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MODERN PHYSICSBUT MOSTLY

QUANTUM MECHANICS

Seeing the invisible, Proving the impossible, Thinking theunthinkable

Modern physics stretches your imagination beyond recognition,pushes the intellectual envelope to the limit, and takes you to the

forefront of human knowledge.

Masayasu AOTANI(青谷正妥)

Kyoto UniversityKyoto, Japan

Anax parthenope （ギンヤンマ）

2013No insects, no life!

Modern Physics

Masayasu AOTANI（青谷正妥）

Spring 2014

Copyright c⃝1991–2014

Masayasu AOTANI

Permission is granted for personal useonly.

Contents

1 Classical Physics vs. Modern Physics 131.1 Extreme Conditions, Precise Measurements, and New Physics . . . . 131.2 Black-Body Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . 151.3 Quantum Theory of Light . . . . . . . . . . . . . . . . . . . . . . . . 15

1.3.1 Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . 171.3.2 Einstein’s Idea of Quantization . . . . . . . . . . . . . . . . . 17

1.4 de Broglie-Einstein Postulates . . . . . . . . . . . . . . . . . . . . . . 18

2 Mathematical Preliminaries 192.1 Linear Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.3 L2-Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.4 The Braket Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.5 Vector Subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.6 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.7 Matrix Representation of Linear Operators . . . . . . . . . . . . . . . 43

2.7.1 Matrix Representations of Operator Products . . . . . . . . . 472.8 The Adjoint of an Operator . . . . . . . . . . . . . . . . . . . . . . . 482.9 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . 502.10 Special Types of Operators . . . . . . . . . . . . . . . . . . . . . . . 53

2.10.1 Hermitian Operators . . . . . . . . . . . . . . . . . . . . . . . 532.10.2 Simultaneous Diagonalization . . . . . . . . . . . . . . . . . . 65

2.11 Active and Passive Transformations . . . . . . . . . . . . . . . . . . . 66

3 Fundamental Postulates and the Mathematical Framework of Quan-tum Mechanics 713.1 The Fundamental Postulates . . . . . . . . . . . . . . . . . . . . . . . 723.2 Unitary Time Evolution . . . . . . . . . . . . . . . . . . . . . . . . . 75

3

3.3 Time-Independent Hamiltonian . . . . . . . . . . . . . . . . . . . . . 773.3.1 Propagator as a Power Series of H . . . . . . . . . . . . . . . 773.3.2 Eigenstate Expansion of the Propagator . . . . . . . . . . . . 81

3.4 The Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . 823.4.1 Uncertainty and Non-Commutation . . . . . . . . . . . . . . . 823.4.2 Position and Momentum . . . . . . . . . . . . . . . . . . . . . 843.4.3 Energy and Time . . . . . . . . . . . . . . . . . . . . . . . . . 85

3.4.3.1 Time rate of change of expectation values . . . . . . 853.4.3.2 Time-energy uncertainty . . . . . . . . . . . . . . . 86

3.5 Classical Mechanics and Quantum Mechanics . . . . . . . . . . . . . 873.5.1 Ehrenfest’s Theorem . . . . . . . . . . . . . . . . . . . . . . . 873.5.2 Exact Measurements and Expectation Values . . . . . . . . . 89

4 Spaces of Infinite Dimensionality 914.1 Two Types of Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . 914.2 Countably Infinite Dimensions . . . . . . . . . . . . . . . . . . . . . . 934.3 Uncountably Infinite Dimensions . . . . . . . . . . . . . . . . . . . . 944.4 Delta Function as a Limit of Gaussian Distribution . . . . . . . . . . 974.5 Delta Function and Fourier Transform . . . . . . . . . . . . . . . . . 974.6 f as a Vector with Uncountably Many Components . . . . . . . . . . 98

4.6.1 Finite Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 984.6.2 Countably Infinite Dimensions . . . . . . . . . . . . . . . . . 984.6.3 Uncountably Infinite Dimensions . . . . . . . . . . . . . . . . 99

4.7 Hermiticity of the Momentum Operator . . . . . . . . . . . . . . . . 1014.7.1 Checking [Pij] = [P ∗

ji] . . . . . . . . . . . . . . . . . . . . . . . 1014.7.2 Hermiticity Condition in Uncountably Infinite Dimensions . . 1024.7.3 The Eigenvalue Problem of the Momentum Operator P . . . . 104

4.8 Relations Between X and P . . . . . . . . . . . . . . . . . . . . . . . 1054.8.1 The Fourier Transform Connecting X and P . . . . . . . . . . 1054.8.2 X and P in the X Basis . . . . . . . . . . . . . . . . . . . . . 1074.8.3 Basis-Independent Descriptions in Reference to the X Basis . 1074.8.4 X and P in the P Basis . . . . . . . . . . . . . . . . . . . . . 1084.8.5 The Commutator [X,P ] . . . . . . . . . . . . . . . . . . . . . 110

5 Schrodinger’s Theory of Quantum Mechanics 1115.1 How was it derived? . . . . . . . . . . . . . . . . . . . . . . . . . . . 1115.2 Born’s Interpretation of Wavefunctions . . . . . . . . . . . . . . . . . 1165.3 Expectation Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

6 The Time-Independent Schrodinger Equation 1256.1 Separation of Time t and Space x . . . . . . . . . . . . . . . . . . . . 1256.2 Conditions on the Wavefunction ψ(x) . . . . . . . . . . . . . . . . . . 127

7 Solutions of Time-Independent Schrodinger Equations in One Di-mension 1317.1 The Zero Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1317.2 The Step Potential (E < V0) . . . . . . . . . . . . . . . . . . . . . . . 1337.3 The Step Potential (E > V0) . . . . . . . . . . . . . . . . . . . . . . . 1397.4 The Barrier Potential (E < V0) . . . . . . . . . . . . . . . . . . . . . 1427.5 The Infinite Square Well Potential . . . . . . . . . . . . . . . . . . . 1437.6 The Simple Harmonic Oscillator Potential . . . . . . . . . . . . . . . 145

7.6.1 Classic Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 1457.6.2 Raising and Lowering Operators . . . . . . . . . . . . . . . . 153

7.6.2.1 Actions of a and a† . . . . . . . . . . . . . . . . . . 159

8 Higher Spatial Dimensions 1698.1 Degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1698.2 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

9 Angular Momentum 1759.1 Angular Momentum Operators . . . . . . . . . . . . . . . . . . . . . 1759.2 Quantum Mechanical Rotation Operator UR . . . . . . . . . . . . . . 1789.3 Rotationally Invariant Hamiltonian . . . . . . . . . . . . . . . . . . . 1829.4 Raising and Lowering Operators: L+ and L− . . . . . . . . . . . . . . 1839.5 Matrix Representations of L2 and Lz . . . . . . . . . . . . . . . . . . 1899.6 Generalized Angular Momentum J . . . . . . . . . . . . . . . . . . . 191

10 The Hydrogen Atom 19510.1 2 Particles Instead of 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 19610.2 Three-Dimensional System . . . . . . . . . . . . . . . . . . . . . . . . 19710.3 The Solutions for Φ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20710.4 The Solutions for Θ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20810.5 Associated Legendre Polynomials and Spherical Harmonics . . . . . . 21410.6 L2, Lz, and the Spherical Harmonics . . . . . . . . . . . . . . . . . . 21610.7 The Radial Function R . . . . . . . . . . . . . . . . . . . . . . . . . . 21710.8 The Full Wavefunction . . . . . . . . . . . . . . . . . . . . . . . . . . 22010.9 Hydrogen-Like Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . 22510.10Simultaneous Diagonalization of H, L2, and Lz . . . . . . . . . . . . 226

10.11Revisiting the Fundamental Postulates . . . . . . . . . . . . . . . . . 230

11 Electron Spin 23511.1 What is the Electron Spin? . . . . . . . . . . . . . . . . . . . . . . . 235

11.1.1 Stern-Gerlach Experiment . . . . . . . . . . . . . . . . . . . . 23611.1.2 Fine Structure of the Hydrogen Spectrum: Spin-Orbit Inter-

action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23711.2 Spin and Pauli Matrices . . . . . . . . . . . . . . . . . . . . . . . . . 23811.3 Sequential Measurements . . . . . . . . . . . . . . . . . . . . . . . . . 24511.4 Spin States in an Arbitrary Direction . . . . . . . . . . . . . . . . . . 248

11.4.1 Spin Matrix Su and Its Eigenvectors . . . . . . . . . . . . . . 24811.4.2 Sequential Measurements Revisited . . . . . . . . . . . . . . . 25111.4.3 Comparison with the Classical Theory . . . . . . . . . . . . . 251

12 Addition of Orbital and Spin Angular Momenta 255

13 Molecular Rotation 25713.1 Rotational Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . 25713.2 Moment of Inertia for a Diatomic Molecule . . . . . . . . . . . . . . . 25813.3 Two-Dimensional Rotation Confined to the x, y-Plane . . . . . . . . . 25913.4 Three-Dimensional Rotation . . . . . . . . . . . . . . . . . . . . . . . 261

Appendix A Matrix Algebra 267A.1 Invertible Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267A.2 Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268A.3 Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271A.4 Multiplicity of an Eigenvalue . . . . . . . . . . . . . . . . . . . . . . 275A.5 Diagonalizability and Simultaneous Diagonalization . . . . . . . . . . 276A.6 Some Useful Definitions, Facts, and Theorems . . . . . . . . . . . . . 286

Appendix B Holomorphic Functional Calculus 287

Appendix C Traveling Waves and Standing Waves 289C.1 The Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 289

C.1.1 Traveling Wave Solutions . . . . . . . . . . . . . . . . . . . . 290C.1.1.1 Real and Complex Expressions for the Traveling Wave290

C.1.2 Standing Wave Solutions . . . . . . . . . . . . . . . . . . . . . 291C.1.2.1 Standing Waves on a String . . . . . . . . . . . . . . 291C.1.2.2 Standing Waves in Quantum Mechanics . . . . . . . 292

Appendix D Continuous Functions 293D.1 Definition of Continuity . . . . . . . . . . . . . . . . . . . . . . . . . 293D.2 Extreme and Intermediate Value Theorems . . . . . . . . . . . . . . . 294D.3 Continuity Condition on the First Derivative . . . . . . . . . . . . . . 294

Appendix E Wronskian and Linear Independence 297

Appendix F Newtonian, Lagrangian, and Hamiltonian Mechanics 301F.1 An Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301F.2 Newtonian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 302F.3 Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 302F.4 Hamiltonian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 303

Appendix G Normalization Schemes for a Free Particle 305G.1 The Born Normalization . . . . . . . . . . . . . . . . . . . . . . . . . 305G.2 The Dirac Normalization . . . . . . . . . . . . . . . . . . . . . . . . . 305

G.2.1 Fourier Transform and the Delta Function . . . . . . . . . . . 306G.2.2 Wavefunctions as Momentum or Position Eigenfunctions . . . 306

G.3 The Unit-Flux Normalization . . . . . . . . . . . . . . . . . . . . . . 309

Appendix H Symmetries and Conserved Dynamical Variables 311H.1 Poisson Brackets and Constants of Motion . . . . . . . . . . . . . . . 311H.2 lz As a Generator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313H.3 Rotation Around The Origin . . . . . . . . . . . . . . . . . . . . . . . 314H.4 Infinitesimal Rotations Around the z-Axis . . . . . . . . . . . . . . . 315

Appendix I Commutators 317I.1 Commutator Identities . . . . . . . . . . . . . . . . . . . . . . . . . . 317I.2 Commutators involving X and P . . . . . . . . . . . . . . . . . . . . 317I.3 Commutators involving X, P , L, and r . . . . . . . . . . . . . . . . 318

Appendix J Probability Current 321

Appendix K Chain Rules in Partial Differentiation 325K.1 One Independent Variable . . . . . . . . . . . . . . . . . . . . . . . . 325K.2 Three Independent Variables . . . . . . . . . . . . . . . . . . . . . . . 325K.3 Reciprocal Relation is False . . . . . . . . . . . . . . . . . . . . . . . 326K.4 Inverse Function Theorems . . . . . . . . . . . . . . . . . . . . . . . . 327

Appendix L Laplacian Operator in Spherical Coordinates 329

Appendix M Legendre and Associated Legendre Polynomials 341M.1 Legendre Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 341M.2 Associated Legendre Polynomials . . . . . . . . . . . . . . . . . . . . 342

Appendix N Laguerre and Associated Laguerre Polynomials 349

Appendix O Fine Structure and Lamb Shift 351O.1 Relativistic Effects: The Dirac Theory . . . . . . . . . . . . . . . . . 351

O.1.1 Variation of the Mass with the Velocity: Hk . . . . . . . . . . 353O.1.2 Spin-Orbit Interaction: HSO . . . . . . . . . . . . . . . . . . . 354O.1.3 The Darwin Term: HD . . . . . . . . . . . . . . . . . . . . . . 355

O.2 The Lamb Shift . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356

Bibliography 358

Answers to Exercises 359Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 376Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379

Subject Index 381

List of Tables

3.1 Physical observables and corresponding quantum operators . . . . . . 733.2 Quantum operators in three dimensions . . . . . . . . . . . . . . . . 73

10.1 Spherical Harmonics for l = 0, 1, and 2 . . . . . . . . . . . . . . . . . 21710.2 Some Normalized Radial Eigenfunctions for the One-Electron Atom . 22610.3 Some Normalized Full Eigenfunctions for the One-Electron Atom . . 227

9

List of Figures

1.1 Black-Body Radiation (Source: Wikimedia Commons; Author: DarthKule) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

10.1 Spherical Coordinate System (Source: Wikimedia Commons; Author:Andeggs) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199

10.2 Spherical Coordinate System Used in Mathematics (Source: Wikime-dia Commons; Author: Dmcq) . . . . . . . . . . . . . . . . . . . . . . 200

10.3 Volume Element in Spherical Coordinates (Source: Victor J. Mon-temayor, Middle Tennessee State University) . . . . . . . . . . . . . . 203

11.1 Stern-Gerlach Experiment (Diagram drawn by en wikipedia TheresaKnott.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236

11.2 Results of Stern-Gerlach Experiment with and without a non-uniformexternal magnetic field (Source: Stern and Gerlach’s original paper[Gerlach and Stern, 1922, p.350]) . . . . . . . . . . . . . . . . . . . . 237

11

Chapter 1

Classical Physics vs. ModernPhysics

Very roughly speaking, Classical Physics consists of Newtonian Mechanics and thetheory of Electricity and Magnetism by Maxwell, while Modern Physics typicallymeans Relativity and Quantum Mechanics.

Classical Physics

• Newtonian Mechanics• Maxwell’s Electricity and Magnetism

Modern Physics

• Relativity• Quantum Mechanics

1.1 Extreme Conditions, Precise Measurements,and New Physics

Quantum Mechanics and the Theory of Relativity emerged early in the 20th cen-tury when technological advances were beginning to make laboratory experimentsunder extreme conditions possible. The same advances also made high-precisionmeasurements readily available. Under these extreme conditions, physicists foundthat Newtonian Mechanics no longer held. One such condition is a microscopicphenomenon such as the behavior of an electron, and another phenomenon involves

13

14 CHAPTER 1. CLASSICAL PHYSICS VS. MODERN PHYSICS

objects moving at speeds comparable to that of light. The former led to the discoveryof Quantum Mechanics, and the latter to Special Relativity1.

Two Types of Extreme Circumstances

1. Microscopic Phenomena (atomic and subatomic)Tunneling: A moving electron can go through a potential barrier which requiresmore energy than the kinetic energy of the electron.Wave-particle duality: Microscopic particles such as an electron sometimesbehaves like a wave showing interference and superposition.

2. At High Speeds (near the speed of light denoted by c)Fast moving particles live longer.Gallilean transformation does not work.

In this book, we will only deal with Quantum Mechanics, and the following listshows some of the key events, discoveries, and theories that contributed to the birthof quantum mechanics.

Historical Events, Discoveries, and Theories

1. Black-body Radiation: (Quantization, Planck’s Postulate)

2. Particle-Like Properties of Radiation: (Wave-Particle Duality)

3. Einstein’s Quantum Theory of Light: A bundle of energy is localized in a smallvolume of space.

4. Atomic spectra indicating discrete energy levels E1, E2, E3, . . . as opposed tocontinuous distribution of energy levels E ∈ [0,∞)2

1There are two distinct theories of relativity; the Special Theory of Relativity and the GeneralTheory of Relativity. It is the Special Theory of Relativity that deals with high-speed phenomena.The General Theory of Relativity is an attempt to explain gravity geometrically. We only concernourselves with the Special Theory, which is a theory of space and time that successfully accountsfor the deviations from Newtonian Mechanics for fast-moving objects.

2We will later see in Section 7.6.1 that E = 0 is impossible in some quantum mechanical systemssuch as a simple harmonic oscillator.

1.2. BLACK-BODY RADIATION 15

1.2 Black-Body RadiationA black body is an idealization of “real-life” black objects in that it does not reflectany light; i.e. perfectly black. When an object is perfectly black, all the radiationcoming from the object is due to emission, and the pattern of emission depends onlyon its temperature. The thermal emission patterns are shown for 3000 K(red), 4000K (green), and 5000 K (blue) in Figure 1.1. The figure also shows the pattern pre-dicted by the classical theory (dark brown). As you can see, the classical theorypredicts that the emitted radiation increases rapidly without bound as the frequencyincreases. This phenomenon is called ultraviolet catastrophe, which does not agreewith the actual observation.

In order to circumvent this problem, in 1901 Max Planck found a mathematicalexpression that fit the data, in which he assumed that energy levels are not contin-uously distributed but are discrete, with each level being an integer multiple of aquantity called Planck constant denoted by h. The Planck constant is a proportion-ality constant between the energy E and the frequency ν such that

E = hν. (1.1)

For curiosity, Planck’s formula for the spectral radiance I(ν, t), the energy per unittime (or the power) radiated per unit area of emitting surface in the normal directionper unit solid angle per unit frequency by a black body at temperature T , is givenby

I(ν, T ) =2hν3

c21

ehνkt − 1

; (1.2)

where ν is the frequency of the radiation, h ≈ 6.62606957×10−34 (Js: joule-seconds)is the Planck constant, c ≈ 299792458 (ms−1: meters per second) is the speed oflight in vacuum, and k ≈ 1.3806488 × 10−23 (m2kg s−2K−1: meter2 kilograms persecond per second per Kelvin) is the Boltzmann constant.

1.3 Quantum Theory of LightWhile it was undeniable that light behaved as a wave, a physical phenomenon calledphotoelectric effect was not consistent with this view.


UV VISIBLE INFRARED

0

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

Sp

ectr

al ra

dia

nce

(kW

· s

r- ¹ ·

m- ²

· n

m- ¹)

Wavelength (m)

5000 K

4000 K

3000 K

Classical theory (5000 K)

0

2

4

6

8

10

12

14

0 0.5 1 1.5 2 2.5 3

Wavelength (m)

5000 K

4000 K

3000 K

Classical theory (5000 K)

Figure 1.1: Black-Body Radiation (Source: Wikimedia Commons; Author: DarthKule)

1.3. QUANTUM THEORY OF LIGHT 17

1.3.1 Photoelectric EffectPhotoelectric effect is a phenomenon where electrons, called photoelectrons, areejected from a metal surface when light shines on it. The following two observa-tions were made.

• The intensity of the incident light had no effect on the maximum kinetic energyof the photoelectrons. The brightness or dimness of the light had no effect onthe maximum kinetic energy if the frequencies are the same.

• No electron was ejected by light with frequencies below a certain cutoff value,called the threshold frequency.

These were not consistent with the wave nature of light. First, the classical the-ory predicted that light with higher intensity ejects electrons with greater energy.Secondly, according to the wave picture, light with lower frequency can also ejectelectrons except that it takes longer.

1.3.2 Einstein’s Idea of QuantizationThese discrepancies were resolved by Einstein’s idea of light quantum, which claimsthat radiation energy is not continuously distributed over the wavefront, but is local-ized forming bundles of energy or particles. They were later called photons. Einsteinpostulated that a single photon interacts with a single electron transferring its energyto the electron instantaneously. The amount of energy E carried by the photon isgiven by the Planck’s formula

E = hν. (1.3)

Another important relation, which is a direct consequence of (1.3) is

p =h

λ; (1.4)

where p is the momentum and λ is the wavelength. 3

3


1.4 de Broglie-Einstein PostulatesEinstein showed that light, which had classically been regarded as an electromagneticwave, can also be viewed as a collection of particles, later called photons. This isknown as wave-particle duality, i.e. light possesses both a wave nature and a particlenature.

A French physicist named de Broglie tried to generalize this idea. As light, classicallya wave, is also particle-like, he conjectured that things such as an electron, so farconsidered to be a particle, could also possess a wave-nature. The wave associatedwith a particle is called de Broglie wave or de Broglie’s matter wave.

In order to compute the frequency and the wavelength of the matter wave, de Broglieassumed that Einstein’s relations (1.3) and (1.4) also apply here.

E = hν p =h

λ(1.7)

The relations (1.7) are known as de Broglie-Einstein relations.

The correctness of these formulas were soon confirmed by two American physicistsDavisson and Germer, who confirmed that the interference pattern for an electronbeam reflected from a single crystal of nickel fitted perfectly to the de Broglie-Einsteinrelations. Note that an interference phenomenon is possible only if electrons exhibita wave nature.

Development of Quantum Mechanics was an attempt to merge this wave-particleduality and classical Newtonian Mechanics in one equation as we will see in Chapter5.

The relation (1.4) is derived from (1.3) and the relativistic energy formula

E2 =(mc2

)2+ (pc)2; (1.5)

where m is the rest mass, and c is the speed of light. Because m = 0 for a photon, we have

E2 = (pc)2 =⇒ E = pc =⇒ E = pνλ =⇒ hν = pνλ =⇒ p =h

λ. (1.6)

Chapter 2

Mathematical Preliminaries

This chapter is a summary of the mathematical tools used in quantum mechanics.It covers various important aspects of basic linear algebra, defines linear operators,and sets convenient notations for the rest of the book.

2.1 Linear Vector SpacesConsider the three dimensional Cartesian space R3 equipped with the x-, y-, and z-axes. We have three unit vectors, perpendicular/orthogonal to one another, denotedby (i, j,k), (ı, ȷ, k), or (x, y, z). In the component notation, we have i = (1, 0, 0),j = (0, 1, 0) and k = (0, 0, 1). Any point p in this space identified by three coordi-nates x, y, and z such that p = (x, y, z) can also be regarded as a vector xi+yj+zkor (x, y, z) in the component notation. Note that we use (x, y, z) both for the co-ordinates and components by abuse of notation. You must be familiar with sucharithmetic operations as addition, subtraction, and scalar multiplication for vectors;which are basically component-wise operations1. This is a canonical example of alinear vector space, and you should keep this example in mind as you study thefollowing definition of a linear vector space. If you are not so inclined, you can safelyskip this definition and still can understand the rest of this book fully.

Definition 2.1 (Linear Vector Space)A linear vector space V is a set v1,v2,v3, . . . of objects called vectors for which“addition” and scalar multiplication are defined, such that

1To be more precise, these operations can be done component-wise though there are otherequivalent definitions/formulations for these arithmetic operations.

19

20 CHAPTER 2. MATHEMATICAL PRELIMINARIES

1. Both addition and scalar multiplication generate another member of V. Thisproperty is referred to as “closure” under addition and scalar multiplication.

2. Addition and scalar multiplication obey the following axioms.

Axioms for Addition: Consider vi, vj, and vk taken from V.(i) vi + vj = vj + vi (commutativity)(ii) vi + (vj + vk) = (vi + vj) + vk (associativity)(iii) There exists a unique null vector, denoted by 0, in V such that

0+ vi = vi + 0 = vi. (existence of the identity element)(iv) For each vi, there exists a unique inverse (−vi) in V such that

vi + (−vi) = 0. (existence of the inverse)a

Axioms for Scalar Multiplication: Consider arbitrary vectors vi, vjand arbitrary scalars α, β.(v) α(vi + vj) = αvi + αvj

(vi) (α + β)vi = αvi + βvi

(vii) α(βvi) = (αβ)vi

Fact 2.1 The following facts can be proved using the axioms.

1. 0v = 0

2. α0 = 0

3. (−1)v = (−v)

aThe axioms (i) through (iv) means that a linear vector space forms an abelian group underaddition.

Definition 2.2 If the allowed values for the scalars α, β, γ, . . . come from somefield F2, we say the linear vector space V is defined over the field F. In particular, ifF is the field of real numbers R, V is a real vector space. Likewise, if F is the field ofcomplex numbers C, V is a complex vector space.

Clearly, R3 is not a complex vector space, but a real vector space. Complex vectorspaces are the canonical vector spaces in quantum mechanics.

2There is an abstract mathematical definition of a field. However, it suffices for elementaryquantum mechanics to know the real numbers and the complex numbers respectively form a field.

2.1. LINEAR VECTOR SPACES 21

Definition 2.3 A set of vectors v1,v2, . . . ,vn is linearly independent (LI) ifn∑i=1

αivi = 0 =⇒ α1 = α2 = . . . = αn = 0. (2.1)

Definition 2.3 is equivalent to saying that no vector in v1,v2, . . . ,vn can beexpressed as a linear combination of the other vectors in the set.

Vectors i, j, and k are linearly independent because

ai + bj + ck = 0 or (a, b, c) = (0, 0, 0) =⇒ a = b = c = 0. (2.2)

Generally speaking, vectors in R3 that are perpendicular to one another are linearlyindependent. However, these are not the only examples of a linearly independent setof vectors. To give a trivial example, vectors i and i + j are linearly independent,though they are not perpendicular to each other. This is because

ai + b(i + j) = 0 =⇒ (a+ b, b) = (0, 0) =⇒ a = b = 0. (2.3)

Definition 2.4 A vector space is n-dimensional if it has at most n vectors that arelinearly independent.

Notation: An n-dimensional vector space over a field F is denoted by Vn(F). Theones we encounter in this course are usually members of Vn(C), including the caseswhere n =∞.

It is often clear what the field F is. In this case, we simply write Vn.

Theorem 2.1 Suppose v1,v2, . . . ,vn are linearly independent vectors in an n-dimensional vector space Vn. Then, any vector v in Vn can be written as a linearcombination of v1,v2, . . . ,vn.

ProofAs Vn is n-dimensional, you can find a set of n + 1 scalars α1, α2, . . . , αn, αn+1,not all zero, such that (

n∑i=1

αivi

)+ αn+1v = 0. (2.4)

Else, v1,v2, . . . ,vn,v are linearly independent, and Vn is at least n+1-dimensional,which is a clear contradiction. Furthermore, αn+1 , 0. If not, at least one of


α1, α2, . . . , αn is not 0, and yet ∑ni=1 αivi = 0, contradicting the assumption that

v1,v2, . . . ,vn are linearly independent. We now have

v =n∑i=1

−αiαn+1

vi. (2.5)

We will next show that there is only one way to express v as a linear combinationof v1,v2, . . . ,vn in Theorem 2.1.

Theorem 2.2 The coefficients of Equation 2.5 are unique.

ProofConsider any linear expression of v in terms of v1,v2, . . . ,vn

v =n∑i=1

βivi. (2.6)

Subtracting Equation 2.6 from Equation 2.5,

0 =n∑i=1

(−αiαn+1

− βi)

vi. (2.7)

However, we know v1,v2, . . . ,vn are linearly independent. Hence,

n∑i=1

(−αiαn+1

− βi)

vi = 0 =⇒ −αiαn+1

− βi = 0 or βi =−αiαn+1

for all i = 1, 2, . . . , n.

(2.8)

In order to understand Definition 2.5 below, just imagine how any vector in R3

can be expressed as a unique linear combination of i, j, and k.

Definition 2.5 A set of linearly independent vectors v1,v2, . . . ,vn, which canexpress any vector v in Vn as a linear combination (Equation 2.6), is called a basisthat spans Vn. The coefficients of the linear combination βi are called the componentsof v in the basis v1,v2, . . . ,vn. We also say that Vn is the vector space spannedby the basis v1,v2, . . . ,vn3.

3This means that we can define Vn as the collection of all objects of the form∑n

i=1 αivi.

2.2. INNER PRODUCT SPACES 23

Once we pick a basis v1,v2, . . . ,vn, and express v as a linear combination of thebasis vectors v =

∑ni=1 βivi, we get a component notation of v as follows.

v =n∑i=1

βivi = (β1, β2, . . . , βn). (2.9)

With this notation, vector additions and scalar multiplications can be done compo-nent by component. For vectors v = (β1, β2, . . . , βn), w = (γ1, γ2, . . . , γn) and ascalar α, we simply have

v + w = (β1 + γ1, β2 + γ2, . . . , βn + γn) and αv = (αβ1, αβ2, . . . , αβn). (2.10)

2.2 Inner Product SpacesThis is nothing but a generalization of the familiar scalar product for V3(R)4. Namely,it is an extension of the inner product defined on R3 to complex vectors. Readersshould be familiar with the component-wise definition of the usual inner product onR3, also called dot product, such that a·b = (a1, a2, a3)·(b1, b2, b3) = a1b1+a2b2+a3b3.We can express this as a matrix product of a row vector a and a column vector b.

[a1 a2 a3

] b1b2b3

=[a1b1 + a2b2 + a3b3

]; (2.11)

where the one-by-one matrix on the right-hand side is a scalar by definition. Thisdefinition of inner product is extended to include complex components by taking thecomplex conjugate of the components of a. So, the extended inner product, whichwe simply call an inner product in the rest of the book, is given by

[a1 a2 a3

]∗ b1b2b3

=[a∗1 a∗

2 a∗3

] b1b2b3

=[a∗1b1 + a∗

2b2 + a∗3b3

]. (2.12)

Definition 2.6 below is an abstract version of the familiar inner product defined on R3.This definition contains the inner product defined on R3, but not all inner productsare of that type. Nevertheless, you should keep the inner product on R3 in mind asyou read the following general definition of inner product. In Definition 2.6, we arewriting a∗ as ⟨a| and b as |b⟩ so that a∗ ·b = ⟨a|b⟩5 in anticipation of Dirac’s braketnotation to be introduced on p.31.

4This is the notation suggested on p.21, which is the same as R3.5Naively, a∗ · b = ⟨a| |b⟩, but we use only one vertical line and write ⟨a|b⟩.


Definition 2.6 (Inner Product) An inner product of vi and vj, denoted by ⟨vi | vj⟩,is a function of two vectors mapping onto C (⟨ | ⟩ : Vn × Vn 7−→ C) satisfying theaxioms below.

(i) ⟨vi | vi⟩ ≥ 0 with the equality holding if and only if vi = 0

(ii) ⟨vi | vj⟩ = ⟨vj | vi⟩∗, where ∗ denotes the complex conjugate

(iii) ⟨vi |αvj + βvk⟩ = α ⟨vi | vj⟩+ β ⟨vi | vk⟩ (linearity in the second argument)

It is convenient to add the fourth property to this list, though it is not an axiom buta consequence of Axioms (i) and (iii).

(iv) ⟨αvi + βvj | vk⟩ = α∗ ⟨vi | vk⟩+ β∗ ⟨vj | vk⟩ (anti-linearity in the first argument)

We say the inner product is linear in the second vector and antilinear in the firstvector.

In order to see how (iv) follows from (ii) and (iii), see the derivation below.

⟨αvi + βvj | vk⟩ = ⟨vk |αvi + βvj⟩∗ by (ii)= (α ⟨vk | vi⟩+ β ⟨vk | vj⟩)∗ by (iii)= α∗ ⟨vi | vk⟩+ β∗ ⟨vj | vk⟩ by (ii)

(2.13)

A moment’s thought would tell us that (iii) and (iv) can be extended to the followingproperty.

⟨n∑i=1

αivi

∣∣∣∣∣∣n∑j=1

βjvj

⟩=

n∑i,j=1

α∗iβj ⟨vi | vj⟩ (2.14)

Definition 2.7 (Inner Product Space) A vector space on which an inner productis defined is called an inner product space.

Our canonical example R3 is naturally an inner product space.

Definition 2.8 (Norm) The norm of a vector v, denoted by either |v| or ∥v∥, isgiven by ⟨v | v⟩1/2. A vector is normalized if its norm is unity. We call such a vectora unit vector.


Definition 2.9 (Metric Space) A metric space is a set where a notion of distance(called a ”metric”) between elements of the set is defined.a In particular, a norm isa metric.

Definition 2.10 (Cauchy Sequence) A Cauchy sequence xnis a sequence forwhich the terms become arbitrarily close to each other if n becomes sufficiently large.For example, a sequence of real numbers xn is Cauchy if for any ε > 0, thereexists a positive integer Nε such that m,n > Nε =⇒ |xm− xn| < ε. More generally,a sequence yn in a metric space is Cauchy if m,n > Nε =⇒ d(ym, yn) < ε.

Definition 2.11 (Complete Metric Space) A metric space M is called completeif every Cauchy sequence of points in M converges in M ; that is, if every Cauchysequence in M has a limit that is also in M.

aThere is a precise notion of a metric which is a real-valued function with two elements of a setas the arguments. Let us use the canonical notation d(x, y). Then, a metric d should satisfy thefollowing conditions.

1. d(x, y) ≥ 0

2. d(x, y) = 0 if and only if x = y

3. d(x, y) = d(y, x)

4. d(x, z) ≤ d(x, y) + d(y, z)

The first condition is redundant as d(x, y) = 12 [d(x, y) + d(y, x)] ≥ 1

2d(x, x) = 0.

Definition 2.12 (Hilbert Space) An inner product space which is complete withrespect to the norm induced by the inner product is called a Hilbert space.6,7

The inner product space R3 is a Hilbert space.

Let us backtrack here and review the relations among a vector space, an innerproduct space, and a Hilbert space.

Vector Space A (linear) vector space is defined by Definition 2.1. Two triv-ial examples are R2 the x, y-plane and R3 the x, y, z-space.

Inner Product Space An inner product space is a vector space on whichan inner product is defined according to Definition 2.6. With the usual dotproduct u ·w, R2 and R3 become inner product spaces.

6A normed vector space is automatically complete if it is finite dimensional.7Parenthetically, a normed complete vector space is called a Banach space.


Hilbert Space A Hilbert space is an inner product space for which all Cauchysequences converge to an element of the space. In particular, all finite dimen-sional inner product spaces are “automatically” Hilbert spaces. Hence, bothR2 and R3 are Hilbert spaces.

Some clarifications are in order about the convergence (inside the Hilbert space H).This means that if ∥vm − vn∥ −→ 0 as both m and n tend to ∞, then there is avector v ∈ H such that ∥vi − v∥ i→∞−−−→ 0. Forgetting about being a vector space fornow, the sequence 1

i∞i=1 in the open interval (0, 1) and closed interval [0, 1] give

an example of converging inside the set and not doing so. Noting that 1i

i→∞−−−→= 0,we can see that 1

i∞i=1 converges inside the set only for [0, 1]. But, as stated above,

(0, 1) and [0, 1] are not vector spaces.

Now, a good example of an inner product space which is not complete, and hence isnot a Hilbert space, is a set of functions defined on an interval [a, b] with k continuousderivatives; denoted by Ck([a, b]).

Example 2.1 (Ck([a, b]) is not complete.) Consider a set Ck([a, b]) of continuousfunctions with k continuous derivatives defined on [a, b]. It is straightforward to showthat Ck([a, b]) is a linear vector space. Furthermore, it is an inner product space withthe following inner product.

⟨f |g⟩ =k∑i=0

∫ b

af i(x)gi(x) dx (2.15)

Here, f i and gi are the i-th derivatives of f and g, respectively. The norm given bythis inner product is

∥f∥ =(

k∑i=0

∫ b

a

∣∣∣f j(x)∣∣∣2)12

. (2.16)

The corresponding Hilbert space, which is the completion of Ck([a, b]) with respect tothe norm (2.16), is called a Sobolev space denoted by Hk((a, b)) or W k,2((a, b)). Itcan be shown that Hk((a, b)) is strictly larger than Ck([a, b]).

Definition 2.13 (Orthogonality) Two vectors u and w are orthogonal to eachother if ⟨u |w⟩ = 0.

Note here that in the familiar 2-dimensional and 3-dimensional cases, namely V2(R)and V3(R), the norm is nothing but the length of the vector, and orthogonality meansthe vectors are geometrically perpendicular to each other.


Definition 2.14 (Orthonormal Set) A set of vectors v1,v2, . . . ,vn form anorthonormal set if

⟨vi | vj⟩ = δij; (2.17)

where the symbol δij is known as Kronecker’s delta with the following straightforwarddefinition.

δij =

1 if i = j0 if i , j (2.18)

Definition 2.15 (Completeness) 8 An orthonormal set of vectors v1,v2, . . . ,vnis complete if

n∑i=1

|vi⟩ ⟨vi| = I. (2.19)

Alternatively, v1,v2, . . . ,vn is complete if any vector v in the Hilbert space canbe expressed as a linear combination of v1, v2, . . ., vn. That is, any v =

∑ni=1 ai |vi⟩

if and only if the set v1,v2, . . . ,vn satisfies ∑ni=1 |vi⟩ ⟨vi| = I.

ProofSuppose ∑n

i=1 |vi⟩ ⟨vi| = I. Then,

v = Iv =

(n∑i=1

|vi⟩ ⟨vi|)|v⟩ =

n∑i=1

⟨vi|v⟩ |vi⟩ . (2.20)

Now, suppose v =∑nj=1 aj |vj⟩ for any v in the Hilbert space. Then,

(n∑i=1

|vi⟩ ⟨vi|)

v =

(n∑i=1

|vi⟩ ⟨vi|) n∑

j=1

aj |vj⟩

=n∑

i,j=1

|vi⟩ ⟨vi| aj |vj⟩

=n∑

i,j=1

aj |vi⟩ ⟨vi|vj⟩ =n∑

i,j=1

aj |vi⟩ δi,j =n∑j=1

aj |vj⟩ = v. (2.21)

Because (2.21) holds for an arbitrary vector v, we have∑ni=1 |vi⟩ ⟨vi| = I by definition.

Definition 2.16 (Orthonormal Basis) If a basis of a vector space Vn forms anorthonormal set, it is called an orthonormal basis.9

8This is also known as the closure condition.9Note that a set of vectors which is orthonormal and complete forms an orthonormal basis.


The unit vectors along the x-, y-, and z-axes, denoted by i, j, and k, form an or-thonormal basis of R3.

Here is a caveat. Note that the representation of a vector v as an n × 1 matrix(column vector) or a 1×n matrix (row vector) can be in terms of any basis. However,an orthonormal basis simplifies computations of inner products significantly. Letu = (u1, u2, . . . , un) and w = (w1, w2, . . . , wn) in the component notation withrespect to some orthonormal basis e1, e2, . . . , en. Then,

⟨u |w⟩ =⟨

n∑i=1

uiei

∣∣∣∣∣∣n∑j=1

wjej

⟩=

n∑i,j=1

u∗iwj ⟨ei | ej⟩ =

n∑i,j=1

u∗iwjδij =

n∑i=1

u∗iwi

= u∗1w1 + u∗

2w2 + . . . + u∗n−1wn−1 + u∗

nwn.a

(2.22)

Theorem 2.3 (Schwarz Inequality) Any inner product satisfies the following in-equality known as Schwarz Inequality.

|⟨vi | vj⟩|2 ≤ |vi|2|vj|2 or |⟨vi | vj⟩| ≤ |vi||vj| (2.23)

ProofConsider the vector

v = vi − ⟨vj | vi⟩vj|vj|2

.

Then, as ⟨v | v⟩ ≥ 0, we get⟨vi − ⟨vj | vi⟩

vj|vj|2

∣∣∣∣∣ vi − ⟨vj | vi⟩ vj|vj|2

⟩= ⟨vi | vi⟩ −

⟨vj | vi⟩ ⟨vi | vj⟩|vj|2

− ⟨vj | vi⟩∗ ⟨vj | vi⟩|vj|2

+⟨vj | vi⟩∗ ⟨vj | vi⟩ ⟨vj | vj⟩

|vj|4

= |vi|2 −⟨vj | vi⟩ ⟨vi | vj⟩

|vj|2− ⟨vj | vi⟩

∗ ⟨vj | vi⟩|vj|2

+⟨vj | vi⟩∗ ⟨vj | vi⟩ |vj|2

|vj|4

= |vi|2 −⟨vj | vi⟩ ⟨vi | vj⟩

|vj|2− ⟨vj | vi⟩

∗ ⟨vj | vi⟩|vj|2

+⟨vj | vi⟩∗ ⟨vj | vi⟩

|vj|2

= |vi|2 −⟨vj | vi⟩ ⟨vi | vj⟩

|vj|2≥ 0 =⇒ |vi|2|vj|2 ≥ |⟨vi | vj⟩|2 . (2.24)

2.3. L2-SPACE 29

We can also see that the equality holds only when

v = vi − ⟨vj | vi⟩vj|vj|2

= 0 or vi = ⟨vj | vi⟩vj|vj|2

. (2.25)

Now, let vi = αvj for some scalar α. Then,

⟨vj | vi⟩vj|vj|2

= ⟨vj |αvj⟩vj|vj|2

= ⟨vj | vj⟩αvj|vj|2

= |vj|2αvj|vj|2

= αvj = vi. (2.26)

Therefore, the equality holds if and only if one vector is a scalar multiple of theother.b

aDon’t make a beginner’s mistake of applying this to vectors expressed in a basis that is notorthonormal. For example, consider the space spanned by e1 and e2; i.e. the x, y-plane if you may.While e1,e2 form an orthonormal basis,

e1,

e1+e2√2

is a basis composed of normalized vectors

which are not mutually orthogonal. Let u = e1 and w = e1+e2√2

. Then, u = (1, 0) and w = (0, 1),

and u∗1w1 + u∗

2w2 = 1× 0 + 0× 1 = 0. But, u ·w = e1 ·(

e1+e2√2

)= e1·e1+e1·e2√

2= 1√

2, 0.

aDon’t make a beginner’s mistake of applying this to vectors expressed in a basis that is notorthonormal. For example, consider the space spanned by e1 and e2; i.e. the x, y-plane if you may.While e1,e2 form an orthonormal basis,

e1,

e1+e2√2

is a basis composed of normalized vectors

which are not mutually orthogonal. Let u = e1 and w = e1+e2√2

. Then, u = (1, 0) and w = (0, 1),

and u∗1w1 + u∗

2w2 = 1× 0 + 0× 1 = 0. But, u ·w = e1 ·(

e1+e2√2

)= e1·e1+e1·e2√

2= 1√

2, 0.

bThis is not surprising as ⟨vj | vi⟩ vj

|vj |2 is the projection of vi along vj . Needless to say, if vi isparallel to vj , its projection in the direction of vj is vi itself.

2.3 L2-SpaceIt was mentioned on p.23 that there are other inner products than the familiar one onR3. And, there is an associated Hilbert space for each of those inner products. Oneimportant class of Hilbert spaces for us is the L2-space. It is a collection of functionswith a common domain, for which an inner product is defined as an integral. Mostof the Hilbert spaces we encounter in this course are L2-spaces. In the following,we will specialize to R as the domain of our functions, but the same definitions anddescriptions apply to a general domain10. In particular, if the domain is R2 or R3, we

have a double integral,+∞!

−∞, or triple integral,

+∞#−∞

, respectively, rather than a single

integral shown here.10To be precise, the domain should be a measure space so that an integral can exist. However,

such purely mathematical details can be omitted as all the domains we consider are measure spaces.


Definition 2.17 (L2-Function) An L2-function f is a square integrable function;that is, f satisfies

∫+∞−∞ |f(x)|2 dx <∞. 11

The next step is to define an inner product on this set of square integrable functions.

Definition 2.18 (L2-Inner Product) Given two square integrable functions f andg, we define an inner product by

⟨f |g⟩ :=∫ +∞

−∞f(x)∗g(x) dx12; (2.27)

where f(x)∗ signifies the complex conjugate.

With this inner product defined on a set of L2-functions, we have a Hilbert space.

As stated in the beginning of this section, the common domain of the functionscan be any other measurable set such as R2, R3, [0, 1], and (−a

2,+a

2) for a > 0. Using

the notation given in Footnote 11, we will denote these L2-spaces by L2(R2), L2(R3),L2[0, 1], and L2(−a

2,+a

2).

The central equation of quantum mechanics, called Schrodinger Equation whichwe will encounter in Chapter 5, is a differential equation, and its solutions are com-plex valued functions on a common domain. Therefore, the L2-space is our canonicalHilbert space.

Some examples of orthonormal bases for L2-spaces are as follows.

1. The trigonometric system ei2πnx+∞n=−∞ is an orthonormal basis for L2[0, 1].

The expansion of a function in this basis is called the Fourier series of thatfunction.

2. The Legendre polynomials, which are obtained by taking the sequence of mono-mials 1, x, x2, x3, . . . and applying the Gram-Schmidt orthogonalization pro-cess to it, form an orthonormal basis for L2[−1, 1]. The first few normalizedpolynomials are as follows.

p1(x) =√1/2

11We often have an interval of finite length as the domain of f . The closed interval [0, 1] is a canon-ical example. In this case, we require

∫ 1

0|f(x)|2 dx < ∞. We use notations such as L2(−∞,+∞)

and L2[0, 1] to make this distinction.12We can easily check that this integral satisfies Definition 2.6.

2.4. THE BRAKET NOTATION 31

p2(x) = x√3/2

p3(x) =(3x2 − 1

) √5/8

p4(x) =(5x3 − 3x

) √7/2

2

p5(x) =(35x4 − 30x2 + 3

) √9/2

8

p6(x) =(63x5 − 70x3 + 15x

) √11/2

8

2.4 The Braket Notation13

This is a notation first used by Dirac. In fact, we have already been using it. You mayhave noticed that the inner product between v and w was denoted by ⟨v |w⟩, and notby the more familiar ⟨v, w⟩. We will see why shortly. Without much ado, let us definea “bra” and a “ket”. On p.23, we have already encountered ⟨a| = a∗ and |b⟩ = bsuch that the inner product between a and b is given by ⟨a|b⟩ = a∗

1b1 + a∗2b2 + a∗

3b3.Here, ⟨a| is called “bra a”, and |b⟩ is called “ket b”. More generally, recall thecomponent notation of a vector v = (v1, v2, . . . , vn). This usually means that wehave picked a particular orthonormal basis14 e1, e2, . . . , en and expressed v asa linear combination of e1, e2, . . . , en such that v = v1e1 + v2e2 + . . . + vnen. Wecan arrange the components either horizontally or vertically, but it is more convenientfor us to arrange it vertically as an n by 1 matrix, often called a column vector. Inquantum mechanics, we denote this column vector by |v⟩ called “ket v”.

|v⟩ =

v1...vn

(2.28)

Now, what is “bra v” denoted by ⟨v|? It is defined as the row vector whose compo-nents are the complex conjugates of the components of v.

⟨v| = [v∗1, . . . , v

∗n] (2.29)

13Note that this is “bra + ket = braket”, and NOT “bracket”.14Actually, it does not have to be an orthonormal basis, but any basis would do. However, we

will specialize to an orthonormal basis here in order to explain why we used the notation ⟨u| |v⟩ forthe inner product. Besides, orthonormal bases are by far the most common and useful bases.


In this notation, a basis vector ei has the following bra and ket. We sometimesdenote ⟨ei| and |ei⟩ by ⟨i| and |i⟩, respectively.

⟨ei| = ⟨i| = [0, 0, . . . , 1, . . . , 0] and |ei⟩ = |i⟩ =

00...1...0

(2.30)

The only nonzero element 1 is in the i-th place. We can express |v⟩ as a linearcombination of |i⟩’s.

|v⟩ =n∑i=1

vi |i⟩ (2.31)

Similarly,

⟨v| =n∑i=1

v∗i ⟨i| . (2.32)

This suggests that we should define |αv⟩ and ⟨αv| by

|αv⟩ = α |v⟩ and ⟨αv| = α∗ ⟨v| = ⟨v|α∗. (2.33)

As shown above, it is customary, for conserving symmetry, to write ⟨v|α∗ ratherthan α∗ ⟨v|. We say α |v⟩ and ⟨v|α∗ are adjoints of each other. More generally, theadjoint of

n∑i=1

αi |vi⟩ (2.34)

isn∑i=1

⟨vi|α∗i , (2.35)

and vice versa. In order to take the adjoint, we can simply replace each scalar by itscomplex conjugate and each bra/ket by the corresponding ket/bra.


Let us now try some manipulations with bras and kets in order to get used totheir properties.

First, consider two vectors u = u1e1 + u2e2 + . . . + unen and w = w1e1 +w2e2 + . . . + wnen in an n-dimensional vector space spanned by an orthonormalbasis e1, e2, . . . , en. We have

⟨u| = [u∗1, . . . , u

∗n] and |w⟩ =

w1...wn

. (2.36)

Hence,

⟨u| |w⟩ = [u∗1, . . . , u

∗n]

w1...wn

= u∗1w1 + u∗

2w2 + . . . + u∗nwn. (2.37)

We drop one vertical bar form ⟨u| |w⟩ and write ⟨u|w⟩, so that

⟨u|w⟩ = [u∗1, . . . , u

∗n]

w1...wn

= u∗1w1 + u∗

2w2 + . . . + u∗nwn. (2.38)

As you can see, the right-hand side is the inner product. Hence, the braket notationis consistent with the notation we have been using for the inner product. In passing,we will make a note of the obvious fact that

⟨ei | ej⟩ = ⟨i | j⟩ = δij. (2.39)

Next, given an orthonormal basis |1⟩ , |2⟩ , . . . , |n⟩, any vector |v⟩ can be ex-panded in this basis so that

|v⟩ =n∑i=1

vi |i⟩ , (2.40)

where vi’s are unique. Taking the inner product of both sides with ⟨j|, we get

⟨j|v⟩ =⟨j∣∣∣ n∑i=1

vi∣∣∣i⟩ = n∑

i=1

vi ⟨j|i⟩ =n∑i=1

viδij = vj. (2.41)

Plugging this back into |v⟩ = ∑ni=1 vi |i⟩,

|v⟩ =n∑i=1

⟨i|v⟩ |i⟩ =n∑i=1

|i⟩ ⟨i|v⟩ . (2.42)


The second equality in Equation 2.42 may not be clear. So, we will show it belowexplicitly by writing out each vector. The term on the left-hand side is as follows.

⟨i|v⟩ |i⟩ =

[0, 0, . . . , 1, . . . , 0]v1...vn

00...1...0

= vi

00...1...0

=

00...vi...0

(2.43)

Note that the only nonzero element 1 is in the i-th place. On the other hand, theterm on the right-hand side is

|i⟩ ⟨i|v⟩ =

00...1...0

[0, 0, . . . , 1, . . . , 0]

v1...vn

=

00...1...0

[vi] =

00...vi...0

. (2.44)

Hence, the second equality in Equation 2.42 indeed holds term by term. In the caseof (2.44), we have a product of an n-by-1 matrix |i⟩, a 1-by-n matrix ⟨i|, and ann-by-1 matrix |v⟩. By the associative law of matrix multiplication, it is possible to


compute |i⟩ ⟨i|15 first.

|i⟩ ⟨i|v⟩ =

00...1...0

[0, 0, . . . , 1, . . . , 0]

v1...vn

=

1 i n

1 0 0 . . . 0 . . . 0

0. . .

......

.... . .

......

i 0 . . . . . . 1 . . ....

......

. . ....

n 0 . . . . . . 0 . . . 0

v1...vn

= [δii]

v1...vn

=

00...vi...0

(2.45)

The symbol δij in (2.45) means an n by n matrix whose only nonzero entry is the 1at the intersection of the i-th row and the j-th column.

[δij] :=

1 j n

1 0 0 . . . 0 . . . 0

0. . .

......

.... . .

......

i 0 . . . . . . 1 . . ....

......

. . ....

n 0 . . . . . . 0 . . . 0

(2.46)

It is easy to see that

|i⟩ ⟨j| = [δij]. (2.47)

15As (2.45) shows, |i⟩ ⟨i| extracts the i-th component of vector v. So, |i⟩ ⟨i| is called the projectionoperator for the ket |i⟩.


And, it is also easy to see thatn∑i=1

|i⟩ ⟨i| =n∑i=1

[δii] = In; (2.48)

where In is the n-dimensional identity operator.16 With this identity, Equation 2.42reduces to triviality.

|v⟩ = I |v⟩ =(

n∑i=1

|i⟩ ⟨i|)|v⟩ =

(n∑i=1

|i⟩ ⟨i| |v⟩)=

n∑i=1

|i⟩ ⟨i|v⟩ (2.49)

Because the identity matrix can be inserted anywhere without changing the outcomeof the computation, we can insert ∑n

i=1 |i⟩ ⟨i| anywhere during our computation. Thisseemingly trivial observation combined with the associativity of matrix multiplica-tion will prove useful later in this book.

Example 2.2 (⟨i|, |i⟩, and |i⟩ ⟨i| in Two Dimensions) In two dimensions, we have

|1⟩ =[10

], |2⟩ =

[01

], ⟨1| =

[1 0

], and ⟨2| =

[0 1

](2.50)

Then,

2∑i=1

|i⟩ ⟨i| =[10

] [1 0

]+

[01

] [0 1

]=

[1 00 0

]+

[0 00 1

]=

[1 00 1

]. (2.51)

Now, consider any vector v = (v1, v2) and compute |1⟩ ⟨1|v and |2⟩ ⟨2|v.

|1⟩ ⟨1|v =

[10

] [1 0

] [ v1v2

]=

[1 00 0

] [v1v2

]=

[v10

](2.52)

|2⟩ ⟨2|v =

[01

] [0 1

] [ v1v2

]=

[0 00 1

] [v1v2

]=

[0v2

](2.53)

16When there is no possibility of confusion, we will simply use I instead of In.


Let us consider the adjoint of (2.40).

⟨v| =n∑i=1

⟨i| v∗i (2.54)

From our previous computation and due to a property of the inner product, we get

⟨v| =n∑i=1

⟨i| (⟨i|v⟩)∗ =n∑i=1

⟨i| ⟨v|i⟩ =n∑i=1

⟨v|i⟩ ⟨i| . (2.55)

Again, recall the identity ∑ni=1 |i⟩ ⟨i| =

∑ni=1[δii] = I. So,

n∑i=1

⟨v|i⟩ ⟨i| = ⟨v|(

n∑i=1

|i⟩ ⟨i|)= ⟨v| I = ⟨v| . (2.56)

It is convenient at this point to have a rule by which we can mechanically arrive atthe adjoint of an expression.

Taking the Adjoint:

1. Reverse the order of all the factors in each term, including numerical coeffi-cients, bras, and kets.

2. Take complex conjugate of the numerical coefficients.

3. Turn bras into kets and kets into bras.

The following example demonstrates what one should do.Consider

α |v⟩ = a |u⟩+ b |w⟩ ⟨s|t⟩ + . . . . (2.57)

What is the adjoint?

After Step 1

|v⟩α = |u⟩ a + |t⟩ ⟨s|w⟩ b + . . . (2.58)

After Step 2

|v⟩α∗ = |u⟩ a∗ + |t⟩ ⟨s|w⟩ b∗ + . . . (2.59)


After Step 3⟨v|α∗ = ⟨u| a∗ + ⟨t|s⟩ ⟨w| b∗ + . . . (2.60)

Therefore, the adjoint of

α |v⟩ = a |u⟩+ b |w⟩ ⟨s|t⟩ + . . . (2.61)

is

⟨v|α∗ = ⟨u| a∗ + ⟨t|s⟩ ⟨w| b∗ + . . . . (2.62)

Theorem 2.4 (Gram-Schmidt Theorem) Given n linearly independent vectors|v1⟩ , |v2⟩ , . . . , |vn⟩, we can construct n orthonormal vectors |1⟩ , |2⟩ , . . . , |n⟩,each of which is a linear combination of |v1⟩, |v2⟩, . . ., |vn⟩. In particular, we canconstruct |1⟩ , |2⟩ , . . . , |n⟩ such that |1⟩ = |v1⟩√

⟨v1 | v1⟩.

ProofWe will construct n mutually orthogonal vectors |1′⟩ , |2′⟩ , . . . , |n′⟩ first. Onceit is done, we will only need to normalize each vector to make them orthonormal.First let

|1′⟩ = |v1⟩ . (2.63)

Next, let|2′⟩ = |v2⟩ −

|1′⟩ ⟨1′ | v2⟩⟨1′ | 1′⟩

, (2.64)

where|1′⟩ ⟨1′ | v2⟩⟨1′ | 1′⟩

is nothing but the projection of |v2⟩ on |1′⟩a. So, in (2.64), the component of |v2⟩ inthe direction of |1′⟩ is subtracted from |v2⟩, making it orthogonal to |1′⟩. Indeed,

⟨1′ | 2′⟩ = ⟨1′ | v2⟩ −⟨1′ | 1′⟩ ⟨1′ | v2⟩⟨1′ | 1′⟩

= 0. (2.66)

The third ket is defined proceeding in the same manner.

|3′⟩ = |v3⟩ −|1′⟩ ⟨1′ | v3⟩⟨1′ | 1′⟩

− |2′⟩ ⟨2′ | v3⟩⟨2′ | 2′⟩

. (2.67)


Then,

⟨1′ | 3′⟩ = ⟨1′ | v3⟩ −⟨1′ | 1′⟩ ⟨1′ | v3⟩⟨1′ | 1′⟩

− ⟨1′ | 2′⟩ ⟨2′ | v3⟩⟨2′ | 2′⟩

= ⟨1′ | v3⟩ − ⟨1′ | v3⟩ − 0 = 0

(2.68)

and

⟨2′ | 3′⟩ = ⟨2′ | v3⟩ −⟨2′ | 1′⟩ ⟨1′ | v3⟩⟨1′ | 1′⟩

− ⟨2′ | 2′⟩ ⟨2′ | v3⟩⟨2′ | 2′⟩

= ⟨2′ | v3⟩ − 0− ⟨2′ | v3⟩ − 0 = 0.

(2.69)

If n ≤ 3, we are done at this point. So, assume n ≥ 4, and suppose we have found imutually orthogonal vectors |1′⟩ , |2′⟩ , . . . , |i′⟩ for i ≤ n − 1. Now, define |(i+ 1)′⟩as follows.

|(i+ 1)′⟩ = |vi+1⟩ −|1′⟩ ⟨1′ | vi+1⟩⟨1′ | 1′⟩

− |2′⟩ ⟨2′ | vi+1⟩⟨2′ | 2′⟩

− . . . − |i′⟩ ⟨i′ | vi+1⟩⟨i′ | i′⟩

(2.70)

For any 1 ≤ j ≤ i,

⟨j′ | (i+ 1)′⟩ = ⟨j′ | vi+1⟩ −i∑

k=1

⟨j′ | k′⟩ ⟨k′ | vi+1⟩⟨k′ | k′⟩

= ⟨j′ | vi+1⟩ −⟨j′ | j′⟩ ⟨j′ | vi+1⟩⟨j′ | j′⟩

= 0.

(2.71)

Hence, by mathematical induction, we can find n mutually orthogonal vectors|1′⟩ , |2′⟩ , . . . , |n′⟩. Finally define |i⟩ for 1 ≤ i ≤ n by

|i⟩ = |i′⟩⟨i′ | i′⟩1/2

, (2.72)

so that

⟨i | i⟩ = ⟨i′ | i′⟩⟨i′ | i′⟩1/2 ⟨i′ | i′⟩1/2

= 1. (2.73)

We have now constructed n orthonormal vectors |1⟩ , |2⟩ , . . . , |n⟩.

Question: Linear IndependenceIf you are a careful reader, you may have noticed that we did not, at least explicitly,


use the fact that the vectors |v1⟩ , |v2⟩ , . . . , |vn⟩ are linearly independent. So, whyis this condition necessary? In order to see this, suppose

|(i+ 1)′⟩ = |vi+1⟩ −|1′⟩ ⟨1′ | vi+1⟩⟨1′ | 1′⟩

− |2′⟩ ⟨2′ | vi+1⟩⟨2′ | 2′⟩

− . . . − |i′⟩ ⟨i′ | vi+1⟩⟨i′ | i′⟩

= 0 (2.74)

for some i ≤ n− 1. Because our construction of |i⟩ requires |vi⟩, this means we haveat most n − 1 orthonormal vectors, and our construction fails. However, we knowthat |j′⟩ is a linear combination of |v1⟩ , |v2⟩ , . . . , and |vj⟩ for any j by construction.Therefore,

−|1′⟩ ⟨1′ | vi+1⟩⟨1′ | 1′⟩

− |2′⟩ ⟨2′ | vi+1⟩⟨2′ | 2′⟩

− . . . − |i′⟩ ⟨i′ | vi+1⟩⟨i′ | i′⟩

(2.75)

in (2.74) is a linear combination of |v1⟩ , |v2⟩ , . . . , and |vi⟩. So, |(i+ 1)′⟩ is a linearcombination of |v1⟩ , |v2⟩ , . . . , and |vi+1⟩ where the coefficient of |vi+1⟩ is 1. But,this means that there is a linear combination of |v1⟩ , |v2⟩ , . . . , and |vi+1⟩ whichequals 0 though not all the coefficients are zero, violating the linear independenceassumption for |v1⟩ , |v2⟩ , . . . , |vn⟩. Turning this inside out, we can conclude that|(i+ 1)′⟩ , 0 for all i ≤ n− 1 if |v1⟩ , |v2⟩ , . . . , |vn⟩ are linearly independent. Thisis why we needed the linear independence of |v1⟩ , |v2⟩ , . . . , |vn⟩ in Theorem 2.4.

aIn the familiar x, y-plane,

|1′⟩ ⟨1′ | v2⟩⟨1′ | 1′⟩

=

(∥1′∥ 1′

)(∥1′∥ ∥v2∥ cos θ)

∥1′∥2= ∥v2∥ cos θ 1′, (2.65)

where θ is the angel between 1′ and v2, and 1′ is the unit vector in the direction of 1′.

Theorem 2.5 The maximum number of mutually orthogonal vectors in an innerproduct space equals the maximum number of linearly independent vectors.

ProofConsider an n-dimensional space Vn, which admits a maximum of n linearly in-dependent vectors by definition. Suppose we have k mutually orthogonal vectors|v1⟩ , |v2⟩ , . . . , |vk⟩, and assume

k∑i=1

αi |vi⟩ = 0. (2.76)

2.6. LINEAR OPERATORS 41

Then,⟨vj

∣∣∣∣∣k∑i=1

αi |vi⟩⟩=

k∑i=1

αi ⟨vj | vi⟩ = αj ⟨vj | vj⟩ = 0 =⇒ αj = 0 for j = 1, 2, . . . , k.

(2.77)

So, the vectors |1⟩ , |2⟩ , . . . , and |k⟩ are linearly independent. This implies thatk ≤ n as Vn can only admit up to n linearly independent vectors. On the other hand,Theorem 2.4 (Gram-Schmidt Theorem) assures that we can explicitly construct nmutually orthogonal vectors. This proves the theorem.

2.5 Vector SubspaceDefinition 2.19 (Vector Subspaces) A subset of a vector space V that form avector space is called a subspace. Our notation for a subspace i of dimensionality niis Vni

i .

Definition 2.20 (Orthogonal Complement) The orthogonal complement of asubspace W of a vector space V equipped with a bilinear form B, of which an innerproduct is the most frequently encountered example, is the set W⊥ of all vectors inV that are orthogonal to every vector in W. It is a subspace of V.

Definition 2.21 (Direct Sum) The direct sum of two vector spaces V and W isthe set V⊕W of pairs of vectors (v,w) in V and W, with the operations:(v,w) + (v′,w′) = (v + v′,w + w′)c(v,w) = (cv, cw)

With these operations, the direct sum of two vector spaces is a vector space.

2.6 Linear OperatorsAn operator Ω operates on a vector |u⟩ and transforms it to another vector |v⟩. Wedenote this as below.

Ω |u⟩ = |v⟩ (2.78)

The same operator Ω can also act on ⟨u| to generate ⟨v|. In this case, we will reversethe order and write

⟨u|Ω = ⟨v| . (2.79)


For an operator Ω to be a linear operator, it should have the following set of prop-erties.

Ω (α |u⟩) = α (Ω |u⟩) (2.80)Ωα |u⟩+ β |v⟩ = α (Ω |u⟩) + β (Ωv) (2.81)

(⟨u|α) Ω = (⟨u|Ω)α (2.82)(⟨u|α + ⟨v| β) Ω = (⟨u|Ω)α + (⟨v|Ω) β (2.83)

Recall that |u⟩ is a column vector and ⟨u| is a row vector. So, the best way tounderstand a linear operator is to regard it as an n × n square matrix. Then, itsaction on |u⟩ and ⟨u| as well as linearity follows naturally. The significance of linear-ity of an operator Ω is that its action on any vector is completely specifiedby how it operates on the basis vectors |1⟩ , |2⟩ , . . . , |n⟩. Namely, if

|v⟩ =n∑i=1

αi |i⟩ , (2.84)

then,

Ω |v⟩ = Ω

(n∑i=1

αi |i⟩)=

n∑i=1

αiΩ |i⟩ . (2.85)

The action of the product of two operators ΛΩ on a ket |u⟩ is defined in an obviousway.

(ΛΩ) |u⟩ = Λ (Ω |u⟩) (2.86)

Ω acts on |u⟩ first followed by Λ. This is an obvious associative law if the operatorsΛ and Ω are regarded as matrices as suggested already. With this view, ΛΩ isnothing but a matrix multiplication, and as such, ΛΩ is not the same as ΩΛ generallyspeaking.

Definition 2.22 (Commutator) The commutator of operators Λ and Ω, denotedby [Λ,Ω] is defined as follows.

[Λ,Ω] := ΛΩ− ΩΛ (2.87)

When the commutator [Λ,Ω] is zero, we say Λ and Ω commute, and the order ofoperation does not affect the outcome. We have the following commutator identities.

[Ω,ΛΓ] = Λ [Ω,Γ] + [Ω,Λ] Γ (2.88)[ΛΩ,Γ] = Λ [Ω,Γ] + [Λ,Γ]Ω (2.89)

2.7. MATRIX REPRESENTATION OF LINEAR OPERATORS 43

Definition 2.23 (Inverse) The inverse Ω−1 of an operator Ω satisfies

ΩΩ−1 = Ω−1Ω = I; (2.90)

where I is the identity operator such that I |u⟩ = |u⟩ and ⟨u| I = ⟨u|. Needless to say,I corresponds to the identity matrix. Note that not all operators have an inverse.

Note that the inverse satisfies (ΩΛ)−1 = Λ−1Ω−1, which is consistent with the matrixrepresentations of Ω and Λ.

2.7 Matrix Representation of Linear OperatorsRecall that the action of a linear operator Ω is completely specified once its actionon a basis is specified. Pick a basis |1⟩ , |2⟩ , . . . , |n⟩ of normalized vectors whichare not necessarily mutually orthogonal; that is, it is not necessarily an orthonormalbasis. The action of Ω on any ket |v⟩ is completely specified if we know the coefficientsΩji for 1 ≤ i, j ≤ n such that

Ω |i⟩ =n∑j=1

Ωji |j⟩ . (2.91)

The coefficients in (2.91) can be computed as follows.

⟨k|Ω |i⟩ =⟨k

∣∣∣∣∣∣n∑j=1

Ωji

∣∣∣∣∣∣ j⟩=

n∑j=1

Ωji ⟨k| |j⟩ =n∑j=1

Ωjiδkj = Ωki (2.92)

So, Ωji = ⟨j|Ω |i⟩ in (2.91). Note that ⟨k| |j⟩ was used instead of ⟨k | j⟩ to make itclear that this matrix product

[ 1 k n

0 0 . . . 1 . . . 0]

00...1...0

1

j

n


is not necessarily an inner product unless our basis happens to be orthonormal.17

There is another way to find Ωji. From (2.48), we have ∑ni=1 |i⟩ ⟨i| = I, and so,

Ω |i⟩ = IΩ |i⟩ =n∑j=1

|j⟩ ⟨j|Ω |i⟩ =n∑j=1

⟨j|Ω |i⟩ |j⟩ =⇒ Ωji = ⟨j|Ω |i⟩. (2.93)

Now, let

|v⟩ =n∑i=1

vi |i⟩ and |v′⟩ = Ω |v⟩ =n∑j=1

v′j |j⟩ .

Then,

|v′⟩ = Ω |v⟩ = Ω

(n∑i=1

vi |i⟩)=

n∑i=1

viΩ |i⟩ =n∑i=1

vin∑j=1

Ωji |j⟩ =n∑j=1

(n∑i=1

Ωjivi |j⟩)

=n∑j=1

(n∑i=1

Ωjivi

)|j⟩ (2.94)

implies

v′j =

n∑i=1

Ωjivi. (2.95)

Consider the matrix [Ωji] = [⟨j|Ω |i⟩], where Ωji is the j-th row and the i-th columnentry. Then, (2.95) implies

v′1......v′n

=

⟨1|Ω |1⟩ · · · ⟨1|Ω |n⟩⟨2|Ω |1⟩ · · · ⟨2|Ω |n⟩

.... . .

...⟨n|Ω |1⟩ · · · ⟨n|Ω |n⟩

v1......vn

(2.96)

or v′1......v′n

=

Ω11 · · · Ω1n

Ω21 · · · Ω2n...

. . ....

Ωn1 · · · Ωnn

v1......vn

. (2.97)

17See the footnote on p.28 for more detail.


Because

Ω11 · · · · · · Ω1k · · · Ω1n...

. . ....

......

. . ....

...

Ωk1 · · · · · · Ωkk · · · Ωkn...

.... . .

...Ωn1 · · · · · · Ωnk · · · Ωnn

0......

1...0

k

=

Ω1k......

Ωkk...

Ωnk

, (2.98)

the k-th column of [Ωij] is nothing but Ω |k⟩.

