℏ𝜔, whichcorresponds to a decrease of atomic energy through the emission of a photon. Ifwe now recall how the above two terms were derived from the correspondingterms of (16.60), then the above conclusions are the same as those we foundearlier using a purely physical argument. If the atom is initially in the groundstate—whence only absorption can take place—we know in advance that thecorresponding transition amplitude will include only the first of the two termsof (16.61), because the second term is negligible for large 𝜏 . We can then write

ai→f (𝜏) = − iℏ𝑈fi 2𝜋𝛿𝜏(𝜔fi − 𝜔), (16.65)

where 𝛿𝜏(Ω) is the function

𝛿𝜏(Ω) =1

2𝜋 ∫𝜏∕2

−𝜏∕2eiΩt dt =

sin(𝜏Ω∕2)𝜋Ω

. (16.66)

The 1∕2𝜋 factor was inserted intentionally in the definition, so that for 𝜏 → ∞ wehave

lim𝜏→∞

𝛿𝜏(Ω) = 𝛿(Ω), (16.67)

where 𝛿(Ω) is the Dirac delta function, according to (16.62)—which is indeedcorrect, since, for any value of the parameter 𝜏 , the function 𝛿𝜏(Ω) has unit totalarea,

∫+∞

−∞𝛿𝜏(Ω) dΩ = 1 ∀𝜏,

while it also becomes increasingly “taller” and “thinner” as 𝜏 goes to infinity.Therefore, 𝛿𝜏(Ω) behaves exactly as Dirac’s delta function (Figure 16.13).

In the limit of a sinusoidal pulse with infinite duration (𝜏 → ∞), (16.65) iswritten as

ai→f (∞) = −2𝜋iℏ

𝑈fi 𝛿(𝜔fi − 𝜔) = −2𝜋iℏ

𝑈fi 𝛿

(Ef − Ei − ℏ𝜔ℏ

)

, (16.68)

where the presence of the delta function expresses, as we noted above, theconservation of energy: the fact that, in this limit, the transition occurs to a


1.5τ / 2π

2π / τ

τ = 10

Ω

–2π / τ

1.0

0.5

1–1–2–3 2 3

Figure 16.13 Graphical sketch of the 𝛿𝜏(Ω) function for a typical value of the parameter 𝜏 . As 𝜏

increases, the bump centered at the origin becomes higher and thinner, while the total areaunder the curve remains equal to unity. A function with these features is an asymptoticrepresentation of Dirac’s delta function.

final state whose energy Ef = Ei + ℏ𝜔 is higher than the initial level Ei by theenergy ℏ𝜔 of the absorbed photon. Of course, in the case of finite 𝜏 , the function𝛿𝜏(𝜔fi − 𝜔) = 𝛿𝜏((Ef − Ei − ℏ𝜔)∕ℏ) is not as “sharp” as the delta function, butit does give a “narrow” distribution of possible values for the final energy Efcentered around the asymptotic result Ei + ℏ𝜔. This is, of course, another wayof looking at the consequences of the time–energy uncertainty principle onthe photon’s energy. A sinusoidal signal sin𝜔t confined to a finite time interval[−𝜏∕2, 𝜏∕2] and assumed zero outside this interval has a frequency uncertaintyΔ𝜔 of the order of 1∕𝜏 . That is, Δ𝜔 ≈ 1∕𝜏 ⇒ Δ𝜖 = Δ(ℏ𝜔) = ℏ∕𝜏 .

Note that the presence of the delta function in (16.68) clearly indicates that weneed to integrate over a set of final states, as we conjectured earlier. We can thusapply formula (16.58) with

Pi→f (𝜏) = |ai→f (𝜏)|2 = 4𝜋2

ℏ2 |𝑈fi|2 𝛿2

𝜏


)

to obtain

Pi→f (𝜏) =4𝜋2

ℏ2 ∫+∞

−∞𝛿2𝜏


)

|𝑈fi|2 𝜌(Ef ) dEf , (16.69)

where, for reasons we explained before, we let the integration over Ef extendthroughout the range −∞ < Ef < +∞, since the sharply focused 𝛿𝜏 and 𝜌

functions ensure that only the regions of Ef values around Ei + ℏ𝜔 contribute tothe integral.

The next step is to assume that the time 𝜏 is large enough that the “focusregion” Δ𝜖 ≈ ℏ∕𝜏 of the function 𝛿𝜏 (i.e., the uncertainty in the photon’s energy)is much smaller than the linewidth ΔE of the final level. This full resonance


condition is crucial for the validity of Fermi’s rule as we shall now see. First,we treat the functions |𝑈fi|

2 and 𝜌(Ef ) as practically constant in the “focusregion”—the full resonance condition is essential here for a sharply localizedfunction such as 𝜌(E)—and substitute them with their value at the center of thisinterval, that is, for Ef = Ei + ℏ𝜔. To facilitate the calculation we perform thefollowing change of variable in (16.69)

Ω =Ef − Ei − ℏ𝜔

ℏ⇒ dEf = ℏ dΩ.

We thus find

Pi→f (𝜏) =4𝜋2

ℏ ∫+∞

−∞𝛿2𝜏 (Ω)|𝑈fi|

2𝜌(Ei + ℏ𝜔 + ℏΩ) dΩ. (16.70)

The center of the “distribution” 𝛿2𝜏 (Ω) will now be at the origin Ω = 0, and (16.70)

gives

Pi→f (𝜏) =4𝜋2

ℏ|𝑈fi|

2𝜌(Ei + ℏ𝜔 + ℏΩ)|Ω=0 ⋅ ∫+∞

−∞𝛿2𝜏 (Ω) dΩ

= 4𝜋2

ℏ|𝑈fi|

2𝜌(Ei + ℏ𝜔)∫+∞

−∞𝛿2𝜏 (Ω) dΩ. (16.71)

The exact value of the last integral can be calculated analytically—or withMathematica, or Matlab—and the result is

∫+∞

−∞𝛿2𝜏 (Ω) dΩ = 𝜏

2𝜋. (16.72)

Consequently, (16.70) takes the final form

Pi→f (𝜏) =4𝜋2

ℏ|𝑈fi|

2𝜌(Ei + ℏ𝜔) ⋅𝜏

2𝜋= 2𝜋

ℏ|𝑈fi|

2𝜌(Ef )𝜏 (Ef = Ei + ℏ𝜔),

where the linear dependence on time 𝜏 has now become evident, and theproportionality coefficient is

Γ = 2𝜋ℏ

|𝑈fi|2𝜌(Ef ), (16.73)

which concludes the proof of Fermi’s rule for resonance transitions.It is important to clarify that the symbol Ef in the final expression (16.73) is

no longer the free integration variable we used throughout but the specific finalenergy Ef = Ei + ℏ𝜔 in the limit of an infinitely long pulse. Note also that thelinear dependence of the (16.72) integral on time 𝜏 is qualitatively evident. We justneed to consider—and consult also Figure 16.13—that since the function 𝛿2

𝜏 (Ω)has a central peak of height 𝜏2∕4𝜋2 and “width” 2 × (2𝜋∕𝜏) = 4𝜋∕𝜏 its area (takenroughly as a triangle around this peak) is approximately equal to

12⋅

4𝜋𝜏

⋅𝜏2

4𝜋2 = 𝜏

2𝜋,

whence the linear dependence on 𝜏 is now obvious. Note, by the way, that thisrough approximation of the “central triangle” area gives the correct result for the


(16.72) integral, as it does also for 𝛿𝜏(Ω) itself, whose area is equal to unity, as wementioned.

