quantum pdf

Lecture notes

PHY3113–Quantum mechanics I

Hubert de Guise

Department of Physics,Lakehead University,

Thunder Bay, Ontario, P7B 5E1

Winter 2010

typeset using LATEX

byLaura Toppozini, Ben Lavoie, Diane Landry

andHubert de Guise

1

Contents

1 Introduction: Where we start, where we are going 11.1 “Shut up and compute!” . . . . . . . . . . . . . . . . . . . . . 11.2 Where did it all go wrong: a naive history of QM . . . . . . . 31.3 How the topics were chosen . . . . . . . . . . . . . . . . . . . 4

2 Heuristic approach to Schrodinger’s equation 62.1 Waves, wave equations and dispersions relations . . . . . . . . 62.2 Schrodinger equation and dispersion relations . . . . . . . . . 92.3 Historical comments . . . . . . . . . . . . . . . . . . . . . . . 112.4 Statistical interpretation . . . . . . . . . . . . . . . . . . . . . 12

3 Expectation values: another look at some known rules 15

I Quantization and boundary conditions 19

4 Basic solution to the Schrodinger equation 204.1 Separation of variables and stationary solutions . . . . . . . . 204.2 Two simple examples . . . . . . . . . . . . . . . . . . . . . . . 23

4.2.1 V (x) = constant = V0. . . . . . . . . . . . . . . . . . . 234.2.2 Infinite discontinuity: the infinite potential step . . . . 24

5 General properties of one-dimensional wave functions 265.1 The no-degeneracy and oscillatory theorems . . . . . . . . . . 265.2 Qualitative features of wave functions. . . . . . . . . . . . . . 29

6 Simple bound state problems in one dimension 316.1 The infinite well . . . . . . . . . . . . . . . . . . . . . . . . . . 316.2 Detour: Symmetric potentials and their solutions . . . . . . . 326.3 The finite well . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6.3.1 Obtaining a transcendental equation . . . . . . . . . . 346.3.2 Example: depth of the deuteron well . . . . . . . . . . 376.3.3 Example: The ammonia molecule . . . . . . . . . . . . 40

6.4 The harmonic oscillator . . . . . . . . . . . . . . . . . . . . . 436.4.1 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . 446.4.2 The harmonic oscillator wavefunctions . . . . . . . . . 48

6.5 δ(x)-well potential . . . . . . . . . . . . . . . . . . . . . . . . 48

2

7 Multidimensional independent-mode problems 51

8 Equivalent one–dimensional problems: the radial equation 548.1 Separation of variables in three dimensions . . . . . . . . . . . 548.2 The case of hydrogen: V (r) = −e2/4πεor. . . . . . . . . . . . . 58

8.2.1 Solving the differential equation . . . . . . . . . . . . . 588.2.2 Hydrogenoid Wave Functions . . . . . . . . . . . . . . 62

8.3 Isotropic oscillator in three dimensions . . . . . . . . . . . . . 648.4 Constant potentials in three dimensions . . . . . . . . . . . . . 68

8.4.1 Spherical Bessel functions . . . . . . . . . . . . . . . . 688.4.2 The Infinite Spherical Well . . . . . . . . . . . . . . . . 758.4.3 The Finite Spherical Well . . . . . . . . . . . . . . . . 76

8.5 The Morse potential . . . . . . . . . . . . . . . . . . . . . . . 79

9 The angular equation 839.1 Separation of the azimuthal and polar variables. . . . . . . . . 839.2 The θ equation . . . . . . . . . . . . . . . . . . . . . . . . . . 849.3 Associated Legendre polynomials . . . . . . . . . . . . . . . . 869.4 Spherical harmonics . . . . . . . . . . . . . . . . . . . . . . . . 879.5 Angular momentum operators . . . . . . . . . . . . . . . . . . 88

II Bras and kets 92

10 Bras, kets and abstract state vectors 9310.1 Introducing the ket vector . . . . . . . . . . . . . . . . . . . . 9310.2 Wavefunctions in the braket notation . . . . . . . . . . . . . . 9610.3 The bra vector . . . . . . . . . . . . . . . . . . . . . . . . . . 9610.4 Projection and closure . . . . . . . . . . . . . . . . . . . . . . 9710.5 |x〉 and ψ(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9810.6 Properties of δ . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

10.6.1 δ(x) in one dimension . . . . . . . . . . . . . . . . . . . 9910.6.2 δ(~r) in three dimensions . . . . . . . . . . . . . . . . . 102

10.7 |p〉 and the Fourier transform . . . . . . . . . . . . . . . . . . 10210.8 The x and p representations . . . . . . . . . . . . . . . . . . . 10510.9 On the use of |k〉 . . . . . . . . . . . . . . . . . . . . . . . . . 10710.10Wave packets and Dirac notation . . . . . . . . . . . . . . . . 108

10.10.1General formulation . . . . . . . . . . . . . . . . . . . 108

3

10.10.2Example: the Gaussian wave packet . . . . . . . . . . . 10910.10.3Qualitative feature of the Gaussian wave packet . . . . 112

11 Scattering states 11511.1 The potential step . . . . . . . . . . . . . . . . . . . . . . . . 11511.2 The potential barrier with E > V0 . . . . . . . . . . . . . . . . 11611.3 The potential hole . . . . . . . . . . . . . . . . . . . . . . . . 11911.4 The potential step with E < V0 . . . . . . . . . . . . . . . . . 12011.5 The S-matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

III Formalism and calculation tools 126

12 Linear stuff 12712.1 Linear spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 12712.2 Inner products and norms . . . . . . . . . . . . . . . . . . . . 12912.3 Linear operators . . . . . . . . . . . . . . . . . . . . . . . . . . 13312.4 Eigenvectors are special . . . . . . . . . . . . . . . . . . . . . 13512.5 The mechanics of matrix diagonalization . . . . . . . . . . . . 13612.6 Application: neutrino mixing . . . . . . . . . . . . . . . . . . 14112.7 The ammonia molecule revisited . . . . . . . . . . . . . . . . . 145

13 Hermitian matrices and their properties 14613.1 Some definitions . . . . . . . . . . . . . . . . . . . . . . . . . . 14613.2 Hermitian operators . . . . . . . . . . . . . . . . . . . . . . . 14713.3 Properties of pairs of hermitian operators . . . . . . . . . . . . 14913.4 Simultaneous eigenvectors of commuting hermitian operators . 149

14 The postulates of Quantum Mechanics 151

15 Algebraic approach to some matrix elements 15415.1 Some properties of commutators . . . . . . . . . . . . . . . . . 154

15.1.1 Linear combination of operators in a commutator . . . 15415.1.2 [x, p] . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

15.2 Harmonic oscillator and the operators a and a† . . . . . . . . 15515.2.1 a and a† . . . . . . . . . . . . . . . . . . . . . . . . . . 15615.2.2 Matrix elements of x, p and their powers . . . . . . . . 16015.2.3 Application: constructing the eigenfunctions of H us-

ing a† . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

4

15.2.4 Application: the harmonic oscillator coherent state . . 16315.3 Angular momentum theory and the operators L+ and L− . . . 167

15.3.1 Commutation relations for angular momentum operators16815.3.2 L+ and L− . . . . . . . . . . . . . . . . . . . . . . . . 17015.3.3 Matrix elements of Lx and Ly . . . . . . . . . . . . . . 173

15.3.4 Constructing the simultaneous eigenfunctions of L2 andLz. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

16 Uncertainty relations 18216.1 〈∆A〉〈∆B〉 ≥ 1

2|〈[A, B]〉| . . . . . . . . . . . . . . . . . . . . 182

16.2 Understanding the assumptions . . . . . . . . . . . . . . . . . 184

A Complex numbers 186A.1 A bit of history . . . . . . . . . . . . . . . . . . . . . . . . . . 186A.2 Adding complex numbers . . . . . . . . . . . . . . . . . . . . . 188A.3 The complex conjugate . . . . . . . . . . . . . . . . . . . . . . 189A.4 Multiplication and division . . . . . . . . . . . . . . . . . . . . 190

A.4.1 Multiplying and dividing in the Cartesian representation190A.4.2 The polar representation . . . . . . . . . . . . . . . . . 191A.4.3 Euler’s theorem . . . . . . . . . . . . . . . . . . . . . . 193A.4.4 Taking powers and roots . . . . . . . . . . . . . . . . . 196

B Numerical integration and Simpson’s rule 199B.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 199B.2 Simpson’s rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 204B.3 Application: quantum particle outside a harmonic well . . . . 205

C Finding zeroes of an equation 209C.1 Bracketing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209C.2 Newton’s method . . . . . . . . . . . . . . . . . . . . . . . . . 209

D Remarks on some second order differential equations 211D.1 The Gauss hypergeometric 2F1(a, b, c; z) . . . . . . . . . . . . 211D.2 The confluent hypergeometric function 1F1(a, c; z) . . . . . . . 214

E Hermite polynomials 216E.1 Series solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 216E.2 Generating function . . . . . . . . . . . . . . . . . . . . . . . . 219E.3 Rodrigues formula and orthogonality . . . . . . . . . . . . . . 219

5

F Eigenvectors: On the importance of being proper 222F.1 General eigenvalue problem . . . . . . . . . . . . . . . . . . . 223F.2 Diagonalizable matrices . . . . . . . . . . . . . . . . . . . . . . 227F.3 Tricks of the trade . . . . . . . . . . . . . . . . . . . . . . . . 228

F.3.1 Powers of a matrix . . . . . . . . . . . . . . . . . . . . 228F.3.2 The determinant and trace conditions . . . . . . . . . . 229

G Collected Problems 2001-2006 230

Introduction 1

1 Introduction: Where we start, where we

are going

1.1 “Shut up and compute!”

Quantum mechanics is an exciting topic: it works and we don’t know why.Physicists don’t solve problems using Newton’s law anymore (they use themore sophisticated Hamilton–Jacobi formulation), but research journals arefull of problems in and problems with quantum mechanics. Quantum me-chanics remains a dynamic topic, rejuvenated by the recent advances in theemerging field of quantum information.

“What is quantum?” The difference between quantum and, say, classicalmechanics is the inherently probabilistic nature of the latter. In quantummechanics, two identical experiments on two systems prepared in exactly thesame manner, can produce different results. What quantum mechanics doesis provide a prescription to enumerate the possible outcome of the experimentand to calculate the relative probabilities of these outcomes.

Before we embark on this course, I must come out clean and dispel someillusions: quantum mechanics is not physics. Quantum mechanics a set ofmathematical rules that allow us to understand the physics of microscopicobjects and to make calculations verified by experiments.

To emphasize the mathematical nature of quantum mechanics, simplynote that, if you walk into a lab where experiments on electricity are done,you will find wires, resistors, capacitors and other such devices. The theoryof electrical circuits has a definite materiality: you can take a resistor in yourhand and (sometimes) feel its heat when a current flows through it. If, onthe other hand, you walk into a quantum optics lab, you will never find aHilbert space, an eigenvector or a wave function: the objects that populate acourse on quantum mechanics are at least once removed from reality. Theyare abstract concepts, mathematical devices and rules that we construct sothat we can extract information and compare with experiments. In timesof despair or confusion (and these times will come), it is always good toremember that quantum mechanics is, ultimately, a mathematical game.

What is unusual is that unreal and abstract mathematical constructsactually work in real practical life. The rules of quantum mechanics havebeen road tested for about 75 years now; they are applicable to a largenumber of disciplines, from solid state to nuclear physics. The whole set of

Introduction 2

rules, however abstract and poorly justified, correctly predicts the behaviorof microscopic objects, and from this, very non-abstract applications havesprung. Semi-conductor devices, which can be explained using quantummechanics, work. NMR works.

The central objective of this course can be summarized by the famousdictum “Shut up and compute!” (attributed to R.P. Feynman).

You can blame Schrodinger and Heisenberg for that. They were ratherdesperate to reproduce some experimental results (spectrum of the hydro-gen and other atoms, in fact): they did not care about shortcuts, fuzzy as-sumptions or whatever else. They found (independently) two methods thatworked, without knowing why they worked.

If you expect to understand ”why” the rules are the way they are, youhave enrolled in the wrong course. This does not imply that the ”why” isunimportant (it is, in fact, absolutely fascinating!); the emphasis on prac-ticality is something that my colleague and I chose to focus on, given thelimited amount of time available in the (often crowded) physics curriculum.Furthermore, it is easier to understand the ”why” if one has had some expo-sure to the mechanics of the calculations. To quote Jean Le Rond d’Alembert,french mathematician of the 18th century: ”Just go on...and faith will soonreturn.” D’Alembert was thus reassuring a friend, who had difficulty under-standing ”why” the rules of calculus were the way they were, this 150 yearsafter the invention of calculus.

The textbooks that we have chosen reflects our choice. Like all textbooks,they will be liked by some, disliked by others. The ”best” textbook onquantum mechanics (or any other subject) is a matter of personal preference;if you do not like our choice, your are strongly encouraged to head for thelibrary: most quantum mechanics texts have a call number starting with QC174. The university has over 150 books on the topic. Most of these dealwith the ”how”; few deal with the ”why”. If you want to understand the”how” better, I do not doubt that one of these 150 books will be ideallysuited for you. If you want to understand the ”why” better, I suggest thatyou buy ”Where does the weirdness go?”, by David Lindley, or consult, forinstance, ”Quantum theory: concepts and methods”, by Asher Peres (QC174.12 P417 1995)

Introduction 3

1.2 Where did it all go wrong: a naive history of QM

The rules of quantum mechanics are not completely arbitrary. Schrodinger, inparticular, was inspired by classical mechanics (what else could have inspiredhim? That was the only game in town at the time!). Central to classicalmechanics is the Hamiltonian function, which, in its simplest conception,represents the total energy of the system.

To integrate Hamilton’s equations of motion, one normally uses a methoddeveloped by Jacobi. Schrodinger (he was trying to solve the hydrogen atom)knew that blunt application of the Hamilton–Jacobi method would not work.He decided on a whim to make some (very inspired) transformation and, bymagic, stumbled on the famous differential equation that now bears his name.(If you want to see the path from Hamilton–Jacobi to Schrodinger’s equation,take the graduate quantum mechanics course.)