Example 2.3 (An Operator on C2) The column vectors[10

]and

[01

]form an

orthonormal basis for C2. If Ω[10

]=

[11

]and Ω

[01

]=

[10

], we should have

Ω =

[1 11 0

](2.99)

because Ω

[10

]is the first column of the matrix representation of Ω, and Ω

[01

]is

the second column. Now consider an arbitrary vector v given by

|v⟩ =[v1v2

]= v1

[10

]+ v2

[01

]. (2.100)

Using the matrix representation

Ω |v⟩ =[1 11 0

] [v1v2

]=

[v1 + v2v1

]. (2.101)

On the other hand,

Ω |v⟩ = Ω

(v1

[10

]+ v2

[01

])= v1Ω

[10

]+ v2Ω

[01

]

= v1

[1 11 0

] [10

]+ v2

[1 11 0

] [01

]

= v1

[11

]+ v2

[10

]=

[v1 + v2v1

]. (2.102)

This checks.


Recall that we denoted α |v⟩ by |αv⟩, and so, it is natural to denote Ω |v⟩ by |Ωv⟩.Then, it is also natural to consider its adjoint ⟨Ωv|. For a scalar α, ⟨αv| = ⟨v|α∗

while |αv⟩ = α |v⟩. Therefore, it is not surprising if ⟨Ωv| , ⟨v| [Ωij] though ⟨v| [Ωij]makes sense as a matrix multiplication. We will next compute the matrix elementsof Ω when it comes out of the ⟨Ωv| and operate on ⟨v| from the right. Looking ahead,let us denote this by Ω† such that ⟨Ωv| = ⟨v|Ω†. We have

|v⟩ =n∑i=1

vi |i⟩ and |v′⟩ = Ω |v⟩ = |Ωv⟩ =n∑j=1

v′j |j⟩ . (2.103)

So,

⟨v| =n∑i=1

⟨i| v∗i and ⟨v′| = adj (Ω |v⟩) = ⟨Ωv| =

n∑j=1

⟨j| v′∗j , (2.104)

where adj means the adjoint. On the other hand, as

Ω |i⟩ =n∑j=1

Ωji |j⟩ ,

⟨v′| = adj (Ω |v⟩) = adj

(Ω

n∑i=1

vi |i⟩)= adj

(n∑i=1

viΩ |i⟩)= adj

n∑i=1

vin∑j=1

Ωji |j⟩

= adj

n∑j=1

n∑i=1

viΩji |j⟩

=n∑j=1

n∑i=1

⟨j| v∗iΩ

∗ji. (2.105)

Hence,

v′∗j =

n∑i=1

v∗iΩ

∗ji for each 1 ≤ j ≤ n. (2.106)

This translates to the following matrix relation.

[v′∗1 . . . v′∗

n

]=[v∗1 . . . v∗

n

]Ω∗

11 Ω∗21 · · · Ω∗

n1

Ω∗12 Ω∗

22 · · · Ω∗n2

......

. . ....

Ω∗1n Ω∗

2n · · · Ω∗nn

(2.107)

or

[v′∗1 . . . v′∗

n

]=[v∗1 . . . v∗

n

]⟨1|Ω |1⟩∗ ⟨2|Ω |1⟩∗ · · · ⟨n|Ω |1⟩∗⟨1|Ω |2⟩∗ ⟨2|Ω |2⟩∗ · · · ⟨n|Ω |2⟩∗

......

. . ....

⟨1|Ω |n⟩∗ ⟨2|Ω |n⟩∗ · · · ⟨n|Ω |n⟩∗

(2.108)


Because

[ k

0 · · · · · · 1 · · · 0]

Ω∗11 · · · · · · Ω∗

k1 · · · Ω∗n1

.... . .

......

.... . .

......

Ω∗1k · · · · · · Ω∗

kk · · · Ω∗nk

......

. . ....

Ω∗1n · · · · · · Ω∗

kn · · · Ω∗nn

=[Ω∗

1k · · · · · · Ω∗kk · · · Ω∗

nk

], (2.109)

the k-th row of[Ω∗ji

]is nothing but ⟨kΩ|.

In summary, if the matrix representation of Ω in |Ωv⟩ = Ω |v⟩ is

Ω = [Ωij] =

Ω11 · · · Ω1n

Ω21 · · · Ω2n...

. . ....

Ωn1 · · · Ωnn

, (2.110)

then, the matrix representation of Ω† such that ⟨Ωv| = ⟨v|Ω† is

Ω† =[Ω∗ji

]Ω∗

11 · · · Ω∗n1

Ω∗12 · · · Ω∗

n2...

. . ....

Ω∗1n · · · Ω∗

nn

. (2.111)

That is, Ω† is the transpose conjugate of Ω.

2.7.1 Matrix Representations of Operator ProductsIf we have matrix representations [Ωij] and [Γkl] for two operators Ω and Γ, it wouldbe convenient if the matrix representation for the operator product ΩΓ is the productof the corresponding matrices. Luckily for us, this is indeed the case as shown below.First, recall that ∑n

k=1 |k⟩ ⟨k| = I. So,

(ΩΓ)ij = ⟨i|ΩΓ |j⟩ =n∑k=1

⟨i|Ω |k⟩ ⟨k|Γ |j⟩ =n∑k=1

ΩikΓkj. (2.112)


Hence, the matrix representation of a product of two operators is the product of thematrix representations for the operators in the same order.

Notation: We often denote [Ωij] by [Ω] for simplicity.

2.8 The Adjoint of an OperatorIn Section 2.7, we have already encountered this, but let us formally define the adjointof an operator and investigate its properties.

Definition 2.24 (The Adjoint) Consider an operator Ω that operates in a vectorspace V.

Ω |v⟩ = |Ωv⟩ (2.113)

If there exists an operator Ω† such that

⟨Ωv| = ⟨v|Ω†. (2.114)

Then, Ω† is called the adjoint of Ω.

We can show that the adjoint indeed exists by explicitly computing the matrix ele-ments in a basis |1⟩ , |2⟩ , . . . , |n⟩.(

Ω†)ij= ⟨i|Ω† |j⟩ = ⟨Ωi | j⟩ = ⟨j |Ωi⟩∗ = ⟨j|Ω |i⟩∗ = (Ωji)

∗ (2.115)

As shown in Section 2.7, the matrix representing Ω† is the transpose conjugate ofthe matrix for Ω.

Example 2.4 (Adjoint of an Operator) Consider

Γ =

[i 10 1

].

Then,

Γ† =

[−i 01 1

].

Now, for v =

[v1v2

],

|Γv⟩ =[i 10 1

] [v1v2

]=

[iv1 + v2v2

]=⇒ ⟨Γv| =

[(iv1 + v2)

∗ v∗2

]

2.8. THE ADJOINT OF AN OPERATOR 49

=[−iv∗

1 + v∗2 v∗

2

]. (2.116)

On the other hand,

⟨v|Γ† =[v∗1 v∗

2

] [ −i 01 1

]=[−iv∗

1 + v∗2 v∗

2

](2.117)

This checks.

Fact 2.2 The adjoint of the product of two operators (ΩΓ)† is the product of theiradjoints with the order switched; that is,

(ΩΓ)† = Γ†Ω†. (2.118)

ProofFirst, regard ΩΓ as one operator. Then,

⟨ΩΓv| = ⟨(ΩΓ)v| = ⟨v| (ΩΓ)†. (2.119)

Next, regard ΩΓv as Ω acting on the vector (Γv) to get

⟨ΩΓv| = ⟨Ω(Γv)| = ⟨Γv|Ω†. (2.120)

But,

⟨Γv| = ⟨v|Γ†

means

⟨ΩΓv| = ⟨v|Γ†Ω†. (2.121)

Therefore,

(ΩΓ)† = Γ†Ω†. (2.122)

Note 2.1 (Taking the Adjoint) In order to take the adjoint:

1. Reverse the order of each element of each term.

2. Take the complex conjugate of the scalars.

3. Switch bras to kets and kets to bras.


4. Take the adjoint of each operator.

For example, consider a linear expression

α |v⟩ = β |u⟩+ γ ⟨w | q⟩ |r⟩+ δΩ |s⟩+ ϵΓΛ |t⟩ .

After Step 1

|v⟩α = |u⟩ β + |r⟩ ⟨w | q⟩ γ + |s⟩Ωδ + |t⟩ΛΓϵ (2.123)

After Step 2

|v⟩α∗ = |u⟩ β∗ + |r⟩ ⟨w | q⟩∗ γ∗ + |s⟩Ωδ∗ + |t⟩ΛΓϵ∗

= |v⟩α∗ = |u⟩ β∗ + |r⟩ ⟨q |w⟩ γ∗ + |s⟩Ωδ∗ + |t⟩ΛΓϵ∗ (2.124)

After Step 3

⟨v|α∗ = ⟨u| β∗ + ⟨r| ⟨q |w⟩ γ∗ + ⟨s|Ωδ∗ + ⟨t|ΛΓϵ∗ (2.125)

After Step 4

⟨v|α∗ = ⟨u| β∗ + ⟨r| ⟨q |w⟩ γ∗ + ⟨s|Ω†δ∗ + ⟨t|Λ†Γ†ϵ∗ (2.126)

However, it will be easier to take the adjoint of each term in one step and add up theterms once you get used to this operation.

2.9 Eigenvalues and EigenvectorsDefinition 2.25 (Eigenvalues, Eigenvectors, and Eigenkets)Consider an operator Ω operating on a vector space V. When a nonzero vector v ora nonzero ket |v⟩ satisfies

Ωv = λv or Ω |v⟩ = λ |v⟩ , (2.127)

v is called an eigenvector, |v⟩ is called an eigenket, and λ is called an eigenvalue.Note that if |v⟩ (v) is an eigenket (eigenvector) with the associated eigenvalue λ, anyscalar multiple of the ket (vector) α |v⟩ (αv), where α , 0, is an eigenket (eigenvector)with the same eigenvalue λ.

2.9. EIGENVALUES AND EIGENVECTORS 51

Example 2.5 (Diagonal Matrices) Diagonal matrices have eigenvalues λi alongthe main diagonal with associated eigenvectors given by

0......

c...0

the i-th entry

; (2.128)

where c is an arbitrary complex number. Here is a 3×3 example. Considerλ1 0 00 λ2 00 0 λ3

. (2.129)

Then, we have λ1 0 00 λ2 00 0 λ3

c00

= λ1

c00

, (2.130)

λ1 0 00 λ2 00 0 λ3

0c0

= λ2

0c0

, (2.131)

andλ1 0 00 λ2 00 0 λ3

00c

= λ3

00c

. (2.132)

In some cases, some of the eigenvalues are the same. For example, the following3×3 matrix has λ1 = λ2 = 1 and λ3 = 2.1 0 0

0 1 00 0 2

(2.133)

When this happens, we say that the multiplicity of the eigenvalue 1 is two.


Example 2.6 (Eigenvalues and Eigenvectors: A two-by-two example) Consider

Ω =

[0.8 0.30.2 0.7

].

Then,

Ω

[0.60.4

]=

[0.8 0.30.2 0.7

] [0.60.4

]= 1 ·

[0.60.4

]and (2.134)

Ω

[1−1

]=

[0.8 0.30.2 0.7

] [1−1

]=

[0.5−0.5

]=

1

2

[1−1

]. (2.135)

Hence, the eigenvalues of Ω are 1 and 12

with corresponding eigenvectors which arescalar multiples of the vectors given above.

There is a standard way to find the eigenvalues of an operator Ω. Here is a sketch.

Suppose λ is an eigenvalue for an operator Ω with a corresponding eigenket |v⟩.Then,

Ω |v⟩ = λ |v⟩ . (2.136)

This means that

(Ω− λI) |v⟩ = |0⟩ , (2.137)

which in turn implies that Ω − λI is not invertible. In order to see this, supposeΩ− λI has the inverse (Ω− λI)−1. Then,

(Ω− λI)−1(Ω− λI) |v⟩ = (Ω− λI)−1 |0⟩ =⇒ |v⟩ = |0⟩ . (2.138)

But, this is a contradiction because |v⟩ , |0⟩ if |v⟩ is an eigenvector. Now considera matrix representation of the operator Ω in some basis. Then, from basic matrixtheory, Ω− λI is invertible if and only if det(Ω− λI) , 0. If Ω is an n× n matrix,det(Ω − λI), called a characteristic polynomial, is an n-th degree polynomial withn rootsa. We can find all the eigenvalues by solving the characteristic equationdet(Ω−λI) = 0. In order to compute det(Ω−λI) we need to pick a particular basis.However, the resulting eigenvalues are basis-independent because the characteristicpolynomial is basis-independent. For a proof, see Theorem A.4 in Appendix A.

aAs described at the end of Example 2.5 on p.51, we do not always have n distinct roots. The

2.10. SPECIAL TYPES OF OPERATORS 53

total number of the roots is n if multiplicity is taken into account. For example, if the polynomialhas (x− 1)2 when factored, 1 is a root with a multiplicity of 2, and it counts as two roots.

Notation: An eigenket associated with the eigenvalue λ is denoted by |λ⟩.

2.10 Special Types of Operators2.10.1 Hermitian Operators18

Definition 2.26 (Hermitian Operators) A Hermitian operator is an operatorwhich is the adjoint of itself; that is, an operator Ω is Hermitian if and only if

Ω = Ω†.

In particular, this means that

⟨Ωv |w⟩ = ⟨v |Ωw⟩ (2.139)

as both sides are equal to⟨v|Ω |w⟩ .

In the linear algebra of real matrices, the matrix representation of a Hermitian op-erator is nothing but a symmetric matrix. But, our matrices are complex, and thematrix representing a Hermitian operator should equal its transpose conjugate.

Definition 2.27 (Anti-Hermitian Operators) a An operator Ω is anti-Hermitianif

Ω† = −Ω.

It is possible to decompose every operator Ω into its Hermitian and anti-Hermitiancomponents; namely if we decompose Ω as

Ω =Ω+ Ω†

2+

Ω− Ω†

2, (2.140)

18These are also called self-adjoint operators. You may hear that self-adjoint is the term used bymathematicians and Hermitian is used by physicists. However, there is a subtle difference between‘self-adjoint” and “Hermitian”. We will not want to get bogged down worrying too much aboutsuch details, but interested readers should check a book titled “Functional Analysis” by MichaelReed and Barry Simon [Reed and Simon, 1980, p.255].


then, the first term Ω+Ω†

2is Hermitian, and the second term Ω−Ω†

2is anti-Hermitian.

aThese are also known as anti-self-adjoint operators.

Hermitian operators play a central role in quantum mechanics due to its specialcharacteristics.Fact 2.3 (Eigenvalues and Eigenkets of a Hermitian Operator)

1) Real Eigenvalues The eigenvalues of a Hermitian operator are real. Mea-surable values in physics such as position, momentum, and energy are realnumbers, and they are obtained as eigenvalues of Hermitian operators in quan-tum mechanics.

2) Orthonormal Eigenkets The eigenkets of a Hermitian operator can always bechosen so that they are mutually orthogonal, and hence linearly independent.19

Because an eigenket can be multiplied by any nonzero number to generate an-other eigenket of any desired “length”, we can always choose the eigenkets whichform an orthonormal set.

3) Basis of Eigenkets The eigenkets of a Hermitian operator is a complete setin the sense that any vector can be expressed as a linear combination of theeigenkets. Therefore, if you know the action of a Hermitian operator Ω onthe eigenkets, you completely understand the action of Ω on any vector in thespace.Verification: Let v =

∑αi |λi⟩ where λi’s are eigenvalues and |λi⟩’s are the

corresponding eigenkets. Then, Ωv = Ω(∑αi |λi⟩) =

∑αi (Ω |λi⟩).

Example 2.7 (A 2-by-2 Hermitian Operator/Matrix) Let

Ω =

[1 i−i 1

].

As the adjoint is the transpose conjugate, we have

Ω† =

[1 i−i 1

]= Ω, (2.141)

19Theorem 2.7 claims that eigenvectors associated with different eigenvalues of a Hermitian op-erator are mutually orthogonal. More generally, according to Fact A.2, eigenvectors associatedwith simple eigenvalues, i.e. eigenvalues of multiplicity one, are linearly independent. Therefore,the significance of this statement is that Hermiticity assures that we can always find l mutuallyorthogonal eigenvectors if the associated eigenvalue is of multiplicity l.


and Ω is Hermitian. Observe that

Ω

[−i1

]=

[1 i−i 1

] [−i1

]=

[00

]= 0 ·

[−i1

](2.142)

and

Ω

[i1

]=

[1 i−i 1

] [i1

]=

[2i2

]= 2 ·

[i1

](2.143)

So, the eigenvalues are indeed real; namely, 0 and 2.We have

|0⟩ =[−i1

]and |2⟩ =

[i1

]. (2.144)

Because

⟨0|2⟩ =[−i1

]† [i1

]=[i 1

] [ i1

]= 0, (2.145)

|0⟩ and |2⟩ are mutually orthogonal.Next note that

|0⟩+ |2⟩2

=1

2

[02

]=

[01

](2.146)

and− |0⟩+ |2⟩

2i=

1

2i

[2i0

]=

[10

]. (2.147)

Therefore, any ket vector v =

[v1v2

]can be expressed as a linear combination of |0⟩

and |2⟩ as follows.

v =

[v1v2

]= v1

[10

]+ v2

[01

]= v1

(|0⟩+ |2⟩

2

)+ v2

(− |0⟩+ |2⟩

2i

)

=(v12− v2

2i

)|0⟩+

(v12

+v22i

)|2⟩ (2.148)

So, any vector v is a linear combination of |0⟩ and |2⟩. If we normalize |0⟩ and |2⟩,we obtain an orthonormal basis for C2. Namely,

1√2

[−i1

]and 1√

2

[i1

](2.149)

form an orthonormal basis.


Example 2.8 (3-by-3 Hermitian and Non-Hermitian Operators) This exam-ple is to demonstrate the claim made by 2) and 3) of Fact 2.3.Consider the following 3×3 Hermitian matrix Ω.

Ω =

1 0 00 1 00 0 2

(2.150)

According to Example 2.5 on p.51 and Fact A.2 on p.276, the eigenvalue 1 is ofmultiplicity 2, and the eigenvalue 2 is simple, i.e. of multiplicity 1.For the eigenvalue 2, we have

Ω

abc

= 2 ·

abc

=⇒

1 0 00 1 00 0 2

abc

=

ab2c

=

2a2b2c

=⇒ a = b = 0. (2.151)

Hence, any nonzero vector of the form 00c

(2.152)

is an eigenvector associated with the eigenvalue 2. In particular,001

(2.153)

is an example of a normalized eigenvector for the eigenvalue 2.Now, for the eigenvalue 1, we have

Ω

abc

= 1 ·

abc

=⇒

1 0 00 1 00 0 2

abc

=

ab2c

=

abc

=⇒ c = 0. (2.154)

So, any nonzero vector of the form ab0

(2.155)


is an eigenvector associated with the eigenvalue 1. In particular,100

and

010

(2.156)

are an example of mutually orthogonal normalized eigenvectors for the eigenvalue 1.At this point, it is trivial to show that the set

001

,100

,010

(2.157)

forms an orthonormal basis of C3. However, no claim is made as to uniqueness ofsuch a set. For example, 1√

10

00

1 + 3i

, 1√2

1i0

, 1√2

1−i0

(2.158)

is another possible choice.Next, consider the following 3×3 non-Hermitian matrix Λ.

Λ =

1 0 01 1 00 0 2

(2.159)

The eigenvalues are 1 of multiplicity 2 and 2 of multiplicity 1 as was the case withΩ.20 For the eigenvalue 2, we have

Λ

abc

= 2 ·

abc

=⇒

1 0 01 1 00 0 2

abc

=

aa+ b2c

=

2a2b2c

=⇒ a = b = 0. (2.160)

Hence, any nonzero vector of the form 00c

(2.161)

20The characteristic equation mentioned on p.52 is (1− λ)2(2− λ) = 0.


is an eigenvector associated with the eigenvalue 2.Next, for the eigenvalue 1,

Λ

abc

= 1 ·

abc

=⇒

1 0 01 1 00 0 2

abc

=

aa+ b2c

=

abc

=⇒ a = c = 0. (2.162)

So, any nonzero vector of the form 0b0

(2.163)

is an eigenvector associated with the eigenvalue 1. Hence, the set of eigenvectors forΛ can be expressed as follows for arbitrary but nonzero complex numbers b and c.

00c

,0b0

(2.164)

Needless to say, we do not have a basis of C3.

Let us prove Fact 2.3 starting with 1).

Theorem 2.6 (Real Eigenvalues) The eigenvalues of a Hermitian operator Ω arereal.

ProofLet λ be an eigenvalue of Ω and |v⟩ be an associated eigenket. Then,

λ∗⟨v∣∣∣ v⟩ = ⟨

λv∣∣∣ v⟩ = ⟨

Ωv∣∣∣ v⟩ = ⟨

v∣∣∣Ω†

∣∣∣ v⟩ = ⟨v∣∣∣Ω∣∣∣ v⟩ = ⟨

v∣∣∣λv⟩ = λ

⟨v∣∣∣ v⟩

=⇒ (λ∗ − λ)⟨v∣∣∣ v⟩ = 0 =⇒ λ∗ = λ (2.165)

because ⟨v | v⟩ is nonzero if |v⟩ is nonzero.21

Theorem 2.7 Eigenkets |v⟩ and |w⟩ of a Hermitian operator Ω associated withdifferent eigenvalues λv and λw are orthogonal.

21Recall that an eigenket/eigenvector is nonzero by Definition 2.25.


Proof

⟨ω |Ω|v⟩ = ⟨w |λv|v⟩ = λv ⟨w | v⟩ (2.166)

On the other hand, as Ω is Hermitian, and λv and λw are real,

⟨w |Ω|v⟩ =⟨w∣∣∣Ω†|v

⟩= ⟨Ωw | v⟩ = ⟨v |Ωw⟩∗ = ⟨v |λw|w⟩∗ = λ∗

w ⟨v |w⟩∗ = λw ⟨w | v⟩ .

(2.167)

So,

λv ⟨w | v⟩ − λw ⟨w | v⟩ = (λv − λw) ⟨w | v⟩ = 0 =⇒ ⟨w | v⟩ = 0 (2.168)

as λv , λw.

We will now prove 2) and 3) of Fact 2.3 and some more.

Lemma 2.1 Let Ω be a linear operator on a vector space Vn and [Ωij]B be itsmatrix representation with respect to a basis B = u1,u2, . . . ,uk . . . ,un. If thek-th vector in the basis B, denoted by uk is an eigenvector with the correspondingeigenvalue λk, such that Ωuk = λkuk, the k-th column of [Ωij]B is given by Ωjk =λkδjk for 1 ≤ j ≤ n. That is, the entries of the k-th column are zero except for thek-th entry, which is λk.

the k-th column of [Ωij]B =

0......

λk...0

k

(2.169)

Proof


From p.45, we know that the k-th column of [Ωij] is nothing but Ω |uk⟩. But,

Ω |uk⟩ = λkuk = λk

0......

1...0

k

=

0......

λk...0

k. (2.170)

In matrix form,

Ω11 · · · · · · Ω1k · · · Ω1n...

. . ....

......

. . ....

...

Ωk1 · · · · · · Ωkk · · · Ωkn...

.... . .

...Ωn1 · · · · · · Ωnk · · · Ωnn

0......

1...0

k

=

Ω1k......

Ωkk...

Ωnk

= λk

0......

1...0

k

=

0......

λk...0

k

⇐⇒ Ωjk = λkδjk. (2.171)

Corollary 2.1 Let Ω be a Hermitian operator on a vector space Vn and [Ωij]B be itsmatrix representation with respect to a basis B = u1,u2, . . . ,uk, . . . ,un. If thek-th vector in the basis B, denoted by uk is an eigenvector with the correspondingeigenvalue λk, such that Ωuk = λkuk, the k-th column of [Ωij]B is given by Ωjk =λkδjk for 1 ≤ j ≤ n. That is, the entries of the k-th column are zero except for thek-th entry, which is λk. Furthermore, the k-th row of [Ωij]B is given by Ωkj = λkδkjfor 1 ≤ j ≤ n.

ProofΩjk = λkδjk follows from Lemma 2.1. Ωkj = λkδkj follows from the definition ofa Hermitian operator and Theorem 2.6. The matrix representation of Ω†, denotedhere by [Ω†

ij]B, is the transpose conjugate of the matrix for Ω denoted by [Ωij]B. As


the k-th column of [Ωij]B is

0......

λk...0

k, (2.172)

the k-th row of [Ω†ij]B is

[ k

0 · · · · · · λ∗k · · · 0

]. (2.173)

But, λ∗k = λk because λk is real. In matrix form, [Ωij] looks like

0......

0 · · · · · · λk · · · 0...0

. (2.174)

Lemma 2.2 (Block Structure of Hermitian Matrices)Consider an ordered list of all the eigenvalues λ1, λ2, . . . , λn of a Hermitian op-erator Ω, where an eigenvalue λ is repeated mλ times if its multiplicity as a rootof the characteristic polynomial is mλ. Without loss of generality, we can assumethat the same eigenvalues appear next to each other without a different eigenvalue inbetween.a Then, the matrix representation of Ω in the basis |λ1⟩ , |λ2⟩ , . . . , |λn⟩,called an eigenbasis, is block-diagonal.

Corollary 2.2 Suppose the eigenvalues λ1, λ2, . . . , λn of a Hermitian operator Ωare all distinct, such that i , j =⇒ λi , λj. Then, the matrix representation of Ωin the eigenbasis |λ1⟩ , |λ2⟩ , . . . , |λn⟩ is diagonal.


aFor simplicity, suppose we have a list of four eigenvalues consisting of two λ’s, one γ, and oneω. Then, the ordered list is λ, λ, ω, γ, λ, λ, γ, ω, ω, λ, λ, γ, γ, λ, λ, ω and not λ, ω, λ.

Theorem 2.8 If an operator Ω on Vn is Hermitian, there is a basis of its orthonor-mal eigenvectors. In this eigenbasis, the matrix representation of Ω is diagonal, andthe diagonal entries are the eigenvalues. Note that this theorem only claims thatthere is at least one such basis. Theres is no claim about uniqueness as it is not true.

ProofConsider the characteristic polynomial which is of n-th degree. If we set it equal tozero, the equation, called the characteristic equation, has at least one root λ1 anda corresponding normalized eigenket |λ1⟩. This can be extended to a full basis Baccording to Theorem A.5. Now, following the Gram-Schmidt procedure, Theorem2.4, we can find an orthonormal basis O = o1,o2,o3, . . . ,on such that o1 = |λ1⟩.In this basis, according to Corollary 2.1, Ω has the following matrix representationwhich we denote by M or [Ω] in keeping with the notation set on p.48.

M = [Ω] =

λ1 0 · · · 00...0

(2.175)

If we denote the shaded submatrix by N , it can be regarded as the matrix represen-tation of an operator acting on the subspace O2 spanned by o2,o3, . . . ,on. Thecharacteristic polynomial for M is given by

det(M − λIn) = (λ1 − λ)× det(N − λIn−1); (2.176)

where In is the n× n identity matrix, In−1 is the (n− 1)× (n− 1) identity matrix,and det(N − λIn−1) is the characteristic polynomial of N . Because det(N − λIn−1)is P n−1(λ), an (n − 1)st degree polynomial in λ, it must have at least one rootλ2 and a normalized eigenket |λ2⟩ unless n = 1. Because we are regarding N asoperating on O2, |λ2⟩ is a linear combination of o2,o3, . . . , and on, and is orthog-onal to |λ1⟩. As before, we can find an orthonormal basis |λ2⟩ ,o′

3,o′4, . . . ,o

′n

of O2. Note that λ2 is an eigenvalue of M , |λ2⟩ is an eigenvector of M , andO1,2 = |λ1⟩ , |λ2⟩ ,o′

3,o′4, . . . ,o

′n is an orthonormal basis of the original vector

space Vn.

Let us take a detour here and see how the relations between M and N work


explicitly. It may help you understand what is going on better.

First, λ2 is clearly an eigenvalue of M as it is a root of the characteristic polyno-mial for M because of the relation (2.176). Next, as |λ2⟩ is a linear combination ofo2,o3, . . . , and on, it is also a linear combination of |λ1⟩ ,o2,o3, . . . , and on. Wecan write

|λ2⟩ = 0 · |λ1⟩+n∑i=2

eioi. (2.177)

So, |λ2⟩ is the following column vector in the original basisO = |λ1⟩ ,o2,o3, . . . ,on.

|λ2⟩ =

0e2...en

(2.178)

Hence,

M |λ2⟩ =

λ1 0 · · · 00...0

0e2...en

=

0

λ2e2...

λ2en

= λ2 |λ2⟩ ; (2.179)

where the second equality in (2.179) follows from

0...0

0e2...en

=

[0

N |λ2⟩

]=

[0

λ2 |λ2⟩

]. (2.180)

Note that |λ2⟩ is regarded as a vector in Vn in (2.179) and as a vector in thesubspace spanned by o2,o3, . . . , and on in (2.180). There is a trade-off betweenavoiding abuse of notation and simplicity, and our choice here is simplicity.


In the basis O1,2, [Ω] takes the following form.

M = [Ω] =

λ1 0 0 · · · 00 λ2 0 · · · 00 0...

...0 0

(2.181)

Repeating the same procedure, we will finally arrive at

M = [Ω] =

λ1 0 0 · · · 00 λ2 0 · · · 00 0 λ3 · · · 0...

......

. . ....

0 0 0 · · · λn

. (2.182)

Note that some eigenvalues may occur more than once, but the above proof stillworks as it is.

Example 2.9 (Diagonalization of a Hermitian Matrix) Consider

Ω =

[1 11 1

].

Ω is clearly Hermitian as it equals its own transpose conjugate. Observe that

Ω

[11

]=

[1 11 1

] [11

]=

[22

]= 2 ·

[11

](2.183)

and

Ω

[1−1

]=

[1 11 1

] [1−1

]=

[00

]= 0 ·

[1−1

]. (2.184)

Define normalized eigenvectors |e1⟩ and |e2⟩ as follows.

|e1⟩ =1√2

[11

]and |e2⟩ =

1√2

[1−1

](2.185)


Then,

Ω11 = ⟨e1|Ω|e1⟩ =1√2

[1 1

] [ 1 11 1

]1√2

[11

]

=1

2

[1 1

] [ 22

]=

1

2· 4 = 2. (2.186)

Similarly, we also get

Ω12 = ⟨e1|Ω|e2⟩ = 0,

Ω21 = ⟨e2|Ω|e1⟩ = 0,

andΩ22 = ⟨e2|Ω|e2⟩ = 0.

(2.187)

Note that the eigenvalues are on the diagonal and all the off-diagonal entries arezero; i.e. we have diagonalized Ω.

Fact 2.4 If Ω is an anti-Hermitian operator:

1. The eigenvalues of Ω are purely imaginary.

2. There is an orthonormal basis consisting of normalized eigenvectors of Ω.

2.10.2 Simultaneous DiagonalizationTheorem 2.9 (Commuting Hermitian Operators) If Ω and Γ are commutingHermitian operators, there exists a basis consisting of common eigenvectors, suchthat both Ω and Γ are diagonal in that basis.

This very useful theorem is actually a corollary of the following more general theo-rems from matrix algebra. For details, check Appendix A.

Theorem 2.10 (Commuting Diagonalizable Matrices) Let A and B be diag-onalizable n×n matrices. Then, AB = BA if and only if A and B are simultaneouslydiagonalizable.

Theorem 2.11 (A Family of Diagonalizable Matrices) Let F be a family ofdiagonalizable n × n matrices. Then, it is a commuting family if and only if it issimultaneously diagonalizable.


2.11 Active and Passive TransformationsWe will mainly deal with active transformations in this book. So, you only need tofamiliarize yourself with the general concept.

In Classical MechanicsIn classical mechanics, it is about a point in space and the coordinate system.

• An active transformation moves the point while leaving the coordinateaxes unchanged/fixed. For example, a rotation of the point (x0, y0) in thex, y-plane by angle θ is effected by the matrix Rθ given by

Rθ =

[cos θ − sin θsin θ cos θ

](2.188)

such that the new position (x′0, y

′0) is obtained as follows.[

x′0

y′0

]= Rθ

[x0y0

]=


] [x0y0

]=

[x0 cos θ − y0 sin θx0 sin θ + y0 cos θ

](2.189)

Under an active transformation, the value of a dynamical variable, often de-noted by ω(q, p), will not necessarily remain the same.

• A passive transformation does not move the point (x0, y0) but only changesthe coordinate system. To take a 2-dimensional rotation as an example, thiscorresponds to rotating the x-y coordinate system without moving the point it-self. Because the only change is in the value of the point’s x- and y-coordinatesrelative to the new rotated axes, and the position itself does not change in thiscase, the physical environment of the point remains the same, and a dynamicalvariable ω(q, p) will have the same value after the transformation.

Note that the new coordinates (x′0, y

′0) are numerically the same whether you rotate

the axes by θ or the point by −θ. Hence, the effect of passive transformation by θcan be computed as below.[

x′0

y′0

]= R−θ

[x0y0

]=

[cos(−θ) − sin(−θ)sin(−θ) cos(−θ)

] [x0y0

]=

[x0 cos θ + y0 sin θ−x0 sin θ + y0 cos θ

](2.190)


Again, despite this relation, the two transformations should not be confused witheach other because the active transformation is the only physically meaningful trans-formation, in the sense that the values of dynamical variables may change.

In Quantum MechanicsIn quantum mechanics, it is about an operator Ω and a basis vector |v⟩. Consider aunitary transformation U which causes a basis change |v⟩ −→ U |v⟩ := |Uv⟩. Underthis transformation, the matrix elements of an operator Ω transforms as follows.

⟨v′|Ω|v⟩ −→ ⟨Uv′|Ω|Uv⟩ = ⟨v′|U †ΩU |v⟩ (2.191)

In (2.191), the change is caused to the vector v in ⟨Uv′|Ω|Uv⟩, and to the operatorΩ in ⟨v′|U †ΩU |v⟩. Nevertheless, the matrix elements of Ω are the same between thetwo formulations or interpretations.

• An active transformation corresponds to

⟨Uv′|Ω|Uv⟩ ; (2.192)

where the vector v is transformed as below.

v −→ Uv (2.193)

• A passive transformation picture is painted by

⟨v′|U †ΩU |v⟩ ; (2.194)

where the operator Ω is transformed as follows.

Ω −→ U †ΩU (2.195)

Because the essence of quantum mechanics is the matrix elements of operators, andall of the physics of quantum mechanics lies in the matrix elements, the equality of(2.192) and (2.194) given in (2.191) indicates that these two views are completelyequivalent. Hence, active and passive transformations in quantum mechanics providetwo equivalent ways to describe the same physical transformation.

68

Exercises1. Consider

Ω =

[0 11 0

].

The eigenvalues are 1 and −1.

(a) Explain why Ω is Hermitian.(b) Find a real normalized eigenvector for each eigenvalue. We will denote

them by |1⟩ and |−1⟩.(c) Verify that the eigenvectors are orthogonal.

(d) Express an arbitrary vector v =

[v1v2

]as a linear combination of |1⟩ and

|−1⟩.(e) Express Ω as a matrix in the basis |1⟩ , |−1⟩.

2. Answer the following questions about the matrix representation of a Hermitianoperator.

(a) Given that the matrix

M =

[6 4α 0

]

is Hermitian, what is α?

(b) The eigenvalues are 8 and −2. Find NORMALIZED and REAL eigen-vectors |8⟩ and |−2⟩, remembering that |λ⟩ means an eigenket associatedwith the eigenvalue λ.

(c) Find the matrix representation of M in the basis consisting of |8⟩ (thefirst basis ket) and |−2⟩ (the second basis ket).

3. LetA =

[0 21 1

]and B =

[−1 21 0

].

(a) Show [A,B] = 0.

69

(b) The eigenvalues forA are 2 and−1. Find two real normalized eigenvectors,|2⟩A and |−1⟩A.

(c) Show that |2⟩A and |−1⟩A are also eigenvectors of B, but the eigenvaluesare not the same.

(d) Verify that |2⟩A , |−1⟩A forms a basis for C2.(e) Find the matrix representations ofA andB relative to the basis |2⟩A , |−1⟩A.

4. Consider two Hermitian matrices

M =

[1 −1−1 1

]and N =

[0 11 0

].

(a) Prove that M and N commute.(b) Find the eigenvalues and real normalized eigenvectors of M and N .(c) Show that M and N are simultaneously diagonalizable.

Chapter 3

Fundamental Postulates and theMathematical Framework ofQuantum Mechanics

This is where quantum mechanics markedly differs from mathematics as you knowit1. The postulates of quantum mechanics cannot be proved or deduced in a purelymathematical fashion. The postulates are hypotheses, that are given to you fromthe beginning. If no disagreement with experiments is found, the postulates are ac-cepted as correct hypotheses. Sometimes, they are called axioms, which means thepostulates are regarded true though not provable.

You should not feel excessively uncomfortable with this. Much of what we knowin physics cannot be proved. For example, consider the celebrated Newton’s SecondLaw of Motion; F = ma. There is no purely mathematical derivation of this relation-ship. In physics, a theory is deemed “correct” if there is a good agreement betweenobservations and theoretical predictions. While classical Newtonian mechanics suf-fices for macroscopic systems, a huge body of experimental evidence supports theview that the quantum postulates provide a consistent description of reality on theatomic scale.

It is possible and is general practice to limit the number of the postulates to aminimum of tow or three. While it may be esthetically pleasing, and mathematically

1I am saying ”mathematics as you know it” because mathematics also relies on many postulatesoften known as axioms. However, most of us are used to encountering definitions and proofs whendoing mathematics and not necessarily postulates.

71

72CHAPTER 3. FUNDAMENTAL POSTULATES AND THE MATHEMATICAL

FRAMEWORK OF QUANTUM MECHANICS

more sound, I believe it helps understand the quantum theory more easily to startwith a greater number of postulates. Necessarily, not all the postulates presentedhere are independent of one another. However, my choice is better understandingfor beginners in exchange for redundancy.

3.1 The Fundamental Postulates1.

An isolated physical system is associated with a topologically separablea,b com-plex Hilbert space H with inner product ⟨ϕ|ψ⟩. Physical states can be iden-tified with equivalence classes of vectors of length 1 in H, where two vectorsrepresent the same state if they differ only by a phase factor.

aA topological space is separable if it contains a countable dense subset; that is, thereexists a sequence xi∞

i=1 of elements of the space such that every nonempty open subset ofthe space contains at least one element of the sequence.

bSeparability means that the Hilbert space H has a basis consisting of countably manyvectors. Physically, this means that countably many observations suffice to uniquely deter-mine the state.

The above is a mathematically correct and rigorous statement of Postulate 1.A more accessible simplified statement would be as follows.

Each physical system has its associated Hilbert space H, so thatthere is one-to-one correspondence between a physical state s ofthe system and a vector vs, or a ket |s⟩, of unit length in H.

2. To every observable q in classical mechanics there corresponds aHermitian operator Q in quantum mechanics. Table 3.1 summarizesthe correspondences.In Table 3.1, I am showing the correspondence between classical observablesand quantum mechanical operators for a single variable x for simplicity. If weinclude y and z, parts of Table 3.1 should be modified as shown in Table 3.2.

2H or H is called Hamiltonian.

3.1. THE FUNDAMENTAL POSTULATES 73

Observable q Q OperationPosition x x or x Mx : multiplicationMomentum p p or p ı

(−i~ ∂

∂x

)Kinetic Energy T T or T − ~2

2m∂2

∂x2

Potential Energy V (x) V (x) or V (x) MV (x) : multiplicationTotal Energy E H or H2 − ~2

2m∂2

∂x2+ V (x)

Angular Momentum lx Lx or Lx −i~(y ∂∂z− z ∂

∂y

)

Table 3.1: Physical observables and corresponding quantum operators

Observable q Q OperationMomentum p p or p −i~

(ı ∂∂x

+ ȷ ∂∂y

+ k ∂∂z

)Kinetic Energy T T or T − ~2

2m

(∂2

∂x2+ ∂2

∂y2+ ∂2

∂z2

)Potential Energy V (r) V (r) or V (r) MV (r) : multiplicationTotal Energy E H or H − ~2

2m

(∂2

∂x2+ ∂2

∂y2+ ∂2

∂z2

)+ V (r)

Angular Momentum lx Lx or Lx −i~(y ∂∂z− z ∂

∂y

)ly Ly or Ly −i~

(z ∂∂x− x ∂

∂z

)lz Lz or Lz −i~

(x ∂∂y− y ∂

∂x

)

Table 3.2: Quantum operators in three dimensions

3. When a measurement is made for an observable q associated withoperator Q, the only possible outcomes are the eigenvalues of Qdenoted by λi.

4. The set of eigenstates/eigenvectors of operator Q, denoted by |λi⟩,forms a complete3 orthonormal set of the Hilbert space H.

3A collection of vectors vαα∈A in a Hilbert space H is complete if ⟨w |vα⟩ = 0 for all α ∈ Aimplies w = 0. Equivalently, vαα∈A is complete if the span of vαα∈A is dense in H, that is,given w ∈ H and ϵ > 0, there exists w′ ∈ spanvα such that ∥w′ −w∥ < ϵ.



5. Suppose a physical system is in state s. If the vector vs ∈ H asso-ciated with the physical state s is given by

vs =n∑i=1

ai |λi⟩ ; (3.1)

where n may go to ∞, the possible outcomes of a measurement ofthe physical quantity q are the eigenvalues λ1, λ2, . . . , λk , . . ., and theprobability that λk is observed is

| ⟨λk |vs⟩ |2 = |ak|2. (3.2)

6. As soon as a measurement is conducted on a physical state s,yielding λk as the outcome, the state vector vs collapses to thecorresponding eigenstate |λk⟩. Therefore, measurement affects thephysical state of the system.4

7. The average value of the observable q after a large number of mea-surements is given by the expectation value

q or ⟨q⟩ = ⟨vs |Q|vs⟩ . (3.3)

8. The state vector |s⟩ associated with a physical state s typically de-pends on time and the spatial coordinates; that is, |s⟩ = |s(r, t)⟩. Thestate evolves in time according to the time-dependent Schrodingerequation.

i~∂

∂t|s(r, t)⟩ = H |s(r, t)⟩ (3.4)

In a typical application, our state vector is a differentiable function, called awavefunction, and the canonical symbol used for the wavefunction is Ψ. Withthis notation, we get

i~∂Ψ(r, t)∂t

= HΨ(r, t). (3.5)

9. We can use the Schrödinger Equation to show that the first deriva-tive of the wave function should be continuous, unless the poten-tial becomes infinite across the boundary. The wavefunction Ψ(r, t)and its spatial derivatives (∂/∂x)Ψ(r, t), (∂/∂y)Ψ(r, t), (∂/∂z)Ψ(r, t) arecontinuous if the potential V (r) = V (x, y, z) is finite.

4This fact is often used in elaborate experimental verifications of quantum mechanics.


10. You can skip this for now.

The total wavefunction must be antisymmetric with respect to theinterchange of all coordinates of one fermiona with those of an-other. Electron spin must be included in this set of coordinates.The Pauli exclusion principle is a direct consequence of this an-tisymmetry principle. Slater determinants provide a convenientmeans of enforcing this property on electronic wavefunctions.

aParticles are classified into two categories, fermions and bosons. Fermions have half-integral intrinsic spins such as 1/2, 3/2, and so forth, while bosons have integral intrinsicspins such as 0, 1, and so forth. Quarks and leptons, as well as most composite particles,like protons and neutrons, are fermions. All the force carrier particles, such as photons, Wand Z bosons, and gluons are bosons, as are those composite particles with an even numberof fermion particles like mesons.

Of the 10 postulates, Postulates 8 is distinctly different from the other nine. Thenine postulates describe the system at a given time t, while Postulate 8 specifies howthe system changes with time.3.2 Unitary Time EvolutionTime evolution of a quantum system is always given by a unitary transformation U ,such that

|s(t)⟩ = U(t) |s(0)⟩ . (3.6)The nature of U(t) depends on the system and the external forces it experiences.However, U(t) does not depend on the state |s⟩. Hence, U(α |s1⟩ + β |s2⟩) =αU |s1⟩ + βU |s2⟩, and the time evolution operator U is linear. Again, the keyhere is that the time evolution operator U is a function of the physical system andnot individual states. Sometimes U(t) is referred to as a propagator.

The unitarity of U follows from the time-dependent Schrodinger equation (3.4)as follows. We have

i~∂

∂t|s(t)⟩ = H(t) |s(t)⟩ or ∂

∂t|s(t)⟩ = −i

~H(t) |s(t)⟩ ; (3.7)

where H is the Hamiltonian of the system and is Hermitian. Suppose

|s(t)⟩ = U(t) |s(0)⟩ (3.8)



for some operator U . Plugging this into the time-dependent Schrodinger equation,we obtain

∂

∂t

(U(t) |s(0)⟩

)=−i~H(t)U(t) |s(0)⟩ =⇒

(∂

∂tU(t)

)|s(0)⟩ =

(−i~H(t)U(t)

)|s(0)⟩

(3.9)

for any physical state s. This means that the actions of

∂

∂tU(t) and −i

~H(t)U(t) (3.10)

on any set of basis vectors are the same. Therefore,

∂

∂tU(t) =

−i~H(t)U(t)a. (3.11)

Taking the adjoint, and noting that H = H†,(∂

∂tU(t)

)†

=(−i~H(t)U(t)

)†=⇒ ∂

∂tU †(t) =

i

~U †(t)H†(t) =

i

~U †(t)H(t). (3.12)

Now, at t = 0, U(0) = I by necessity, and this gives U †(0)U(0) = I. On the otherhand,

∂

∂t

(U †(t)U(t)

)=

(∂

∂tU †(t)

)U(t) + U †(t)

(∂

∂tU(t)

)

=

(∂

∂tU †(t)

)U(t) + U †(t)

(∂

∂tU(t)

)=i

~U †(t)H(t)U(t) + U †(t)

−i~H(t)U(t)

=i

~U †(t)

(H(t)− H(t)

)U(t) = 0. (3.13)

Hence, U †(t)U(t) = I at all times t, and U(t) is unitary. Because U(t) is unitary,

∥ |s(t)⟩∥2 = ⟨s(t)|s(t)⟩ =⟨s(0)

∣∣∣ U †(t)U(t)|s(0)⟩= ⟨s(0)|I|s(0)⟩ = ∥ |s(0)⟩ ∥2,

(3.14)

and the magnitude of the state vector is preserved. One way to interpret timeevolution is to regard it as a rotation of the state vector in Hilbert space.


3.3 Time-Independent HamiltonianSo far, we have considered a Hamiltonian with explicit time-dependence as is clearfrom the notation H(t). However, for many physical systems, the Hamiltonian isnot time-dependent. Fro example, consider a typical Hamiltonian given by

aOne way to understand what this equality means is to think in terms of their matrix represen-tations with respect to a basis.

H = − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+ V (r) (3.15)

in Table 3.2. In the most general case, the potential energy term V can have explicittime dependence so that V = V (r, t). However, if V does not depend on t, andneither does H as shown above, there are simple relations among the time evolutionoperator U(t), the system Hamiltonian H, and the eigenstates of H.

3.3.1 Propagator as a Power Series of HWe start with the differential relation (3.11)

d

dtU(t) =

−i~H(t)U(t); (3.16)

where we switched to total time derivative as we will only focus on t here, holdingall the other variables fixed. Let us now remove time-dependence from H in (3.16).

d

dtU(t) =

−i~HU(t); (3.17)

We will compare this with the familiar case of a real-valued function f : R → Rsatisfying the differential equation

d

dtf(t) =

−i~Hf(t); (3.18)

where H is some scalar. As the answer to this differential equation is

f(t) = f(0) exp[−iHt~

], (3.19)



you may think the answer to the analogous operator differential equation (3.17) issomething like

U(t) = U(0) exp[−iHt~

]= exp

[−iHt~

]; (3.20)

where the second equality holds as U(0) is the identity operator by necessity. As itturns out, this is the right answer with the interpretation of the right-hand side as apower series in operator H. This is explained in more detail in Appendix B. But, wewill only check it formally here to convince ourselves that the solution really works.Maclaurin series of the exponential function gives

exp[−iHt~

]=

∞∑n=0

[−iH~

]nn!

tn = I +∞∑n=1

[−iH~

]nn!

tn; (3.21)

where[

−iH~

]0= I by definition. Then,

d

dt

(exp

[−iHt~

])=

d

dt

I + ∞∑n=1

[−iH~

]nn!

tn

=d

dt

∞∑n=1

[−iH~

]nn!

tn

=

∞∑n=1

[−iH~

]nn(n− 1)!

ntn−1 =−iH~

∞∑n=1

[−iH~

]n−1

(n− 1)!tn−1 =

−iH~

∞∑n=0

[−iH~

]nn!

tn

=−i~H exp

[−iHt~

]. (3.22)

So,

U(t) = exp[−iHt~

](3.23)

satisfies the differential equation (3.17).

In Section 3.2, we showed that the time evolution operator U(t) is unitary forany Hamiltonian H. When H does not depend on time, we have an alternativeproof that U(t) is unitary. However, note that our discussion below is somewhatheuristic and not completely rigorous mathematically speaking.


First, taking the term-by-term adjoint of the right-hand side of (3.21), we get

(exp

[−iHt~

])†=

∞∑n=0

[−iH~

]nn!

tn

†

=∞∑n=0

[(−iH~

)†]n

n!tn

=∞∑n=0

[+iH~

]nn!

tn = exp[iHt

~

]. (3.24)

It suffices to show

U(t)U †(t) = exp[−iHt~

]exp

[iHt

~

]= I. (3.25)

Recall from Theorem 2.8 that a matrix representing a Hermitian operator is diago-nalizable with its eigenvalues on the diagonal. Because H is Hermitian, there exitsan invertible matrix J such that

JHJ−1 = D =

λ1

. . .

λm

or H = J−1DJ = J−1

λ1

. . .

λm

J ;(3.26)

where λ1, . . . , λm are the eigenvalues of H. We also have

Hn =(J−1DJ

)n= J−1DnJ = J−1

λ1

. . .

λm

n

J = J−1

λn1

. . .

λnm

J.(3.27)

Therefore,

U(t) = e− i~Ht =

∞∑n=0

(−i~

)nHntn

1

n!=

∞∑n=0

(−i~

)nJ−1DnJtn

1

n!

=∞∑n=0

J−1(−i~

)nDntn

1

n!J = J−1

( ∞∑n=0

(−i~

)nDntn

1

n!

)J

= J−1

∞∑n=0

(−i~

)n λ1

. . .

λm

n

tn1

n!

J



= J−1

∞∑n=0

(−i~

)n λn1

. . .

λnm

tn 1

n!

J

= J−1

∑∞n=0

(−i~

)nλn1 t

n 1n!

. . . ∑∞n=0

(−i~

)nλnmt

n 1n!

J

= J−1

∑∞n=0

(−i~λ1t

)n1n!

. . . ∑∞n=0

(−i~λmt

)n1n!

J

= J−1

exp

(−i~λ1t

). . .

exp(

−i~λmt

) J. (3.28)

Similarly,

U †(t) = ei~Ht = J−1

exp

(i~λ1t

). . .

exp(i~λmt

) J. (3.29)

We now have

U(t)U †(t)

= J−1

exp

(−i~λ1t

). . .

exp(

−i~λmt

) JJ−1

exp

(i~λ1t

). . .

exp(i~λmt

) J

= J−1

exp

(−i~λ1t

). . .

exp(

−i~λmt

)

exp(i~λ1t

). . .

exp(i~λmt

) J

= J−1

1

. . .

1

J = I. (3.30)


Therefore, U(t) is indeed unitary for all t.

3.3.2 Eigenstate Expansion of the PropagatorHere, we will discuss how the propagator U(t) can be expressed using the normalizedeigenkets of H. The eigenvalue problem for H is

H |E⟩ = E |E⟩ ; (3.31)

where E is an eigenvalue representing the total energy, and |E⟩ is an associatednormalized eigenket. From (2.48),∑

|E⟩ ⟨E| = I. (3.32)

Hence,

|s(t)⟩ =∑|E⟩ ⟨E|s(t)⟩ , (3.33)

and ⟨E|s(t)⟩ gives the coefficients when |s(t)⟩ is expressed as a linear combinationof |E⟩. For simplicity of notation, let cE(t) = ⟨E|s(t)⟩, so that

|s(t)⟩ =∑

cE(t) |E⟩ . (3.34)

From Postulate 8,

i~Ψ(r, t)∂t

= HΨ(r, t) =⇒ i~∂ |s(t)⟩∂t

= H |s(t)⟩

=⇒ i~∂∑cE(t) |E⟩∂t

= H∑

cE(t) |E⟩ =⇒∑

i~∂cE(t)

∂t|E⟩ =

∑cE(t)E |E⟩

=⇒∑(

i~∂cE(t)

∂t− EcE(t)

)|E⟩ = 0. (3.35)

As |E⟩’s are linearly independent,

i~∂cE(t)

∂t− EcE(t) = 0 for each value of E. (3.36)

This implies that cE(t) is an exponential function.

i~∂cE(t)

∂t− EcE(t) = 0 =⇒ ∂cE(t)

∂t=−iE~

cE(t) =⇒ cE(t) = cE(0)e−iEt/~. (3.37)



Remembering that cE(t) = ⟨E|s(t)⟩, and in particular cE(0) = ⟨E|s(0)⟩, we have

|s(t)⟩ =∑⟨E|s(0)⟩ e−iEt/~ |E⟩ =

∑|E⟩ ⟨E|s(0)⟩ e−iEt/~. (3.38)

Comparing this with (3.6), we get

U(t) =∑E

|E⟩ ⟨E| e−iEt/~. (3.39)

This is the eigenstate expansion of the propagator U(t).

Let us double check and see if this expression of U(t) is unitary. Following thesteps described in Note 2.1,

U †(t) =

(∑E

|E⟩ ⟨E| e−iEt/~)†

=∑E

(|E⟩ ⟨E| e−iEt/~

)† Step 1−−−→∑E

e−iEt/~ ⟨E| |E⟩

Step 2−−−→∑E

eiEt/~ ⟨E| |E⟩ Step 3−−−→ U †(t) =∑E

eiEt/~ |E⟩ ⟨E| . (3.40)

Hence,

U(t)U †(t) =

(∑E

|E⟩ ⟨E| e−iEt/~)(∑

E′eiE

′t/~ |E ′⟩ ⟨E ′|)

=∑E,E′|E⟩ ⟨E|E ′⟩ ⟨E ′| e−i(E−E′)t/~ =

∑E=E′

|E⟩ ⟨E ′| e−i(E−E′)t/~ =∑E

|E⟩ ⟨E| = I.

(3.41)

This checks.

3.4 The Uncertainty PrincipleThe uncertainty principle is one of the celebrated consequences of quantum mechan-ics, whose influence has gone beyond the realm of physics into humanities such asphilosophy.

3.4.1 Uncertainty and Non-CommutationTheorem 3.1 (The Uncertainty Relation) Consider two physical observables rep-resented by Hermitian operators A and B. We denote the expectation value of an

3.4. THE UNCERTAINTY PRINCIPLE 83

operator Ω by ⟨Ω⟩; that is, if the state of the system is |u⟩ after normalization,

⟨Ω⟩ = ⟨u|Ω |u⟩ . (3.42)

In addition, define A by

A =√⟨(A− ⟨A⟩I)2⟩5,6. (3.43)

Then, [A,B] = iα implies AB ≥ |α|2

; where α is a real scalar.

ProofWe will first define two Hermitian operators CA and CB as follows.

CA = A− ⟨A⟩I and CB = B − ⟨B⟩I (3.44)

Then,

[CA, CB] = [A− ⟨A⟩I, B − ⟨B⟩I] = [A− ⟨A⟩I, B]− [A− ⟨A⟩I, ⟨B⟩I] = [A,B] = iα.(3.45)

Next define two vectors by

|v⟩ = CA |u⟩ and |w⟩ = CB |u⟩ . (3.46)

Then,

⟨v|v⟩ = ⟨u|C†ACA |u⟩ = ⟨u|CACA |u⟩ = ⟨u|C2

A |u⟩= ⟨u| (A− ⟨A⟩I)2 |u⟩ = ⟨(A− ⟨A⟩)2⟩ (3.47)

and

⟨w|w⟩ = ⟨u|C†BCB |u⟩ = ⟨u|CBCB |u⟩ = ⟨u|C2

B |u⟩= ⟨u| (B − ⟨B⟩I)2 |u⟩ = ⟨(B − ⟨B⟩)2⟩. (3.48)

5From the formula, you can see that A is the standard deviation of the observable A in thestate |u⟩.

6This definition means (A)2 = ⟨(A− ⟨A⟩)2⟩ = ⟨(A2−2A⟨A⟩+⟨A⟩2)⟩ = ⟨A2⟩−2⟨A⟩⟨A⟩+⟨A⟩2 =

⟨A2⟩ − ⟨A⟩2 or A =√⟨A2⟩ − ⟨A⟩2.



Therefore,

⟨v|v⟩ = (A)2 and ⟨w|w⟩ = (B)2 . (3.49)

Now, by Schwarz Inequality (Theorem 2.3),

(A)2 (B)2 = ⟨v|v⟩ ⟨w|w⟩ ≥ | ⟨v|w⟩ |2 = | ⟨u|CACB|u⟩ |2

= | ⟨u|(CACB − CBCA + CACB + CBCA)/2|u⟩ |2 =1

4| ⟨u|[CA, CB] + CA, CB|u⟩ |2;

(3.50)

where CA, CB = CACB + CBCA is called an anti-commutator. Because CA andCB are Hermitian,

⟨u|CA, CB|u⟩ = ⟨u|CACB + CBCA|u⟩ = ⟨u|CACB|u⟩+ ⟨u|CBCA|u⟩= ⟨u|C†

ACB|u⟩+ ⟨u|C†BCA|u⟩ = ⟨CAu|CBu⟩+ ⟨CBu|CAu⟩

= ⟨CAu|CBu⟩+ ⟨CAu|CBu⟩∗ = 2Re ⟨CAu|CBu⟩ . (3.51)

Fro simplicity of notation, let β = 2Re ⟨CAu|CBu⟩. Then, we have

(A)2 (B)2 ≥ 1

4|iα+ β|2 = 1

4(α2 + β2) ≥ 1

4α2. (3.52)

Therefore,AB ≥ |α|

2. (3.53)

3.4.2 Position and Momentum

From Table 3.1, p = −i~ ddx

and x =Mx. In order to compute the commutator [x, p],we will let [x, p] act on a function f(x).

[x, p]f(x) = x(−i~ ddx

)f(x)− (−i~ ddxx)f(x) = −i~x d

dxf(x) + i~

d

dx(xf(x))

= −i~x ddxf(x) + i~f(x) + i~x

d

dxf(x) = i~f(x) =⇒ [x, p] = i~ (3.54)


Therefore,

xp ≥ ~2. (3.55)

3.4.3 Energy and TimeThe momentum-position uncertainty principle xp ≥ ~

2has an energy-time ana-

log, Et ≥ ~2. However, this must be a different kind of relationship because t is

not a dynamical variable. This uncertainty cannot have anything to do with lack ofcommutation.

3.4.3.1 Time rate of change of expectation values

How do expectation values change in time? That is, how can we compute

d

dt⟨A⟩(t) = d

dt⟨u(t)|A|u(t)⟩ ; (3.56)

where |u(t)⟩ is the state of the system at time t. From (3.4),

d

dt|u(t)⟩ = 1

i~H |u(t)⟩ and d

dt⟨u(t)| = − 1

i~⟨u(t)|Ha. (3.57)

So,

d

dt⟨A⟩(t) =

(d

dt⟨u(t)|

)A |u(t)⟩+ ⟨u(t)|A

(d

dt|u(t)⟩

)

= − 1

i~⟨u(t)|HA |u(t)⟩+ ⟨u(t)|A 1

i~H |u(t)⟩

=1

i~⟨u(t)| −HA+ AH |u(t)⟩ = 1

i~⟨[A,H]⟩(t). (3.58)

We haved

dt⟨A⟩(t) = 1

i~⟨[A,H]⟩(t). (3.59)

The indication here is that dynamical evolution of an observable depends on its levelof commutativity with H, at least on the average.



3.4.3.2 Time-energy uncertainty

Our derivation of the general form of the uncertainty principle (3.53) concerns twoobservables. However, as mentioned already, time t is not a dynamical variable. Weneed a way to make a connection between t and an observable A if we are to usethe general inequality (3.53) to formulate time-energy uncertainty.

We will use an observable A to characterize the change in the system in time.As A and the energy E are observables, we have

AE ≥ 1

2|⟨[A,H]⟩| (3.60)

from (3.53). Now, from (3.59),

1

2|⟨[A,H]⟩| = 1

2

∣∣∣∣∣i~ ddt⟨A⟩∣∣∣∣∣ = ~2

∣∣∣∣∣ ddt⟨A⟩∣∣∣∣∣ . (3.61)

Hence,

AE ≥ ~2

∣∣∣∣∣ ddt⟨A⟩∣∣∣∣∣ . (3.62)

The ratio of the amount of uncertainty in A, denoted by A, to the rate of changein the expected value of A,

∣∣∣ ddt⟨A⟩

∣∣∣, is one candidate for t. With this definition,we have

t = A∣∣∣ ddt⟨A⟩

∣∣∣ . (3.63)

Then,

Et = EA∣∣∣ ddt⟨A⟩

∣∣∣ ≥~2

∣∣∣ ddt⟨A⟩

∣∣∣∣∣∣ ddt⟨A⟩

∣∣∣ =~

2=⇒Et ≥ ~

2. (3.64)

Again this inequality is completely different from that for two observables. It de-pends on how you devise your way to measure time which is different from ”the”usual time t. This may be the reason why time-energy uncertainty is so notoriouslycontroversial.


3.5 Classical Mechanics and Quantum MechanicsAccording to the correspondence principle advocated by Niels Bohr, every new phys-ical theory must contain as a limiting case the old theory. So, quantum mechanicshas to be a superset of classical mechanics. Indeed, classical mechanics is a limitingcase of quantum mechanics as h or ~→ 0. It was also shown by Ehrenfest that theexpectation values of quantum mechanical observables satisfy the same equationsas the corresponding variables in classical mechanics.

3.5.1 Ehrenfest’s TheoremOne embodiment of the correspondence principle are the two relations proved byEhrenfest that connect classical mechanics and quantum mechanics.

Theorem 3.2 (Ehrenfest Theorem) If Ω is a quantum mechanical operator and⟨Ω⟩ is its expectation value, we have

d

dt⟨Ω⟩ = 1

i~⟨[Ω, H]⟩+

⟨∂Ω

∂t

⟩. (3.65)

ProofThis is a generalized version of (3.59); where the operator is possibly time-dependent.We will write out the integral for the expectation values explicitly in order to seethe reasons for the total derivative d

dtand the partial derivative ∂

∂tclearly. We will

specialize to a one-dimensional case without loss of generality and let Ψ(x, t) be anormalized wavefunction representing the state. Then,

d

dt⟨Ω⟩ = d

dt

∫Ψ(x, t)∗Ω(x, t)Ψ(x, t) dx =

∫ ∂

∂t

(Ψ(x, t)∗Ω(x, t)Ψ(x, t)

)dxb

=∫ ∂

∂t

(Ψ(x, t)∗

)Ω(x, t)Ψ(x, t) dx

+∫

Ψ(x, t)∗ ∂

∂t

(Ω(x, t)

)Ψ(x, t) dx

+∫Ψ(x, t)∗Ω(x, t)

∂

∂t

(Ψ(x, t)

)dx. (3.66)



Now, from Postulate 8 on p.74, we have

i~∂

∂t|s(r, t)⟩ = H |s(r, t)⟩

=⇒ ∂

∂t|s(r, t)⟩ = 1

i~H |s(r, t)⟩ and ∂

∂t⟨s(r, t)| = 1

−i~⟨s(r, t)| Hc. (3.67)

In our case (3.67) becomes

∂

∂t

(Ψ(x, t)

)=

1

i~HΨ(x, t) and ∂

∂t

(Ψ(x, t)∗

)=−1i~

Ψ(x, t)∗H. (3.68)

Substituting these into (3.66),

d

dt⟨Ω⟩ =

∫ −1i~

Ψ(x, t)∗HΩ(x, t)Ψ(x, t) dx+

⟨∂Ω

∂t

⟩

+∫Ψ(x, t)∗Ω(x, t)

1

i~HΨ(x, t) dx

=1

i~

∫Ψ(x, t)∗(ΩH −HΩ)Ψ(x, t) dx+

⟨∂Ω

∂t

⟩

=1

i~⟨[Ω, H]⟩+

⟨∂Ω

∂t

⟩(3.69)

In particular, if the Hamiltonian is time-independent as in Section 3.3, EhrenfestTheorem implies the following.

Corollary 3.1 (Time-Independent Hamiltonian) If the Hamiltonian H doesnot have explicit time dependence, the time rate of change of the expectation valueof a variable Ω represented by operator Ω, < Ω >, is proportional to the expectationvalue of the commutator between Ω and H, [Ω, H], with a proportionality constant −i

~.

Therefore, < Ω > is conserved if Ω commutes with a time-independent HamiltonianH.

aOne way to understand this is to regard |u(t)⟩ as an n × 1 matrix (a column vector) and⟨u(t)| as its transpose conjugate (a row vector). With this interpretation, d

dt |U(t)⟩ becomes entry


by entry differentiation of a matrix, and(

ddt |u(t)⟩

)†= d

dt ⟨u(t)| and(1i~H |u(t)⟩

)†= − 1

i~ ⟨u(t)|Hnaturally.

aOne way to understand this is to regard |u(t)⟩ as an n × 1 matrix (a column vector) and⟨u(t)| as its transpose conjugate (a row vector). With this interpretation, d

dt |U(t)⟩ becomes entryby entry differentiation of a matrix, and

(ddt |u(t)⟩

)†= d

dt ⟨u(t)| and(1i~H |u(t)⟩

)†= − 1

i~ ⟨u(t)|Hnaturally.

bWe need to switch from ddt to ∂

∂t because the independent variables are x and t before theintegration with respect to x removes x as a variable.

cNote that ∂∂t ⟨s(r, t)| =

1−i~ ⟨s(r, t)| H follows from

(∂∂t |s(r, t)⟩

)†=(

1i~H |s(r, t)⟩

)†.

3.5.2 Exact Measurements and Expectation ValuesWe have seen in Tables 3.1 and 3.2 that each classical observable has its associatedquantum mechanical operator. This is an exact one-to-one correspondence.

However, in terms of actual measurements, a quantum mechanical operator doesnot generate one exact and fixed value under repeated measurements as in the classi-cal measurement. Instead, each quantum mechanical operator specifies a probabilitydistribution of possible outcomes of the measurement by way of a sate function orwavefunction. This necessitates a probabilistic interpretation of classical laws ofphysics.

As we saw in Ehrenfest Theorem, the general strategy is to use expectation valuesrather than exact measured values. In the quantum mechanical formulationof classical laws of physics, the expectation values play the role of theclassical variables.

90

Exercises1. Let Q be a quantum mechanical operator representing a physical observableq. An example of Q is the momentum operator P = −i~ ∂

∂xin one dimension,

representing the classical momentum p. For simplicity, we will only considerthe cases with distinct eigenvalues.

(a) What can you say about Q? In other words, what is the name of the classof operators/matrices Q belongs to?

(b) Now consider the set |λi⟩ of all normalized eigenkets7 of Q.i. What is ⟨λi|λj⟩? This property makes λi an orthonormal set.ii. The set λi is complete. What does this mean?

2. Let |λi⟩ and |λj⟩ be two distinct eigenstates of Q, i.e. i , j. Note that |λi⟩ and|λj⟩ are normalized as they are states.

(a) If α |λi⟩ + β |λj⟩ is a state, not necessarily an eigenstate, where α and βare scalars, what is the relation between α and β?

(b) With what probabilities will λi and λj be observed if a measurement ismade?

(c) A measurement has been made for q and the value obtained was λi. Whatis the state the system is in after the measurement? You can only answerthis up to an arbitrary phase eiθ.

(d) A second measurement is made on the resulting state above. What value/values will be observed with what probability/probabilities?

7Recall we use eigenket, eigenvector, and eigenfunction interchangeably.

Chapter 4

Spaces of Infinite Dimensionality

So far, we have focused mostly on finite-dimensional spaces. This was because finitedimensions make it easy to explain mathematics and physics for the author and alsoeasy to understand mathematics and physics for the readership. However, the at-tentive reader may have wondered why finite dimensions will suffice.

The simplest example is the position x and its quantum mechanical representation(operator) X or Mx. The position observable is continuously distributed over theinterval (−∞,+∞), and hence, there are infinitely many distinct eigenvalues forthe operator X, which in turn implies that there are infinitely many eigenvectorswhich are mutually orthogonal according to Theorem 2.7. Needless to say, the di-mensionality of the Hilbert space in which the operator X operates has to be infinite.

Another example is the discrete energy distribution as indicated in 4 on p.14. Thisstates that the total energy E can only take discrete values E1, E2, E3, . . . and notany value E ∈ [0,∞) as in classical mechanics where E is continuously distributed.However, it does not place an upper limit on E, and infinitely many discrete valuesof E are possible.

4.1 Two Types of InfinityTwo kinds of infinite dimensional vector spaces are encountered in quantum me-chanics. One infinity is known as countable or denumerable infinity, and the otherinfinity is called uncountable/non-denumerable infinity. Discrete energy level dis-tribution is an example of countable infinity, and the position x is an example ofuncountable infinity. A canonical example of countable infinity is the number of

91

92 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY

natural numbers N = 1, 2, 3, . . ., and a good example of uncountable infinity isthe number of points in the interval (0, 1). In fact, countable infinity is far “smaller”than uncountable infinity. Namely, countably infinite number of points, when putnext to each other, fit in a line segment of length 0, while uncountably many pointsform a line of strictly positive length. In order to understand the difference, let ussee how we can show N is “of length 0”.