For the rigorous readers, there is one more question that we need to address: Isthe inequality (16.56),Γ𝜏 ≪ 1, compatible with the spirit of our present analysis,where 𝜏 appears to be large enough so that the “full resonance condition”

Δ𝜖 ≈ ℏ

𝜏≪ ΔE (16.74)

holds? The answer is not trivial. The inequality Γ𝜏 ≪ 1—which, as you mayrecall, is necessary to obtain the linear dependence as an approximation of anexponential law, and also for the perturbative treatment of the problem to bevalid—requires 𝜏 to be relatively small. Specifically, we must have

𝜏 ≪ Γ−1 = 𝜏Γ, (16.75)where 𝜏Γ is the characteristic time of the process. On the other hand,(16.74)—taking into account that ℏ∕ΔE = 𝜏sp = Γ−1

sp —requires that

𝜏 ≫ Γ−1sp = 𝜏sp (16.76)

so 𝜏 should obey the double inequality𝜏sp ≪ 𝜏 ≪ 𝜏Γ, (16.77)

which is only possible if the interval [𝜏sp, 𝜏Γ] exists and is wide enough. That is,𝜏sp ≪ 𝜏Γ or Γsp ≫ Γ. (16.78)

The inequality Γ≪ Γsp is thus the crucial test for the validity of Fermi’s rule. Solet us see what we obtain if we take Γ from that rule and require that Γ≪ Γsp. Foran order-of-magnitude comparison we omit 2𝜋 from Fermi’s rule and use thetime–energy uncertainty principle (𝜏spΔE ≈ ℏ) to estimate Γsp = 𝜏−1

sp . We thenhave

Γ ≈ 1ℏ|𝑈fi|

2𝜌(Ef ) ≈1ℏ|𝑈fi|

2 1ΔE

≪ Γsp ≈ 𝜏−1sp ≈ ΔE

ℏ

⇒1ℏ|𝑈fi|

2 1ΔE

≪ΔEℏ

⇒ |𝑈fi|2 ≪ (ΔE)2

⇒ |𝑈fi|≪ ΔE, (16.79)which is exactly the condition for the validity of Fermi’s rule we announced at thebeginning of this section (Eq. (16.53)). Given that 𝑈fi (= e 0zfi∕2) is on the orderof ea00 (0 is the intensity of the electric field), the validity of (16.79) dependscrucially on the intensity of the incident EM wave. So it transpires that Fermi’srule survives in almost all practical cases except for very intense laser beams, inwhich case (16.79) is violated and interesting new physical phenomena—such asRabi oscillations—arise (See OS16.2.).

Problems

16.8 A capacitor is in the process of discharging and the electric field(t) = (t)z between its plates decays exponentially as (t) = 0e−𝜆t

(t > 0). A hydrogen atom is placed between the capacitor plates at timet0 = 0. If the atom is initially in its ground state, what is the probability

16.8 The Light Itself: Polarized Photons and Their Quantum Mechanical Description 511

of finding it in each of the states 2s, 2px, 2py, or 2pz after time t? Is theprobability non-negligible for realistic values of 0 and 𝜆?

16.9 A harmonic oscillator in its ground state is subject to the followingGaussian pulse of a force parallel to the oscillation axis

f (t) = f0e−(t∕𝜏)2.

Show that, after the pulse has “passed,” the probability for the transition0 → 1 is given by

P0→1 = 𝜋

2f 20 𝜏

2e−𝜏2∕2 (ℏ = m = 𝜔 = 1).

What about the probabilities 0 → n, for n ≥ 2? Also, recast the aboveresult in ordinary units.

16.10 Use Fermi’s rule to prove that the rates for the two resonanceprocesses—absorption and emission—are always the same. What isthe general mathematical reason for this?

16.11 In a resonance absorption from the ground state of an atom, thelinewidth of the excited state is ΔE ≈ 10−8 eV. Check the validity ofFermi’s rule when the intensity of the incident photon beam is equal to(a) I = 10−4 W∕cm2, (b) I = 103 W∕cm2.

16.8 The Light Itself: Polarized Photons and TheirQuantum Mechanical Description

So far in this chapter, we have discussed the interaction of photons with matter. Inthis last section of the book we take a look at the photons themselves, focusing ontheir own properties, such as polarization—the analog of spin for electrons—andits quantum mechanical description.

16.8.1 States of Linear and Circular Polarization for Photons

As we know, attributing corpuscular nature to light has not led us to reject itswave nature. We should thus regard photons as EM waves and particles at thesame time. As such, photons must carry the basic properties of EM waves, such aslinear or circular polarization. In this spirit, the following definition is plausible.

Definition (Polarized photons): A photon is called linearly or circularly polarizedif its associated EM wave is similarly polarized.

We remind the readers that a plane EM wave is linearly polarized, say, along thex axis, if its electric field always points along this axis, while its intensity variesharmonically with time. In other words, E = xE0 sin(𝜔t + 𝜙), where 𝜔 is theangular frequency of the wave. Note that the standard convention is to consider


Ey Ex

z

Figure 16.14 Linear polarization states of an EM field. Since EMwaves are transverse, an EM wave that propagates along the z axishas its electric field on the x–y plane.

a wave propagating along the z axis, which we picture in the horizontal direction(see Figure 16.14). In this case, the linear polarization vectors point along the xand y axes, that is, perpendicular to the z axis, since EM waves are transverse.

For a circularly polarized EM wave, its electric field does not have a constantdirection but rotates on the x–y plane with constant angular velocity that is equalto the angular frequency 𝜔 of the wave. Being a rotating vector of magnitude E0,the electric field of a (clockwise) circularly polarized light is written in complexform as

ER(t) = E0ei𝜔t = E0(cos𝜔t + i sin𝜔t)= E0 cos𝜔t + i(E0 sin𝜔t) = Ex(t) + iEy(t),

which is a superposition, with relative phase 𝜋∕2, of the linearly polarized fieldsEx = E0 cos𝜔t and Ey = E0 sin𝜔t. Similarly, for a counterclockwise circularlypolarized light, we have

EL(t) = E0e−i𝜔t = E0(cos𝜔t − i sin𝜔t)= Ex(t) − iEy(t),

whence the relation between the circularly and linearly polarized fields is

ER = Ex + iEy, EL = Ex − iEy (16.80)

or, conversely (i.e., solving for Ex and Ey),

Ex =12(ER + EL), Ey =

12i(ER − EL). (16.81)