The next task was to actually solve the differential equation. (There is nophysics up to that point, just a really sneaky transformation that could havelead nowhere.) The differential equation is of a type known as an eigenvalueequation of the Sturm–Liouville type. Well, Schrodinger knew how to solvethis, so he plowed forward and found that some solutions produced the correctenergy for the hydrogen atom while others did not. The final trick of his paper(first of a series of three) was to invoke some boundary condition to removeall the unwanted solutions. Voila! We now have found the spectrum of thehydrogen atom without really knowing what we are doing.

This is apparently not a serious advance. After all, Bohr had alreadycomputed that spectrum. However, Bohr’s theory was known to be falsebecause it did not correctly predict the angular momentum of each energylevel nor did it correctly count the number of states having a given energy.(This can be determined by Zeeman splitting, something easily done at thetime.) Schrodinger’s solution did agree with all the experimental data fromthe hydrogen atom, and also was reasonably good at predicting the spectrumof the helium atom, something at which the Bohr model failed completely.Schrodinger had indeed found something new.

The solutions to the differential equation Schrodinger called wave func-tions. The name apparently comes because Schrodinger once overheard aninnocent remark at the end of a talk on Compton scattering. The remarkwas to the effect that, if particles really behaved like waves, there should bea wave equation associated with this behavior.

Schrodinger did not really know what to make of those beastly wave func-

Introduction 4

tions. They are, in general, complex (in the mathematical sense), which pre-cludes directly associating a physical meaning to them. Schrodinger thoughtthe wave functions were somehow waves that “guided” the particles alongsome kind of classical trajectories. That idea rapidly fell apart: under “Shutup and compute”, others had taken up his prescription and found that thewave function for two electrons (in a helium atom) was actually a functionof six variables (three for each electrons). Maybe the wave was guided insix–dimensional space??

It was up to Born, a few years later, to make some sense of it all. Heproposed that the meaningful quantity was not the wave function itself butits magnitude squared (which is always a positive number). Born then af-firmed that this magnitude squared was a probability density, meaning thatit provided information about the probability distribution of a particle (orparticles) in space. This is just another example of “Shut up and compute”:we don’t know what the wave function is (in fact, according to Born, the wavefunction itself has no meaning) but we know how to use it in calculations.

I tell this story because it contains two essential elements for this course.First, quantum mechanics tells us about the possible outcome of an experi-ment designed to measure some physical quantity (here, the possible energylevels). By solving the Schrodinger equation, we answer the question “Whatare possible levels that can be measured experimentally?”. Second, quantummechanics tells us how to calculate the probability of each outcome. Theseare two central objectives of the course.

To finish this brief history we must allow Dirac to enter. He realizedthat Schrodinger and Heisenberg were really going the same thing. He ac-tually generalized the concept of wave function to that of state vector, andintroduced the bra and ket notation that is now commonly used in the field.Dirac also realized that, most of the time, we only really need properties ofwave functions rather than the wave functions themselves: he pioneered thecalculation of matrix elements (particularly for the harmonic oscillator) Thesecond half of the course is implicitly devoted to quantum mechanics “a laHeisenberg and Dirac”, but we have some work to do before we get there.

1.3 How the topics were chosen

We cannot cover everything in one term: there is just too much material forthat. The course will focus on simple quantum systems: square wells andharmonic oscillators in one and three dimensions, and hydrogenoid atoms.

Introduction 5

Pretty much any type of problem can be approximated by one of these prob-lems. These problems can all be understood by solving the Schrodinger dif-ferential equation. We will mostly be interested in enumerating and countingenergy levels.

Next, we will deal with the generalization to state vectors. This is muchmore useful and much more “quantum” than the wave function approach.The question of what is a state vector “in real life” is clearly moot in thisformalism. The state vector formulation relies on properties of the solutionsto the Schrodinger equation much more than on the solution themselves. Itleads to the calculation of matrix elements of operators and the recasting ofSchrodinger’s equation from a differential eigenvalue problem into a matrixeigenvalue problem. We will only cover the simplest of examples, whichwill allow us to understand, for instance, the ammonia molecule, and theprobability of finding the molecule in one of its two forms.

Heuristic Schrodinger equation 6

2 Heuristic approach to Schrodinger’s equa-

tion

Schrodinger’s equation is a wave equation, describing wave–like properties ofparticles. Schrodinger’s equation describes how the derivative w/r to timeof a function (called in the jargon a wave–function) is connected to the valueof this function and of its spatial derivatives.

2.1 Waves, wave equations and dispersions relations

A wave is a disturbance in space and time. Normally, a physical wave (like asound wave or an electromagnetic wave) will carry energy. Mathematically,a wave (in one dimension, to simplify the analysis for the moment) movingin the +x direction is any function of the combination x − vt, where x is aposition, t is time and v is the velocity of the wave.

The combination x− vt has units of length, and it mathematically moreconvenient to consider the dimensionless quantity

k (x− vt) = kx− ωt, (2.1)

where k, a quantity called the wave vector, has units of inverse length.Clearly, we have

kv = ω (2.2)

where ω is the angular frequency.An example is given in Fig., where we have four plots of

Ψ(x, t) =1

1 + (2x− t)2 , (2.3)

for t = −2,−1, 0 and 1. It is clear that the wave travels “to the right”. Awave traveling to the left would be a function of the combination kx + ωt.

Note that it is not the overall sign of kx−ωt that determines the directionof motion but rather the relative sign of kx and ωt. Thus, a function of ωt−kxstill describes a wave traveling to the right. Note also that the functionΨ(x, t) that mathematically describes the wave need not be periodic.

One very common type of wave is the so–called harmonic, or free, wave.Mathematically, it is described by the general form

Ψ(x, t) = a cos (kx− ωt + ϕ) , (2.4)


where ϕ is an angle called the initial phase of the wave. This type of waveis obviously periodic. Its amplitude is a.

For a variety of reason, it is often more covenient to use complex numbersand represent Ψ(x, t) as

Ψ(x, t) = Aei(kx−ωt), (2.5)

where A is now a complex number.The mathematical function Ψ(x, t) describing the disturbance is usu-

ally the solution to a wave equation. For instance, the function Ψ(x, t)of Eqn.(2.5) is a solution of the differential equation

∂2

∂x2Ψ(x, t)− 1

v2

∂2

∂t2Ψ(x, t) = 0. (2.6)

Eqn.(2.6) is so common that it is sometimes referred to as “the wave equa-tion”. However, this is by no means unique. For instance, the much morecomplicated partial differential equation

∂

∂tΨ(x, t) +

∂3

∂x3Ψ(x, t) + 6Ψ(x, t)

∂

∂xΨ(x, t) = 0 (2.7)

is known as the Korteweg-de Vries wave equation. It has as a solution thefunction

Ψ(x, t) =v

2

1

cosh2[√

v2

(x− vt)] , (2.8)

and it describes shallow water waves. This is still a wave, owing to the x−vtcombination, but it is not harmonic.

There are fundamental differences between Eqns.(??) and (2.7). First,the form of the differential equation is completely different: Eqn.(??) it con-tains only second order derivatives whereas Eqn.(2.7) contains derivatives ofvarious order. An extremely important property for quantum mechanicsis that Eqn.(??) is linear. This means that, if Ψ1(x, t) and Ψ2(x, t) aresolutions to Eqn.(??), then

Ψ(x, t) = αΨ1(x, t) + βΨ2(x, t) (2.9)

is also solution for constant α and constant β. You can verify this directly bysubstituting Eqn.(2.9) into Eqn.(2.6). However, this does not hold for thesolutions to Eqn.(2.7), i.e. Ψ(x, t) is not, in general, solution to Eqn.(2.7).


The linearity property of solutions is crucial because it is closely andintimately related to the possibility of producing interference between waves,something that was experimentally verified for particles. In other words,experiment suggests that, in order to describe the known wave propertiesof particles, we should primarily consider linear wave equations. (Nota:this does not exclude the possibility of non–linear wave equations; it simplymeans that such non–linear equations would not be good at describing someof the known wavelike properties of particles.)

When a differential equation is linear, like Eqn.(??), it usually followsfrom the differential equation that some kind of subsidiary condition mustbe met. This condition usually takes the form of a relation between ω andk and is known as the dispersion relation for the wave.

For instance, the dispersion relation for a real physical piano string is ofthe form

ω2

k2=

To

ρo

+ αk2, (2.10)

where To is the tension in the string, ρo is the mass density of the string, andα is a small positive constant that would be zero if the string were perfectlyflexible.

Normally, it is possible to recover the dispersion relation from the waveequation. Consider, for instance, the modification of Eqn.(??)

∂2

∂x2Ψ(x, t)− 1

c2

∂2

∂t2Ψ(x, t) =

m2c2

~2Ψ(x, t), (2.11)

where an extra term has been added and the velocity is taken to be the speedof light c. (Remember that ~ has units of [energy]×[time], which means m2c2

~2has the correct units of [length]−2). Inserting the ansatz Ψ(x, t) = Aei(kx−ωt),we find that we have a solution provided that the dispersion relation

−k2 +ω2

c2=

m2c2

~2(2.12)

is satisfied. This is easily rearranged into

(~ω)2 = (~k)2 c2 + m2c4. (2.13)

If one remembers from quantum physics the basic rules

E = hν = ~ω, (2.14)

p = ~k, (2.15)


then Eqn.(2.13) is an expression for the relativistic energy of a particle withmomentum p and rest mass m. Eqn.(2.11) is candidate for a wave equationappropriate for the description of a quantum relativistic particle. (Whetheror not the wave properties of a relativistic particle is actually described bythe solutions of Eqn.(2.11) is another matter.)

Notice how Eqns.(2.10) and (2.13) contains the physical parameters thatthe mathematical description of the wave is suppose to embody. Anotherexample of the physics contained in a dispersion relation is provided by theequation

ω2 =4πNe2

me

+ c2k2, (2.16)

= ω2p + c2k2 (2.17)

found to describe the propagation of waves in a plasma. Here, e is theelectronic charge, me is the electronic mass and N is the electronic density ofthe plasma. Eqn.(2.16) shows that, if ω2

p > ω2, it is possible for k to becomeimaginary, in which case the form of Eqn.(2.5) becomes

Ψ(x, t) = Ae−κxe−iωt, (2.18)

where

κ =1

c

√ω2

p − ω2. (2.19)

In such a case, the wave attenuates with x and there is no oscillation. Again,the dispersion relation describes the physics of the plasma through the ad-ditional term 4πNe2

me, which itself depends on various physical constants and

the electronic density appropriate to the particular situation at hand.

2.2 Schrodinger equation and dispersion relations

The phenomenological starting points of Schrodinger equation are the re-lations giving wavelike properties to particles. More specifically, to a freeparticle of momentum p we assign a free wave with wave vector k related top by the de Broglie relation

p = ~k. (2.20)

The energy of a free particle is completely kinetic, so

E =p2

2m=~2k2

2m. (2.21)


To transform this into a dispersion relation, we observe that, for photon, therelation

E = hν = ~ω (2.22)

holds. Combining these last two equations yields

~ω =~2k2

2m(2.23)

As a free wave is described by Eqn.(2.5):

Ψ(x, t) = Aei(kx−ωt),

we need to find a differential equation (i.e. a wave equation) that will re-produce the dispersion relation of Eqn.(2.23) when we input Eqn.(2.5) as asolution. We also want our differential equation to be linear so that we canreproduce interference effects. The simplest thing around is

i~∂

∂tΨ(x, t) = − ~

2

2m

∂2

∂x2Ψ(x, t). (2.24)

If the particle is not free but evolves in a potential V (x), then the dispersionrelation becomes

~ω =~2k2

2m+ V (x), (2.25)

and the wave equation is postulated to change to

i~∂

∂tΨ(x, t) = − ~

2

2m

∂2

∂x2Ψ(x, t) + V (x)Ψ(x, t). (2.26)

This is the time–dependent Schrodinger equation. Formally, it amount to atransformation of the equation E = p2/2m+V (x) into a differential equationfor Ψ(x, t) obtained by replacing the physical quantities E, p and x by the“operators”

E 7→ H = i~∂

∂t, (2.27)

p 7→ p = −i~∂

∂x, (2.28)

x 7→ x = x . (2.29)

The simple harmonic wave of Eqn.(2.5) is no longer a solution of Eqn.(2.26),but the harmonic wave is suppose to describe something oscillating freely, notsomething subject to a potential.


2.3 Historical comments

If the “derivation” of Eqn.(2.26) looks very fake, it is because it is fake andvery handwavy. One cannot derive the Schrodinger equation from first prin-ciples. More interestingly, this is NOT how Schrodinger ever approached theproblem. His argument was based on classical mechanics. Modern classicalmechanics has nothing to do with Newton and is based instead on a theHamiltonian function H, which is essentially the total energy of the system.In our case, the classical Hamiltonian looks like

H =p2

2m+ V (x). (2.30)

The motion of a particle can be obtained from the Hamiltonian using Hamil-ton’s equations of motions. These equations are not always easy to solve,so there is a refinement called Hamilton-Jacobi theory, especially powerfulwhen the motion is periodic. In the Hamilton–Jacobi equation, momentumgets replaced by a partial derivative of the action integral S:

S =

∫ (p2

2m− V (x)

)dt (2.31)

p 7→ ∂S

∂x. (2.32)

The Hamilton–Jacobi equation then looks like

H(x,∂S

∂x, t) +

∂S

∂t= 0, (2.33)

1

2m

(∂S

∂x

)2

+ V (x) +∂S

∂t= 0. (2.34)

If H does not depend on this explicitly (as we are assuming in this discussion),then H is the total energy E of the system so that all of this can be simplifiedto

1

2m

(∂S

∂x

)2

+ V (x)− E = 0. (2.35)

Schrodinger then postulated the form

S = ξ ln Ψ(x, t), (2.36)


where ξ is some constant. Inserting Eqn.(2.36) into Eqn.(??) yields (afterminor tinkering) :

ξ1

Ψ(x, t)

∂

∂xΨ(x, t) =

∂S

∂x(2.37)

(∂

∂xΨ(x, t)

)2

− 2m

ξ2(E − V (x))Ψ2(x, t) = 0. (2.38)

This is not linear, so a further massaging must be done. Suffice it to saythat he eventually got what he wanted, i.e. a differential equation. Schro-dinger was gunning for the energy levels of the hydrogen atom; as we willsee, it is not the easiest problem to start with. After (again quite artificially)imposing boundary conditions on his solution, he found that the got preciselythe energy levels of the hydrogen atom if ξ was set to ~.