ProofLet 0 < ε < 1 be an arbitrary number. Then, we have

1 ∈(1− ε(1− ε)

2, 1 +

ε(1− ε)2

), 2 ∈

(2− ε2(1− ε)

2, 2 +

ε2(1− ε)2

),

. . . , n ∈(n− εn(1− ε)

2, n+

εn(1− ε)2

), . . . . (4.1)

This means n is contained in an interval of length εn(1−ε). Therefore, N is containedin an interval of length

∞∑n=1

εn(1− ε). (4.2)

Noting that a geometric series ∑∞n=1 ar

n for 0 < r < 1 converges to a1−r and we have

(a, r) = (ε(1− ε), ε),∞∑n=1

εn(1− ε) = ε(1− ε)1− ε

= ε. (4.3)

Therefore, N can be contained in an interval of arbitrary length ε, which implies thelength of an “interval” containing all natural numbers is 0.

You may wonder why the same argument cannot be used for uncountably manypoints, but the reason is in the name itself. If there are uncountably many points,you cannot include all of them even if you do ∑∞

n=1 εn(1− ε) because this is nothing

but counting, and you cannot count the uncountable. Here is an additional quali-tative description of what is happening. Each point is, by definition, of length 0. Ifthere are only countably many points, the total length is 0 even if there are infinitelymany points. However, if there are uncountably many points, the total length is

93

strictly positive as uncountable infinity overwhelms the length of 0 of each point.

Think of the following two situations as analogues of the above example.

1. Let f(x) = 1/x and g(x) =√x. Then, lim

x→∞f(x) = 0 and lim

x→∞g(x) =∞, but

limx→∞

f(x)g(x) = limx→∞

1/√x = 0.

2. Let f(x) = 1/x and g(x) = x2. Then, limx→∞

f(x) = 0 and limx→∞

g(x) = ∞, butlimx→∞

f(x)g(x) = limx→∞

x =∞.

In Case 1, g(x) behaves like countable infinity, while g(x) is like uncountable infinityin Case 2.

It is now clear that countable infinity and uncountable infinity should be treatedseparately.

4.2 Countably Infinite DimensionsIf the Hilbert space is of countably infinite dimensions, we have a complete orthonor-mal basis

|e1⟩ , |e2⟩ , . . . , |en⟩ , . . . or |ek⟩ ; k = 1, 2, . . . , ∞, (4.4)

satisfying

⟨ei|ej⟩ = δij (orthonormality) (4.5)

and∞∑i=1

|ei⟩ ⟨ei| = I (completeness). (4.6)

In this case, aside from the infinitely many components for vectors and infinitelymany elements matrices carry, everything is the same as for finite dimensionalHilbert spaces.


4.3 Uncountably Infinite DimensionsLet us consider the set of position eigenkets |x⟩, each labeled by the positioneigenvalue x. As x is an observable and X is a Hermitian operator, |x⟩ forms abasis. In particular, it is an orthonormal basis if the set |x⟩ satisfies

⟨x|x′⟩ = δ(x− x′) (orthonormality) (4.7)

and ∫ +∞

−∞|x⟩ ⟨x| dx = I (completeness); (4.8)

where the right-hand side of (4.7) is a real-valued function called the Dirac deltafunction with the following definition and properties. Delta function is necessarywhenever the basis kets are labeled by a continuous index including the positionvariable x.

Heuristic DescriptionQualitatively, the delta function can be described by

δ(x) =

∞ x = 00 x , 0

(4.9)

and ∫ +∞

−∞δ(x) dx = 1. (4.10)

Needless to say, this is an “extended” function, and no usual function has these prop-erties. Rather than regarding the delta function as a function as we know them, weshould think of it as a device to make quantum mechanics as well as other theoriesof physics work in a consistent fashion.

Properties of the Dirac Delta FunctionLet us look at the properties of δ(x− x′) found in (4.7).

1. As before, we have

δ(x− x′) = 0 if x , x′

95

and (4.11)∫ +∞

−∞δ(x− x′) dx′ = 1.

2. The delta function is sometimes called the sampling function as it samples thevalue of the function f(x′) at one point x; namely,∫ +∞

−∞δ(x− x′)f(x′) dx′ = f(x). (4.12)

3. The delta function is even as shown below.

δ(x− x′) = ⟨x|x′⟩ = ⟨x′|x⟩∗ = δ(x′ − x)∗ = δ(x′ − x) (4.13)

The last equality holds because the delta function is a real-valued function asstated on p.94.

Now, let f(x) be a differentiable function. Then,

d

dxf(x− x′) = lim

h→0

f(x− x′)− f((x− h)− x′)

h(4.14)

andd

dx′f(x− x′) = lim

h→0

f(x− (x′ + h))− f(x− x′)

h

= limh→0

f(x− h− x′)− f(x− x′)

h

= − limh→0

f(x− x′)− f((x− h)− x′)

h

= − d

dxf(x− x′) (4.15)

imply the following relation for the “formal” derivatives of the delta function.a

δ′(x− x′) =d

dxδ(x− x′) = − d

dx′ δ(x− x′) (4.16)

4. We have δ′(x− x′) = δ(x− x′) ddx′ , and more generally,

dnδ(x− x′)

dxn= δ(x− x′)

dn

dx′n . (4.17)


Therefore, ∫δ′(x− x′)f(x′) dx′ =

∫δ(x− x′)

d

dx′f(x′) dx′

=∫δ(x− x′)f ′(x′) dx′ = f ′(x), (4.18)

and more generally,∫δn(x− x′)f(x′) dx′ =

∫δ(x− x′)

dn

dxnf(x′) dx′

=∫δ(x− x′)fn(x′) dx′ = fn(x). (4.19)

Next, consider δ(ax) for any real number a , 0. If we let y = ax, then, dydx

= a =⇒dx = dy

a, y : −∞ −→ +∞ if a > 0, and y : +∞ −→ −∞ if a < 0. So,

∫ +∞

−∞f(x)δ(ax) dx =

(∫ +∞

−∞f(y

a

)δ(y) dy

)1

a

=1

af(0

a

)=

1

af(0) if a > 0 (4.20)

and∫ +∞

−∞f(x)δ(ax) dx =

(∫ −∞

+∞f(y

a

)δ(y) dy

)1

a=(∫ +∞

−∞f(y

a

)δ(y) dy

)1

−a

=1

−af(0

a

)=

1

−af(0) if a < 0. (4.21)

5. The following rescaling property holds for the delta function.

δ(ax) =δ(x)

|a|for any a , 0 (4.22)

97

4.4 Delta Function as a Limit of Gaussian Distri-bution

The Gaussian distribution gσ(x − x′) labeled by the standard deviation σ is givenby

gσ(x− x′) =1

(πσ2)12

exp[−(x− x′)2

σ2.

](4.23)

The Gaussian function gσ is symmetric about x = x′, has widthb σ, and its peakheight is (πσ)− 1

2 at x = x′. Furthermore, the area under the curve is unity irre-spective of the value of the standard deviation σ or variance σ2. Hence, it is atleast qualitatively obvious that gσ provides a better and better approximation ofthe delta function as σ approaches zero.

4.5 Delta Function and Fourier TransformFor a function f(x), its Fourier transform is

f(k) =1√2π

∫ +∞

−∞e−ikxf(x) dx, (4.24)

while the inverse Fourier transform is

f(x′) =1√2π

∫ +∞

−∞eikx

′f(k) dk. (4.25)

Substituting (4.24) into (4.25), we obtain the following relation.

f(x′) =∫ +∞

−∞

(1

2π

∫ +∞

−∞eik(x

′−x) dk)f(x) dx (4.26)

Now, compare (4.26) with (4.12). Then, we can see

1

2π

∫ +∞

−∞eik(x

′−x) dk = δ(x′ − x) (4.27)


or1

2π

∫ +∞

−∞eik(x−x′) dk = δ(x− x′). (4.28)

This is another characterization of the delta function.

4.6 f as a Vector with Uncountably Many Com-ponents

In order to understand a function f and its value at x, normally denoted by f(x),we will consider vectors in a finite dimensional space, a space of countably infinitedimension, and a space spanned by uncountably many basis vectors, in this order.Without loss of generality, we will consider orthonormal bases.

4.6.1 Finite DimensionsConsider a vector |v⟩ in Vn (an n-dimensional vector space). If a particular or-thonormal basis B = |e1⟩ , |e2⟩ , . . . , |en⟩ is chosen, we can express v as a columnvector with respect to B.

|v⟩ =

31

32...3n

B

(4.29)

We have

|v⟩ =n∑i=1

3i |ei⟩ (4.30)

and3i = ⟨ei|v⟩ . (4.31)

4.6.2 Countably Infinite DimensionsIf the vector space has a countably infinite dimension, and an orthonormal basisB = |e1⟩ , |e2⟩ , . . . , |en⟩ , . . ., we can express |v⟩ as an infinitely long column

99

vector with respect to B.

|v⟩ =

31

32...3n......

B

(4.32)

We have

|v⟩ =∞∑i=1

3i |ei⟩ (4.33)

and3i = ⟨ei|v⟩ . (4.34)

4.6.3 Uncountably Infinite DimensionsWhen the vector space has an uncountably infinite dimension and an orthonormalbasis B = |ex⟩x∈R or x∈(−∞,+∞); where the kets in the basis B are indexed bya continuous variable x, each vector v has uncountably many components 3xx∈Rwith a continuous index x ∈ R , as opposed to a discrete index i ∈ N. We can stillregard |v⟩ as a column vector.

|v⟩ =

3−∞↑...3x...↓3+∞

B

(4.35)

By abuse of notation, we can write

|v⟩ =∑x∈R3x |ex⟩ (4.36)

and, as before,


3x = ⟨ex|v⟩ . (4.37)

Let us switch to a more familiar notation. Instead of |ex⟩ and |v⟩, we write |x⟩ and|f⟩. Then, we have

|f⟩ =

f−∞↑...fx...↓

f+∞

B

, (4.38)

|f⟩ =∑x∈R

fx |x⟩ , (4.39)

andfx = ⟨x|f⟩ . (4.40)

Now let us write f(x) instead of fx to get

|f⟩ =∑x∈R

f(x) |x⟩ , (4.41)

andf(x) = ⟨x|f⟩ . (4.42)

Notationally more appropriate expression for |f⟩ than (4.41) is obtained using thecompleteness relation (4.8).

|f⟩ = I |f⟩ =∫ +∞

−∞|x⟩ ⟨x|f⟩ dx =

∫ +∞

−∞f(x) |x⟩ dx (4.43)

In this view, a function |f⟩ is a vector which is a linear combination of orthonormalbasis vectors |x⟩x∈R, and its coefficients, components of the column vector with

101

respect to the basis |x⟩x∈R, are our familiar f(x). We can rewrite (4.38) as follows.

|f⟩ =

f(−∞)↑...

f(x)...↓

f(+∞)

|x⟩x∈R

(4.44)

Finally, you can see that the inner product is the one defined for L2-functions inSection 2.3. All we need to do is to use the completeness property (4.8).

⟨f |g⟩ = ⟨f |I|g⟩ =⟨f∣∣∣∣∫ +∞

−∞|x⟩ ⟨x| dx

∣∣∣∣ g⟩ =∫ +∞

−∞⟨f |x⟩ ⟨x|g⟩ dx

=∫ +∞

−∞⟨x|f⟩∗ ⟨x|g⟩ dx =

∫ +∞

−∞f(x)∗g(x) dx (4.45)

4.7 Hermiticity of the Momentum OperatorRecall that the momentum operator P in one dimension is given by

P = −i~ ∂∂x

= −i~ ddx. (4.46)

This operator acts on differentiable L2 functions f(x) and has to be Hermitian asthe momentum p is a physical observable. Let us verify this. We have the followingin our notation.

P |f⟩ =∣∣∣∣∣−i~ dfdx

⟩(4.47)

⟨x|P |f⟩ =⟨x

∣∣∣∣∣− i~ ddx∣∣∣∣∣f⟩=

⟨x

∣∣∣∣∣ −i~ dfdx⟩= −i~ df

dx(x) = −i~df(x)

dx(4.48)

4.7.1 Checking [Pij] = [P ∗ji]

In this section, we will check if P is represented by a matrix which is equal to itstranspose conjugate. Because our basis consists of uncountably many orthonormal


vectors |x⟩x∈R, it is actually more appropriate to write [Pxx′ ] = [P ∗x′x]; where

Pxx′ = ⟨x|P |x′⟩. Using the completeness property (4.8), we can rewrite (4.48).

⟨x|P |f⟩ = ⟨x|PI|f⟩ =⟨x

∣∣∣∣∣P∫ b

a|x′⟩ ⟨x′| dx′

∣∣∣∣∣ f⟩

=∫ b

a⟨x|P |x′⟩ ⟨x′|f⟩ dx′ =

∫ b

a⟨x|P |x′⟩ f(x′) dx′ = −i~df(x)

dx(4.49)

Note that we used a and b as our limits of integration because the domain of thewavefunction is not always (−∞,+∞).

Now compare (4.18) and (4.49) to obtain

⟨x|P |x′⟩ = −i~δ′(x− x′) = −i~δ(x− x′)d

dx′ . (4.50)

As shown in the footnote on p.95, the first derivative of the real-valued δ-functionδ′(y) is an odd function. Hence,

P ∗x′x = ⟨x′|P |x⟩∗ = (−i~δ′(x′ − x))∗

= (−i~)∗(δ′(x′ − x))∗

= i~δ′(x′ − x) = i~(−δ′(x− x′)) = −i~δ′(x− x′) = Pxx′ . (4.51)

Therefore, the matrix representation of P = −i~ ddx

, denoted by [Pxx′ ], is equal to itstranspose conjugate. This was a necessary and sufficient condition for an operatorto be Hermitian if we have a finite dimensional Hilbert space (See Definition 2.26).However, this condition alone is not sufficient when the dimension is uncountablyinfinite.

4.7.2 Hermiticity Condition in Uncountably Infinite Dimen-sions

We will go back to the definition of Hermiticity. Definition 2.26 states that anoperator Ω is Hermitian if and only if

Ω = Ω†, (4.52)

which is equivalent to

⟨Ωv |w⟩ = ⟨v |Ωw⟩ (4.53)

103

for all v and w. As we are interested in P = −i~ ddx

, we would like to see if⟨−i~ df

dx

∣∣∣∣∣ g⟩=

⟨f

∣∣∣∣∣− i~dgdx⟩⇐⇒ i~

⟨df

dx

∣∣∣∣∣ g⟩= −i~

⟨f

∣∣∣∣∣ dgdx⟩

⇐⇒⟨df

dx

∣∣∣∣∣ g⟩= −

⟨f

∣∣∣∣∣ dgdx⟩. (4.54)

Let us first work on the left-hand side. Completeness condition and integration byparts give us the following.⟨

df

dx

∣∣∣∣∣ g⟩=

⟨df

dx

∣∣∣∣∣ I∣∣∣∣∣ g⟩=

⟨df

dx

∣∣∣∣∣∫ b

a|x⟩ ⟨x| dx

∣∣∣∣∣ g⟩

=∫ b

a

⟨df

dx

∣∣∣∣∣x⟩⟨x| g⟩ dx =

∫ b

a

(df(x)

dx

)∗

g(x) dx =∫ b

a

df ∗(x)

dxg(x) dx

= f ∗(x)g(x)∣∣∣+∞

−∞−∫ b

af ∗(x)g′(x) dx. (4.55)

On the other hand, the right-hand side is

−⟨f

∣∣∣∣∣ dgdx⟩= −

⟨f

∣∣∣∣∣∫ b

a|x⟩ ⟨x| dx

∣∣∣∣∣ dgdx⟩= −

∫ b

a⟨f |x⟩

⟨x

∣∣∣∣∣ dgdx⟩dx

= −∫ b

af ∗(x)g′(x) dx. (4.56)

The expressions (4.55) and (4.56) are equal if and only if

f ∗(x)g(x)∣∣∣ba= 0. (4.57)

Needless to say, this is not satisfied by all functions f and g. However, there area couple of notable cases where (4.57) holds.

Single-Valuedness : When the spherical coordinates (r, θ, ϕ) are used as inChapter 10, (r, θ, 0) and (r, θ, 2π) are the same point in the three-dimensionalspace. Therefore, we require Ψ(r, θ, 0, t) = Ψ(r, θ, 2π, t) due to the single-valuedness condition, Condition 2, on p.128. In this case, a = 0 and b = 2π,and we indeed have

f ∗(x)g(x)∣∣∣ba= f ∗(ϕ)g(ϕ)

∣∣∣2π0

= Ψ∗(r, θ, ϕ, t)Ψ(r, θ, ϕ, t)∣∣∣2π0

= 0 (4.58)


with respect to the ϕ-integral.c One class of functions that satisfy (4.57)consists of periodic functions. For example, if a = 0 and b = 2π, trigonometricfunctions such as sin x and cos x and their complex counterparts includingeix satisfy this condition.

Functions Vanishing at Infinity : This is when a = −∞ and b = +∞,and we need

f ∗(x)g(x)∣∣∣+∞

−∞= 0. (4.59)

In particular, this equality holds for L2-functions, described on p.30 of Section2.3, as f ∈ L2(−∞,+∞) implies lim

|x|→∞f(x) = 0.

4.7.3 The Eigenvalue Problem of the Momentum OperatorP

The eigenvalue problem for P in an arbitrary basis is

P |λ⟩ = λ |λ⟩ . (4.60)

Taking the projection on ⟨x|,

⟨x|P |λ⟩ = λ ⟨x|λ⟩ . (4.61)

Using completeness,

⟨x|P |λ⟩ = ⟨x|PI|λ⟩ = ⟨x|P(∫|x′⟩ ⟨x′| dx′

)|λ⟩ =

∫⟨x|P |x′⟩ ⟨x′|λ⟩ dx′

= λ ⟨x|λ⟩ . (4.62)

Now, from (4.50), we have

⟨x|P |x′⟩ = −i~δ(x− x′)d

dx′ . (4.63)

Hence,

⟨x|P |λ⟩ =∫⟨x|P |x′⟩ ⟨x′|λ⟩ dx′ =

∫−i~δ(x− x′)

d

dx′ ⟨x′|λ⟩ dx′

105

= −i~ ddx⟨x|λ⟩ (4.64)

If we denote ⟨x|λ⟩ by fλ(x), we have

−i~ ddxfλ(x) = λfλ(x). (4.65)

Had we started with the X-basis made up of the eigenvectors of the position oper-ator X, we would have obtained (4.65) immediately as P = −i~ d

dxin this basis.

Solving the ordinary differential equation (4.65), we get

fλ(x) = Ceiλx/~; (4.66)

where C is an arbitrary constant. It is now clear that P takes any real number λ asits eigenvalue with the corresponding eigenfunction Ceiλx in the X-basis. In orderto normalize the eigenfunctions, we let C = 1√

2π~. Then,

⟨x|λ⟩ = fλ(x) =1√2π~

eiλx/~, (4.67)

and, due to (4.28) and (4.22),

⟨λ|λ′⟩ =∫ +∞

−∞⟨λ|x⟩ ⟨x|λ′⟩ dx =

1√2π~

1√2π~

∫ +∞

−∞e−iλx/~eiλ

′x/~ dx

=1

2π~

∫ +∞

−∞ei

1~(λ′−λ)x dx =

1

~

(1

2π

∫ +∞

−∞ei

1~(λ′−λ)x dx

)=

1

~δ(1

~(λ− λ′)

)=

1

~

δ(λ− λ′)

|1/~|= δ(λ− λ′). (4.68)

So, |λ⟩λ∈R forms an orthonormal basis.

4.8 Relations Between X and P

4.8.1 The Fourier Transform Connecting X and P

Given a physical system S , there is an associated Hilbert space H according toPostulate 1 on p.72. This Hilbert space is spanned either by the eigenkets of P(the P basis) or the eigenkets of X (the X basis) as both X and P are Hermitian


operators. Given a ket |f⟩, it can be expanded either in the X basis |x⟩ or in theP basis |p⟩.

Note here that we switched the notation for the P basis from |λ⟩ to |p⟩ forclarity.

Recall that we have

|f⟩ =∑x∈R⟨x|f⟩ |x⟩ and |f⟩ =

∑p∈R⟨p|f⟩ |p⟩ . (4.69)

So, f(x) = ⟨x|f⟩ is the “x-th” coefficient/component of |f⟩, and f(p) = ⟨p|f⟩ is the“p-th” coefficient/component of |f⟩. They are coefficients in view of the expansion(4.69), but they are also components of the column vector |f⟩. Let us compare f(p)and f(x) using (4.67).

f(p) = ⟨p|f⟩ =∫ +∞

−∞⟨p|x⟩ ⟨x|f⟩ dx =

∫ +∞

−∞⟨x|p⟩∗ ⟨x|f⟩ dx

=⇒ f(p) =1√2π~

∫ +∞

−∞e−ipx/~f(x) dx (4.70)

f(x) = ⟨x|f⟩ =∫ +∞

−∞⟨x|p⟩ ⟨p|f⟩ dp

=⇒ f(x) =1√2π~

∫ +∞

−∞eipx/~f(p) dp (4.71)

These are nothing but the Fourier transform and the Fourier inverse transform.Therefore, the expressions of a function |f⟩ in this Hilbert space in terms of thecomplete X basis and complete P basis are related through the familiar Fouriertransform.

You may note that the more familiar form of the Fourier transform can beobtained if we set ~ = 1.d Indeed, if we consider the operator K = P/~ andits orthonormal eigenbasis |k⟩k∈R, which satisfies ⟨x|k⟩ = 1√

2πeikx as opposed to

⟨x|p⟩ = 1√2π~eipx/~, f(k) and f(x) are related as follows.

f(k) = ⟨k|f⟩ =∫ +∞

−∞⟨k|x⟩ ⟨x|f⟩ dx =

∫ +∞

−∞⟨x|k⟩∗ ⟨x|f⟩ dx

=1√2π

∫ +∞

−∞e−ikxf(x) dx (4.72)

107

f(x) = ⟨x|f⟩ =∫ +∞

−∞⟨x|k⟩ ⟨k|f⟩ dk =

1√2π

∫ +∞

−∞eikxf(k) dk (4.73)

As advertised, these are the Fourier transform and its inverse in their more familiarforms.

4.8.2 X and P in the X BasisWe already know that the position operator X in the X basis is x or Mx, which is ascalar multiplication. We also know that the momentum operator P in the X basisis −i~ d

dx. We can check the action of X in the framework of this chapter as follows.

X |x⟩ = x |x⟩ =⇒ ⟨x′|X|x⟩ = ⟨x′|x|x⟩ = x ⟨x′|x⟩ = xδ(x′ − x) (4.74)

=⇒ ⟨x|X|f⟩ = ⟨x|XI|f⟩ = ⟨x|X(∫|x′⟩ ⟨x′| dx′

)|f⟩ =

∫⟨x|X|x′⟩ ⟨x′|f⟩ dx′

=∫x′δ(x− x′)f(x′) dx′ = xf(x) (4.75)

In order to express this action in a basis-independent manner, we can adopt Shankar’snotation [Shankar, 1980, p.74].

4.8.3 Basis-Independent Descriptions in Reference to the XBasis

Recall that |f⟩ is a basis-independent expression for a vector (function) in our Hilbertspace. One way to specify f is by giving ⟨x|f⟩ = f(x) for the orthonormal basis|x⟩x∈R. Differently put, we cannot really specify the nature of f unless we go tosome basis such as the position basis xx∈R. With this specification, it is now possi-ble to express f as a basis-independent ket in reference to the position orthonormalbasis |x⟩x∈R. This potentially confusing strategy reduces to writing the basis-independent ket |f⟩ as |f(x)⟩. What does this mean? It means the following.

1. |f⟩ itself is basis-independent.

2. However, we know |f⟩ is fully characterized by the coefficients f(x)x∈R.

3. So, we write |f(x)⟩ for an abstract basis-independent vector whose coefficientswith respect to |x⟩x∈R are f(x)x∈R.


For example, from (4.75), the coefficients of X |f⟩ with respect to |x⟩x∈R, namely⟨x|X|f⟩x∈R, are xf(x)x∈R. Therefore, the action of X can be expressed ina basis-independent manner, but in reference to the orthonormal position basis|x⟩x∈R, as below.

X |f(x)⟩ = |xf(x)⟩ (4.76)

Similarly, a basis-independent description of the momentum operator P in referenceto |x⟩x∈R is

P |f(x)⟩ =∣∣∣∣∣−i~df(x)dx

⟩. (4.77)

4.8.4 X and P in the P BasisFrom (4.60), we have

P |p⟩ = p |p⟩ (4.78)

in the P basis. Hence, the matrix element Ppp′ = ⟨p|P |p′⟩ is given by

⟨p|P |p′⟩ = ⟨p|p′|p′⟩ = p′ ⟨p|p′⟩ = p′δ(p− p′). (4.79)

It remains to show how X operates in this basis. The matrix element Xpp′ = ⟨p|X|p′⟩can be obtained as follows.

First, note that

i~d

dp

(e−ipx/~

)= i~(−ix/~)e−ipx/~ = xe−ipx/~. (4.80)

Using (4.67) and (4.80), we get

⟨p|X|p′⟩ = ⟨p|XI|p′⟩ = ⟨p|X(∫|x⟩ ⟨x| dx

)|p′⟩ =

∫⟨p|X|x⟩ ⟨x|p′⟩ dx

=∫⟨p|x|x⟩ ⟨x|p′⟩ dx =

∫x ⟨p|x⟩ ⟨x|p′⟩ dx =

∫x ⟨x|p⟩∗ ⟨x|p′⟩ dx

=∫x

(1√2π~

eipx/~)∗ (

1√2π~

eip′x/~

)dx =

1

2π~

∫xe−ipx/~eip

′x/~ dx

109

=1

2π~

∫i~d

dp

(e−ipx/~eip

′x/~)dx = i

d

dp

(1

2π

∫ei(p

′−p)x/~ dx)

= i~d

dpδ(p− p′) = i~δ′(p− p′). (4.81)

The ~ in front of the delta function arises in the following manner. If we let y = x/~,dx = ~dy, and y : −∞ −→ +∞ as x : −∞ −→ +∞. So,

1

2π

∫ +∞

−∞ei(p

′−p)x/~ dx =1

2π

∫ +∞

−∞ei(p

′−p)y ~dy = ~(

1

2π

∫ +∞

−∞ei(p

′−p)y dy)

= ~δ(p′ − p) = ~δ(p− p′). (4.82)

You may realize that this can be modified to generate an alternative proof of Prop-erty 5 or (4.22) on p.96.

Hence, the action of X in the K basis is described by

Xf(k) = i~df(k)

dk. (4.83)

As before, the basis independent form of this with our notational convention is

X |f(k)⟩ =∣∣∣∣∣i~ ddkf(k)

⟩. (4.84)

Let us summarize what we have shown in Sections 4.8.2 and 4.8.4.

• In the X basis:

X ←→ x or Mx (4.85)

P ←→ −i~ ddx

(4.86)

• In the P basis:

X ←→ i~d

dp(4.87)

P ←→ p or Mp (4.88)


4.8.5 The Commutator [X,P ]

Let [X,P ]X and [X,P ]P be commutators of X and P in the X and P bases, respec-tively. Then,

[X,P ]Xf(x) =

[Mx

(−i~ d

dx

)−(−i~ d

dxMx

)]f(x)

= −i~xdf(x)dx

+ i~d

dx(xf(x)) = −i~xdf(x)

dx+ i~

(f(x) + x

df(x)

dx

)= i~f(x) =⇒ [X,P ]X = i~I (4.89)

and

[X,P ]Pf(p) =

[(i~d

dp

)Mp −Mp

(i~d

dp

)]f(p)

= i~d

dp(pf(p))− i~pdf(p)

dp= i~

(f(p) + p

df(p)

dp

)− i~pdf(p)

dp

= i~f(p) =⇒ [X,P ]p = i~I. (4.90)

Because the actual expression for the identity operator I is basis-independent, wedo not write IX or IP . We conclude that

[X,P ] = i~I (4.91)

in either basis.aAs the delta function is even, its “derivative” is an odd function. Generally speaking, let f(x)

be a differentiable even function and y = −x, then, for an arbitrary value x0,

d

dxf(x)

∣∣∣∣x=x0

=d

dxf(−x)

∣∣∣∣x0

= − d

d(−x)f(−x)

∣∣∣∣x0

= − d

dyf(y)

∣∣∣∣y=−x0

= − d

dxf(x)

∣∣∣∣x=−x0

.

So, we have f ′(x0) = −f ′(−x0) for any x0.bThe “width” is defined as the distance between the two points of inflection of gσ.cWe will see in Chapter 10 that we can separate the four variables (r, θ, ϕ, t) and write Ψ as a

product of functions of r, θ, ϕ, and t; namely Ψ(r, θ, ϕ, t) = R(r)Θ(θ)Φ(ϕ)T (t).dThis is the reason why we left 2π~ as it was even though 2π~ = 2π h

2π = h.

Chapter 5

Schrodinger’s Theory of QuantumMechanics

In 1925 an Austrian physicist Erwin Rudolf Josef Alexander Schrodinger proposedthe Schrodinger Equation which is a differential equation whose solutions are wave-functions. He chose a differential equation to describe new physics since a differen-tial equation was the most common type of equation known to the physicists witha function for a solution. In Chapter 3, the postulates of quantum mechanics werepresented. However, these postulates were put together in a post hoc fashion. In thischapter, we will learn how quantum mechanical framework was initially constructed,drawing on the analogy to classical mechanics and incorporating the emerging ideasof wave-particle duality.

In the language of Chapter 3, and in retrospect, Schrodinger’s work was an attemptto associate a Hilbert space of L2-functions, now known as wavefunctions, to a givenphysical system. The functions are called wavefunctions as they are supposed toreflect the wave nature of particles including superposition and interference.

5.1 How was it derived?In his derivation, he was guided by the following.

1. The classical traveling wave: a simple sinusoidal traveling wave

Ψ(x, t) = sin 2π(x

λ− νt

)1 (5.1)

111

112 CHAPTER 5. SCHRODINGER’S THEORY OF QUANTUM MECHANICS

2. The de Broglie-Einstein postulates

λ =h

p, ν =

E

h(5.2)

3. The Newtonian energy equation

E =P 2

2m+ V (5.3)

4. The equation should be linear in terms of its solutions.If

Ψ1(x, t), Ψ2(x, t) (5.4)

are two solutions of the Schrodinger Equation, then any linear combination ofthe two

Ψ(x, t) = c1Ψ1(x, t) + c2Ψ2(x, t) (5.5)

is also a solution; where c1 and c2 are arbitrary constants.=⇒ This makes superposition and interference possible.

Here is what we want.

• A differential equation, later named the Schrodinger Equation, which looks likethe energy equation.

• Wave functions, which are the solutions of the Schrodinger Equation and be-have like classical waves, including interference and superposition.

Let us first consider the energy equation and observe the following.

1You may be more familiar with Ψ(x, t) = sin(kx− ωt). The equivalence of sin 2π(

xλ − νt

)and

sin(kx−ωt) is a direct consequence of the definition of the wavenumber k and the relation betweenthe frequency ν and the angular frequency ω; namely, k = 2π

λ and ω = 2πν. This is as shown onp.113.

5.1. HOW WAS IT DERIVED? 113

Let k = 2πλ

2 and ω = 2πν. Then, we have

Ψ(x, t) = sin(kx− ωt). (5.6)

On the other hand, the energy equation is

E =P 2

2m+ V. (5.7)

Plug P = hλ

and E = hν into (5.7) to obtain

h2

2mλ2+ V (x, t) = hν. (5.8)

Recall that k = 2πλ

and ω = 2πν. So, we have the following relations.(1

λ

)2

=

(k

2π

)2

ν =ω

2π(5.9)

We now substitute these back into the energy equation (5.7).

E =P 2

2m+ V =⇒ h2

2mλ2+ V (x, t) = hν =⇒ h2k2

(2π)2· 1

2m+ V (x, t) = h

(ω

2π

)

=⇒(h

2π

)2

k2 · 1

2m+ V (x, t) =

(h

2π

)ω (5.10)

Now we introduce a new notation.

~ =h

2π(5.11)

When you read it aloud, it is pronouced either h bar or h cross, with h bar seemingmore dominant these days. With this notation, (5.10) becomes

~2k2

2m+ V (x, t) = ~ω. (5.12)

We will next look at the classical sinusoidal traveling wave given by

Ψ(x, t) = sin(kx− ωt). (5.13)2This k is called the wavenumber or angular/circular wavenumber to make a clear distinction

from another wavenumber 1/λ, which is also called the spectroscopic wavenumber. By definition,k is the number of wavelengths per 2π units of distance.


Here are three of its partial derivatives which are of interest to us.

∂Ψ

∂x= k cos(kx− ωt) (5.14)

∂2Ψ

∂x2= −k2 sin(kx− ωt) (5.15)

∂Ψ

∂t= −ω cos(kx− ωt) (5.16)

From these, we arrive at the implications below.

=⇒

∂2

∂x2gets us k2.

∂∂t

gets us ω.

(5.17)

Note that this argument is all qualitative and of strory-telling type. So is the rest ofthe argument.

By abuse of notation, let Ψ(x, t) be the desired wave function which acts like thetraveling wave.

As the energy equation, whose solution is Ψ(x, t), contains k2 and ω, we consider thecorrespondences below.

~2k2

2m⇐⇒ ∂2

∂x2Ψ(x, t) (5.18)

~ω ⇐⇒ ∂

∂tΨ(x, t) (5.19)

We have used guiding assumptions (hypotheses) 1 through 3 to arrive at

α∂2

∂x2Ψ(x, t) + V (x, t) = β

∂Ψ(x, t)

∂t. (5.20)

Why α and β? A very short answer is “because it works”. But, a slightly longer an-swer would be that our derivation is largely qualitative, and we need to incorporatesome quantitative degrees of freedom in our trial formula. At any rate, whateverworks!

How about the fourth assumption; the linearity assumption?

5.1. HOW WAS IT DERIVED? 115

Let Ψ1 and Ψ2 be two solutions of the above equation. Then we have

αΨ1,xx + V = βΨ1,t and αΨ2,xx + V = βΨ2,t

=⇒ α(Ψ1,xx +Ψ2,xx) + 2V = β(Ψ1,t +Ψ2,t). (5.21)

We are unfortunately off by the factor 2 in front of the potential V .

And one way around this problem is to modify the equation in the following way.

α∂2

∂x2Ψ(x, t) + V (x, t)Ψ(x, t) = β

∂Ψ(x, t)

∂t(5.22)

With a little more work, we can determine α and β. The easiest way is to considerV = 0 which gives us a free particle or a free traveling wave

Ψ(x, t) = ei(kx−ωt). (5.23)

Substituting V = 0 and (5.23) into (5.22),

α∂2

∂x2Ψ(x, t) + V (x, t)Ψ(x, t) = β

∂Ψ(x, t)

∂t

=⇒ α∂2

∂x2ei(kx−ωt) = β

∂

∂tei(kx−ωt) =⇒ −αk2ei(kx−ωt) = −iωβei(kx−ωt)

=⇒ −αk2 = −iωβ (5.24)

Comparing (5.24) with (5.12) when V (x, t) = 0, we get

−αk2 = −iωβ ←→ ~2k2

2m= ~ω. (5.25)

Therefore,

α =−~2

2mand β =

~

−i= i~. (5.26)

And our end product is

− ~2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂t. (5.27)

This is known as the Schrodinger Equation or the Time-Dependent SchrodingerEquation. We often write it as below.[

− ~2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t(5.28)


The solution set Ψ(x, t) represents the wave functions which are to be associatedwith the motion of a particle of mass m under the influnece of the force F describedby V (x, t); namely

F (x, t) = −∂V (x, t)

∂x. (5.29)

Agreement with experiment? How do you check it? We first need a physical inter-pretation of Ψ to do so.

5.2 Born’s Interpretation of WavefunctionsWe now have

− ~2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂tor[

− ~2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t.

(5.30)

So, if V (x, t) is given, a wavefunction Ψ(x, t) can be obtained at least in principle.But, what is Ψ?

Since Ψ is a complex function, Ψ itself can not be a real physical wave. How can weextract physical reality from this?

Max Born realized that Ψ∗Ψ = |Ψ|2 is real while Ψ itself is not and postulated thatP (x, t) = Ψ(x, t)∗Ψ(x, t) can be regarded as a probability. He postulated that theprobability P (x, t)dx of finding the particle at a coordinate between x and x+ dx isequal to Ψ(x, t)∗Ψ(x, t)dx.

However, there is a problem with this view as it is. It is too simplistic. To understandthe nature of the problem simply note that 2Ψ(x, t) is also a solution if Ψ(x, t) is asolution of

− ~2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂t. (5.31)

This means that the probability is not unique, which of course is not acceptable.Furthermore, the total probability is not necessarily one if we used an arbitrary

5.3. EXPECTATION VALUES 117

solution. Our necessary condition is∫ +∞

−∞P (x, t)dx = 1. (5.32)

In other words, we need to find Ψ such that∫ +∞

−∞P (x, t)dx =

∫ +∞

−∞Ψ∗(x, t)Ψ(x, t)dx = 1. (5.33)

This is the physically meaningful Ψ called a normalized wavefunction. This processis called normalization. You should recognize that this normalized wavefunction is“the state” first mentioned in Postulate 1 on p.72 of Chapter 3.

5.3 Expectation ValuesPostulate 7 on p.74, Chapter 3, gives

q or ⟨q⟩ = ⟨vs |Q|vs⟩ . (5.34)

For a Hilbert space of L2-functions, we have

⟨q⟩ = ⟨vs |Q|vs⟩ =∫

vs(x)∗Qvs(x) dx; (5.35)

where the state vector vs(x) is an L2-function of unit norm, i.e.∫vs(x)

∗vs(x) dx = 1. (5.36)

In this section, we will start from scratch and see how (5.35) is “derived”, so to speak,for the momentum P and the total energy E.

Recall that for a die∑

value× probability = 1× 1

6+ 2× 1

6+ 3× 1

6+ 4× 1

6+ 5× 1

6+ 6× 1

6

= (1 + 2 + 3 + 4 + 5 + 6) · 16

(5.37)

gives the expectation value.


Question : How do we calculate the “average” position of a particle?

Answer : By computing the expectation value of the position x denoted by ⟨x⟩ orx.

Analogously to the die above, with x as the value, we have

⟨x⟩ =∑

x× probability. (5.38)

The continuous version of the above is

⟨x⟩ = x =∫ +∞

−∞xΨ∗(x, t)Ψ(x, t)dx =

∫ +∞

−∞Ψ∗(x, t)xΨ(x, t)dx. (5.39)

How about P and E? We are tempted to write

P =∫ +∞

−∞Ψ∗(x, t)PΨ(x, t)dx (5.40)

and

E =∫ +∞

−∞Ψ∗(x, t)EΨ(x, t)dx. (5.41)

It turns out this is what we should indeed do.

Recall we are always to be guided by the classical wave. Instead of a sine wave, letus look at a particular solution of the Schrodinger Equation called a free particlesolution. The free-ness of a free partilce derives from the lack of any external force.In other words, V (x, t) = 0 for a free particle.[

− ~2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t

=⇒ − ~2

2m

∂2

∂x2Ψ(x, t) = i~

∂Ψ(x, t)

∂t(5.42)

We will solve this in detail in the next chapter, but the answer is the classical trav-eling wave.

Previously, we used

Ψ(x, t) = sin(kx− ωt). (5.43)


Now we will use its complex version.

Ψ(x, t) = cos(kx− ωt) + i sin(kx− ωt) = ei(kx−ωt) (5.44)

Taking the first partial derivative with respect to x,

∂Ψ

∂x= ikΨ(x, t) = i

P

~Ψ(x, t); (5.45)

where the last equality follows from

k =2π

λ=

2π

h/p=

P

h/2π=P

~. (5.46)

Therefore,

P [Ψ(x, t)] = −i~ ∂∂x

[Ψ(x, t)] . (5.47)

Similarly,

∂Ψ(x, t)

∂t= −iωΨ(x, t) = −iE

~Ψ(x, t); (5.48)

where the last equality follows from

ω = 2πν = 2πE

h=

E

h/2π=E

~. (5.49)

Therefore,

E [Ψ(x, t)] = i~∂

∂t[Ψ(x, t)] . (5.50)

On the other hand, let us compare

P 2

2m+ V (x, t) = E (5.51)

with [− ~

2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t(5.52)

term by term.


The first terms give us

−~2

2m

∂2

∂x2=P 2

2m. (5.53)

Working formally,

√−~2

√∂2

∂x2=√P 2 =⇒ ±i~ ∂

∂x= P (5.54)

We will choose the minus sign because of our previous work.How about E? Quite straightforwardly, we obtain

E = i~∂

∂t. (5.55)

Hence, it is consistent, so to speak, with the Schrodinger Equation.

We have P [Ψ(x, t)] = −i~ ∂

∂x[Ψ(x, t)]

E [Ψ(x, t)] = i~ ∂∂t[Ψ(x, t)]

. (5.56)

So,

⟨P ⟩ =∫ +∞

−∞Ψ∗(x, t)PΨ(x, t)dx =

∫ +∞

−∞Ψ∗(x, t)

(−i~ ∂

∂x

)Ψ(x, t)dx

= −i~∫ +∞

−∞Ψ∗ ∂

∂xΨdx (5.57)

and

⟨E⟩ =∫ +∞

−∞Ψ∗(x, t)EΨ(x, t)dx = i~

∫ +∞

−∞Ψ∗ ∂

∂tΨdx. (5.58)

Incidentally, this is the same as

⟨E⟩ =∫ +∞

−∞Ψ∗(x, t)

(− ~

2

2m

∂2

∂x2+ V (x, t)

)Ψ(x, t)dx. (5.59)


An Infinite Square Well Potential

Consider a particle with total energy E in the potential well given below.

V (x, t) =

0 −a

2< x < a

2

(a > 0)∞ |x| ≥ a

2

(5.60)

One normalized solution to the Schrodinger Equation is

Ψ(x, t) =

A cos πx

ae−iEt/~ −a

2< x < a

2

(a > 0)0 |x| ≥ a

2

; (5.61)

where A is known as a normalization constant.Let us conduct a consistency check with this function. Since

∂Ψ

∂t=−iE~

e−iEt/~A cos πxa

=−iE~

Ψ(x, t), (5.62)

we get

⟨E⟩ = i~∫ +∞

−∞Ψ∗(x, t)

∂Ψ(x, t)

∂tdx = i~

∫ +∞

−∞Ψ∗(x, t)

−iE~

Ψ(x, t)

= E∫ +∞

−∞Ψ∗Ψdx. (5.63)

However, since we have a normalized wavefunction,∫ +∞

−∞Ψ∗Ψdx = 1 (5.64)

and

⟨E⟩ = E. (5.65)

I said A was a normalization constant, but I never told you what its value was. Let’sfind out.∫ +∞

−∞Ψ∗Ψdx =

∫ +a2

− a2

(A cos πx

ae−iEt/~

)∗A cos πx

ae−iEt/~dx

=∫ +a

2

− a2

A∗ cos πxae+iEt/~A cos πx

ae−iEt/~dx =

∫ +a2

− a2

A∗A cos2 πxae+iEt/~e−iEt/~dx


= |A|2∫ +a

2

− a2

cos2 πxadx = 2|A|2

∫ +a2

0cos2 πx

adx = 1 (5.66)

If we let y = πxa

, we get dx = aπdy and y : 0 → π

2as x : 0 → a

2. So, integration by

this subsitution gives us∫ +a

2

0cos2 πx

adx =

∫ +π2

0

(a

π

)cos2 ydy =

(a

π

) [y

2+

sin 2y

4

]π/20

=a

π· π4=a

4. (5.67)

Therefore,

2|A|2a4= |A|2a

2= 1 =⇒ |A| =

√2

a. (5.68)

Note here that the normalization constant A can not be detrmined uniquely. Indeed,the above relation indicates that there are infinitely many values of A that servesthe purpose.

|A| =√2

a=⇒ A =

√2

aeiθ; (5.69)

where θ is any real number. This fact that the normalization constant A can onlybe specified up to the argument is another reason why the wavefunction itself doesnot represent physical reality.

123

Exercises1. Consider a classical sinusoidal traveling wave given by the equation below;

where λ is the wavelength and ν is the frequency so that νλ = 3 (the speed ofits propagation).

Ψ(x, t) = A sin[2π

λ(x− 3t)

]Show that

Ψ(x, t) = Ψ(x+ 3t0, t+ t0),

and briefly explain what this means.3

2. Write down, but do not derive, the time-dependent Schrodinger equation forV (x, t). This is a one-dimensional case, and the only variables are the spatialvariable x and the temporal variable t.

3. If Ψ1(x, t) and Ψ2(x, t) are both solutions of the Schrodinger equation, showthat any linear combination αΨ1(x, t) + βΨ2(x, t) is also a solution; where αand β are scalars.

4. Define eiθ by eiθ = cos θ + i sin θ and prove the following relationships.

(a) eiθeiϕ = ei(θ+ϕ)

(b)(eiθ)n

= einθ

5. This problem is the same as Chapter 7 Problem 1.Consider a particle of mass m and total energy E which can move freely alongthe x-axis in the interval [−a

2,+a

2], but is strictly prohibited from going outside

this region. This corresponds to what is called an infinite square well potentialV (x) given by V (x) = 0 for −a

2< x < +a

2and V (x) = ∞ elsewhere. If we

solve the Schrodinger equation for this V (x), one of the solutions is

Ψ(x, t) =

A cos

(πxa

)e−iEt/~ −a

2< x < a

2

(a > 0)0 |x| ≥ a

2

.

3Note thatνλ = 3 =⇒ ω = 2πν =

2π

λ3.

Therefore, we haveΨ(x, t) = A sin

[2π

λx− ωt

].

124

(a) Find A so that the function Ψ(x, t) is properly normalized.(b) In the region where the potential V (x) = 0, the Schrodinger equation

reduces to− ~

2

2m

∂2

∂x2Ψ(x, t) = i~

∂Ψ(x, t)

∂t.

Plug Ψ(x, t) = A cos(πxa

)e−iEt/~ into this relation to show E = π2~2

2ma2.

(c) Show that ⟨P ⟩ = 0.(d) Evaluate ⟨x2⟩ for this wavefunction. You will probably need an integral

table for this. (Of course, you can always try contour integration or somesuch. But, that is beyond this course.)

(e) Evaluate ⟨P 2⟩ for the same wavefunction. You will not need an integraltable if you use the fact that the function is normalized. Of course, abrute force computation will yield the same result as well.

6. This problem is the same as Chapter 7 Problem 2.If you are a careful student who pays attention to details, you may have realizedthat A can only be determined up to the argument, or up to the sign if youassume A is a real number.

(a) What is the implication of this fact as to the uniqueness of wavefunction?(b) What is the implication of this fact as to the probability density Ψ∗(x, t)Ψ(x, t)?

How about ⟨P ⟩, ⟨P 2⟩, and ⟨x2⟩?

Chapter 6

The Time-IndependentSchrodinger Equation

6.1 Separation of Time t and Space x

In the Schrodinger Equation[− ~

2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t(6.1)

the potential often does not depend on time; i.e. V (x, t) = V (x). An example of thiswould be the infinite square well potential.

V (x, t) =

0 −a

2< x < a

2

(a > 0)∞ |x| ≥ a

2

(6.2)

When V (x, t) = V (x), we can express the multivariable function Ψ(x, t) as a productof a function of x and a function of t as follows.

Ψ(x, t) = ψ(x)ϕ(t) (6.3)

This technique is called “separation of variables”. Let us substitute the above Ψ intothe Schrodinger Equation.[

− ~2

2m

∂2

∂x2+ V (x)

]ψ(x)ϕ(t) = i~

∂ψ(x)ϕ(t)

∂t

=⇒ − ~2

2m

∂2

∂x2ψ(x)ϕ(t) + V (x)ψ(x)ϕ(t) = i~

∂ψ(x)ϕ(t)

∂t

125

126 CHAPTER 6. THE TIME-INDEPENDENT SCHRODINGER EQUATION

=⇒ − ~2

2mϕ(t)

∂2

∂x2ψ(x) + V (x)ψ(x)ϕ(t) = ψ(x)i~

∂ϕ(t)

∂t

=⇒ ϕ(t)

[− ~

2

2m

∂2

∂x2ψ(x) + V (x)ψ(x)

]= ψ(x)i~

∂ϕ(t)

∂t(6.4)

Dividing through by ψ(x)ϕ(t), we get

1

ψ(x)

[− ~

2

2m

d2

dx2ψ(x) + V (x)ψ(x)

]︸︷︷︸

H(x)

= i~1

ϕ(t)

dϕ(t)

dt︸︷︷︸K(t)

; (6.5)

where we changed the partial derivatives to total derivatives as we now have one-variable functions.

Generally speaking,

H(x) = K(t) for all pairs (x, t) (6.6)

means that both H(x) and K(t) are a constant. Let H(x) = K(t) = G (a constant).Then,

K(t) = G⇐⇒ i~1

ϕ(t)

dϕ(t)

dt= G⇐⇒ dϕ(t)

dt=

1

i~Gϕ(t) =

−i~Gϕ(t). (6.7)

Therefore,

ϕ(t) = Ae−iGt/~. (6.8)

Compare this with the time-dependent part of the travelling wave

ei(kx−ωt) = eikxe−iωt. (6.9)

We realize that G~= 2πG

h“should be” ω.

2πG

h= ω =⇒ G =

ωh

2π=

2πνh

2π= hν = E. (6.10)

We now have G = E, and

ϕ(t) = e−iEt/~. (6.11)

6.2. CONDITIONS ON THE WAVEFUNCTION ψ(X) 127

This also implies

H(x) =1

ψ(x)

[− ~

2

2m

d2

dx2ψ(x) + V (x)ψ(x)

]= E

=⇒ − ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x); (6.12)

where E is the total energy.

− ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x) (6.13)

or[− ~

2

2m

d2

dx2+ V (x)

]ψ(x) = Eψ(x) (6.14)

is the Time-Independent Schrodinger Equation.

Note here that the full wavefunction is given by

Ψ(x, t) = ψ(x)e−iEt/~. (6.15)

In the rest of this course, we will solve the Time-Independent Schrodinger Equationfor various systems, culminating in a solution for the hydrogen atom.

6.2 Conditions on the Wavefunction ψ(x)

Previously, I mentioned that only normalized solutions are the physically meaningfulsolutions in order for the Born’s interpretation to make sense. However, this restric-tion alone would not eliminate all the mathematically correct solutions which arenot acceptable on physical grounds. Hence, we need to put further restrictions onthe nature of the solutions. We will place three types of restrictions on both ψ(x)

and dψ(x)dx

, which can be extended naturally to higher dimensions such as ψ(x, y) andψ(x, y, z). The following are ad-hoc and post-hoc conditions albeit physically veryreasonable.

Physical Sense Revisited:1. Finiteness and Integrability: The solution and its first derivative have to take

a finite value for all x. Furthermore, the wavefunction has to be integrablein order to be normalizable; i.e.

∫+∞−∞ ψ∗ψdx < ∞. This is due to Born’s

probabilistic interpretation.

128 CHAPTER 6. THE TIME-INDEPENDENT SCHRODINGER EQUATION

2. Single-valuedness: The solution and the first derivative should have one valueat each point in space.

3. Continuity: In principle, both the solution and its first derivative have to becontinuous everywhere.

We will find solutions of this type. Incidentally, Condition 3 on the derivativecan not be imposed if the potential V (x) does not remain finite. We will see such anexample in Section 7.5 where the potential blows up to +∞. Else, both ψ(x) anddψ(x)dx

are continuous. In particular, this means that ψ(x) and dψ(x)dx

are continuouseven if the potential V (x) itself is discontinuous. However, the second derivatived2ψ(x)dx2

is discontinuous if V (x) is, which is clear from the Schrodinger equation itselfas shown in (6.16). Note that the right-hand side of (6.16) is discontinuous if V (x)is discontinuous because ψ(x) is continuous.

− ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x) =⇒ − ~

2

2m

d2

dx2ψ(x) = (E − V (x))ψ(x)

=⇒ d2

dx2ψ(x) =

2m

~2(V (x)− E)ψ(x) (6.16)

One can also see from (6.16) that dψ(x)dx

has to be continuous if V (x) is finite becausea function has an infinite first derivative at a discontinuity and the first derivativein this case is d

dx

(dψ(x)dx

)= d2ψ(x)

dx2.

In case V (x) is continuous, we can prove continuity of dψ(x)dx

as follows. Let δ be anarbitrary positive number. We will show continuity of dψ(x)

dxat x = x0.

∫ x0+δ

x0

d2

dx2ψ(x) dx =

2m

~2

∫ x0+δ

x0(V (x)− E)ψ(x) dx

=⇒∫ x0+δ

x0

d

dx

(dψ(x)

dx

)dx =

2m

~2

∫ x0+δ


=⇒[dψ(x)

dx

]x0+δx0

=2m

~2

∫ x0+δ


=⇒[dψ(x)

dx

]x0+δ

−[dψ(x)

dx

]x0

=2m

~2

∫ x0+δ

x0(V (x)− E)ψ(x) dx (6.17)

6.2. CONDITIONS ON THE WAVEFUNCTION ψ(X) 129

Now, when the potential V (x) is continuous, |V (x)− E| and |ψ(x)| are bounded bya finite number M from above on the interval [x0, x0 + δ].a And hence, we have∣∣∣∣∣

[dψ(x)

dx

]x0+δ

−[dψ(x)

dx

]x0

∣∣∣∣∣ = 2m

~2

∣∣∣∣∣∫ x0+δ


∣∣∣∣∣≤ 2m

~2

∫ x0+δ

x0|V (x)− E| |ψ(x)| dx ≤ 2m

~2·M2 · δ. (6.18)

Now let δ → 0. Noting that the relation 6.18 holds for smaller δ with the same M ,we can conclude

dψ(x)

dx

∣∣∣∣∣x0+δ

− dψ(x)

dx

∣∣∣∣∣x0

δ→0−−→ 0. (6.19)

This means dψ(x)dx

is continuous from the right or right-continuous at x = x0. Like-wise,

∫ x0

x0−δ

d2

dx2ψ(x) dx =

2m

~2

∫ x0

x0−δ(V (x)− E)ψ(x) dx

=⇒ dψ(x)

dx

∣∣∣∣∣x0

− dψ(x)

dx

∣∣∣∣∣x0−δ

δ→0−−→ 0. (6.20)

And, dψ(x)dx

is also left-continuous at x = x0. Hence, the first derivative is continuousat x = x0.

aThis is from Theorem D.1 called the Extreme Value Theorem.

See Appendix D for a mathematical definition of continuity and a mathematicallymore rigorous proof of continuity of dψ(x)

dx.

130

Exercises1. Write down, but do not derive, the time-independent Schrodinger equation forV (x), that is, V is now a function of x only. This is a one-dimensional case,and the only spatial variable is x.

2. The time-dependent Schrodinger Equation is

− ~2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂t.

Consider a case where the potential does not depend on time; i.e. V (x, t) =V (x). Assume Ψ(x, t) = ψ(x)e−iEt/~ and derive the time-independent SchrodingerEquation.

3. Discuss in concrete terms and sufficient detail why Φ(ϕ) = sinϕ is an accept-able wavefunction while Φ(ϕ) = ϕ is not. We are using the spherical polarcoordinates here.

Chapter 7

Solutions of Time-IndependentSchrodinger Equations in OneDimension

Though you may, perhaps naively, think our world is three-dimensional, there aremany physical systems which are of lower dimensions. For example, a thin film formsa two-dimensional system, and a particle moving freely without any external forceis the simplest one-dimensional system. Furthermore, the Schrodinger equations forone-dimensional systems are easier to solve. Hence, it makes good mathematical andphysical sense to start with one dimensional cases.

7.1 The Zero PotentialThis is the case where V (x) = 0 for all x.

Our Time-Independent Schrodinger Equation is

−~2

2m

d2ψ(x)

dx2= Eψ(x). (7.1)

The most general solution for this second order linear ordinary differential equationis of the form

ψ(x) = A1 sin√2mE

~x+ A2 cos

√2mE

~x = C1e

ikx + C2e−ikx; (7.2)

131

132CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRODINGER EQUATIONS IN

ONE DIMENSION

where A1, A2, C1, and C2 are arbitrary constants, and k =√2mE~

.

On the other hand, we always have

ϕ(t) = e−iEt/~ = e−ihνt/(h/2π) = e−i2πνt = e−iωt (7.3)

for time dependence.

Therefore, the full time-dependent wavefunction is

Ψ(x, t) = ψ(x)ϕ(t) =(C1e

ikx + C2e−ikx

)e−iωt

= C1ei(kx−ωt) + C2e

−i(kx+ωt). (7.4)

Let us take

Ψ(x, t) = C1ei(kx−ωt)

(= C1e

ikxe−iEt/~). (7.5)

Then,

P = ⟨P ⟩ =∫ +∞

−∞Ψ∗PΨdx =

∫ +∞

−∞C∗

1e−ikxeiEt/~(−i~) ∂

∂xC1e

ikxe−iEt/~dx

=∫ +∞

−∞C∗

1e−ikxeiEt/~(−i~)(ik)C1e

ikxe−iEt/~dx =(∫ +∞

−∞Ψ∗Ψdx

)= ~k

∫ +∞

−∞ψ∗ψdx = ~k =

h

2π· 2mE~

=√2mE. (7.6)

This makes sense as

P 2

2m= E =⇒ P =

√2mE (7.7)

classically, and (7.6) serves as yet another consistency checking device.

As simple as the concept and the wavefunctions of a free particle are, there is oneserious complication regarding normalization. To see this, consider an eigenfunctionψ(x) = Aeikx. In order for this function to be normalizable, it has to be integrableto begin with. However,∫ +∞

−∞

(Aeikx

)∗ (Aeikx

)dx = |A|2

∫ +∞

−∞e−ikxeikx dx = |A|2

∫ +∞

−∞1 dx =∞, (7.8)

7.2. THE STEP POTENTIAL (E < V0) 133

and ψ(x) is not normalizable in the usual way.

In order to circumvent this problem, three normalization schemes have been pro-posed; namely, the Born normalization, Dirac normalization, and unit-flux normal-ization. As these normalization procedures are not crucial for the rest of this book,their discussions are relegated to Appendix G.

7.2 The Step Potential (E < V0)We now consider a step potential given by

V (x) =

V0 x > 00 x ≤ 0

(7.9)

for E < V0.

In Newtonian (classical) mechanics, the particle can not enter the region (0,∞). Wehave

E = K.E.+ P.E. = K.E.+ V0 in (0,∞) =⇒ K.E. = E − V0 < 0. (7.10)

This does not make any sense of any kind classically. Is this also the case in quantummechanics? Let us just plunge in and solve it!

The time-independent Schrodinger Equation is

− ~2

2m

d2ψ(x)

dx2+ V (x)ψ(x) = Eψ(x). (7.11)

In our case, we have the following.

−~2

2m

d2ψ(x)

dx2= Eψ(x) x ≤ 0 (7.12)

−~2

2m

d2ψ(x)

dx2+ V0ψ(x) = Eψ(x) x > 0 (7.13)

Equation 7.12 is that for a free particle. Therefore, the general solution is

ψ(x) = Aeik1x +Be−ik1x; (7.14)


ONE DIMENSION

where k1 =√2mE~

, and A and B are arbitrary constants.

On the other hand, (7.13) can be rewritten as follows.

−~2

2m

d2ψ(x)

dx2= (E − V0)ψ(x)⇐⇒

~2

2m

d2ψ(x)

dx2= (V0 − E)ψ(x)

⇐⇒ d2ψ(x)

dx2=

2m(V0 − E)~2

ψ(x) (7.15)

The general solution for x > 0 is

ψ(x) = Cek2x +De−k2x; (7.16)

where k2 =

√2m(V0−E)

~, and C and D are arbitrary constants. Note that V0 − E is

positive in this region.

As stated above, these are indeed the solutions, or the (most) general solutions, sincewe have linear second order ordinary differential equations.

We have ψ(x) = Aeik1x +Be−ik1x (k1 =√2mE~

) x ≤ 0

ψ(x) = Cek2x +De−k2x (k2 =

√2m(V0−E)

~) x > 0

. (7.17)

Recall the conditions on (ψ, dψdx).

1. Finite

2. Single-valued

3. Continuous

We will determine A, B, C, and D using these conditions.

Let us begin with the region x > 0. In this region,

ψ(x) = Cek2x +De−k2x; where k2 =

√2m(V0 − E)~

. (7.18)

Therefore,

ψ∗(x)ψ(x) =(C∗ek2x +D∗e−k2x

) (Cek2x +De−k2x

)


= |C|2e2k2x +2ReC∗D︷︸︸︷

D∗C + C∗D+|D|2e−2k2x. (7.19)

Since the wavefunction has to be normalizable, it has to satisfy the finiteness condi-tion ∫ +∞

−∞ψ∗(x)ψ(x)dx =

∫ +∞

−∞|ψ(x)|2dx <∞. (7.20)

Noting that this ψ is the wavefuntion only in the region x > 0, we actually need∫ +∞

0|ψ(x)|2dx <∞. (7.21)

Hence, we have to have

C = 0. (7.22)

Though this may seem intuitively obvious, let us conduct a complete verification ofthis for once. We will start with (7.19) and note that∫ ∞

0|D|2e−2k2x dx = |D|2 1

−2k2

[e−2k2x

]∞0

= |D|2 1

−2k2(−1) = |D|

2

2k2<∞. (7.23)

So, we need only to deal with the first two terms |C|2e2k2x +

2ReC∗D︷︸︸︷D∗C + C∗D. Set

R = 2ReC∗D and consider |C|2e2k2x +2ReC∗D︷︸︸︷

D∗C + C∗D = |C|2e2k2x +R. If R ≥ 0,∫ ∞

0|C|2e2k2x +Rdx (7.24)

is clearly not finite unless C = R = 0. So, consider R < 0 and C , 0, and letL = −R > 0. We have |C|2e2k2x + R = |C|2e2k2x − L, and the following inequalityholds.

x >ln 2L

|C|2

2k2=⇒ 2k2x > ln 2L

|C|2=⇒ e2k2x > e

ln 2L|C|2 =

2L

|C|2=⇒ |C|2e2k2x > 2L

=⇒ |C|2e2k2x − L > L > 0 =⇒ |C|2e2k2x +R > L > 0 (7.25)

Therefore,

∫ ∞

0|C|2e2k2x +

2ReC∗D︷︸︸︷D∗C + C∗D dx =

∫ ∞

0|C|2e2k2x +Rdx


ONE DIMENSION

=∫ ln 2L

|C|22k2

0|C|2e2k2x +Rdx+

∫ ∞ln 2L

|C|22k2

|C|2e2k2x +Rdx

≥∫ ln 2L

|C|22k2

0|C|2e2k2x +Rdx+

∫ ∞ln 2L

|C|22k2

Ldx (7.26)

is not finite and we need to require C = 0. Note that this also means R =2ReC∗D︷︸︸︷

D∗C + C∗D = 0.At this point, we have ψ(x) = Aeik1x +Be−ik1x k1 =

√2mE~

x ≤ 0

ψ(x) = De−k2x (k2 =

√2m(V0−E)

~) x > 0

. (7.27)

Next, we will use the continuity condition on ψ(x) and dψ(x)dx

.Continuity of ψ(x) at x = 0 gives

De−k2·0 = Aeik1·0 +Be−ik1·0 =⇒ D = A+B. (7.28)

Since the first derivative isdψ(x)dx

= A(ik1)eik1x +B(−ik1)e−ik1x k1 =

√2mE~

x ≤ 0dψ(x)dx

= D(−k2)e−k2x (k2 =

√2m(V0−E)

~) x > 0

, (7.29)

continuity of dψ(x)dx

at x = 0 gives

−k2De−k2·0 = ik1Aeik1·0 − ik1Be−ik1·0 =⇒ −k2D = ik1(A−B)

=⇒ −k2ik1

D = A−B. (7.30)

We now have a simultaneous equationik2k1D = A−BD = A+B

(7.31)

for A, B, and D with the solution

A = 12

(1 + ik2

k1

)D

B = 12

(1− ik2

k1

)D. (7.32)


This gives

ψ(x) =

D2(1 + ik2/k1)e

ik1x + D2(1− ik2/k1)e−ik1x x ≤ 0

De−k2x x > 0. (7.33)

So, the full solution is

Ψ(x, t) = ψ(x)ϕ(t) = ψ(x)e−iEt/~

=

D2(1 + ik2/k1)e

i(k1x−Et/~) + D2(1− ik2/k1)e−i(k1x+Et/~) x ≤ 0

De−k2xe−iEt/~ x > 0. (7.34)

Recall that

e−iEt/~ = e−iωt. (7.35)

This is because

Et/~ = hνt/~ = hνt/(h

2π) = 2πνt = ωt. (7.36)

Therefore, in the region x ≤ 0,

Ψ(x, t) = Aei(k1x−ωt) +Bei(−k1x−ωt). (7.37)

Note that

Aei(k1x−ωt) (7.38)

is traveling to the right while

Bei(−k1x−ωt) (7.39)

is traveling to the left. In other words,

Aei(k1x−ωt) (7.40)

is the incident wave, incident on the step form the left, while

Bei(−k1x−ωt) (7.41)

is the wave reflected by the step.


ONE DIMENSION

What is the reflection coefficient? Since the reflection coefficient is the ratio of theamplitude of the reflected wave and the amplitude of the incident wave, we have

the reflection coefficient = the amplitude of the reflected wavethe amplitude of the incident wave

=|Bei(−k1x−ωt)|2

|Aei(k1x−ωt)|2=|B|2|ei(−k1x−ωt)|2

|A|2|ei(k1x−ωt)|2=|B|2

|A|2=B∗B

A∗A

=

D∗

2

(1− ik2

k1

)∗D2

(1− ik2

k1

)D∗

2

(1 + ik2

k1

)∗D2

(1 + ik2

k1

) =

(1 + ik2

k1

) (1− ik2

k1

)(1− ik2

k1

) (1 + ik2

k1

) = 1 (7.42)

And, we have a total reflection.

Indeed, we have a standing wave in the region x ≤ 0. To see this, plug eik1x =cos k1x+ i sin k1x into

ψ(x) =

D2

[(1 + ik2/k1)e

ik1x + (1− ik2/k1)e−ik1x]x ≤ 0

De−k2x x > 0(7.43)

to obtain

ψ(x) =

D cos k1x−D k2

k1sin k1x x ≤ 0

De−k2x x > 0. (7.44)

So, for x ≤ 0,

Ψ(x, t) = D

(cos k1x−

k2k1

sin k1x)e−iEt/~. (7.45)

Remember the following from high school math,

α cos kx− β sin kx =

(α√

α2 + β2cos kx− β√

α2 + β2sin kx

) √α2 + β2

=√α2 + β2 cos(kx+ θ); (7.46)

where θ is such that

cos θ = α√α2 + β2

and sin θ = β√α2 + β2

. (7.47)

7.3. THE STEP POTENTIAL (E > V0) 139

Therefore, the nodes of Ψ(x, t)∗Ψ(x, t) do not move, and we indeed have a standingwave.

Now, in the region x > 0,

Ψ∗(x, t)Ψ(x, t) = D∗De−2k2x. (7.48)

While

limx→∞

Ψ∗Ψ = 0, (7.49)

Ψ∗Ψ , 0 (7.50)

for ∀x > 0. This means that the probability of finding the particle to the right of thestep is not zero even if the total energy E is smaller than the step height V0. Thisphenomenon is definitely non-classical and is known as “barrier penetration”.

7.3 The Step Potential (E > V0)We will now consider the case where the total energy E is greater than the stepheight.

Classically, we have the following.

1. Not a total reflection

2. Has to be a total penetration into (0,∞)

But, quantum mechanically, the Schrodinger Equation predicts the following.

1. Not a total reflection (agreement with the classical theory)

2. There is partial reflection at the boundary x = 0 (disagreement)

Let us see how it works.

As usual, Ψ(x, t) = ψ(x)e−iEt/~, and the Time-Independent Schrodinger Equation is−~22m

d2ψ(x)dx2

= Eψ(x) x < 0−~22m

d2ψ(x)dx2

= (E − V0)ψ(x) x > 0(7.51)


ONE DIMENSION

As E − V0 > 0 this time, the two equations are basically the same. They are bothfree particles, and the solutions are

ψ(x) =

Aeik1x +Be−ik1x x < 0Ceik2x +De−ik2x x > 0

; (7.52)

where k1 =√2mE~

and k2 =

√2m(E−V0)~

.

Suppose the particle is in the region x < 0 or (−∞, 0) at t = 0. In other words, theparticle is incident on the step from the left. Then, since the wave is travelling to theright in the region (0,∞), we should set D = 0. This is simply because e−ik2xe−iEt/~

is a wave travelling to the left.

Imposing the continuity condition on ψ and dψdx

at x = 0, we can express the constantsB and C in terms of A as follows.

ψ(x)|x=0− = ψ(x)|x=0+dψ(x)dx

∣∣∣x=0−

= dψ(x)dx

∣∣∣x=0+

=⇒

A+B = Ck1(A−B) = k2C

=⇒B = Ak1−k2

k1+k2

C = A 2k1k1+k2

(7.53)

Therefoere,

ψ(x) =

Aeik1x + Ak1−k2

k1+k2e−ik1x

A 2k1k1+k2

eik2x(7.54)

The reflection coefficient R is given by

R =B∗B

A∗A=

(k1 − k2k1 + k2

)2

. (7.55)

Since the transmission coefficient T is related to R via R + T = 1,

T = 1−R = 1−(k1 − k2k1 + k2

)2

=(k1 + k2 + k1 − k2)(k1 + k2 − k1 + k2)

(k1 + k2)2

=4k1k2

(k1 + k2)2. (7.56)

7.3. THE STEP POTENTIAL (E > V0) 141

As advertised, we do not have a total reflection, and we do not have a total penetra-tion into (0,∞), either.