16.8.2 Linear and Circular Polarizers

Just as in classical EM waves, we systematically produce and measure linearlyor circularly polarized photons by using polarizers. In their simplest form, theseare plates or filters of some material that allows only the suitably polarizedcomponent of incident light to pass through. So, as (non-polarized) photonspass through a linear polarizer along the x axis, only those that are linearlypolarized along the same axis will exit. The same polarizers can also be used tomeasure the polarization of a light beam and thus of its photons. So, the photonsof the beam will be, say, linearly polarized along a certain direction if the beamsuffers no loss as it passes through the corresponding linear polarizer. If no lightgets transmitted at all, then surely the beam’s polarization is perpendicular tothe direction of the polarizer. If the two directions form an angle 𝜃, then theintensity of the beam is reduced by a factor cos2𝜃, since only the component of


Figure 16.15 Schematic diagram of linear (a)and circular (b) polarizers.

x y R(b)(a)

L

the initial electric field along the direction of the polarizer can pass, while thecomponent perpendicular to it is blocked. It is customary to denote linear andcircular polarizers by the simple graphical representations of Figure 16.15.

16.8.3 Quantum Mechanical Description of Polarized Photons

What is the form of the wavefunctions that describe a particular photon polar-ization state? Not having discussed at all up to now what kind of wavefunctionsare needed for this purpose—and it is not a simple matter—we shall have to relyon an abstract formalism for quantum mechanical states introduced by Dirac,which only utilizes their basic property as vectors, that is, as elements of a vectorspace. The standard notation for such a vector is |𝛼⟩, which is called a ket vectoror, more specifically, the ket vector corresponding to the state with physicalfeatures described by the letter 𝛼.

The key concept here is vector space. What is a vector space? It is a set whoseelements can be added or multiplied by (real or complex) numbers such that theresult is still an element of the same set. To put it more succinctly, a set is a vectorspace if every linear combination of its elements is also an element of the set.The connection to quantum mechanics stems from the following fact. The set ofsquare integrable wavefunctions—and hence the set of all physically realizablestates of a quantum system—is a vector space. Here is why. A wavefunction issquare integrable if it decays sufficiently fast at infinity. If 𝜓1 and 𝜓2 are two suchwavefunctions, then every linear combination of them, 𝜓 = c1𝜓1 + c2𝜓2, is alsoa square integrable wavefunction, since it decays at infinity at least as fast as theslowest decaying of the two functions in the linear combination. So the squareintegrable wavefunctions satisfy the definition of a vector space and can thusbe regarded as vectors in this space. The Dirac formalism offers a handy way oftreating quantum mechanical states (i.e., their corresponding wavefunctions) byusing only their most abstract general property: The fact that they are vectors ina suitable vector space. Even though we shall not be needing this for the limitedpurposes of the present discussion, let us also note that the Dirac formalismincludes, apart from the vectors |𝛼⟩—the ket vectors— the so-called bra vectorsthat are denoted as ⟨𝛼| and are the hermitian conjugates—or adjoints—of the ketvectors, in the same manner that the line vectors X† are the hermitian conjugatesof the column vectors X. That is, (|𝛼⟩)† = ⟨𝛼| and also |𝛼⟩ = (⟨𝛼|)†. In this spirit,just like the inner product (X,Y ) =

∑Ni=1 x∗

i yi of two column vectors

X =⎛⎜⎜⎝

x1⋮

xN

⎞⎟⎟⎠

, Y =⎛⎜⎜⎝

y1⋮

yN

⎞⎟⎟⎠

(16.82)

is equivalently written as

(X,Y ) ≡ X†Y , (16.83)


the inner product (|𝛼⟩, |𝛽⟩) of two ket vectors is also written as

(|𝛼⟩, |𝛽⟩) = (|𝛼⟩)†|𝛽⟩ = ⟨𝛼| ⋅ |𝛽⟩ ≡ ⟨𝛼|𝛽⟩, (16.84)

whence the names bra and ket for the two kinds of Dirac vectors. Theycorrespond to the left and right bracket respectively, in the traditional notationof the inner product.

In the context of this formalism, the states of linear polarization of photonsalong the x or y axes are written as |x⟩ or |y⟩, while for the states of circularpolarization we write |R⟩ and |L⟩ for the clockwise and counterclockwisepolarization, respectively. In addition, and according to (16.80), the relationbetween the two kinds of photon polarization states is written—taking alsonormalization into account— as

|R⟩ = 1√

2(|x⟩ + i|y⟩), |L⟩ = 1

√2(|x⟩ − i|y⟩) (16.85)

⇒ |x⟩ = 1√

2(|R⟩ + |L⟩), |y⟩ = 1

i√

2(|R⟩ − |L⟩). (16.86)

Being quantum superposition states, each one of (16.85), say, the first one, tellsus that, if the photon described by this state is measured by an x polarizer, it hasa probability of 50% to go through—in which case the photon is detected in thestate |x⟩− and a probability of 50% not to go through, in which case the photonis clearly in state |y⟩. This interpretation justifies the factor 1∕

√2 in (16.85). The

sum of the probabilities for the photon to be in either state must equal unity.The photon polarization states are, of course, assumed normalized and

orthogonal to each other, so the following relations hold:

⟨x|x⟩ = ⟨y|y⟩ = ⟨R|R⟩ = ⟨L|L⟩ = 1, ⟨x|y⟩ = ⟨R|L⟩ = 0. (16.87)

Here is a pertinent exercise for the readers: Show that if |x⟩ and |y⟩ are nor-malized and orthogonal, then the same must hold for |R⟩ and |L⟩ as definedby (16.85), and vice versa. The following examples should also help readersfamiliarize themselves with these concepts.

Example 16.5 A linearly polarized photon along the x axis is transmittedfirst through a clockwise circular polarizer, and immediately after that, througha counterclockwise polarizer. What is the probability for the photon to passthrough the first polarizer and to pass through both polarizers? Are theprobability values shown in the figure correct?

R50% 0%x

L

Solution: To answer the question, we must write the state |x⟩ of the impingingphoton as a linear combination of the states of circular polarization |R⟩ and|L⟩, since the polarizers in our experimental device measure this kind of polar-ization. Thus, according to (16.86), the state |x⟩ of the incoming photon willbe a superposition of the two possible circular polarization states with a 50%


probability for each one. Therefore, the probability that the first measurementyields the state |R⟩ (i.e., the probability that the photon gets transmitted throughthe first polarizer) is 50%. According to the principle of quantum measurement,the state of the photon that went through the first detector must now be |R⟩and thus the probability for the photon to go through the second detectoris zero. (Since we already know that the photon is clockwise polarized, thesecond detector cannot observe anything different, namely, counterclockwisepolarization.)

Example 16.6 Calculate the probabilities that a circularly polarized photongoes through each one of the linear polarizers arranged as in the following figureand verify the probability values noted there.