2.4 Statistical interpretation

What now of the so–called wave function Ψ(x, t)? Let us assume V = 0, sowe already know Ψ(x, t) = Aei(kx−ωt).

Let P (x, t) = Ψ∗(x, t)Ψ(x, t) ≥ 0, and consider

∂

∂tP (x, t) =

(∂

∂tΨ∗(x, t)

)Ψ(x, t) + Ψ∗(x, t)

∂

∂tΨ(x, t) (2.39)

Using the Schrodinger equation to change the time derivative into spatialderivatives:

∂

∂tΨ(x, t) = − i

~

(− ~

2

2m

∂2

∂x2

)Ψ(x, t) ,

∂

∂tΨ∗(x, t) =

i

~

(− ~

2

2m

∂2

∂x2

)Ψ∗(x, t) ,

we obtain

∂

∂tP (x, t) = Ψ(x, t)

(i~2m

∂2

∂x2Ψ∗(x, t)

)+ Ψ∗(x, t)

(− i~

2m

∂2

∂x2Ψ(x, t)

).

(2.40)We can clean up the expressions to get

∂

∂tP (x, t) =

i~2m

[Ψ∗(x, t)

∂2

∂x2Ψ(x, t)−Ψ(x, t)

∂2

∂x2Ψ∗(x, t)

](2.41)

=∂

∂x

[i~2m

(Ψ∗(x, t)

∂

∂xΨ(x, t)−Ψ(x, t)

∂

∂xΨ∗(x, t)

)](2.42)


This is a continuity equation:

∂

∂t(density of “stuff”) +

∂

∂x(current of “stuff”) = 0 . (2.43)

The ”stuff” has density P (x, t) = Ψ∗(x, t)Ψ(x, t), which is always ≥ 0.This is interpreted in Quantum Mechanics as a probability density, meaningthat

P (x0, t0)dx = Ψ∗(x0, t0)Ψ(x0, t0)dx (2.44)

is the probability of finding a system described by Ψ(x, t) is a small region

between x0 and x0 + dx at time t0, and Pab =∫ b

aP (x, t)dx is the probability

of finding the system described by Ψ(x, t) in the interval a ≤ x ≤ b.Note that the probability of finding the system exactly at x0 is

Px0x0 =

∫ x0

x0

P (x, t)dx = 0 , (2.45)

contrary to one of the numerous incorrect examples found in Beiser.Obviously, a requirement is that the total probability of find the particle

somewhere in space, i.e. in the interval −∞ < x < ∞, be 1, i.e.

1 =

∞∫

−∞

P (x, t)dx =

∞∫

−∞

Ψ∗(x, t)Ψ(x, t)dx . (2.46)

This is known as the normalization condition.Let us come back to Ψ(x, t) = Aei(kx−Et/~) . Note that Ψ∗(x, t)Ψ(x, t) =

AA∗ ≥ 0 so

1 = |A|2∞∫

−∞

dx = ∞! (2.47)

This is a conflict between the physical interpretation and the mathemat-ical apparatus: we’ll resolve (but not completely solve) that bug later. Theconflict is similar to calculating the total energy of a plane wave in electro-magnetic theory: since this is proportional to the integral of | ~E|2, which isconstant for a plane wave propagating in vacuum, integrating over the entirespace gives an infinite value for the total energy, an obvious nonsense. Likein electromagnetic theory, a plane wave is really just an idealization, and weshouldn’t be too surprised if, in such cases, things take a left turn.


Anyways, with P (x0, t) identified as a probability density, we now notethat the probability current is

J(x, t) = − i~2m

(Ψ∗(x, t)

(∂

∂xΨ(x, t)

)−Ψ(x, t)

(∂

∂xΨ∗(x, t

))(2.48)

What about problems with potential energies? There, we simply extendthe procedure by directly adding V (x)

+i~∂

∂tΨ(x, t) =

~2

2m

∂2

2x2Ψ(x, t) + V (x0, t)Ψ(x, t) (2.49)

Provided that V (x) is real and does not depend on t , the interpretationof P (x, t) as a probability density and J(x, t) as a probability current stillholds.

Note that, since∞∫−∞

P (x, t)dx = 1 , the probability of finding the par-

ticle somewhere is independent of t, as expected. Finally, observe that, if∞∫∞

P (x, t)dx = 1 , we necessarily have, for continuous Ψ(x, t) (which is rea-

sonable),lim

x→±∞Ψ(x, t) = 0 . (2.50)

Solutions for which Eq.(2.50) is applicable are known as bound states.As the name implies, such wave functions describe “bound” particles whichcannot escape to infinity. Solutions for which Eq.(2.50) does not apply areknown as free states (or scattering states).

Expectation values 15

3 Expectation values: another look at some

known rules

Suppose that we want to calculate the average weight of the students in theclass. If Nwi

students have weight wi , then the average weight is

〈w〉 =∑

i

wi Nwi(3.1)

Now suppose that we want to compute the average w as a function oftime. Then,

〈w(t)〉 =∑

i

wiNwi(t) (3.2)

What we mean by this is that the average weight may change in time becausethe number of student with a given weight will change in time. The weightwi that occurs on the R.H.S. does not change in time. For instance, if todaythere are 5 students weighing 60kg, 3 students weighing 65kg and 7 studentsweight 70kg, then the average weight is

〈w(today)〉 =60.5 + 65.3 + 70.7

5 + 3 + 7(3.3)

If tomorrow there are 4 students weighing 60kg, 5 weighing 65kg and 6weighing 70, then the average weight is

〈w(tomorrow)〉 =60.4 + 65.5 + 70.6

4 + 5 + 6(3.4)

Even if the average weight 〈w(t)〉 varies with t , the individual values ofweights, 60kg, 65kg, 70kg, do not. Going now to the continuous case, thesummation becomes an integral and the discrete distribution becomes thedistribution P (w, t)

〈w(t)〉 =

∫P (w, t)w dw∫P (w, t)dw

(3.5)

and

〈w(t)〉 =

∫O(w, t)w dw∫P (w, t)dw

(3.6)


The variation in time of the average weight is due to the variation in timein the distribution of these weights: the weight w = 65kg is the same at allt. Let us apply this to

〈x(t)〉 =

∞∫

−∞

P (x, t) x dx , (3.7)

where

P (x, t) = ψ∗(x, t)ψ(x, t) ,

∫ ∞

−∞P (x, t)dx = 1 . (3.8)

Then,

〈x(t)〉 =

∞∫

−∞

P (x, t)xdx ,

=

∞∫

−∞

(∂

∂xJ(x, t)

)xdx = xJ(x, t)

∣∣∣∞

−∞−

∞∫

−∞

J(x, t)dx (3.9)

= xi~2m

(Ψ∗(x, t)

∂


∂

∂xΨ∗(x, t)

) ∣∣∣∞

−∞

− i~2m

∞∫

−∞

[Ψ∗(x, t)

∂


(∂

∂xΨ∗(x, t)

)]dx. (3.10)

The boundary term is zero if we assume limx→±∞ ψ(x, t) = 0 . In the lastterm, integrate by parts

−∞∫

−∞

Ψ(x, t)∂

∂xΨ∗(x, t)dx = −Ψ(x, t)Ψ∗(x, t)

∣∣∣∞

−∞+

∞∫

−∞

Ψ∗(x, t)

(∂

∂xΨ(x, t)

)

=

∞∫

−∞

Ψ∗(x, t)

(∂

∂xΨ(x, t)

)dx (3.11)

so that

〈x(t)〉 = −i~m

∞∫

−∞

Ψ∗(x, t)

(∂

∂xΨ(x, t)

)dx (3.12)


The momentun p = mx , thus 〈p〉 = m〈x〉 is therefore given by

〈p(t)〉 = −i~∞∫

−∞

Ψ∗(x, t)∂

∂xΨ(x, t)dx (3.13)

=

∞∫

−∞

ψ∗(x, t)p Ψ(x, t)dx (3.14)

with

p = −i~∂

∂x. (3.15)

Thus, the rule of Eq.(3.15), which had initially been guessed at using planewaves, is now shown to be consistent with our previous assumptions aboutthe nature of Ψ(x, t) and of Ψ(x, t)∗Ψ(x, t) even in cases where the solutionis not a plane wave.

Now, in the Schrodinger equation,

i~∂

∂tΨ(x, t) = − ~

2

2m

∂2

∂x2Ψ(x, t) + V (x)Ψ(x, t) (3.16)

the term

− ~2

2m

∂2

∂x2⇐⇒ p2

2m, (3.17)

i.e. the kinetic energy. The remains to see if we can interpret V (x) as apotential energy. Consider a hypothetical case where there is no keneticenergy. Then all the energy is potential, and P (x, t) is constant (the particle

does not move). Setting∂2

∂x2Ψ(x, t) = 0 in the Schrodinger equation, we

have

i~∂

∂tΨ(x, t) = V (x)Ψ(x, t) =⇒ Ψ(x, t) = e−iV (x)t/~Ψ(x, 0) (3.18)

and

P (x, t) = Ψ∗(x, t)Ψ(xt)− eiV (x)t/~Ψ∗(x, 0)e−iV (x)t/~Ψ(x, 0) (3.19)

= Ψ∗(x, 0)Ψ(x, 0) = P (x, 0) (3.20)

independent of time. Thus, V (x) can be properly thought of as a potentialenergy.


The Schrodinger equation

i~∂

∂tΨ(x, t) =

~2

2m

∂2

∂x2Ψ(x, t) + V (x)Ψ(x, t) (3.21)

is therefore an operation transcription of a statement about energy:

E ←→ i~∂

∂t, (3.22)

p2

2m←→ − ~

2

2m

∂2

∂x2, (3.23)

V (x) ←→ V (x) . (3.24)

19

Part I

Quantization and boundaryconditionsIn this part of the course, we will look first at energy quantization.

It is an experimentally verifiable fact that energy quantization occurs onlywhen the motion of a particle or group of particles is spatially confined orbounded. If the motion is not confined, as is the motion of a free chargedparticle, there is no energy quantization.

Mathematically, the confinement appears by imposing appropriate bound-ary solutions of the Schrodinger equation. Energy quantization in boundstates and boundary conditions are closely linked. The properties of theresulting solutions will also be examined.

In Chapter 8, we will extend the analysis to the quantization of angularmomentum.

Basic solutions 20

4 Basic solution to the Schrodinger equation

4.1 Separation of variables and stationary solutions

The (full, time–dependent) Schrodinger equation reads

i~∂

∂tΨ(x, t) = − ~

2

2m

∂2

∂t2Ψ(x, t) + V (x)Ψ(x, t) (4.1)

We look for solutions of this partial differential equation using the tech-nique of separation of variables, i.e. we assume

Ψ(x, t) = Φ(t)ψ(x) (4.2)

Plugging this ansatz into Schrdinger’s equation, we find

i~(

d

dtΦ(t)

)ψ(x) = Φ(t)

[−~2

2m

d2

dx2ψ(x) + V (x)ψ(x)

](4.3)

This is a functional identity valid ∀t, ∀x so dividing by Φ(t)ψ(x), we get

i~1

Φ(t)

(d

dtΦ(t)

)=

1

ψ(x)

[−~

−2

2m

d2

dx2ψ(x) + V (x)ψ(x)

](4.4)

The l.h.s. depends on t only, and is always equal to the right h.s., whichdepends on x only. Thus, both sides must be equal to the same constantwhich we call E. Hence,

i~1

Φ(t)dΦ(t) = E dt =⇒ Φ(t) = Ae−iEt/~ (4.5)

− ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x) (4.6)

withΨ(x, t) = e−iEt/~ψ(x) (4.7)

Eqn.(4.6) is known as the time–independent, or stationary Schrodingerequation. It depends explicitly on the potential V (x) and so the solutions toEqn.(4.6) vary from one problem to the next. The entire time dependenceof Ψ(x, t) is contained in the complex exponential e−iEt/~; provided that wefind E (via the solution Eqn.(4.6), we do not really need to worry about the

Basic solutions 21

time dependence: this is why a good deal of this section is devoted to solvingEqn.(4.6).

Let us look more closely at one of the essential properties of this type ofseparable solution: the probability density

Ψ(x, t)∗Ψ(x, t) = ψ∗(x)ψ(x) , (4.8)

does not depend on time. Thus, solutions of the form of Eqn.(4.7) are knownas stationary solutions.

This does not mean that the probability density is always time–independent.For instance, if Ψ1(x, t) and Ψ2(x, t) are two solutions of the time–dependentSchrodinger equation having different time dependence:

Ψ1(x, t) = e−iE1t/~ψ1(x) , Ψ2(x, t) = e−iE2t/~ψ2(x) , E1 6= E2 , (4.9)

then the linear combination

Ψ(x, t) = Ψ1(x, t) + Ψ2(x, t) (4.10)

is also a solution of Eqn(4.1) but Ψ(x, t)∗Ψ(x, t) will depend on t.We have explicitly restricted our wave functions to separable functions.

Is this a serious restriction? The answer is no, but must be interpretedcarefully. Not every solution to the Schrodinger equation is separable; forinstance, the solution Ψ(x, t) of Eqn.(4.10) is not separable: it is not possibleto write Ψ(x, t) as a function of x only multiplied by a function of t only.However, it can be shown that every solution of the Schodinger equation canbe written as a sum of separable solutions.

This can be compared to the following statement for ordinary functions ofone variable: not every function is a polynomial function, but (barring patho-logical cases) every function can be expanded as a sum of polynomial (thinkTaylor series). In this statement, ”polynomial” plays the role of ”separablesolution” in the previous paragraph.

It is also important to realize that not all solutions of the Schodingerequation are acceptable as wavefunctions: Eqn.(2.50) is an example of arestriction on the possible solutions to the Schrodinger equation if a solutionis to describe a bound system. More generally, the ”legal” solutions mustsatisfy a certain number of conditions. We impose on the wave functionsthe extra conditions that ψ(x) be everywhere continuous, finite and single–valued.