We now solve the same problem for a wave incident on the potential boundary atx = 0 from the right. In other words, the particle is in the region (0,∞) at t = 0and traveling to the left.

Our general solution is

ψ(x) =

Aeik1x +Be−ik1x x < 0Ceik2x +De−ik2x x > 0

; (7.57)

where k1 =√2mE~

and k2 =

√2m(E−V0)~

.

Since there is no boundary to reflect the wave in the region (−∞, 0), the wave shouldbe traveling to the left in this region. We set A = 0.

ψ(x) =

Be−ik1x x < 0Ceik2x +De−ik2x x > 0

(7.58)

The first derivative with respect to x is

ψ′(x) =

−ik1Be−ik1x x < 0ik2Ce

ik2x + (−ik2)De−ik2x x > 0. (7.59)

Imposing the continuity condition on ψ(x) and dψ(x)dx

at x = 0, we getB = C +D

−ik1B = ik2(C −D)=⇒

C +D = BC −D = −k1

k2B

=⇒C = k2−k1

2k2B

D = k2+k12k2

B. (7.60)

What are the reflection coefficient and the transmission coefficient in this case?

R =C∗C

D∗D=

(k2−k12k2

)2B∗B(

k2+k12k2

)2B∗B

=

(k1 − k2k1 + k2

)2

(7.61)

and

T = 1−R =4k1k2

(k1 + k2)2(7.62)


ONE DIMENSION

Note that these are the same as before when we computed R and T for the particlemoving in from the left.

Incidentally, B , 0 in (7.52) can be proved on purely mathematical gorunds as fol-lows.

If B = 0,

ψ(x) =

Aeik1x x < 0Ceik2x x > 0

=⇒

A = Ck1A = k2C

. (7.63)

But, it is obvious that this can not be solved for A and C unless we accept A = C = 0which is physically meaningless. Hence, B , 0 on this ground.

7.4 The Barrier Potential (E < V0)The potential V (x) is V0 in the region (0, a) and 0 elsewhere.

V (x) =

V0 0 < x < a0 x < 0, a < x

(7.64)

Our Time-Independent Schrodinger Equations are−~22m

d2ψ(x)dx2

= Eψ(x) x < 0−~22m

d2ψ(x)dx2

+ V0ψ(x) = Eψ(x) 0 < x < a−~22m

d2ψ(x)dx2

= Eψ(x) a < x

, (7.65)

and the general solutions are

ψ(x) =

Aeik1x + Be−ik1x x < 0Ce−k2x + Dek2x 0 < x < aFeik1x + Ge−ik1x a < x

; (7.66)

where k1 =√2mE~

and k2 =

√2m(V0−E)

~.

Assuming x < 0 at t = 0, that is, the particle originated at x = −∞ and is movingto the right, we have to set G = 0 as there is nothing to reflect the wave in the regiona < x.

7.5. THE INFINITE SQUARE WELL POTENTIAL 143

On the other hand, our boundary conditions are

ψ(x)|x=0− = ψ(x)|x=0+

ψ(x)|x=a− = ψ(x)|x=a+dψ(x)dx

∣∣∣x=0−

= dψ(x)dx

∣∣∣x=0+

dψ(x)dx

∣∣∣x=a−

= dψ(x)dx

∣∣∣x=a+

. (7.67)

We have the above four equations and the five unknowns A, B, C, D, and F . But,we also have a normalization condition. So, we can solve for the five unknowns.

As it turned out, F , 0, and the probability density Ψ∗(x, t)Ψ(x, t) = ψ∗(x)ψ(x) hasa (nonzero) tail in the region x > a. Of course, it is nonzero also inside the barrier.Therefore, we have both barrier penetration and tunneling.

7.5 The Infinite Square Well PotentialThe potential is given by

V (x) =

∞ |x| > a

2

0 |x| < a2

, (7.68)

and we get, as we have already seen,

ψ(x) =

0 |x| > a

2

A sin kx+B cos kx |x| < a2

; (7.69)

where k =√2mE~

. As usual, the full wavefunction is

Ψ(x, t) = ψ(x)e−iEt/~. (7.70)

It turned out the continuity condition of the first derivative is too restrictive for thissystem as the potential blows up to ∞ unlike any known physical system.The continuity condition on ψ(x) at |x| = a

2gives

A sin ka2

+ B cos ka2

= 0−A sin ka

2+ B cos ka

2= 0

=⇒

2B cos ka2

= 02A sin ka

2= 0

=⇒A = 0 and cos ka

2= 0

B = 0 and sin ka2= 0

. (7.71)


ONE DIMENSION

Therefore, ψ(x) = B cos kx where cos ka

2= 0

orψ(x) = A sin kx where sin ka

2= 0

. (7.72)

We have two families of solutions.ψn(x) = Bn cos knx where kn = nπ

an = 1, 3, 5, . . .

ψn(x) = An sin knx where kn = nπa

n = 2, 4, 6, . . .. (7.73)

Note that n is a positive integer in both cases because kn > 0. We can also notethat negative n’s give redundant solutions and n = 0 gives a physically meaninglesstrivial solution. Indeed,

n = 0 =⇒ ψ0(x) = A0 sin 0 = 0. (7.74)

Let us compute the normalization constants An and Bn. We will obtain real andpositive constants.∫ +a

2

− a2

(Bn cos knx)∗ (Bn cos knx) dx = |Bn|2∫ +a

2

− a2

cos2 knx dx

= |Bn|2∫ +a

2

− a2

1 + cos 2knx2

dx =|Bn|2

2

[x+

1

2knsin 2knx

]+a2

− a2

=|Bn|2

2

[x+

a

2nπsin 2nπ

ax]+a

2

− a2

=|Bn|2

2

[(a

2+

a

2nπsinnπ

)−(−a2− a

2nπsinnπ

)]=|Bn|2

2· a = 1

=⇒ Bn =

√2

a(7.75)

∫ +a2

− a2

(An sin knx)∗ (An sin knx) dx = |An|2∫ +a

2

− a2

sin2 knx dx

= |An|2∫ +a

2

− a2

1− cos 2knx2

dx =|An|2

2

[x− 1

2knsin 2knx

]+a2

− a2

=|An|2

2

[x− a

2nπsin 2nπ

ax]+a

2

− a2

7.5. THE INFINITE SQUARE WELL POTENTIAL 145

=|An|2

2

[(a

2− a

2nπsinnπ

)−(−a2+

a

2nπsinnπ

)]=|An|2

2· a = 1

=⇒ An =

√2

a(7.76)

Hence, (7.73) becomes ψn(x) =√

2a

cos knx where kn = nπa

n = 1, 3, 5, . . .

ψn(x) =√

2a

sin knx where kn = nπa

n = 2, 4, 6, . . .. (7.77)

We can also write (7.77) as follows.

ψn(x) =

√2

asin

nπ(x+ a

2

)a

=

√2

asin

[nπ

(x

a+

1

2

)](7.78)

Furthermore, if we shift the x-axis so that V (x) = 0 for 0 < x < a, we will get

ψn(x) =

√2

asin

(nπx

a

). (7.79)

Because kn =√2mEn/~, the total energy En can only take discrete values.

kn =

√2mEn~

=⇒ 2mEn = ~2k2n

=⇒ En =~2k2n2m

=π2~2n2

2ma2n = 1, 2, 3, 4, 5, . . . (7.80)

Note that En ∝ n2 and the spacing between neighboring energy levels

En+1 − En =π2~2(n+ 1)2

2ma2− π2~2n2

2ma2=π2~2(2n+ 1)

2ma2(7.81)

is proportional to 2n+ 1. This is the reason for the widening of the gap as n→∞.More significant implication of this is that the particle can not have zero total energy.Indeed, the lowest energy level

E1 =π2~2

2ma2> 0. (7.82)

This is one reason why absolute zero degree temperature can not be achieved. Whilethe infinite square well potential is a mathematical idealization, it approximates thepotential well experienced by atoms or nuclei in a crystal for example.

Another type of bound system frequently encountered in nature is a diatomic molecule.


ONE DIMENSION

7.6 The Simple Harmonic Oscillator Potential7.6.1 Classic SolutionConsider a typical spring encountered in high school physics with a spring constantk such that the restoring force F is given by F = −kx for the displacement x fromthe equilibrium position. For no other reason than convention, we will use c insteadof k for the spring constant.

Because

V (x) =1

2cx2, (7.83)

the Time-Independent Schrodinger Equation is

−~2

2m

d2ψ(x)

dx2+c

2x2ψ(x) = Eψ(x). (7.84)

We will need some fancy footwork to solve this equation. As it turns out, it isconvenient to introduce a new variable ν analogously to the classical theory. Hence,

ν =1

2πω =

1

2π

√c

m1. (7.85)

After substituting c = (2π)2ν2m into the equation and dividing through by −~22m

, weget

d2ψ

dx2+

[2mE

~2−(2πmν

~

)2

x2]ψ = 0. (7.86)

1To see this, we start with the classical equation F = mα. For us,

−cx = md2x

dt2=⇒ d2x

dt2= − c

mx =⇒ x(t) = A sin

(√c

mt

)+B cos

(√c

mt

)

=√A2 +B2 sin

(√c

mt+ θ

);

where cos θ = A√A2+B2

and sin θ = B√A2+B2

. Therefore, we have

ω =

√c

m.

7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 147

Now, let α = 2πmν/~ and β = 2mE/~2 to simplify the equation to

d2ψ

dx2+ (β − α2x2)ψ = 0. (7.87)

Here are further manipulations.

u =√αx =

[2πm

~2π(c

m)1/2

]1/2x =

(cm)1/4

~1/2x (7.88)

dψ

dx=dψ

du

du

dx=√αdψ

du(7.89)

d2ψ

dx2=

d

dx

(dψ

dx

)=d(dψ

dx)

dx=d(√αdψdu)

dx=du

dx

d

du

(√αdψ

du

)

=√α√αd

du

(dψ

du

)= α

d2ψ

du2(7.90)

You may be more comfortable with the following.

du =√αdx =⇒ dx =

1√αdu (7.91)

d2ψ

dx2=

d1√αdu

dψ1√αdu

= αd2ψ

du2(7.92)

Either way is fine, and we now have

d2ψ

dx2+ (β − α2x2)ψ = α

d2ψ

du2+ (β − αu2)ψ = 0

=⇒ d2ψ

du2+

(β

α− u2

)ψ = 0. (7.93)

Here comes a very inexact argument.

As |u| → ∞, the differential equation behaves like

d2ψ

du2− u2ψ = 0 (7.94)


ONE DIMENSION

or

d2ψ

du2= u2ψ (7.95)

as β/α is a constant.The (mathematical) general solution of this differential equation is approximately2

ψ = Ae−u2/2 +Beu2/2 as |u| → ∞. (7.96)

But, we have to set

B = 0 (7.97)

to satisfy the finiteness condition. So,

ψ(u) = Ae−u2/2 as |u| → ∞. (7.98)

We will look for a solution of the form

ψ(u) = Ae−u2/2H(u); (7.99)

where H(u) must be slowly varying in comparison to e−u2/2.Compute d2ψ(u)

du2, then plug it back into the equation

d2ψ(u)

du2+

(β

α− u2

)ψ(u) = 0 (7.100)

to obtain

d2H

du2− 2u

dH

du+

(β

α− 1

)H = 0 (7.101)

2

d2

du2

(e−u2/2

)=

d

du

(−ue−u2/2

)= −e−u2/2 − u

(−ue−u2/2

)= (u2 − 1)e−u2/2

−−−−−→|u|−→∞

u2e−u2/2

d2

du2

(eu2/2

)=

d

du

(ueu2/2

)= eu2/2 + u

(ueu2/2

)= (u2 + 1)eu2/2

−−−−−→|u|−→∞

u2eu2/2

Hence, (7.96) holds.


as follows.

Without loss of generality, set A = 1, and consider

ψ(u) = e−u2/2H(u). (7.102)

Then, we have

dψ

du= −ue−u2/2H(U) + e−u2/2H ′(u) (7.103)

andd2ψ

du2= −e−u2/2H(u)− u

(−ue−u2/2

)H(u)− ue−u2/2H ′(u)

−ue−u2/2H ′(u) + e−u2/2H ′′(u)

= −e−u2/2H(u) + u2e−u2/2H(u)− 2ue−u2/2H ′(u) + e−u2/2H ′′(u). (7.104)

Therefore,

d2ψ(u)

du2+

(β

α− u2

)ψ(u) = −e−u2/2H(u) + u2e−u2/2H(u)− 2ue−u2/2H ′(u)

+e−u2/2H ′′(u) +

(β

α− u2

)e−u2/2H(u)

= −e−u2/2H(u)− 2ue−u2/2H ′(u) + e−u2/2H ′′(u) +β

αe−u2/2H(u) = 0. (7.105)

Dividing through by e−u2/2, we get

−H(u)− 2uH ′(u) +H ′′(u) +β

αH(u)

= H ′′(u)− 2uH ′(u) +

(β

α− 1

)H(u) = 0, (7.106)

which is (7.101).

We will try a power series solution for H(u).Let

H(u) =∞∑l=0

al · ul = a0 + a1u+ a2u2 + a3u

3 + · · · . (7.107)


ONE DIMENSION

Then,

udH(u)

du= (a1 + 2a2u+ 3a3u

2 + 4a4u3 + · · · )u =

( ∞∑l=0

(l + 1)al+1ul

)· u

=∞∑l=0

(l + 1)al+1ul+1 =

∞∑l=1

lalul (7.108)

and

d2H(u)

du2= 2a2 + 2 · 3a3u+ 3 · 4a4u2 + · · · =

∞∑l=0

(l + 1)(l + 2)al+2ul. (7.109)

Plugging the above into

d2H

du2− 2u

dH

du+

(β

α− 1

)H = 0 (7.110)

gives us∞∑l=0

(l + 1)(l + 2)al+2ul − 2

∞∑l=0

lalul +

(β

α− 1

) ∞∑l=0

alul = 0 (7.111)

or∞∑l=0

[(l + 1)(l + 2)al+2 − 2lal +

(β

α− 1

)al

]ul = 0. (7.112)

For a power series to be zero, each term has to be zero; that is to say all the coefficientshave to be zero.We now have the following recursion relation.

al+2 = −(β/α− 1− 2l)

(l + 1)(l + 2)al (7.113)

The general solution is a sum of even and odd series; namely,

H(u) = a0

(1 +

a2a0u2 +

a4a0u4 +

a6a0u6 + · · ·

)+a1

(u+

a3a1u3 +

a5a1u5 +

a7a1u7 + · · ·

)(7.114)

= a0(1 + b1u2 + b2u

4 + b3u6 + . . .) + a1u(1 + c1u

2 + c2u4 + c3u

6 + . . .)

= a0∑l even

b l2ul + a1u

∑l even

c l2ul; (7.115)


where

b l2=ala0

and c l2=al+1

a1for l = 2, 4, 6, . . . , (7.116)

and the reason for the awkward notations b l2

and c l2

for the coefficients will becomeclear when compared with (7.119). What (7.114) means is that one can get all evencoefficients using (7.113) if a0 is given3, and all odd coefficients can be computed ifa1 is chosen. However, there is no relation between the even and odd coefficients,and the even series and odd series of (7.114) or (7.115) are completely independent.

As l→∞,

al+2

al= −(β/α− 1− 2l)

(l + 1)(l + 2)→ 2l

l2→ 2

l. (7.117)

So, we can see immediately that

b l2+1

b l2

=al+2

al

l→∞−−−−−→ 2

land

c l2+1

c l2

=al+3

al+1

l→∞−−−−−→ 2

l. (7.118)

Now compare this with the Taylor series expansion (about 0) of eu2 where we changethe running index from i to l = 2i and denote 1

( l2)!

by d l2.

eu2

=∞∑i=0

((u2)i)

i!=

∞∑i=0

u2i

i!=

even∑l=0

ul(l2

)!=

even∑l=0

d l2ul (7.119)

Observe that

d l2+1

d l2

=

1

( l2+1)!1

( l2)!

=

(l2

)!(

l2+ 1

)!=

(l2

)!(

l2+ 1

) (l2

)!

=1

l2+ 1

l→∞−−−−−→ 1

l/2=

2

l. (7.120)

Therefore, both the even series and the odd series of (7.115) behave like the Taylorexpansion of eu2 for higher order terms, or as l → ∞. Guided by this, we assumeH(u) behaves like

H(u) = a0Keu2 + a1K

′ueu2

. (7.121)3For example, a6

a0= a6

a4

a4

a2

a2

a0.


ONE DIMENSION

So, e−u2/2H(u) of (7.102) and (7.99) behaves like

a0Keu2/2 + a1K

′ueu2/2 as |u| → ∞4. (7.122)

In order to satisfy the integrability condition, Condition 1 on p.127, H(u) has toterminate after some n to prevent the exponential terms in (7.122) from blowing up.This means we have to set β/α = 2n+ 1 for n = 0, 1, 2, 3, 4, 5, . . . due to (7.113).

When β/α = 2n + 1 we get a Hermite polynomial denoted by Hn(u). This gives usa series of eigenfunctions

ψn(u) = Ane−u2/2Hn(u). (7.123)

The first six functions look like this.n = 0 ψ0 = A0e

−u2/2

1 ψ1 = A1ue−u2/2

2 ψ2 = A2(1− 2u2)e−u2/2

3 ψ3 = A3(3u− 2u3)e−u2/2

4 ψ4 = A4(3− 12u2 + 4u4)e−u2/2

5 ψ5 = A5(15u− 20u3 + 4u5)e−u2/2

(7.124)

Recalling that u =√αx, α = 2πmν/~, and β = 2mE/~2, we get

β/α =2mE/~2

2πmν/~=

E

~πν=

Eh2π· πν

=2E

hν(7.125)

Setting this equal to 2n+ 1,

β/α = 2n+ 1 =⇒ 2Enhν

= 2n+ 1 =⇒ En =(n+

1

2

)hν

n = 0, 1, 2, 3, 4, 5, . . . . (7.126)

Recall that

ν =1

2π

√c

m. (7.127)

Therefore, if the force constant c and the mass m are known, we know the energylevels of this harmonic oscillator.

These are the two important facts about the simple harmonic oscillator.4As |u| → ∞, higher order terms with large values of l become dominant, and this is where the

series in (7.122) begin to behave like the Taylor expansion (7.119).


1. The energy levels are equally spaced. This is markedly different from theinfinite square well. This means many higher energy levles can be achievedmore easily.

2. The lowest energy posssible is not zero but E0 =12hν for n = 0. Since vibrations

of diatomic molecules can be closely approximated by the simple harmonicoscillator, this is another reason why the absolute zero degree temperature cannot be achieved.

When a molecule drops from one energy level to a lower level, a photon carryingthat much energy is released. On the other hand, a molecule moves to a higherenergy level if it absorbs a photon carrying the energy corresponding to the differ-ence between the two levels. These phenomena are called emission and absorptionrespectively.

7.6.2 Raising and Lowering OperatorsOur primary goal in this section is to simplify the computation of the energy levels fora simple harmonic oscillator. The beauty of this alternative approach is that we willbe able to obtain the energy levels without explicitly computing the wavefunctions.The concept of raising and lowering operators has applications to angular momen-tum discussed in Chapter 9 and electron spin, a special type of angular momentum,described in Chapter 11.

From (7.84), we have

−~2

2m

d2ψ(x)

dx2+c

2x2ψ(x) = Eψ(x). (7.128)

This equation is often expressed in terms of ω = 2πν.

ν =1

2π

√c

m=⇒ ω = 2πν =

√c

m=⇒ c = mω2 (7.129)

Substituting mω2 for c, we get


ONE DIMENSION

[−~2

2m

d2

dx2+

1

2mω2x2

]ψ(x) = Eψ(x). (7.130)

Our first step is to modify and rescale the variables. Let

X =

√mω

~x (7.131)

and

P =1√m~ω

p; (7.132)

where x is the usual multiplication operator Mx and p = −i~ ∂∂x

as given in Table3.1. We will not use capital letters X and P in this section in order to make aclear distinction between the derived (X, P ) and the original (X,P ). X and P aredimensionless variables satisfying a simpler commutation relation. Indeed,

XP − P X =

√mω

~x

1√m~ω

p− 1√m~ω

p

√mω

~x =

1

~(xp− px), (7.133)

and we have

[X, P ] = i. (7.134)

We also introduce a rescaled Hamiltonian H defined by

H =1

~ωH. (7.135)

Now note that

X2 + P 2 =mω

~x2 +

1

m~ω

(−i~ d

dx

)(−i~ d

dx

)=mω2

~ωx2 − ~2

m~ω

d2

dx2

=2

~ω

(−~2

2m

d2

dx2+

1

2mω2x2

)= 2

1

~ωH. (7.136)

Therefore,


H =1

2

(X2 + P 2

). (7.137)

We will now define a new operator a and its adjoint a†, which will turn out to be”raising” and ”lowering” operators in the title of this section.

a =1√2

(X + iP

)(7.138)

a† =1√2

(X − iP

)(7.139)

Going backwards, we get the following if the above relations are solved for X and P .

X =1√2

(a† + a

)(7.140)

P =i√2

(a† − a

)(7.141)

Next, we will compute the commutator of a and a†.[1√2

(X + iP

),

1√2

(X − iP

)]=

1

2

[X + iP , X − iP

]=i

2

[P , X

]−[X, P

]=i

2(−i− i) = 15 (7.142)

Therefore,

[a, a†] = 1. (7.143)

Now,

a†a =1

2

(X − iP

) (X + iP

)=

1

2

(X2 + P 2 + iXP − iP X

)=

1

2

(X2 + P 2 − 1

)= H − 1

2. (7.144)

So,


ONE DIMENSION

H = a†a+1

2. (7.145)

On the other hand,

aa† = [a, a†] + a†a = 1 + H − 1

2= H +

1

2. (7.146)

And,

H = aa† − 1

2. (7.147)

At this point, let us introduce another self-adjoint operator N defined by

N = a†a. (7.148)

Note that(a†a

)†= a†

(a†)†

= a†a, which proves that N is self-adjoint. The operatorN satisfies the following commutation relations.

[N, a] = [a†a, a] = a†aa+ (a†aa− a†aa)− aa†a = [a†, a]a = −a (7.149)[N, a†] = [a†a, a†] = a†aa† + (a†a†a− a†a†a)− a†a†a = a†[a, a†] = a† (7.150)

We have now replaced the initial eigenvalue problem for H, which is a function of xand p, with that of N , which is a product of a and a†. Because we have

H = ~ωH, H = N +1

2=⇒ H = ~ω

(N +

1

2

), (7.151)

N |n⟩ = n |n⟩ =⇒ H |n⟩ = ~ω(n+

1

2

)|n⟩ ; (7.152)

where we are looking ahead and already using suggestive notations n and |n⟩, asopposed to the usual λ and |λ⟩. However, it is only on p.158 that we actually showthat the eigenvalues of N are nonnegative integers. Relations (7.151) and (7.152)show that the eigenvalue problem for N is equivalent to the eigenvalue problem forH. When the eigenvalue problem for N is solved, we will get En =

(n+ 1

2

)~ω as


the total energy for the simple harmonic oscillator.

We will need the following three facts before solving the eigenvalue problem for N .

Fact 7.1 Each eigenvalue n of the operator N is nonnegative.

ProofSuppose N |n⟩ = n |n⟩. Then,

⟨n|N |n⟩ = n ⟨n|n⟩ . (7.153)

On the other hand,

⟨n|N |n⟩ = ⟨n|a†a|n⟩ = ∥a |n⟩ ∥2. (7.154)

Therefore,

n ⟨n|n⟩ = ∥a |n⟩ ∥2 =⇒ n =∥a |n⟩ ∥2

⟨n|n⟩≥ 0. (7.155)

Fact 7.2 If the eigenvalue n = 0 for N , a |0⟩ = 0. Conversely, if a |v⟩ = 0,N |v⟩ = 0 |v⟩. For other values of n, i.e. n > 0 from Fact 7.1, a |n⟩ is an eigenvectorof N with the eigenvalue n− 1.

ProofSuppose N |0⟩ = 0 |0⟩. Then,

∥a |0⟩ ∥2 = ⟨0|a†a|0⟩ = ⟨0|N |0⟩ = ⟨0|0|0⟩ = 0 =⇒ a |0⟩ = 0. (7.156)

Now suppose a |v⟩ = 0, then,

a |v⟩ = 0 =⇒ a†a |v⟩ = N |v⟩ = 0 |v⟩ . (7.157)

So, any nonzero vector |v⟩ with a |v⟩ = 0 is an eigenvector of N with the eigenvaluen = 0.Next, consider a strictly positive eigenvalue n of the operator N . Then, from (7.149),

[N, a] |n⟩ = Na |n⟩ − aN |n⟩ = Na |n⟩ − an |n⟩ = Na |n⟩ − na |n⟩ = −a |n⟩


ONE DIMENSION

=⇒ N (a |n⟩) = na |n⟩ − a |n⟩ = (n− 1) (a |n⟩) . (7.158)

Therefore, a |n⟩ is an eigenvector of N with the eigenvalue n− 1.

Fact 7.3 If |n⟩ is an eigenvector of N with the eigenvalue n. Then, a† |n⟩ is aneigenvector of N with the eigenvalue n+ 1.

ProofWe will first show a† |n⟩ is nonzero. From (7.143),

∥a† |n⟩ ∥2 = ⟨n|aa†|n⟩ = ⟨n|[a, a†] + a†a|n⟩ = ⟨n|1 +N |n⟩= ⟨n|n⟩+ n ⟨n|n⟩ = (n+ 1)∥ |n⟩ ∥2. (7.159)

As |n⟩ is an eigenvector of N , it is nonzero, and a† |n⟩ is nonzero.Now from (7.143),

[N, a†] |n⟩ = Na† |n⟩ − a†N |n⟩ = Na† |n⟩ − a†n |n⟩ = Na† |n⟩ − na† |n⟩ = a† |n⟩=⇒ N

(a† |n⟩

)= (n+ 1)

(a† |n⟩

). (7.160)

Therefore, a† |n⟩ is an eigenvector of N with the eigenvalue n+ 1.

We are now ready to show that the eigenvalues of N are non-negative integers.Suppose an eigenvalue λ of N is not an integer, and let |λ⟩ denote an associatedeigenvector. From Fact 7.1, we already know that λ is strictly positive. From Fact7.2, we know that a |λ⟩ is an eigenvector of N with the eigenvalue λ − 1. Afterrepeated applications of a, on |λ⟩, k times, we get ak |λ⟩ which is an eigenvector ofN with the associated eigenvalue λ−k. As k can be as large as we wish it to be, thisimplies that N has strictly negative eigenvalues, contradicting Fact 7.1. Therefore,the eigenvalues of N are all nonnegative integers.

Consider an eigenvalue n, which is a positive integer, and an associated eigenvectordenoted by |n⟩. Then, due to Fact 7.2, ak |n⟩ > is an eigenvector for k = 1, 2, . . . , n.When k = n, the eigenvalue associated with an |n⟩ is 0. Then, from Fact 7.2,a (an |n⟩) = 0, terminating the chain of eigenvectors |n⟩ , a |n⟩ , a2 |n⟩ , . . . at k = n.So, an eigenvalue n of N can only be a nonnegative integer.


On the other hand, given an eigenvector |0⟩ of N , we can apply a† to the vector anynumber of times, with each application generating another eigenvector whose asso-ciated eigenvalue is greater than the previous eigenvalue by 1. This is guaranteed byFact 7.3. Therefore, the eigenvalues of N , H, and H are all nonnegative integers.

We have

En =(n+

1

2

)~ω for n = 0, 1, 2, . . . (7.161)

as before.

7.6.2.1 Actions of a and a†

Let |0⟩ , |1⟩ , |2⟩ , . . . be an orthonormal basis6 for the Hilbert space for H, H, orN . Recall that the eigenvectors are the same for these three operators. We wouldlike to find exactly what a and a† do to these basis vectors.

We already know

a |n⟩ = cn |n− 1⟩ (7.162)

for some constant cn. Taking the adjoint,

⟨n| a† = ⟨n− 1| c∗n. (7.163)

So,

⟨n|a†a|n⟩ = ⟨n− 1|n− 1⟩ c∗ncn = c∗

ncn =⇒ ⟨n|N |n⟩ = c∗ncn

=⇒ ⟨n|n|n⟩ = c∗ncn =⇒ n = |cn|2. (7.164)

We have

cn =√neiϕ. (7.165)

By convention, we normally set ϕ = 0. Then,

cn =√n and a |n⟩ =

√n |n− 1⟩ . (7.166)

6Because these are eigenvectors of a Hermitian operator H, they form a basis automatically.The only additional condition imposed here is normalization of each ket.


ONE DIMENSION

Similarly, we also know

a† |n⟩ = dn |n+ 1⟩ (7.167)

for some constant dn. Taking the adjoint,

⟨n| a = ⟨n+ 1| d∗n. (7.168)

This gives us

⟨n|aa†|n⟩ = ⟨n+ 1|d∗ndn|n+ 1⟩ . (7.169)

The left-hand side can be transformed as below using the relation (7.143).

⟨n|aa†|n⟩ = ⟨n|[a, a†] + a†a|n⟩ = ⟨n|1 +N |n⟩ = 1 + n (7.170)

Therefore,

|dn|2 = n+ 1. (7.171)

Setting the phase equal to 0 as before, we get

dn =√n+ 1 and a† |n⟩ =

√n+ 1 |n+ 1⟩ . (7.172)

We can represent the actions of a and a† as matrices with infinitely many entries.

a ↔

0√1 0 0 . . .

0 0√2 0 . . .

0 0 0√3 . . .

0 0 0 0 . . ....

......

.... . .

(7.173)

a† ↔

0 0 0 0 . . .√1 0 0 0 . . .

0√2 0 0 . . .

0 0√3 0 . . .

......

....... . .

(7.174)

I hope it is now clear to you why a is the lowering operator and a† is the raisingoperator7.

7Note that a and a† are also called destruction and creation operators, respectively.

161

Exercises1. This problem is the same as Chapter 5 Problem 5.

Consider a particle of mass m and total energy E which can move freely alongthe x-axis in the interval [−a

2,+a

2], but is strictly prohibited from going outside

this region. This corresponds to what is called an infinite square well potentialV (x) given by V (x) = 0 for −a

2< x < +a

2and V (x) = ∞ elsewhere. If we

solve the Schrodinger equation for this V (x), one of the solutions is

Ψ(x, t) =

A cos

(πxa

)e−iEt/~ −a

2< x < a

2

(a > 0)0 |x| ≥ a

2

.

(a) Find A so that the function Ψ(x, t) is properly normalized.(b) In the region where the potential V (x) = 0, the Schrodinger equation

reduces to− ~

2

2m

∂2

∂x2Ψ(x, t) = i~

∂Ψ(x, t)

∂t.

Plug Ψ(x, t) = A cos(πxa

)e−iEt/~ into this relation to show E = π2~2

2ma2.

(c) Show that ⟨P ⟩ = 0.(d) Evaluate ⟨x2⟩ for this wavefunction. You will probably need an integral

table for this. (Of course, you can always try contour integration or somesuch. But, that is beyond this course.)

(e) Evaluate ⟨P 2⟩ for the same wavefunction. You will not need an integraltable if you use the fact that the function is normalized. Of course, abrute force computation will yield the same result as well.

2. This problem is the same as Chapter 5 Problem 6.If you are a careful student who pays attention to details, you may have realizedthat A in Problem 1 can only be determined up to the argument, or up to thesign if you assume A is a real number.

(a) What is the implication of this fact as to the uniqueness of wavefunction?(b) What is the implication of this fact as to the probability density Ψ∗(x, t)Ψ(x, t)?

How about ⟨P ⟩, ⟨P 2⟩, and ⟨x2⟩?

162

3. Solve the time-independent Schrodinger equation for a free particle. (Note thata free particle is a particle with no force acting on it. In terms of the potentialV (x), this means that it is a constant. For simplicity, assume that V (x) = 0in this problem.)

4. Consider a particle of mass m which can move freely along the x-axis anywherefrom x = −a/2 to x = +a/2 for some a > 0, but is strictly confined to theregion [−a/2,+a/2]. This corresponds to what is called an infinite square wellpotential V (x) given by V (x) = 0 for −a/2 < x < +a/2 and V (x) = ∞elsewhere. Where the potential is infinite, the wavefunction is zero. Use thecontinuity condition on ψ(x) at |x| = a/2, but not on ψ′(x), to find at leasttwo of the solutions.

5. Give a rigorous mathematical proof that “H(x) = K(t) for all pairs of realnumbers (x, t)” implies that “H(x) and K(t) are a constant”.

6. Given a potential V (x) with the following profile

V (x) =

0 x < 0V0 x > 0

solve the Time-Independent Schrodinger Equation for a particle with the totalenergy E < V0 traveling from x = −∞ to x = +∞. In fact, we have alreadysolved this problem in this chapter. I just want you to review the procedure.

7. Show that we have total reflection in 6 above by direct computation usingprobability current described in Appendix J.

8. Consider the potential V (x) given by V (x) = 0 for x < 0 and V (x) = V0 > Efor x > 0, and show that we have a standing wave in the region x < 0 for aparticle incident on the discontinuity at x = 0 from the left, that is, from theregion characterized by negative values of x.

9. Consider the potential V (x) given by V (x) = 0 for x < 0 and V (x) = V0 < E.

(a) Study the complete solution of the time-independent Schrodinger equationwe solved in this chapter.

(b) Sow that the transmission coefficient is given by 4k1k2(k1+k2)2

.

163

10. Consider a step potential given by

V (x) =

0 x < 0V0 x > 0

.

(a) Write down, but do not derive or solve, the time-independent Schrodingerequation for a particle of mass m and total energy E > V0 for each region.

(b) For a particle moving toward the step from the left, the solutions are

ψ(x) = Aeik1x +Be−ik1x (x < 0)

andψ(x) = Ceik2x (x > 0),

where k1 =√2mE~

and k2 =√

2m(E−V0)~

. Use the two boundary conditionsat x = 0 to express C in terms of A, k1, and k2.

(c) As it turns out, we also get B = k1−k2k1+k2

A. Compute the reflection coefficientR; that is, express it in terms of k1 and k2.

11. In order to solve the Time-Independent Schrodinger Equation for a particleincident on the potential barrier given by

V (x) =

V0 > E 0 < x < a0 elsewhere

from the left, that is, x < 0 at t = 0, you need to consider the wavefunctiongiven by

ψ(x) =

Aeik1x + Be−ik1x x < 0Fe−k2x + Gek2x 0 < x < aCeik1x + De−ik1x x > a

; (7.175)

with the following set of boundary conditions.

ψ(x)|x=0− = ψ(x)|x=0+

ψ(x)|x=a− = ψ(x)|x=a+dψ(x)dx

∣∣∣x=0−

= dψ(x)dx

∣∣∣x=0+

dψ(x)dx

∣∣∣x=a−

= dψ(x)dx

∣∣∣x=a+

(7.176)

(a) What are k1 and k2 in (7.175)?

164

(b) Why should D in (7.175) be 0?(c) Write down the boundary conditions (7.176) after setting D = 0.

12. Consider a particle of energy E > V0 moving from x = +∞ to the left in thestep potential:

V (x) =

V0 x > 0

0 x < 0.

(a) Write the time-independent Schrodinger equations in the regions x < 0and x > 0.

(b) The general solutions are of the formA× (wave traveling to the right) +B × (wave traveling to the left) x < 0

C × (wave traveling to the right) +D × (wave traveling to the left) x > 0.

Complete the general solutions; i.e. substitute actual formulas for (wavetraveling to the right) and (wave traveling to the left). For simplicity ofnotation, let k1 =

√2mE~

and k2 =

√2m(E−V0)~

.(c) Determine the value of one of the constants based on the fact that there

is no wave coming back from x = −∞ and traveling to the right.(d) Use the boundary conditions at x = 0 to get the relations among the

constants.(e) Compute the reflection coefficient R.

13. Consider a barrier potential given by

V (x) =

0 (x < −a)V0 (−a < x < 0)0 (x > 0)

.

A stream of particles are originating at x = −∞ and traveling to the right.The total energy E is less than the barrier height V0.

(a) Write down, but do not derive or solve, the time-independent Schrodingerequation for a particle of mass m and total energy E < V0 for each region.

(b) For a particle moving toward the step from the left, the solutions are

ψ(x) =

Aeik1x +Be−ik1x (x < −a)Cek2x +De−k2x (−a < x < 0)Geik1x (x > 0)

165

where k1 =√2mE~

and k2 =

√2m(V0−E)

~.

i. Explain in words why we have ψ(x) = Geik1x and not ψ(x) = Geik1x+He−ik1x in the region x > 0.

ii. Give the two continuity coonditions at x = 0.iii. On physical grounds, we also need continuity of ψ∗ψ as well as its

first derivative at the boundaries. Use this continuity condition ondψ∗ψdx

at x = 0 to show neither C nor D is 0.

14. Show the following for the infinite square well potential described in the class.

V (x) =

0 −a

2< x < a

2

∞ elsewhere

(a) Two general solutions ψ(x) = A sin(kx) + B cos(kx) and ψ(x) = Ceikx +De−ikx are equivalent. Here, k =

√2mE/~, and A, B, C, and D are

arbitrary constants.(b) The usual boundary conditions at x = 0 and a allow only discrete values

of k, which in turn implies that the total energy is discretized.

15. The potential for the infinite square well is given by

V (x) =

∞ |x| > a

2

0 |x| < a2

,

and the solutions for the time-independent Schrodinger equation given in theclass were

ψn(x) = Bn cos knx where kn = nπa

n = 1, 3, 5, . . .ψn(x) = An sin knx where kn = nπ

an = 2, 4, 6, . . .

.

(a) Find a normalized solution for n = 2 for which the normalization constantA2 is purely imaginary.

(b) What is the expectation value of the position x for this solution? Use thefull wavefunction to solve this problem.

(c) Plug the above solution into the lefthand side of the time-independentSchrodinger equation to determine E2, the total energy associated withthis state.

166

16. The potential for the infinite square well is given by

V (x) =

∞ |x| > a

2

0 |x| < a2

.

(a) Write down, but do not derive or solve, the time-independent Schrodingerequation for a particle of mass m and total energy E in the region −a

2<

x < a2.

(b) Find a solution of the form ψ(x) = A sin(Bx) for −a2< x < a

2, where A

and B are constants, and compute the total energy following the stepsbelow.

i. Express B as a function of E so that ψ(x) = A sin(Bx) is a solutionof the Shcrodinger equation. (Choose the positive square root.)

ii. Find the smallest value of B that satisfies the boundary conditions.iii. Compute the total energy corresponding to the B above.iv. Determine A so that the wavefunction is properly normalized.

17. Consider an infinite square well whose potential is given by

V (x) =

∞ x < 00 0 < x < L∞ L < x

.

(a) Write down, but do not derive or solve, the time-independent Schrodingerequation for a particle of mass m and total energy E in the region 0 <x < L.

(b) The most general solution of the time-independent Schrodinger equationin the region 0 < x < L is of the form

ψ(x) = A sin kx+B cos kx;

where A and B are arbitrary constants and k =√2mE~

.i. Using one of the boundary conditions on ψ(x), determine the value

of B.ii. Find all possible positive values k can take by applying one of the

boundary conditions on ψ(x). Express k in terms of π, L, and anarbitrary natural number n. We will refer to each of these values askn. With this, we can write our wavefunction as

ψn(x) = An sin knx+Bn cos knxin order to make (potential) dependence on n more explicit.

167

iii. Impose the normalization condition to find the positive real value ofAn given the above kn.

(c) Now, use the relation

kn =

√2mEn~

to determine all possible values the total energy can take. In other words,express En in terms of m, h, L, and n.

18. Try a series solutionf(x) =

∞∑j=0

ajxj

for the differential equationdf(x)

dx+ f(x) = 3x2 + 8x+ 3,

and show thataj + (j + 1)aj+1 = 0 for j ≥ 3.

19. For a simple harmonic oscillator, show that the wavefunction corresponding ton = 1 is given by

ψ1 = A1ue−u2/2,

and the wavefunction for n = 2 takes the form

ψ2 = A2(1− 2u2)e−u2/2.

20. A time-independent ground state wavefunction of the simple harmonic oscil-lator is given by Φ0 = A0exp[−u2

2], where u = [(Cm)

14/~

12 ]x. Find the expec-

tation value of x as a function of the normalization constant A0. Use the fullwavefunction with the total energy E0.

21. This question concerns the simple harmonic oscillator.

(a) The full ground state wavefunction Ψ0 can be expressed as

ψ0e−iEt/~

with

ψ0 = A0e−u2/2;

where A0 is a normalization constant, and u =[(Cm)

14~− 1

2

]x. Compute

the expectation value < x > of the position for this state.

168

(b) Show by explicit computation that the first two wavefunctions of a simpleharmonic oscillator are orthogonal. Use the functions given below, whereA0 and A1 are normalization constants.

ψ0 = A0e−u2/2

ψ1 = A1ue−u2/2

(c) Why do you know the third and the fifth eigenfunctions, denoted by ψ2

and ψ4 as the first eigenfunction is ψ0, are orthogonal to each other withoutconducting explicit integration?

ψ2 = A2(1− 2u2)e−u2/2

ψ4 = A4(3− 12u2 + 4u4)e−u2/2

(d) Explain in fewer than 50 words why a solid made of diatomic molecules cannot be cooled to absolute zero. Here, I am only asking you to reproducethe crude argument I gave you in the class as incomplete as it was.

22. Suppose |n⟩ is an eigenket of a†a with the eigenvalue n. Show that a† |n⟩ is aneigenket of a†a with the eigenvalue n+ 1.

23. Describe, in a sentence or two, one of the differences between classical mechan-ics and quantum mechanics. Your answer should be about an actual physicalphenomenon or its interpretation. Hence, for example, the difference in formbetween the Schrodinger equation and the Newton’s equation is not an accept-able answer.

Chapter 8

Higher Spatial Dimensions

So far, we have only dealt with one dimensional cases partly for simplicity. However,most physical systems have multiple spatial dimensions, and there are a few newphenomena which you can observe only if you have more than one dimension. Wewill discuss two of such new features; degeneracy and angular momentum.

8.1 DegeneracyIn a typical dictionary, degeneracy is defined as a state of being degenerate, and theprimary or ordinary meanings of the adjective degenerate are “Having declined, asin function or nature, from a former or original state” and “Having fallen to an in-ferior or undesirable state, especially in mental or moral qualities.” But, degeneracyin quantum mechanics does not have much to do with declining or becoming infe-rior. According to Wikipedia [Wikipedia contributors, nd], degeneracy in quantummechanics has the following definition.

In quantum mechanics, a branch of physics, two or more different states of asystem are said to be degenerate if they are all at the same energy level. It isrepresented mathematically by the Hamiltonian for the system having morethan one linearly independent eigenstate with the same eigenvalue. Con-versely, an energy level is said to be degenerate if it contains two or moredifferent states. The number of different states at a particular energy levelis called the level’s degeneracy, and this phenomenon is generally known as aquantum degeneracy.From the perspective of quantum statistical mechanics, several degeneratestates at the same level are all equally probable of being filled.

169

170 CHAPTER 8. HIGHER SPATIAL DIMENSIONS

One example of this is the hydrogen atom discussed in Chapter 10. In this chapter,we will give a general proof that there is no degeneracy for “bound” states in onedimension, leaving discussions of degeneracy for specific cases to other chapters.

Definition 8.1 (Bound States) Bound states occur whenever the particle cannotmove to infinity. That is, the particle is confined or bound at all energies to movewithin a finite and limited region of space which is delimited by two classical turningpoints.

As the particle cannot move to infinity, we need to have

ψ(x −→ ±∞) −→ 0 (8.1)

for any bound state.

Theorem 8.1 There is no degeneracy for one-dimensional bound states.ProofSuppose there is degeneracy for total energy E. This means there are two unitvectors |u⟩ and |w⟩, representing distinct physical states, such that

H |u⟩ = E |u⟩ and H |w⟩ = E |w⟩ . (8.2)

Let x be the only position variable in this one-dimensional space. Then, writingψu(x) and ψw(x) for |u⟩ and |w⟩ respectively, we get

− ~2

2m

d2

dx2ψu(x) + V ψu(x) = Eψu(x) (1)

and

− ~2

2m

d2

dx2ψw(x) + V ψw(x) = Eψw(x). (2)

Multiplying (1) by ψw(x) and (2) by ψu(x) gives us(− ~

2

2m

d2

dx2ψu(x)

)ψw(x) + V ψu(x)ψw(x) = Eψu(x)ψw(x) (3)

and(− ~

2

2m

d2

dx2ψw(x)

)ψu(x) + V ψw(x)ψu(x) = Eψw(x)ψu(x). (4)

8.1. DEGENERACY 171

Now, subtracting (4) from (3) leads to

− ~2

2m

(d2

dx2ψu(x)

)ψw(x)−

(− ~

2

2m

)(d2

dx2ψw(x)

)ψu(x) = 0

=⇒(d2

dx2ψu(x)

)ψw(x)−

(d2

dx2ψw(x)

)ψu(x) = 0. (5)

Noting that (d2

dx2ψu(x)

)ψw(x)−

(d2

dx2ψw(x)

)ψu(x)

=

(d2

dx2ψu(x)

)ψw(x) +

d

dxψu(x)

d

dxψw(x)

− d

dxψu(x)

d

dxψw(x)−

(d2

dx2ψw(x)

)ψu(x)

=

(d2

dx2ψu(x)

)ψw(x) +

d

dxψu(x)

d

dxψw(x)

−(d

dxψu(x)

d

dxψw(x) +

(d2

dx2ψw(x)

)ψu(x)

)

=d

dx

[(d

dxψu(x)

)ψw(x)

]− d

dx

[(d

dxψw(x)

)ψu(x)

]

=d

dx

[(d

dxψu(x)

)ψw(x)−

(d

dxψw(x)

)ψu(x)

], (6)

(5) implies (d

dxψu(x)

)ψw(x)−

(d

dxψw(x)

)ψu(x) = c (7)

where c is some constant. Because ψu and ψw are bound states,

ψu(x)x→±∞−−−−−→ 0 and ψw(x)

x→±∞−−−−−→ 0. (8)Therefore, the constant c in (7) is 0.1 Naively, i.e. assuming ψu and ψw are positivefor now, we have

1

ψu(x)

d

dxψu(x) =

1

ψw(x)

d

dxψw(x) =⇒ d lnψu(x)

dx=d lnψw(x)

dx

1A more rigorous proof also requires an examination of the behavior of ddxψu(x) and d

dxψw(x)as x tends to ±∞. For now, just think of exponential functions e∓kx with k > 0 and their firstderivatives as x→ ±∞.


=⇒∫ d lnψu(x)

dxdx =

∫ d lnψw(x)dx

dx =⇒ lnψu(x) = lnψw(x) + k

=⇒ lnψu(x) = ln ekψw(x) =⇒ ψu(x) = ekψw(x). (9)

So, in this case, ψu(x) is a scalar multiple of ψw(x), and they represent the samephysical state. This in turn implies that there is no degeneracy. If you think thereis too much hand-waving in this “proof”, you are extremely right! I gave this naiveargument first in order to make this proof more accessible for, say, freshmen.

Having said that, a more rigorous proof can be given resorting to the Wronskianargument described in Appendix E. According to Appendix E, two solutions, ψu andψw, of a linear second-order ordinary differential equation of the form

ψ′′(x) + p(x)ψ′(x) + q(x)ψ(x) = 0 (10)

where p(x) and q(x) are continuous, are linearly dependent if and only if the Wron-skian, W (ψu, ψw), given by

W (ψu, ψw)(x) = ψu(x)ψ′w(x)− ψ′

u(x)ψw(x) (11)

is 0 for all x in the domain. As this is the case for us, we have ψu(x) = Kψw(x) forsome constant K, and this proves the theorem.

We have now shown that degeneracy, which requires two linearly independent eigen-functions with the same energy eigenvalue E, is a phenomenon specific to more thanone dimension.

In passing, let us make a note of the following theorem found on p.217 of “Quan-tum Mechanics: Concepts and Applications (second edition)” by Nouredine Zettili[Zettili, 2009, p.217], which includes discreteness of energy levels for bound states.

Theorem 8.2 (Discrete and Nondegenerate) In a one-dimensional problem theenergy levels of a bound state system are discrete and not degenerate.

The condition (8.1) together with Condition 3 (continuity conditions) on p.128 can besatisfied only for certain special values of energy, making allowed total energy valuesdiscrete. We saw how this works repeatedly in Chapter 7. Zettili also provides thefollowing theorem.

Theorem 8.3 The wave function ψn(x) of a one-dimensional bound state systemhas n nodes (i.e., ψn(x) vanishes n times) if n = 0 corresponds to the ground stateand (n− 1) nodes if n = 1 corresponds to the ground state.

8.2. ANGULAR MOMENTUM 173

Remark 8.1 (Unbound One-Dimensional States) Note that we do have de-generacy if we have unbound states. A good example is a free particle traveling wavewhere we have

ei(kx−ωt) and e−i(kx+ωt) (8.12)

according to (7.4).

These are linearly independent states corresponding to particles/waves traveling tothe right ei(kx−ωt) and to the left e−i(kx+ωt). Because k =

√2mE~

for both waves, theyhave the same energy, and we have two-fold degeneracy.

8.2 Angular MomentumThe only momentum we encounter in a one-dimensional system is the linear mo-mentum p = mv as we only have a linear motion in one-dimension. However, inhigher dimensions, rotational motions are also possible, and we have another varietyof momentum called the angular momentum. The classical definition of the angularmomentum L of a particle of mass m with respect to a chosen origin is given by

L = r × p; (8.13)

where r is the position vector connecting the origin and the particle, p = mv is thelinear momentum, and × is the usual vector cross product. If we denote the angleformed by r and p by θ, we have

∥L∥ = ∥p∥∥r∥ sin θ = m∥v∥∥r∥ sin θ or L = mvr sin θ.a (8.14)

aWe can choose either angle for this purpose as sin θ = sin (π − θ).

The angular momentum is subject to the fundamental constraints of the conser-vation of angular momentum principle if there is no external torque, rotation-causingforce, on the object. Using the usual substitutions


xi ←→Mxiand pi ←→ −i~

∂

∂xi; (8.15)

where x1 = x, x2 = y, x3 = z, and likewise for pi’s, we can “derive” quantummechanical angular momentum operators Lx, Ly, and Lz drawing on the classicalrelation (8.13).

Because the study of the angular momentum is a large and thriving industry,we will devote the entire Chapter 9 to a detailed discussion of quantum mechanicalangular momentum.

Chapter 9

Angular Momentum

As we move from one-dimensional systems to higher dimensionality, more and moreinteresting physics emerges. In this chapter, we will discuss how angular momentumis dealt with in quantum mechanics and the consequences of quantum mechanicalcomputations of angular momenta. The angular momentum relations are very im-portant in the study of atoms and nuclei. Like the total energy E, the magnitudeof angular momentum, usually denoted by L, as well as its z-component, Lz areconserved. Furthermore, angular momentum is quantized like energy levels.

9.1 Angular Momentum OperatorsQuantum mechanical operators representing the x-, y-, and z-component of angularmomentum were already presented in Table 3.2 without derivation. Now, we willsee how we can “derive” quantum mechanical counterpart drawing on the classicaltheory of angular momentum.

The angular momentum of a particle about the origin is given by

L = r × p =⇒ lxı+ ly ȷ+ lzk =

∣∣∣∣∣∣∣ı ȷ kx y zpx py pz

∣∣∣∣∣∣∣ =⇒

lxlylz

=

ypz − zpyzpx − xpzxpy − ypx

; (9.1)

where the expression L = r × p is coordinate-independent, but the component-by-component expressions on the right are in Cartesian coordinates. The quantummechanical operator L = Lxı+ Ly ȷ+ Lzk is obtained as follows.

175

176 CHAPTER 9. ANGULAR MOMENTUM

First, noting that px ↔ −i~ ∂∂x , py ↔ −i~ ∂∂y , and pz ↔ −i~ ∂∂z , we get:

Lx = y

(−i~ ∂

∂z

)− z

(−i~ ∂

∂y

)= −i~

(y∂

∂z− z ∂

∂y

)

Ly = z

(−i~ ∂

∂x

)− x

(−i~ ∂

∂z

)= −i~

(z∂

∂x− x ∂

∂z

)(9.2)

Lz = x

(−i~ ∂

∂y

)− y

(−i~ ∂

∂x

)= −i~

(x∂

∂y− y ∂

∂x

)

This book culminates in a full solution of the Schrodinger equation for the hydro-gen atom; where we only have a central force, and the spherical coordinate systemis employed to simplify the computation. In spherical coordinates, we get:

Lx = i~

(sinϕ ∂

∂θ+ cot θ cosϕ ∂

∂ϕ

)

Ly = i~

(− cosϕ ∂

∂θ+ cot θ sinϕ ∂

∂ϕ

)(9.3)

Lz = −i~∂

∂ϕ.

Note that Lz is particularly simple. As for the magnitude of angular momentum,we will work with the squared magnitude, denoted by L2.

L2 = L2x + L2

y + L2z = −~2

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

](9.4)

We have the following commutation relations.

[Lx, Ly] = i~Lz (9.5)[Ly, Lz] = i~Lx (9.6)[Lz, Lx] = i~Ly (9.7)

[L2, Lx] = [L2, Ly] = [L2, Lz] = 0 (9.8)

The relations (9.5), (9.6), and (9.7) can be summarized as

L×L = i~L. (9.9)

9.1. ANGULAR MOMENTUM OPERATORS 177

Let us verify (9.5) only, as (9.6) and (9.7) can be verified in the same manner. We willverify (9.5) in three different ways for the sake of comparison. The first verificationuses (9.2). For simplicity of notation, we will denote ∂

∂x, ∂∂y

, and ∂∂z

by ∂x, ∂y, ∂z,respectively. You might as well familiarize yourself with this notational scheme atthis point, unless you are already comfortable with it, because it is widely used inphysics and engineering.

[Lx, Ly] = (−i~)2 [(y∂z − z∂y) (z∂x − x∂z)− (z∂x − x∂z) (y∂z − z∂y)]= (−i~)2 [y∂zz∂x − y∂zx∂z − z∂yz∂x + z∂yx∂z

− (z∂xy∂z − z∂xz∂y − x∂zy∂z + x∂zz∂y)]

= (−i~)2(y∂x + yz∂z∂x − yx∂2z − z2∂y∂x + zx∂y∂z − zy∂x∂z

+z2∂x∂y + xy∂2z − x∂y − xz∂z∂y)

= i~(−i~) (x∂y − y∂x) = i~(−i~)(x∂

∂y− y ∂

∂x

)= i~Lz (9.10)

Next, we will verify (9.5) in spherical coordinates.

[Lx, Ly] = LxLy − LyLx= i~ (sinϕ∂θ + cot θ cosϕ∂ϕ) i~ (− cosϕ∂θ + cot θ sinϕ∂ϕ)− i~ (− cosϕ∂θ + cot θ sinϕ∂ϕ) i~ (sinϕ∂θ + cot θ cosϕ∂ϕ)

= (i~)2 [sinϕ∂θ (− cosϕ∂θ) + sinϕ∂θ (cot θ sinϕ∂ϕ)+ cot θ cosϕ∂ϕ (− cosϕ∂θ) + cot θ cosϕ∂ϕ (cot θ sinϕ∂ϕ)]− (i~)2 [− cosϕ∂θ(sinϕ∂θ)− cosϕ∂θ(cot θ cosϕ∂ϕ)+ cot θ sinϕ∂ϕ(sinϕ∂θ) + cot θ sinϕ∂ϕ(cot θ cosϕ∂ϕ)]

= (i~)2(− sinϕ cosϕ∂2θ + sin2 ϕ

−1sin2 θ

∂ϕ + sin2 ϕ cot θ∂θ∂ϕ

+ cot θ cosϕ sinϕ∂θ − cot θ cos2 ϕ∂ϕ∂θ + cot2 θ cos2 ϕ∂ϕ+ cot2 θ sinϕ cosϕ∂2ϕ

)− (i~)2

(− sinϕ cosϕ∂2θ − cos2 ϕ −1sin2 θ

∂ϕ − cos2 ϕ cot θ∂θ∂ϕ

+ cot θ sinϕ cosϕ∂θ + sin2 ϕ cot θ∂ϕ∂θ − cot2 θ sin2 ϕ∂ϕ

+ cot2 θ sinϕ cosϕ∂2ϕ)

= (i~)2(

sin2 ϕ−1

sin2 θ∂ϕ + cot2 θ cos2 ϕ∂ϕ + cos2 ϕ −1sin2 θ

∂ϕ

+ cot2 θ sin2 ϕ∂ϕ)


= (i~)2( −1

sin2 θ∂ϕ + cot2 θ∂ϕ

)= i~(−i~)

(1

sin2 θ− cot2 θ

)∂ϕ

= i~(−i~)1− cos2 θsin2 θ

∂ϕ = i~(−i~) ∂∂ϕ

= i~Lz (9.11)

Finally, we will use the commutator identity (I.5).

[Lx, Ly] = [ypz − zpy, zpx − xpz]= [ypz, zpx]− [ypz, xpz]− [zpy, zpx] + [zpy, xpz]

= y[pz, z]px + z[y, px]pz + zy[pz, px] + [y, z]pzpx

− (y[pz, x]pz + x[y, pz]pz + xy[pz, pz] + [y, x]pzpz)

− (z[py, z]px + z[z, px]py + zz[py, px] + [z, z]pypx)

+ z[py, x]pz + x[z, pz]py + xz[py, pz] + [z, x]pypz

= y(−i~I)px + z · 0 · pz + zy · 0 + 0 · pzpx− (y · 0 · pz + x · 0 · pz + xy · 0 + 0 · pzpz)− (z · 0 · px + z · 0 · py + zz · 0 + 0 · pypx)+ z · 0 · pz + x(i~I)py + xz · 0 + 0 · pypz

= i~(xpy − ypx) = i~Lz (9.12)

Now, we will verify (9.8), using (I.4) of Appendix I; namely, [AB,C] = [A,C]B +A[B,C].

[L2, Lx] = [L2x + L2

y + L2z, Lx] = [L2

x, Lx] + [L2y, Lx] + [L2

z, Lx]

= 0 + [Ly, Lx]Ly + Ly[Ly, Lx] + [Lz, Lx]Lz + Lz[Lz, Lx]

= −i~LzLy − i~LyLz + i~LyLz + i~LzLy = 0 (9.13)

Likewise for Ly and Lz.

9.2 Quantum Mechanical Rotation Operator URLet R(ϕk) be a physical rotation, a counterclockwise rotation around the z-axis byϕ, and UR be the quantum mechanical operator associated with the physical rotationR. Then, the action of UR is characterized by

UR |x, y, z⟩ = |x cosϕ− y sinϕ, x sinϕ+ y cosϕ, z⟩ (9.14)

In order to derive UR, we will first consider an infinitesimal physical rotation

9.2. QUANTUM MECHANICAL ROTATION OPERATOR UR 179

R(εk) and its quantum mechanical counterpart URε given to first order in ε by

URε = I + εΩ. (9.15)

As it turns out, the correct form is

URε = I − iε

~Lz; (9.16)

where Lz is the quantum mechanical angular momentum operator around the z-axis(9.2). We can convince ourselves that (9.16) is indeed the desired quantum mechan-ical operator as follows. Note that this is not a very rigorous proof, but a roughsketch of how it should be shown.

By definition, we have

URε |x, y, z⟩ = |x− yε, y + xε, z⟩ . (9.17)

It follows from (9.17) that

URε |ψ⟩ = URε

+∞$−∞

|x, y, z⟩ ⟨x, y, z|ψ⟩ dxdydz

=

+∞$−∞

URε |x, y, z⟩ ⟨x, y, z|ψ⟩ dxdydz

=

+∞$−∞

|x− yε, y + xε, z⟩ ⟨x, y, z|ψ⟩ dxdydz. (9.18)

Now, let x′ = x− yε, y′ = y + xε, and z′ = z. Then, the Jacobian is∣∣∣∣∣∣∣∣∂x′

∂x∂y′

∂x∂z′

∂x∂x′

∂y∂y′

∂y∂z′

∂y∂x′

∂z∂y′

∂z∂z′

∂z

∣∣∣∣∣∣∣∣−1

=

∣∣∣∣∣∣∣1 ε 0−ε 1 00 0 1

∣∣∣∣∣∣∣−1

=1

1 + ε2≈ 1− ε2 = 1 (9.19)


to first order in ε. So,

URε |ψ⟩ =+∞$

−∞

|x′, y′, z′⟩ ⟨x, y, z′|ψ⟩ 1

1 + ε2dx′dy′dz′

=

+∞$−∞

|x′, y′, z′⟩ ⟨x, y, z′|ψ⟩ dx′dy′dz′

=

+∞$−∞

|x′, y′, z′⟩ ⟨x′ + yε, y′ − xε, z′|ψ⟩ dx′dy′dz′ (9.20)

to first order in ε. Hence,

⟨x, y, z|URε |ψ⟩ = ⟨x, y, z|+∞$

−∞

|x′, y′, z′⟩ ⟨x′ + yε, y′ − xε, z′|ψ⟩ dx′dy′dz′

=

+∞$−∞

⟨x, y, z|x′, y′, z′⟩ ⟨x′ + yε, y′ − xε, z′|ψ⟩ dx′dy′dz′

=

+∞$−∞

δ(x− x′)δ(y − y′)δ(z − z′) ⟨x′ + yε, y′ − xε, z′|ψ⟩ dx′dy′dz′

= ⟨x+ yε, y − xε, z|ψ⟩ . (9.21)

So, we have

=⇒ ⟨x, y, z|URε |ψ⟩ = (URε |ψ⟩) (x, y, z) = ψ(x+ yε, y − xϵ, z). (9.22)

Expanding ψ(x+ yε, y − xε, z) in a Taylor series to order ε, we obtain

ψ(x+ yε, y − xε, z) = ψ(x, y, z) +∂ψ

∂x(yε) +

∂ψ

∂y(−xε). (9.23)

On the other hand,

⟨x, y, z|URε |ψ⟩ = ⟨x, y, z|I + εΩ|ψ⟩ = ⟨x, y, z|ψ⟩+ ε ⟨x, y, z|Ω|ψ⟩= ψ(x, y, z) + ε ⟨x, y, z|Ω|ψ⟩ . (9.24)

9.2. QUANTUM MECHANICAL ROTATION OPERATOR UR 181

Therefore,

⟨x, y, z|URε |ψ⟩ = ψ(x+ yε, y − xϵ, z)

=⇒ ψ(x, y, z) + ε ⟨x, y, z|Ω|ψ⟩ = ψ(x, y, z) +∂ψ

∂x(yε) +

∂ψ

∂y(−xε)

=⇒ ε ⟨x, y, z|Ω|ψ⟩ = ε

[y∂ψ

∂x+ (−x)∂ψ

∂y

]

=⇒ (Ω |ψ⟩) (x, y, z) =[(y∂

∂x− x ∂

∂y

)ψ

](x, y, z)

=⇒ Ω = y∂

∂x− x ∂

∂y=⇒ i~Ω = i~

(y∂

∂x− x ∂

∂y

)

=⇒ −i~(x∂

∂y− y ∂

∂x

)= Lz =⇒ Ω =

Lzi~

=−i~Lz (9.25)

This verifies (9.16).

URε = I − iε

~Lz (9.26)

The quantum mechanical operator UR for a finite rotation R(ϕ0k) can be constructedfrom (9.16) as follows.

UR = limN→∞

(I − i

~

ϕ0

NLz

)N= exp (−iϕ0Lz/~) (9.27)

The exponential here is to be regarded as a convergent power series.

exp (−iϕ0Lz/~) =∞∑n=0

(−iϕ0/~)n

n!Lnz (9.28)

Now, from (9.3),

Lz = −i~∂

∂ϕ. (9.29)

So, in spherical coordinates,

UR = exp (−iϕ0Lz/~) = exp(−iϕ0

(−i~ ∂

∂ϕ

)/~

)= exp

(−ϕ0

∂

∂ϕ

). (9.30)


9.3 Rotationally Invariant HamiltonianSuppose the only force in the system is a central force F (r), which derives from acentral potential V (r) where r is the distance from the origin, so that

F = −∇V (r) = −dV (r)

drr. (9.31)

Then, the Hamiltonian

H =p2

2m+ V (r) (9.32)

is invariant under arbitrary rotations, and it commutes with Lx, Ly, Lz, and L2.From (I.20) and (I.21), we have the following.

[H,Lx] = [H,Ly] = [H,Lz] = 0 and [H,L2] = 0 (9.33)Therefore, L2, Lx, Ly, and Lz are constants of motion and conserved.1 We know fromTheorem 2.9 that commuting Hermitian operators Ω and Γ have common eigenvec-tors which diagonalize Ω and Γ simultaneously.

Because Lx, Ly, and Lz do not commute with each other, we cannot diagonalizeH, L2, Lx, Ly, and Lz simultaneously. The best we can do is to diagonalize H, L2,and only one of the components of L. It is customary to choose Lz. So, we will diag-onalize H, L2, and Lz simultaneously 2, which amounts to finding all eigenfunctionsψ(r, θ, ϕ) which are, by definition, stationary under H, L2, and Lz.

The explicit forms of the eigenfunctions will be found in Chapter 10. But, thereis a way to solve the eigenvalue problem of L2 and Lz using raising and loweringoperators as for the simple harmonic oscillator discussed in Section 7.6.2.

1 We can show that dLdt = 0 if V = V (r) within the framework of classical mechanics as well.

Namely, we have

L = r × p =⇒ dL

dt=dr

dt× p + r × dp

dt=dr

dt× p + r ×mdv

dt= v × p + r ×mα = v × p + r × F

= v ×mv + r ×(−dV (r)

dr

r

∥r∥

)= m(v × v)− dV (r)

dr

1

∥r∥(r × r) = 0. (9.34)

2 We can draw a contradiction if we assume there is a common eigenstate |l⟩ for Lz and Lx suchthat

Lx |l⟩ = lx |l⟩ and Lz |l⟩ = lz |l⟩ .

9.4. RAISING AND LOWERING OPERATORS: L+ AND L− 183

9.4 Raising and Lowering Operators: L+ and L−

In this section, we will find the eigenvalues of L2 and Lz without finding the eigen-functions explicitly. Looking ahead, we will denote the eigenvalues of L2 by ~2l(l+1)and those of Lz by ml~. We will write |l,ml⟩ for the orthonormal eigenstates withassociated eigenvalues ~2l(l + 1) and ml~.

L2 |l,ml⟩ = ~2l(l + 1) |l,ml⟩Lz |l,ml⟩ = ml~ |l,ml⟩ (9.35)⟨l,ml|l′m′

l⟩ = δll′δml,m′l

We will show that l is a nonnegative integer and ml = −l,−l+1, . . . , 0, . . . , l− 1, l.The number l is referred to as an orbital quantum number, and ml is called either amagnetic3 quantum number or an azimuthal4 quantum number. We first define theraising and lowering operators as follows. The reason for the naming will becomeclear shortly.

L+ = Lx + iLy (9.36)L− = Lx − iLy (9.37)

Because Lx and Ly are Hermitian, it follows that

L†+ = (Lx + iLy)

† = L†x − iL†

y = Lx − iLy = L− (9.38)

This implies

[Lz, Lx] |l⟩ = i~Ly |l⟩ = LzLx |l⟩ − LxLz |l⟩ = lzlx |l⟩ − lxlz |l⟩ = 0.

This in turn implies

[Lx, Ly] |l⟩ = i~Lz |l⟩ = LxLy |l⟩ − LyLx |l⟩ = 0− Ly(lx |l⟩) = 0− 0 = 0.

We now haveLy |l⟩ = Lz |l⟩ = 0 =⇒ Lx |l⟩ = 0.

Therefore, only possible common states are those for zero angular momentum. Needless to say, wecannot find a set of orthonormal eigenstates that simultaneously diagonalizes Lx, Ly, and Lz.

3As we will see later, the energy levels among the states with different ml’s are the same in theabsence of an external magnetic field.

4It is so called as the angle θ in spherical coordinate system is called an azimuth.


andL†

− = (Lx − iLy)† = L†x + iL†

y = Lx + iLy = L+. (9.39)

Incidentally, using (9.3), we can express L+ and L− in spherical coordinates.

L± = ~e±iϕ(± ∂

∂θ+ i cot θ ∂

∂ϕ

)(9.40)

As we already know L2 commutes with Lx and Ly, we have

[L2, L+] = [L2, L−] = 0. (9.41)

Let us also compute other commutators.

[L+, L−] = [Lx + iLy, Lx − iLy] = [Lx, Lx] + [Lx,−iLy]+ [iLy, Lx] + [iLy,−iLy]

= 0 + (−i)i~Lz + i(−i~Lz) + 0 = 2~Lz (9.42)

[L+, Lz] = [Lx + iLy, Lz] = [Lx, Lz] + i[Ly, Lz] = −i~Ly + i(i~Lx)

= −~Lx − ~Lx = −~(Lx + iLy) = −~L+ (9.43)

[L−, Lz] = [Lx − iLy, Lz] = [Lx, Lz]− i[Ly, Lz] = −i~Ly − i(i~Lx)= ~Lx − i~Ly = ~(Lx − iLy) = ~L− (9.44)

We also have the following.