Rx y45° 45°

50% 25% 12.5%

Solution: Since |R⟩ = (|x⟩ + i|y⟩)∕√

2, the probability that the photon goesthrough the first polarizer—the x polarizer—is clearly 50%. As it comes out ofthis polarizer, it is polarized along the x axis and is in the |x⟩ state. To see whathappens next, it is necessary to observe—can you explain why?—that the stateof linear polarization at an angle 𝜃 with respect to the x axis is written as

|𝜃⟩ = cos 𝜃|x⟩ + sin 𝜃|y⟩.

Therefore, the probability for a photon to pass through a linear polarizer at anangle 𝜃 with the photon’s polarization is equal to cos2 𝜃. (This result is the sameas in the corresponding classical experiment, where the passage of a linearlypolarized beam through a polarizer at an angle 𝜃 with respect to the directionof the beam’s polarization causes a reduction of its intensity by a factor cos2𝜃.)Returning to our problem, the probability that the photon going through thefirst polarizer goes through the second one as well is cos2 45∘ = (1∕

√2)2 = 1∕2,

so the combined probability that the photon goes through both of them is 25%.In the same manner, we find that the probability for the photon to go through allthree polarizers is 12.5%.

We conclude this section with a brief discussion of the problem of spinmeasurement—that is analogous to polarization—and the Stern–Gerlach deviceused for it. The analogy is warranted, because both the photon polarizers andthe Stern–Gerlach device for particles with spin (say, spin 1∕2) are consideredas model measurement devices. As such, they are pertinent to all discussionson the foundations of quantum mechanics, and in particular, the problem ofmeasurement. The schematic representation of the Stern–Gerlach device isgiven in Figure 16.16.

But the experiments of interest—just as with polarizers—are those involving aserial arrangement of Stern–Gerlach devices, each with a different orientation ingeneral.


+

SGz–

Figure 16.16 Schematic representation of a Stern–Gerlach device where the fieldis inhomogeneous along the z axis: The symbols + and − on the right denote theexit points of the particles that registered with spin up and down, respectively.

We leave it as a last exercise for the readers to describe what happens inthe following arrangement of Stern–Gerlach devices and verify whether theprobability values are correct as stated.

36%

X3

4i

+ z+ z+

X––

+

–

+

–

100% 50%

SGz SGz SGx

Note that in the first two devices the − exit is blocked, while in the third devicethe+ exit is blocked. Here, the symbols x±, y±, z± denote particles with positive ornegative projection—that is, spin “up” or “down”—onto the corresponding axis.

But let us focus our attention on the entry to the third device. All particlesarriving to it are identical. They each have spin up along the z axis and are

described by the same state vector X =(

10

)

. There is no way whatsoever to

distinguish one particle from the other. And yet, half of these particles leave thethird device from one exit and half from the other, the two populations havingopposite spin orientations with regard to the x axis. The particles were identicalas they entered, they were subjected to exactly the same process—they movedunder the same magnetic field—and yet they were different when they exitedthe device! What is the cause for this difference? None, according to quantummechanics, as presented in this book.

“Whoever has not been shocked by quantum mechanics, has not understood it,”Bohr had once said. The shock is indeed necessary, but does not automaticallylead to a genuine understanding of quantum mechanics, unless it is accompaniedby the reverse shock—the shock caused by the spectacular failure of classicalphysics to explain the mystery of the stability of atoms and molecules and theirunique form. “A pure miracle when considered from the standpoint of classicalphysics,” as Bohr also said.

So we reiterate a suggestion that we already made back in Chapter 1, as a“remedy” vis-a-vis the paradoxical nature of the quantum world. We suggest thatthe readers revisit from time to time the mystery of atomic stability, and oncethey allow themselves to be “shocked” afresh, to retrace the thought process

Stability → Quantization → Wavelike behavior → Wave= Probability wave → Collapse of the wavefunction upon measurement

to ensure that quantum mechanics and its statistical interpretation are ratherinescapable. Of course, no physical theory is eternal. But one thing is certain.Whatever theory might one day replace quantum mechanics as more funda-mental, it will undoubtedly incorporate it as a limiting case that will continueto hold in all those scales and at the level of precision where the present theoryhas been experimentally tested. When it comes to the nuclear and atomic world,the scientists and engineers of the future—however distant that future is—will


continue to apply today’s quantum mechanics, in precisely the same way wecontinue to apply Newtonian mechanics when we want to send a vehicle to Marsor calculate how a tsunami propagates. Quantum mechanics is here to stay.

Problems

16.12 Consider a photon that is linearly polarized along the x axis. The photonenters the configuration of polarizers depicted below. Calculate the prob-ability that the photon passes through all three polarizers. How wouldyour answer change if the photon were linearly polarized at an angle 𝜃with respect to the x axis?

Rx

y L

16.13 A circularly polarized photon enters the configuration of polarizersdepicted in the following figure. Calculate the probability that thephoton passes through all three polarizers. If the photon were linearlypolarized at an angle 𝜃 with respect to the x axis, determine the value of𝜃 that maximizes the probability for the photon to pass through all threepolarizers.

Rx yθ

16.14 Consider a photon whose polarization state is described by the followingstate vector:

|𝜓⟩ = 1√

3|R⟩ +

√23|L⟩.

What is the probability that the photon passes through a linear polarizerplaced along the x axis?

Further Problems

16.15 Use the formulas for 𝜎R and Γsp given in the text to show that 𝜎R ∼ 𝜆2. Inother words, show that, at resonance, the atom “appears” to the incidentphoton as having a size comparable to the photon’s wavelength.

16.16 Calculate the spontaneous emission rates for the decays 3pz → 1s and3pz → 2s, and check whether the “cubic-power law” is indeed valid inthis case.


16.17 Prove the selection rule Δn = ±1 for a harmonic oscillator in resonancewith an incident EM field that is polarized along the oscillation axis.

16.18 Explain why the electric dipole transitions in many-electron atoms obeythe selection rule ΔS = 0, where S is the total-spin quantum number ofthe atom. Spin-changing transitions can take place only via magneticdipole interactions of the form 𝑈 = −𝝁 ⋅B(t). Explain why magneticdipole transitions are weaker than electric dipole ones by a factor of theorder of 𝛼2.

16.19 A harmonic oscillator is subject to the time-dependent perturbations(a) V = 1

2k(t)x2

(b) V = g(t)𝛿(x).Which are the selection rules in each case?

16.20 The same question as before for a hydrogen atom when the followingperturbations are applied to it(a) V = g(t)x(b) V = g(t)F(r)(c) V = g(t)F(r) cos 𝜃

16.21 A particle in the ground state of an infinite potential well at t = 0 is subjectto the time-dependent perturbation

V (x, t) = gxe−𝜆t.

What is the probability for the particle to be in the first excited state ofthe well at time t? What are the selection rules in this case?