Basic solutions 22

There are also conditions on the derivative dψ(x)/dx. Consider

− ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x) (4.11)

Push V (x)ψ(x) to the right hand side and integrate from −ε to ε:

− ~2

2m(ψ′(ε)− ψ′(−ε)) =

∫ ε

−ε

(E − V (x))ψ(x)dx , (4.12)

where

ψ′(x) ≡ dψ(x)

dx. (4.13)

Now recall some geometrical properties of integrals of general functionsf(ξ). Let m and M denote, respectively, the min and max of a function f(ξ)over the interval [a, b] . Then

m(b− a) ≤∫ b

a

f(ξ)dξ ≤ M(b− a) (4.14)

Use this with a = −ε, b = ε, ξ = x and f = V −E, and take the limit ε → 0:

limε→0

2εm ≤ limε→0

ε∫

−ε

(V (x)− E)ψ(x)dx ≤ limε→0

2Mε . (4.15)

Thus, unless m → −∞ or M → ∞, we have, by the sandwich theorem ofcalculus

limε→0

(ψ′(ε)− ψ′(−ε)) → 0 , (4.16)

i.e. ψ′ is continuous at x = 0. A similar argument can be used to prove theresult for any value x = x0.

The proof fails if m → −∞ or M →∞, meaning that the product (V (x)−E)ψ(x) blows up somewhere. This can happen if ψ(x) blows up (which willonly happen in some very specific and very rare circumstances), if E becomesinfinite (also rather pathological) or if V (x) has an infinite discontinuitysomewhere. Otherwise, ψ′(x) will be continuous. Curiously enough, infinitediscontinuities in V (x) are rather frequent: the prime example is the infinitelydeep well, where V suddenly goes from 0 to +∞ at the ends of the well. Wewill see other examples of useful infinite discontinuity later.

Having dealt with the continuity of ψ′(x), we now impose that ψ′(x)be finite everywhere: indeed, the calculation of 〈p〉 involves ψ′(x), and itdoesn’t make much sense to have infinite momentum, as this would implyinfinite kinetic energy. Finally, ψ′(x) should also be single–valued.

Basic solutions 23

4.2 Two simple examples

Before we consider problems with complicated potentials, let us gain someexperiences with two very simple but illustrative examples.

4.2.1 V (x) = constant = V0.

By separation of variables, we have

Ψ(x, t) = e−iEt/~ψ(x) (4.17)

− ~2

2m

d2

dx2ψ(x) + V0ψ(x) = Eψ(x) (4.18)

=⇒ − ~2

2m

d2

dx2ψ(x) = (E − V0)ψ(x) (4.19)

Assume E − V0 ≥ 0. Then the general solution is expressed in terms oftwo constants A,B, with

ψ(x) = Aeikx + Be−ikx

~2k2

2m= E − V0 ⇒ k = ±

√2m(E − V0)

~2. (4.20)

Therefore,Ψ(x, t) = Aei(kx−ωt) + Be−i(kx+ωt) . (4.21)

The term is kx− ωt describes a (free) particle moving to the right

〈p〉 ∼∫

dxAe−i(kx−ωt)

(−i~

∂

∂xAei(kx−ωt)

)(4.22)

∼ +~k (4.23)

Furthermore, if t increases, then x must also increase to keep kx−ωtconstant. Similarly, Be−i(kx+ωt) describes a particle moving to the left.

Assume now E − V0 < 0. Then, the solution is of the form

ψ(x) = Aeκx + Be−κx

~2κ2

2m= V0 − E ⇒ κ = ±

√2m(V0 − E)

~2. (4.24)

In this case,Ψ(x, t) = Ae(κx−iωt) + Be−(κx+iωt) . (4.25)

Basic solutions 24

4.2.2 Infinite discontinuity: the infinite potential step

Consider another simple example:

V (x) =

0 if x < 0∞ if x > 0

(4.26)

Suppose some source is placed at x = −∞. Then we have, for x < 0

ψ(x) = Aeikx + Be−ikx , ~k =√

2mE : (4.27)

On the right hand side, we have E − V (x) < 0 and thus

~κ = ±√

2m(E − V ) = ±∞ . (4.28)

Oops!Well, on the right of the barrier, let us write

ψ(x) = Ceikx + De−ikx , (4.29)

and try to determine the two constants C and D. Substituting k = −i∞,we get

ψ(x) = Ce+∞x + De−∞x . (4.30)

Since C and D are constants, they can be determined by choosing anypoint x > 0. Choose any finite x, say x = x0. Then, as ψ(x) must be finiteeverywhere and, in particular, at x0, (since ψ∗(x)ψ(x) must itself be finiteeverywhere), the term e+∞x must be removed from the solution since it is ∞at x0: loosely, this amount to setting C = 0 in the expression for ψ(x). Onecan imagine being nasty and choosing C = e−∞x0 , so that the product Ce∞x

is finite at x0. That doesn’t do a lot of mileage, as this product is infinite forany point to the right of x0: there is no way around it: we have to removethe term in e∞x.

Let us now look at the term De−∞x. For any finite D, this is zero ev-erywhere on the right of the barrier: this term disappears by itself. Again,one can play games and choose D to be, say, D = e∞x0 so that the produceDe−∞x0 is finite at x0. The problem is as before: if x′ < x0, then De−∞x′

would blow up. Hence, we conclude that we must also remove the term con-taining D from the solution: we have effectively argued that, on the right ofthe barrier, ψ(x) ≡ 0.

Basic solutions 25

By continuity at x = 0, we therefore have

ψ(0) = A + B = 0 ⇒ A = −B (4.31)

This gives, for x < 0, the solution

ψ(x) = A(eikx − e−ikx) = A sin kx . (4.32)

Note that, if the barrier is at x = L rather than x = 0, we can simplytranslate the solution so that it becomes

ψ(x) = A sin k(x− L) . (4.33)

There is so particular restriction on k, so all energies are permitted.

General properties of ψ(x) 26

5 General properties of one-dimensional wave

functions

In this section, we will discuss some general properties of wave functions.Some of these properties are formal, others are less formal, but all are usefulto have a better qualitative understanding of wavefunctions.

5.1 The no-degeneracy and oscillatory theorems

Suppose ψ (ξ) is solution to the Schrodinger equation describing the n’thbound state of a system, having energy En, i.e. ψ (ξ) is solution to theSchrodinger equation

d2

dξ2ψ (ξ) +

2m

~2[En − V (ξ)] ψ (ξ) = 0. (5.1)

Then, there is no other function, other than multiple of ψ (ξ) , that has thisenergy.

This is a proof by contradiction. Indeed assume ψ (ξ) and φ (ξ) are bothsolutions, with ψ (ξ) 6= cφ (ξ) for an arbitrary complex constant c. Then:

1

ψ (ξ)

d2

dξ2ψ (ξ) = −2m

~2[En − V (ξ)] =

1

φ (ξ)

d2

dξ2φ (ξ) , (5.2)

which can be rearranged into

0 = φ (ξ)d2

dξ2ψ (ξ)− ψ (ξ)

d2

dξ2φ (ξ) ,

=d

dξ

(φ (ξ)

(d

dξψ (ξ)

)− ψ (ξ)

(d

dξφ (ξ)

)), (5.3)

so that

φ (ξ)

(d

dξψ (ξ)

)− ψ (ξ)

(d

dξφ (ξ)

)= γ, (5.4)

where γ is a constant.However, for the wave functions to be normalizable, we need

limξ→∞

φ (ξ) = limξ→∞

ψ (ξ) = 0. (5.5)


Therefore, evaluating Eqn.(5.4) at infinity gives γ = 0 and thus

φ (ξ)

(d

dξψ (ξ)

)= ψ (ξ)

(d

dξφ (ξ)

), (5.6)

1

ψ (ξ)

d

dξψ (ξ) =

1

φ (ξ)

d

dξφ (ξ) . (5.7)

Integrating gives ψ (ξ) = cφ (ξ) , with c a constant, in contraction with theassumption that ψ (ξ) and φ (ξ) have the same energy but ψ (ξ) 6= cφ (ξ) .

This result is obviously applicable to one-dimensional problems where, forinstance, ξ = x, but it is also applicable to a much larger class of problems.For instance, the radial part of the Schrodinger equation can be written inthe form of a one-dimensional equation:

1

χ (r)

d2

dr2χ (r) +

2m

~2[En` − V (r)]− `(` + 1)

r2, (5.8)

where the radial wave function R(r) is related to the auxiliary wave functionχ (r) by R(r) = χ (r) /r. Thus, Eqn.(5.8) is the Schrodinger equation de-scribing a one-dimensional particle restricted to r ≥ 0 subject to an effectivepotential

Veff (r) = V (r) +~2

2m

`(` + 1)

r2, (5.9)

with the extra term known as the ”centrifugal term”, which depends on theangular quantum number `. Because the effective potential depends on `, wehave added this extra subscript to En`. Our result applies to χ (r) and thusto R(r), meaning that, for specified `, only one solution χ (r) will have energyEn`. Note that, since Veff depends on `, there may be more than one wavefunction that will have energy E = En`, but these wave functions will belabeled by different values of `, i.e. these wave functions will be solutions fordifferent effective potentials. (This is precisely the situation of the hydrogenatom.)

Next, we will prove what is called the oscillatory theorem. For conve-nience of notation, recall

H = − ~2

2m

d2

dξ2+ V (ξ). (5.10)

V (ξ) may or may not be an effective potential: this is irrelevant to ourargument.


The oscillatory theorem states that, if

Hψ1(ξ) = E1ψ1(ξ) , Hψ2(ξ) = E2ψ2(ξ) , E1 < E2, (5.11)

then ψ2(ξ) has more zeroes (nodes) than ψ1(ξ).To show this, manipulate the Schrodinger equation so that

1

ψ1(ξ)

~2

2m

d2

dξ2ψ1(ξ) + E1 = V (ξ) =

1

ψ2(ξ)

~2

2m

d2

dξ2ψ2(ξ) + E2, (5.12)

which one then reorganizes as

ψ2(ξ)d2

dξ2ψ1(ξ)− ψ1(ξ)

d2

dξ2ψ2(ξ) =

d

dξ

(ψ2(ξ)

d

dξψ1(ξ)− ψ1(ξ)

d

dξψ2(ξ)

)

=~2

2m(E2 − E1) ψ1(ξ)ψ2(ξ). (5.13)

Integrating from ξ = a to ξ = b, we find

~2

2m(E2 − E1)

∫ b

a

dξψ1(ξ)ψ2(ξ)

=

(ψ2(b)

d

dξψ1(b)− ψ1(b)

d

dξψ2(b)

)

−(

ψ2(a)d

dξψ1(a)− ψ1(a)

d

dξψ2(a)

)(5.14)

If one chooses the points a, b to be consecutive zeroes of ψ1, and if ψ1 iscontinuous, then ψ1does not change sign over the interval [a, b] . We canassume, without loss of generality, that ψ1 is positive on this interval, so thatψ′1(a) ≥ 0 and ψ′1(b) ≤ 0. Then, using these with ψ1(a) = ψ1(b) = 0 inEqn.(5.14), we obtain

~2

2m(E2 − E1)

∫ b

a

dξψ1(ξ)ψ2(ξ) = ψ2(b)d

dξψ1(b)− ψ2(a)

d

dξψ1(a). (5.15)

Assume now that ψ2(ξ) is always positive on [a, b] . Then the integrandis positive, so the integral will be positive (the area under a positive curveis necessarily positive!) On the right hand side, however, the first term isnegative by virtue of ψ′1(b) ≤ 0 while the second is positive. Overall, the


right hand side is therefore negative, while the left side is negative, a con-tradiction : ψ2(ξ) cannot be always positive on [a, b] , meaning that betweentwo consecutives zeroes of ψ1(ξ), the function ψ2(ξ) must itself go throughzero. In other words, ψ2(ξ) has more zeroes than ψ1(ξ). This proves thetheorem.

The theorem is beautifully illustrated with the harmonic oscillator wavefunction. By inspection, the wave function for the lowest energy has nozeroes, the wave functions for the next lowest energies have one, two, three,etc. nodes, the number of nodes increasing with the energy. It is also easilyverified that, between two nodes of a solution, another solution of higherenergy will go through zero at least once.

The theorem can also be verified for hydrogen wave functions, althoughit must be done a little more carefully: for fixed ` (since the effective one-dimensional potential depends on ` and we must compare solutions in thesame potential), solutions with higher energies have a larger number of zeroes.

5.2 Qualitative features of wave functions.

Is there a way to understand ”physically” the contents of the oscillatorytheorem? Consider the geometrical interpretation of

d2

dξ2ψ (ξ) = −2m

~2[En − V (ξ)] ψ (ξ) . (5.16)

at a point where En − V (ξ) ≥ 0. This is the ”classical” region where thetotal energy is greater than the potential energy. Obviously, this must holdin classical physics, as the kinetic energy if always positive and the totalenergy is the sum of kinetic and potential energy. The left hand side ofEqn.(5.16) is the ”curvature” of the function ψ (ξ) . Evaluated at a pointξ0 where ψ (ξ0) is positive, we see that the curvature must be negative, i.e.the function ψ (ξ) must curve downwards when it is positive. Likewise, ifψ (ξ0) is negative, then the curvative is positive, so that ψ (ξ) must curveupwards when it is negative. This is typical of oscillatory functions, and canbe verified using the solutions of the infinite well or the harmonic oscillator.

Note that, for points where En − V (ξ) < 0, the curvature of ψ (ξ) ispositive for positive ψ (ξ) and negative for negative ψ (ξ) , a behaviour typicalof the exponential function.

Next, the magnitude of the curvature at ξ0 is, basically, given by En −V (ξ0). If En − V (ξ0) is large, then the function has a large curvature at


x

x

Figure 1: Left: In the classically allowed region, a positive wave function musthave a negative curvature. Right: in the same region, a negative wave functionmust have a positive curvature.

that point; if En − V (ξ0) is small, ψ (ξ) will be flatter than elsewhere atξ0. In a harmonic oscillator, for instance, the curvature of the solution isalways maximal at the center of the well ξ0 = 0, and this is the point whereEn − V (ξ0) is largest.