L+L− = (Lx + iLy)(Lx − iLy) = L2x − iLxLy + iLyLx + L2

y

= L2 − L2z − i[Lx, Ly] = L2 − L2

z − i(i~Lz) = L2 − L2z + ~Lz (9.45)

L−L+ = (Lx − iLy)(Lx + iLy) = L2x + iLxLy − iLyLx + L2

y

= L2 − L2z + i[Lx, Ly] = L2 − L2

z + i(i~Lz) = L2 − L2z − ~Lz (9.46)

Because the operators Lx, Ly, and Lz are all Hermitian,

⟨l,ml|L2|l,ml⟩ = ⟨l,m|L2x + L2

y + L2z|l,ml⟩

= ⟨l,m|L2x|l,ml⟩+ ⟨l,m|L2

y|l,ml⟩+ ⟨l,m|L2z|l,ml⟩

= ⟨l,m|L†xLx|l,ml⟩+ ⟨l,m|L†

yLy|l,ml⟩+ ⟨l,m|L†zLz|l,ml⟩


= ⟨Lx(l,ml)|Lx(l,ml)⟩+ ⟨Ly(l,ml)|Ly(l,ml)⟩ (9.47)+ ⟨Lz(l,ml)|Lz(l,ml)⟩

= ∥ |Lx(l,ml)⟩ ∥2 + ∥ |Ly(l,ml)⟩ ∥2 + ∥ |Lz(l,ml)⟩ ∥2 ≥ 0

On the other hand,

⟨l,ml|L2|l,ml⟩ = ~2l(l + 1) ⟨l,ml|l,ml⟩ = ~2l(l + 1). (9.48)

Therefore,

~2l(l + 1) ≥ 0 =⇒ l ≥ 0 or l ≤ −1. (9.49)

Because the parabola y = x(x + 1) is symmetric about the axis x = −12, we can

choose l ≥ 0 and still get all possible nonnegative values l(l + 1) can take.

Now, let

|+⟩ := L+ |l,ml⟩ ; (9.50)

where |l,ml⟩ is a normalized eigenstate of L2 with the eigenvalue ~2l(l + 1) and alsoa normalized eigenstate of Lz with the eigenvalue ml from (9.35). Then,

L2 |+⟩ = L2L+ |l,ml⟩ = L+L2 |l,ml⟩ = L+

(~2l(l + 1) |l,ml⟩

)= ~2l(l + 1)L+ |l,ml⟩ = ~2l(l + 1) |+⟩ . (9.51)

Hence, |+⟩ is still an eigenstate of L2 with the same eigenvalue ~2l(l + 1). On theother hand, due to (9.35) and (9.43),

Lz |+⟩ = LzL+ |l,ml⟩ = ([Lz, L+] + L+Lz) |l,ml⟩= (~L+ +ml~L+) |l,ml⟩ = (ml + 1)L+ |l,ml⟩ = (ml + 1)~ |+⟩ . (9.52)

Hence, |+⟩ is still an eigenstate of Lz but with an eigenvalue which is ~ greater thanthe original ml~. Likewise, if we let

|−⟩ := L− |l,ml⟩ , (9.53)

we get


L2 |−⟩ = L2L− |l,ml⟩ = L−L2 |l,ml⟩ = L−

(~2l(l + 1) |l,ml⟩

)= ~2l(l + 1)L− |l,ml⟩ = ~2l(l + 1) |−⟩ (9.54)

and

Lz |−⟩ = LzL− |l,ml⟩ = ([Lz, L−] + L−Lz) |l,ml⟩= (−~L− +ml~L−) |l,ml⟩ = (ml − 1)L− |l,ml⟩

= (ml − 1)~ |−⟩ . (9.55)

This shows that |−⟩ is an eigenstate of L2 with an eigenvalue ~2l(l + 1) and is aneigenstate of Lz with an eigenvalue that is ~ smaller than the original ml~. In sum,we have shown that L+/L− raises/lowers the magnetic quantum number ml by one,but preserves the total angular momentum l. In other words, L+ |l,ml⟩ is propor-tional to |l,ml + 1⟩, and L− |l,ml⟩ is proportional to |l,ml − 1⟩.

Let us now compute the square of the norm for |+⟩ and |−⟩, respectively.

∥ |+⟩ ∥2 = ⟨+|+⟩ = ⟨L+(l,ml)|L+(l,ml)⟩= ⟨l,ml|L−L+|l,ml⟩ = ⟨l,ml|L2 − L2

z − ~Lz|l,ml⟩=(~2l(l + 1)− ~2m2

l − ~2ml

)⟨l,ml|l,ml⟩ = ~2(l(l + 1)−ml(ml + 1))

= ~2(l2 + l −m2l −ml) = ~

2((l +ml)(l −ml) + (l −ml))

= ~2(l −ml)(l +ml + 1) (9.56)

∥ |−⟩ ∥2 = ⟨−|−⟩ = ⟨L−(l,ml)|L−(l,ml)⟩= ⟨l,ml|L+L−|l,ml⟩ = ⟨l,ml|L2 − L2

z + ~Lz|l,ml⟩=(~2l(l + 1)− ~2m2

l + ~2ml

)⟨l,ml|l,ml⟩ = ~2(l(l + 1)−ml(ml − 1))

= ~2(l2 + l −m2l +ml) = ~

2((l +ml)(l −ml) + (l +ml))

= ~2(l +ml)(l −ml + 1) (9.57)

Because we have to have ∥ |+⟩ ∥2 ≥ 0 and ∥ |−⟩ ∥2 ≥ 0, from (9.56) and (9.57), wecan see that

~2(l −ml)(l +ml + 1) ≥ 0 (9.58)and


~2(l +ml)(l −ml + 1) ≥ 0. (9.59)

The inequality (9.58) requires either

I. l ≥ ml and l ≥ −ml − 1

or

II. l ≤ ml and l ≤ −ml − 1;

while the other inequality (9.59)implies either

III. l ≥ −ml and l ≥ ml − 1

or

IV. l ≤ −ml and l ≤ ml − 1.

Therefore, we should have either condition I or condition II holding true, and at thesame time, one of conditions III and IV to be true.

First, suppose ml ≥ 0. As specified on p.185, we chose l ≥ 0. If ml is nonnegative,l ≤ −ml − 1 is impossible. Hence, condition I should be satisfied. In particular, weneed l ≥ ml as l ≥ −ml−1 is automatically satisfied in this case. Consider conditionsIII and IV next. Condition III reduces to l ≥ ml− 1, and condition IV is impossiblebecause l ≤ −ml and l ≤ ml − 1 implies 2l ≤ −1 which contradicts l ≥ 0. So, wehave to satisfy l ≥ ml and l ≥ ml − 1 simultaneously. Therefore,

ml ≤ l if ml ≥ 0. (9.60)

Next, suppose ml < 0. Condition I reduces to l ≥ −ml − 1, and condition II isimpossible as l > 0 > ml. Now, condition III reduces to l ≥ −ml as l ≥ ml − 1 isautomatically satisfied, and condition IV is impossible because l ≥ 0 > ml > ml− 1.Hence, we require l ≥ −ml − 1 and l ≥ −ml. Therefore,

ml ≥ −l if ml < 0. (9.61)

The inequalities (9.60) and (9.61) together with the fact that L+/L− raises/lowersthe magnetic quantum number ml by one imply


ml = −l,−l + 1, . . . , l − 1,+l ; where l ≥ 0. (9.62)

As the spacing for each pair of neighboring ml values is 1, it is necessary thatl − (−l) = 2l is a nonnegative integer. But, this is possible only for an integralor half-integral l. It turns out l is an integer for the orbital angular momentum5.

Before closing this section, we will briefly discuss a simple implication of (9.51),(9.52), and (9.56), which will be useful later. A similar result follows from (9.54),(9.55), and (9.57). From (9.51), we can see that L+ |l,ml⟩ is an eigenvector of L2

with the eigenvalue ~2l(l + 1). Furthermore, (9.52) indicates that L+ |l,ml⟩ is aneigenvector of Lz with the eigenvalue (ml + 1)~. Hence, we have

L+ |l,ml⟩ = C |l,ml + 1⟩ (9.63)

for some scalar C. However, we already know from (9.56) that

∥L+ |l,ml⟩ ∥2 = ⟨l,ml + 1|C∗C|l,ml + 1⟩ = |C|2

= ~2(l −ml)(l +ml + 1), (9.64)

and so,

C = eiθ~√(l −ml)(l +ml + 1). (9.65)

According to standard convention, we choose θ = 0 or eiθ = 1 [Shankar, 1980, p.336]to get

C = ~√(l −ml)(l +ml + 1) =⇒

L+ |l,ml⟩ = ~√(l −ml)(l +ml + 1) |l,ml + 1⟩ . (9.66)

Similarly, we can also show

L− |l,ml⟩ = ~√(l +ml)(l −ml + 1) |l,ml − 1⟩ . (9.67)

Finally, noting that |l,−l⟩, |l,−l + 1⟩, . . . , |l, l − 1⟩, and |l, l⟩ are the only eigen-vectors of Lz from (9.62), we have to have

5Half-integral l’s correspond to an internal or intrinsic angular momentum called the spin of aparticle. We will discuss electron spin in Chapter 11.

9.5. MATRIX REPRESENTATIONS OF L2 AND LZ 189

L+ |l, l⟩ = 0 (9.68)and

L− |l,−l⟩ = 0 (9.69)

as in Section 7.6.2.

9.5 Matrix Representations of L2 and Lz

In the infinite-dimensional basis of the orthonormal eigenstates |l,ml⟩, where l =0, 1, 2, 3, . . . and ml = −l,−l + 1,−l + 2, . . . , 0, . . . , l − 2, l − 1, l for each l, L2 andLz are simultaneously diagonal. Needless to say, the matrices have infinitely manyrows and columns.

L2 =

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . . .

0 2~2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . . .

0 0 2~2 0 0 0 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 2~2 0 0 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 0 6~2 0 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 0 0 6~2 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 0 0 0 6~2 0 0 0 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 6~2 0 0 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 6~2 0 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 12~2 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 0 12~2 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 0 0 12~2 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 0 0 0 12~2 0 0 0 . . .

0 0 0 0 0 0 0 0 0 0 0 0 0 12~2 0 0 . . .

0 0 0 0 0 0 0 0 0 0 0 0 0 0 12~2 0 . . .

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 12~2 . . .

.

.....

.

.....

.

.....

.

.....

.

.....

.

.....

.

.....

.

.....

. . .

(9.70)


Lz =

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . . .

0 −~ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 +~ 0 0 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 0 −2~ 0 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 0 0 −~ 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 +~ 0 0 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 +2~ 0 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 −3~ 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 0 −2~ 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 0 0 −~ 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . . .

0 0 0 0 0 0 0 0 0 0 0 0 0 +~ 0 0 . . .

0 0 0 0 0 0 0 0 0 0 0 0 0 0 +2~ 0 . . .

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 +3~ . . .

.

.....

.

.....

.

.....

.

.....

.

.....

.

.....

.

.....

.

.....

. . .

(9.71)

The diagonal blocks correspond to l = 0, l = 1, l = 2, . . . starting from the trivial1-by-1 block in the upper left-hand corner. For L2, the diagonal entries for the l-thblock are the same and given by l(l + 1)~2, while they range from −l~ to +l~ forLz. Here, L2 is exhibiting “degeneracy” discussed in Section 8.1 on p.169. We have(2l + 1)-fold degeneracy for each diagonal. To the contrary, Lz is not degenerate ineach diagonal as the eigenvalues range from −l~ to +l~. For example, consider thel = 2 block. The eigenstates, |l,ml⟩’s = |2,ml⟩’s with ml = −2,−1, 0,+1,+2, aregiven by

|2,−2⟩ =

000010000...

, |2,−1⟩ =

000001000...

, |2, 0⟩ =

000000100...

, |2,+1⟩ =

000000010...

, |2,+2⟩ =

000000001...

. (9.72)

9.6. GENERALIZED ANGULAR MOMENTUM J 191

Fro these states, we have

L2 |2,−2⟩ = L2 |2,−1⟩ = L2 |2, 0⟩ = L2 |2,+1⟩ = L2 |2,+2⟩ = 2(2 + 1)~2 = 6~2,(9.73)

and

Lz |2,−2⟩ = −2~, Lz |2,−1⟩ = −~, Lz |2, 0⟩ = 0, Lz |2,+1⟩ = +~,

and Lz |2,+2⟩ = +2~. (9.74)

9.6 Generalized Angular Momentum J

As shown on p.176, the three operators associated with the components of an arbi-trary classical angular momentum, Lx, Ly, and Lz, satisfy the commutation relations(9.5), (9.6), and (9.7). These relations originate from the geometric properties of ro-tations in three-dimensional space. However, in quantum mechanics it is sometimesnecessary to consider more abstract “angular momentum” with no counterpart inclassical mechanics. What we do is to go backwards, from the commutation rela-tions to “rotations”, and define quantum mechanical angular momentum J as anyset of three observables/Hermitian operators Jx, Jy, and Jz which satisfies the samecommutation relations as Lx, Ly, and Lz. Namely,

[Jx, Jy] = i~Jz, (9.75)[Jy, Jz] = i~Jx, (9.76)[Jz, Jx] = i~Jy, (9.77)

and[J2, Jx] = [J2, Jy] = [J2, Jz] = 0; (9.78)

where (9.78) follows from (9.75), (9.76), and (9.77) as in (9.13) on p.178. Note thatJ2 = J2

x + J2y + J2

z (a scalar) by definition, and (9.78) implies [J2, J ] = 0. With thistheoretical framework, quantum mechanical angular momentum is based entirely onthe commutation relations (9.75), (9.76), and (9.77).

If you examine the computations we conducted for L, L2, Lx, Ly, and Lz, you cansee that we only used the commutation relations (9.5), (9.6), and (9.7), which areequivalent to (9.75), (9.76), and (9.77). Therefore, all the results we obtainedfor L apply to J without modification. In particular, we can define raising andlowering operators by


J+ = Jx + iJy (9.79)and

J− = Jx − iJy, (9.80)

which satisfy

J+ |j,mj⟩ = ~√(j −mj)(j +mj + 1) |j,mj + 1⟩ (9.81)

and

J− |j,mj⟩ = ~√(j +mj)(j −mj + 1) |j,mj − 1⟩ (9.82)

as well as

J+ |j, j⟩ = 0 (9.83)and

J− |j,−j⟩ = 0. (9.84)

So, we have

−j ≤ mj ≤ +j for j ≥ 0, (9.85)J2 |j,mj⟩ = ~2j(j + 1) |j,mj⟩ , (9.86)

andJz |j,mj⟩ = mj~ |j,mj⟩ . (9.87)

Finally, recall that one immediate implication of this theory for experimentalphysics is that simultaneous measurements of two components of J is impossiblewhile J2 and one component of J , typically chosen to be Jz, can be determined atthe same time. We will revisit the concept of generalized angular momentum inChapter 11, where we discuss the electron spin.

193

Exercises1. Show that

[Lz, Lx] = i~Ly.

Chapter 10

The Hydrogen Atom

This is the first “real” system with multiple variables. So, let us take a deep breathhere and briefly summarize what we have established in terms of the SchrodingerEquation with one variable.

So far1. TDSE (the Time-Dependent Schrodinger Equation)[

− ~2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t(10.1)

2. TISE (the Time-Independent Schrodinger Equation)[− ~

2

2m

d2

dx2+ V (x)

]ψ(x) = Eψ(x); (10.2)

where ψ(x) is an eigenfunction, and E is an eigenvalue.

3. The Relation between Ψ and ψ

Ψ(x, t) = ψ(x)e−iEt/~ (10.3)

We have not really tried any real system yet. But, we need to carry out a realcomputation for a real system for an experimental confirmation of the correctness ofthe Schrodinger Equation

One of the simplest systems is a hydrogen atom. However, this poses a couple ofdifficulties since (1) two particles are involved now and also since (2) it is no longerone-dimensional, but three-dimensional.

195

196 CHAPTER 10. THE HYDROGEN ATOM

10.1 2 Particles Instead of 1Let M be the mass of the nucleus and m the mass of the electron. The nucleus andthe electron are both moving about their fixed center of mass. But, we definitelyprefer a system where the nucleus is standing still and the electron is the only objectin motion. Introduction of the reduced mass µ given by

µ =Mm

M +m(10.4)

solves this problem. It literally reduces the original two-body problem to a one-body problem, where a body of mass µ experiences the same force that is at workbetween the two bodies. 1 This force is the Coulombic force for us and is inverselyproportional to the square of the distance between the two bodies. For the hydrogenatom, M = 1.67262178 × 10−27 kilograms and m = 9.10938291 × 10−31 kilograms.So,

1The reduced mass µ is an effective inertial mass in the following sense. First, let F be the forceexerted on the electron by the nucleus. Then, we know from Newton’s third law that the forceexerted by the electron on the nucleus is −F . Now, denote the accelerations of the nucleus andelectron by aM and am. We have

F = mam and − F =MaM =⇒ mam = −MaM . (10.5)

Hence, the relative acceleration a, the acceleration of the electron relative to the nucleus, is givenby

a = am − aM =F

m− −F

M=

(1

m+

1

M

)F =

M +m

MmF . (10.6)

Dividing (10.6) through by M+mMm to make the expression look like Newton’s second law, we get

F =Mm

M +ma. (10.7)

This can be interpreted as force F acting on an object of mass µ = MmM+m . Finally, note that the

relation between µ, M , and m can be expressed as follows.

1

µ=

1

M+

1

m(10.8)

10.2. THREE-DIMENSIONAL SYSTEM 197

µ =(

M

m+M

)m ≈ 0.999455679m ≈ m, (10.9)

and we can simply use m instead of µ in this case.

10.2 Three-Dimensional SystemThe classical energy equation is

p2

2µ+ V = E (10.10)

or

1

2µ

(p2x + p2y + p2z

)+ V (x, y, z) = E. (10.11)

We want to transform this into a three-dimensional Schrodinger Equation.Our solution, the wavefunction, now depends on four variables x, y, z, and t.

Ψ(x, t) =⇒ Ψ(x, y, z, t) (10.12)

Here are the operator correspondences we are going to use.

px ⇐⇒ −i~∂

∂xpy ⇐⇒ −i~

∂

∂ypz ⇐⇒ −i~

∂

∂zE ⇐⇒ i~

∂

∂t(10.13)

Of course, this means we have the following.

p2x = −~2∂2

∂x2p2y = −~2

∂2

∂y2p2z = −~2

∂2

∂z2(10.14)

Therefore,

p2

2µ=p2x + p2y + p2z

2µ=

1

2µ

[−~2 ∂

2

∂x2+

(−~2 ∂

2

∂y2

)+

(−~2 ∂

2

∂z2

)]


=−~2

2µ

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

). (10.15)

On the other hand, the potential energy is now a function of x, y, and z as below.

V = V (x, y, z) =−e2

4πε0√x2 + y2 + z2

(10.16)

We now have the following Time-Dependent Schrodinger Equation in three di-mensions.

−~2

2µ

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ(x, y, z, t) + V (x, y, z)Ψ(x, y, z, t)

= i~∂Ψ(x, y, z, t)

∂t(10.17)

Let us introduce a Laplacian operator or “del squared” ∇2 defined by

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2. (10.18)

This simplifies the equation to the following form.

−~2µ∇2Ψ+ VΨ = i~

∂Ψ

∂t; (10.19)

where

V = V (x, y, z) and Ψ = Ψ(x, y, z, t). (10.20)

Now let

Ψ(x, y, z, t) = ψ(x, y, z)e−iEt/~. (10.21)

Then, we get the Time-Independent Schrodinger Equation.


−~2

2µ∇2ψ(x, y, z) + V (x, y, z)ψ(x, y, z) = Eψ(x, y, z) (10.22)

As it turns out, it is easier to solve this equation in spherical (polar) coordinates,where (x, y, z) are replaced by (r, θ, ϕ).2 Since a hydrogen atom possesses sphericalsymmetry, this is a better coordinate system. Figure 10.1 shows the spherical co-ordinates used in physics, while Figure 10.2 is the convention used in mathematics.It is unfortunate that the roles of θ and ϕ are switched between the two systems,but there is not much chance of confusion once you become used to the sphericalcoordinate system. In this class, we will use the physicists’ convention.

x

y

z

(r, θ, φ)

φ

θ

r

Figure 10.1: Spherical Coordinate System (Source: Wikimedia Commons; Author:Andeggs)

Right away, V (x, y, z) simplifies as follows.

V (x, y, z) =−e2

4πε0√x2 + y2 + z2

=−e2

4πε0r= V (r, θ, φ) (10.23)

In fact, the potential energy only depends on the distance from the origin.

2Note that in pure math books θ and ϕ are usually switched.


x

y

z

(r, θ, φ)

φ

θ

r

Figure 10.2: Spherical Coordinate System Used in Mathematics (Source: WikimediaCommons; Author: Dmcq)

V (r, θ, φ) =−e2

4πε0r= V (r) (10.24)

Of course,

ψ(x, y, z) =⇒ ψ(r, θ, φ). (10.25)

This means that we have to convert ∇2 = ∂2

∂x2+ ∂2

∂y2+ ∂2

∂z2to another equivalent

operator expressed as derivatives with respect to r, θ, and φ.Then, we will have

−~2

2µ∇2ψ(r, θ, φ) + V (r)ψ(r, θ, φ) = Eψ(r, θ, φ). (10.26)

The answer is

∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2, (10.27)


and we can see that ∇2 appears far more complicated in the spherical coordinates.

Here is how it is done. We begin with the three relations between (x, y, z) and(r, θ, φ).

x = r sin θ cosφy = r sin θ sinφz = r cos θ

(10.28)

We can see how ∂2

∂x2etc. can be converted. What we should do is to keep using chain

rules and relations like ∂x∂r

= sin θ cosφ. But, the actual operation is quite tedious.

A full and complete proof of (10.27) can be found in Appendix L. But, let usprove only a part of this conversion here. Incidentally, it will be worth your while toread Sections K.1, K.2, and K.3 of Appendix K because the rules involving partialdifferentiation is often confusing.

Consider a one-variable function ψ(r) for r =√x2 + y2 + z2.

As (K.8) gives

∂ψ

∂x=∂ψ

∂r

∂r

∂x+∂ψ

∂θ

∂θ

∂x+∂ψ

∂ϕ

∂ϕ

∂x=∂ψ

∂r

∂r

∂x=

x√x2 + y2 + z2

∂ψ

∂r=x

r

∂ψ

∂r(10.29)

and∂r

∂x=

1

2

(x2 + y2 + z2

)−1/2· 2 =

x

r, (10.30)

we have∂2ψ

∂x2=

∂

∂x

(x

r

∂ψ

∂r

)=

∂

∂x

(x · 1

r

∂ψ

∂r

)=∂x

∂x

(1

r

∂ψ

∂r

)+ x

∂

∂x

(1

r

∂ψ

∂r

)

=1

r

∂ψ

∂r+ x

∂r

∂x

∂

∂r

(1

r

∂ψ

∂r

)=

1

r

∂ψ

∂r+ x · x

r

∂

∂r

(1

r

∂ψ

∂r

)

=1

r

∂ψ

∂r+x2

r

∂

∂r

(1

r

∂ψ

∂r

). (10.31)

Similarly,

∂2ψ

∂y2=

1

r

∂ψ

∂r+y2

r

∂

∂r

(1

r

∂ψ

∂r

)(10.32)


and∂2ψ

∂z2=

1

r

∂ψ

∂r+z2

r

∂

∂r

(1

r

∂ψ

∂r

). (10.33)

Therefore,

∇2ψ =3

r

∂ψ

∂r+x2 + y2 + z2

r

∂

∂r

(1

r

∂ψ

∂r

)=

3

r

∂ψ

∂r+ r

∂

∂r

(1

r

∂ψ

∂r

)

=3

r

∂ψ

∂r+ r

(− 1

r2∂ψ

∂r+

1

r

∂2ψ

∂r2

)=

2

r

∂ψ

∂r+∂2ψ

∂r2=

1

r2∂

∂r

(r2∂ψ

∂r

). (10.34)

In order to get other parts, try ψ = ψ(φ) and ψ = ψ(θ).

In preparation for the rest of the course as well as your career as a physicist orjust as someone who needs to understand and use physics, please do get used to thespherical coordinates. For example, the volume element

dxdydz

becomes

r2dr sin θdθdφ or − r2drd(cos θ)dφ (10.35)

in spherical coordinates as shown in Figure 10.3. Needless to say, the limits ofintegration should be adjusted accordingly. For example, the normalization integralsare

+∞∫−∞

+∞∫−∞

+∞∫−∞

Ψ∗(x, y, z, t)Ψ(x, y, z, t) dxdydz, (10.36)

∞∫0

π∫0

2π∫0

Ψ∗(r, θ, ϕ, t)Ψ(r, θ, ϕ, t) r2dr sin θdθdφ, (10.37)

and∞∫0

+1∫−1

2π∫0

Ψ∗(r, θ, ϕ, t)Ψ(r, θ, ϕ, t) r2drd(cos θ)dφ; (10.38)

where the last integral (10.38) is particularly useful when we have a wavefunctionwhose θ-dependence is via cos θ or sin θ. As you can see in Table (10.3), this is indeed


the case for the wavefunctions of the hydrogen atom. We sometimes write dV forthe volume element and dΩ for sin θdθdφ, so that

dV = r2drdΩ. (10.39)

The quantity Ω is known as solid angle.

Figure 10.3: Volume Element in Spherical Coordinates (Source: Victor J. Mon-temayor, Middle Tennessee State University)

Let us pause for a minute here, take a deep breath, and summarize what wehave achieved so far to clearly understand where we are now, as well as, “hopefully”,where we are heading.

1. Due to the spherical symmetry inherent in the hydrogen atom, we made adecision to use the spherical coordinates as opposed to the familiar Cartesiancoordinates. In particular, this simplifies the expression for the potential en-ergy greatly. It is now a function only of the radial distance between the


electron and the proton nucleus.a

V (x, y, z) = V (r) =−e2

4πε0r(10.40)

2. However, a rather heavy trade-off was the conversion of ∇2 to the equivalentexpression in the spherical coordinates. The conversion process is cumbersome,and the resulting expression is far less palatable.

∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2(10.41)

aIf the charge on the nucleus is Ze rather than just e, we have V (x, y, z) = V (r) = −Ze2

4πε0rinstead. This makes extending the results obtained for the hydrogen atom to hydrogen-like atomsquite straightforward as we will see in Section 10.9.

We are now ready to solve the Time-Independent Schrodinger Equation in sphericalcoordinates given by

−~2

2µ∇2ψ(r, θ, φ) + V (r)ψ(r, θ, φ) = Eψ(r, θ, φ). (10.42)

We will resort to the separation of variables technique yet one more time. So, considerψ as a product of three single variable functions of r, θ, and φ respectively.

ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ) (10.43)

Then, the Time-Independent Schrodinger Equation is

[−~2

2µ∇2

(r,θ,φ) + V (r)

]R(r)Θ(θ)Φ(φ) = ER(r)Θ(θ)Φ(φ). (10.44)

Let us write it out in its full glory and start computing!

−~2

2µ

[1

r2∂

∂r

(r2∂(RΘΦ)

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ∂(RΘΦ)

∂θ

)+

1

r2 sin2 θ

∂2(RΘΦ)

∂φ2

]

+V (r)R(r)Θ(θ)Φ(φ) = ER(r)Θ(θ)Φ(φ)


=⇒ −~2

2µ

[1

r2∂

∂r

(r2∂R

∂rΘΦ

)+

1

r2 sin θ∂

∂θ

(sin θ∂Θ

∂θRΦ

)+

1

r2 sin2 θ

∂2Φ

∂φ2RΘ

]

+V (r)RΘΦ

=−~2

2µ

[ΘΦ

1

r2∂

∂r

(r2∂R

∂r

)+

1

r2 sin θRΦ∂

∂θ

(sin θ∂Θ

∂θ

)+

1

r2 sin2 θRΘ

∂2Φ

∂φ2

]

+V (r)RΘΦ

=−~2

2µ

[ΘΦ

1

r2d

dr

(r2dR

dr

)+

1

r2 sin θRΦd

dθ

(sin θdΘ

dθ

)+

1

r2 sin2 θRΘ

d2Φ

dφ2

]

+V (r)RΘΦ = ER(r)Θ(θ)Φ(φ) (10.45)

Now, multiply through by−2µ~2· r2 sin2 θ

1

RΘΦ.

−~2

2µ· −2µ~2

r2 sin2 θ1

RΘΦ

[ΘΦ

1

r2d

dr

(r2dR

dr

)+

1

r2 sin θRΦd

dθ

(sin θdΘ

dθ

)

+1

r2 sin2 θRΘ

d2Φ

dφ2

]+ V (r)RΘΦ

−2µ~2

r2 sin2 θ1

RΘΦ

= ERΘΦ · −2µ~2

r2 sin2 θ1

RΘΦ(10.46)

=⇒ sin2 θ

R

d

dr

(r2dR

dr

)+

sin θΘ

d

dθ

(sin θdΘ

dθ

)+

1

Φ

d2Φ

dφ2+−2µ~2

r2 sin2 θV (r)

=−2µ~2

r2 sin2 θE

=⇒ 1

Φ

d2Φ

dφ2= −sin2 θ

R

d

dr

(r2dR

dr

)− sin θ

Θ

d

dθ

(sin θdΘ

dθ

)− 2µ

~2r2 sin2 θ [E − V (r)]

(10.47)


The LHS (lefthand side) is a function of φ, and the RHS (righthand side) is afunction of r and θ. And the equality holds for all values of r, θ, and φ. Therefore,both sides have to equal a constant. For later convenience, we denote this by −m2

l .So, we have

1

Φ

d2Φ

dφ2= −m2

l (10.48)

or

d2Φ

dφ2= −m2

lΦ. (10.49)

According to (10.47), this also means

−sin2 θ

R

d

dr

(r2dR

dr

)− sin θ

Θ

d

dθ

(sin θdΘ

dθ

)− 2µ

~2r2 sin2 θ [E − V (r)] = −m2

l . (10.50)

Dividing through by sin2 θ,

− 1

R

d

dr

(r2dR

dr

)− 1

Θ sin θd

dθ

(sin θdΘ

dθ

)− 2µ

~2r2 [E − V (r)] = − m2

l

sin2 θ

=⇒ 1

R

d

dr

(r2dR

dr

)+

2µ

~2r2 [E − V (r)] =

m2l

sin2 θ− 1

Θ sin θd

dθ

(sin θdΘ

dθ

). (10.51)

The LHS of (10.51) depends on r, and the RHS is a function of θ. As the equalityholds for any values of r and θ, both sides have to be a constant. Denote that constantby l(l + 1)3, for the reason that will become clear shortly, to get

−1sin θ

d

dθ

(sin θdΘ

dθ

)+m2lΘ

sin2 θ= l(l + 1)Θ (10.52)

and

3We are not placing any restriction on the values l can take. Because l can be a complex number,any scalar can be represented as l(l+1). However, we will later show that l is in fact a nonnegativeinteger.

10.3. THE SOLUTIONS FOR Φ 207

1

r2d

dr

(r2dR

dr

)+

2µ

~2[E − V (r)]R = l(l + 1)

R

r2. (10.53)

We have now reduced the problem to that of solving the following three equations,each in one variable.

d2Φ

dφ2= −m2

lΦ (10.54)

−1sin θ

d

dθ

(sin θdΘ

dθ

)+m2lΘ

sin2 θ= l(l + 1)Θ (10.55)

1

r2d

dr

(r2dR

dr

)+

2µ

~2[E − V (r)]R = l(l + 1)

R

r2(10.56)

10.3 The Solutions for Φ

The easiest to solve is equation (10.54).

d2Φ

dφ2= −m2

lΦ (10.57)

So,

Φ(φ) = eimlφ. (10.58)

To be precise, we have

Aeimlφ (10.59)

for an arbitrary scalar A. However, we can adjust the constant when we normalizethe total wavefunction Ψ(r, φ, θ, t). So, we have set A = 1 for now.

Consider single-valuedness.


Φ(0) = Φ(2π) =⇒ ei·ml·0 = ei·ml·2π

=⇒ 1 = eiml·2π = cos(ml2π) + i sin(ml2π) (10.60)

Now, let ml = a+ bi for (a, b ∈ R).

|1| =∣∣∣eiml2π

∣∣∣ = ∣∣∣ei(a+bi)2π∣∣∣ = ∣∣∣eia2π∣∣∣ ∣∣∣e−b2π∣∣∣ (10.61)∣∣∣e−b2π

∣∣∣ = e−b2π = 1 (10.62)

2πb = 0 =⇒ b = 0 (10.63)

It is now obvious that ml is a real number and

|ml| = 0, 1, 2, 3, 4, . . . . . . . (10.64)

For each value of ml, there is a corresponding solution

Φml(φ) = eimlφ. (10.65)

The numbers ml are known as magnetic quantum numbers.

Now we need to solve equations (10.55) and (10.56).

10.4 The Solutions for Θ

Equation (10.55) is known as the angular equation.

− 1

sin θd

dθ

(sin θdΘ

dθ

)+m2lΘ

sin2 θ= l(l + 1)Θ (10.66)

Note that

d cos θ = − sin θdθ. (10.67)

So,

−1sin θ

d

dθ=

d

d cos θ , (10.68)

10.4. THE SOLUTIONS FOR Θ 209

and

sin θ ddθ

= sin θ dd cos θ− sin θ

= − sin2 θd

d cos θ = −(1− cos2 θ) d

d cos θ . (10.69)

This suggests that (10.55) can be simplified if we let z = cos θ. We have

d

d cos θ

(−(1− cos2 θ) dΘ

d cos θ

)+

m2lΘ

1− cos2 θ = l(l + 1)Θ, (10.70)

which leads to

d

dz

(−(1− z2)dΘ

dz

)+

m2lΘ

1− z2= l(l + 1)Θ =⇒

d

dz

((1− z2)dΘ

dz

)+

(l(l + 1)− m2

l

1− z2

)Θ = 0 (10.71)

At the time Schrodinger was solving his equation for the hydrogen atom, this wasalready a well-known differential equation, and the solutions are called the associ-ated Legendre functions denoted by Θlml

(z). This is related to a better known set offunctions called the Legendre polynomials denoted by Pl(z). The relation betweenΘlml

(z) and Pl(z) is as follows.

Θlml(z) = (1− z2)|ml|/2d

|ml|Pl(z)

dz|ml|(10.72)

Our first job is to prove this.

Now Pl(z) is a solution to

(1− z2)d2Pldz2− 2z

dPldz

+ l(l + 1)Pl = 0. (10.73)

Differentiate both sides with respect to z to obtain

−2z d2Pldz2

+ (1− z2)d3Pldz3− 2

dPldz− 2z

d2Pldz2

+ l(l + 1)dPldz

= (1− z2)d3Pldz3− 4z

d2Pldz2

+

((l(l + 1)− 2)

dPldz

)= 0. (10.74)


Try ddz

again.

−2d2Pldz2− 2z

d3Pldz3− 2z

d3Pldz3

+ (1− z2)d4Pldz4− 2

d2Pldz2− 2

d2Pldz2

−2z d3Pldz3

+ l(l + 1)d2Pldz2

= (1− z2)d4Pldz4− 6z

d3Pldz3

+ (l(l + 1)− 6)d2Pldz2

= 0 (10.75)

Suppose we get

(1− z2)dk+2Pldzk+2

− 2(k + 1)zdk+1Pldzk+1

+ (l(l + 1)− k(k + 1))dk

dzkPl = 0 (10.76)

after differentiating k times. Relation (10.76) holds for k = 0, 1, and 2 according to(10.73), (10.74), and (10.75).Now apply d

dzone more time.

−2z dk+2Pldzk+2

+ (1− z2)dk+3Pldzk+3

− 2(k + 1)dk+1Pldzk+1

− 2(k + 1)zdk+2Pldzk+2

+(l(l + 1)− k(k + 1))dk+1Pldzk+1

= (1− z2)d(k+1)+2Pldz(k+1)+2

− 2((k + 1) + 1)zd(k+2)+1Pldz(k+2)+1

+(l(l + 1)− (k + 1)((k + 1) + 1)dk+1Pldzk+1

= 0 (10.77)

So, by mathematical induction,

(1− z2) dk+2

dzk+2Pl − 2(k + 1)z

dk+1Pldzk+1

+ (l(l + 1)− k(k + 1))dkPldzk

= 0 (10.78)

when we apply dk

dzk to both sides of (10.73).

On the other hand, consider

Θlml= (1− z2)|ml|/2Γ(z) (10.79)

and substitute this into


d

dz

((1− z2)dΘ

dz

)+

(l(l + 1)− m2

l

1− z2

)Θ = 0. (10.80)

We have the following lengthy computation.

d

dz

((1− z2) d

dz(1− z2)k/2Γ

)+

(l(l + 1)− k2

1− z2

)(1− z2)k/2Γ

= −2z ddz

((1− z2)k/2Γ) + (1− z2) d2

dz2((1− z2)k/2Γ)

+

(l(l + 1)− k2

1− z2

)(1− z2)k/2Γ

= −2z(k

2(1− z2)(k/2−1)(−2z)Γ + (1− z2)k/2dΓ

dz

)

+(1− z2) ddz

(k

2(1− z2)(k/2−1)(−2z)Γ + (1− z2)k/2dΓ

dz

)

+

(l(l + 1)− k2

1− z2

)(1− z2)k/2Γ

= −2z(kz(1− z2)(k/2−1)Γ + (1− z2)k/2dΓ

dz

)

+(1− z2)[k

2(k/2− 1)(1− z2)(k/2−2)(−2z)(−2z)Γ +

k

2(1− z2)(k/2−1)(−2)Γ

+k

2(1− z2)(k/2−1)(−2z)dΓ

dz+k

2(1− z2)(k/2−1)(−2z)dΓ

dz+ (1− z2)k/2d

2Γ

dz2

]

+

(l(l + 1)− k2

1− z2

)(1− z2)k/2Γ

= 2kz2(1− z2)(k/2−1)Γ− 2z(1− z2)k/2dΓdz

+(1− z2)[2kz2

(k

2− 1

)(1− z2)(k/2−2)Γ− k(1− z2)(k/2−1)Γ


−kz(1− z2)(k/2−1)dΓ

dz− kz(1− z2)(k/2−1)dΓ

dz+ (1− z2)k/2d

2Γ

dz2

]

+

(l(l + 1)− k2

1− z2

)(1− z2)k/2Γ = 0 (10.81)

Multiplying through by (1− z2)−k/2,2kz2

1− z2Γ− 2z

dΓ

dz+ (1− z2)

[2kz2

(k

2− 1

)(1− z2)−2Γ− k(1− z2)−1Γ

−kz(1− z2)−1dΓ

dz− kz(1− z2)−1dΓ

dz+d2Γ

dz2

]+

(l(l + 1)− k2

1− z2

)Γ

=2kz2

1− z2Γ− 2z

dΓ

dz+ 2kz2

(k

2− 1

)(1− z2)−1Γ− kΓ− kzdΓ

dz− kzdΓ

dz

+(1− z2)d2Γ

dz2+

(l(l + 1)− k2

1− z2

)Γ

= (1− z2)d2Γ

dz2+ (−2kz − 2z)

dΓ

dz+

l(l + 1) +2kz2 + 2kz2

(k2− 1

)− k2

1− z2− k

Γ

= (1− z2)d2Γ

dz2− 2(k + 1)z

dΓ

dz+

[l(l + 1) +

k2z2 − k2

1− z2− k

]Γ

= (1− z2)d2Γ

dz2− 2(k + 1)z

dΓ

dz+[l(l + 1)− k2 − k

]Γ = 0. (10.82)

So, we have shown

(1− z2)d2Γ

dz2− 2(k + 1)z

dΓ

dz+ [l(l + 1)− k(k + 1)] Γ = 0 (10.83)

or

(1− z2)d2Γ

dz2− 2(|ml|+ 1)z

dΓ

dz+ [l(l + 1)− |ml| (|ml|+ 1)] Γ = 0. (10.84)

Comparing this with (10.76), which is given below again for reference,

(1− z2) dk+2

dzk+2Pl − 2(k + 1)z

dk+1Pldzk+1

+ (l(l + 1)− k(k + 1))dkPldzk

= 0, (10.85)


we conclude

Θlml= (1− z2)|ml|/2d

|ml|Pldz|ml|

. (10.86)

It remains to solve (10.73)

(1− z2)d2Pldz2− 2z

dPldz

+ l(l + 1)Pl = 0 (10.87)

for Pl. Try

Pl(z) =∞∑k=0

akzk. (10.88)

We get a recursion relation as we did before on p.150 in Section 7.6.1 for the simpleharmonic oscillator.

aj+2 =j(j + 1)− l(l + 1)

(j + 2)(j + 1)aj (10.89)

Also as before, we do not want an infinite series; that is, we want the series to termi-nate after a finite number of terms, so that the function is integrable/normalizable.We need

j(j + 1)− l(l + 1) = j2 + j − l2 − l = (j + l)(j − l) + j − l= (j − l)(j + l + 1) = 0 =⇒ l = j or − j − 1. (10.90)

As j is a nonnegative integer, l can be any integer, positive, 0, or negative. However,the only meaningful quantity is l(l + 1). Noting that the parabola

y(l) = l(l + 1) =(l +

1

2

)2

−(1

2

)2

(10.91)

is symmetric about l = −12, it is sufficient to consider l = 0, 1, 2, 3, . . . . So, l =

0, 1, 2, 3, . . . give all acceptable solutions. In fact, y(−l− 1) = (−l− 1)(−l− 1+1) =(−l−1)(−l) = l(l+1) proves that the pairs (0,−1), (1,−2), (2,−3), (3,−4) . . . givethe same value for l(l + 1), and negative l values are clearly redundant.We get


P0 = 1, P1 = z, P2 = 1− 3z2, P3 = 3z − 5z3, . . . . (10.92)

So, the corresponding Θlml’s are

Θ00 = 1, Θ10 = z, Θ1±1 = (1− z2)1/2, Θ20 = 1− 3z2,

Θ2±1 = (1− z2)1/2z, Θ2±2 = 1− z2, . . . (10.93)

Note that Pl(z) is an lth degree polynomial. Therefore, for each l, we have

ml = −l, − l + 1, . . . , 0, . . . , l − 1, + l. (10.94)

10.5 Associated Legendre Polynomials and Spher-ical Harmonics

Recall that Pl(z)’s = Pl(cos θ)’s are called the Legendre polynomials, and Θlml’s are

called associated Legendre functions/polynomials as explained on p.209. The asso-ciated Legendre polynomials are also denoted by Pml

l (cos θ) or simply by Pml (cos θ).

So,

Θlml(cos θ) = Pml

l (cos θ). (10.95)

We now introduce a new set of functions known as the spherical harmonics, whichare the products of Φ and Θ up to the normalization constant. It is customary todenote each spherical harmonic by Y ml

l or simply by Y ml . So, we have

Y mll (θ, ϕ) ∝ Φml

(ϕ)Θlml(cos θ) = Φml

(ϕ)Pmll (cos θ) = eimlϕPml

l (cos θ). (10.96)

With the appropriate normalization constant4,

4Different definitions for spherical harmonics are used in different fields such as geodesy, mag-netics, quantum mechanics, and seismology. The definition given here is commonly adopted in thequantum mechanics community.

10.5. ASSOCIATED LEGENDRE POLYNOMIALS AND SPHERICAL HARMONICS

Y mll (θ, ϕ) = (−1)ml

√√√√2l + 1

4π

(l −ml)!

(l +ml)!Pmll (cos θ)eimlϕ; (10.97)

where the normalization is chosen such that∫ 2π

0

∫ π

0(Y ml

l (θ, ϕ))∗ Y mll (θ, ϕ) sin θdθdϕ

=∫ 2π

0

∫ +1

−1(Y ml

l (θ, ϕ))∗ Y mll (θ, ϕ)d(cos θ)dϕ = 1, (10.98)

or

=

"(Y ml

l (θ, ϕ))∗ Y mll (θ, ϕ) dΩ = 1. (10.99)

In fact, Y mll ’s are orthonormal such that∫ 2π

0

∫ π

0(Y ml

l (θ, ϕ))∗ Ym′

ll′ (θ, ϕ) sin θdθdϕ

=∫ 2π

0

∫ +1

−1(Y ml

l (θ, ϕ))∗ Ym′

ll′ (θ, ϕ)d(cos θ)dϕ

=

"Y ml∗l Y ml

l dΩ = δll′δmlm′l. (10.100)

We will see how this comes about in Section 10.6. In passing let us make a note of thefact that the normalization constant from the ϕ-dependent portion of the integral is√

12π

, and the normalization constant from the θ-dependent portion of the integral is√(2l+1)(l−m)!

2(l+ml)!.

Spherical harmonics for ml < 0 has the following alternative formula

Y mll (θ, ϕ) =

√√√√2l + 1

4π

(l − |ml|)!(l + |ml|)!

P|ml|l (cos θ)eimlϕ. (10.101)

This is a direct consequence of the following identity satisfied by the associatedLegendre polynomials.5 We will write m instead of ml for simplicity.

P−ml (x) = (−1)m (l −m)!

(l +m)!Pml (x) (10.102)

5See Fact M.1 of Appendix M on p.342 for a proof.


Indeed, for m = −|m| < 0, we have

Y ml (θ, ϕ) = Y

−|m|l (θ, ϕ) = (−1)−|m|

√√√√2l + 1

4π

(l + |m|)!(l − |m|)!

P−|m|l (cos θ)e−i|m|ϕ

= (−1)−|m|

√√√√2l + 1

4π

(l + |m|)!(l − |m|)!

[(−1)|m| (l − |m|)!

(l + |m|)!P

|m|l (cos θ)

]ei(−|m|)ϕ

=

√√√√2l + 1

4π

(l − |m|)!(l + |m|)!

P|m|l (cos θ)eimϕ. (10.103)

We can combine (10.97) and (10.103) in the following manner.

Y mll (θ, ϕ) = (−1)(ml+|ml|)/2

√√√√2l + 1

4π

(l − |ml|)!(l + |ml|)!

P|ml|l (cos θ)eimlϕ (10.104)

Note that the phase factor (−1)ml or (−1)(ml+|ml|)/2 does not change the physicalobservables, and it is more or less a mathematical book keeping device.

10.6 L2, Lz, and the Spherical HarmonicsThe orthonormality relation (10.100) is a consequence of the following fact.

Fact 10.1 The spherical harmonics are simultaneous eigenstates of L2 and Lz cor-responding to the eigenvalues l(l + 1)~2 and ml~.

L2Y mll (θ, ϕ) = l(l + 1)~2Y ml

l (θ, ϕ) (10.105)and

LzYmll (θ, ϕ) = ml~Y

mll (θ, ϕ) (10.106)

Because L2 and Lz are Hermitian operators, and also because Y mll ’s are normalized,

the set

Y mll

−l≤ml≤+ll=0,1,2,... (10.107)

forms an orthonormal basis. Hence, Y mll ’s are not only orthonormal, but also com-

plete, and any function of θ and ϕ can be represented as a superposition (linearcombination) of spherical harmonics.

10.7. THE RADIAL FUNCTION R

Recall that for each of l = 0, 1, 2, 3, . . . we haveml = −l,−l+1,−l+2, . . . 0, . . . , l−2, l − 1, l. All the spherical harmonics for l = 0, 1, and 2 are listed in Table 10.1.

Spherical Harmonicsl = 0 ml = 0 Y 0

0 (θ, ϕ) =12

√1π

l = 1ml = −1 Y −1

1 (θ, ϕ) = 12

√32πe−iϕ sin θ = 1

2

√32π

x−iyr

ml = 0 Y 01 (θ, ϕ) =

12

√3π

cos θ = 12

√3πzr

ml = +1 Y 11 (θ, ϕ) =

−12

√32πeiϕ sin θ = −1

2

√32π

x+iyr

l = 2

ml = −2 Y −22 (θ, ϕ) = 1

4

√152πe−i2ϕ sin2 θ = 1

4

√152π

(x−iy)2r2

ml = −1 Y −12 (θ, ϕ) = 1

2

√152πe−iϕ sin θ cos θ = 1

2

√152π

(x−iy)zr2

ml = 0 Y 02 (θ, ϕ) =

14

√5π(3 cos2 θ − 1) = 1

4

√5π2z2−x2−y2

r2

ml = +1 Y 12 (θ, ϕ) =

−12

√152πeiϕ sin θ cos θ = −1

2

√152π

(x+iy)zr2

ml = +2 Y 22 (θ, ϕ) =

14

√152πei2ϕ sin2 θ = 1

4

√152π

(x+iy)2

r2

Table 10.1: Spherical Harmonics for l = 0, 1, and 2

Next on our agenda is the radial function R.

10.7 The Radial Function R

The general radial equation for a one-electron atom is

1

r2d

dr

(r2dR

dr

)+

2µ

~2

[E +

Ze2

4πε0r

]R = l(l + 1)

R

r2; (10.108)

where Ze is the charge on the nucleus. We will only solve this for Z = 1 or for thehydrogen atom.


Let us first invoke some substitutions to put the radial equation in a more man-ageable form.

β2 =2µE

~(β > 0) (10.109)

ρ = 2βr (10.110)

γ =µe2

4πε0~2β(10.111)

With these substitutions, we get

1

ρ2d

dρ

(ρ2dR

dρ

)+

[−1

4− l(l + 1)

ρ2+γ

ρ

]R = 0. (10.112)

As ρ→∞R(ρ) ≈ e−ρ/2. (10.113)

So, let us write

R(ρ) = e−ρ/2F (ρ), (10.114)

and substitute this into the radial equation to get

d2F

dρ2+

(2

ρ− 1

)dF

dρ+

[γ − 1

ρ− l(l + 1)

ρ2

]F = 0. (10.115)

Once again, we will try a series solution with

F (ρ) = ρs∞∑k=0

akρk (a0 , 0, s ≥ 0); (10.116)

where the condition on s is in place to prevent F from blowing up at the origin, whilea0 , 0 assures that our solution is not trivial as we will see in (10.119).

After substituting this trial series solution into the differential equation (10.115),we get

[s(s+ 1)− l(l + 1)]a0ρs−2 +

∞∑j=0

[(s+ j + 1)(s+ j + 2)− l(l + 1)]aj+1

10.7. THE RADIAL FUNCTION R 219

−(s+ j + 1− γ)ajρs+j−1 = 0. (10.117)

For the left-hand side to be zero for any value of ρ, the following two relations shouldhold.

s(s+ 1)− l(l + 1) = 0

aj+1 =s+ j + 1− γ

(s+ j + 1)(s+ j + 2)− l(l + 1)aj

(10.118)

(10.119)

The first equality (10.118) gives s = l and s = −(l+1). But, s = −(l+1) should berejected as s ≥ 0. The second relation (10.119)

aj+1 =j + l + 1− γ

(j + l + 1)(j + l + 2)− l(l + 1)aj (10.120)

indicates that

γ = j + l + 1 = n (Call this n.) (10.121)

assures termination of the series after a finite number of terms; namely, after the jthterm. Hence,

n = l + 1, l + 2, l + 3, · · · (10.122)

as j runs from 0 to ∞ with l = 0, 1, 2, 3, . . ..

Now recall from (10.109) on p.218

β2 =2µE

~2(10.123)

and from (10.111) on p.218

γ

=

n

=µe2

4πε0~2β(10.124)

So,


En = − µe4

(4πε0)22~2n2= −

(µe4

32π2ε20~2

)1

n2= −

(~2

2µa20

)1

n2

for n = 1, 2, 3, · · · . (10.125)

Here,

a0 =4πε0~

2

µe2(10.126)

is analogous to the Bohr radius aB defined by

aB =4πε0~

2

mee2, (10.127)

where me is the rest mass of an electron. The numerical value of aB is about5.29177× 10−11 m, 52.9177 pm, 0.0529177 nm, or 0.529177 angstroms.

Because the radial function R(r) depends on two parameters, n and l, we writeRnl(r) for the radial part of the wavefunction.

10.8 The Full WavefunctionPutting it all together, we get

ψnlml(r, θ, φ) = Rnl(r)Y

mll (θ, φ) = Rnl(r)Θlml

(θ)Φml(φ); (10.128)

where

Φml(φ) = eimlφ |ml| = 0, 1, 2, 3, · · · , (10.129)

Θlml(θ) = Θlml

(cos θ) = Pmll (cos θ) = sin|ml| θFl|ml|(cos θ), (10.130)

and

10.8. THE FULL WAVEFUNCTION 221

Rnl(r) = e−r/na0(r

a0

)2

Gnl

(r

a0

). (10.131)

The functions F and G are both polynomials;

Fl|ml|(cos θ) (10.132)

is a polynomial in cos θ, and

Gnl

(r

a0

)(10.133)

is a polynomial in r/a0.

More specifically, the normalized radial wavefunctions are6,7

Rnl(r) =

√√√√( 2

na0

)3 (n− l − 1)!

2n[(n+ l)!]3e−r/na0

(2r

na0

)lL2l+1n−l−1

(2r

na0

)(10.134)

and the normalized full wavefunctions are

ψnlml= Rnl(r)Y

mll (θ, ϕ)

=

√√√√( 2

na0

)3 (n− l − 1)!

2n[(n+ l)!]3e−r/na0

(2r

na0

)lL2l+1n−l−1

(2r

na0

)×

√√√√2l + 1

4π

(l −ml)!

(l +ml)!Pmll (cos θ)eimlϕ

. (10.135)

Let us now summarize the relations among n, l, and ml. We have obtained

|ml| = 0, 1, 2, 3, · · · , (10.136)l = |ml| , |ml|+ 1, · · · , |ml|+ 3, · · · , (10.137)

and

6The appropriate normalization scheme for Rnl is discussed in Note 10.1.7L in (10.134) is an associated Laguerre polynomial described in Appendix N.


n = l + 1, l + 2, l + 3, · · · (10.138)

in this order, but this can be reorganized as follows.

n = 1, 2, 3, 4, · · · (10.139)l = 0, 1, 2, · · · , n− 1 (10.140)

ml = −l,−l + 1, · · · , 0, · · · ,+l (10.141)

The first number n is associated with R(r) and is called the principal quantum num-ber or the radial quantum number. The second number l is associated with Θ(θ) andis called the orbital quantum number or the angular momentum quantum number.Finally, the number ml is called the magnetic quantum number which is associatedwith Φ(φ). In addition to these, we also have a spin quantum number ms, which wewill discuss in Chapter 11.

Remark 10.1 (Degeneracy for the Hydrogen Atom) The total energy E de-pends only on the principal quantum number n for the hydrogen atom. This meansall the distinct states ψnlml

for l = 0, 1, . . . , n − 1 and −l ≤ ml ≤ +l share thesame energy level. Noting that there are 2l + 1 distinct states for each l, we have

n−1∑l=0

2l + 1 = 2n−1∑l=0

l +n−1∑l=0

1 = 2 · (n− 1 + 0) · n2+ n = n2 − n+ n = n2. (10.142)

So, the n-th energy level of the Hydrogen atom is n2-fold degenerate.

As it turned out, this is a special property of a 1r

potential. In order to see thespecial nature of the 1

rpotential, we will compare the expectation values

⟨rk⟩

fork = −3,−2,−1, 1, and 2. According to the normalization scheme described in Note10.1, we have the following equality.*

ψnlml(r, θ, ϕ)∗rkψnlml

(r, θ, ϕ) dV

=

*(Rnl(r)Θlml

(θ)Φml(φ))∗ rk (Rnl(r)Θlml

(θ)Φml(φ)) dV

=∫ ∞

0

(Rnl(r)Θlml

(θ)Φml(φ))∗ rk (Rnl(r)Θlml

(θ)Φml(φ)) r2drdΩ

=∫ ∞

0rk+2Rnl(r)

∗Rnl(r) dr∫ π

0Θlml

(θ)∗Θlml(θ) sin θdθ

∫ 2π

0Φml

(ϕ)∗Φml(ϕ) dϕ

=∫ ∞

0rk+2Rnl(r)

∗Rnl(r) dr × 1× 1 =∫ ∞

0rk+2Rnl(r)

∗Rnl(r) dr (10.143)

10.8. THE FULL WAVEFUNCTION 223

Therefore,⟨rk⟩=∫ ∞

0rk+2Rnl(r)

∗Rnl(r) dr or∫ ∞

0rk+2 |Rnl(r)|2 dr. (10.144)

Actual computations tend to be quite tedious, but the following are the results.⟨r2⟩=a20n

2

2

[5n2 + 1− 3l(l + 1)

](10.145)

⟨r⟩ = a02

[3n2 − l(l + 1)

](10.146)⟨

1

r

⟩=

1

n2a0(10.147)⟨

1

r2

⟩=

1(l + 1

2

)n3a20

(10.148)

⟨1

r3

⟩=

1

l(l + 1

2

)(l + 1)n3a30

(10.149)

One can note that only⟨1r

⟩does not depend on the angular momentum quantum

number l.

The first 6 normalized radial wavefunctions for a hydrogen-like atom, where thepositive charge on the nucleus is Ze rather than e, are presented in Table 10.2.Likewise, the first 14 normalized full wavefunctions are listed in Table 10.3.

Note 10.1 (Individual Normalization for Rnl, Θlml, and Φml

) We know that anormalized time-independent wavefunction ψ(r, θ, φ) and time-dependent wavefunc-tion Ψ(r, θ, φ, t) = ψ(r, θ, φ)e−iEt/~ should satisfy∫

Ψ(r, θ, φ, t)∗Ψ(r, θ, φ, t) dV =∫ (

ψ(r, θ, φ)e−iEt/~)∗ψ(r, θ, φ)e−iEt/~ dV

=∫ψ(r, θ, φ)∗ψ(r, θ, φ) dV = 1; (10.150)

where dV = r2dr sin θdθdϕ is the volume element. More specifically, we have∫ψnlml

(r, θ, φ)∗ψnlml(r, θ, φ) dV =

∞∫0

∫ψnlml

(r, θ, φ)∗ψnlml(r, θ, φ) r2drdΩ

=

∞∫0

π∫0

2π∫0

(Rnl(r)Θlml(θ)Φml

(φ))∗ (Rnl(r)Θlml(θ)Φml

(φ)) r2dr sin θdθdφ


=

∞∫0

Rnl(r)∗Rnl(r) r

2dr

π∫0

Θlml(θ)∗Θlml

(θ) sin θdθ2π∫0

Φml(φ)∗Φml

(φ) dφ = 1.

(10.151)

Now, it is absolutely NOT necessary that we have∞∫0

Rnl(r)∗Rnl(r) r

2dr =

π∫0

Θlml(θ)∗Θlml

(θ) sin θdθ =2π∫0

Φml(φ)∗Φml

(φ) dφ = 1

(10.152)

separately for the r-, θ-, and φ-dependent parts of the wavefunction ψ. However, itis often convenient to adjust the scaling factors so that (10.152) is indeed satisfied.Normalization for Φ is assured by

Φml(φ) =

1√2πeimlφ (10.153)

because∫ 2π

0Φ(φ)∗Φ(φ) dφ =

∫ 2π

0

1√2πe−imlφ

1√2πeimlφ dφ =

1

2π

∫ 2π

01 dφ = 1. (10.154)

On the other hand, from p.215, the normalization constant for Θ is√√√√(2l + 1)(l −m)!

2(l +ml)!, (10.155)

and, from (10.97), we have the following normalized Θlmlas well as Y ml

l (θ, φ) =Θlml

(θ)× Φml(φ).

Θlml= (−1)ml

√√√√(2l + 1)(l −ml)!

2(l +ml)!Pmll (cos θ) (10.156)

Finally, we need∞∫0

Rnl(r)∗Rnl(r) r

2dr = 1. (10.157)

This is satisfied by the Rnl(r) given in (10.134).

Rnl(r) =

√√√√( 2

na0

)3 (n− l − 1)!

2n[(n+ l)!]3e−r/na0

(2r

na0

)lL2l+1n−l−1

(2r

na0

)(10.158)

10.9. HYDROGEN-LIKE ATOMS 225

Definition 10.1 (Radial Probability Density P (r)) The radial probability den-sity P (r) is defined by

P (r) =∫ψ∗(r, θ, ϕ)ψ(r, θ, ϕ) r2dΩ =

∫ π

0

∫ 2π

0Ψ∗(r, θ, ϕ)Ψ(r, θ, ϕ) r2 sin θdθdϕ

=∫ +1

−1

∫ 2π

0ψ∗(r, θ, ϕ)ψ(r, θ, ϕ) r2d cos θdϕ, (10.159)

so that P (r0)dr gives the probability of finding the electron in the range [r0, r0 + dr],which is the spherical shell of thickness dr bounded by the surfaces of the spheres ofradii r0 and r0 + dr. Needless to say, P (r) satisfies∫ ∞

0P (r) dr = 1. (10.160)

10.9 Hydrogen-Like AtomsA hydrogen-like atom, or hydrogen-like ion, is an atomic nucleus with one electron.Some examples other than the hydrogen atom itself are He+, Li2+, Be3+, and B4+.If we denote the atomic number by Z, the charge carried by hydrogen-like ions aree(Z − 1). With this notation, we can derive all the wavefunctions for the hydrogen-like ions from the wavefunctions for the hydrogen atom by replacing the charge e onproton by Ze. You can see this from the time-independent Hamiltonian.

−~2

2µ

[1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)

+1

r2 sin2 θ

∂2

∂φ2

]− Ze2

4πε0r

ψ(r, θ, φ) = Eψ(r, θ, φ) (10.161)

The only difference is Z appearing once in the potential term. The resulting wave-functions are

ψnlml,Z = Rnl,Z(r)Ymll (θ, ϕ)

=

√√√√( 2Z

na0

)3 (n− l − 1)!

2n[(n+ l)!]3e−Zr/na0

(2Zr

na0

)lL2l+1n−l−1

(2Zr

na0

)×

√√√√2l + 1

4π

(l −ml)!

(l +ml)!Pmll (cos θ)eimlϕ

. (10.162)

And, the total energy EZ is given by


EZ,n = − Z2µe4

(4πε0)22~2n2= −

(Z2µe4

32π2ε20~2

)1

n2= −

(Z2~2

2µa20

)1

n2

for n = 1, 2, 3, · · · . (10.163)

Quantum Numbersn l Eigenfunctions Rnl(r)

1 0 R10 = 2(Za0

)3/2e−Zr/a0

2 0 R20 =1

2√2

(Za0

)3/2 (2− Zr

a0

)e−Zr/2a0

2 1 R21 =1

4√6

(Za0

)3/2Zra0e−Zr/2a0

3 0 R30 =1

9√3

(Za0

)3/2 (6− 4Zr

a0+ 4Z2r2

9a20

)e−Zr/3a0

3 1 R31 =1

27√6

(Za0

)3/2 (4− 2Zr

3a0

)2Zra0e−Zr/3a0

3 2 R32 =1

81√30

(Za0

)3/24Z2r2

a20e−Zr/3a0

Table 10.2: Some Normalized Radial Eigenfunctions for the One-Electron Atom

10.10 Simultaneous Diagonalization of H, L2, andLz

Let us examine the big picture now. In Section 10.6, we saw that the sphericalharmonics

Y mll

−l≤ml≤+ll=0,1,2,... (10.164)

form an orthonormal basis that simultaneously diagonalizes L2 and Lz. Likewise,the set of normalized eigenfunctions or eigenkets

ψnlml or |n, l,ml⟩ (10.165)

10.10. SIMULTANEOUS DIAGONALIZATION OF H, L2, AND LZ 227

Quantum Numbersn l ml Eigenfunctions ψnlml

(r, θ, ϕ)

1 0 0 ψ100 =1√π

(Za0

)3/2e−Zr/a0

2 0 0 ψ200 =1

4√2π

(Za0

)3/2 (2− Zr

a0

)e−Zr/2a0

2 1 0 ψ210 =1

4√2π

(Za0

)3/2Zra0e−Zr/2a0 cos θ

2 1 ±1 ψ21±1 =1

8√π

(Za0

)3/2Zra0e−Zr/2a0 sin θ e±iϕ

3 0 0 ψ300 =1

18√3π

(Za0

)3/2 (6− 4Zr

a0+ 4Z2r2

9a20

)e−Zr/3a0

3 1 0 ψ310 =1

27√2π

(Za0

)3/2 (4− 2Zr

3a0

)Zra0e−Zr/3a0 cos θ

3 1 ±1 ψ31±1 =1

54√π

(Za0

)3/2 (4− 2Zr

3a0

)Zra0e−Zr/3a0 sin θ e±iϕ

3 2 0 ψ320 =1

81√6π

(Za0

)3/2Z2r2

a20e−Zr/3a0(3 cos2 θ − 1)

3 2 ±1 ψ32±1 =1

81√π

(Za0

)3/2Z2r2

a20e−Zr/3a0 sin θ cos θ e±iϕ

3 2 ±2 ψ32±2 =1

162√π

(Za0

)3/2Z2r2

a20e−Zr/3a0 sin2 θ e±i2ϕ

Table 10.3: Some Normalized Full Eigenfunctions for the One-Electron Atom


for n = 1, 2, 3, . . ., l = 0, 1, . . . , n − 1, and ml = −l,−l + 1, . . . 0 . . . , l − 1,+lform an orthonormal basis that simultaneously diagonalizes the Hamiltonian H, themagnitude of the orbital angular momentum squared L2, and the z-component of theorbital angular momentum Lz. We have the following set of simultaneous eigenvalueproblems with a common solution |n, l,ml⟩.

H |ψ⟩ = En |ψ⟩L2 |ψ⟩ = l(l + 1)~2 |ψ⟩ (10.166)Lz |ψ⟩ = ml~ |ψ⟩

The reason why such common eigenkets can be found is that the commutators [H,L2]and [H,Lz] are both zero in addition to [L2, Lz] = 0 which we know from (9.8). Asyou can see in Appendix L, there are alternative forms of ∇2. The form given in(10.27) is

∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2, (10.167)

and one alternative form is

∇2 =∂2

∂r2+

2

r

∂

∂r+

1

r2∂2

∂θ2+

cos θr2 sin θ

∂

∂θ+

1

r2 sin2 θ

∂2

∂ϕ2(10.168)

or

∇2 =∂2

∂r2+

2

r

∂

∂r+

1

r2

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

]. (10.169)

From (9.4) and (10.169), we have

∇2 =∂2

∂r2+

2

r

∂

∂r+

1

r2

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

]

=∂2

∂r2+

2

r

∂

∂r− 1

~2r2(−~2)

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

]

=∂2

∂r2+

2

r

∂

∂r− L2

~2r2. (10.170)

We will use this form of ∇2 in order to show [H,L2] = 0 and [H,Lz] = 0.8

8Both [H,L2] = 0 and [H,Lz] = 0 are proved in a different manner in Appendix I.

10.10. SIMULTANEOUS DIAGONALIZATION OF H, L2, AND LZ 229

Fact 10.2 ([H, L2] = 0) The full Hamiltonian

H =−~2

2µ∇2 + V (r) =

−~2

2µ

[∂2

∂r2+

2

r

∂

∂r− L2

~2r2

]+ V (r) (10.171)

commutes with the squared magnitude of the angular momentum

L2 = −~2[

1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

]. (10.172)

ProofThe important point here is that the expression of H contains only the radial variabler and L2, while the expression of L2 contains the azimuthal and polar variables ϕand θ. With this in mind, the commutator [H,L2] can be computed as follows.

[H,L2

]=

[−~2

2µ

[∂2

∂r2+

2

r

∂

∂r− L2

~2r2

]+ V (r), L2

]

=

[−~2

2µ

[∂2

∂r2+

2

r

∂

∂r− L2

~2r2

], L2

]+[V (r), L2

]=

[−~2

2µ

[∂2

∂r2+

2

r

∂

∂r

], L2

]+

[L2

2µr2, L2

]+[V (r), L2

]. (10.173)

It is now clear that each term of (10.173) is zero, and we have verified[H,L2

]= 0. (10.174)

Fact 10.3 ([H, Lz] = 0) The full Hamiltonian

H =−~2

2µ∇2 + V (r) =

−~2

2µ

[∂2

∂r2+

2

r

∂

∂r− L2

~2r2

]+ V (r) (10.175)

commutes with the z-component of the angular momentum

Lz = −i~∂

∂ϕ. (10.176)


Proof

[H,Lz] =

[−~2

2µ

[∂2

∂r2+

2

r

∂

∂r− L2

~2r2

]+ V (r), Lz

]

=

[−~2

2µ

[∂2

∂r2+

2

r

∂

∂r

], Lz

]+

[L2

2µr2, Lz

]+ [V (r), Lz] = 0. (10.177)

10.11 Revisiting the Fundamental PostulatesLet us recall the Fundamental Postulates presented in Section 3.1. The followingis the set of guiding principles that connects the physical reality with mathematicalabstraction.

• Each physical system S has an associated Hilbert space H.

• Each physical state s of the physical system S has an associated normalizedket |s⟩, called a state ket, in H.

• Every physical observable q is represented by a Hermitian Operator Q whosedomain is H.

• Measurement of q in the physical system S is mirrored in the abstract quantummechanical system by the action of the corresponding operator Q on the stateket |s⟩.

• The only possible outcomes of quantum mechanical measurements are theeigenvalues of the corresponding operator Q.

• A state ket is initially a linear combination of normalized eigenkets of Q. How-ever, after the measurement, the state ket becomes the eigen ket associated withthe measured eigenvalue.

You can note here that the quantum mechanical system is composed of the threesome(H, |s⟩, Q), which are purely mathematical and abstract constructs. In particular,no reference is made to any coordinate system or a function defined relative to such

10.11. REVISITING THE FUNDAMENTAL POSTULATES 231

coordinates.