519

Appendix

Definitions and unit conversions

angström 1 Å = 10−8 cm = 10−10 mfermi 1 F = 10−13 cm = 10−15 melectron volt 1 eV = 1.602 × 10−12 erg = 1.602 × 10−19 Jtesla 1 tesla = 104 gaussjoule 1 J = 107 erg

Useful order-of-magnitude values

Electron rest energy ∶ mec2 ≈ 0.5 MeVProton rest energy ∶ mpc2 ≈ 1 GeV(kT) room

temperature(≈300 K)≈ 1

40eV

(kT)T=12000 K ≈ 1 eVPhoton frequency of energy 1 eV ∶ f ≈ 2.4 × 1014 HzPhoton wavelength of energy 1 eV ∶ 𝜆 ≈ 12000 ÅBohr magneton in practicalunits (eV/gauss) ∶ 𝜇B ≈ 0.5 × 10−8 ≈ 10−8 eV∕gaussℏc = 12400 eVÅ ≈ 12000 eVÅ

Approximate formulas

Photon energy in eV as a functionof its wavelength in Å ∶ 𝜖(eV) ≈ 12000

𝜆(Å)de Broglie wavelength of an electronin Å, as a function of its energy in eV ∶ 𝜆(Å) ≈ 12

√E(eV)

Wavelength of maximum emissionfor a blackbody of temperature T ∶ 𝜆max(cm) ≈ 0.3

T(K)


520 Appendix

Physical constants in cgs units

Planck’s constant⎧⎪⎨⎪⎩

h = 6.626 × 10−27 erg s

ℏ = h2𝜋

= 1.054 × 10−27 erg s

Speed of light c = 2.997 × 1010 cm∕s

Electron charge e = 4.803 × 10−10 esu

Electron mass me = 9.109 × 10−28 g

Proton mass mp = 1.672 × 10−24 g

Proton mass/electron massmp

me= 1836.151

Fine structure constant 𝛼 = e2

ℏc= 1

137.036

Bohr radius a0 = ℏ2

me2 = 0.529 Å

Ionization energy ofhydrogen atom (with mp → ∞)

WI(H) =mee4

2ℏ2 = 13.605 eV

Compton wavelengthof electron

Classical electron radius r0 = e2

mec2 = 2.817 fermi

Bohr magneton 𝜇B = eℏ2mec

= 9.274 × 10−21 erg∕gauss

= 5.788 × 10−9 eV∕gauss

Nuclear magneton 𝜇N = eℏ2mpc

= 5.050 × 10−24 erg∕gauss

= 3.152 × 10−12 eV∕gauss

Boltzmann’s constant k = 1.380 × 10−16 erg∕K

Avogadro’s number NA = 6.022 × 1023 mol−1

Universal gravitational constant G = 6.672 × 10−8 cgs

Appendix 521

Trigonometric identities

sin2x = 12(1 − cos 2x)

cos2x = 12(1 + cos 2x)

sin3x = 14(3 sin x − sin 3x)

cos3x = 14(3 cos x + cos 3x)

sin 2x = 2 sin x cos x

cos 2x = cos2x − sin2x

sin 3x = 3 sin x − 4sin3x

cos 3x = 4cos3x − 3 cos x

sin(x + y) = sin x cos y + sin y cos x

cos(x + y) = cos x cos y − sin x sin y

sin x cos y = 12(sin(x + y) + sin(x − y))

sin x sin y = 12(cos(x − y) − cos(x + y))

cos x cos y = 12(cos(x + y) + cos(x − y))

522 Appendix

Useful integrals

∫∞

0xne−𝜆x dx = n!

𝜆n+1

∫+∞


√𝜋

𝜆, Re𝜆 > 0

∫+∞

−∞x2 e−𝜆x2 dx = 1

2𝜆

√𝜋

𝜆

∫+∞

−∞x4 e−𝜆x2 dx = 3

4𝜆2

√𝜋

𝜆

∫+∞

−∞x2n e−𝜆x2 dx = 1 ⋅ 3 · · · (2n − 1)

(2𝜆)n

√𝜋

𝜆= (2n)!

n!(4𝜆)n

√𝜋

𝜆

∫+∞

−∞e−𝜆x2+𝜇x dx = e𝜇2∕4𝜆

√𝜋

𝜆, Re𝜆 > 0

∫∞

0

x3

ex − 1dx = 𝜋4

15

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

∫ sin2 kx dx = x2− 1

4ksin 2kx

∫ x sin kx dx = sin kxk2 − x cos kx

k

∫ x cos kx dx = x sin kxk

+ cos kxk2

∫ x2sin2kx dx = x3

6− x cos 2kx

4k2 − (2k2x2 − 1) sin 2kx8k3

∫ xe−𝜆x dx = −( x𝜆+ 1𝜆2

)

e−𝜆x

∫ x2e−𝜆x dx = −(

x2

𝜆+ 2x𝜆2 + 2

𝜆3

)

e−𝜆x

∫ x3e−𝜆x dx = −(

x3

𝜆+ 3x2

𝜆2 + 6x𝜆3 + 6

𝜆4

)

e−𝜆x

∫dx

(x2 + a2)2 = x2a2(x2 + a2)

+ 12a3 tan−1(x∕a)

523

Bibliography

A: Introductory undergraduate level

Feynman, R. (1971) Lectures on Physics, Vol. III, Addison-Wesley.French, A. (1978) An Introduction to Quantum Mechanics, W. W. Norton.Wichmann, E.H. (1971) Quantum Physics, Berkeley Physics Course, Vol. 4,

McGraw-Hill.

B: Intermediate undergraduate level

Abers, E.S. (2003) Quantum Mechanics, Prentice Hall.Bohm, D. (1951) Quantum Theory, Prentice Hall.Bransden, B.H. and Joachain, C.J. (2000) Quantum Mechanics, 2nd edn,

Prentice Hall.Cohen-Tannoudji, C., Diu, B., and Laloë, F. (1992) Quantum Mechanics I & II,

Wiley-VCH.Eisberg, R. (1960) Fundamentals of Modern Physics, John Wiley & Sons.Eisberg, R. and Resnick, R. (1985) Quantum Physics of Atoms, Molecules, Solids,

Nuclei and Particles, 2nd edn, John Wiley & Sons.Gasiorowitz, S. (2003) Quantum Physics, 3rd edn, John Wiley & Sons.Griffiths, D.J. (2016) Introduction to Quantum Mechanics, 2nd edn, Cambridge

University Press.Liboff, R.L. (2002) Introductory Quantum Mechanics, Addison-Wesley.Miller, D. (2008) Quantum Mechanics for Scientists and Engineers, Cambridge

University Press.McIntyre, D. and Manogue, C. (2012) Quantum Mechanics: A Paradigms Approach,

Pearson.Shankar, R. (2011) Principles of Quantum Mechanics, 2nd edn, Plenum Press.Singh, J. (1997) Quantum Mechanics (Fundamentals & Applications to Technology),

John Wiley & Sons.


524 Bibliography

Susskind, L. and Friedman, A. (2015) Quantum Mechanics: The TheoreticalMinimum, Basic Books.

Townsend, J. (2012). A Modern Approach to Quantum Mechanics, 2nd edn,University Science Books.

Yariv, A. (1982) Theory and Applications of Quantum Mechanics, John Wiley& Sons.