Finally, since

− ~2

2m

d2

dξ2∼ p2

2m(5.17)

is interpreted as the kinetic energy, we expect that intervals where p2/2mis large correspond to intervals where the particles moves ”fast”, whereasintervals where p2/2m is small correspond to intervals where the particlesmoves ”slowly”. Hence, the probability of finding the particle in an intervalwhere it is moving fast should be less than the probability of finding theparticle in an interval where it moves slowly. Where therefore expect thatψ∗ (ξ) ψ (ξ) and thus ψ (ξ) itself will have a smaller amplitude in the ”fast”region than in the ”slow” region. Again, this can be verified for the knownwave functions.

Simple bound states 31

6 Simple bound state problems in one dimen-

sion

6.1 The infinite well

Suppose next that we place two infinite steps at x = 0 and x = L, so that

V (x) =

∞ , if x ≤ 0,0 , if 0 < x < L,∞ , if x ≥ L .

(6.1)

Inside this infinite well, we have

ψ(x) = Aeikx + Be−ikx (6.2)

At x = 0, we have ψ(0) = 0 since the potenial step is infinite there:

ψ(0) = 0 = A + B ⇒ A = −B ⇒ ψ(x) = A sin kx (6.3)

Next, at x = L , we also need to have ψ(L) = 0 :

ψ(L) = A sin kL = 0 (6.4)

Clearly if A = 0 , then we have accomplished nothing because ψ(x) = 0everywhere: ψ∗(x)ψ(x) = 0 everywhere and we are describing nothing. Thus,the only way to satisfy simultaneously ψ(0) = ψ(L) = 0 is to choose thosevalues of k for which

sin kL = 0 ⇒ kn =nπ

L(6.5)

The subscript n on kn has been introduced to distinguish between the variousdiscrete values kn that will make kn satisfy sin knL = 0.

There are two things to note. First, because the potential steps are∞ , we did not use any conditions on the continuity of dψ

dx; in fact dψ

dxis

discontinuous at x = 0 or x = L since V (x) has an infinite discontinuity atx = 0 and x = L. The second and most crucial observation is that only adiscrete (or quantized) set of values of k can make this work. Since ~k = p,we have

~kn = pn =√

2mEn . (6.6)


Thus we find that the possible values of E are also quantized, i.e. they forma discrete set indexed by the quantum number n :

~kn =~nπ

L=

√2mEn ⇒ En =

~2n2π2

2mL2, kn =

nπ

L, n ∈ Z+ (6.7)

We can distinguish the various acceptable solutions by their energies, orby the label n. Thus, the full time–dependent solution with energy En is

Ψn(x, t) = A e−iEnt/~ sin(knx) (6.8)

The last job to do is to normalize ψn(x) and, by the same token, normalizeΨn(x, t). This is easily done

1 =

∫ ∞

−∞dxψ∗n(x)ψn(x) =

∫ L

0

dxψ∗n(x)ψn(x) , (6.9)

since ψn(x) = 0 for x < 0 and x > L Thus:

1 = A · A∗∫ L

0

dx sin2(knx) (6.10)

= A · A∗ ·(

+1

2x− 1

4kn

sin 2knx

)L

0

(6.11)

Now recall that knL = nπ, sin 2nπ = 0 Thus, we end up with

1 = A · A∗ · L

2. (6.12)

Choosing A =√

2L

will normalized the wave functions, so that we have

ψn(x) =

√2

Lsin knx (6.13)

Ψn(x, t) =

√2

Le−i~n2π2t/2mL2

sin knx (6.14)

6.2 Detour: Symmetric potentials and their solutions

Some potentials have the property that V (−x) = V (x). These are known assymmetric potentials. Furthermore, although some potential are not explic-itly symmetric, it is often possible to “trick” the problem into believing that


the potential is symmetric. For instance, the infinite well of section 6.1 can betransformed into a symmetric well under the change of variables ξ = x−L/2,so that the potential well now extends from −ξ to +ξ. Other examples ofsymmetric potentials include the harmonic oscillator V (x) = 1

2kx2 of section

6.4 and the square well of finite depth of section 6.3.We now prove the following important claim. Suppose the potential is

symmetric: V (x) = V (−x). Then it is always possible to choose the solu-tions ψ(x) of the corresponding time–independent Schrodinger equation tobe either even functions, ψ(−x) = ψ(x), or odd functions, ψ(−x) = −ψ(x).

To prove this, suppose ψ(x) is solution to

− ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x) , (6.15)

and make the change x → −x to get

− ~2

2m

d2

dx2ψ(−x) + V (−x)ψ(−x) = Eψ(−x) . (6.16)

Note that, under this change, d2/d(−x)2 = d2/dx2.As the potential is symmetric, V (x) = V (−x), we see that ψ(−x) is

also solution to the time–independent Schrodinger equation with the sameenergy. Now, recall that any linear combination of two solutions of the time–independent Schrodinger equation with the same energy is also a solution ofthe time–independent Schrodinger equation. Thus, we can combine ψ(x) andψ(−x) into two new linear combinations

ψ+(x) =1√2

(ψ(x) + ψ(−x)) , ψ−(x) =1√2

(ψ(x)− ψ(−x)) . (6.17)

which are also solutions to the Schrodinger equation for this potential. Thesolutions ψ±(x) also have energy E. The functions ψ+(x) and ψ−(x) are,respectively, even and odd. This proves the claim.

What this doesn’t mean is that, if there is an even solution for a givenenergy E, there is also an odd solution with that energy. In fact, we willshow that this is never the case in general. What it does say is that, if youset up your problem so that V (x) = V (−x), then you can find solutions thatare even or odd. This will be clarified by examples below.


6.3 The finite well

6.3.1 Obtaining a transcendental equation

Consider a well of finite depth, described by a symmetric potential of theform

V (x) =

0 if x < −L ,−V0 if − L ≤ x ≤ L ,0 if x > L .

(6.18)

where V0 > 0. This is a symmetric potential, so we expect that, inside thewell at least, solutions will be of the form ψ(x) = Aeikx + Be−ikx. The evenand odd combinations of these are, in fact, ψ(x) = A cos(kx) and ψ(x) =B sin(kx), respectively. The bound states will have energies < 0, so we writeE = −ε, with E < 0 and ε > 0.

Let us look for even solutions. Once again, we have three regions, so wehave three regions for the solutions:

ψ(x) =

Ceκx if x < −L ,A cos(kx) if − L ≤ x ≤ L ,Ce−κx if x > L .

(6.19)

As always,

k =

√2m(E − V )

~2=

√2m(V0 − ε)

~2,

κ =

√2m(V − E)

~2=

√2mε

~2. (6.20)

In writing the form of ψ(x), we have explicitly used the symmetry of V (x) toinstantly reduce the number of unknown amplitudes: the amplitude of thedecreasing exponentials must be identical by symmetry.

We now proceed along the familiar lines. By continuity of ψ(x) at x = L,and by continuity of dψ/dx at x = L, we find

A cos(kL) = Ce−κL ,

−kA sin(kL) = −κCe−κL . (6.21)

Taking now the ratio of these equation to eliminate A and B, we find, aftermultiplication by L,

kL tan(kL) = κL . (6.22)


It is not possible to solve the equation algebraically: it must be solved nu-merically. To do this, we need to write explictly Eqn.(6.22) in the form

√2mL2V0

~2

(1− ε

V0

)tan

[√2mL2V0

~2

(1− ε

V0

) ]=

√2mL2V0

~2

ε

V0

, (6.23)

and use the variables

ξ =

√2mL2V0

~2, z =

ε

V0

≤ 1 . (6.24)

so that Eqn.(6.22) becomes

ξ√

1− z tan(ξ√

1− z)

= ξ√

z ,

tan(ξ√

1− z)

=

√z

1− z. (6.25)

Before we actually solve this, it is worth observing that the parameter ξhas no units and is completely determined by the properties of the potential(its depth V0, its half-width L) and by the mass of the “particle” trapped inthis potential.

For instance, in problems where the particle is an electron, typical energiesare in eV and typical lenghts are measured in units of the Bohr radius a0.Thus, writing L = ` a0, we have typical values of ξ given in the table below(using the mass of the electron):

V0 (in eV) ` (no units) ξ (no units)1 1 0.2708

2 0.54173 0.8125

3 1 0.46915 1 0.605610 1 0.8565

For nuclear–type problems, with masses of the other of proton or neutronmasses, V0 is most often measured in MeV and L in F = 10−15m. Typicalvalues of ξ are comparable if somewhat a little lower.

Returning now to Eqn.(6.25), we solve using a graphical method. Let usassume, for the sake of argument, that ξ = 4. On the same graph, plot of


the left hand side and the right hand side of Eqn.(6.25). You get somethinglike Fig. 2.

The intersections of the two curves are the points where tan(4√

1− z) =√z/(1− z) and therefore represent the values of z that are solutions to

Eqn.(6.25). Note that, for plotting purposes, 0 < z ≤ 1.

0.2 0.4 0.6 0.8 1

1

2

3

4

5

Figure 2: The graphical solutions to Eqn.(6.25) with ξ = 4.

First, we see that there are two intersection points and thus two solutions.They can be located precisely by numerically solving Eqn.(6.25). This gives:z2 = 0.192 , z1 = 0.902.

Thus, assuming that we know the depth V0 of the potential, we canrecover ε1 = z1 × V0 = 0.902 × V0 and so forth. Given the energies, wecan recover k1, k2, κ1 and κ2. Using Eqn.(6.21), we can obtain the ratiosC1/A1 and C2/A2 in our two solutions ψ1(x) and ψ2(x). Finally, using thenormalization condition, we can determine, say, A1 and A2. This closes theproblem of solving for even wave functions.

Now, the same tactics must be applied to odd wave functions. This time,however, our solution looks like

ψ(x) =

−Ceκx if x < −L ,A sin(kx) if − L ≤ x ≤ L ,Ce−κx if x > L .

(6.26)


Note the way in which the sign of C changes with the region, so as to makethe wave function odd.

It is quite easy to work out that, after the same manipulations as before,we eventually get

tan(ξ√

1− z)

= −√

(1− z)

z. (6.27)

–4

–3

–2

–1

00.2 0.4 0.6 0.8 1

z

Figure 3: The graphical solutions to Eqn.(6.27) with ξ = 4.

Clearly, there is only one bound state, with energy obtained from z3 =0.617. It is worth observing that the lowest energy for the even solution islower than the lowest energy of the odd solution.

6.3.2 Example: depth of the deuteron well

As a first approximation, the force between a neutron and a proton can bedescribed by an attractive square-well potential, given in Eqn.(G.132) andillustrated in Fig. 33.

V (r) =

∞ for r ≤ 0 ,−V0 for r ≤ R0 ,0 for r > R0 .

(6.28)

Here, r is the relative radius between the two nucleons, and V0 > 0.


V=0

r=R o

V=−Vo MeV

Figure 4: An approximation to the nuclear well.

The range R0 of the nuclear force is known to be of the order of theCompton wavelength of the pion:

R0 =~

mπc, (6.29)

where mπ = 139MeV/c2 is the mass of the pionKnowing that the proton–neutron has a only one bound state with energy

−2.22MeV, and knowing this state has angular momentum ` = 0, we canfind the depth of the potential V0 as follows.

Because the region r ≤ 0 is inaccessible, the wave function for a boundstate in this potential must be 0 at r ≤ 0. Thus, we have

ψ(r) =

0 if r ≤ 0 ,A sin kr , if 0 < r ≤ R0 ,Be−κr if r > R0 .

(6.30)

This form in each region is the only one that will satisfy the boundary con-ditions: ψ(0) = 0 and ψ(∞) → 0.

The energy of the bound state is −2.22MeV , which we write −ε, withε = 2.22MeV . Inside the well, the time–independent Schrodinger equationis

− ~2

2m

d2ψ(r)

dr2− V0ψ(r) = −εψ(r) , (6.31)

which can be written more simply as

~2

2m

d2ψ(r)

dr2= − (V0 − ε) ψ(r) . (6.32)


Clearly, V0 > ε, i.e. the well must be deeper than 2.22MeV to accommodatea bound state of that energy, so the r.h.s. of Eq.(6.32) is negative, justifyingthe oscillatory form of ψ(r) inside the well, with k2 = 2µ(V0 − ε)/~2 and µthe reduced mass of the system. The cosine term cannot appear as it doesnot satisfy the boundary condition ψ(0) = 0.

For r > R0, outside the well in the classically forbidden region, we have

~2

2m

d2ψ(r)

dr2= εψ(r) , (6.33)

which must be again of the form suggested for the wavefunction to be nor-malizable. Here, κ2 = 2µε/~2.

The continuity of the wavefunction at r = 0 is guaranteed by the choiceof the sine term for the wavefunction inside the well. Because there is aninfinite jump the potential at r = 0, there is no condition on the derivativeof ψ at that point.

The continuity of the wavefunction and of its derivative at r = R0 yields

A sin(kR0) = Be−κR0 , (6.34)

kA cos(kR0) = −κBe−κR0 . (6.35)

We can divide the first equation by the second to eliminate the factors Aand B and obtain

tan(kR0) = −k

κ, (6.36)

tan

(√2µ(V0 − ε)R2

0

~2

)= −

√2µ(V0−ε)

~2√2µε/~2

, (6.37)

= −√

V0 − ε

ε(6.38)

We know ε and we are looking for V0. So, to solve this equation, we setz = V0/ε and rearrange to

tan

√2µε(V0

ε− 1)R2

0

~2

= −

√V0

ε− 1 , (6.39)

tan

(√2µεR2

0

~2

√z − 1

)= −√z − 1 . (6.40)


We now compute the numerical factor

√2µεR2

0

~2=

√2µ ε

~2

~2

m2πc2

=

√2µ ε

m2πc2

=

√2µ c2ε

m2πc4

. (6.41)

The reduced mass is given by 1/µ = 1/mp+1/mn. Taking mass of the protonand the neutron to be equal to 940MeV/c2, we get µ c2 = 1

2940. Thus, we

have √2µε R2

0

~2=

√940× 2.22

1392= 0.329 (6.42)

and thus, we are looking for solutions to

tan(0.329

√z − 1

)= −√z − 1 , (6.43)

10 20 30 40

-10

-8

-6

-4

-2

Figure 5: The graphical intersection of the curves tan(0.329√

z − 1) and −√z − 1.