Therefore, the generic simultaneous eigenvalue problems (10.166) and their solutions(10.165) are actually coordinate free. In other words, the eigenkets |n, l,ml⟩ “live”in some abstract unspecified Hilbert space H. From Section 4.6, we can expand|n, l,ml⟩ in the coordinate basis |r, θ, ϕ⟩ to get

|n, l,ml⟩ = I |n, l,ml⟩ =

∑0≤r<+∞

∑0≤θ≤π

∑0≤ϕ<2π

|r, θ, ϕ⟩ ⟨r, θ, ϕ|

|n, l,ml⟩

=∑

0≤r<+∞

∑0≤θ≤π

∑0≤ϕ<2π

⟨r, θ, ϕ|n, l,ml⟩ |r, θ, ϕ⟩ (10.178)

=∑

0≤r<+∞

∑0≤θ≤π

∑0≤ϕ<2π

ψnlml(r, θ, ϕ) |r, θ, ϕ⟩ (10.179)

or more appropriately

|n, l,ml⟩ =

∞∫0

π∫0

2π∫0

|r, θ, ϕ⟩ ⟨r, θ, ϕ| r2dr sin θdθdϕ |n, l,ml⟩

=

∞∫0

π∫0

2π∫0

⟨r, θ, ϕ|n, l,ml⟩ |r, θ, ϕ⟩ r2dr sin θdθdϕ (10.180)

=

∞∫0

π∫0

2π∫0

ψnlml(r, θ, ϕ) |r, θ, ϕ⟩ r2dr sin θdθdϕ; (10.181)

where we used the identities∑0≤r<+∞

∑0≤θ≤π

∑0≤ϕ<2π

|r, θ, ϕ⟩ ⟨r, θ, ϕ| = I

and∞∫0

π∫0

2π∫0

|r, θ, ϕ⟩ ⟨r, θ, ϕ| r2dr sin θdθdϕ = I.

When we write

ψnlml(r, θ, ϕ) = ⟨r, θ, ϕ|n, l,ml⟩ (10.182)

for the coefficients in (10.178) and (10.180), we are taking the projection of |n, l,ml⟩onto the infinite-dimensional coordinate system whose axes are defined by |r, θ, ϕ⟩,


to use a geometric language. In a more common language, this amounts to solv-ing the set of eigenvalue problems (10.166) using the spherical coordinate system(r, θ, ϕ) in order to obtain the eigenvectors as functions of r, θ, and ϕ. The operatorsH, L2, and Lz are also abstract and coordinate-independent generic objects whoserepresentations in the spherical coordinate system are

H =−~2

2µ

[1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2

]+ V (r), (10.183)

L2 = −~2[

1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

], (10.184)

and

Lz = −i~∂

∂ϕ. (10.185)

Similarly, we can project |n, l,ml⟩ onto the Cartesian system defined by the familiar|x, y, z⟩.9

ψnlml(x, y, z) = ⟨x, y, z|n, l,ml⟩ (10.186)

In this case, we already know

H =−~2

2µ

[∂2

∂x2+

∂2

∂y2+

∂2

∂z2

]+ V (x, y, z) (10.187)

and

Lz = −i~(x∂

∂y− y ∂

∂x

). 10 (10.188)

9This is not to be confused with the usual projection onto the x-, y-, and z-axes of a vectorv, which allows us to write v = (vx, vy, vz). Our coordinate system here is not the usual three-dimensional system, but it is of infinite dimensions, each axis of which is defined by one of |x, y, z⟩;where −∞ < x, y, z < +∞.

10As we know Lx = −i~(y ∂

∂z − z∂

∂y

)and Ly = −i~

(z ∂

∂x − x∂

∂z

)as well as L2 = L2

x + L2y + L2

z,we can also express L2 in terms of x, y, z and ∂

∂x ,∂

∂y ,∂

∂z . However, the resulting messy formula isnot of much instructional value or other use to us. That is the reason why L2 is not included inthis list.

233

Exercises1. Consider a one-variable function ψ(r) for r =

√x2 + y2 + z2, and derive the

relation below.∂2ψ

∂x2=

1

r

∂ψ

∂r+x2

r

∂

∂r

(1

r

∂ψ

∂r

)

2. Without showing detailed computations, simply sketch the proof that

Θlml= (1− z2)|ml|/2d

|ml|Pldz|ml|

.

3. Hydrogen, deuterium, and singly ionized helium are all examples of one-electronatoms. The deuterium nucleus has the same charge as the hydrogen nucleus,and almost exactly twice the mass. The helium nucleus has twice the chargeof the hydrogen nucleus, and almost exactly four times the mass. Make anapproximate prediction of the ratios of the ground state energies of these atoms.

4. Consider n = 3 state of a hydrogen atom.

(a) What l values are possible?(b) For each value of l, what ml values are possible?(c) Finally, how many degenerate states do we have for n = 3?

5. What is the energy of a photon emitted when the electron drops from the 3rdhighest energy level (n = 3) to the ground state (n = 1)? Leave µ, e, π, ε0,and ~ as they are.

6. Answer the following questions about the hydrogen atom.

(a) The differential equation of the ϕ-dependent function Φ(ϕ) is given byd2Φdϕ2

= −k2Φ with a solution Φ(ϕ) = eikϕ. Find all values of k = a + bi

(a, b real) consistent with the single-valuedness condition. Show all yourwork.

(b) For each principal quantum number n, there are n2 degenerate energylevels. Prove this.

7. Answer the following questions about the hydrogen atom.

(a) Consider n = 4.

234

i. What is the largest allowed value of l?ii. What is the magnitude of the corresponding angular momentum?

Leave ~ as it is.iii. How many different z-components may this angular momentum vec-

tor have?iv. What is the magnitude of the largest z component? Leave ~ as it is.

(b) Consider an electron in the ground state of the hydrogen atom charac-terized by n = 0. The normalized ground state wavefunction is givenby

ψ000 =1√π

(1

a0

)3/2

e−r/a0 .

You may use∫x2ebx dx = ebx

(x2

b− 2x

b2+

2

b3

)+ C

and∫x3ebx dx = ebx

(x3

b− 3x2

b2+

6x

b3− 6

b4

)+ C.

i. What is the probability P (0 ≤ r < a0) that the electron lies inside asphere of radius a0 centered at the origin? Leave e as it is.

ii. Find the average distance < r > of the electron from the nucleus.(c) For a hydrogen-like atom with Z = 3, what is the energy of a photon

emitted when the electron drops from the 3rd highest energy level (n = 3)to the ground state (n = 1)? Leave µ, e, π, ε0, and ~ as they are. Do notuse a0 in your answer.

Chapter 11

Electron Spin

There are quantum mechanical phenomena that do not fit in the framework based onthe postulates given in Section 3.1 in a straightforward manner. While the postulatesdraw heavily on the analogy based on the corresponding classical mechanical system,these phenomena have no classical analogues. They require an additional variablecalled intrinsic spin and a special Hilbert space on which the spin operators act. Theintrinsic spin S is a very important signature for many particles. We will devote thischapter to the discussion of electron spin as it is the most representative example.Intrinsic spin can be characterized and understood best if we regard it as anotherkind of angular momentum that requires a separate treatment different from thatfor the familiar orbital angular momentum represented by Lx, Ly, and Lz.

11.1 What is the Electron Spin?

The electron was discovered in 1897 by Joseph John Thompson. It was the firstsubatomic and fundamental particle discovered. Physicists knew from the beginningthat an electron had charge and mass. But, the concept of electron spin was proposedby Samuel Abraham Goudsmit and George Uhlenbeck much later in 1925. Becausethe electron was usually regarded as a point particle with no internal structure andno physical dimension, it was actually impossible for the electron to spin like a top.Therefore, it was necessary to assume that the electron was born with intrinsic angu-lar momentum, called electron spin, which is not associated with its orbital motion.The first sign of electron spin came from an experiment conducted by Otto Sternand Walther Gerlach in 1921.

235

236 CHAPTER 11. ELECTRON SPIN

11.1.1 Stern-Gerlach ExperimentIn this experiment, electrically neutral silver atoms passed through a nonuniformmagnetic field B = Bz, oriented along the z-axis, which deflected the atoms. Theatoms, after passing through the magnetic field, hit a photographic plate generatingsmall visible dots. The deflection is along the z-axis and the magnitude of the forceon the atoms is proportional to dB

dz. In the classical picture, the orbiting electron

produces magnetic dipole moments whose magnitudes are continuously distributed.If this is the case, the extent of vertical deflection should also be continuously dis-tributed forming a “blob” on the plate. On the other hand, in the quantum mechan-ical picture, there are 2l + 1 z-components, ml’s, for l = 1, 2, 3, . . .. In particular,note that ml = 0 is always possible. When ml = 0, there is no force on the atomand the atoms with ml = 0 will not be deflected. Hence, if l = 0, no atom will bedeflected.

Figure 11.1: Stern-Gerlach Experiment (Diagram drawn by en wikipedia TheresaKnott.)

The results of Stern-Gerlach experiment presented in their original paper arereproduced in Figure 11.2 [Gerlach and Stern, 1922, p.350]. The picture on the leftshows only one line as there was no external magnetic field. When an external non-uniform magnetic field was turned on, there were two lines shown on the right due tothe force proportional to dB

dz. However, this was inconsistent with their expectations

in two ways.

11.1. WHAT IS THE ELECTRON SPIN? 237

1. There was no line at the center corresponding to ml = 0.

2. There should be an odd number of lines; namely 2l + 1. But, they observedtwo lines.

Figure 11.2: Results of Stern-Gerlach Experiment with and without a non-uniform external magnetic field (Source: Stern and Gerlach’s original paper[Gerlach and Stern, 1922, p.350])

Later in 1927, T. E. Phipps and J. B. Taylor reproduced this effect using hydrogenatoms in the ground state (l = 0 andml = 0). This was a convincing piece of evidencethat there was “something” other than the orbital angular momentum because l = 0implies the orbital angular momentum is 0.

11.1.2 Fine Structure of the Hydrogen Spectrum: Spin-OrbitInteraction

Generally speaking, the term fine structure refers to the splitting of spectral linesdue to first order relativistic effects. These relativistic effects add three fine structureterms to the Hamiltonian known as kinetic, spin-orbit interaction, and Darwin terms.And it is the spin-orbit splitting that provided one of the first pieces of evidence forelectron spin.

We relegate the discussion of these relativistic effects to Appendix O.


11.2 Spin and Pauli MatricesOne way to account for these apparent discrepancies is to assume the existence of afourth quantum number known as a spin quantum number denoted by s.1 We caneasily see that if we assume the magnitude of the spin S is given by S = ~

√s(s+ 1)

and s = 1/2, the number of values its z-component Sz = ms~ can take is 2s+ 1 = 2,and we can successfully explain the two-lines observed. In a nutshell, this is whatwe are going to do. While we will specialize to the electron shortly, spin itself is anintrinsic property possessed by any fundamental particle. Every fundamental parti-cle has its own unique signature value of s that characterizes the particle. While ms

varies between −s and +s, s itself does not.

As mentioned on p.235, spin is best understood as intrinsic angular momentum.According to Section 9.6, this means that the three components of the spin satisfythe same commutation relations as Lx, Ly, and Lz; or their generalized counterpartsJx, Jy, and Jz. Because spin is a generalized angular momentum, we can understandspin if we understand the quantum mechanical properties of angular momentum.Here are the basic properties of spin as a quantum mechanical angular momentum.

S = Sxx + Syy + Szz (11.1)S2 = S2

x + S2y + S2

z (11.2)[Sx, Sy] = i~Sz (11.3)[Sy, Sz] = i~Sx (11.4)[Sz, Sx] = i~Sy (11.5)

[S2, Sx] = [S2, Sy] = [S2, Sz] = [S2, S] = 0 (11.6)

From these properties and the results of actual measurements, we can construct thefollowing framework to describe and understand the electron spin. We will find an or-thonormal basis that diagonalizes S2 and Sz simultaneously and find (non-diagonal)matrix representations for Sx and Sy in that basis.

First, based on the outcomes of various experiments, we know s = 1/2 andms = ±1/2. This means S2 = s(s+ 1)~2I = 1

2

(12+ 1

)~2I = 3

4~2I, where I is the

1In the framework of our quantum theory based on the Schrodinger equation, spin is an a poste-riori consideration. However, spin appears naturally when the relativistic analog of the Schrodingerequation, known as the Dirac equation, is used. Spin is a consequence of demanding that the Diracequation be first order in time and space.

11.2. SPIN AND PAULI MATRICES 239

identity operator, and the eigenvalues of Sz are ± ~2. Now, due to Properties 2) and

3) on p.54 as well as Theorem 2.7 on p.58, two normalized eigenvectors of Sz form anorthonormal basis for the corresponding Hilbert space. We will denote a normalizedeigenvector for the eigenvalue + ~

2by |↑⟩ and that for the eigenvalue − ~

2by |↓⟩. Using

the notation of Section 9.6 on generalized angular momentum, |j,mj⟩, we have

|↑⟩ = |s = 1/2,ms = 1/2⟩ = |1/2, 1/2⟩ (11.7)and

|↓⟩ = |1/2,−1/2⟩ . (11.8)

Because our Hilbert space is two-dimensional, all operators can be represented by2-by-2 matrices. Let us compute the matrix elements for S2, Sz, Sx, and Sy.

Let us arrange |↑⟩ and |↓⟩ in this order, so that

|↑⟩ =[10

](11.9)

and

|↓⟩ =[01

], (11.10)

and express the i-th row and j-th column component of the (2-by-2) matrix repre-sentation of an operator Ω by Ωij. For S2,

S211 = ⟨↑ |S2| ↑⟩ =

⟨↑∣∣∣∣34~2I

∣∣∣∣ ↑⟩ =3

4~2 ⟨↑ | ↑⟩ = 3

4~2 (11.11)

S212 = ⟨↑ |S2| ↓⟩ =

⟨↑∣∣∣∣34~2I

∣∣∣∣ ↓⟩ =3

4~2 ⟨↑ | ↓⟩ = 0 (11.12)

S221 = ⟨↓ |S2| ↑⟩ =

⟨↓∣∣∣∣34~2I

∣∣∣∣ ↑⟩ =3

4~2 ⟨↓ | ↑⟩ = 0 (11.13)

S222 = ⟨↓ |S2| ↓⟩ =

⟨↓∣∣∣∣34~2I

∣∣∣∣ ↓⟩ =3

4~2 ⟨↓ | ↓⟩ = 3

4~2 (11.14)

⇓


S2 =

[34~2 00 3

4~2

]=

3

4~2[1 00 1

]. (11.15)

For Sz,

Sz11 = ⟨↑ |Sz| ↑⟩ =⟨↑∣∣∣∣+~2

∣∣∣∣ ↑⟩ = +~

2⟨↑ | ↑⟩ = +

~

2(11.16)

Sz12 = ⟨↑ |Sz| ↓⟩ =⟨↑∣∣∣∣−~2

∣∣∣∣ ↓⟩ = −~2⟨↑ | ↓⟩ = 0 (11.17)

Sz21 = ⟨↓ |Sz| ↑⟩ =⟨↓∣∣∣∣+~2

∣∣∣∣ ↑⟩ = +~

2⟨↓ | ↑⟩ = 0 (11.18)

Sz22 = ⟨↓ |Sz| ↓⟩ =⟨↓∣∣∣∣−~2

∣∣∣∣ ↓⟩ = −~2⟨↓ | ↓⟩ = −~

2(11.19)

⇓

Sz =

[+ ~

20

0 − ~2

]=~

2

[1 00 −1

]. (11.20)

In order to compute the matrix representations of Sx and Sy, it is convenient to useraising and lowering operators S+ = Sx + iSy and S− = Sx − iSy. Solving for Sx andSy, we get

Sx =1

2(S+ + S−) (11.21)

and

Sy =1

2i(S+ − S−). (11.22)

From Section 9.6, we have

S+ |s,ms⟩ = ~√(s−ms)(s+ms + 1) |s,ms + 1⟩ (11.23)

and

S− |s,ms⟩ = ~√(s+ms)(s−ms + 1) |s,ms − 1⟩ .a (11.24)


aRecall that these relations hold for any generalized momentum, including, of course, electronspin. All we require are the commutation relations

[Jx, Jy] = i~Jz,

[Jy, Jz] = i~Jx,

and[Jz, Jx] = i~Jy;

where J is S for us.

These relations give the following.

S+ |↑⟩ = S+ |s = 1/2,ms = 1/2⟩ = 0 (11.25)

S+ |↓⟩ = S+ |1/2,−1/2⟩ = ~√(

1

2−(−1

2

))(1

2+(−1

2

)+ 1

)|1/2, 1/2⟩

= ~ |1/2, 1/2⟩ = ~ |↑⟩ (11.26)

S− |↑⟩ = S− |1/2, 1/2⟩ = ~√(

1

2+

1

2

)(1

2− 1

2+ 1

)|1/2,−1/2⟩

= ~ |1/2,−1/2⟩ = ~ |↓⟩ (11.27)S− |↓⟩ = S− |1/2,−1/2⟩ = 0 (11.28)

For Sx = 12(S+ + S−),

Sx11 = ⟨↑ |Sx| ↑⟩ =⟨↑∣∣∣∣12 (S+ + S−)

∣∣∣∣ ↑⟩ =1

2(⟨↑ |S+| ↑⟩+ ⟨↑ |S−| ↑⟩)

=1

2(0 + ⟨↑ |~| ↓⟩) = 1

2(0 + 0) = 0 (11.29)

Sx12 = ⟨↑ |Sx| ↓⟩ =⟨↑∣∣∣∣12 (S+ + S−)

∣∣∣∣ ↓⟩ =1

2(⟨↑ |S+| ↓⟩+ ⟨↑ |S−| ↓⟩)

=1

2(⟨↑ |~| ↑⟩+ 0) =

~

2(11.30)

Sx21 = ⟨↓ |Sx| ↑⟩ =⟨↓∣∣∣∣12 (S+ + S−)

∣∣∣∣ ↑⟩ =1

2(⟨↓ |S+| ↑⟩+ ⟨↓ |S−| ↑⟩)

=1

2(0 + ⟨↓ |~| ↓⟩) = ~

2(11.31)

Sx22 = ⟨↓ |Sx| ↓⟩ =⟨↓∣∣∣∣12 (S+ + S−)

∣∣∣∣ ↓⟩ =1

2(⟨↓ |S+| ↓⟩+ ⟨↓ |S−| ↓⟩)


=1

2(⟨↓ |~| ↑⟩+ 0) =

1

2(0 + 0) = 0 (11.32)

⇓

Sx =

[0 ~

2~2

0

]= ~

2

[0 11 0

]. (11.33)

For Sy = 12i(S+ − S−),

Sy11 = ⟨↑ |Sy| ↑⟩ =⟨↑∣∣∣∣ 12i (S+ − S−)

∣∣∣∣ ↑⟩ =1

2i(⟨↑ |S+| ↑⟩ − ⟨↑ |S−| ↑⟩)

=1

2i(0− ⟨↑ |~| ↓⟩) = 1

2i(0− 0) = 0 (11.34)

Sy12 = ⟨↑ |Sy| ↓⟩ =⟨↑∣∣∣∣ 12i (S+ − S−)

∣∣∣∣ ↓⟩ =1

2i(⟨↑ |S+| ↓⟩ − ⟨↑ |S−| ↓⟩)

=1

2i(⟨↑ |~| ↑⟩+ 0) =

~

2i= −i~

2(11.35)

Sy21 = ⟨↓ |Sy| ↑⟩ =⟨↓∣∣∣∣ 12i (S+ − S−)

∣∣∣∣ ↑⟩ =1

2i(⟨↓ |S+| ↑⟩ − ⟨↓ |S−| ↑⟩)

=1

2i(0− ⟨↓ |~| ↓⟩) = − ~

2i= i~

2(11.36)

Sy22 = ⟨↓ |Sy| ↓⟩ =⟨↓∣∣∣∣ 12i (S+ − S−)

∣∣∣∣ ↓⟩ =1

2i(⟨↓ |S+| ↓⟩ − ⟨↓ |S−| ↓⟩)

=1

2i(⟨↓ |~| ↑⟩+ 0) =

1

2i(0 + 0) = 0 (11.37)

⇓

Sy =

[0 −i ~

2

i ~2

0

]= ~

2

[0 −ii 0

]. (11.38)

Definition 11.1 (Pauli Matrices) The Pauli matrices σ is defined by

S =~

2σ; (11.39)

where S = (Sx, Sy, Sz) and σ = (σx, σy, σz). Component by component, we havethe following matrix representations of σi’s in the orthonormal basis consisting ofnormalized eigenvectors of Sz.


σx =

[0 11 0

], σy =

[0 −ii 0

], and σz =

[1 00 −1

]. (11.40)

In addition, we often define σ0 as follows; namely, σ0 is the identity matrix.

σ0 =

[1 00 1

](11.41)

Noting that

iσy =

[0 1−1 0

], (11.42)

we can see

1

2(σx + iσy) =

1

2

([0 11 0

]+

[0 1−1 0

])=

[0 10 0

], (11.43)

Similarly,

1

2(σx − iσy) =

1

2

([0 11 0

]−[

0 1−1 0

])=

1

2

[0 02 0

]=

[0 01 0

]. (11.44)

1

2(σ0 + σz) =

1

2

([1 00 1

]+

[1 00 −1

])=

[1 00 0

], (11.45)

and1

2(σ0 − σz) =

1

2

([1 00 1

]−[1 00 −1

])=

[0 00 1

]. (11.46)

Therefore, σ0, σx, σy, σz forms a basis for the vector space spanned by

[1 00 0

],

[0 10 0

],

[0 01 0

], and

[0 00 1

]. (11.47)

Properties of Pauli Matrices

1. |↑⟩ and |↓⟩ are normalized eigenvectors of σz, and also Sz. Similarly, |↑⟩xand |↓⟩x are normalized eigenvectors of σx, and |↑⟩y and |↓⟩y are normalized


eigenvectors of σy. Here, |↑⟩x and |↓⟩x are the states corresponding to spin upand spin down along the x-axis; and likewise for |↑⟩y and |↓⟩y.2

2. Pauli spin matrices are Hermitian. For example,

σ†y =

[0 −ii 0

]†

=

[0 −ii 0

]= σy, (11.48)

and similarly for σx and σz.

3. Pauli matrices are unitary. For example,

σ†yσy =

[0 −ii 0

]† [0 −ii 0

]=

[0 −ii 0

] [0 −ii 0

]

=

[(−i)i 0

0 i(−i)

]=

[1 00 1

]. (11.49)

Likewise for σx and σz.

4. Pauli matrices anti-commute; i.e. the anti-commutator

[σi, σj]+ = σiσj + σjσi = 0 if i , j and i, j , 0.3 (11.50)

5. We have the following cyclic relations.

σxσy = iσz, σyσz = iσx, and σzσx = iσy. (11.51)

6. Pauli matrices are traceless.

Tr σx = Tr σy = Tr σz = 0 (11.52)

7. Pauli matrices are idempotent, which is a direct consequence of Properties 2and 3.

σ2i = I. (11.53)

2The eigenvalues of Sx, Sy, and Sz are ± ~2 ; while the eigenvalues are ±1 for the rescaled σx, σy,and σz.

3Some authors use for the anti-commutator. However, is used for the Poisson bracket inclassical mechanics. So, we will use [ ]+ instead.

11.3. SEQUENTIAL MEASUREMENTS 245

11.3 Sequential MeasurementsWe will now consider consecutive Stern-Gerlach type experiments. This will demon-strate how Postulates 3, 5, and 6 from Chapter 3 work.

We need to do some preliminary work first. In the Sz basis |↑⟩ , |↓⟩, we have

σx|↑⟩+ |↓⟩√

2=

1√2

[0 11 0

]([10

]+

[01

])=

1√2

[0 11 0

] [11

]

=1√2

[11

]= 1 · 1√

2

([10

]+

[01

])= 1 · |↑⟩+ |↓⟩√

2(11.54)

and

σx|↑⟩ − |↓⟩√

2=

1√2

[0 11 0

]([10

]−[01

])=

1√2

[0 11 0

] [1−1

]

=1√2

[−11

]= −1 · 1√

2

([10

]−[01

])= −1 · |↑⟩ − |↓⟩√

2. (11.55)

Therefore,

|↑⟩x =1√2(|↑⟩+ |↓⟩) (11.56)

and

|↓⟩x =1√2(|↑⟩ − |↓⟩). (11.57)

Similarly,

σy|↑⟩+ i |↓⟩√

2=

1√2

[0 −ii 0

]([10

]+

[0i

])=

1√2

[0 −ii 0

] [1i

]

=1√2

[1i

]= 1 · 1√

2

([10

]+

[0i

])= 1 · |↑⟩+ i |↓⟩√

2(11.58)

and

σy|↑⟩ − i |↓⟩√

2=

1√2

[0 −ii 0

]([10

]−[0i

])=

1√2

[0 −ii 0

] [1−i

]

=1√2

[−1i

]= −1 · 1√

2

([10

]−[0i

])= −1 · |↑⟩ − i |↓⟩√

2. (11.59)


Therefore,

|↑⟩y =1√2(|↑⟩+ i |↓⟩) (11.60)

and

|↓⟩y =1√2(|↑⟩ − i |↓⟩). (11.61)

Note that (11.56), (11.57), (11.60), and (11.61) also mean

|↑⟩ = 1√2(|↑⟩x + |↓⟩x) =

1√2(|↑⟩y + |↓⟩y) (11.62)

and

|↓⟩ = 1√2(|↑⟩x − |↓⟩x) =

1√2i(|↑⟩y − |↓⟩y) =

−i√2(|↑⟩y − |↓⟩y). (11.63)

Now we are ready to embark on a study of sequential Stern-Gerlach experiments/measurements. Stern-Gerlach device does not only allow us to detect two compo-nents, spin up and spin down, but also enables us to isolate one variety of electrons,be it spin up or spin down. This is the feature we will need.

We start with a state

α |↑⟩+ β |↓⟩ with |α|2 + |β|2 = 1. (11.64)

The condition on α and β assures that the linear combination α |↑⟩+ β |↓⟩ is of unitlength; i.e. it is indeed a state. The fact that we can only observe Sz = + ~

2or Sz = ~

2

is a consequence of Postulate 3 and the matrix representation of Sz given by (11.20),which implies that the eigenvalues of Sz are indeed + ~

2and − ~

2. After the electrons

in this state go through the first Stern-Gerlach device, you will observe spin up |↑⟩and spin down |↓⟩ states with probabilities |α|2 and |β|2, respectively according toPostulate 5.

If an electron with the spin wavefunction (11.64) is found to have an up-spin,Postulate 6 forces its wavefunction to be |↑⟩ after the measurement. So, we have

11.3. SEQUENTIAL MEASUREMENTS 247

α |↑⟩+ β |↓⟩ first measurement−−−−−−−−−−−−−−−−−−→ |↑⟩ . (11.65)

Having isolated |↑⟩, we now have a pure eigenstate with an up-spin, and the secondStern-Gerlach device find |↑⟩ with a probability 1.0. This obvious consequence isdescribed in (11.66).

α |↑⟩+ β |↓⟩ first measurement−−−−−−−−−−−−−−−−−−→ |↑⟩ second measurement−−−−−−−−−−−−−−−−−−−−→ |↑⟩ (11.66)

Next, let us see what will happen if the second Stern-Gerlach device is rotated by90 so that it is now in the x-direction. With what probabilities will we observe |↑⟩xand |↓⟩x? The answer lies in (11.62), which gives

|↑⟩ = 1√2(|↑⟩x + |↓⟩x). (11.67)

According to Postulate 5, the probabilities of observing each eigenstate is given by∣∣∣ 1√2

∣∣∣2. Therefore, we will observe the up-spin 50% of the time and the down-spin50% of the time.

α |↑⟩+ β |↓⟩ first measurement−−−−−−−−−−−−−−−−−−−−−−−−−−→to choose the spin-up state

|↑⟩

second measurement−−−−−−−−−−−−−−−−−−−−→in the x-direction

|↑⟩x (50%) and |↓⟩x (50%) (11.68)

Similarly, we have

α |↑⟩+ β |↓⟩ first measurement−−−−−−−−−−−−−−−−−−−−−−−−−−−−→to choose the spin-down state

|↓⟩

second measurement−−−−−−−−−−−−−−−−−−−−→in the x-direction

|↑⟩x (50%) and |↓⟩x (50%) (11.69)

as

|↓⟩ = 1√2(|↑⟩x − |↓⟩x) (11.70)


from (11.63).

What if |↑⟩x goes through a Stern-Gerlach device oriented in the y-direction? Bysymmetry, we should observe the up-spin and down-spin states with a probability of0.5, respectively. Let us double check this by an explicit computation. By (11.56),(11.62), and (11.63),

|↑⟩x =1√2(|↑⟩+ |↓⟩) = 1√

2

(1√2(|↑⟩y + |↓⟩y) +

1√2i(|↑⟩y − |↓⟩y)

)

=1

2(|↑⟩y + |↓⟩y) +

1

2i(|↑⟩y − |↓⟩y) =

1

2(|↑⟩y + |↓⟩y)−

1

2i(|↑⟩y − |↓⟩y)

=1− i2|↑⟩y +

1 + i

2|↓⟩y . (11.71)

Because

∣∣∣∣1− i2

∣∣∣∣2 =( √

2

2

)2

=1

2and

∣∣∣∣1 + i

2

∣∣∣∣2 =( √

2

2

)2

=1

2, (11.72)

our intuition is supported mathematically according to Postulate 5.

11.4 Spin States in an Arbitrary DirectionAs our system possesses spherical symmetry, the choice of the z-axis is arbitrary.In other words, we should observe the same physics despite different choices of thez-direction. The following is a mathematical treatment of this fact.

11.4.1 Spin Matrix Su and Its EigenvectorsRecall the spin operator S given in (11.1).

S = Sxx + Syy + Szz. (11.73)

This is so designed that the following “inner products” provide the spin matrices inthe x-, y-, and z-directions.

S · x = Sx, S · y = Sy, S · z = Sz (11.74)

11.4. SPIN STATES IN AN ARBITRARY DIRECTION 249

Analogously, the spin matrix in the direction defined by a unit vector u extendingfrom the origin to the point (r, θ, ϕ), u = (sin θ cosϕ, sin θ sinϕ, cosϕ), is given by

S · u = (Sx, Sy, Sz) · (sin θ cosϕ, sin θ sinϕ, cos θ)= sin θ cosϕSx + sin θ sinϕSy + cos θ Sz

= sin θ cosϕ[0 ~

2~2

0

]+ sin θ sinϕ

[0 −i ~

2

i ~2

0

]+ cos θ

[+ ~

20

0 − ~2

]

=~

2

[cos θ sin θ(cosϕ− i sinϕ)

sin θ(cosϕ+ i sinϕ) − cos θ

]=~

2

[cos θ sin θe−iϕ

sin θeiϕ − cos θ

](11.75)

Let us compute the eigenvalues and eigenvectors of (11.75). It suffices to consider[cos θ sin θe−iϕ


](11.76)

instead. If we denote the eigenvalues by λ and λ′, we have, by (A.8),

λ+ λ′ = trace[

cos θ sin θe−iϕ


]= 0 =⇒ λ′ = −λ (11.77)

and

λλ′ = −λ2 = det[



]= − cos2 θ − sin2 θ = −1. (11.78)

So, the two eigenvalues of[



]are +1 and −1, which in turn im-

plies that the two eigenvalues of ~2



]are + ~

2and − ~

2. You can

clearly see here that the electron spin is nothing like the orbital angular momentum,either classically or quantum mechanically, as the values along any axis are ± ~

2,

irrespective of θ and ϕ. Recall that quantum mechanical angular momentum has2l+1 values for Lz. This is an odd number while the electron spin has an even num-ber of components; namely, two. Also, compare this with a classical orbital angularmomentum vector L, for which Lz depends on the angle θ formed by L and the z-axis.

Next, we will determine the eigenvectors. As usual, this is only up to an arbitrary

scaling factor. Let[ab

]be an eigenvector of



]. Then,



] [ab

]=

[a cos θ + b sin θe−iϕ

a sin θeiϕ − b cos θ

]= ±1 ·

[ab

]. (11.79)


Consider [a cos θ + b sin θe−iϕ


]=

[ab

](11.80)

first. a cos θ + b sin θe−iϕ = aa sin θeiϕ − b cos θ = b

=⇒b sin θe−iϕ = (1− cos θ)aa sin θeiϕ = (1 + cos θ)b =⇒

ab

= sin θe−iϕ

1−cos θab

= 1+cos θsin θeiϕ

=⇒

ab

=2 sin θ

2cos θ

2e−iϕ

2 sin2 θ2

ab

=2 cos2 θ

2

2 sin θ2

cos θ2eiϕ

=⇒ a

b=

cos θ2

sin θ2

e−iϕ or a

b=

cos θ2e−iϕ/2

sin θ2eiϕ/2

(11.81)

Next, consider[a cos θ + b sin θe−iϕ


]= −1 ·

[ab

]=

[−a−b

]. (11.82)

a cos θ + b sin θe−iϕ = −aa sin θeiϕ − b cos θ = −b

=⇒b sin θe−iϕ = −(1 + cos θ)aa sin θeiϕ = −(1− cos θ)b =⇒

ab

= sin θe−iϕ

−(1+cos θ)

ab

= −(1−cos θ)sin θeiϕ

=⇒

ab

=2 sin θ

2cos θ

2e−iϕ

−2 cos2 θ2

ab

=−2 sin2 θ

2

2 sin θ2

cos θ2eiϕ

=⇒ a

b=

sin θ2

− cos θ2

e−iϕ or a

b=

sin θ2e−iϕ/2

− cos θ2eiϕ/2

(11.83)

This gives us the following set of eigenvectors.

|↑⟩u =[cos θ

2e−iϕ

sin θ2

]or

[cos θ

2e−iϕ/2

sin θ2eiϕ/2

](11.84)

|↓⟩u =[sin θ

2e−iϕ

− cos θ2

]or

[sin θ

2e−iϕ/2

− cos θ2eiϕ/2

](11.85)

11.4. SPIN STATES IN AN ARBITRARY DIRECTION 251

11.4.2 Sequential Measurements RevisitedRecalling that we are in the Sz-basis, (11.84) and (11.85) mean

|↑⟩u = cos θ2e−iϕ |↑⟩+ sin θ

2|↓⟩ (11.86)

and

|↓⟩u = sin θ2e−iϕ |↑⟩ − cos θ

2|↓⟩ . (11.87)

Suppose we have made a measurement of the spin in the u direction and found anup-spin state, the ket after the measurement is, according to Postulate 6 on p.74,Chapter 3,

|↑⟩u = cos θ2e−iϕ |↑⟩+ sin θ

2|↓⟩ . (11.88)

If we further measure the spin in the z-direction, the probability of finding an up-spinis

| ⟨↑ | ↑⟩u |2 =

∣∣∣∣∣cos θ2e−iϕ ⟨↑ | ↑⟩+ sin θ

2⟨↑ | ↓⟩

∣∣∣∣∣2

=

∣∣∣∣∣cos θ2e−iϕ

∣∣∣∣∣2

= cos2 θ2, (11.89)

and the probability for the down-spin is

| ⟨↓ | ↑⟩u |2 =

∣∣∣∣∣cos θ2e−iϕ ⟨↓ | ↑⟩+ sin θ

2⟨↓ | ↓⟩

∣∣∣∣∣2

=

∣∣∣∣∣sin θ2∣∣∣∣∣2

= sin2 θ

2, (11.90)

according to Postulate 5 on p.74, Chapter 3.

11.4.3 Comparison with the Classical TheoryAs mentioned on p.89 in Chapter 3, the expectation values play the role of the clas-sical variables in the quantum mechanical formulation of classical laws of physics.Let us see how this plays out for electron spin.

If you have the state |↑⟩u, the angular momentum vector of length ~2

is pointing inthe direction of u = (r, θ, ϕ). Hence, the z-component is ~

2cos θ in the classical pic-

ture. Let us compute the expectation value of Sz for the state |↑⟩u. Our computationwill be in the Sz-basis.

u ⟨↑ |Sz| ↑⟩u =[cos θ

2eiϕ sin θ

2

] [+ ~2

00 − ~

2

] [cos θ

2e−iϕ

sin θ2

]


=[cos θ

2eiϕ sin θ

2

] [+ ~2

cos θ2e−iϕ

− ~2

sin θ2

]=~

2

(cos2 θ

2− sin2 θ

2

)=~

2cos

(2 · θ

2

)=~

2cos θ

This confirms that the quantum mechanical expectation value indeed plays the roleof the corresponding classical variable in this case as well.

253

Exercises1. Consider an electron in the initial spin state

1√2|↑⟩+ 1√

2|↓⟩ .

(a) If two Stern-Gerlach devices are sequentially arranged so that the first de-vice measures the spin in the z-direction and the second in the y-direction,what is the probability that the first device measures an up-spin in thez-direction and the second device measures a down-spin in the y-direction?

(b) In the above, what is the final spin state of the electron?

Chapter 12

Addition of Orbital and SpinAngular Momenta

255

Chapter 13

Molecular Rotation

There is only one three-dimensional motion for a point mass such as an electron;namely translation or simply the motion of its center of mass, i.e. its position, in thethree-dimensional space. However, for entities with more complex structures suchas molecules, there are three types of motion; translation, vibration, and rotation.Translation is the change of the position of its center of mass and vibration is as ina simple harmonic oscillator. Fro rotation, think of a top spinning while remainingat the same spot. The point is that translation is the only motion that involves achange in the center of gravity, and three dimensional motions of molecules can bedecomposed into translation, vibration, and rotation. As we have already discussedtranslational motion of particles and the simple harmonic oscillator as an example ofvibration of diatomic molecules, rotation is our final frontier. We will only considera rigid body in this chapter.

13.1 Rotational Kinetic Energy1

The classical formula for the rotational kinetic energy E of a freely rotating moleculeis

E =1

2Ixω

2x +

1

2Iyω

2y +

1

2Izω

2z , (13.1)

where x, y, and z are principal axes of rotation, and Ix, Iy, and Iz are the momentsof inertia about the x-, y-, and z-axes. As the angular momenta about the principal

1There is no such thing as rotational potential energy, and we only concern ourselves withrotational kinetic energy.

257

258 CHAPTER 13. MOLECULAR ROTATION

axes are given by

Lx = Ixωx, Ly = Iyωy, and Lz = Izωz, (13.2)

we can also express the rotational kinetic energy as follows in terms of the angularmomenta about the principal axes.

E =L2x

2Ix+L2y

2Iy+L2z

2Iz(13.3)

The quantum mechanical Hamiltonian is obtained by the usual substitution of clas-sical observables by the corresponding quantum mechanical operators. Namely,

H =L2x

2Ix+L2y

2Iy+L2z

2Iz; (13.4)

where

Lx = −i~(y∂

∂z− z ∂

∂y

)= i~

(sinϕ ∂

∂θ+ cot θ cosϕ ∂

∂ϕ

)

Ly = −i~(z∂

∂x− x ∂

∂z

)= i~

(− cosϕ ∂


∂ϕ

)(13.5)

Lz = −i~(x∂

∂y− y ∂

∂x

)= −i~ ∂

∂ϕ.

from (9.2) and (9.3).

13.2 Moment of Inertia for a Diatomic MoleculeMoment of inertia is defined relative to a chosen axis of rotation. For a point mass,the moment of inertia is defined as the product of the mass and the square of per-pendicular distance to the axis of rotation. Therefore,

I = mr2, (13.6)

where m is the mass and r is the distance between the point mass and the axisof rotation. To a first approximation, a diatomic molecule can be regraded as twopint masses M and m connected by a rigid massless rod of length r. Now introduce

13.3. TWO-DIMENSIONAL ROTATION CONFINED TO THE X,Y -PLANE 259

a canonical coordinate axes, call them the x- and y-axes, so that the massless rodlies on the x-axis with the center of mass of the molecule at the origin. Supposethe coordinate of the point mass M is (r1, 0), and that of m is (−r2, 0), where thepositive numbers r1 and r2 satisfy r1+ r2 = r. Our axis of rotation is the y-axis thatgoes through the center of mass. Then, the moment of inertia I for this diatomicmolecule is the sum of the moment for the two point masses to a first approximation.

I =Mr21 +mr22 (13.7)

Now, because the center of mass is at the origin, we haveMr1 −mr2M +m

= 0 =⇒Mr1 = mr2. (13.8)

We can use this relation to express r1 and r2 in terms of r.

Mr1 = mr2 =⇒ r1 =m

Mr2 =

m

M(r − r1) =⇒

(1 +

m

M

)r1 =

m

Mr

M +m

Mr1 =

m

Mr =⇒ r1 =

m

M +mr =⇒ r2 =

M

mr1 =

M

M +mr (13.9)

Plugging these into (13.7),

I =Mr21 +mr22 =Mm2

(M +m)2r2 +m

M2

(M +m)2r2

=Mm(m+M)

(M +m)2r2 =

Mm

M +mr2. (13.10)

Recall that µ = MmM+m

is the reduced mass. Therefore, we get

I = µr2. (13.11)

13.3 Two-Dimensional Rotation Confined to thex, y-Plane

The angular momentum vector is always parallel to the z-axis; i.e. J = Jzk2. But,

the difference from the classical counterpart is the quantization of Jz. Without lossof generality, we will specialize to rotations about the z-axis. As Lx and Ly are 0 in(13.4) and (13.5), we get

2In the literature about molecular rotation, the angular momentum vector is typically denotedby J rather than L.


H =L2z

2I=

(−i~ ∂

∂ϕ

) (−i~ ∂

∂ϕ

)2I

= − ~2

2I

∂2

∂ϕ2; (13.12)

where we dropped the subscript z from Iz because Iz is the only nonzero moment ofinertia in our situation. Our Schrodinger equation is

Hψ = Eψ =⇒ − ~2

2I

∂2ψ

∂ϕ2= Eψ. (13.13)

Because ψ here is a function of ϕ alone, and also because it is basically the same asΦ encountered in Chapter 10 “The Hydrogen Atom”, it is customary to denote thiswavefunction by Φ. With this convention, we now have

− ~2

2I

∂2Φ(ϕ)

∂ϕ2= EΦ(ϕ). (13.14)

The solutions are

Φm(ϕ) =

√1

2πeimϕ for m = 0,±1,±2, . . . . (13.15)

As in the case of the hydrogen atom, possible values for m are determined by thesingle-valuedness condition; namely, Φ(ϕ) = Φ(ϕ + 2nπ) for any integer n. For thetotal energy, we have

m = ±√2IE

~=⇒ E =

m2~2

2Ifor m = 0,±1,±2, . . . . (13.16)

The term m2 in the expression for the energy implies that we have two-fold degen-eracy for each value of m except for m = 0, which means there is no rotation. In theclassical picture, we can regard this as deriving from the sense of rotation, clockwiseor counter-clockwise.

13.4. THREE-DIMENSIONAL ROTATION 261

13.4 Three-Dimensional RotationIn this section, we still consider a rotational motion confined to a plane. However,unlike in Section 13.3 where the rotation was confined to the x, y-plane, we nowallow the plane of rotation to be tilted relative to the z-axis. In other words, it isnecessary to give the magnitude of angular momentum ∥J∥ and its tilt from thez-axis, measured by Jz as before, to completely specify the rotation.

From (10.41) the full Hamiltonian in spherical coordinates is

−~2

2µ

[1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2

]

=−~2

2µr2

[∂

∂r

(r2∂

∂r

)+

1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

]. (13.17)

Because we are dealing with a rigid molecule, r is fixed, and there is no kinetic energyof radial motion. So, the term

1

r2∂

∂r

(r2∂

∂r

)(13.18)

vanishes. Also, we know from (13.11) that I = µr2. Therefore, the Hamiltonian forthree-dimensional rotation of a rigid body is given by

−~2

2I

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

]. (13.19)

This is basically the same as the θ- and φ-dependent parts of the Hamiltonian for thehydrogen atom given in Chapter 10. The time-independent Schrodinger equation is

−~2

2I

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

]ψ(θ, φ) = Eψ(θ, φ). (13.20)

Separation of variables work here as well. So, let

ψ(θ, φ) = Θ(θ)Φ(φ). (13.21)

Then, we will get


d2Φ

dφ2= −m2

JΦ (13.22)

and~2

2I

[1

sin θd

dθ

(sin θdΘ

dθ

)− m2

JΘ

sin2 θ

]+ EΘ = 0. (13.23)

Due to single-valuedness and normalizability conditions on the wavefunctions, we get

E =~2

2IJ(J + 1) for J = 1, 2, 3, . . . (13.24)

andmJ = −J,−J + 1, . . . , 0, . . . , J − 1, J. (13.25)

The wavefunctions are

ΦmJ(φ) =

1

2πeimJφ (13.26)

and

ΘmJJ(θ) =

[(2J + 1)(J − |mJ |)!

2(J + |mJ |)!

] 12

P|mJ |J (cos θ); (13.27)

where P |mJ |J (cos θ)’s denote the associated Legendre polynomials first encountered in

(10.95) on p.214. We usually use the symbol B for ~22I

, and so, we have

E = BJ(J + 1). (13.28)

As for mJ , it is proportional to the z-component Jz of the rotational angular mo-mentum; namely,

Jz = mJ~. (13.29)

This is another example of degeneracy in higher dimensions discussed in Section 8.1.The total rotational energy depends only on J , but there are 2J + 1 distinct states

13.4. THREE-DIMENSIONAL ROTATION 263

with mJ = −J,−J + 1, . . . , 0, . . . , J − 1, J which share the same J , and hence, havethe same total energy E.

In the most general case, we have Ix, Iy, and Iz which are different from each other.For a rigid body like this, we need to introduce three constants A = ~2

2Ix, B = ~2

2Iy,

and C = ~2

2Izrather than just B = ~2

2I. Interested readers should consult “Microwave

Molecular Spectra (3rd edition)” by Gordy and Cook [Gordy and Cook, 1984].

Appendices

265

Appendix A

Matrix Algebra

A.1 Invertible MatricesFact A.1 (Invertible Matrix) Let A be a square n×n matrix. Then the followingstatements are equivalent. Hence, they are either all true or all false.

1. A is an invertible matrix.

2. A is bijective; that is, A is one-to-one and onto.

3. A is row equivalent to the n × n identity matrix.

4. A has n pivot positions.

5. The equation Ax = 0 has only the trivial solution 0.

6. The null space of A is 0.

7. The dimension of the null space of A is 0.

8. The columns of A form a linearly independent set.

9. The linear transformation x 7→ Ax is one-to-one.

10. The equation Ax = b has at least one solution for each b ∈ Rn.

11. The columns of A span Rn.

12. The linear transformation x 7→ Ax maps Rn onto Rn.

13. There is an n× n matrix L such that LA = I.

267

268 APPENDIX A. MATRIX ALGEBRA

14. There is an n× n matrix R such that AR = I.

15. At is an invertible matrix.

16. The columns of A form a basis of Rn.

17. The column space C(A) = Rn.

18. The dimension of C(A) = n.

19. rank A = n

20. The number 0 is not an eigenvalue of A.

21. det A , 0

A.2 Rank of a MatrixDefinition A.1 (Rank of a Matrix) The column rank of a matrix A is the max-imum number of linearly independent column vectors of A. The row rank of A is themaximum number of linearly independent row vectors of A. Equivalently, the columnrank of A is the dimension of the column space of A, while the row rank of A isthe dimension of the row space of A. The column rank and the row rank are alwaysequal, and we call this number simply the rank of A, denoted by rank A.1

Theorem A.1 (Column Rank = Row Rank) The column rank and the row rankof a matrix A are equal.

The following proofs are basically from Wikipedia.

ProofConsider an m×n matrix A whose i-th row and j-th column entry is Aij and columnrank is r. Pick an arbitrary basis v1,v2, . . . ,vr of the column space of A. Eachbasis vector is an m× 1 column vector, and we can form an m× r matrix

C = [v1 v2 . . . vr] = [vkl]1≤l≤r1≤k≤m. (A.1)

1Alternatively, the row rank of A is the number of non-zero rows in the row reduced from of A.

A.2. RANK OF A MATRIX 269

Note that we are writing the k-th component of the m × 1 column vector vl as vklwith 1 ≤ k ≤ m such that

vl =

v1lv2l...vml

(A.2)

for bookkeeping purposes. The j-th column [Aj] of A can be expressed as a linearcombination of v1,v2, . . . ,vr, or vi for short, with coefficients bij such that

[Aj] =r∑p=1

bpjvp for 1 ≤ j ≤ n. (A.3)

In its full glory, (A.3) can be written as follows.A1j

A2j...

Amj

=

∑rp=1 v1pbpj∑rp=1 v2pbpj

...∑rp=1 vmpbpj

(A.4)

Hence,

Aij =r∑p=1

vipbpj =⇒ A = CB; (A.5)

where A = [Aij]1≤j≤n1≤i≤m is the original m×n matrix, C = [vkl]

1≤l≤r1≤k≤m is the m×r matrix

whose columns are the basis vectors v1,v2, . . . ,vr, and B = [bpj]1≤j≤n1≤p≤r is the r×n

coefficient matrix defined by (A.3). Now consider the i-th row of A denoted here by[Ai].

[Ai] = [Ai1 Ai2 . . . Ain] =

r∑p=1

vipbp1r∑p=1

vipbp2 . . .r∑p=1

vipbpn

=

r∑p=1

vip [bp1 bp2 . . . bpn] (A.6)

Because [bp1 bp2 . . . bpn] is the p-th row of B, the rows Ai of the m × n matrixA are linear combinations of the rows of B, we have row rank A ≤ row rank B ≤


r = column rank A. We have now shown that row rank A ≤ column rank A. Fur-thermore, if we subject At to the same procedure, we will arrive at row rank At ≤column rank At =⇒ column rank A ≤ row rank A. This proves the theorem.

An Alternative Proof for Real MatricesLet A be an m × n real matrix whose row rank is r; i.e. the dimension of the rowspace of A is r. Choose a basis v1,v2, . . . ,vr of the row space of A. Note thatvi’s are 1× n row vectors, (vi)t’s are n× 1 column vectors, and A(vi)

t’s are m× 1column vectors. Now, consider a linear combination

r∑i=1

ciA(vi)t = A

[r∑i=1

ci(vi)t

]= Awt = 0; (A.7)

where ci’s are some scalars, and w :=r∑i=1

civi is some vector in the row space of A.

Because Aw = 0 and the i-th component of the m × 1 column vector Awt is givenby ∑n

j=1Aijwj, which is the inner product between the i-th row of A and w, (A.7)implies that w is orthogonal to each row vector of A. Hence, w is orthogonal to allthe vectors in the row space of A. But, we know w belongs to the row space of A.Therefore, w =

r∑i=1

civi = 0, which in turn implies that all ci’s are 0 because vi’s

are linearly independent. It now follows that A(v1)t, A(v2)

t, . . . , A(vr)t are linearly

independent. Noting that [A(vi)t]k =

∑nl=1Akl(vi)

tl , we have

A(vi)t =

∑nl=1A1l(vi)

tl∑n

l=1A2l(vi)tl

...∑nl=1Aml(vi)

tl

=

A11(vi)

t1 + A12(vi)

t2 + . . .+ A1n(vi)

tn

A21(vi)t1 + A22(vi)

t2 + . . .+ A2n(vi)

tn

...Am1(vi)

t1 + Am2(vi)

t2 + . . .+ Amn(vi)

tn

= (vi)t1

A11

A21...

Am1

+ (vi)t2

A12

A22...

Am2

+ . . .+ (vi)tn

A1n

A2n...

Amn

, (A.8)

which is a linear combination of the column vectors of A. Hence, we have r lin-early independent vectors A(v1)

t, A(v2)t, . . . , A(vr)

t in the column space of A. Weconclude that the dimension of the column space is at least r, and we have shownthat column rank A ≥ row rank A. We next consider At and go through the sameprocedure to show that column rank At ≥ row rank At. But, column rank At =

A.3. CHANGE OF BASIS 271

row rank A and row rank At = column rank A give us row rank A ≥ column rank A.This concludes the proof.

A.3 Change of BasisTheorem A.2 (Change of Basis: Transition Matrix) [Anton, 2010, p.218] Con-sider two bases B = u1,u2, . . . ,un and B′ = u′

1,u′2, . . . ,u

′n of an n-dimensional

vector space Vn; where we regard ui’s and u′i’s as n× 1 column vectors. If vB is the

n × 1 column matrix representation of a vector v ∈ Vn in B, and vB′ the same inB′, then, the relation between vB and vB′ can be described by

vB = PB′→BvB′ ; (A.9)

where PB′→B, the transition matrix from B′ to B, is an n×n matrix whose columnsare u′

1B,u′2B, . . . ,u

′nB. Note that with our notation u′

jB is the j-th basis vector ofB′ expressed as a column vector in the basis B. Similarly, due to symmetry, withanother transition matrix PB→B′ whose columns are ujB′, we have

vB′ = PB→B′vB. (A.10)

In fact, we can show

PB′→BPB→B′ = PB→B′PB′→B = I (A.11)

and PB′→B is the inverse of PB→B′.

ProofAs B and B′ are bases, each vector in B can be expressed as a unique linear combi-nation of the vectors in B′, and each vector in B′ can be expressed as a unique linearcombination of the vectors in B. Hence, we can find the unique coefficients Sij andTij, for 1 ≤ j, i ≤ n, or differently put, unique matrices S = [Sij] and T = [Tij] suchthat

ui =n∑j=1

Sjiu′j and u′

i =n∑j=1

Tjiuj (A.12)

or in matrix form

[ u1 u2 · · · un ] = [ u′1 u′

2 · · · u′n ]

S11 · · · S1n...

. . ....

Sn1 · · · Snn

(A.13)


and

[ u′1 u′

2 · · · u′n ] = [ u1 u2 · · · un ]

T11 · · · T1n...

. . ....

Tn1 · · · Tnn

. (A.14)

If we combine the two relations of (A.12), we get

ui =n∑j=1

Sjiu′j =

n∑j=1

Sjin∑k=1

Tkjuk =n∑k=1

n∑j=1

TkjSji

uk (A.15)

and

u′i =

n∑j=1

Tjiuj =n∑j=1

Tjin∑k=1

Skju′k =

n∑k=1

n∑j=1

SkjTji

u′k. (A.16)

However, because B = u1,u2, . . . ,un and B′ = u′1,u

′2, . . . ,u

′n are bases, ui’s

and u′i’s are linearly independent. Hence,

n∑j=1

TkjSji =n∑j=1

SkjTji = δki. (A.17)

Recall that

δki =

1 i = k0 i , k

. (A.18)

If we form the matrices S = [Sij] and T = [Tij],∑nj=1 SkjTji is the k-th row and the

i-th column entry of the matrix product ST , and ∑nj=1 TkjSji is the k-th row and the

i-th column entry of the matrix product TS. Therefore,

ST = TS = I, (A.19)

and T is the inverse of S. Getting back to (A.12),

ui =n∑j=1

Sjiu′j = S1iu

′1 + S2iu

′2 + . . . + Sniu

′n =

S1i

S2i...Sni

B′

(A.20)

A.3. CHANGE OF BASIS 273

and

u′i =

n∑j=1

Tjiuj = T1iu1 + T2iu2 + . . . + Tniun =

T1iT2i...Tni

B

; (A.21)

where []B and []B′ mean they are with respect to B and B′, respectively. This provesthat the j-th column of [Sij] is ujB′, and that the j-th column of [Tij] is u′

jB. Itremains to show that

[Tij] = PB′→B and [Sij] = PB→B′ (A.22)

as described in the theorem. Take any vector v ∈ Vn and express it as linearcombinations of ui’s and u′

i’s, respectively.

v =

v1v2...vn

B

=n∑i=1

viui and v =

v′1

v′2...v′n

B′

=n∑i=1

v′iu

′i (A.23)

We get

v =n∑i=1

v′iu

′i =

n∑i=1

v′i

n∑j=1

Tjiuj =n∑j=1

(n∑i=1

Tjiv′i

)uj

=

(n∑i=1

T1iv′i

)u1 +

(n∑i=1

T2iv′i

)u2 + . . . +

(n∑i=1

Tniv′i

)un =

∑ni=1 T1iv

′i∑n

i=1 T2iv′i

...∑ni=1 Tniv

′i

B

=⇒ [v]B = [Tij][v]B′ (A.24)

and

v =n∑i=1

viui =n∑i=1

vin∑j=1

Sjiu′j =

n∑j=1

(n∑i=1

Sjivi

)u′j

=

(n∑i=1

S1ivi

)u′

1 +

(n∑i=1

S2ivi

)u′

2 + . . . +

(n∑i=1

Snivi

)u′n =

∑ni=1 S1ivi∑ni=1 S2ivi...∑n

i=1 Snivi

B′


=⇒ [v]B′ = [Sij][v]B (A.25)

This proves (A.22), and we have now proved the theorem by an explicit constructionof the transition matrices PB′→B and PB→B′.

Theorem A.3 (Change of Basis and Linear Transformations) Let Ω be a lin-ear operator operating on a vector space Vn. In other words, Ω is a linear map fromVn into Vn. Consider two bases B = u1,u2, . . . ,un and B′ = u′

1,u′2, . . . ,u

′n

of Vn and the corresponding matrix representations of Ω in those bases, denoted by[Ωij] and [Ω′

ij] respectively. Then,

Ω′ij =

n∑k,l=1

T−1ik ΩklTlj or [Ω′

ij] = [T−1ij ][Ωij][Tij] or Ω′ = T−1ΩT ; (A.26)

where T = PB′→B.

ProofFrom (2.91) on p.43,

Ωui =n∑j=1

Ωjiuj (A.27)

andΩu′

i =n∑j=1

Ω′jiu

′j. (A.28)

Next, from (A.12), we have

u′i =

n∑j=1

Tjiuj. (A.29)

Substituting (A.29) into the right-hand side of (A.28),

n∑j=1

Ω′jiu

′j =

n∑j=1

Ω′ji

(n∑k=1

Tkjuk

)=

n∑k=1

n∑j=1

TkjΩ′ji

uk. (A.30)

Now, substituting (A.29) and (A.27) successively into the left-hand side of (A.28),

Ωu′i = Ω

n∑j=1

Tjiuj

=n∑j=1

TjiΩuj =n∑j=1

Tji

(n∑k=1

Ωkjuk

)=

n∑k=1

n∑j=1

ΩkjTji

uk.

(A.31)

A.4. MULTIPLICITY OF AN EIGENVALUE 275

Because uk’s are linearly independent, (A.30) and (A.31) imply thatn∑j=1

TkjΩ′ji =

n∑j=1

ΩkjTji (A.32)

for all 1 ≤ i, k ≤ n. This means that the (k, i)-entry of the matrix product TΩ′

equals the (k, i)-entry of the matrix product ΩT for all 1 ≤ i, k ≤ n. Therefore,

TΩ′ = ΩT =⇒ Ω′ = T−1ΩT ; (A.33)

where we know T = PB′→B from Theorem A.2.

Theorem A.4 (Basis Independence of Characteristic Polynomials) The char-acteristic polynomial of a linear transformation Ω is well defined. That is, it isbasis-independent and does not depend on the choice of a basis.

ProofConsider two bases B = ui and B′ = v′

i, and denote the matrix representationsof a linear transformation OMEGA with respect to these bases by Ω and Ω′. IfT := PB′→B so that Tv′

k = uk, then, from Theorem A.3 we have Ω′ = T−1ΩT . Thisgives us

det(Ω′ − λI) = det(T−1ΩT − λI) = det(T−1ΩT − T−1(λI)T )

= det(T−1(Ω− λI)T ) = det(T−1)det(Ω− λI))detT =1

detTdet(Ω− λI)detT

= det(Ω− λI), (A.34)

which is the desired result.

Theorem A.5 If there is a linearly independent set S, it can be extended to a basisof the vector space V; that is, if u1,u2, . . . ,um (m < n) are linearly independent,we can find um+1 . . . ,un such that u1,u2, . . . ,um+1 . . . un is a basis for V.

Proof by Zorn’s Lemma

A.4 Multiplicity of an EigenvalueIf M is an n × n matrix, then, det(A − λI) is an n-th degree polynomial, calledthe characteristic polynomial, and the eigenvalues are the roots of the characteristicpolynomial. If we have a factor of the form (λ−c)l in the factorization of det(A−λI),we say the eigenvalue c is of multiplicity l.


Fact A.2 (Multiplicity and Linear Independence) Let λ1, λ2, . . . , λk be thelist of all eigenvalues, including repeated roots of the characteristic polynomial, i.e.λ0 of multiplicity l appears in the list λ1, λ2, . . . , λk l-times.

1. An eigenvalue/a root that occurs only once in the list λ1, λ2, . . . , λk is calledsimple.

2. If λ0 occurs l times, λ0 is of multiplicity l.

3. Eigenvectors associated with simple eigenvalues are linearly independent.

4. An eigenvalue of multiplicity l may have 1,2, ,. . . , or l linearly independenteigenvectors.

A.5 Diagonalizability and Simultaneous Diagonal-ization

In the following, we will denote the set of all n× n matrices by Mn.

Definition A.2 (Similarity) A matrix B ∈ Mn is similar to another matrix A ∈Mn if there exists an invertible matrix S ∈Mn such that

B = S−1AS. (A.35)

Definition A.3 (Diagonalizability) A matrix A ∈ Mn is diagonalizable if it issimilar to a diagonal matrix.

Theorem A.6 A matrix A ∈ Mn is diagonalizable if and only if there exists a setof n linearly independent eigenvectors of A.

Proof(=⇒)Let S be an invertible matrix such that S−1AS = D, where D is a diagonal matrix.We have S−1AS = D =⇒ S (S−1AS) = SD =⇒ AS = SD. Now, denote the j-thcolumn of S by sj, which can also be regarded as an n× 1 column vector such that

sj = [sij] with 1 ≤ i ≤ n, (A.36)

A.5. DIAGONALIZABILITY AND SIMULTANEOUS DIAGONALIZATION 277

where sij is the i-th row and j-th column entry of S. With this notation,

Asj =

∑i

A1isij

...∑i

Akisij

...∑i

Anisij

.

(A.37)

So, Asj is the j-th column of AS, and we have

AS = [As1 As2 . . . Asn]. (A.38)

Next, let D = [dii], where dii is the i-th diagonal entry of D. Then, the i-th row andthe j-th column entry of SD denoted by (SD)ij is given by

(SD)ij =∑k

sikdkj = sijdjj = djjsij. (A.39)

Hence, the j-th column of SD is given bydjjs1j...

djjskjdjjsnj

= djjsj, (A.40)

which in turn indicates

SD = [d11s1 d22s2 . . . dnnsn]. (A.41)

Putting these all together,

AS = SD =⇒ [As1 As2 . . . Asn] = [d11s1 d22s2 . . . dnnsn] =⇒ Asj = djjsj

for 1 ≤ j ≤ n. (A.42)

Therefore, each sj is an eigenvector of A with corresponding eigenvalue djj. BecauseS is invertible, rank S = n, and sj’s are linearly independent.(=⇒)Conversely, suppose there is a linearly independent set s1 s2 . . . sn of eigenvectorsof A with associated eigenvalues s1 s2 . . . sn, and form S = [s1 s2 . . . sn]. As before,


AS = [As1 As2 . . . Asn] = [λ1s1 λ2s2 . . . λnsn]. If we let D be the diagonal matrixwhose k-th diagonal entry is λk. Then, SD = [λ1s1 λ2s2 . . . λnsn] = AS. Becausesj is a linearly independent set, S is of full rank, and S−1 exists. We now haveAS = SD =⇒ S−1AS = S−1SD = D, and A is diagonalizable.

Definition A.4 (Simultaneous Diagonalization) Two matrices A,B ∈ Mn aresaid to be simultaneously diagonalizable if there exists an invertible matrix S suchthat both S−1AS and S−1BS are diagonal matrices.

Theorem A.7 (Simultaneous Diagonalization and Commutativity) Let A,B ∈Mn be diagonalizable. Then AB = BA if and only if A and B are simultaneouslydiagonalizable.

In order to prove Theorem A.7, we need the following two lemmas.

Lemma A.1 Two matrices A ∈ Mn and B ∈ Mm are diagonalizable if and only ifthe composite block matrix

C =

[A 00 B

]∈Mn+m (A.43)

is diagonalizable.

Proof(=⇒)Suppose A ∈ Mn and B ∈ Mm are diagonalizable. Then, we can find two invertiblematrices SA ∈ Mn and SB ∈ Mm such that S−1

A ASA and S−1B BSB are diagonal

matrices. Define S ∈Mn+m by

S :=

[SA 00 SB

]. (A.44)

Then,

S−1 =

[S−1A 00 S−1

B

], (A.45)

and

S−1CS = S−1

[A 00 B

]S =

[S−1A 00 S−1

B

] [A 00 B

] [SA 00 SB

]


=

[S−1A ASA 00 S−1

B BSB

](A.46)

is a diagonal matrix, and C is diagonalizable.(⇐=)Conversely, assume that C is diagonalizable. Then, there exists an invertible matrixS ∈ Mn+m such that D := S−1CS is diagonal. Now, denote the j-th column of Sby sj ∈ Cn+m. We can write S = [s1 s2 . . . sn+m], and because CS = SD, with theusual representation D = [di], where di is the i-th diagonal entry, we have

[Cs1 Cs2 . . . Csn+m] = [d1s1 d2s2 . . . dn+msn+m] (A.47)

as in (A.42). This shows that Csk = dksk, and sk’s are eigenvectors of C with theassociated eigenvalues dk. Now, we write each sk in composite form

sk =

[xkyk

], (A.48)

where xk ∈ Cn and yk ∈ Cm. Due to the block structure of S, Csk = dksk requiresAxk = dkxk and Byk = dkyk. Denoting [x1 x2 . . . xn+m] by X and [y1 y2 . . . yn+m]by Y , we have

S =

[XY

]. (A.49)

Because X is an n× (n+m) matrix and Y is an m× (n+m) matrix,

rank X ≤ n and rank Y ≤ m. (A.50)

On the other hand,

rank S = n+m (A.51)

as S in invertible. We have the following implications.

n+m = rank S ≤ rank X + rank Y ≤ n+m

=⇒ rank X + rank Y = n+m =⇒ rank X = n, rank Y = m (A.52)

Therefore, X contains n linearly independent columns and Y contains m linearlyindependent columns. Recalling that A ∈ Mn operates on vectors in Cn and thecolumns of X are eigenvectors of A, we can now see that A is diagonalizable byTheorem A.6. Likewise, B is also diagonalizable as B ∈ Mm operates on vectors inCm and the columns of Y are eigenvectors of B.


Lemma A.2 For a diagonal matrix

D =

d1 0 . . . 0 00 d2 . . . 0 0...

.... . .

......

0 0 . . . dn−1 00 0 . . . 0 dn

∈Mn, (A.53)

there exists an invertible matrix P ∈Mn such that P−1DP switches the positions ofdi and dj for any 1 ≤ i, j ≤ n. In fact, such P can be obtained by switching the i-throw and the j-th row of the identity matrix I ∈Mn.

Before embarking on a general proof of this lemma. Let us see how P works with a4× 4 matrix

D =

d1 0 0 00 d2 0 00 0 d3 00 0 0 d4

. (A.54)

We will try to switch d1 and d3 to obtain

Dd1↔d3 =

d3 0 0 00 d2 0 00 0 d1 00 0 0 d4

. (A.55)

According to our program, P is obtained by switching the first and the third rowsof I ∈M4. Hence,

P =

0 0 1 00 1 0 01 0 0 00 0 0 1

and P−1 =

0 0 1 00 1 0 01 0 0 00 0 0 1

. (A.56)

With this P , we have

P−1DP =

0 0 1 00 1 0 01 0 0 00 0 0 1

d1 0 0 00 d2 0 00 0 d3 00 0 0 d4

0 0 1 00 1 0 01 0 0 00 0 0 1


=

0 0 1 00 1 0 01 0 0 00 0 0 1

0 0 d1 00 d2 0 0d3 0 0 00 0 0 d4

=

d3 0 0 00 d2 0 00 0 d1 00 0 0 d4

= Dd1↔d3 (A.57)

Proof of Lemma A.2Our proof is by way of an explicit computation. Let us see how the matrix P givenby (A.58) switches di and dj. In (A.58), “. . .”, “...”, and a blank space signify 0, while“. . .” stands for 1.

i j

1 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0

0 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0...

.... . .

......

0 0 . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0

i 0 0 . . . . . . 0 . . . . . . . . . . . . 1 . . . . . . . . . 0 0

0 0 . . . . . . 0 1 . . . . . . . . . 0 . . . . . . . . . 0 0...

... . . . . . .... . . .

. . . . . . . . .... . . . . . . . . .

......

...... . . . . . .

... . . . . . .. . . . . .

... . . . . . . . . ....

...

0 0 . . . . . . 0 . . . . . . . . . 1 0 . . . . . . . . . 0 0

j 0 0 . . . . . . 1 . . . . . . . . . . . . 0 . . . . . . . . . 0 0

0 0 . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . 0 0...

.... . .

......

......

. . ....

...

0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 0

0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 1

(A.58)

Note first that P has the following block diagonal structure; where Ii−1 is the (i −1)× (i− 1) identity matrix, In−j is the (n− j)× (n− j) identity matrix, “. . .” stands


for 0, and “. . .” denotes 1.

P =

i j

Ii−1 . . . . . . . . . . . . . . . . . . . . .

i . . . 0 . . . . . . . . . . . . 1 . . .

. . . 0 1 . . . . . . . . . 0 . . .

. . .... . . .

. . . . . . . . .... . . .

. . .... . . . . . .

. . . . . .... . . .

. . . 0 . . . . . . . . . 1 0 . . .

j . . . 1 . . . . . . . . . . . . 0 . . .