Zettili, N. (2009) Quantum Mechanics: Concepts and Applications, 2nd edn, JohnWiley & Sons.

C: Advanced undergraduate and graduate level

Baym, G. (1969) Lectures on Quantum Mechanics, Benjamin.Bethe, H.A. and Jackiw, R. (1986) Intermediate Quantum Mechanics,

Addison-Wesley.Dirac, P.A.M. (1967) The Principles of Quantum Mechanics, Oxford University Press.Landau, L. and Lifshitz, E. (1958) Quantum Mechanics, Pergamon Press.Merzbacher, E. (1970) Quantum Mechanics, 2nd edn, 1977, John Wiley & Sons.Messiah, A. (1965) Quantum Mechanics I & II, North-Holland Publishing Co.Sakurai, J.J. (1993) Modern Quantum Mechanics, revised edn, Addison Wesley.Schiff, L. (1955) Quantum Mechanics, McGraw-Hill.

D: Structure of matter and light-matter interaction

Atkins, P. and Friedman, R. (2010) Molecular Quantum Mechanics, OxfordUniversity Press.

Bethe, H. and Salpeter, E. (1977) Quantum Mechanics of One and Two ElectronAtoms, 2nd edn, Plenum Publishing Corporation.

Dresselhaus, M.S. (1996) Science of Fullerenes and Carbon Nanotubes: TheirProperties and Applications, Academic Press.

Economou, E. N. (2010) The Physics of Solids: Essentials and Beyond, Springer.Gasiorowitz, S. (1979) The Structure of Matter, Addison-Wesley.Kaxiras, E. (2003) Atomic and Electronic Structure of Solids, Cambridge University

Press.Loudon, R. (1973) The Quantum Theory of Light, Oxford University Press.Morrison, M.A. (1976) Quantum States of Atoms, Molecules and Solids,

Prentice Hall.Murrel, J.N., Kettle, S.F., and Tedder, J.M. (1985) The Chemical Bond, John Wiley

& Sons.Read, F.H. (1980) Electromagnetic Radiation, John Wiley & Sons.Svelto, O. (2010) Principles of Lasers, 5th edn, Springer-Verlag.Woodgate, G.K. (1980) Elementary Atomic Structure, Oxford University Press.Yariv, A. (1975) Quantum Electronics, 2nd edn, John Wiley & Sons.

Bibliography 525

E: Problems

Constantinescu, F. and Magyari, E. (1971) Problems in Quantum Mechanics,Pergamon Press.

Flügge, S. (1971) Practical Quantum Mechanics I & II, Springer-Verlag.Gol'dman, I.I. and Krivchenkov, V.D. (1961) Problems in Quantum Mechanics,

Pergamon Press.Squires, G. (1995) Problems in Quantum Mechanics, Cambridge University Press.

527

Index

aacetylene molecule 406adjoint operator OS2.2, OS10.1CAiry equation 195, 206algebraic theory of

harmonic oscillator OS6.2orbital angular momentum

OS10.2spin OS10.2

alkali metals (or alkalis) 331, 341alpha decay 160, 162, OS5.2ammonia maser 374ammonia molecule 373–377

and nitrogen inversion 373, 374angular momentum 122

addition law 296commutation relations for 124conservation 231–232eigenstates 237eigenstates of total 300eigenvalues 237, 242orbital 231peculiar properties of quantum

261–262spin (see spin)total 295uncertainty relations OS10.2

annihilation operator OS6.2antibonding orbital 359antisymmetric states 307Archimedean solids 431 see also

semi-regular polyhedraaromatic hydrocarbons 418, 421

and Hückel’s rule 418stabilization energy 421

associated Legendreequation 240, OS9.3polynomials 242, OS9.3

atomic energy scale 15atomic length scale 19atomic magnetism 329atomic orbitals 258

1s, 2s orbitals 259px, py, pz orbitals 260–261

atomic polarizability OS12.2atomic stability 13

according to Bohr 39under collisions 13under external EM radiation 17under internal radiation 21

atomic units 216of energy 249of length 249

bBalmer’s formula 25band

gap 439spectrum 441structure 440

Bell inequalities 229benzene molecule 409, 414

delocalization in 423the free-electron model in 417the resonance concept 424

Bessel’s equation 205bidimensional equations 194

of the first (second, third) kind 205blackbody radiation 44


528 Index

Blochoscillations 451theorem 450

Bohrfrequency 24, 93magneton 272radius 28theory 28

Bohr’s quantization condition 28and de Broglie waves 28for power-law potentials 50

bond(s)delocalized 414double 364hybridized 393, 395 (see also

hybridization)ionic 337polar 337metallic 431nonpolar covalent 337𝜋 bond 364polar covalent 337𝜎 bond 364

bonding and antibonding orbitals 359bosons 308, 314box

one-dimensional 132three-dimensional 148two-dimensional 148

Brillouin zone 449bulk modulus OS15.3butadiene molecule, (C4H6) 425

ccarbon nanotubes 434central potential 231Chandrasekhar limit OS15.4chemical bond 331

directionality of 331double-well model 352elementary theory 352molecular orbital 331valence 331valence electrons 332valence shell 331

circular polarization 514cold emission 165

collapse of the wavefunction uponmeasurement 35, 86

and action at a distance 36and the double-slit experiment 36

commutation relations OS3.2commutator 121compatible physical quantities 119Compton effect 5, 8Compton wavelength 8conductivity

Drude formula 442the “mystery” of 443

conductor 443confinement, resistance to 182conformation

eclipsed 407staggered 407

conjugated hydrocarbons 408cyclic chains 418Hückel’s theory 421hybridization theory for 408LCAO theory for 408linear chains 424stabilization energy 421

conservation ofangular momentum 231–232, 485,

OS9.2probability 62, OS2.3

continuity conditions 130 see alsomatching conditions

continuous spectrum 86, 378and non-localized solutions 132

Copenhagen interpretation 229correspondence principle 496Coulomb potential 210

bare 324screened 324

cross section 473, 475as a function of photon energy 482scattering 476

crystal momentum 450

dDavisson–Germer experiment 33de Broglie hypothesis 12debye (unit) 368degeneracy 250

Index 529

accidental 255exchange 317order of 255pressure see Fermi pressureremoval of 257rotational 256and symmetry 256

delocalized bonds 414in two-dimensional arrays 429–433

delta function see Dirac delta functiondensity of states 457

experimental determination 460fractional 492in free-electron model 457in one, two and three dimensions

458typical experimental curve 459

diamagnetic (atoms) 331dibromobenzene molecule 415diffraction 38, 111dimensional analysis (or method) 41

and blackbody radiation 44the fundamental theorem of 41and the hydrogen atom 47and ultraviolet catastrophe 46

dipole moment (atoms) 485discrete spectrum 82, 86

and localized solutions 132dispersion relation 59, 451Dirac

bra vectors 513, OS11.2delta function 89, OS2.1Bformalism 513, OS11.2ket vectors 513, OS11.2