A plot of this shows there is a single intersection near z = 30. A refinedsearch yields z = 29.49, and thus V0 = ε×29.49 = 65.5MeV . This completesthe calculation of V0.

6.3.3 Example: The ammonia molecule

The ammonia molecule, NH3, is in the form of pyramid, with the nitrogenatom at vertex and three hydrogen atoms at the base. The nitrogen atom


may be at one of two symmetric equilibrium positions: “above” or “below”the plane of the hydrogen atoms. The molecule is illustrated in Fig.G. (Bothfigures taken from Wikipedia at http://en.wikipedia.org/wiki/Ammonia ).

Figure 6: The ammonia molecule, showing the nitrogen atom in relation to thethree planar hydrogen atoms.

The potential energy for the motion of the nitrogen atom is illustratedin Fig.7 as the smooth “W”–shape curve. The figure also illustrates theapproximation of this potential by a infinite one–dimensional square wellwith a central bump of height V0 = 0.245eV. The infinite walls are locatedat ±b = ±0.5548× 10−10m; the central bump extends from −a to +a, witha = 0.2298× 10−10m.

V

a

b

o

b−a

Figure 7: The potential for the nitrogen atom of the ammonia molecule. The exactpotential (wavy line) is approximated by an infinite square well with a bump inthe middle (heavy line).

We will look for the lowest energy states of this molecule. The energy will


be denoted E. It is measured from the bottom of the well and thus positive.Let x denotes the position of the nitrogen atom. If x > 0, then the

nitrogen atom is “above” the hydrogen plane, while the nitrogen atom isbelow this plane when x < 0.

We expect on general grounds that an even wave function will have lesscurvature than the odd wavefunction, so we will assume our wavefunction tobe even. Our candidate wavefunction is

ψ(x) =

B sin(k(b + x)) if − b ≤ x ≤ a ,

A (eκx + e−κx) if − a < x < a ,

B sin(k(b− x)) if a ≤ x ≤ b .

(6.44)

It is certainly even under the transformation x → −x, and it satisfies theboundary condition that ψ(±b) = 0 because the wavefunction outside thewell must be 0. Both the increasing and decreasing exponentials appear asboth as solution of the Schrodinger equation when one assumes E < V0. (Ifit turns out there are no solutions with E < V0, this assumption will have todropped and the form of the wavefunction in the region of the central bumpwill have to be modified.)

As always, we set

κ =

√2m(V0 − E)

~2, k =

√2mE

~2. (6.45)

The continuity of ψ and dψ/dx at x = a gives two equations:

A(eκa + e−κa

)= B sin(k(b− a)) , (6.46)

Aκ(eκa − e−κa

)= −B k cos(k(b− a)) . (6.47)

The continuity of ψ and its derivative at x = −a gives the same two equations.Dividing these so as to eliminate A and B, we obtain the consistency

equation:1

κ

(eκa + e−κa

eκa − e−κa

)= −1

ktan(k(b− a)) . (6.48)

To bring this to a solvable form, we set

ξ =

√2mV0 a2

~2= 3.91 , z =

E

V0

, (6.49)


where

mNH3 : 2300 MeV/c2 ~c : 197.3 MeV× 10−15mb : 0.5548× 10−10m a : 0.2298× 10−10m

2mV0 a2

~2: 15.3182

√2mV0 a2

~2: 3.9138

−1 + b/a : 1.4143

and so obtain the transcendental equation

− (e3.91√

1−z + e−3.91√

1−z)

(e3.91√

1−z − e−3.91√

1−z)√

1− z=

tan(5.54√

z)√z

. (6.50)

The solution is obtained graphically by using the graph of Fig.8.

0.2 0.4 0.6 0.8 1

-3

-2.5

-2

-1.5

-1

-0.5

Figure 8: Graphical solutions to − (e3.91√

1−z + e−3.91√

1−z)(e3.91

√1−z − e−3.91

√1−z)

√1− z

=tan(5.54

√z)√

z.

Visibly, there are two solutions, at z = 0.228 and z = 0.844. Thus, fromz = E/V0, we see that the lowest energy state is located at E = 0.228×V0 =0.056eV. The second even level is located at E = 0.206eV. There is, of course,an odd level with energy in between those of the two even levels. There aremore even levels, but their energy is above the bump, so the form of thewavefunction must be modified before a search for these solutions can occur.

6.4 The harmonic oscillator

The harmonic oscillator is important as many potentials can be approximatedby a harmonic potential for sufficiently small displacements. For a harmonic


oscillator, we haveV (x) = 1

2kx2 . (6.51)

The time–independent Schrodinger equation for ψ(x) is now

− ~2

2m

d2

dx2ψ(x) + 1

2kx2ψ(x) = Eψ(x) . (6.52)

6.4.1 Solution

To simplify the notation, let us introduce

ω =

√k

m, ξ =

√mω

~x , λ =

2E

~ω, (6.53)

so that we can rewrite the differential equation in x as

d2

dξ2ψ(ξ) + (λ− ξ2)ψ(ξ) = 0 . (6.54)

To proceed, we examine the boundary conditions of the problem. V (x)(or V (ξ)) is continuous everywhere, so we get nothing from that. On theother hand, ψ(ξ) must be finite everywhere, and, in particular, ψ(x) mustremain finite at ξ = ±∞ . Thus, for ξ →∞, we can write

d2

dξ2ψ(ξ)− ξ2ψ(ξ) ≈ 0 , ξ2 À λ . (6.55)

This is an approximate equality, and we can verify that solutions of the form

ψ±(ξ) = ξpe±ξ2/2 (6.56)

are approximately solutions provided that p is finite. To show this, we con-centrate on ψ−(ξ) and work out

d

dξξpe−ξ2/2 = pξp−1 e−ξ2/2 − ξ ξpe−ξ2/2 (6.57)

d2

dξ2ξp e−ξ2/2 = p(p− 1) ξp−2 e−ξ2/2 − p ξp−1 ξ e−ξ2/2

−(p + 1)ξp e−ξ2/2 + ξ2 ξp e−ξ2/2 (6.58)

= ξp+2 e−ξ2/2

(p(p− 1)

ξ4− (2p + 1)

ξ2+ 1

)(6.59)

' ξ2ξpe−ξ2/2 (6.60)


for ξ →∞ ifp/ξ2 → 0 . (6.61)

The same argument holds for ξpe+ξ2/2 . However, we can throw out thissolution because the wave function must remain finite everywhere, and ξpeξ2/2

diverges for large ξ.Since we know approximately how ψ(ξ) behaves for large ξ, let us write

ψ(ξ) = e−ξ2/2H(ξ) (6.62)

and try to determine H(ξ).Plugging this form for ψ(x) into

d2

dξ2ψ(ξ) + (λ− ξ2)ψ(ξ) = 0 (6.63)

gives:

d

dξψ(ξ) =

d

dξ

(e−ξ2/2H(ξ)

)

= −ξe−ξ2/2H(ξ) + e−ξ2/2H ′(ξ) (6.64)

d2

dξ2ψ(ξ) =

d

dξ

(−ξe−ξ2/2H(ξ) + e−ξ2/2H ′(ξ)

)(6.65)

= −e−ξ2/2H(ξ) + ξ2e−ξ2/2H(ξ)− ξe−ξ2/2H ′(ξ)

−ξe−ξ2/2H ′(ξ) + e−ξ2/2H ′′(ξ) (6.66)

and thus,

[ξ2H(ξ)− 2ξH ′(ξ) + H ′′(ξ) + (λ− 1)H(ξ)− ξ2H(ξ)]e−ξ2/2 = 0 (6.67)

which can be simplified to

H ′′(ξ)− 2ξH ′(ξ) + (λ− 1)H(ξ) = 0 . (6.68)

Now the equationy′′ − 2ξy′ + 2ny = 0 (6.69)

is known as Hermite’s differential equation. By the condition of Eqn.(6.61),we are restricted to solutions of finite degree; these are the so–called Hermite


polynomials Hn(ξ):

H0(ξ) = 1 H1(ξ) = 2ξ

H2(ξ) = 4ξ2 − 2 H3(ξ) = 8ξ3 − 12ξ

H4(ξ) = 16ξ4 − 48ξ2 + 12 H5(ξ) = 32ξ5 − 160x3 + 120

(6.70)

It is shown in Appendix E how to solve explicitly Eqn.(6.69). It is alsoshown in Eqs. (E.16) and (E.17) that, unless the factor n in Eq.(6.69) is apositive integer, the solution to Eq.(6.69) eventually grows like e2ξ2

and thusmakes the whole wavefunction

ψ(x) ∼ e−ξ2/2e2ξ2

(6.71)

divergent. Thus, we rediscover the restriction that H(ξ) be a polynomial offinite degree in ξ.

With the identification 2n = λ− 1 , we have

2n =2En

~ω− 1 ⇒ En = (n + 1

2)~ω , (6.72)

with n a positive integer.Going back to the old variables, we obtain the n’th allowed solutions

ψn(x) = NnHn

(√mω

~x

)e−mωx2/2~ , (6.73)

Ψn(x, t) = ψn(x)e−i(n+ 12)ωt (6.74)

where Nn is a normalization constant: it is tricky to work out but one ulti-mately finds:

Nn =(mω

π~

)1/4

× 1√2nn!

. (6.75)

The first few solutions are plotted in Fig. 6.4.1, with the wavefunctionon the left and the corresponding probability density on the right.


-4 -2 2 4

-0.6

-0.4

-0.2

0.2

0.4

0.6

-4 -2 2 4

0.05

0.1

0.15

0.2

0.25

0.3

0.35

-4 -2 2 4

-0.4

-0.2

0.2

0.4

0.6

-4 -2 2 4

0.05

0.1

0.15

0.2

0.25

0.3

0.35

-4 -2 2 4

-0.6

-0.4

-0.2

0.2

0.4

0.6

-4 -2 2 4

0.1

0.2

0.3

0.4

-4 -2 2 4

0.1

0.2

0.3

0.4

0.5

0.6

0.7

-4 -2 2 4

0.1

0.2

0.3

0.4

0.5

Figure 9: The harmonic oscillator wave functions for n = 0, 1, 2, 3 and their re-spective probability densities. Note the eveness of the solutions.


6.4.2 The harmonic oscillator wavefunctions

Here’s an explicit list of the first few harmonic oscillator wavefunctions:

n ψn(x)

0(mω

~π

)1/4

e−mωx2/2~

1(mω

~

)3/4 √2

π1/4 x e−mωx2/2~

2(mω

~π

)1/4 (−1+2x2 mω~ )√

2π1/4 e−mωx2/2~

3(mω

~

)3/4

x(−3+2x2 mω

~ )√3π1/4 e−mωx2/2~

4(mω

~π

)1/4(3−12x2 mω

~ +4x4(mω~ )

2)

2√

6π1/4 e−mωx2/2~

(6.76)

6.5 δ(x)-well potential

We now introduce one final type of potential: the so–called δ–well. TheDirac δ(x) function will be discussed at greater length in Sec.10.6. For themoment, it is useful to think of the δ–well as a infinitely deep well that isinfinitely thin.

First, imagine the function f(x, ε)

f(x, ε) =

0 if x < −ε ,1

εif − ε ≤ x ≤ ε ,

0 if x > ε .

(6.77)

in the limit ε → 0. Figure 6.5 gives an idea of f(x, ε) for ε = 10, 100 and1000. One can see that the function becomes a very very thin spike withdecreasing ε.

In the limit where ε → 0, the spike becomes a Dirac δ, i.e.

δ(x) = limε→0

f(x, ε) , (6.78)


-0.4 -0.2 0.2 0.4

2

4

6

8

10

-0.4 -0.2 0.2 0.4

20

40

60

80

100

-0.4 -0.2 0.2 0.4

200

400

600

800

1000

Figure 10: A sequence of f(x, ε) converging to a δ–function. Here, we havef(x, 1/10), f(x, 1/100), f(x, 1/1000).

where f(x, ε) is defined in Eqn.(6.77).Thus, we consider a potential of the form

V (x) = −Ω δ(x) , (6.79)

where Ω > 0. This is a discontinuous potential well with no width.The Schrodinger equation for this potential takes the form

− ~2

2m

d2

dx2ψ(x)− Ωδ(x)ψ(x) = Eψ(x) . (6.80)

Because this is an infinite step, there is a discontinuity in ψ′(x):

ψ′(0+)− ψ′(0−) = −2m

~2Ωψ(0) . (6.81)

Since V (x) = 0 for x 6= 0, and V (0) = −Ω∞, the expression for wavefunctionon the left and on right of the potential is

ψ(x) =

Ae−κx + Beκx , x < 0 ,Ce−κx + Deκx , x > 0 ,

(6.82)

where, as always, κ =√

2mε~2 , ε = −E > 0.

Now, if ψ(x) is to remain finite at x = ±∞, we must have A = 0 andD = 0. The condition of continuity of ψ(x) at x = 0 yields

B = C . (6.83)

Using this in Eqn.(6.81) for the discontinuity of ψ′(x) provides another equa-tion: when evaluated at the discontinuity, the jump in ψ′(x) gives

−Cκ− Cκ = −2m

~2Ω C , (6.84)


with solution

κ =mΩ

~2. (6.85)

This is the only possible solution for κ, and thus there is only one boundstate with energy

κ2 =m2Ω2

~4= −2mE

~2⇒ E = −mΩ2

2~2. (6.86)

The δ–well is useful in many approximations where it replaces the exactpotential due to atomic nucleus. Examples include the Kronig–Penney modelin solid state physics and simple models bonding and antibonding in the H2

molecule.

Independent modes 51

7 Multidimensional independent-mode prob-

lems

We now consider a class of multidimensional problems that can be reducedto a number of independent problems. The results are applicable to someproblems describing a single particles in two or three dimensions, to someproblems dealing with many particles in one or several dimensions, or tocombinations of several particles in more than one dimensions.

Suppose that one can find variables ξ, η so that the Schrodinger equationcan be written as

i~∂

∂tΨ(ξ, η, t) =

(− ~2

2mξ

∂2

∂ξ2+ V (ξ)− ~2

2mη

∂2

∂η2+ U(η)

)Ψ(ξ, η, t). (7.1)

Under the usual identifications

i~∂

∂t↔ E , −i~

∂

∂ξ↔ pξ , −i~

∂

∂η↔ pη, (7.2)

Eqn.(7.1) is nothing than a statement about the total energy of the system,which is now equal to the sum of total energies for the variables ξ, η. Notethat the masses mξ and mη need not be the same.