. . . . . . . . . . . . . . . . . . . . . In−j

=

Ii−1 . . . . . .. . . Q . . .. . . . . . In−j

; (A.59)

where Q ∈Mj−i+1 is defined by

Q =

0 . . . . . . . . . . . . 10 1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . 1 01 . . . . . . . . . . . . 0

. (A.60)

It is easy to see Q2 = I or Q−1 = Q. Due to the block diagonal structure of P , itsuffices to consider the action of Q on the following submatrix D of D.

D :=

di . . . . . . . . . . . . 00 di+1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . dj−1 00 . . . . . . . . . . . . dj

(A.61)

We have

Q−1DQ =

0 . . . . . . . . . . . . 10 1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . 1 01 . . . . . . . . . . . . 0


di . . . . . . . . . . . . 00 di+1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . dj−1 00 . . . . . . . . . . . . dj

0 . . . . . . . . . . . . 10 1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . 1 01 . . . . . . . . . . . . 0

=

0 . . . . . . . . . . . . 10 1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . 1 01 . . . . . . . . . . . . 0

0 . . . . . . . . . . . . di0 di+1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . dj−1 0dj . . . . . . . . . . . . 0

=

dj . . . . . . . . . . . . 00 di+1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . dj−1 00 . . . . . . . . . . . . di

. (A.62)

This proves Lemma A.2.

There is an obvious corollary to Lemma A.2.

Corollary A.1 Given a diagonal matrix D ∈ Mn and D′permutation ∈ Mn, where

D′permutation is another diagonal matrix whose diagonal entries are a permuted

version of the original D, there exists an invertible matrix Q ∈ Mn such thatD′

permutation = Q−1DQ. In other words, one can permute the diagonal entriesby some similarity transformation Q−1DQ.

Proof of Theorem A.7(simultaneous diagonalizability =⇒ commutativity)As S−1AS and S−1BS are both diagonal, and diagonal matrices commute with eachother, we get

AB = S(S−1AS

) (S−1BS

)S−1 = S

(S−1BS

) (S−1AS

)S−1 = BA. (A.63)


(commutativity =⇒ simultaneous diagonalizability)As A is diagonalizable, there is an invertible matrix S ∈Mn such that D = S−1AS isdiagonal. By Corollary A.1, we can permute the diagonal entries of D by multiplyingS by an appropriate invertible matrix. Hence, we may assume that

D =

λ1Im1 0 . . . 00 λ2Im2 0...

. . ....

0 . . . 0 λrImr

; (A.64)

where λi’s are distinct, mj is the multiplicity of λj, and Imjis the mj ×mj identity

matrix. As AB = BA,(S−1AS

) (S−1BS

)= S−1ABS = S−1BAS =

(S−1BS

) (S−1AS

). (A.65)

Denoting (S−1BS) by C, we have DC = CD. Now, let cij, dij, (cd)ij, and (dc)ijsignify the i-th row and the j-th column entries of C, D, CD, and DC respectively.As D is diagonal, we can write dij = δijdjj. So,

(cd)ij =∑k

cikdkj =∑k

cik(δkjdjj) = cijdjj (A.66)

and(dc)ij =

∑k

dikckj =∑k

(δikdii)ckj = diicij. (A.67)

Because CD = DC means (cd)ij = (dc)ij for all 1 ≤ i, j ≤ n, we have

(cd)ij = (dc)ij =⇒ cijdjj = diicij =⇒ (dii − djj)cij = 0 for 1 ≤ i, j ≤ n. (A.68)

Therefore, cij = 0 unless dii = djj. But, dii = djj only if dii and djj belong to thesame diagonal block of (A.64). This implies that cij = 0 unless dii and djj belong tothe same block of (A.64). Hence, the matrix C is block diagonal whose i-th block isan mi×mi submatrix. In other words, C has the same block structure as D, namely

C =

C1 0 . . . 00 C2 0...

. . ....

0 . . . 0 Cr

; where Cl ∈Mmlfor each l. (A.69)

As B is also diagonalizable, there is an invertible matrix R such that R−1BR isdiagonal. So,

R−1SCS−1R = R−1SS−1BSS−1R = R−1BR, (A.70)


and C is diagonalizable. From Lemma A.1, all the blocks of C are diagonalizable;that is, for each j, there exists an invertible matrix Tj ∈ Mmj

such that T−1j CjTj is

diagonal. Next, define a block diagonal matrix T ∈Mn by

T =

T1 0 . . . 00 T2 0...

. . ....

0 . . . 0 Tr

. (A.71)

Then, in a straightforward manner,

T−1 =

T−11 0 . . . 00 T−1

2 0...

. . ....

0 . . . 0 T−1r

. (A.72)

We now have

(ST )−1B(ST ) = T−1(S−1BS)T = T−1CT

=

T−11 C1T1 0 . . . 0

0 T−12 C2T2 0

.... . .

...0 . . . 0 T−1

r CrTr

(A.73)

and(ST )−1A(ST ) = T−1(S−1AS)T = T−1DT

=

T−11 λ1Im1T1 0 . . . 0

0 T−12 λ2Im2T2 0

.... . .

...0 . . . 0 T−1

r λrImrTr

=

λ1Im1 0 . . . 00 λ2Im2 0...

. . ....

0 . . . 0 λrImr

= D. (A.74)

As both (A.73) and (A.74) are diagonal, A and B are simultaneously diagonalizableby ST ∈Mn.


A.6 Some Useful Definitions, Facts, and TheoremsDefinition A.5 (The Characteristic Polynomial) The characteristic polynomialof a square matrix M is the function

f(λ) = det (λI −M). (A.75)

Fact A.3 If a square matrix M has eigenvalues λi (i = 1, . . . , n), we can write thecharacteristic polynomial f(λ) as the following product.

f(λ) = det (λI −M) =∏i

(λ− λi) (A.76)

Theorem A.8 (Trace, Determinant, and Eigenvalues) If M is an n×n squarematrix with eigenvalues λi (i = 1, 2, . . . n), the following equalities hold.

traceM =n∑i=1

λi (A.77)

detM =n∏i=1

λi (A.78)

ProofWe can expand the product (A.76) as follows.

f(λ) =∏i

(λ− λi) = λn − λn−1n∑i=1

λi + (−1)nn∏i=1

λi (A.79)

Appendix B

Holomorphic Functional Calculus

Consider a function f : C→ C and an operator Ω. If f can be expanded in a Taylorseries about z = 01 such that

f(z) =∞∑n=0

f (n)(0)

n!zn, (B.1)

we can define a map f : Ω 7→ f(Ω) by

f(Ω) =∞∑n=0

f (n)(0)

n!Ωn. (B.2)

One way to understand this formalism is to consider a matrix representation of Ω.Then, the sum is also a square matrix and convergence of the series is nothing butan entry-by-entry convergence.

Fact B.1

Fact B.2 (Basis Independence of the Characteristic Polynomial) The char-acteristic polynomial is basis independent. Hence, the characteristic equation, theeigenvalues, and their multiplicities are basis-independent.Proof

1The general form of Taylor series about point z0 is

f(z) =∞∑

n=0

f (n)(z0)

n!(z − z0)n.

When the Taylor expansion is about 0, it is also called a Maclaurin series.

287

288 APPENDIX B. HOLOMORPHIC FUNCTIONAL CALCULUS

Consider two bases B = b1, b2, . . . bn and D = d1,d2, . . . dn, and let M = [Mkl]be the matrix of basis change, such that

bi =n∑j=1

Mijdj (B.3)

Appendix C

Traveling Waves and StandingWaves

When a wave is not confined to a finite portion of space, we observe a traveling wave.On the other hand, it is possible to have a wave confined to a given space producing aregular pattern, and we call this a standing wave. Traveling waves transport energyfrom one area of space to another, whereas standing waves can not transport energy.

In this section again, we will straddle over the line between classical mechanicsand quantum mechanics.

C.1 The Wave Equation

The classical wave equation for a plane wave y(x, t) traveling on a string in thex-direction is

∂2y

∂x2=

1

v2∂2y

∂t2. (C.1)

In (C.1), v is the propagation speed (the phase velocity1) of the wave satisfyingνλ = v, where ν is the frequency and λ is the wavelength.

1 There are phase velocity, group velocity, and signal velocity. For example, you can find anextensive discussion at http://www.mathpages.com/home/kmath210/kmath210.htm.

289

290 APPENDIX C. TRAVELING WAVES AND STANDING WAVES

C.1.1 Traveling Wave SolutionsFor an ideal string extending from x = −∞ to x = +∞, two useful solutions are

y(x, t) = A sin 2π

λ(x± vt) or A sin k(x± vt)

or A sin(kx± ωt) or A sin 2π(x

λ± t

T

); (C.2)

where A is an arbitrary amplitude, k = 2πλ

is the wave number, ω = 2πν is theangular frequency, and T is the period.

In (C.2), sin k(x− vt) represents a wave traveling to the right, and sin k(x+ vt)is a wave moving to the left. In order to see this, just consider why the propagationspeed v is known as the “phase” velocity. The angle k(x ± vt) is the phase whichspecifies each point of the traveling wave, Hence, we should follow the points withthe same phase as the time t progresses in order to trace the movement of the wave.

Consider a point on the traveling wave located at position x0 at time t0 denoted by(x0, t0). The phase that identifies this point is k(x0 ± vt0). Now, consider a latertime t > t0. Where is the point? Let the point be at x. Because it is the same point,the phase indicated by (x0, t0) should equal the phase for (x, t). Therefore, we have

x0 ± vt0 = x± vt =⇒ x = x0 ± vt0 ∓ vt =⇒ x = x0 ∓ v(t− t0) (C.3)or

x0 − vt0 = x− vt =⇒ x = x0 + v(t− t0)and

x0 + vt0 = x+ vt =⇒ x = x0 − v(t− t0)

(C.4)

Now, x0+v(t−t0) means the point has moved to the right by v(t−t0), and x0−v(t−t0)indicates that the point is now displaced by v(t− t0) to the left. Therefore, the wavey(x, t) = A sin(kx − ωt) travels to the right, and the wave y(x, t) = A sin(kx + ωt)travels to the left as claimed above.

C.1.1.1 Real and Complex Expressions for the Traveling Wave

In Section C.1.1, we discussed representative solutions y(x, t) = A sin(kx ± ωt).However, cos(kx ± ωt) are also solutions, and the most general solution for a givenk or λ is of the form2

A sin(kx+ ωt) +B cos(kx+ ωt) + C sin(kx− ωt) +D cos(kx− ωt). (C.5)2We fixed the speed v when we wrote down the wave equation (C.1).

C.1. THE WAVE EQUATION 291

A complex form

Aei(kx−ωt) +Bei(kx+ωt) (C.6)

can also be used. With the definition

eiθ = cos θ + i sin θ, (C.7)

we can show that (C.5) and (C.6) are equivalent.

We generally use sin(kx ± ωt) or cos(kx ± ωt) for classical waves which have to bereal. On the other hand, the complex form ei(kx±ωt) may be preferable in quantummechanics as it often simplifies the computation.

C.1.2 Standing Wave SolutionsGenerally speaking, standing waves can be produced by two identical waves travelingin opposite directions. Due to the superposition principle, these waves form aninterference pattern that does not seem to move.

C.1.2.1 Standing Waves on a String

For a string of length L fixed at both ends, the solutions can be expressed as follows.

y(x, t) = A sinωnt sin nπxL

(C.8)

If T represents a constant horizontal tension, we have

ωn =

√T

ρ

nπ

Lfor n = 1, 2, 3, . . . . (C.9)

For the string, the propagation speed v =√

Tρ, and hence the frequency

νn = ωn/2π =

√T

ρ

nπ

L

1

2π= v

nπ

L

1

2π=nv

2L. (C.10)

292 APPENDIX C. TRAVELING WAVES AND STANDING WAVES

C.1.2.2 Standing Waves in Quantum Mechanics

A typical standing wave in quantum mechanics is generated when an incident waveinterferes with a reflected wave. Such reflection occurs at a potential boundary. Forexample, on p.138 in Section 7, an incident wave originating at x = −∞ interfereswith the reflected wave bouncing back from the potential discontinuity at x = 0.The full wavefunction given in (7.45)

Ψ(x, t) = D

(cos k1x−

k2k1

sin k1x)e−iEt/~ (C.11)

can be rewritten as follows.

Ψ(x, t) =√α2 + β2 cos(kx+ θ)e−iEt/~ (C.12)

Therefore, as claimed on p.139, the wavefunction Ψ(x, t) has fixed nodes, which inturn implies that it is a standing wave by definition.

Appendix D

Continuous Functions

It is neither necessary nor feasible to conduct all derivations in physics in a math-ematically rigorous manner. This is particularly so when a new theory is beingdeveloped. More often than not, however, sloppy mathematics when learning an es-tablished theory becomes an obstruction to going beyond the theory and doing ownphysics. It is for this reason that mathematically rigorous definitions and proofs arepresented here.

D.1 Definition of ContinuityWe specialize to a one-variable function f(x) with x ∈ R here.

Definition D.1 (Continuous Functions: δ-ϵ) A function f(x) is continuous atx = c ∈ R if for an arbitrarily chosen ε > 0 there exists δε > 0 such that |x − c| <δε =⇒ |f(x)− f(c)| < ε.

In words, this means that f(x) is arbitrarily close to f(c) if x is sufficiently close to c.Sometimes this is referred to as the delta-epsilon (δ-ε) definition and is one of severalequivalent definitions of continuity. Some readers may prefer another definition interms of the convergence of a sequence.

Definition D.2 (Convergence of a Sequence) A sequence xii∈N converges toc if for an arbitrary ε > 0 there exists nε ∈ N such that i > nε =⇒ |xi − c| < ε.When this happens, we write

limi→∞

xi = c. (D.1)

293

294 APPENDIX D. CONTINUOUS FUNCTIONS

This means that xi is arbitrarily close to c if i is sufficiently large.

Definition D.3 (Continuous Functions: Convergence of Sequences) A func-tion f(x) is continuous at x = c ∈ R if for any sequence xii∈N converging to c thesequence f(xi)i∈N converges to f(c). That is,

limi→∞

xi = c =⇒ limi→∞

f(xi) = f(c). (D.2)

D.2 Extreme and Intermediate Value TheoremsThere are two powerful and convenient theorems for a continuous function definedon a closed interval [a, b].

Theorem D.1 (Extreme Value Theorem) If a continuous function f is definedon a closed interval [a, b], then the function attains both its maximum and minimumvalues on the interval. That is to say there are numbers c, d ∈ [a, b] such thatf(c) ≥ f(x) and f(d) ≤ f(x) for all x ∈ [a, b].

Note that, this does not necessarily apply to open endpoints. For example, considerex on the interval (0, 1]. While the maximum value of e1 = e is indeed taken, theminimum does not even exist. All we can say is that ex is bounded from below by1, but ex is always strictly greater than 1 on (0, 1].

Theorem D.2 (Intermediate Value Theorem) If a continuous function f is de-fined on a closed interval [a, b] and f(a) ≤ k ≤ f(b) or f(a) ≥ k ≥ f(b), one canfind c ∈ [a, b] such that f(c) = k.

This means that any value between f(a) and f(b) is actually taken by f at least onceon the interval [a, b].

D.3 Continuity Condition on the First DerivativeWe are now ready to give a more rigorous proof of the continuity condition shown onp.128 when the potential V (x) is continuous. We will show that the first derivativedψ(x)dx

is continuous at x = x0.

First note that it suffices to consider δ < 1 in the δ − ϵ definition (Definition D.1) of continuity. Because both V (x) and ψ(x) are continuous, |V (x)− E| and |ψ(x)|

D.3. CONTINUITY CONDITION ON THE FIRST DERIVATIVE 295

are continuous and bounded from above by some finite number N on the interval[x0 − 1, x0 + 1]. Now, given ϵ > 0, let δ = min1, ~2ϵ

2mN2. Then,

|x− x0| < δ =⇒∣∣∣∣∣[dψ(x)

dx

]x

−[dψ(x)

dx

]x0

∣∣∣∣∣ = 2m

~2

∣∣∣∣∫ x


∣∣∣∣≤ 2m

~2

∫ x

x0|V (x)− E| |ψ(x)| dx (x ≥ x0) or 2m

~2

∫ x0

x|V (x)− E| |ψ(x)| dx (x ≤ x0)

<2m

~2·N2 · δ ≤ 2m

~2·N2 · ~

2ϵ

2mN2= ϵ. (D.3)

Therefore, we have shown

|x− x0| < δ =⇒∣∣∣∣∣[dψ(x)

dx

]x

−[dψ(x)

dx

]x0

∣∣∣∣∣ < ϵ. (D.4)

And, dψ(x)dx

is continuous at x = x0.

296 APPENDIX D. CONTINUOUS FUNCTIONS

Appendix E

Wronskian and LinearIndependence

The following theorems were taken from lecture notes prepared by Kiam Heong Kwa[Kwa, nd] except for the last one.

Definition E.1 (The Wronskian) Let y1 and y2 be two differentiable functions ofx. Then, the Wronskian W (y1, y2), associated to y1 and y2, is the function

W (y1, y2)(x) =

∣∣∣∣∣ y1(x) y2(x)y′1(x) y′

2(x)

∣∣∣∣∣ = y1(x)y′2(x)− y′

1(x)y2(x). (E.1)

Theorem E.1 [Bocher, 1901] If f(x) and g(x) are differentiable functions on anopen interval I such that W (f, g)(x0) , 0 for some point x0 in I, then f(x) and g(x)are linearly independent on I. Equivalently, if f(x) and g(x) are linearly dependenton I, then W (f, g)(x) = 0 for all x ∈ I.

Remark E.1 The converse of Theorem E.1 does not hold without additional condi-tions on the pair of functions involved. That is to say, it is not true in general thattwo differentiable functions f(x) and g(x) with a vanishing Wronskian on an openinterval must be linearly dependent on the same interval.

Theorem E.2 If f(x) and g(x) are differentiable functions on an open interval Isuch that f(x) , 0 and W (f, g)(x) = 0 for all x ∈ I, then f(x) and g(x) are linearlydependent. In particular, g(x) = kf(x) for a constant k.

297

298 APPENDIX E. WRONSKIAN AND LINEAR INDEPENDENCE

Theorem E.3 (Abel’s Theorem) Consider a linear second-order ordinary differ-ential equation

y′′ + p(x)y′ + q(x)y = 0; (E.2)

where the coefficients p(x) and q(x) are continuous on an open interval I.The Wronskian W (y1, y2) is given by the following integral on I for some constantthat depends on y1(x) and y2(x).

W (y1, y2)(x) = c exp[−∫p(x) dx

](E.3)

To be more explicit, if x0 is any given point in I,

W (y1, y2)(x) = W (y1, y2)(x0) exp[−∫ x

x0p(s) ds

](E.4)

for all x ∈ I.

Remark E.2 By (E.3) and (E.4), the Wronskian of any two solutions of (E.2) iseither always negative, always zero, or always positive on the interval I.

Theorem E.4 [Bostan and Dumas, 2010] Let y1(x) and y2(x) be solutions of (E.2).Then y1(x) and y2(x) are linearly independent if and only if W (y1, y2)(x) , 0for all x ∈ I. Equivalently, y1(x) and y2(x) are linearly dependent if and only ifW (y1, y2)(x) = 0 for all x ∈ I.

Definition E.2 (Analytic Functions) A function f(z) is said to be analytic abouta point z0 if it has a convergent power series representation in a neighborhood of z0;that is,

f(z) =∞∑k=0

fk(z − z0)k (E.5)

is convergent in an open set, or interval if z is real, containing z0. In addition, thecoefficients fk are uniquely determined by

fk =f (k)(z0)

k!(E.6)

for each k = 0, 1, 2, . . . .Therefore, a function is (globally) analytic if and only if its Taylor series about z0converges to the function in some neighborhood for every z0 in its domain.

299

Theorem E.5 Let f(x) and g(x) be analytic functions on an open interval I. IfW (f, g)(x) = 0 on I, then f(x) and g(x) are linearly dependent.

Fact E.1 (Principle of Superposition) Let y1 and y2 be solutions of a homoge-neous linear differential equation of the general form

A0(x)dny(x)

dxn+ A1(x)

dn−1y(x)

dxn−1+ . . .

+An−1(x)dy(x)

dx+ An(x)y(x) = 0 (E.7)

orn∑i=0

Ai(x)dn−iy(x)

dxn−i =

[n∑i=0

Ai(x)dn−i

dxn−i

]y(x) = 0; (E.8)

where d0

dx0y(x) is to be regarded as y(x) by definition. Then,

c1y1(x) + c2y2(x) (E.9)

is a solution of the same differential equation for any constants c1 and c2.

Theorem E.6 (Fundamental Set) If y1(x) and y2(x) are two linearly independentsolutions of the equation

y′′(x) + p(x)y′(x) + q(x)y(x) = 0, (E.10)

then, any solution y(x) is given by

y(x) = c1y1(x) + c2y2(x) (E.11)

for some constants c1 and c2. In this case, the set y1(x), y2(x) is called a funda-mental set of solutions.

Corollary E.1 Two polynomials p(x) and q(x) are linearly dependent if and only ifW (p, q)(x) = 0 everywhere.

300 APPENDIX E. WRONSKIAN AND LINEAR INDEPENDENCE

Appendix F

Newtonian, Lagrangian, andHamiltonian Mechanics

Newtonian mechanics, Lagrangian mechanics, and Hamiltonian mechanics are threeequivalent formulations of classical mechanics. Therefore, any of these formulationscan be used to solve a problem in classical mechanics. However, depending on thenature of the problem, one of these formulations leads to equations which are mucheasier to solve. One major difference between Newtonian mechanics and the other twois that Newtonian mechanics describes the system in terms of force, while Lagrangianand Hamiltonian mechanics do so in terms of energy. The following is a very briefreview and description of the three formulations.

F.1 An OverviewHere is a summary of the qualitative differences among Newtonian, Lagrangian, andHamiltonian formulations.

Newtonian formulation considers all the forces acting in/on the system. If youknow the forces and the mass, you can get a differential equation for the position ofeach particle.

Lagrangian formulation is based on an action principle. In a conservative field,action should be minimized for the actual trajectory (path of motion) of a particle.The action in this case is what is known as the Lagrangian L = T − V , total energyminus potential energy, and its minimization leads to a description of the behaviorof each particle by way of Lagrange’s (differential) equations. The variables involved

301

APPENDIX F. NEWTONIAN, LAGRANGIAN, AND HAMILTONIAN MECHANICS

are generalized positions (often denoted by q), generalized velocities (often denotedby q), and time t; i.e. L = L(q, q, t).

Hamiltonian formulation is usually derived from the Lagrangian. When the totalenergy is conserved, which is the case in this book, the Hamiltonian H is nothing butthe total energy of the system, H = T + V . The variables are generalized positionsq, generalized momenta p, and time t; i.e. H = H(p, q, t). This is the formulationemployed almost exclusively in quantum mechanics.

More quantitative/mathematical descriptions will follow.

F.2 Newtonian MechanicsNewton’s second law is

Fj = maj or Fj = md2xjdt2

for j = 1, . . . , n. (F.1)

The position xj(t) is obtained by solving this second-order ordinary differential equa-tion with some boundary condition such as the initial position.

The Lagrangian and Hamiltonian are formulated in an n-dimensional configura-tion space.

F.3 Lagrangian MechanicsThe Lagrangian, L, is defined as the difference between the kinetic energy T and thepotential energy V .

L = T − V (F.2)

The motion is described by Lagrange’s equations of motion

d

dt

(∂L

∂qj

)=∂L

∂qjfor j = 1, . . . , n; (F.3)

where qj’s are generalized coordinates (positions) and qj’s are generalized velocities1.1Generalized coordinates refer to the parameters that describe the configuration of the system

relative to some reference configuration. Their time derivatives are the generalized velocities.

F.4. HAMILTONIAN MECHANICS 303

F.4 Hamiltonian MechanicsThe Hamiltonian H is defined as the sum of the kinetic and potential energies.

= T + V (F.4)

Hamilton’s equations of motion are as follows.

∂H

∂qj= −pj and ∂H

∂pj= qj for j = 1, . . . , n; (F.5)

where qj’s are generalized coordinates (positions) and pj’s are generalized momenta.We also have the following time derivatives connecting H and L.

∂H

∂t= −∂L

∂t(F.6)

One apparent difference between the Lagrangian formulation and Hamiltonianformulation is that Lagrange’s equations describe the system by a set of n second-order differential equations, while Hamilton’s equations are n sets of coupled first-order differential equations. Hence, Hamiltonian mechanics requires 2n first-orderdifferential equations while Lagrangian mechanics requires n second-order differentialequations. As mentioned in the beginning on p.301, one can also note that “force”enters the equation of motion explicitly only for Newtonian mechanics.

APPENDIX F. NEWTONIAN, LAGRANGIAN, AND HAMILTONIAN MECHANICS

Appendix G

Normalization Schemes for a FreeParticle

We will only consider ψ(x) = Aeikx for k =√2mE~

. Computations for ψ(x) = Ae−ikx

are completely analogous.

G.1 The Born NormalizationThis highly controversial but widely used procedure involves restricting the rangeof integration to a finite length L and letting L to infinity afterwards. If we onlyconsider real and positive normalization constants,∫ +L/2

−L/2

(Aeikx

)∗ (Aeikx

)dx = |A|2

∫ +L/2

−L/2e−ikxeikx dx = |A|2L = 1

=⇒ A =1√L. (G.1)

Hence, the Born-normalized wavefunction is

ψ(x) =1√Leikx. (G.2)

G.2 The Dirac NormalizationDirac normalization is based on the Dirac delta function δ(x−x′) described in Section4.3. In particular, we will use the relation between Fourier transform and the deltafunction.

305

APPENDIX G. NORMALIZATION SCHEMES FOR A FREE PARTICLE

G.2.1 Fourier Transform and the Delta FunctionThe derivations found in Section 4.5 is reproduced below with slightly different no-tations more fitting for this argument.1 For a function f(x), we have

F (k) =1√2π

∫ +∞

−∞f(x)e−ikx dx the Fourier transform (G.3)

and

f(x′) =1

2π

∫ +∞

−∞F (k)eikx

′dk the inverse Fourier transform (G.4)

Substituting (G.3) for F (k) in (G.4),

f(x′) =∫ +∞

−∞f(x)

[1

2π

∫ +∞

−∞eik(x−x′) dk

]dx. (G.5)

This leads to one representation of Dirac delta function

δ(x′ − x) = 1

2π

∫ +∞

−∞eik(x−x′) dk =

1

2π~

∫ +∞

−∞eip(x

′−x)/~ dp; (G.6)

where we used the relation k =√2mE~

=

√2m p2

2m

~= p/~. Similarly,

1

2π

∫ +∞

−∞e−ik′xeikx dx =

1

2π

∫ +∞

−∞eix(k−k′) dx = δ(k − k′). (G.7)

G.2.2 Wavefunctions as Momentum or Position Eigenfunc-tions

For a free particle, we have

H =P 2

2m(G.8)

and the commutator

[H,P ] = 0. (G.9)1In the author’s modest and personal opinion, seeing/understanding the same derivation and/or

relation expressed with different notations is one good way to solidify your grasp of a mathematicaland physical concept.

G.2. THE DIRAC NORMALIZATION 307

Therefore, the stationary solutions, or the eigenfunctions, of the Hamiltonian H arealso eigenfunctions of the momentum operator P . Indeed,

PAeikx = −i~ ddxAei

p~x = −i~

(ip

~

)Aei

p~x = pAeikx. (G.10)

In the case of a free particle, the distribution of the momentum eigenvalues p iscontinuous ranging from −∞ to +∞ just like the position x. This is the situationdescribed in Section 4.8. We will label each normalized eigenket of P by its associatedmomentum p to obtain |p⟩ or Apei

p~x; where Ap’s are normalization constants.

Now, from (4.87), the representation of X in the P basis is given by

X = i~d

dp. (G.11)

Noting that

XAe−ikx = i~d

dpAe−i p

~x = i~

(−ix~

)Ae−i p

~x = xAe−ikx, (G.12)

we will label each normalized eigenket of X by its associated position eigenvalue xto obtain |x⟩ or Axe−ix

~p; where Ax’s are normalization constants.

It is important to understand the reason behind writing |p⟩ as Apeip~x and |x⟩ as

Axe−ix~p. When we interpret p as a fixed value and x as the index in the X space, we

write ei p~x, while we write e−ix

~p when x is fixed and p is the index in the P space. It is

all a matter of what is interpreted as fixed and what is interpreted as a running index.

In the braket notation, we get

⟨x|p⟩ = Apei p~x (G.13)

and⟨p|x⟩ = Axe

−ix~p. (G.14)

Needless to say, Ax = A∗p is a necessary condition because ⟨p|x⟩ = ⟨x|p⟩∗ gives

Axe−ix~p = A∗

pe−i p~x =⇒ Ax = A∗

p. (G.15)

Based on (4.7), the appropriate normalization scheme suggests itself at this point.


Definition G.1 (The Dirac Normalization) The free particle eigenfunctions |p⟩are normalized a la Dirac if

⟨p′|p⟩ = ⟨p|p′⟩ =∫ +∞

−∞

(Ape

i p′~x)∗ (

Apei p~x)dx = δ(p− p′). (G.16)

We have ∫ +∞

−∞

(Ape

ik′x)∗ (

Apeikx)dx = |Ap|2

∫ +∞

−∞e−ik′xeikx dx

= |Ap|2∫ +∞

−∞e−i p′

~xei

p~x dx = |Ap|2

∫ +∞

−∞e−ix(p′−p)/~ dx

= 2π~|Ap|2[1

2π

∫ +∞

−∞e−iy(p′−p) dy

]= 2π~|Ap|2δ(p− p′); (G.17)

where we let y = x/~. The expression (G.17) becomes δ(p− p′) if we let Ap = 1√2π~

.This gives us |p⟩ = ψp with

⟨x|p⟩ = ψp(x) =1√2π~

eikx or 1√2π~

eip~x. (G.18)

with the proposed normalization (G.16)

⟨p|p′⟩ =∫ +∞

−∞ψp(x)

∗ψp′(x) dx =1

2π

∫ +∞

−∞e−iy(p−p′) dy = δ(p− p′). (G.19)

We also have ∫ +∞

−∞

(Axe

−ik′x)∗ (

Axe−ikx

)dp = |Ax|2

∫ +∞

−∞eik

′xe−ikx dp

= |Ax|2∫ +∞

−∞ei

x′~pe−ix

~p dp = |Ax|2

∫ +∞

−∞eip(x

′−x)/~ dp

= 2π~|Ax|2[1

2π

∫ +∞

−∞eiy(x

′−x) dy]= 2π~|Ax|2δ(x− x′); (G.20)

where we let y = p/~. The expression (G.20) becomes δ(x− x′) if we let Ax = 1√2π~

.This gives us |x⟩ = ψx with

⟨p|x⟩ = ψx(p) =1√2π~

eikx or 1√2π~

eix~p (G.21)

with the normalization

⟨x|x′⟩ =∫ +∞

−∞ψx(p)

∗ψx′(p) dp =1

2π

∫ +∞

−∞eiy(x

′−x) dy = δ(x− x′). (G.22)

G.3. THE UNIT-FLUX NORMALIZATION 309

We have made sure Ap = A∗x as required by (G.15).

Finally, note again that there is symmetry between the Dirac normalization con-dition ⟨p|p′⟩ = δ(p − p′) for the eigenkets |p⟩ and the completeness (or closure)condition ⟨x|x′⟩ = δ(x− x′) for the position kets |x⟩.

G.3 The Unit-Flux NormalizationThis is based on the concept of particle current density discussed in Appendix J. Awavefunction is normalized if it has a unit current density. Namely, the followingfunctions are normalized in the unit flux sense.

ψ±(x) =1√~k/m

e±ikx ; where k =

√2mE

~(G.23)

Let us verify that we indeed have a unit current density.

j =i~

2m

(ψ∂

∂xψ∗ − ψ∗ ∂

∂xψ

)

=i~

2m

1√~k/m

e±ikx ∂

∂x

1√~k/m

e∓ikx − 1√~k/m

e∓ikx ∂

∂x

1√~k/m

e±ikx

=

i~

2m

1

(~k/m)

(e±ikx(∓ik)e∓ikx − e∓ikx (±ik) e±ikx

)=

i

2k(∓2ik) = ±1 (G.24)

The ± sign signifies the direction of the flow relative to the way the x-axis is set up.Now note that

~k/m =(~P

~

)/m =

P

m=mv

m= v. (G.25)

Hence, the probability density is given by

ψ±(x)∗ψ±(x) =

1√~k/m

e±ikx

∗1√~k/m

e±ikx =1

~k/m=

1

v, (G.26)

which provides a classical explanation for the unit flux because the “amount” ofprobability that flows through any point x = x0 per unit time, say one second, is1v× v = 1.

Appendix H

Symmetries and ConservedDynamical Variables

The description and consequences of symmetry are important and beautiful compo-nents of both classical mechanics and quantum mechanics. In the context of physics,symmetry means that a certain observable remains constant under some transfor-mation. In other words, symmetry is always in reference to a given transformation.

H.1 Poisson Brackets and Constants of MotionDefinition H.1 (Poisson Bracket) Let p and q denote sets of n generalized mo-menta and generalized coordinates; p = p1, p2, . . . , pn and q = q1, q2, . . . , qn.The Poisson bracket between ω(p, q) and λ(p, q), denoted by ω, λ, is defined asbelow.

ω, λ =n∑i=1

(∂ω

∂qi

∂λ

∂pi− ∂ω

∂pi

∂λ

∂qi

)(H.1)

Poisson bracket arises naturally as the total time derivative of a function ω(p, q),which does not depend on time t explicitly, as follows.

dω

dt=

n∑i=1

(∂ω

∂qiqi +

∂ω

∂pipi

)

=n∑i=1

(∂ω

∂qi

∂H

∂pi+∂ω

∂pi

∂H

∂qi

)= ω,H (H.2)

311

312 APPENDIX H. SYMMETRIES AND CONSERVED DYNAMICAL VARIABLES

Fact H.1 From (H.2), we can see that a variable γ whose Poisson bracket with Hvanishes, γ,H = 0, is a constant of motion (COM), or preserved.

Theorem H.1 (Invariant H and Conservation of a Dynamical Variable) Considera dynamical variable g(p, q) which generates1 the following infinitesimal transforma-tion

qi → qi = qi + ε ∂g∂pi

= qi + δqi

pi → pi = pi − ε ∂g∂qi= pi + δpi

. (H.3)

If H is invariant under the infinitesimal transformation (H.3), then g is conserved;i.e. g is a constant of motion (COM).

ProofThe Taylor series Tf (x1, x2, . . . , xn) of a multivariate function f(x1, x2, . . . , xn)about a point (a1, a2, . . . , an) is given by

Tf (x1, x2, . . . , xn) =∞∑i1=0

∞∑i2=0

· · ·∞∑in=0

(x1 − a1)i1(x2 − a2)i2 · · · (xn − an)ini1!i2! · · · in!(

∂i1+i2+···+inf

∂xi11 ∂xi22 · · · ∂xinn

)(a1, a2, . . . , an). (H.4)

Therefore, the Taylor series expansion TH (p1, p2, . . . , pn, q1, q2, . . . , qn) of our Hamil-tonian

H (p, q) = H (p1, p2, . . . , pn, q1, q2, . . . , qn) (H.5)

about the point (a1, a2, . . . , a2n) is given by

TH (p1, p2, . . . , pn, q1, q2, . . . , qn) =∞∑i1=0

∞∑i2=0

· · ·∞∑

i2n=0

(p1 − a1)i1 · · · (qn − a2n)i2n

i1!i2! · · · (i2n)!(∂i1+i2+···+i2nH

∂pi11 ∂pi22 · · · ∂qi2n

n

)(a1, a2, . . . , a2n). (H.6)

Now we let pi = pi, qi = qi, ai = pi for 1 ≤ i ≤ n, and aj = qj for n+ 1 ≤ j ≤ 2n in(H.6).

TH (p1, p2, . . . , pn, q1, q2, . . . , qn) =∞∑i1=0

∞∑i2=0

· · ·∞∑

i2n=0

(p1 − p1)i1 · · · (qn − qn)i2n

i1!i2! · · · (i2n)!

1We say g is the generator of the infinitesimal transformation (H.3).

H.2. LZ AS A GENERATOR 313

(∂i1+i2+···+i2nH

∂pi11 ∂pi22 · · · ∂qi2n

n

)(p1, p2, . . . , qn). (H.7)

Note that pi − pi = −ε ∂g∂qiand qi − qi = ε ∂g

∂pifrom (H.3). Substituting these into (H.

7), we get

TH (p1, p2, . . . , pn, q1, q2, . . . , qn) =∞∑i1=0

∞∑i2=0

· · ·∞∑

i2n=0

(−ε ∂g

∂qi

)i1 · · · (ε ∂g∂pi

)i2n

i1!i2! · · · (i2n)!

×(∂i1+i2+···+i2nH

∂pi11 ∂pi22 · · · ∂qi2n

n

)(p1, p2, . . . , qn)

= H (p1, p2, . . . , qn) +n∑i=1

[∂H

∂pi

(−ε ∂g

∂qi

)+∂H

∂qi

(ε∂g

∂pi

)]+O(ε2). (H.8)

As H is invariant under (H.3), we have

H (p1, p2, . . . , qn)−H (p1, p2, . . . , qn)

=n∑i=1

[∂H

∂pi

(−ε ∂g

∂qi

)+∂H

∂qi

(ε∂g

∂pi

)]+O(ε2)

= εn∑i=1

(−∂H∂pi

∂g

∂qi+∂H

∂qi

∂g

∂pi

)+O(ε2) = εH , g+O(ε2)

= ε [H , g+O(ε)] = 0 (H.9)

Suppose H , g(p, q), which is H , g evaluated at (p, q), is γ > 0 or δ < 0.As O(ε) ε→0−−→ 0, |H , g+O(ε)| > 0 if ε is sufficiently small. But, this contradicts(H.9). Therefore, H , g = 0, and this in turn implies that the dynamical variableg is conserved due to (H.2).

H.2 lz As a GeneratorLet g in (H.3) be the z-component of angular momentum; i.e. g = lz = xpy − ypxfrom (9.1). Then, we have the following.

x→ x = x+ ε∂g

∂px= x− yε (H.10)


y → y = y + ε∂g

∂py= xε+ y (H.11)

z → z = z + ε∂g

∂pz= z (H.12)

Comparing this with (H.20), we can see that this is an infinitesimal rotation aroundthe z-axis. Hence, the angular momentum around the z-axis is the generator ofrotations around the z-axis. Therefore, lz is conserved if H is invariant underrotations of a physical system around the z-axis. Fro example, if we only havea central force F (r), and hence a central potential V (r), there is no force in thetangential direction as we move around a circle ∥r∥ = r0, θ = θ0, whose center is onthe z-axis. Because we have no torque in this case, it follows that lz is a constant ofmotion (COM).

H.3 Rotation Around The OriginOne simple example of such an observable-transformation pair is the distance fromthe origin, f(x, y, z) =

√x2 + y2 + z2, of a particle under a rotation about the origin

denoted by R. In this case, if the rotation is by ϕ about the x-axis, followed by arotation by θ about the y-axis, followed by a rotation by ψ about the z-axis, all in thecounter-clockwise direction, the transformation R is given by the following matrix.

R =

[cos θ cosϕ cosϕ sinψ + sinϕ sin θ cosψ sinϕ sinψ − cosϕ sinψ − cosϕ sin θ cosψ

− cos θ sinψ cosϕ cosψ − sinϕ sin θ sinψ sinϕ cosψ + cosϕ sin θ sinψsin θ − sinϕ cos θ cosϕ cos θ

](H.13)

As it is very cumbersome to carry on a computation with the full three-dimensionalrotation matrix, we will only consider a counter-clockwise rotation Rz about thez-axis given by the matrix below.

Rz =

cos θ − sin θ 0sin θ cos θ 00 0 1

(H.14)

Under the transformation, (x, y, z) becomes (x, y, z). xyz

= Rz

xyz

=

cos θ − sin θ 0sin θ cos θ 00 0 1

xyz

=

x cos θ − y sin θx sin θ + y cos θ

z

(H.15)

So,

f(Rz[x, y, z]) = f(x, y, z) = f(x cos θ − y sin θ, x sin θ + y cos θ, z)

H.4. INFINITESIMAL ROTATIONS AROUND THE Z-AXIS 315

=√(x cos θ − y sin θ)2 + (x sin θ + y cos θ)2 + z2

=√x2 cos2 θ − 2xy sin θ cos θ + y2 sin2 θ + x2 sin2 θ + 2xy sin θ cos θ + y2 cos2 θ + z2

=√x2(sin2 θ + cos2 θ) + y2(sin2 θ + cos2 θ)− 2xy sin θ cos θ + 2xy sin θ cos θ + z2

=√x2 + y2 + z2 = f(x, y, z), (H.16)

and the distance from the origin f is invariant under a rotation about the z-axis.Similarly, we can also show

f(Rx[x, y, z]) = f(x, y, z) and f(Ry[x, y, z]) = f(x, y, z); (H.17)

where Rx and Ry are rotations about the x- and y-axes, respectively. As any rotationis a combination of Rx, Ry, and Rz, this proves that the distance form the origin isinvariant under any rotation about the origin as it should be.

H.4 Infinitesimal Rotations Around the z-AxisWe will first specialize to rotations about the origin in two dimensions for computa-tional simplicity. The results can be extended to rotations about the z-axis in threedimensions in a straightforward manner.

In two dimensions, we only have Rz. The effect of Rz by θ is as follows.[xy

]= Rz

[xy

]=


] [xy

]=

[x cos θ − y sin θx sin θ + y cos θ

](H.18)

Next, consider an infinitesimal transformation which differs from the identity trans-formation I by angle ε. In the limit as ε approaches 0, cos ε approaches 1 and sin εtends to ε. Hence, [

xy

]=

[x cos ε− y sin εx sin ε+ y cos ε

]ε → 0−−−→

[x− yεxε+ y

]. (H.19)

This extends to three dimensions as below. xyz

=

x cos ε− y sin εx sin ε+ y cos ε

z

ε → 0−−−→

x− yεxε+ yz

. (H.20)

Appendix I

Commutators

I.1 Commutator IdentitiesHere are some basic and useful commutator identities.

[A,B + C] = [A,B] + [A,C] (I.1)[A+B,C] = [A,C] + [B,C] (I.2)[A,BC] = [A,B]C +B[A,C] (I.3)[AB,C] = [A,C]B + A[B,C] (I.4)

[AB,CD] = A[B,C]D + C[A,D]B + CA[B,D] + [A,C]BD (I.5)[A, [B,C]] + [C, [A,B]] + [B, [C,A]] = 0 (I.6)

Let us verify (I.3), (I.4), and (I.5). It is straightforward for (I.3) and (I.4).[A,BC] = ABC −BCA = ABC −BAC +BAC −BCA= (AB −BA)C +B(AC − CA) = [A,B]C +B[A,C] (I.7)

Likewise for (I.4). Verification of (I.5) is similar.[AB,CD] = [(AB), CD] = [AB,C]D + C[AB,D]

= ([A,C]B + A[B,C])D + C([A,D]B + A[B,D])

= A[B,C]D + C[A,D]B + CA[B,D] + [A,C]BD (I.8)

I.2 Commutators involving X and P

From Tables 3.1 and 3.2, we have Mxi(multiplication by xi) and xi

(−i~ ∂

∂xi

)for

(x1, x2, x3) = (x, y, z) and (x1, x2, x3) =(i, j, k

). Of 18 possible combinations for the

317

318 APPENDIX I. COMMUTATORS

commutator, there are only three that are nonzero. We have:

[Xi, Xj] = [Mxi,Mxj

] = 0 for all (i, j) (I.9)

[Pi, Pj] =

[−i~ ∂

∂xi,−i~ ∂

∂xj

]= 0 for all (i, j) (I.10)

[Pi, Xj] = −i~δij for all (i, j). (I.11)

These relations are very useful in computing other commutators such as those amongLx, Ly, Lz, L2, and H.

I.3 Commutators involving X, P , L, and r

[Lz, x] = i~y (I.12)[Lz, y] = −i~x (I.13)[Lz, z] = 0 (I.14)

[Lz, px] = i~py (I.15)[Lz, py] = −i~px (I.16)

[Lz, pz] = 0 (I.17)

[Lz, r2] = 0 (I.18)

[Lz, p2] = 0 (I.19)

[H,Lx] = [H,Ly] = [H,Lz] = 01 when V = V (r) (I.20)[H,L2] = 0 (I.21)

Verification

[Lz, x] = [xpy − ypx, x] = [xpy, x]− [ypx, x] = [x, x]py + x[py, x]− ([y, x]px + y[px, x])

= 0 + 0− 0− y[px, x] = −y(−i~) = i~y (verification of I.12)

[Lz, px] = [xpy − ypx, px] = [xpy, px]− [ypx, px] = [x, px]py + x[py, px]

1One can write these as [H,L] = 0.

I.3. COMMUTATORS INVOLVING X, P , L, AND R 319

−([y, px]px + y[px, px]) = i~py + 0− 0− 0 = i~py (verification of I.15)

Recall from (9.3) that

Lz = −i~∂

∂ϕ. (I.22)

Because Lz is a partial derivative with respect to ϕ, it commutes with multiplicationby any function of r, including r2. This proves (I.18).

[Lz, p2] = [Lz, p

2x + p2y + p2z] = [Lz, p

2x] + [Lz, p

2y] + [Lz, p

2z]

= [Lz, px]px + px[Lz, px] + [Lz, py]py + py[Lz, py] + [Lz, pz]pz + pz[Lz, pz]

= i~pypx + i~pxpy − i~pxpy − i~pypx + 0 + 0 = 0 (verification of I.19)

[H,Lz] =

[p2

2m+ V (r), Lz

]=

1

2m[p2, Lz] + [V (r), Lz] = 0 + 0 = 0 (verification of I.20)

[H,L2] = [H,L2x + L2

y + L2z] = [H,L2

x] + [H,L2y] + [H,L2

z]

= [H,Lx]Lx + Lx[H,Lx] + [H,Ly]Ly + Ly[H,Ly] + [H,Lz]Lz + Lz[H,Lz]

= 0 + 0 + 0 + 0 + 0 + 0 (verification of I.21)

320 APPENDIX I. COMMUTATORS

Appendix J

Probability Current

The following is based on a lecture note of Professor Steven Carlip at UC Davis[Carlip, 2013].

Let R be a region in R3 with a boundary ∂R, and consider a particle whosewavefunction is Ψ(x, t). Then, the probability P (R) that the particle is found in Ris given by

P (R) =∫RΨ∗(x, t)Ψ(x, t) dxdydz. (J.1)

If the particle is moving in the classical sense, or the magnitude of the wavefunctionis changing with time in the quantum mechanical sense, P (R) changes with time,and the first derivative is given by

d

dtP (R) =

∫R

∂

∂t(Ψ(x, t)Ψ(x, t)) dxdydz =

∫R

(∂Ψ∗

∂tΨ+Ψ∗∂Ψ

∂t

)dxdydz. (J.2)

The time-dependent Schrodinger equation gives

i~∂Ψ

∂t= − ~

2

2m∇2Ψ+ VΨ (J.3)

as well as its complex conjugate

−i~∂Ψ∗

∂t= − ~

2

2m∇2Ψ∗ + VΨ∗. (J.4)

Therefore,

∂Ψ∗

∂tΨ =

1

−i~

(− ~

2

2m∇2Ψ∗ + VΨ∗

)Ψ =

~

i2m

(∇2Ψ∗

)Ψ− V

i~Ψ∗Ψ (J.5)

321

322 APPENDIX J. PROBABILITY CURRENT

and

Ψ∗∂Ψ

∂t= Ψ∗ 1

i~

(− ~

2

2m∇2Ψ+ VΨ

)= − ~

i2mΨ∗∇2Ψ+

V

i~Ψ∗Ψ. (J.6)

So, the integrand of (J.2) is

∂Ψ∗

∂tΨ+Ψ∗∂Ψ

∂t=~

i2m

(Ψ∇2Ψ∗ −Ψ∗∇2Ψ

)= − i~

2m

(Ψ∇2Ψ∗ −Ψ∗∇2Ψ

). (J.7)

Now we need the following identity.

Fact J.1 For differentiable functions f and g we have

∇ · (f∇g − g∇f) = f∇2g − g∇2f. (J.8)

ProofFor simplicity, we will introduce the following notation, which is used widely in thephysics literature.

fxi:=

∂f

∂xiand fxixj

:=∂2f

∂xj∂xi; where xi, xj = x, y, or z (J.9)

Then,

∇ · (f∇g − g∇f) = ∇ ·[f(gxı+ gy ȷ+ gzk

)− g

(fxı+ fy ȷ+ fzk

)]=

(ı∂

∂x+ ȷ

∂

∂y+ k

∂

∂z

)·[f(gxı+ gy ȷ+ gzk

)− g

(fxı+ fy ȷ+ fzk

)]=

∂

∂x(fgx − gfx) +

∂

∂y(fgy − gfy) +

∂

∂z(fgz − gfz)

= fxgx + fgxx − gxfx − gfxx + fygy + fgyy − gyfy − gfyy+ fzgz + fgzz − gzfz − gfzz

= fgxx − gfxx + fgyy − gfyy + fgzz − gfzz= f(gxx + gyy + gzz)− g(fxx + fyy + fzz)

= f∇2g − g∇2f. (J.10)

Now, let f = Ψ and g = Ψ∗ in (J.8) to obtain

Ψ∇2Ψ∗ −Ψ∗∇2Ψ = ∇ · (Ψ∇Ψ∗ −Ψ∗∇Ψ) . (J.11)

323

If we define a vector j by

j :=i~

2m(Ψ∇Ψ∗ −Ψ∗∇Ψ) , (J.12)

then, from (J.11) we have

∇ · j =i~

2m

(Ψ∇2Ψ∗ −Ψ∗∇2Ψ

), (J.13)

and (J.2) becomes

d

dtP (R) =

∫R

(∂Ψ∗

∂tΨ+Ψ∗∂Ψ

∂t

)dxdydz

= −∫R

i~

2m

(Ψ∇2Ψ∗ −Ψ∗∇2Ψ

)dxdydz

= −∫R∇ · j dxdydz. (J.14)

Next, remember Stokes’ Theorem from multivariable calculus.

Theorem J.1 (Stokes’ Theorem: the Divergence Theorem)∫R∇ · j d3x =

∫∂R

j · n d2x; (J.15)

where n is the unit normal to the boundary or surface ∂R.

We finally have

d

dtP (R) = −

∫∂R

j · n d2x. (J.16)

What (J.16) is telling us is that the amount of j escaping through the surface ∂Requals, in magnitude, the change in the amount of probability we have in the regionR. This is why j is called the probability current.

Definition J.1 (Probability Current) Given a wavefuntion Ψ(x, t) we define theprobability current j by

j :=i~

2m(Ψ∇Ψ∗ −Ψ∗∇Ψ) . (J.17)

324 APPENDIX J. PROBABILITY CURRENT

In one dimension, this reduces to

j :=i~

2m

(Ψ∂

∂xΨ∗ −Ψ∗ ∂

∂xΨ

). (J.18)

In particular, when the potential, and hence the Hamiltonian, does not have explicittime dependence, j reduces to an expression in the spatial part of the wavefunctionψ(x) rather than the full wave function Ψ(x, t)1 such that

j =i~

2m

(ψ∂

∂xψ∗ − ψ∗ ∂

∂xψ

). (J.19)

Note that this is completely analogous in form and substance to the continuity equa-tion of fluid dynamics and electromagnetism.

dQ

dt= −

∫∂R

j · n dS (dS is the surface integral over ∂R, the boundary of R.)2

(J.20)In fluid dynamics, Q is the amount of the fluid, and j is the flow of the fluid. Inelectromagnetism, Q is the amount of electric charge, and j is the electric current.Finally, let us make a note of the fact that the equality does not hold if there isa “sink” or “source” in the region R, which is the difference between the classicalcontinuity equation and our quantum mechanical equation.

1 The derivation is straightforward as below.

j :=i~

2m

(Ψ∂

∂xΨ∗ −Ψ∗ ∂

∂xΨ

)=

i~

2m

(ψ(x)e−iωt ∂

∂x

(ψ(x)e−iωt

)∗

−(ψ(x)e−iωt

)∗ ∂

∂x

(ψ(x)e−iωt

))=

i~

2m

(ψ(x)e−iωt ∂

∂xψ(x)∗eiωt − ψ(x)∗eiωt ∂

∂xψ(x)e−iωt

)=

i~

2m

(ψ∂

∂xψ∗ − ψ∗ ∂

∂xψ

)2This form is known as the integral form. Its differential counterpart, called the differential form

quite appropriately, is given by

∂ρ(x, t)

∂t= −∇ · j(x, t) or ∂ρ(x, t)

∂t= − ∂

∂xj(x, t) in one dimension;

where ρ is the density.

Appendix K

Chain Rules in PartialDifferentiation

K.1 One Independent VariableConsider a function f = f(x, y, z), where x = x(t), y = y(t), and z = z(t). Then, wehave

df

dt=∂f

∂x

dx

dt+∂f

∂y

dy

dt+∂f

∂z

dz

dt. (K.1)

K.2 Three Independent VariablesConsider a function f = f(x, y, z), where x = x(r, θ, ϕ), y = y(r, θ, ϕ), and z =z(r, θ, ϕ). Then, we have

∂f

∂r=∂f

∂x

∂x

∂r+∂f

∂y

∂y

∂r+∂f

∂z

∂z

∂r, (K.2)

∂f

∂θ=∂f

∂x

∂x

∂θ+∂f

∂y

∂y

∂θ+∂f

∂z

∂z

∂θ, (K.3)

and∂f

∂ϕ=∂f

∂x

∂x

∂ϕ+∂f

∂y

∂y

∂ϕ+∂f

∂z

∂z

∂ϕ. (K.4)

Actually, the following operator forms may be more convenient in actual use.∂

∂r=∂x

∂r

∂

∂x+∂y

∂r

∂

∂y+∂z

∂r

∂

∂z(K.5)

325

326 APPENDIX K. CHAIN RULES IN PARTIAL DIFFERENTIATION

∂

∂θ=∂x

∂θ

∂

∂x+∂y

∂θ

∂

∂y+∂z

∂θ

∂

∂z(K.6)

∂

∂ϕ=∂x

∂ϕ

∂

∂x+∂y

∂ϕ

∂

∂y+∂z

∂ϕ

∂

∂z(K.7)

Of course, we also have

∂f

∂x=∂f

∂r

∂r

∂x+∂f

∂θ

∂θ

∂x+∂f

∂ϕ

∂ϕ

∂x, (K.8)

∂f

∂y=∂f

∂r

∂r

∂y+∂f

∂θ

∂θ

∂y+∂f

∂ϕ

∂ϕ

∂y, (K.9)

and∂f

∂z=∂f

∂r

∂r

∂z+∂f

∂θ

∂θ

∂z+∂f

∂ϕ

∂ϕ

∂z, (K.10)

or

∂

∂x=∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ, (K.11)

∂

∂y=∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂ϕ

∂y

∂

∂ϕ, (K.12)

and∂

∂z=∂r

∂z

∂

∂r+∂θ

∂z

∂

∂θ+∂ϕ

∂z

∂

∂ϕ. (K.13)

What we do not have is the following.

K.3 Reciprocal Relation is FalseUnlike the total derivative, we do not have

∂r

∂x=

(∂x

∂r

)−1

. FALSE (K.14)

Instead, we generally have

∂r

∂x,

(∂x

∂r

)−1

. TRUE (K.15)

K.4. INVERSE FUNCTION THEOREMS 327

This is easy to understand because ∂r∂x

means the derivative/change of r with respectto x where y and z are held constant, while ∂x

∂ris the derivative/change of x with

respect to r where θ and ϕ are held fixed. An explicit computation follows.(∂r

∂x

)y,z: fixed

=

(∂√x2 + y2 + z2

∂x

)y,z: fixed

=1

2

2x√x2 + y2 + z2

=x

r= sin θ cosϕ

(K.16)(∂x

∂r

)θ,ϕ: fixed

=

(∂r sin θ cosϕ

∂r

)θ,ϕ: fixed

= sin θ cosϕ (K.17)

Hence, the inequality (K.15) holds in this case. Needless to say, (K.15) is generallytrue and is not limited to spherical and Cartesian coordinates.

Compare (K.15) with the following example where we have total derivatives.y = x2 or x = ±√y (K.18)

dy

dx= 2x and dx

dy=d± √ydy

= ±1

2

1√y= ±1

2

1

±x=

1

2x(K.19)

dy

dx=

(dx

dy

)−1

(K.20)

Note here that (K.20) really means

dy

dx

∣∣∣∣∣at x=x0

=

dxdy

∣∣∣∣∣at y=x20

−1

. (K.21)

See Theorem K.1 for a more general description.

K.4 Inverse Function TheoremsThe reciprocal relation (K.20) is nothing but a straightforward application of theInverse Function Theorem.Theorem K.1 (Inverse Function Theorem: Single Variable) If f is a contin-uously differentiable function of x with nonzero derivative at the point a, then, f isinvertible in a neighborhood of a, the inverse f−1 is continuously differentiable, and(

f−1)′(b) =

1

f ′(a), (K.22)

where b = f(a).

328 APPENDIX K. CHAIN RULES IN PARTIAL DIFFERENTIATION

Theorem K.2 (Inverse Function Theorem: Multiple Variables) If the totalderivative of a continuously differentiable multivariable function F defined on anopen subset O of Rn and into Rn is invertible at a point p, that is, the Jacobian Jof F at p is non-zero, then, F is an invertible function with the inverse F−1 nearthe point p. This means that the inverse F−1 is defined in some neighborhood ofF (p). Moreover, the inverse function F−1 is continuously differentiable. As for theJacobian, we have

JF−1 (F (p)) = [JF (p)]−1 ; (K.23)

where JG(q) is the Jacobian of the function G at the point q, and [JF (p)]−1 is the

matrix inverse.

Appendix L

Laplacian Operator in SphericalCoordinates

The Laplacian operator in Cartesian coordinates is

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2. (L.1)

We will use the following relations

r =√x2 + y2 + z2 x = r sin θ cosϕ

y = r sin θ sinϕ z = r cos θ

cos θ = z√x2 + y2 + z2

sin θ =√x2 + y2√

x2 + y2 + z2

cosϕ =x√

x2 + y2sinϕ =

y√x2 + y2

,

along with the chain rules below from p.325 of Appendix K.∂

∂x=∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ, (L.2)

∂

∂y=∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂ϕ

∂y

∂

∂ϕ, (L.3)

∂

∂z=∂r

∂z

∂

∂r+∂θ

∂z

∂

∂θ+∂ϕ

∂z

∂

∂ϕ. (L.4)

∂

∂r=∂x

∂r

∂

∂x+∂y

∂r

∂

∂y+∂z

∂r

∂

∂z(L.5)

329

330 APPENDIX L. LAPLACIAN OPERATOR IN SPHERICAL COORDINATES

∂

∂θ=∂x

∂θ

∂

∂x+∂y

∂θ

∂

∂y+∂z

∂θ

∂

∂z(L.6)

∂

∂ϕ=∂x

∂ϕ

∂

∂x+∂y

∂ϕ

∂

∂y+∂z

∂ϕ

∂

∂z(L.7)

The partial derivatives of r with respect to x, y, and z are as follows.

∂r

∂x=

1

2

2x√x2 + y2 + z2

=x

r= sin θ cosϕ (L.8)

∂r

∂y=

1

2

2y√x2 + y2 + z2

=y

r= sin θ sinϕ (L.9)

∂r

∂z=

1

2

2z√x2 + y2 + z2

=z

r= cos θ (L.10)

Next, we will use z = r cos θ to compute ∂θ∂x

, ∂θ∂y

, and ∂θ∂z

. 1

∂z

∂x=∂r

∂xcos θ + r

∂ cos θ∂x

= sin θ cosϕ cos θ + r

(∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ

)(cos θ)

= sin θ cosϕ cos θ + r∂θ

∂x(− sin θ) =⇒ 0 = sin θ cosϕ cos θ − r sin θ ∂θ

∂x

=⇒ r sin θ ∂θ∂x

= sin θ cosϕ cos θ

1 There is a simpler, but possibly less comfortable way to compute ∂θ∂x , ∂θ

∂y , and ∂θ∂z using the

equalities

cos θ = z

rand sin θ =

√x2 + y2√

x2 + y2 + z2.

Namely,

cos θ = z

r=⇒ − sin θdθ = z(−x)

r3dx =⇒ ∂θ

∂x=

cos θ cosϕr

,

cos θ = z

r=⇒ − sin θdθ = z(−y)

r3dy =⇒ ∂θ

∂y=

cos θ sinϕr

,

and

sin θ =√x2 + y2√

x2 + y2 + z2=⇒ cos θdθ =

√x2 + y2(−z)

r3dz =⇒ ∂θ

∂z= − sin θ

r.

331

=⇒ ∂θ

∂x=

cos θ cosϕr

(L.11)

∂z

∂y=∂r

∂ycos θ + r

∂ cos θ∂y

= sin θ sinϕ cos θ + r

(∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂ϕ

∂y

∂

∂ϕ

)(cos θ)

= sin θ sinϕ cos θ + r∂θ

∂y(− sin θ) =⇒ 0 = sin θ sinϕ cos θ − r sin θ∂θ

∂y

=⇒ r sin θ∂θ∂y

= sin θ sinϕ cos θ

=⇒ ∂θ

∂y=

cos θ sinϕr

(L.12)

∂z

∂z=∂r

∂zcos θ + r

∂ cos θ∂z

= cos θ cos θ + r

(∂r

∂z

∂

∂r+∂θ

∂z

∂

∂θ+∂ϕ

∂z

∂

∂ϕ

)(cos θ)

= cos θ cos θ + r∂θ

∂z(− sin θ) =⇒ 1 = cos2 θ − r sin θ∂θ

∂z

=⇒ r sin θ∂θ∂z

= − sin2 θ

=⇒ ∂θ

∂z= −sin θ

r(L.13)

Now, use x = r sin θ cosϕ to compute ∂ϕ∂x

, ∂ϕ∂y

, and ∂ϕ∂z

. 2

∂x

∂x=

(∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ

)(r sin θ cosϕ)

2 There is an alternative way using the relations

cosϕ =x√

x2 + y2and sinϕ =

y√x2 + y2

.

Namely,

cosϕ =x√

x2 + y2=⇒ − sinϕdϕ =

(1√

x2 + y2+ x

(−12

)1

(x2 + y2)3/2(2x)

)dx

=

(1√

x2 + y2− x2

(x2 + y2)3/2

)dx =

x2 + y2 − x2

(x2 + y2)3/2dx =

y2

(x2 + y2)3/2dx


= sin θ cosϕ sin θ cosϕ+cos θ cosϕ

rr cos θ cosϕ− r sin θ sinϕ∂ϕ

∂x

= sin2 θ cos2 ϕ+ cos2 θ cos2 ϕ− r sin θ sinϕ∂ϕ∂x

= cos2 ϕ− r sin θ sinϕ∂ϕ∂x

=⇒ 1 = cos2 ϕ− r sin θ sinϕ∂ϕ∂x

=⇒ r sin θ sinϕ∂ϕ∂x

= − sin2 ϕ

=⇒ ∂ϕ

∂x= − sinϕ

r sin θ (L.14)

∂x

∂y=

(∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂ϕ

∂y

∂

∂ϕ

)(r sin θ cosϕ)

= sin θ sinϕ sin θ cosϕ+cos θ sinϕ

rr cos θ cosϕ− r sin θ sinϕ∂ϕ

∂y

= sin2 θ sinϕ cosϕ+ cos2 θ sinϕ cosϕ− r sin θ sinϕ∂ϕ∂y

= sinϕ cosϕ− r sin θ sinϕ∂ϕ∂y

=⇒ 0 = sinϕ cosϕ− r sin θ sinϕ∂ϕ∂y

=⇒ r sin θ sinϕ∂ϕ∂y

= sinϕ cosϕ

=⇒ ∂ϕ

∂y=

cosϕr sin θ (L.15)

∂x

∂z=

(∂r

∂z

∂

∂r+∂θ

∂z

∂

∂θ+∂ϕ

∂z

∂

∂ϕ

)(r sin θ cosϕ)

=(r sin θ sinϕ)2

(r sin θ)3 dx =sin2 ϕ

r sin θdx =⇒ ∂ϕ

∂x=

sin2 ϕ

r sin θ1

(− sinϕ) = − sinϕr sin θ ,

and

sinϕ =y√

x2 + y2=⇒ cosϕdϕ =

(1√

x2 + y2+ y

(−12

)1

(x2 + y2)3/2(2y)

)dy

=

(1√

x2 + y2− y2

(x2 + y2)3/2

)dy =

x2 + y2 − y2

(x2 + y2)3/2dy =

x2

(x2 + y2)3/2dy

=(r sin θ cosϕ)2

(r sin θ)3 dy =cos2 ϕr sin θ dy =⇒ ∂ϕ

∂y=

cos2 ϕr sin θ

1

cosϕ =cosϕr sin θ .

333

= cos θ sin θ cosϕ− sin θrr cos θ cosϕ− r sin θ sinϕ∂ϕ

∂z

= −r sin θ sinϕ∂ϕ∂z

=⇒ ∂ϕ

∂z= 0 (L.16)

Incidentally, we already knew that ∂ϕ∂z

= 0 from geometric considerations; that is, ϕis determined completely by the x- and y-coordinates.

Let us take a deep breath at this point and summarize what we have found sofar.

∂r∂x

= sin θ cosϕ ∂r∂y

= sin θ sinϕ ∂r∂z

= cos θ∂θ∂x

= cos θ cosϕr

∂θ∂y

= cos θ sinϕr

∂θ∂z

= − sin θr

∂ϕ∂x

= − sinϕr sin θ

∂ϕ∂y

= cosϕr sin θ

∂ϕ∂z

= 0

(L.17)

Our next step is to compute ∂∂x

, ∂∂y

, and ∂∂z

.

∂

∂x=∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ= sin θ cosϕ ∂

∂r+

cos θ cosϕr

∂

∂θ− sinϕr sin θ

∂

∂ϕ(L.18)

∂

∂y=∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂ϕ

∂y

∂

∂ϕ= sin θ sinϕ ∂

∂r+

cos θ sinϕr

∂

∂θ+

cosϕr sin θ

∂

∂ϕ(L.19)

∂

∂z= cos θ ∂

∂r− sin θ

r

∂

∂θ+ 0 · ∂

∂ϕ= cos θ ∂

∂r− sin θ

r

∂

∂θ(L.20)

We are ready to compute ∇2 = ∂2

∂x2+ ∂2

∂y2+ ∂2

∂z2= ∂

∂x∂∂x

+ ∂∂y

∂∂y

+ ∂∂z

∂∂z

.