Dopplerbroadening of spectral lines 492shifts of spectral lines 492

double-slit experiment 34double-well model 352

asymmetric double-well 356symmetric double-well 356

Drude formula 442dynamical symmetry 257

eEhrenfest theorem OS3.4effective charge 324, 335

effective cross section 473 see alsocross section

effective mass 447, 449effective potential 235eigenfunction(s)

completeness 98orthogonality 97

electric dipole transitions 486electron affinity 337electronegativity 337electronic specific heat 463electron microscope 32electron mobility 447energy

eigenfunctions 80eigenvalues 80, 81spectrum 86, 131energy bands 439, 443

energy gaps 439energy quantization

mathematical mechanism 85and the stability of matter 13

equationsassociated tridimensional 205bidimensional 194unidimensional 195

equipartition theorem 463exactly solvable potentials 203, OS7.2,

OS7.3Kratzer potential 197Morse potential 197

exchange degeneracy 317exclusion principle see Pauli exclusion

principleexpectation value see mean valueEuler equations see unidimensional

equations

fFermi

pressure 462sea 463temperature 464wavenumber 456

Fermi energy 453, 456qualitative understanding 461

Fermi’s (golden) rule 478

530 Index

fermions 308, 314ferromagnetism 310fine structure constant 29Floquet’s theorem OS15.2forbidden reflection 53forbidden transition 263forbidden transmission see tunneling

effectfree-electron model

for benzene 417for C60 431density of states in 457for a solid 454

fullerenes 429C60 fullerene 431free-electron model for 431

gGaussian integral 68

and its derivatives 69generalized 503

Gaussian wavefunction 69gedanken experiment 111graphene 430gravitational collapse 463, OS15.4

in neutron stars OS15.4in white dwarfs OS15.4

gyromagnetic ratio 269, 272

hHall experiment 446halogens 336Hamiltonian

classical 56quantum 56

harmonic oscillator 167algebraic solution OS6.2allowed and forbidden transitions

186anisotropic 188classical limit 179correspondence principle 185creation and annihilation operators

OS6.2eigenfunctions 174, 178eigenvalues 174emission of radiation by 184

penetration into classicaly forbiddenregions 181

selection rules 185solution method 169three-dimensional 188Stark effect for the 187zero-point energy 183

Heisenberg’s uncertainty principle seeuncertainty principle

heliumatom 328excited states 349 see also

orthohelium; parahelium“molecule” 363

helium-like systems 343Hermite polynomials 174Hermite equation 174Hermitian conjugate (or adjoint)

OS2.2Hermitian matrix 287Hermitian operator 95, 99

properties of 98hexatriene molecule 409hidden variables 229Hückel’s rule 418Hückel’s theory 421Hund’s (first) rule 330hybridization 393, 396

for inorganic molecules 409nontypical 436partial 401, 405sp1 402, 403sp2 402, 403sp3 402total 401

hydrogen atom 207, 246accidental (or hydrogenic) degeneracy

255allowed and forbidden transitions

263, 264atomic orbitals 258atomic units 216conceptual questions 227–228eigenenergies (or allowed energies)

25, 214eigenfunctions 213, 251energy-level diagram 26, 254

Index 531

energy spectrum 25fine structure 299forbidden regions 224hyperfine structure 304ionization energy 25, 219Lentz vector 256lifetime of 2p state 494minimum excitation energy 218penetration in the classicaly

forbidden region 224principal quantum number 249quantum mechanics vs. Bohr’s

theory 225radial functions Rn𝓁(r) 253selection rules 263spectroscopic notation of states

258and the uncertainty principle 218,

221hydrogen bond(s) 370

in DNA molecule 372hydrogen-like atoms (or ions) 30hydrogen molecule 360hyperfine interaction 304hypergeometric equation 202, 205

confluent 202, 205

iidentical particles 305

interacting pairs of OS11.1Bprinciple of indistinguishability of

305incompatible physical quantities 119indeterminacy principle see

uncertainty principleinfinite potential well 132

classical limit (correspondenceprinciple) 138

eigenfunctions 135eigenvalues 135

information waves 35inner product 99

properties of 100interference of quantum waves

33–34insulator 443insulator vs. conductor 443

ionic bond 337ionic crystal 336ionization of atoms by light 472ionization energy 334, 335

kkinetic energy

and resistance to confinement 112Klein–Gordon equation 56Kronecker delta 398

lLaguerre’s equation 202Lande factor 281Larmor frequency 276laser

amplification factor 490cavity 488continuous wave 491directionality 491four-level 493intensity 491monochromaticity 492operation principle 487phase coherence 491population inversion 487pulsed 491pumping 488threshold condition 488, 489

lattice constant 449LCAO (Linear Combination of Atomic

Orbitals) 352, OS14.2, OS14.3Legendre equation 240

polynomials 241–242Lentz vector 256line shape (or line profile) 492line-width 115

natural 117linear operator 58linear polarization 512lithium

atom 329, 336molecule 364

local conservation of probabilityOS2.3

Lyman series 277

532 Index

mmagnetic dipole 270magnetic moment 270, 271magnetic quantum number 239, 269magnetic spin anomaly 283many-electron atoms 324

arrangement of energy levels 328maser 374matching conditions 130matter waves see wave-particle duality

of mattermatrix

elements of an operator 102, 354mechanics OS14.2representation of an operator 101

mean free path 14, 473, 475mean lifetime 473mean value

of a quantum mechanical quantity71

of a statistical quantity 63measurement

collapse of the wavefunction upon86, 87, 104

complete OS9.2principle 104in quantum mechanics 86

methane molecule 394, 398mirror symmetry 139, 178mobility (of electrons in crystalline

solids) 447momentum

crystal momentum 447eigenvalue equation 88momentum-position uncertainty

principle 108quantum mechanical operator

72molecular spectra 377

moleculeC2H2 (acetylene) 406C2H4 406C2H6 407C4H4 429C4H6 (butadiene) 427C6H8 (hexatriene) 409CH4 (methane) 394, 398

HCl (hydrogen chloride) 388Morse potential 200

application to vibrational spectrum201

exact solution 200

nneutron star

and Chandrasekhar’s limit OS15.4and gravitational collapse OS15.4

nitrogen inversion 373, 374nitrogen molecule 366noble gases 329nodal lines 26

nodal surfaces 26node(s) 26

theorem 178normalization

condition 61factor (constant) 62

normal modes 12norm (or length) of a wavefunction

101nuclear energy scale 15nuclear length scale 19nuclear magnetic resonance (NMR)