Assuming Ψ(ξ, η, t) = Φ(t)ψ(ξ, η) leads to the separation of the timecomponent:

Ψ(ξ, η, t) = e−iEt/~ψ(ξ, η) , (7.3)

with E the separation constant and where the time–independent wave func-tion ψ(ξ, η) satisfies

(− ~2

2mξ

∂2

∂ξ2+ V (ξ)− ~2

2mη

∂2

∂η2+ U(η)

)ψ(ξ, η) = Eψ(ξ, η). (7.4)

To convert this last equation from a partial to a pair of ordinary differentialequations, we look once again for a separable solution:

ψ(ξ, η) = X (ξ) N (η) . (7.5)

This yields

N(η)

(− ~2

2mξ

d2

dξ2X (ξ) + V (ξ)X (ξ)

)

+X(ξ)

(− ~2

2mη

d2

dη2N (η) + U(η)N (η)

)= E X(ξ) N (η) . (7.6)


Dividing through by X(ξ) N(η) and reorganizing, we obtain

1

X(ξ)

(− ~2

2mξ

d2

dξ2X (ξ) + V (ξ)X (ξ)

)

= E − 1

N(η)

(− ~2

2mη

d2

dη2N (η) + U(η)N (η)

). (7.7)

Since both sides must be equal for all values of ξ and all values of η, theymust be both equal to a common constant Eξ. Denoting Eη = E − Eξ andrearranging yields

(− ~2

2mξ

d2

dξ2+ V (ξ)

)X (ξ) = EξX (ξ) , (7.8)

(− ~2

2mη

d2

dη2+ U(η)

)N (η) = EηN (η) , (7.9)

with the constraint Eξ + Eη = E. Each of the previous two equations canthen be solved separately, producing solutions of the form

Ψ(ξ, η, t) = e−iEt/~X (ξ) N (η) . (7.10)

Note that we have not specified if ξ, η refer to a single particle in two dimen-sions or to two particles in one dimension. In the former case, one couldidentify ξ = x, η = y. In the latter, one would have ξ = x1 for the firstparticle and η = x2 for the second.

This technique is directly applicable to, for instance, the problem of find-ing the energy levels of a single particle in a two-dimensional infinite wellhaving length Lx and Ly in the x and y directions, respectively. Clearlythen, Eqn.(7.8), with V = 0 and the boundary condition X (0) = X(Lx) = 0will give

X(x) =

√2

Lx

sin

(nxπx

Lx

)(7.11)

while Eqn.(7.9), with U = 0 and the boundary conditions N (0) = N(Ly) =0, will give

N(y) =

√2

Ly

sin

(nyπx

Ly

). (7.12)


Furthermore,

Enx = ~2n2xπ

2/(2mL2

x

),

Eny = ~2n2yπ

2/(2mL2

y

),

E = Enx + Eny =~2π2

2m

(n2

x

L2x

+n2

y

L2y

). (7.13)

Another example is the isotropic two-dimensional harmonic oscillator,with Hamiltonian

H =1

2m

(p2

x + p2y

)+

1

2mω2(x2 + y2), (7.14)

which can be separated in two one-dimensional oscillators, one along x andthe other along y. The possible energies are

E =(nx + 1

2

)~ω +

(ny + 1

2

)~ω

= (N + 1) ~ω , (7.15)

where N = nx + ny.

The radial equation 54

8 Equivalent one–dimensional problems: the

radial equation

We now consider a class of problems for which we can use our results fromone dimensions. When the potential depends only on the combinaison r =√

x2 + y2 + z2 of the x, y, z variables in three dimensions, a change of co-ordinates from Cartesian to spherical allows one to seek solutions of thetime–independent Schodinger equation in the form of a product

ψ(x, y, z) 7→ ψ(r, θ, φ) = R(r)f(θ, φ) , (8.1)

where R(r) and f(θ, φ) are solution to the radial and angular Schrodingerequations, respectively.

The radial solutions R(r) can be considered as one–dimensional functionsprovided that the radius r is properly restricted to non–negative values.

8.1 Separation of variables in three dimensions

If

px 7→ px = −i~d

dx(8.2)

in one dimension, we can pretty much guess that

px 7→ px = −i~∂

∂x, py 7→ py = −i~

∂

∂y, pz 7→ pz = −i~

∂

∂z, (8.3)

and the time–independent Schrodinger equation in three dimensions becomes

[− ~

2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+ V (

√x2 + y2 + z2)

]ψ(x, y, z) = Eψ(x, y, z) .

(8.4)We now take advantage of the special form of V : V (

√x2 + y2 + z2) and

introduce the spherical coordinates

z = r cos θ , y = r sin θ sin ϕ , x = r sin θ cos ϕ , (8.5)

where

r = (x2 + y2 + z2)1/2 , θ = arctan

√x2 + y2

z2, ϕ = arctan

(y

x

). (8.6)


Thus,

∂

∂z=

∂

∂r

∂r

∂z+

∂

∂θ

∂θ

∂z+

∂

∂ϕ

∂ϕ

∂z(8.7)

=2z

2√

x2 + y2 + z2

∂

∂r+

1

1 +(

x2+y2

z2

)√x2 + y2

(− 1

z2

)∂

∂θ(8.8)

= cos θ∂

∂r− 1

rsin θ

∂

∂θ(8.9)

∂

∂y=

∂

∂r

∂r

∂y+

∂

∂θ

∂θ

∂y+

∂

∂ϕ

∂ϕ

∂y(8.10)

=2y

2√

x2 + y2 + z2

∂

∂r

+1

1 +(

x2+y2

z2

) 1

z

y

(x2 + y2)1/2

∂

∂θ+

1(1 + y2

x2

) · 1

x

∂

∂ϕ(8.11)

= sin θ sin ϕ∂

∂r+

1

1 + tan2 θ· tan θ sin ϕ

r sin θ

∂

∂θ

+1

1 + tan2 ϕ

1

r sin θ cos ϕ(8.12)

= sin θ sin ϕ∂

∂r+ cos2 θ · sin θ

cos θ

sin ϕ

r sin θ

∂

∂θ+

cos2 ϕ

r sin θ cos ϕ

∂

∂ϕ(8.13)

= sin θ sin ϕ∂

∂r+

cos θ sin ϕ

r

∂

∂θ+

cos ϕ

r sin θ

∂

∂ϕ(8.14)

After manipulations of a similar caliber, one finds

∂

∂x=

x√x2 + y2 + z2

∂

∂r+

1(1 + x2+y2

z2

) 1

z

x√x2 + y2

∂

∂θ

+1(

1 + x2

y2

)y

(− 1

x2

)· ∂

∂ϕ(8.15)

= sin θ cos ϕ∂

∂r+

1

1 + tan2 θ· tan θ cos ϕ

r sin θ

∂

∂θ

− 1

1 + tan2 ϕ

(r sin θ sin ϕ

r2 sin2 θ cos2 ϕ

)∂

∂ϕ(8.16)

= sin θ cos ϕ∂

∂r+

cos θ

rcos ϕ

∂

∂θ− sin ϕ

r sin θ

∂

∂ϕ(8.17)


We actually need terms like ∂2/∂x2. More straightforward but tediousmanipulations eventually produce

∂2

∂x2+

∂2

∂y2+

∂2

∂z2=

1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)

+1

r2 sin2 θ∂2

∂ϕ2(8.18)

Note that this form is “set up” for separation of variables, as derivatives areneatly separated out. Hence, ψ(r, θ, ϕ) satisfies

− ~2

2m

[1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

∂2

∂ϕ2

]ψ(r, θ, ϕ)

+V (r)ψ(r, θ, ϕ) = Eψ(r, θ, ϕ) (8.19)

Assumeψ(r, θ, ϕ) = R(r)Y (θ, ϕ) . (8.20)

Inserting this into the time–independent Schrodinger equation, dividingthrough by R(r)Y (θ, ϕ), and multiplying by−2mr2/~2 gives, after rearrange-ment:

1

R(r)

d

dr

(r2 d

dr

)R(r) +

2mr2

~2(E − V (r)) (8.21)

= − 1

Y (θ, ϕ)

(1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

)Y (θ, ϕ) .(8.22)

The left hand side depends only on r, while the right hand side dependsonly on the angular variables (θ, ϕ). If this equality is to hold for every(r, θ, ϕ), both sides must be equal to a common constant, denoted for conve-nience by −`(` + 1) with ` ≥ 0.

Thus, we now have the two differential equation

1

r2

d

dr

(r2 d

dr

)R(r) +

2m

~2(E − V (r))R(r) = −`(` + 1)

r2R(r) (8.23)

(1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

)Y (θ, ϕ) = `(` + 1)Y (θ, ϕ) (8.24)

The eigenvalue equation for r depends explicitly on the potential V (r).The solution to Eqn.(G.43) will therefore depend on V (r); nothing further


can be said until we know what V (r) is. On the other hand, the eigenvalueequation for the angular part is always the same and does not depend onthe details of V (r), provided that V (r) is a function of r only (this wasthe original assumption that allowed the separation of variables). Thus, theangular equation can be solved once and for all: this will be done later.

Eqn.(G.43) can be further simplified, since

1

r2

d

dr

(r2 d

dr

)R(r) =

1

r2

d

dr

(r2R′(r)

)

=1

r2

(2rR′(r) + r2R′′(r)

)

=2

rR′(r) + R′′(r) (8.25)

Observing that1

r

d2

dr2rR(r) =

d2

dr2R(r) +

2

r

dR

dr, (8.26)

we obtain1

r2

d

drr2

(d

drR(r)

)=

1

r

d2

dr2rR(r) (8.27)

The radial equation now takes the form

1

r

d2

dr2rR(r) +

2m

~2(E − V (r)) R(r)− `(` + 1)

r2R(r) = 0 . (8.28)

Multiplying through by r, we obtain, using χ(r) = rR(r),

d2

dr2χ(r) +

2m

~2(E − V (r))χ(r)− `(` + 1)

r2χ(r) = 0 ,

− ~2

2m

d2

dr2χ(r) +

(V (r) +

~2

2m

`(` + 1)

r2

)χ(r) = Eχ(r) . (8.29)

Eqn.(8.29) is nothing but the Schrodinger equation for the potential

Veff(r) = V (r) +~2

2m

`(` + 1)

r2. (8.30)

The extra term is a “centrifugal” barrier, that imposes conservation of an-gular momentum on the radial motion of the particle.

We see that, in order to solve Eqn.(8.29), it is not sufficient to know V (r)but one must also specify `. This parameter will be later identified as the


angular momentum quantum number; for the moment, it just enters via theseparation constant `(` + 1).

In spite of the similarities between Eqn.(8.29) and the one–dimensionalSchrodinger equation, we note that the boundary conditions on χ(r) are cer-tainly different from those of a 1-d problem. The variable r is, by geometricalconsiderations, restricted to r > 0, so we must have, by continuity of the wavefunction, χ(r) = 0 for r < 0, and thus χ(r) = 0 at r = 0. [Nota: Typo inSchaum 8.13]

8.2 The case of hydrogen: V (r) = −e2/4πεor.

8.2.1 Solving the differential equation

Let us define the Bohr radius:

ao =4π2~2εo

πme2=

4π~2εo

me2(8.31)

and the dimensionless quantity ρ = r/ao. Then we get

d

dr=

d

dρ

dρ

dr=

1

ao

d

dρ(8.32)

and thus

−~2

2m

m2e4

(4πεo)2~4

d2

dρ2χ(ρ) +

[V (ρ) +

~2`(` + 1)

2m

m2e4

(4πεo)2~4

1

ρ2

]χ(ρ)

=−me4

2(4πεo)2~2

d2

dρ2χ(ρ) +

[V (ρ) +

me4

2(4πεo)2~2

`(` + 1)

ρ2

]χ(ρ) = Eχ(ρ)(8.33)

The combinationme4

2(4πεo)2~2≈ 13.6eV (8.34)

is a convenient energy scale, so we define

E ≡ me4

2(4πεo)2~2(8.35)

and rearrange to get the much cleaner expression

− d2

dρ2χ(ρ) +

[V (ρ) +

`(` + 1)

ρ2

]χ(ρ) = −νχ(ρ) (8.36)


with

V (ρ) =V (ρ)

E, ν = −E

E, χ(ρ) = ρR(ρ) . (8.37)

Let

V (ρ) = − e2

4πεor

me2

(4πεo)~2

1

ρ= − me4

(4πεo)2~4

1

ρV (ρ) = −2

ρ. (8.38)

The radial part of Schrodinger for the Coulomb problem now reads like

− d2

dρ2χ(ρ) +

[−2

ρ+

`(` + 1)

ρ2

]χ(ρ) = −νχ(ρ) (8.39)

In spherical coordinates, the radial probability density is ρ2|R(ρ)|2. If∫ ∞

0

ρ2|R(ρ)|2 dρ =

∫ ∞

0

|χ(ρ)|2 dρ (8.40)

is to remain finite, we require χ(ρ) → 0 as ρ →∞ and χ(ρ) → 0 as ρ → 0.In the limit ρ →∞, the differential equation becomes

− d2

dρ2χ(ρ) = −νχ(ρ) . (8.41)

The solution to this depends on the sign of ν. If ν > 0, then χ(ρ) ∼ e±√

νρ.If ν < 0, then χ(ρ) ∼ e±i

√νρ, which is oscillatory. As we want χ(ρ) → 0 for

ρ →∞, we choose χ(ρ) ∼ e−√

νρ, so that E = −νE < 0.To find the solution for finite ρ, we write χ(ρ) = e−

√νρξ(ρ). The function

ξ(ρ) must now satisfy with ξ(ρ) → 0 as ρ → 0 (so that χ(ρ) → 0 as ρ → 0).Taking the ansatz χ(ρ) = ξ(ρ)e−

√νρ in the full differential equation we

obtain,

d

dρχ(ρ) = −√νe−

√νρξ(ρ) + e−

√νρξ′(ρ) (8.42)

d2

dρ2χ(ρ) =

(−√ν)2

e−√

νρξ(ρ) + 2(−√νe−

√νρξ′(ρ)