∂2

∂x2=

∂

∂x

∂

∂x=

(sin θ cosϕ ∂

∂r+

cos θ cosϕr

∂


∂

∂ϕ

)·(

sin θ cosϕ ∂∂r

+cos θ cosϕ

r

∂


∂

∂ϕ

)

=

(sin θ cosϕ ∂

∂r

)(sin θ cosϕ ∂

∂r

)+

(sin θ cosϕ ∂

∂r

)(cos θ cosϕ

r

∂

∂θ

)


−(

sin θ cosϕ ∂∂r

)(sinϕr sin θ

∂

∂ϕ

)

+

(cos θ cosϕ

r

∂

∂θ

)(sin θ cosϕ ∂

∂r

)+

(cos θ cosϕ

r

∂

∂θ

)(cos θ cosϕ

r

∂

∂θ

)

−(

cos θ cosϕr

∂

∂θ

)(sinϕr sin θ

∂

∂ϕ

)

−(

sinϕr sin θ

∂

∂ϕ

)(sin θ cosϕ ∂

∂r

)−(

sinϕr sin θ

∂

∂ϕ

)(cos θ cosϕ

r

∂

∂θ

)

+

(sinϕr sin θ

∂

∂ϕ

)(sinϕr sin θ

∂

∂ϕ

)

= sin2 θ cos2 ϕ ∂2

∂r2+ sin θ cos θ cos2 ϕ ∂

∂r

(1

r

∂

∂θ

)

− sinϕ cosϕ ∂∂r

(1

r

∂

∂ϕ

)

+cos θ cos2 ϕ

r

∂

∂θ

(sin θ ∂

∂r

)+

cos θ cos2 ϕr2

∂

∂θ

(cos θ ∂

∂θ

)

− cos θ sinϕ cosϕr2

∂

∂θ

(1

sin θ∂

∂ϕ

)

− sinϕr

∂

∂ϕ

(cosϕ ∂

∂r

)− cos θ sinϕ

r2 sin θ∂

∂ϕ

(cosϕ ∂

∂θ

)

+sinϕ

r2 sin2 θ

∂

∂ϕ

(sinϕ ∂

∂ϕ

)


∂r2+ sin θ cos θ cos2 ϕ

[− 1

r2∂

∂θ+

1

r

∂2

∂r∂θ

]

− sinϕ cosϕ[− 1

r2∂

∂ϕ+

1

r

∂2

∂r∂ϕ

]

+cos θ cos2 ϕ

r

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ

]

+cos θ cos2 ϕ

r2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2

]

− cos θ sinϕ cosϕr2 sin θ

[−cos θ

sin θ∂

∂ϕ+

∂2

∂θ∂ϕ

]

335

− sinϕr

[− sinϕ ∂

∂r+ cosϕ ∂2

∂r∂ϕ

]

− cos θ sinϕr2 sin θ

[− sinϕ ∂

∂θ+ cosϕ ∂2

∂θ∂ϕ

]

+sinϕ

r2 sin2 θ

[cosϕ ∂

∂ϕ+ sinϕ ∂2

∂ϕ2

]


∂r2+

cos2 θ cos2 ϕr2

∂2

∂θ2+

sin2 ϕ

r2 sin2 θ

∂2

∂ϕ2+

cos2 θ cos2 ϕ+ sin2 ϕ

r

∂

∂r

+ sin θ cos θ cos2 ϕ[− 1

r2∂

∂θ+

1

r

∂2

∂r∂θ

]− sinϕ cosϕ

[− 1

r2∂

∂ϕ+

1

r

∂2

∂r∂ϕ

]

+cos θ cos2 ϕ

r

[+ sin θ ∂2

∂r∂θ

]

+cos θ cos2 ϕ

r2

[− sin θ ∂

∂θ

]

− cos θ sinϕ cosϕr2 sin θ

[−cos θ

sin θ∂

∂ϕ+

∂2

∂θ∂ϕ

]

− sinϕr

[+ cosϕ ∂2

∂r∂ϕ

]


[− sinϕ ∂

∂θ+ cosϕ ∂2

∂θ∂ϕ

]

+sinϕ

r2 sin2 θ

[cosϕ ∂

∂ϕ

](L.21)

∂2

∂y2=

∂

∂y

∂

∂y=

(sin θ sinϕ ∂

∂r+

cos θ sinϕr

∂

∂θ+

cosϕr sin θ

∂

∂ϕ

)·(

sin θ sinϕ ∂∂r

+cos θ sinϕ

r

∂

∂θ+

cosϕr sin θ

∂

∂ϕ

)

=

(sin θ sinϕ ∂

∂r

)(sin θ sinϕ ∂

∂r

)+

(sin θ sinϕ ∂

∂r

)(cos θ sinϕ

r

∂

∂θ

)

+

(sin θ sinϕ ∂

∂r

)(cosϕr sin θ

∂

∂ϕ

)

+

(cos θ sinϕ

r

∂

∂θ

)(sin θ sinϕ ∂

∂r

)+

(cos θ sinϕ

r

∂

∂θ

)(cos θ sinϕ

r

∂

∂θ

)


+

(cos θ sinϕ

r

∂

∂θ

)(cosϕr sin θ

∂

∂ϕ

)

+

(cosϕr sin θ

∂

∂ϕ

)(sin θ sinϕ ∂

∂r

)+

(cosϕr sin θ

∂

∂ϕ

)(cos θ sinϕ

r

∂

∂θ

)

+

(cosϕr sin θ

∂

∂ϕ

)(cosϕr sin θ

∂

∂ϕ

)

= sin2 θ sin2 ϕ∂2

∂r2+ sin θ cos θ sin2 ϕ

∂

∂r

(1

r

∂

∂θ

)

+ sinϕ cosϕ ∂∂r

(1

r

∂

∂ϕ

)

+cos θ sin2 ϕ

r

∂

∂θ

(sin θ ∂

∂r

)+

cos θ sin2 ϕ

r2∂

∂θ

(cos θ ∂

∂θ

)

+cos θ sinϕ cosϕ

r2∂

∂θ

(1

sin θ∂

∂ϕ

)

+cosϕr

∂

∂ϕ

(sinϕ ∂

∂r

)+

cos θ cosϕr2 sin θ

∂

∂ϕ

(sinϕ ∂

∂θ

)

+cosϕr2 sin2 θ

∂

∂ϕ

(cosϕ ∂

∂ϕ

)

= sin2 θ sin2 ϕ∂2

∂r2+ sin θ cos θ sin2 ϕ

[− 1

r2∂

∂θ+

1

r

∂2

∂r∂θ

]

+ sinϕ cosϕ[− 1

r2∂

∂ϕ+

1

r

∂2

∂r∂ϕ

]

+cos θ sin2 ϕ

r

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ

]

+cos θ sin2 ϕ

r2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2

]

+cos θ sinϕ cosϕ

r2 sin θ

[−cos θ

sin θ∂

∂ϕ+

∂2

∂θ∂ϕ

]

+cosϕr

[cosϕ ∂

∂r+ sinϕ ∂2

∂r∂ϕ

]

+cos θ cosϕr2 sin θ

[cosϕ ∂

∂θ+ sinϕ ∂2

∂θ∂ϕ

]

337

+cosϕr2 sin2 θ

[− sinϕ ∂

∂ϕ+ cosϕ ∂2

∂ϕ2

](L.22)

∂2

∂z2=

∂

∂z

∂

∂z=

(cos θ ∂

∂r− sin θ

r

∂

∂θ

)·(

cos θ ∂∂r− sin θ

r

∂

∂θ

)

=

(cos θ ∂

∂r

)(cos θ ∂

∂r

)−(

cos θ ∂∂r

)(sin θr

∂

∂θ

)−(

sin θr

∂

∂θ

)(cos θ ∂

∂r

)

+

(sin θr

∂

∂θ

)(sin θr

∂

∂θ

)

= cos2 θ ∂2

∂r2− sin θ cos θ ∂

∂r

(1

r

∂

∂θ

)− sin θ

r

∂

∂θ

(cos θ ∂

∂r

)

+sin θr2

∂

∂θ

(sin θ ∂

∂θ

)

= cos2 θ ∂2

∂r2− sin θ cos θ

[− 1

r2∂

∂θ+

1

r

∂2

∂r∂θ

]− sin θ

r

[− sin θ ∂

∂r+ cos θ ∂2

∂r∂θ

]

+sin θr2

[cos θ ∂

∂θ+ sin θ ∂

2

∂θ2

](L.23)

Finally, we can compute ∇2, combining (L.21), (L.22), and (L.23). There are somenice cancellations and simplifications. But, we have to be methodical and systematicin deciding on our computational strategy, that is, which terms to combine in whatorder.

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2

=[sin2 θ

(sin2 ϕ+ cos2 ϕ

)+ cos2 θ

] ∂2∂r2

+[sin θ cos θ

(sin2 ϕ+ cos2 ϕ

)− sin θ cos θ

] [− 1

r2∂

∂θ+

1

r

∂2

∂r∂θ

]

+ (sinϕ cosϕ− sinϕ cosϕ)[− 1

r2∂

∂ϕ+

1

r

∂2

∂r∂ϕ

]

+cos θ(sin2 ϕ+ cos2 ϕ)

r

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ

]

+cos θ(sin2 ϕ+ cos2 ϕ)

r2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2

]


+

(cos θ sinϕ cosϕ

r2 sin θ − cos θ sinϕ cosϕr2 sin θ

)[−cos θ

sin θ∂

∂ϕ+

∂2

∂θ∂ϕ

]The following three terms derive from ∂2

∂x2.

− sinϕr

[− sinϕ ∂

∂r+ cosϕ ∂2

∂r∂ϕ

]


[− sinϕ ∂

∂θ+ cosϕ ∂2

∂θ∂ϕ

]

+sinϕ

r2 sin2 θ

[cosϕ ∂

∂ϕ+ sinϕ ∂2

∂ϕ2

]The following three terms derive from ∂2

∂y2.

+cosϕr

[cosϕ ∂

∂r+ sinϕ ∂2

∂r∂ϕ

]

+cos θ cosϕr2 sin θ

[cosϕ ∂

∂θ+ sinϕ ∂2

∂θ∂ϕ

]

+cosϕr2 sin2 θ

[− sinϕ ∂

∂ϕ+ cosϕ ∂2

∂ϕ2

]The following two terms derive from ∂2

∂z2.

− sin θr

[− sin θ ∂

∂r+ cos θ ∂2

∂r∂θ

]

+sin θr2

[cos θ ∂

∂θ+ sin θ ∂

2

∂θ2

]

=∂2

∂r2+

cos θr

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ

]+

cos θr2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2

]

+(sin2 ϕ+ cos2 ϕ+ sin2 θ

) 1r

∂

∂r+

(cos θr2 sin θ

)(sin2 ϕ+ cos2 ϕ

) ∂

∂θ+

sin θ cos θr2

∂

∂θ

+1

r2 sin2 θ(sinϕ cosϕ− sinϕ cosϕ) ∂

∂ϕ+

sin2 θ

r2∂2

∂θ2− sin θ cos θ

r

∂2

∂r∂θ

+sinϕ cosϕ− sinϕ cosϕ

r

∂2

∂r∂ϕ+

cos θr2 sin θ (sinϕ cosϕ− sinϕ cosϕ) ∂2

∂θ∂ϕ

+sin2 ϕ+ cos2 ϕ

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

cos θr

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ

]+

cos θr2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2

]

339

+(1 + sin2 θ

) 1r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

sin θ cos θr2

∂

∂θ

+sin2 θ

r2∂2

∂θ2− sin θ cos θ

r

∂2

∂r∂θ

+1

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

cos θr

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ− sin θ ∂2

∂r∂θ

]

+cos θr2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2+ sin θ ∂

∂θ

]

+(1 + sin2 θ

) 1r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ

+sin2 θ

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

cos θr

[cos θ ∂

∂r

]+

cos θr2

[cos θ ∂

2

∂θ2

]

+(1 + sin2 θ

) 1r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

sin2 θ

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

cos θr2

[cos θ ∂

2

∂θ2

]+(1 + sin2 θ + cos2 θ

) 1r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ

+sin2 θ

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

2

r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

sin2 θ + cos2 θr2

∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

2

r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

1

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2(L.24)

Therefore, we have shown

∇2 =∂2

∂r2+

2

r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

1

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2. (L.25)

Alternatively, we can also write

∇2 =1

r2∂

∂rr2∂

∂r+

1

r2 sin θ∂

∂θsin θ ∂

∂θ+

1

r2 sin2 θ

∂2

∂ϕ2. (L.26)


Verification is straightforward as below.

1

r2∂

∂rr2∂

∂r+

1

r2 sin θ∂

∂θsin θ ∂

∂θ

=1

r2

(2r

∂

∂r+ r2

∂2

∂r2

)+

1

r2 sin θ

(cos θ ∂

∂θ+ sin θ ∂

2

∂θ2

)

=∂2

∂r2+

2

r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

1

r2∂2

∂θ2(L.27)

Appendix M

Legendre and Associated LegendrePolynomials

M.1 Legendre PolynomialsThe n-th degree Legendre polynomial Pn(x) is a solution to the following Legendre’sdifferential equation, where n = 0, 1, 2, . . ..

d

dx

[(1− x2) d

dxPn(x)

]+ n(n+ 1)Pn(x) = 0 (M.1)

Each Pn(x) can be obtained using Rodrigues’ formula

Pn(x) =1

2nn!

dn

dxn

[(x2 − 1)n

], (M.2)

which can be verified by differentiating both sides of the identity

(x2 − 1)d

dx(x2 − 1)n = 2nx(x2 − 1)n (M.3)

(n+ 1) times. When the condition Pn(1) = 1 is imposed, we have

P0(x) = 1

P1(x) = x

P2(x) =1

2(3x2 − 1)

P3(x) =1

2(5x3 − 3x)

341

342 APPENDIX M. LEGENDRE AND ASSOCIATED LEGENDRE POLYNOMIALS

P4(x) =1

8(35x4 − 30x2 + 3)

P5(x) =1

8(63x5 − 70x3 + 15x)

P6(x) =1

16(231x6 − 315x4 + 105x2 − 5)

M.2 Associated Legendre PolynomialsThe associated Legendre polynomial of degree l and order m, denoted by Pm

l (x),satisfies the following general Legendre equation.

(1− x2) d2

dx2Pml (x)− 2x

d

dxPml (x) +

[l(l + 1)− m2

1− x2

]Pml (x) = 0 (M.4)

Pml (x) can be expressed in the following Rodrigues type form.

Pml (x) =

(−1)l

2ll!(1− x2)m/2 d

l+m

dxl+m(1− x2)l (M.5)

In particular, for m > 0, we have

Pml (x) = (1− x2)m/2 d

m

dxmPl(x). (M.6)

Fact M.1 We have the following identity.

P−ml (x) = (−1)m (l −m)!

(l +m)!Pml (x) (M.7)

Proof (taken from a lecture note prepared by Professor Hitoshi Murayama of UCBerkeley)We will first compute how

dl+m

dxl+m(1− x2)l is related to dl−m

dxl−m(1− x2)l.

We need two lemmas for this proof.

Lemma M.1 (Combinatorial Identity) We have the following identity.(k + 1r

)=

(k

r − 1

)+

(kr

)(M.8)

M.2. ASSOCIATED LEGENDRE POLYNOMIALS 343

Proof of Lemma M.1It is nothing but a straightforward computation.(

kr − 1

)+

(kr

)=

k!

(k − r + 1)!(r − 1)!+

k!

(k − r)!r!

=k!r

(k + 1− r)!r(r − 1)!+

k!(k − r + 1)

(k − r + 1)(k − r)!r!=k!(r + k − r + 1)

(k + 1− r)!r!

=k!(k + 1)

(k + 1− r)!r!=

(k + 1)!

(k + 1− r)!r!=

(k + 1r

)(M.9)

Lemma M.2 Let F (x) and G(x) be two functions which are at least n-times differ-entiable. Then,

dn

dxnF (x)G(x) =

n∑r=0

(nr

)[dr

dxrF (x)

] [dn−r

dxn−rG(x)

]. (M.10)

Proof of Lemma M.2Let us introduce a short-hand notation

drF :=dr

dxrF (x); (M.11)

where d0

dx0F (x) = F (x) by definition. Our proof is by mathematical induction. First,

when n = 1, we have

d(FG) = (dF )G+ F (dG) =

(10

)(dF )G+

(11

)F (dG)

=1∑r=0

(1r

)[dr

dxrF (x)

] [d1−r

dx1−rG(x)

], (M.12)

and (M.10) holds for n = 1. Now, suppose (M.10) holds for n = k ≥ 1, so that

dk

dxkF (x)G(x) =

k∑r=0

(kr

)[dr

dxrF (x)

] [dk−r

dxk−rG(x)

]=

k∑r=0

(kr

)drFdk−rG.

(M.13)

Then, for n = k + 1, we get

dk+1

dxk+1F (x)G(x) =

d

dx

[dk

dxkF (x)G(x)

]


=d

dx

k∑r=0

(kr

)[dr

dxrF (x)

] [dk−r

dxk−rG(x)

]

=d

dx

[k∑r=0

(kr

)drFdk−rG

]=

k∑r=0

(kr

)d(drFdk−rG

)

=k∑r=0

(kr

)(dr+1Fdk−rG+ drFdk−r+1G

)

=k∑r=0

(kr

)dr+1Fdk−rG+

k∑r=0

(kr

)drFdk−r+1G. (M.14)

Now, let u = r + 1. Then,k∑r=0

(kr

)dr+1Fdk−rG =

k+1∑u=1

(k

u− 1

)duFdk+1−uG

=k+1∑r=1

(k

r − 1

)drFdk+1−rG. (M.15)

Plug this into (M.14) and use Lemma M.1.

dk+1

dxk+1F (x)G(x) =

k∑r=0

(kr

)dr+1Fdk−rG+

k∑r=0

(kr

)drFdk−r+1G

=k+1∑r=1

(k

r − 1

)drFdk+1−rG+

k∑r=0

(kr

)drFdk−r+1G

=

(kk

)dk+1Fd0G+

k∑r=1

(k

r − 1

)drFdk+1−rG+

(k0

)d0Fdk+1G

+k∑r=1

(kr

)drFdk+1−rG

=

(k + 10

)d0Fdk+1G+

k∑r=1

[(k

r − 1

)+

(kr

)]drFdk+1−rG

+

(k + 1k + 1

)dk+1Fd0G

=

(k + 10

)d0Fdk+1G+

k∑r=1

(k + 1r

)drFdk+1−rG+

(k + 1k + 1

)dk+1Fd0G

=k+1∑r=0

(k + 1r

)drFdk+1−rG (M.16)


So, M.10 holds for n = k + 1, and we have proved the lemma by mathematicalinduction.

Let us now resume the proof of Fact M.1. We have

dl+m

dxl+m(1− x2)l = dl+m

dxl+m(1− x)l(1 + x)l

=l+m∑r=0

(l +mr

)(dr

dxr(1− x)l

)(dl+m−r

dxl+m−r (1 + x)l). (M.17)

However, because (1− x)l and (1 + x)l are both l-th degree polynomials,dr

dxr(1− x)l = 0 if r > l (M.18)

anddl+m−r

dxl+m−r (1 + x)l = 0 if r < m. (M.19)

Therefore, the only nonzero terms are for m ≤ r ≤ l, and

dl+m

dxl+m(1− x2)l =

l∑r=m

(l +mr

)(dr

dxr(1− x)l

)(dl+m−r

dxl+m−r (1 + x)l). (M.20)

Now, observe the following.

d

dx(1− x)l = d(1− x)l

d(1− x)d(1− x)dx

= l(1− x)l−1(−1),

d2

dx2(1− x)l = d

dx

[d

dx(1− x)l

]= (−1)l d

dx(1− x)l−1

= (−1)l d(1− x)l−1

d(1− x)d(1− x)dx

= (−1)l(l − 1)(1− x)l−2(−1)

= (−1)2l(l − 1)(1− x)l−2, (M.21)and more generally

dk

dxk(1− x)l = (−1)kl(l − 1) . . . (l − k + 1)(1− x)l−k = (−1)k l!

(l − k)!(1− x)l−k

(M.22)

Similarly, we also have

dk

dxk(1 + x)l =

l!

(l − k)!(1 + x)l−k. (M.23)


Therefore, (M.20) can be expressed as follows.

dl+m

dxl+m(1− x2)l =

l∑r=m

(l +mr

)(−1)rl!(l − r)!

(1− x)l−r l!

(r −m)!(1 + x)r−m (M.24)

At this point, let s = r −m to obtain

dl+m

dxl+m(1− x2)l =

l−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!(1− x)l−m−s l!

s!(1 + x)s. (M.25)

Next, we will multiply the right-hand side by (1−x2)m

(1−x2)m .

dl+m

dxl+m(1− x2)l =

l−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!(1− x)l−m−s l!

s!(1 + x)s

=(1− x2)m

(1− x2)ml−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!(1− x)l−m−s l!

s!(1 + x)s

=(1− x)m(1 + x)m

(1− x2)ml−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!

(1− x)l−m−s l!

s!(1 + x)s

=1

(1− x2)ml−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!

(1− x)l−m−s(1− x)m l!s!(1 + x)s(1 + x)m

=1

(1− x2)ml−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!

(1− x)l−s l!s!(1 + x)m+s

=1

(1− x2)ml−m∑s=0

(l +m)!

(l − s)!(m+ s)!

(−1)m+sl!

(l −m− s)!

(1− x)l−s l!s!(1 + x)m+s

=1

(1− x2)ml−m∑s=0

(l +m)!

s!(l −m− s)!(−1)m+sl!

(l − s)!

(1− x)l−s l!

(m+ s)!(1 + x)m+s


=(l +m)!

(l −m)!

1

(1− x2)ml−m∑s=0

(l −m)!

s!(l −m− s)!(−1)m+s

[(−1)s(−1)s l!

(l − s)!(1− x)l−s

][

l!

(l − (l −m− s))!(1 + x)l−(l−m−s)

]

=(l +m)!

(l −m)!

1

(1− x2)ml−m∑s=0

(l −m)!

s!(l −m− s)!(−1)m+s

[(−1)s(−1)s l!

(l − s)!(1− x)l−s

] [l!

(m+ s)!(1 + x)m+s

]

=(l +m)!

(l −m)!

(−1)m

(1− x2)ml−m∑s=0

(l −ms

)[(−1)s l!

(l − s)!(1− x)l−s

] [l!

(m+ s)!(1 + x)m+s

]

=(l +m)!

(l −m)!

(−1)m

(1− x2)ml−m∑s=0

(l −ms

)[ds

dxs(1− x)l

] [dl−m−s

dxl−m−s (1 + x)l]

=(l +m)!

(l −m)!

(−1)m

(1− x2)mdl−m

dxl−m(1− x2)l; (M.26)

where the last equality follows from (M.17) with l +m replaced by l −m.Then, from (M.5) and (M.26), we get

Pml (x) =

(−1)l

2ll!(1− x2)m/2 d

l+m

dxl+m(1− x2)l

=(−1)l

2ll!(1− x2)m/2 (l +m)!

(l −m)!

(−1)m

(1− x2)mdl−m

dxl−m(1− x2)l

=(l +m)!

(l −m)!(−1)m (−1)l

2ll!(1− x2)−m/2 d

l−m

dxl−m(1− x2)l

=(l +m)!

(l −m)!(−1)mP−m

l (x)

=⇒ P−ml (x) = (−1)m (l −m)!

(l +m)!Pml (x) (M.27)

Appendix N

Laguerre and Associated LaguerrePolynomials

For a nonnegative integer m, i.e. m = 0, 1, 2, 3, . . ., Laguerre’s equation is the linearsecond order ordinary differential equation

xy′′(x) + (1− x)y′(x) +my(x) = 0, (N.1)

whose polynomial solutions, denoted by Lm(x) are known as Laguerre polynomials.The general term is given by the Rodrigues formula

Ln(x) =ex

n!

dn

dxn

(e−xxn

), (N.2)

or the closed form1

Lk(x) =k∑i=0

(k

i

)(−1)i

i!xi or

k∑i=0

kCi(−1)i

i!xi. (N.3)

The first few polynomials are given below.

L0(x) = 1

L1(x) = −x+ 1

L2(x) =1

2(x2 − 4x+ 2)

L3(x) =1

6(−x3 + 9x2 − 18x+ 6)

1Note that(

ki

)and kCi mean the same thing. Depending on where you went to school, you may

be familiar with either.

349

350 APPENDIX N. LAGUERRE AND ASSOCIATED LAGUERRE POLYNOMIALS

L4(x) =1

24(x4 − 16x3 + 72x2 − 96x+ 24)

L5(x) =1

120(−x5 + 25x4 − 200x3 + 600x2 − 600x+ 120)

L6(x) =1

720(x6 − 36x5 + 450x4 − 2400x3 + 5400x2 − 4320x+ 720)

The Laguerre polynomials satisfy a three-term recurrence relation

(k + 1)Lk+1(x) = (2k + 1− x)Lk(x)− kLk−1(x). (N.4)

According to Favard’s theorem, the Laguerre polynomials form a family of orthogonalpolynomials.

Theorem N.1 (Favard’s Theorem) If a system of polynomials Pn(x) of degreen = 0, 1, 2, . . . satisfies

xPn(x) = anPn+1(x) + bnPn(x) + cnPn−1(x) n ≥ 1, (N.5)xP0(x) = a0P1(x) + b0P0(x) (N.6)

where an, bn (n ≥ 0), and cn (n ≥ 1) are real constants, and ancn+1 > 0, then,there exists a positive measure µ such that the polynomials Pn(x) are orthogonal withrespect to µ.2

Indeed, Lk’s are orthogonal with respect to an inner product defined by

⟨f |g⟩ =∫ ∞

0f(x)g(x)e−x dx. (N.7)

2The following holds for µ.

1. The measure µ may not be unique even if a scaling factor is ignored.

2. If µ is unique, the polynomials are dense in L2 with respect to µ.

3. If there is a µ with bounded support, then, µ is unique.

Appendix O

Fine Structure and Lamb Shift

O.1 Relativistic Effects: The Dirac TheoryThe Hamiltonian H0 for the hydrogen atom used in Chapter 10 was:

H0 =P2

2µ+ V (r) =

P2

2µ− q2

4πε0

1

r=

P2

2µ− e2

r; (O.1)

where q is the electron charge and e2 := q2

4πε0.1 The first term is the kinetic energy in

the center of mass frame, and the second term is the electrostatic interaction energybetween the electron and the proton.

Now, the problem is that this Hamiltonian is not exact as it does not includerelativistic effects. For example, the magnetic effects involving the electron spin aswell as magnetic interactions related to the proton spin are ignored. In fact, theserelativistic effects are typically about (1/137)2 the size of H0. Still, spectroscopicmeasurements are accurate enough to detect these effects. In 1928, Charles GaltonDarwin2 and Walter Gordon found the following expressions for the energy of an

1Nothing is carved in stone. However, it appears customary to express the Hamiltonian in termsof e as defined here. Note that what we called e in Chapter 10 is denoted by q here. For those“old-timers” who are familiar with the CGS (centimeter-gram-second) System of Units, this is adevice to connect CGS with SI (the International System of Units) if you may.

2Sir Charles Galton Darwin (1987-1962), a British physicist, was a grandson of the CharlesDarwin of evolution fame. Sir Charles was the first, with Gordon, to work out the exact energylevels of Hydrogen according to the Dirac equation and thereby discovered the eponymous term inthe levels. He worked out the Lagrangian and Hamiltonian for classical motion of several interactingcharges correct to O(v2/c2). He also worked on statistical mechanics (Darwin-Fowler method).Later in life he took part in the Manhattan project.

351

352 APPENDIX O. FINE STRUCTURE AND LAMB SHIFT

electron in a hydrogen atom.

W = mc21 +

γ2

(k′ + n′)2

− 12

; (O.2)

where γ = 2π2

c~is the fine structure constant, k(= 0, 1, 2, . . .) is the azimuthal quantum

number, k′ =√(k + 1)2 − γ2, and n′ is a nonnegative integer [Darwin, 1928, p.672].

E =mc2√

1 + α2

(√K2−α2+n)

2

; (O.3)

where α is the fine structure constant, K = ±1,±2,±3, . . ., and n = 0, 1, 2, . . .[Gordon, 1928, p.14][Chen, nd, p.6].

E

mc2=

1√αZ

n−k+√k2−α2Z2

2 (O.4)

[Ketterle, nd, p.2]

E = mc2[1 +

γ2

(s+ n′)2

]− 12

; (O.5)

where γ = Ze2

~c, k = ±1,±2, . . ., s = (k2 − γ2)

12 , and n′ = 0, 1, 2, . . . [Schiff, 1968, p.

486].

E = µc2[1 +

Z2α2

n+√λ2 − Z2α2

]− 12

; (O.6)

where µ is the reduced mass, α = e2

c~≈ 1/137, Z is the atomic number of the nucleus,

and λ is an integer [Chongming and Dianyan, 1984, p.171].

Enjlm = mc2

1 + Zα

n− (j + 1/2) +√(j + 1/2)2 − (Zα)2

2

− 12

; (O.7)

where n = 1, 2, . . . are the principal quantum numbers, and j+1/2 ≤ n [Murayama, nd,p.17].

O.1. RELATIVISTIC EFFECTS: THE DIRAC THEORY 353

Because the hydrogen atom is a weakly relativistic system3, it is useful to ap-proximate the Hamiltonian to the lowest order in v/c. We can write

H = mc2 +p2

2m− e2

r+HFS = mc2 +H0 +HFS; (O.8)

where mc2 is the electron’s rest energy, p2

2m− e2

r= H0 is the non-relativistic Hamil-

tonian, and HFS is the fine structure interaction given by

HFS = − p4

8m3c2+

e2

2m2c21

r3L · S − ~2

8m2c2∇2 e

2

r.4 (O.9)

The three terms of HFS are known as the kinetic, spin-orbit interaction, and Dar-winian terms, denoted by Hk, HSO, and HD, respectively. Contribution made byeach of the three terms is of the order

(vc

)2.

HFS = − p4

8m3c2︸︷︷︸Hk

+e2

2m2c21

r3L · S︸︷︷︸

HSO

− ~2

8m2c2∇2 e

2

r︸︷︷︸HD

(O.10)

O.1.1 Variation of the Mass with the Velocity: Hk

The relativistic expression for the energy E of a classical particle of rest-mass m andmomentum p is given by

E = c√p2 +m2c2, (O.11)

and the first three terms of its expansion in powers of p/mc gives

E = mc2 +p2

2m− p4

8m3c2︸︷︷︸Hk

+ . . . (O.12)

= mc2︸︷︷︸rest-mass energy

+p2

2m︸︷︷︸non-relativistic kinetic energy

− p4

8m3c2︸︷︷︸first correction

+ . . . . (O.13)

This is the origin of Hk.3In the Bohr model, the velocity v of an electron in the orbit characterized by n = 1 is given by

v = (e2/~)c ≈ (1/137)c≪ 1.4Some schools define S and L such that ∥S∥~ and ∥L∥~ are the magnitudes of the corresponding

spin and orbital angular momenta. When this convention is adopted, the second term becomes~2e2

2m2c21r3 L · S.


O.1.2 Spin-Orbit Interaction: HSO

The intrinsic spin dipole magnetic moment of an electron µS is given by

µS = −gSµBS

~; (O.14)

where the spin g-factor gS for an electron is approximately 2 (gS ≈ 2), µB is aphysical constant known as Bohr magneton5, and S is the electron’s spin angularmomentum whose magnitude S is given by S2 = s(s+1)~2 = 3

4~2. As the electron is

a spin one-half particle, S = ~2, and the magnitude µS of µS is approximately equal

to one Bohr magneton.

µS ≈ 2e~

2m

~

2

1

~= µB (O.15)

The spin-orbit interaction energy stems from the interaction between this magneticdipole and the motional magnetic field B experienced by the electron as it movesthrough the electric field of the proton. Naively, the motional magnetic field Bnaiveis given by

Bnaive = −v

c×E = −v

c× e

r3r =

e

mc

1

r3L. (O.16)

However, the relativistic transformation of a vector between two coordinate framesmoving with different velocities, which are not collinear, involves a rotation in addi-tion to a dilation. That rotation is known as the Thomas precession, which effectivelycuts the motional magnetic field in half. Therefore, the true motional magnetic fieldBmotion experienced by the electron is given by

Bmotion =1

2

e

mc

1

r3L.6 (O.17)

5The Bohr magneton is given by

µB =e~

2m

in SI units and by

µB =e~

2mc

in Gaussian CGS units.

O.1. RELATIVISTIC EFFECTS: THE DIRAC THEORY 355

Therefore, the total spin-orbit interaction HSO is given by

HSO = −µS ·Bmotion = −(−gSµB

S

~

)·(1

2

e

mc

1

r3L)= 2 · e~

2mc

1

~

1

2

e

mc

1

r3L · S

=e2

2m2c21

r3L · S. (O.18)

O.1.3 The Darwin Term: HD

A particle has what is known as the Compton wavelength λC given by

λC =h

mc; (O.19)

where the capital C in λC stands for “Compton”, and the small letter c in thedenominator of the right-hand side is the speed of light. Often, the reduced Comptonwavelength λrC defined by

λrC =λC2π

=~

mc(O.20)

is used instead. For the electron, the reduced Compton wavelength is roughly 386 fm(femtometers) or 3.86× 10−13 meters7. The Compton wavelength can be thought ofas a fundamental limitation on measuring the position of a particle, taking quantummechanics and special relativity into account.

What this means is that we should use an effective potential Veffective, whichis V (r) averaged over a small ball whose radius is the Compton wavelength. Inother words, rather than V (r), we should average V over all the points r + ε with∥ε∥ ≤ λC or λrC . Typically, we resort to the Taylor expansion of V (r + ε) asfollows and conduct term-by-term integration to get the average. This fluctuationor uncertainty is sometimes called Zitterbewegung (trembling motion).

V (r + ε) = V (r) + ε · ∇V (r) +3∑

i,j=1

1

2εiεj

∂

∂xi

∂

∂xjV (r) + . . . (O.21)

When it is all said and done, we get

HD =~2

8m2c24π

(Ze2

4πε0

)δ3(r) =

Ze2~2

8m2c2ε0δ3(r); (O.22)

6More detail can be found in Ketterle’s lecture notes for example [Ketterle, nd].71 nanometer = 106 femtometers


where the identity

∇21

r= −4πδ3(r) (O.23)

was used.

Now, even the exact solution of the Dirac equation still has some shortcomings.One notable problem is that it predicts degeneracy for such pairs of states as (2S 1

2,

2P 12), (3S 1

2, 3P 1

2), (3P 3

2, 3D 3

2), and (3D 5

2, 3F 5

2)8, when indeed these states are not

degenerate. This problem arises because states of the same values of n and j aredegenerate according to the Dirac theory as shown in (O.7). We need to consideran additional quantum mechanical effect sometimes called the vacuum interaction toremove this degeneracy. The effect is the largest for n = 2, and the energy splittingbetween the 2S 1

2state and the 2P 1

2state is called the Lamb Shift.

O.2 The Lamb ShiftSome physicists say ”Vacuum is noisy.” Others say ”Vacuum is not really empty.”What they typically mean is that there is quantum fluctuation which allows nonzeroelectromagnetic fields in vacuum. We have zero-point fluctuation of the electromag-netic modes of free space because they behave like harmonic oscillators with a strictlypositive zero-point energy hν/2, and the time averages of the squared amplitudes ofelectric and magnetic fields are strictly positive. This electric field causes the averagepotential experienced by the electron to deviate from V (r), and this is the reasonbehind the Lamb Shift. The observed value is 1,058 MHz.

Interested readers can find a very readable exposition in a lecture note by Wolf-gang Ketterle of MIT [Ketterle, nd].

8Spectroscopic NotationThe following explanation was adopted with slight modifications from Jim Branson’s notes[Branson, nd]. Spectroscopic notation is widely used in atomic physics, which provides a stan-dard way to describe the angular momentum quantum numbers of a state. The general form isN2s+1Lj ; where N is the principal quantum number, s signifies the total spin quantum numberso that there are 2s + 1 spin states, L refers to the orbital angular momentum quantum numberl customarily written as S, P,D, F, . . . for l = 0, 1, 2, 3, . . ., and j is the total angular momentumquantum number. Note that N is often dropped. For the hydrogen atom, we have 1 2S 1

2, 2 2S 1

2,

2 2P 32, 2 2P 1

2, 3 2S 1

2, 3 2P 3

2, 3 2P 1

2, 3 2D 5

2, 3 2D 3

2, 4 2S 1

2, 4 2P 3

2, 4 2P 1

2, 4 2D 5

2, 4 2D 3

2, 4 2F 7

2, 4 2F 5

2,

and so on.

Bibliography

[Anton, 2010] Anton, H. (2010). Elementary Linear Algerbra. Wiley, Hoboken, NJ,10th edition.

[Bocher, 1901] Bocher, M. (1901). Certain cases in which the vanishing of the wron-skian is a sufficient condition for linear dependence. Transactions of the AmericanMathematical Society, 2(2):139–149.

[Bostan and Dumas, 2010] Bostan, A. and Dumas, P. (2010). Wronskians and linearindependence. American Mathematical Monthly, 117(8):722–727.

[Branson, nd] Branson, J. (n.d.). Spectroscopic notation.

[Carlip, 2013] Carlip, S. (2013). Probability current.

[Chen, nd] Chen, R. (n.d.). Equivalent solution for radial dirac coulomb equation.

[Chongming and Dianyan, 1984] Chongming, X. and Dianyan, X. (1984). Diracequation and energy levels of hydrogenlike atoms in robertson-walker metric. IlNuovo Cimento B Series 11, 83(2):162–172.

[Darwin, 1928] Darwin, C. G. (1928). The wave equations of the electron. Proceed-ings of the Royal Society of London. Series A, 118(780):654–680.

[Gerlach and Stern, 1922] Gerlach, V. W. and Stern, O. (1922). Der experimentellenachweis der richtungsquantelung im magnetfeld. Zeitschrift fur Physik, 9:349–352.

[Gordon, 1928] Gordon, W. (1928). Die energieniveaus des wasserstoffatoms nachder diracschen quantentheorie des elektrons. Zeitschrift für Physik, 48(1-2):11–14.

[Gordy and Cook, 1984] Gordy, W. and Cook, R. L. (1984). Microwave molecularspectra. Wiley, New York, 3rd edition.

357

358 BIBLIOGRAPHY

[Ketterle, nd] Ketterle, W. (n.d.). Fine structure and lamb shift.

[Kwa, nd] Kwa, m. (n.d.). Linear independence and the wronskian.

[Wikipedia contributors, nd] Wikipedia contributors (n.d.). Degenerate energy lev-els.

[Murayama, nd] Murayama, H. (n.d.). 221b lecture notes: Relativistic quantummechanics.

[Reed and Simon, 1980] Reed, M. and Simon, B. (1980). Functional analysis: Re-vised and enlarged edition. Methods of modern mathematical physics. AcaemicPress, New York.

[Schiff, 1968] Schiff, L. I. (1968). Quantum mechanics. McGraw-Hill, New York.

[Shankar, 1980] Shankar, R. (1980). Principles of quantum mechanics. Plenum, NewYork.

[Zettili, 2009] Zettili, N. (2009). Quantum mechanics: Concepts and applications.John Wiley & Sons, West Sussex, UK, second edition.

Answers to Exercises

This section is a work in progresswith some dummy place-holders atthis point.Chapter 2

1. (a) Ω is Hermitian because (Ωt)∗= Ω as shown below.[ 0 1

1 0

]t∗

=

([0 11 0

])∗

=

[0 11 0

]

(b) [0 11 0

] [ab

]= 1 ·

[ab

]=⇒

[ba

]=

[ab

]=⇒ a = b[

0 11 0

] [ab

]= −1 ·

[ab

]=⇒

[ba

]=

[−a−b

]=⇒ a = −b

Hence, |1⟩ = 1√2

[11

], and |−1⟩ = 1√

2

[1−1

].

(c)

⟨1| − 1⟩ = 1√2

[1 1

] 1√2

[1−1

]= 0

(d) Noting that

|1⟩+ |−1⟩√2

=

[10

]

359

360 ANSWERS TO EXERCISES

and|1⟩ − |−1⟩√

2=

[01

],

we get

v =

[v1v2

]= v1

[10

]+ v2

[01

]= v1

(|1⟩+ |−1⟩√

2

)+ v2

(|1⟩ − |−1⟩√

2

)

=v1 + v2√

2|1⟩+ v1 − v2√

2|−1⟩

(e)

⟨1|Ω|1⟩ = 1√2

[1 1

] [ 0 11 0

]1√2

[11

]=

1

2

[1 1

] [ 11

]= 1

⟨1|Ω| − 1⟩ = 1√2

[1 1

] [ 0 11 0

]1√2

[1−1

]=

1

2

[1 1

] [ −11

]= 0

⟨−1|Ω|1⟩ = 1√2

[1 1

] [ 0 11 0

]1√2

[1−1

]=

1

2

[1 1

] [ −11

]= 0

⟨−1|Ω| − 1⟩ = 1√2

[1 −1

] [ 0 11 0

]1√2

[1−1

]=

1

2

[1 −1

] [ −11

]= −1

Hence,

[Ωij] =

[1 00 −1

].

2. (a) As M is Hermitian, we have to have M t∗ =M . So,[ 6 4α 0

]t∗

=

[6 α4 0

]∗

=

[6 α∗

4 0

]=

[6 4α 0

]=⇒ α = 4.

(b) [6 44 0

] [ab

]=

[6a+ 4b4a

]= 8

[ab

]=⇒ a = 2b =⇒ |8⟩ = 1√

5

[21

][6 44 0

] [ab

]=

[6a+ 4b4a

]= −2

[ab

]=⇒ b = −2a =⇒ |−2⟩ = 1√

5

[1−2

]

CHAPTER 2 361

(c)

M11 =1√5

[2 1

] [ 6 44 0

]1√5

[21

]=

1

5

[2 1

] [ 168

]=

1

5· 40 = 8

M12 =1√5

[2 1

] [ 6 44 0

]1√5

[1−2

]=

1

5

[2 1

] [ −24

]= 0

M21 =1√5

[1 −2

] [ 6 44 0

]1√5

[21

]=

1

5

[1 −2

] [ −24

]= 0

M22 =1√5

[1 −2

] [ 6 44 0

]1√5

[1−2

]=

1

5

[1 −2

] [ −24

]= −2

Therefore,

M = [Mij] =

[8 00 −2

].

3. (a)

[A,B] = AB −BA =

[0 21 1

] [−1 21 0

]−[−1 21 0

] [0 21 1

]

=

[2 00 2

]−[2 00 2

]= 0

(b) [0 21 1

] [ab

]= 2

[ab

]=⇒

[2ba+ b

]=

[2a2b

]=⇒ a = b

So,

|2⟩A =1√2

[11

].

[0 21 1

] [cd

]= −1

[cd

]=⇒

[2dc+ d

]=

[−c−d

]=⇒ c = −2d

So,

|−1⟩A =1√5

[2−1

]. (13.24)


(c)

B |2⟩A =

[−1 21 0

]1√2

[11

]= 1 · 1√

2

[11

]

B |−1⟩A =

[−1 21 0

]1√5

[2−1

]=

1√5

[−42

]= −2 · 1√

5

[2−1

]

Therefore, |2⟩A is an eigenvector of B with the associated eigenvalue 1,and |−1⟩ is an eigenvector of B with the associated eigenvalue −2.

(d)√5 |−1⟩A +

√2 |2⟩A

3=

1

3

([2−1

]+

[11

])=

1

3

[30

]=

[10

]√5 |−1⟩A − 2

√2 |2⟩A

−3=

1

−3

([2−1

]−[22

])=

1

−3

[0−3

]=

[01

]

Therefore, |2⟩A , |−1⟩A indeed forms a basis for C2.(e) Remembering that Ωij = ⟨i|Ω|j⟩, we can compute the entries as follows.

A11 =A ⟨2|A|2⟩A =[

1√2

1√2

] [ 0 21 1

] [ 1√2

1√2

]

=[

1√2

1√2

] [ √2√2

]= 2; where A ⟨2| is the adjoint of |2⟩A

A12 =[

1√2

1√2

] [ 0 21 1

] [ 2√5

−1√5

]=[

1√2

1√2

] [ −2√5

1√5

]=−1√10

A21 =[

2√5

−1√5

] [ 0 21 1

] [ 1√2

1√2

]=[

2√5

−1√5

] [ √2√2

]=

√2

5

A22 =[

2√5

−1√5

] [ 0 21 1

] [ 2√5

−1√5

]=[

2√5

−1√5

] [ −2√5

1√5

]= −1

B11 =[

1√2

1√2

] [ −1 21 0

] [ 1√2

1√2

]=[

1√2

1√2

] [ 1√2

1√2

]= 1

B12 =[

1√2

1√2

] [ −1 21 0

] [ 2√5

−1√5

]=[

1√2

1√2

] [ −4√5

2√5

]=−2√10

B21 =[

2√5

−1√5

] [ −1 21 0

] [ 1√2

1√2

]=[

2√5

−1√5

] [ 1√2

1√2

]=

1√10

CHAPTER 2 363

B22 =[

2√5

−1√5

] [ −1 21 0

] [ 2√5

−1√5

]=[

2√5

−1√5

] [ −4√5

2√5

]= −2

4. (a) As M = I −N , [M,N ] = [I −N,N ] = 0 is obvious.(b)

det

(λI −

[0 11 0

])= det

[λ −1−1 λ

]= λ2 − 1 = (λ+ 1)(λ− 1) = 0

=⇒ λ = ±1

So, the eigenvalues of N are ±1.[0 11 0

] [ab

]=

[ba

]= ±1 ·

[ab

]=⇒ a = ±b

=⇒ |1⟩N =1√2

[11

]and |−1⟩N =

1√2

[1−1

]

As we already know M and N commute, they share the same eigenvectors.[1 −1−1 1

] [11

]=

[00

]= 0 ·

[11

][

1 −1−1 1

] [1−1

]=

[2−2

]= 2 ·

[1−1

]

So, the eigenvalues of M are 0 and 2, and corresponding normalized eigen-vectors are

|0⟩M =1√2

[11

]and |2⟩M =

1√2

[1−1

].

(c) ( 1√2

[1 1−1 1

])t∗ [1 −1−1 1

](1√2

[1 1−1 1

])

=1

2

[1 −11 1

] [1 −1−1 1

] [1 1−1 1

]

=1

2

[1 −11 1

] [2 0−2 0

]=

1

2

[4 00 0

]=

[2 00 0

]


and( 1√2

[1 1−1 1

])t∗ [0 11 0

](1√2

[1 1−1 1

])

=1

2

[1 −11 1

] [0 11 0

] [1 1−1 1

]

=1

2

[1 −11 1

] [−1 11 1

]=

1

2

[−2 00 2

]=

[−1 00 1

]

Chapter 31. [3 pts]

(a) 0.4772 [1 pt](b) 0.5− 0.4495 = 0.0505 [1 pt](c) P (|z| > 1.96) = 2× (0.5− 0.4750) = 0.05 [1 pt]

Chapter 51.

Ψ(x+ vt0, t+ t0) = A sin[2π

λ(x+ vt0 − v(t+ t0))

]= A sin

[2π

λ(x+ vt0 − vt− vt0)

]= A sin

[2π

λ(x− vt)

]= Ψ(x, t)

Ψ(x, t) is the displacement at location x at time t. As the speed is v, this movesto the position “x+ vt0” time “t0” later or at time “t+ t0”.

2.− ~

2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂t

3. We have

− ~2

2m

∂2Ψ1(x, t)

∂x2+ V (x, t)Ψ1(x, t) = i~

∂Ψ1(x, t)

∂tand

CHAPTER 5 365

− ~2

2m

∂2Ψ2(x, t)

∂x2+ V (x, t)Ψ2(x, t) = i~

∂Ψ2(x, t)

∂t.

Noting that ∂2

∂x2, multiplication by V (x, t), and ∂

∂tare all linear such that

∂2

∂x2[αΨ1(x, t) + βΨ2(x, t)] = α

∂2

∂x2Ψ1(x, t) + β

∂2

∂x2Ψ2(x, t)

V (x, t) [αΨ1(x, t) + βΨ2(x, t)] = αV (x, t)Ψ1(x, t) + βV (x, t)Ψ2(x, t)

∂

∂t[αΨ1(x, t) + βΨ2(x, t)] = α

∂

∂tΨ1(x, t) + β

∂

∂tΨ2(x, t),

we get

− ~2

2m

∂2 [αΨ1(x, t) + βΨ2(x, t)]

∂x2+ V (x, t) [αΨ1(x, t) + βΨ2(x, t)]

= − ~2

2m

[α∂2

∂x2Ψ1(x, t) + β

∂2

∂x2Ψ2(x, t)

]+ αV (x, t)Ψ1(x, t) + βV (x, t)Ψ2(x, t)

= α

[− ~

2

2m

∂2

∂x2Ψ1(x, t) + V (x, t)Ψ1(x, t)

]

+ β

[− ~

2

2m

∂2

∂x2Ψ2(x, t) + V (x, t)Ψ2(x, t)

]

= α

[i~∂Ψ1(x, t)

∂t

]+ β

[i~∂Ψ2(x, t)

∂t

]

= i~∂ [αΨ1(x, t) + βΨ2(x, t)]

∂t

=⇒ − ~2

2m

∂2 [αΨ1(x, t) + βΨ2(x, t)]

∂x2+ V (x, t) [αΨ1(x, t) + βΨ2(x, t)]

= i~∂ [αΨ1(x, t) + βΨ2(x, t)]

∂t.

Therefore, αΨ1(x, t) + βΨ2(x, t) is also a solution.

4. (a) eiθeiϕ = (cos θ+i sin θ)(cosϕ+i sinϕ) = (cos θ cosϕ−sin θ sinϕ)+i(sin θ cosϕ+cos θ sinϕ) = cos(θ + ϕ) + i sin(θ + ϕ) = ei(θ+ϕ)

(b) Our proof is by mathematical induction.Consider (cos θ+i sin θ)k. For k = 1, we have (cos θ+i sin θ)1 = ei·1·θ = eiθ.Hence, (cos θ + i sin θ)k = eikθ holds fro k = 1.


Now suppose (cos θ + i sin θ)k = eikθ = cos(kθ) + i sin(kθ) for some k.Then, (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) = [cos(kθ) +i sin(kθ)](cos θ + i sin θ) = cos(kθ) cos θ − sin(kθ) sin θ + i[sin(kθ) cos θ +sin θ cos(kθ)] = cos[(k + 1)θ] + i sin[(k + 1)θ] = ei(k+1)θ.Therefore,

(eiθ)n

= einθ by induction.

5. (a) ∫ +a2

− a2

Ψ∗Ψ dx =∫ +a

2

− a2

A∗ cos(πx

a

)e+iEt/~A cos

(πx

a

)e−iEt/~ dx

=∫ +a

2

− a2

A∗A cos2(πx

a

)dx = |A|2

∫ +a2

− a2

1 + cos(2πxa

)2

dx

= |A|2[x

2+

a

2 · 2πsin

(2πx

a

)]+a2

− a2

= |A|2[a

4+

a

4πsinπ − −a

4− a

4πsin(−π)

]= |A|2a

2= 1

=⇒ A = eiθ√2

a

(b)

− ~2

2m

∂2

∂x2

[A cos

(πx

a

)e−iEt/~

]= i~

∂[A cos

(πxa

)e−iEt/~

]∂t

=⇒ − ~2

2mA

[−(π

a

)2]

cos(πx

a

)e−iEt/~

= i~A cos(πx

a

)(−iE~

)e−iEt/~

=⇒ − ~2

2m

[−(π

a

)2]= i~

(−iE~

)= E =⇒ E =

π2~2

2ma2(13.25)

(c)

⟨P ⟩ =∫ +a

2

− a2

A∗ cos(πx

a

)e+iEt/~

(−i~ ∂

∂x

)A cos

(πx

a

)e−iEt/~ dx

=2

a(−i~)

∫ +a2

− a2

cos(πx

a

) −πa

sin(πx

a

)dx

=iπ~

a2

∫ +a2

− a2

2 sin(πx

a

)cos

(πx

a

)dx

CHAPTER 5 367

=iπ~

a2

∫ +a2

− a2

sin(2πx

a

)dx =

iπ~

a2

[−a2π

cos(2πx

a

)]+a2

− a2

=iπ~

a2−a2π

[cos π − cos (−π)] = iπ~

a2−a2π

[−1− (−1)] = 0

(d)

⟨x2⟩ =∫ +a

2

− a2

A∗ cos(πx

a

)e+iEt/~x2A cos

(πx

a

)e−iEt/~ dx

=2

a

∫ +a2

− a2

x2 cos2(πx

a

)dx

Let y =π

ax, so that dx =

a

πdy.

=2

a

∫ +π2

− π2

(a

π

)2

y2 cos2 y aπdy =

2

a

(a

π

)3 ∫ +π2

− π2

y2 cos2 y dy

=2

a

(a

π

)3[y3

6+

(y2

4− 1

8

)sin 2y +

y cos 2y4

]+π2

− π2

=2

a

(a

π

)3[y3

6+y cos 2y

4

]+π2

− π2

=2

a

(a

π

)3

2(π2

)36

+ 2π2

cos π4

=2a2

π3

[π3

24− π

4

]

=a2

12

(1− 6

π2

);

where we used the formula∫y2 cos2 y dy =

y3

6+

(y2

4− 1

8

)sin 2y +

y cos 2y4

+ C.

(e)

⟨P 2⟩ =∫ +a

2

− a2

A∗ cos(πx

a

)e+iEt/~

(−i~ ∂

∂x

)2

A cos(πx

a

)e−iEt/~ dx

= −~2[−(π

a

)2] ∫ +a

2

− a2

A∗ cos(πx

a

)e+iEt/~A cos

(πx

a

)e−iEt/~ dx.


But, the normalization condition gives∫ +a2

− a2

A∗ cos(πx

a

)e+iEt/~A cos

(πx

a

)e−iEt/~ dx = 1.

Therefore, we get

⟨P 2⟩ =(π~

a

)2

.

6. (a) As eiθA for any real number θ can replace A, the wavefunction is notunique in that sense.

(b) The normalization constant eiθA enters into the computations of Ψ∗(x, t)Ψ(x, t),⟨P ⟩, ⟨P 2⟩, and ⟨x2⟩ as

(eiθA

)∗ (eiθA

). However,

(eiθA

)∗ (eiθA

)= e−iθA∗eiθA =

A∗A, and the value of θ does not affect these physical observables.

Chapter 61.

− ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x)

2. Plugging V (x, t) = V (x) and Ψ(x, t) = ψ(x)e−iEt/~ into the time-dependentSchrodinger Equation, we get

− ~2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂t

=⇒ − ~2

2m

∂2ψ(x)e−iEt/~

∂x2+V (x)ψ(x)e−iEt/~ = i~

∂ψ(x)e−iEt/~

∂t

=⇒ − ~2

2me−iEt/~∂

2ψ(x)

∂x2+V (x)ψ(x)e−iEt/~ = i~(−iE/~)e−iEt/~ψ(x)

=⇒ − ~2

2m

∂2ψ(x)

∂x2+V (x)ψ(x) = Eψ(x).

3. The first derivative of sinϕ is cosϕ. Both sinϕ and cosϕ are continuous, andsin2 ϕ is obviuosly integrable over the interval [0, 2π]. Furthermore, sin(ϕ +2mπ) = sinϕ and cos(ϕ+2mπ) = cosϕ for all integers m. Hence, the conditionof single-valuedness is also satisfied.On the other hand, if ψ(ϕ) = ϕ, we have ψ(0) = 0 and ψ(2π) = 2π , 0 = ψ(0)despite the fact that ϕ = 0 and ϕ = 2π signify the same position in space.Hence, this function is not single-valued and cannot be a wavefunction as such.

CHAPTER 7 369

Chapter 71. Same as Chapter 5 Problem 5. So, see the answers to Chapter 5

Problem 5.

2. Same as Chapter 5 Problem 6. So, see the answers to Chapter 5Problem 6.

3.

− ~2

2m

d2

dx2ψ(x) = Eψ(x) =⇒ d2

dx2ψ(x) = −2mE

~2ψ(x) =⇒ ψ(x) = Ae±i

√2mE~

x

4. Let k =√2mE~

. Then, the general solution is

ψ(x) = C sin kx+D cos kx.

Taking the boundary conditions into account,

ψ(±a2

)= 0 =⇒

C sin ka2+D cos ka

2= 0

−C sin ka2+D cos ka

2= 0

=⇒D cos ka

2= 0

C sin ka2= 0

.

Because ka2> 0, we have

D cos ka2

= 0⇐⇒ D = 0 or ka

2=(n+

1

2

)π for n = 0, 1, 2, . . .

and

C sin ka2

= 0⇐⇒ C = 0 or ka

2= nπ for n = 1, 2, 3, . . . .

Note that C = D = 0 leads to a zero wavefunction, and ka2

=(n+ 1

2

)π and

ka2= nπ cannot hold simultanepously

Hence, we have either

ka

2=(n+

1

2

)π (n = 0, 1, 2, . . .) and C = 0

orka

2= nπ (n = 1, 2, 3, . . .) and D = 0.


When, C = n = 0, k = πa

and

ψ(x) = D cos πxa.

When D = 0 and n = 1, k = 2πa

and

ψ(x) = C sin 2πx

a.

Let us now calculate the real and positive normalization constants.For D,

∫ +a2

− a2

ψ∗(x)ψ(x) dx =∫ +a

2

− a2

|D|2 cos2 πxadx = |D|2

∫ +a2

− a2

1 + cos 2πxa

2dx

=|D|2

2

[x+

a

2πsin 2πx

a

]+a2

− a2

=|D|2

2· a = 1 =⇒ D =

√2

a.

For C,∫ +a

2

− a2

ψ∗(x)ψ(x) dx =∫ +a

2

− a2

|C|2 sin2 2πx

adx = |C|2

∫ +a2

− a2

1− cos 4πxa

2dx

=|C|2

2

[x− a

4πsin 4πx

a

]+a2

− a2

=|C|2

2· a = 1 =⇒ C =

√2

a.

So, the functions we have found are

ψ1(x) =

√2

acos πx

aand ψ2(x) =

√2

asin 2πx

a.

5. Because H(x) = K(t) for all (x, t) ∈ R2 := R × R, H(0) = K(t) and H(x) =K(0) for all (x, t) ∈ R2. Combining this with H(x) = K(t), we get H(0) =K(0), which in turn implies H(x) = K(t) = H(0)(= K(0)) for all (x, t) ∈ R2.

6. Here is a summary of the full solution given in Section 7.2. The Schrodingerequations are

−~2

2m

d2ψ(x)

dx2= Eψ(x) x ≤ 0

−~2

2m

d2ψ(x)

dx2+ V0ψ(x) = Eψ(x) x > 0

CHAPTER 7 371

with the corresponding general solutions

ψ−(x) = Aeik1x +Be−ik1x where k1 =√2mE

~and x < 0

and

ψ+(x) = Cek2x +De−k2x where k2 =

√2m(V0 − E)~

and x > 0.

As ψ+ has to be normalizable/integrable in the region x > 0, we set C = 0 toobtain

ψ+(x) = De−k2x where k2 =

√2m(V0 − E)~

and x > 0.

Now use two continuity conditions

ψ+(0) = ψ−(0) and ψ′+(0) = ψ′

−(0)

to obtain ik2k1D = A−BD = A+B

with the solution

A = 12

(1 + ik2

k1

)D

B = 12

(1− ik2

k1

)D.

This gives

ψ(x) =

D2(1 + ik2/k1)e

ik1x + D2(1− ik2/k1)e−ik1x x ≤ 0

De−k2x x > 0.

7.

8.

9.


10. (a)

− ~2

2m

d2ψ(x)

dx2= Eψ(x) x < 0

− ~2

2m

d2ψ(x)

dx2+ V0ψ(x) = Eψ(x) x > 0

(b) Continuity of the wavefunction ψ(x) at x = 0 gives

Aeik1x +Be−ik1x = Ceik2x (at x = 0) or A+B = C.

Continuity of the first derivative dψ(x)dx

gives

ik1Aeik1x − ik1Be−ik1x = ik2Ce

ik2x (at x = 0) or k1(A−B) = k2C.

The rest is Arithmetic, and we get

C =2k1

k1 + k2A.

(c)

R =B∗B

A∗A=

(k1 − k2k1 + k2

)2

11.

12. (a) −~22m

d2ψ(x)dx2

= Eψ(x) x < 0−~22m

d2ψ(x)dx2

+ V0ψ(x) = Eψ(x) x > 0

(b) Aeik1x +Be−ik1x x < 0

Ceik2x +De−ik2x x > 0

(c) A = 0

(d) ψ(x)|0− = ψ(x)|0+ψ′(x)|0− = ψ′(x)|0+

=⇒

B = C +D−ik1B = ik2(C −D)

=⇒C +D = BC −D = −k1

k2B

=⇒C = k2−k1

2k2B

D = k2+k12k2

B.

CHAPTER 7 373

(e)

R =C∗C

D∗D=

(k2−k12k2

)2B∗B(

k2+k12k2

)2B∗B

=

(k1 − k2k1 + k2

)2

13. (a)

− ~2

2m

d2ψ(x)

dx2= Eψ(x) x < −a and x > 0

− ~2

2m

d2ψ(x)

dx2+ V0ψ(x) = Eψ(x) − a < x < 0

(b) i. Since there is nothing that reflects the wave in the region x > 0, thereshould not be a wave traveling to the left in this region. This meansH = 0 as e−ik1xe−iωt = e−i(kx+ωt) represents a wave traveling towardthe left.

ii. Continuity of the wavefunction ψ(x) at x = 0 gives

Cek2x +De−k2x = Geik1x (at x = 0) =⇒ C +D = G.

Continuity of the first derivative dψ(x)dx

at x = 0 gives

k2Cek2x−k2De−k2x = ik1Ge

ik1x (at x = 0) =⇒ k2(C−D) = ik1G.

iii. In the region x > 0, ψ∗(x)ψ(x) = G∗e−ik1xGeik1x = |G|2, and dψ∗(x)ψ(x)dx

∣∣∣0+

=

0. Hence, we also need dψ∗(x)ψ(x)dx

∣∣∣0−

= 0.Suppose C = 0. Then, we have ψ(x) = De−k2x in the region−a < x <

0. This gives dψ∗(x)ψ(x)dx

∣∣∣0−

= dD∗De−2k2x

dx

∣∣∣0−

= (−2k2)|D|2e−2k2x∣∣∣0−

=

−2k2|D|2 = 0 =⇒ D = 0, and both C and D are zero. Similarly,D = 0 forces C to be zero.If ψ(x) is zero inside the barrier, the continuity conditions on ψ(x) anddψ(x)dx

at x = −a, 0 implies ψ(x) = 0 everywhere, which is a physicallymeaningless solution. Therefore, neither C nor D can be zero.

14.

15. (a) ∫ +a/2

−a/2ψ∗ψdx = 2|A|2

∫ a/2

0sin2 kx dx = 1 =⇒ A = ±

√2

ai


=⇒ ψ2(x) = ±√2

ai sin 2πx

a(one of these)

(b) ∫ +a/2

−a/2Ψ∗xΨ dx =

∫ +a/2

−a/2A∗

2 sin (k2x)eiE2t/~ · x · A2 sin (k2x)e

−iE2t/~ dx

= |A2|2∫ +a/2

−a/2x sin2 (k2x) dx = 0 (We have an odd function.)

(c)−~2

2m

d2

dx2A2 sin (k2x) =

2π2~2

ma2A2 sin (k2x) =⇒ E2 =

2π2~2

ma2

16. (a) − ~22m

d2ψ(x)dx2

= Eψ(x)

(b) i. d2ψ(x)dx2

= ddx

(AB cos(Bx)) = −AB2 sin(Bx)=⇒ − ~2

2m· (−AB2) sin(Bx) = AE sin(Bx)

=⇒ B2 = 2mE~2

=⇒ B =√2mE~

ii. sin(±a

2B)= 0 =⇒ a

2B = π =⇒ B = 2π

a

iii. B2 = 4π2

a2= 2mE

~2=⇒ E = 2π2~2

ma2

iv.∫ a/2

−a/2A sin(Bx) · A sin(Bx)dx = 1 =⇒ A =√2/a

17. (a) − ~22m

d2ψ(x)dx2

= Eψ(x)

(b) i. ψ(0) = A sin 0 +B cos 0 = B = 0

ii. ψ(L) = A sin kL = 0 =⇒ kL = nπ =⇒ k = nπL

iii.∫ L0 ψ

∗ψdx =∫ L0 |A|2 sin2 knxdx =

[x2− sin(2knx)

4kn

]L0= |A|2L

2= 1 =⇒

An =√

2L

(c) kn = nπL

=√2mEn

~=⇒ En =

(~nπL

)212m

= n2h2

8mL2

18.df(x)

dx=

∞∑j=1

jajxj−1 =

∞∑j=0

(j + 1)aj+1xj

=⇒ df(x)

dx+ f(x) =

∞∑j=0

[aj + (j + 1)aj+1]xj = 3x2 + 8x+ 3

=⇒ aj + (j + 1)aj+1 = 0 for j ≥ 3.

CHAPTER 7 375

19.

20. The full wave function is ψ0e−iE0t/~.

⟨x⟩ =∫ +∞

−∞(ψ0e

−iE0t/~)∗x(ψ0e−iE0t/~)dx

=∫ +∞

−∞ψ∗0xψ0dx

=∫ +∞

−∞xA∗

0e−u2/2A0e

−u2/2dx

= |A0|2∫ +∞

−∞xe−[(Cm)1/2/~]x2︸︷︷︸

an odd functiondx

= 0

21. (a) The full ground state wavefuntion Ψ0 is given by Ψ0 = ψ0e−iEt/~. There-

fore,

< x >=∫ +∞

−∞Ψ∗

0xΨ0 dx =∫ +∞

−∞ψ∗0eiEt/~xψ0e

−iEt/~ dx =∫ +∞

−∞ψ∗0xψ0 dx∫ +∞

−∞A∗

0e−u2/2xA0e

−u2/2 dx = A∗0A0

∫ +∞

−∞xe−u2 dx

= A∗0A0

∫ +∞

−∞xe−[(Cm)1/2~−1]x2 dx = 0

as the integrand is an odd function.(b) ∫ +∞

−∞ψ∗0ψ1 dx =

∫ +∞

−∞A∗

0e−u2/2A1ue

−u2/2 dx = A∗0A1

∫ +∞

−∞ue−u2 dx

= A∗0A1

1√α

∫ +∞

−∞ue−u2 du = 0

because ue−u2 is an odd function.(c) Because the Hamiltonian is a Hermitian operator, its eigenvectors/func-

tions/kets associated with different eigenvalues are orthogonal to eachother.

(d) Vibration of diatomic molecules can be approximated by the simple har-monic oscillator. As the lowest allowed energy is strictly positive, thereis always some residual motion. This nonzero kinetic energy prevents thesystem from reaching zero degrees kelvin.


22.

23. Some examples are as follows.

• Energy levels are continuous in classical mechanics, while they are discretein quantum mechanics.• There is barrier penetration only in quantum mechanics.• The position of the particle is probabilistically determined in quantum

mechanics. But, it takes one unique value in classical mechanics.

Chapter 81.

Chapter 91. Proceeding as in (9.12),

[Lz, Lx] = [xpy − ypx, ypz − zpy]= [xpy, ypz]− [xpy, zpy]− [ypx, ypz] + [ypx, zpy]

= x[py, y]pz + y[x, pz]py + yx[py, pz] + [x, y]pypz

− (x[py, z]py + z[x, py]py + zx[py, py] + [x, z]pypy)

− (y[px, y]pz + y[y, pz]px + yy[px, pz] + [y, y]pxpz)

+ y[px, z]py + z[y, py]px + zy[px, py] + [y, z]pxpy

= x(−i~I)pz + y · 0 · py + yx · 0 + 0 · pypz− (x · 0 · py + z · 0 · py + zx · 0 + 0 · pypy)− (y · 0 · pz + y · 0 · px + yy · 0 + 0 · pxpz)+ y · 0 · py + z(i~I)px + zy · 0 + 0 · pxpy

= i~(zpx − xpz) = i~Ly.

In spherical coordinates as in (9.11),

[Lz, Lx] = LzLx − LxLz= −i~∂ϕi~ (sinϕ∂θ + cot θ cosϕ∂ϕ)− i~ (sinϕ∂θ + cot θ cosϕ∂ϕ) (−i~)∂ϕ

CHAPTER 10 377

= −(i~)2(cosϕ∂θ + sinϕ∂ϕ∂θ − cot θ sinϕ∂ϕ + cot θ cosϕ∂2ϕ

)+ (i~)2

(sinϕ∂θ∂ϕ + cot θ cosϕ∂2ϕ

)= i~ [i~ (− cosϕ∂θ + cot θ sinϕ∂ϕ)]

= i~

[i~

(− cosϕ ∂


∂ϕ

)]= i~Ly.

Chapter 101.

2.

3.

4. (a)l = 0, 1, 2

(b)ml = 0 for l = 0,

ml = −1, 0, 1 for l = 1,

ml = −2,−1, 0, 1, 2 for l = 2

(c) 9 dgenerate states

5.

E3 − E1 = −µe4

(4πε0)22~232−(− µe4

(4πε0)22~212

)=

8µe4

(4πε0)22~29=

µe4

36(πε0)2~2

6. (a) eik(2πm) = eik(2πn) for all (m,n). Consider (m,n) = (1, 0). Then, we have

|eik2π| = |ei(a+bi)2π| = |eia2π|e−b2π = 1 =⇒ b = 0.

Now consider m = n+ 1.

eik2π = 1 =⇒ k = 0,±1,±2, · · · (necessary condition).

But, this is also sufficient.


(b) For each n, l = 0, 1, 2, · · · , n− 1, and ml = −l,−l + 1, · · · , l − 1, l︸︷︷︸2l+1

n−1∑l=0

(2l + 1) = 2n−1∑l=0

l +n−1∑l=0

1 = 2 · n(n− 1)

2+ n = n2 Q.E.D.

7. (a) i. l = 3

ii. ∥L∥ =√3(3 + 1)~ = 2

√3~

iii. 2l + 1 = 2 · 3 + 1 = 7

iv. 3~

(b) i.

P (0 ≤ r < a0) =

a0∫0

π∫0

2π∫0

(1√π

(1

a0

)3/2

e−r/a0

)∗

(1√π

(1

a0

)3/2

e−r/a0

)r2dr sin θdθdϕ

=1

π

1

a30

a0∫0

π∫0

2π∫0

e−2r/a0r2dr sin θdθdϕ =1

π

1

a30(4π)

a0∫0

r2e(−2/a0)r dr

=4

a30

e−2r/a0

[r2

−2/a0− 2r

(−2/a0)2+

2

(−2/a0)3

]a00

=4

a30

e−2

[−1

2a30 −

1

2a30 −

1

4a30

]−[−1

4a30

]= 4e−2

[−2− 2− 1

4

]+ 1 = 1− 5e−2

ii.

< r >=

∞∫0

π∫0

2π∫0

(1√π

(1

a0

)3/2

e−r/a0

)∗

r

(1√π

(1

a0

)3/2

e−r/a0

)r2dr sin θdθdϕ

=1

πa30

∞∫0

π∫0

2π∫0

e−2r/a0r3dr sin θdθdϕ =1

πa30(4π)

∞∫0

r3e(−2/a0)r dr

CHAPTER 11 379

=4

a30

e−2r/a0

[r3

−2/a0− 3r2

(−2/a0)2+

6r

(−2/a0)3− 6

(−2/a0)4

]∞

0

=4

a30· 1 · 6a

40

4 · 4=

3

2a0

(c)

EZ,n = − Z2µe4

(4πε0)22~2n2= −

(Z2µe4

32π2ε20~2

)1

n2=⇒ En,3 = −

(9µe4

32π2ε20~2

)1

n2

Hence,

E3,3 − E1,3 = −(

9µe4

32π2ε20~2

)(1

9− 1

)= −

(9µe4

32π2ε20~2

)(−8

9

)=

µe4

4π2ε20~2

Chapter 111. (a) The probability of the first device getting an up-spin is

(1√2

)2= 0.5. After

the measurement, the state is

|↑⟩ = 1√2|↑⟩y +

1√2|↓⟩y .

Hence, the probability that the second device measures a down-spin in they-direction is also

(1√2

)2= 0.5. Therefore, the answer is 0.5× 0.5 = 0.25.

(b) The final spin state is |↓⟩y due to the fundamental postulates of quantummechanics.

Chapter 121.

Subject Index

This is a work in progress. A very slow progress it is.

E

Erwin Schrodinger, 111

N

Newtonian Mechanics, 13

Q

Quantum Mechanics, 13

R

Relativity, 13

T

traveling wave, 111

381

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