377, OS16.3nuclear magneton 273

ooperators 58

commutativity of 59linear 58

orthogonality 82, 99, 100orthohelium 319, 349oxygen molecule 364

pparabolic approximation 168parahelium 319, 349paramagnetic (atoms) 331parity, 484

operator OS6.1BPauli exclusion principle 310Pauli matrices 288, 290Pauli principle 305, 308, 317

and atomic magnetism 329

Index 533

penetrationin the classical limit 144into classicaly forbidden regions

143length 144

periodic potentials 439, OS15.2periodic table 327, 332

chemical periodicity 332shell model 328small periodic table 327

perturbation theory (time-dependent)and Fermi’s rule 499

perturbation theory (time-independent)342

application to helium-like systems343–345

application to Stark effect OS12.2first order 343for degenerate levels OS12.1Bsecond order OS12.2systematic theory OS12.2

phase factor 91photoelectric effect 4

cutoff potential 6Einstein’s photoelectric equation 5experimental facts 6and measurement of Planck’s

constant 5photon

as a particle (Compton experiment)8

as the quantum of light (photoelectriceffect) 4

Planck’s constant 3reduced 24

Planck’s radiation formula 45polarized photons 511, 513

circularly 511linearly 511polarizers 513

polynomial method 191polynomial solutions, existence

theorem 196population inversion 487position

eigenvalue equation 89

position-momentum uncertaintyprinciple 108

probability density 61quantum mechanical operator 72

positronium 230postulates of quantum mechanics 104potential barrier

rectangular 156transmission probability 158tunneling effect see tunneling effect

potential wellsquare 140general shape 131infinite 132

power-series method 191precession of spin in a magnetic field

292, OS16.3principal quantum number 249probability waves (see quantum

waves)principle of

wave-particle duality 3wave-particle duality of light 4wave-particle duality of matter 11

probabilistic interpretation of matterwaves 21

probability current density 149,OS2.3

probability density 22, 64

qquantum discontinuities

in the classical limit 137quantum jump (or leap) 13quantum number

angular momentum 249magnetic 249principal 249spin 278

quantum transitions, the problem of470

quantum wavesDavisson–Germer experiment 33the double-slit experiment 34standing 12

quasi-momentum see crystalmomentum

534 Index

rRabi oscillations 510, OS16.2radial

function 245probability density 222wavefunction 245

radial Schrödinger equationin an arbitrary central potential 235in the hydrogen atom 246

Rayleigh–Jeans law 46reaction (chemical)

addition 422substitution 422

recombination 444rectangular potential barrier see square

potential barrierrectangular potential well see square

potential wellrecurrence relation 192, 194reduced mass 207, 379reflection coefficient

for a potential step 151for a square potential barrier 157

reflection symmetry see mirrorsymmetry

resonance cross section 481, 482resonant absorption 472

cross section 481and laser operation 487

rigid rotor 378rotational spectrum 378

selection rule 380Rydberg constant 25Rydberg states 496

sscattering

boundary conditions 151cross section 476from a delta function barrier 164of light by atoms 472from a potential step 150from a rectangular potential barrier

157states 149

Schrödinger equation 53, 103–104bound states 85

general solution 78–80history 53–54scattering states 149stationary states 91statistical interpretation 61time-dependent 79, 103time-independent 79

selection rulesfor light-atom interactions 483for rotational transitions 380for vibrational transitions 383

semiclassical theory of light-atominteractions 469

angular distribution of emittedphotons 498

the “cubic-power law” 496polarization of emitted photons 498selection rules 483

semiconductor 441doped 445effective mass 449energy bands 445Hall experiment 446holes 445recombination process 444valence band 445

semi-regular polyhedra 431separation

constant 79, 234of variables 78, 233

shell model (of the atom) 328singular points 240solid angle (definition) 253spectrum

absorption 24continuous 86discrete 86electronic 384emission 24line 24mixed 86rotational 384vibrational 384vibrational–rotational 387, 388

specific resistance 441spectroscopic notation 258spherical harmonics 243, 252

Index 535

spherical potential well OS8.1Bspin

elementary theory 278, 285experimental confirmation 278and the Pauli principle 307spin magnetic anomaly 283spin matrices 289spin-orbit coupling 298spin quantum number 278wavefunction (or state vector) 286time evolution (or spin precession)

in a magnetic field 292, 294what spin really is 284, 291

spontaneous decay see spontaneousemission

spontaneous emission 472, 494correspondence principle 496the “cubic-power law” 495, 496Rydberg states 496semiclassical treatment 497

spring constant 42square potential(s) 129square potential barrier see potential

barriersquare potential well 140, OS4.2,

OS4.3stability of matter 13stabilization energy 421standard deviation (or uncertainty) 65standing quantum waves 12

and energy quantization 12and the stability of matter 13

Starkeffect 267problem 268shift 268shift for hydrogen OS12.1B, OS12.2

state vector 286, 500, 517stationary state 91statistical interpretation 61, 71statistical moments 66statistical interpretation of

quantum mechanics 60the wavefunction 61, 83

Stefan–Boltzmannconstant 45, 49law 45

step potential 150classical limit 154forbidden reflection 153reflection coefficient 153scattering from 151

Stern–Gerlach device 280, 515–516,OS10.1B

experiment 280stimulated emission 472strong quantum limit 136

superposition states 81–82symmetry

and degree of degeneracy 257dynamical 257spherical 256

system(s) of units 10atomic units 216cgs vs. SI 10

tthermal light see blackbody radiationthermal neutrons 33thermal speed 14threshold frequency 5time evolution of

mean values OS3.4probability OS2.3wavefunctions 77

time evolution operator OS10.1Ctrace (of a matrix) 289transition

probabilities 471, 502, 505–506rate 473

transmission coefficientin a potential barrier 160in a potential step 151

transmission resonance 159, OS5.1Btime-dependent perturbations 499

Fermi’s rule 503–504transition probabilities 502

time–energy uncertainty principle 114tunneling effect

and alpha decay of nuclei 161,OS5.2

in an arbitrary potential barrier 162exponential sensitivity of 159, 162in a square potential barrier 156

536 Index

two-photon decay 264two-state systems OS12.1B, OS16.2

uultraviolet catastrophe 46unbound states see scattering statesuncertainty principle 107

classical 111in the classical limit 118generalized 119, OS3.3position–momentum 108time–energy 114

unidimensional equations 195unitary operators OS10.1Cunits see systems of units

vvalence 331valence shell 331van der Waals forces 361variational

helium-like systems 346, 348method 346parameter 346theorem 349

vector space 100, 513, OS2.1Bvibrational–rotational spectrum

385, 387selection rules 386

vibrational spectrum 382selection rule 383

visible spectrum 384

wwater molecule 332, 367, 411

dipole moment of 368shape 332, 367, 411

wave(s)plane 60

sinusoidal 60information 35

wavefunction(s) 22antisymmetric 312, 316in the continuous spectrum 89–91normalized(-able) 61square integrable 61statistical interpretation 61symmetric 312, 316

wavepacket propagation OS2.1Bwave–particle duality 3wave–particle duality of light 4

experimental confirmation 5, 8wave–particle duality of matter 11

and the double-slit experiment 34experimental confirmation 33practical formulas and applications

31probabilistic interpretation 21and the problem of atomic stability

13, 21and the problem of energy scales

15and the problem of length scales

19weak quantum limit 136Wien’s law 45work function 4Wronskian OS15.2

zZeeman

anomalous Zeeman effect 277effect 267, 274, 276normal Zeeman effect 277shift 268spectrum 269, 276splitting 269, 275

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