)+ e−

√νρξ′′(ρ)

= νe−√

νρξ(ρ)− 2√

νe−√

νρξ′(ρ) + e−√

νρξ′′(ρ) (8.43)

and thus

−νe−√

νρξ(ρ) + 2√

νe−√

νρξ′(ρ)− e−√

νρξ′′(ρ)

−(

2

ρ− `(` + 1)

ρ2

)e−

√νρξ(ρ) = −νe−

√νρξ(ρ) (8.44)


or

ξ′′(ρ)− 2√

νξ′(ρ) +

(2

ρ− `(` + 1)

ρ2

)ξ(ρ) = 0. (8.45)

Next we consider the small ρ behaviour of this equation: as ρ → 0 we get

ξ′′(ρ)− 2√

νξ′(ρ)− `(` + 1)

ρ2ξ(ρ) = 0 (8.46)

Since χ(ρ) → 0 as ρ → 0, and χ(ρ) = e−√

νρξ(ρ) → 0 as ρ → 0, we musthave ξ(ρ) ∼ ρs with s > 0 for small ρ. To determine s, take ξ(ρ) ∼ ρs to get:

ξ′′(ρ) = s(s− 1)ρs−2 , (8.47)

ξ′(ρ) = sρs−1 . (8.48)

and insert this in Eqn.(8.46) to obtain

s(s− 1)ρs−2 − 2√

νsρs−1 − `(` + 1)ρs−2 = 0

ρs−2[s(s− 1)− 2

√νsρ− `(` + 1)

]= 0 (8.49)

To solve, we observe that, as ρ → 0, the most pathological term is of theform

ρs−2 (s(s− 1)− `(` + 1)) . (8.50)

To eliminate this term, we need s(s − 1) = `(` + 1) ⇒ s = ` + 1, so ξ(ρ)behaves like ρ`+1 for small ρ. Note that another solution to Eqn.(8.49) iss = −`, but this violates the condition that s ≥ 0. Furthermore, with thischoice, we have, as ρ → 0,

ξ(ρ) ∼ ρ`−1(−2√

ν (` + 1)ρ) ∼ ρ` , (8.51)

which always remains finite as ` ≥ 0: the choice s = `+1 save the wavefunc-tion for small ρ.

To interpolate between ρ → 0 and finite ρ, we write ξ(ρ) = ρ`+1L(ρ).Since χ(ρ) = ρ`+1L(ρ)e−

√νρ must remain finite as ρ →∞, the function L(ρ)

is restricted to functions such that

limρ→∞

ρ`+1L(ρ)e−√

νρ → 0 . (8.52)

Using ξ(ρ) = ρ`+1L(ρ) in Eqn.(8.45), we obtain the derivatives

ξ′(ρ) = (` + 1)ρ`L(ρ) + ρ`+1L′(ρ) (8.53)

ξ′′(ρ) = `(` + 1)ρ`−1L(ρ) + 2(` + 1)ρ`L′(ρ) + ρ`+1L′′(ρ) (8.54)


So our differential equation Eqn.(8.45) is turned into a differential equationfor L(ρ):

ρ`+1L′′(ρ) + 2(` + 1)ρ`L′(ρ) + `(` + 1)ρ`−1L(ρ)− 2√

ν(` + 1)ρ`L(ρ)

−2√

νρ`+1L′(ρ) + 2ρ`L(ρ)− `(` + 1)ρ`−1L(ρ) = 0 , (8.55)

or, after factoring a common non–zero factor of ρ`:

ρL′′(ρ) +(2` + 2− 2

√νρ

)L′(ρ)− 2

(√ν(` + 1)− 1

)L(ρ) = 0 (8.56)

Let

σ = 2√

νρ,d

dρ=

d

dσ

dσ

dρ= 2

√ν

d

dσ. (8.57)

Then

ρL′′(ρ) =σ

2√

ν4νL′′(σ) = 2

√νσL′′(σ) (8.58)

L′(ρ) = 2√

νL′(σ) . (8.59)

This transforms the second order differential equation Eqn.(8.56) into

2√

ν

[σL′′(σ) + (2` + 2− σ) L′(σ)− (` + 1− 1√

ν)L(σ)

]= 0 (8.60)

The term in bracket is an example of Laguerre’s differential equation:

xy′′ + (k + 1− x)y′ + qy = 0 , (8.61)

itself a special case of the confluent hypergeometric differential equation,Eqn.(G.44):

zu′′(z) + (c− z)u′(z)− au(z) = 0 .

Here we have

x = σ, c = 2(` + 1), k = 2` + 1, a = (` + 1− 1√ν) = −q, (8.62)

The solutions to this are generalized Laguerre functions Lkq(x). These

functions remain finite as x → ∞ only if a is a negative integer, in whichcase these functions are generalized Laguerre polynomials. Hence we need

a = (` + 1− 1√ν) ∈ Z− . (8.63)


For this to occur, we must have

1√ν∈ Z , (8.64)

so set ν = 1/n2r to get 1/

√ν = nr. This, in turn, implies that

Eν = −νE = − E

n2r

7→ En = − E

n2r

. (8.65)

Since a ≤ 0, we have the extra condition

` + 1− nr ≤ 0 , (8.66)

or nr ≥ ` + 1, which restricts the possible values of `.

8.2.2 Hydrogenoid Wave Functions

A solution to the confluent hypergeometric equation

xy′′(x) + (c− x)y′(x)− ay(x) = 0

that remains finite at x = 0 and x →∞ is the function

y(x) = 1F1(a, c; x) = 1 +a

c

x

1!+

a(a + 1)

c(c + 1)

x2

2!+

a(a + 1)(a + 2)

c(c + 1)(c + 2)

x3

3!+ . . .

with a ∈ Z−.Comparing with our equation

σL′′(σ) + (2(l + 1)− σ) L′(σ)− (` + 1− nr)L(σ) = 0 ,

we see thaty(σ) = 1F1(` + 1− nr, 2(` + 1); σ). (8.67)

The associated Laguerre polynomials are related to 1F1 via

Lka(σ) =

(k + a)!

k!a!1F1(a, k + 1; σ). (8.68)

Going back to the original variable

σ = 2√

νρ =2r

nrao

,


we have thus found

χnr,`(r) =

(2r

nrao

)`+1

e−r/nrao1F1(` + 1− nr, 2(` + 1);

2r

aonr

) (8.69)

and

Rnr`(r) =

[(2

nrao

)3(nr − `− 1)!

2nr(nr + `)!

] 12

e−r/nrao

(2r

nrao

)`

L2`+1`+1−nr

(2r

nrao

),

(8.70)where the radial wave function Rnr`(r) has been properly normalized.

The first few L2`+1`+1−nr

(x) are given below. They follow Bateman’s con-vention, and are those produced by Mathematica c©.

nr ` L2`+1`+1−nr

(x)

1 0 1

2 0 2− x

1 1

3 0 12(6− 6x + x2)

1 4− x

2 1

4 0 16(24− 36x + 12x2 − x3)

1 12(20− 10x + x2)

2 6− x

3 1

5 0 124

(120− 240x + 120x2 − 20x3 + x4)

1 16(120− 90x + 18x2 − x3)

2 12(42− 14x + x2)

3 8− x

4 1

The first few normalized complete hydrogen radial wave functions Rnr` (r)are thus


nr ` Rnr` (r)

1 0 2e−r/a01

a3/20

2 0 e−r/2a0(2− r/a0)

2√

2a3/20

1 e−r/2a0r

2√

6a5/20

3 0 2e−r/3a0(2r2 − 18ra0 + 27a2

0)

81√

3a7/20

1 e−r/3a0

√2r (4− 2r/3a0)

27√

3a5/20

2 e−r/3a02√

2r2

81√

15a7/20

These wave functions agree with those in Schaum’s.

8.3 Isotropic oscillator in three dimensions

We next consider the case of a three–dimensional isotropic oscillator, with

V (r) =1

2mω2 r2 . (8.71)

The radial equation for χ(r) = rR(r) becomes

d2

dr2χ(r) +

2m

~2

(E − 1

2mω2r2

)χ(r)− `(` + 1)

r2χ(r) = 0 (8.72)

For ` = 0, this is just the ordinary 1-d harmonic oscillator. For ` 6= 0, weare working in an effective potential V (r) = 1

2mω2r2 + ~

2mr2 `(` + 1), whichdrives the particle away from the origin as r → 0.

Set2mE

~2= k2,

mω

~= λ,

k2

2λ=

E

~ω= µ (8.73)


5 10 15 20 25

0.1

0.2

0.3

nr=3,L=0

5 10 15 20 25

0.02

0.04

0.06

0.08

0.1

0.12

0.14

nr=3,L=0

2 4 6 8 10

0.02

0.04

0.06

0.08

0.1

0.12

0.14

nr=2,L=1

2 4 6 8 10

0.025

0.05

0.075

0.1

0.125

0.15

0.175

0.2nr=2,L=1

2 4 6 8 10

0.2

0.4

0.6

nr=2,L=0

2 4 6 8 10

0.025

0.05

0.075

0.1

0.125

0.15

0.175

0.2nr=2,L=0

1 2 3 4 5

0.5

1

1.5

2nr=1,L=0

1 2 3 4 5

0.1

0.2

0.3

0.4

0.5

nr=1,L=0

Figure 11: The first few hydrogenoıd wave functions and their respective proba-bility densities.


to getd2

dr2χ(r) +

[k2 − λ2r2 − `(` + 1)

r2

]χ(r) = 0 (8.74)

For large r, we have

d2

dr2χ(r) ' −λ2r2χ(r) , (8.75)

which, in a manner similar to the one–dimensional oscillator, produces theasymptotic behavior

χ(r) ∼ rse−λr2/2 , (8.76)

as long as s is finite.For small r, we have

d2

dr2χ(r) ' `(` + 1)

r2χ(r) , (8.77)

which produces, as in manner similar to the hydrogen atom solution,

χ(r) ∼ r`+1 (8.78)

Thus, set χ(r) = r`+1f(r)e−λr2/2, ` finite. Now

d

drχ(r) =

[(` + 1− λr2)f(r) + rf ′(r)

] 1

2r`e−λr2/2 (8.79)

d2

dr2χ(r) =

1

2r`−1e−λr2/2

[`(` + 1) + λr2

(−3 + λr2 − 2`)]

f(r)

+2r(` + 1− λr2

)f ′(r) + r2f ′′(r) , (8.80)

so that we get the differential equation

1

2r`e−λr2/2

[r f ′′(r) + 2(` + 1− λr2)f ′(r) + r

(k2 − (2` + 3)λ

)f(r)

]= 0 .

(8.81)To proceed, we need to make a change of variables similar to the one–dimensional case: set t = λr2 and obtain

d

dr=

d

dt

dt

dr= 2λr

d

dt= 2

√λt

d

dt, (8.82)

d2

dr2= 2

√λt

d

dt

(2√

λtd

dt

)= 2

√λt

(2√

λ

2√

t

d

dt+ 2

√λt

d2

dt2

),

= 2λd

dt+ 4λt

d2

dt2. (8.83)


This converts our differential equation to

√t

λ(2λf ′(t) + 4λt f ′′(t)) + 2(` + 1− t)2λ

√t

λf ′(t)

+2λ

√t

λ

1

2λ

(k2 − (2` + 3) λ

)f(t) = 0 (8.84)

Thus:

tf ′′(t) +

(` +

3

2− t

)f ′(t) +

(k2

4λ− (2` + 3)

4

)f(t) = 0 . (8.85)

We now compare Eqn.(8.85) with the standard form of the confluenthypergeometric differential equation:

xy′′ + (c− x)y′ − ay = 0 (8.86)

and findf(t) = 1F1(−nr, ` + 3

2; t) , (8.87)

where

a = −(

k2

4λ− (2` + 3)

4

)= −nr , nr ∈ Z+ . (8.88)

This last requirement must be enforced if the series solution

1F1(a, c; x) = 1 +a

cx +

a(a + 1)

c(c + 1)

x2

2!+

a(a + 1)(a + 2)

c(c + 1)(c + 2)

x3

3!+ . . .

is to terminate at some finite degree, thereby guaranteeing that χ(r) → 0 asr →∞.

Alternately comparing with the associated Laguerre equation

xLkn(x)′′ + (k + 1− x)Lk

n(x)′ + nLkn(x) = 0 (8.89)

we find f(t) = L`+1/2nr (t)

Going back to the original variables, we find

E

2~ω= nr +

2` + 3

4⇒ E =

(2nr + ` +

3

2

)~ω =

(n +

3

2

)~ω , (8.90)


with n = 2nr+`. Since nr ∈ Z+, i.e. nr = 0, 1, 2, 3, . . . whereas ` = 0, 1, 2, . . .,we find n = 0, 1, 2, . . . subject to the restrictions

` =

n, n− 2, n− 4, . . . , 0 if n is evenn, n− 2, n− 4, . . . , 1 if n is odd

(8.91)

The solutions, properly normalized, work out to

Rn`(r) =

√(` + 1/2)!nr!

(nr + ` + 1/2)!Γ(` + 3

2

)√

2λ`+3/2e−λr2/2r`L`+1/2nr

(λr2) (8.92)

nr l n L`+1/2nr (λr2) Rn`(r)

0 0 0 1 2λ3/4

π1/4 e−λr2/2

0 1 1 1 2λ5/4

π1/4

√23r e−λr2/2

1 0 2 32− λr2 2

√23

λ3/4

π1/4

(32− λr2

)e−λr2/2

0 2 2 1 4√15

λ7/4

π1/4 r2 e−λr2/2

1 1 3 52− λr2 4√

15λ5/4

π1/4 r(

52− λr2

)e−λr2/2

0 3 3 1 4√

2105

λ9/4

π1/4 r3 e−λr2/2

2 0 4 18(15− 20λr2 + 4λ2r4) 1√

30λ3/4

π1/4 (15− 20λr2 + 4λ2r4)e−λr2/2

1 2 4 72− λr2 4

√2

105λ7/4

π1/4 r2(

72− λr2

)e−λr2/2

0 4 4 1 83√

105λ11/4

π1/4 r4 e−λr2/2

8.4 Constant potentials in three dimensions

8.4.1 Spherical Bessel functions

Suppose V = Vo, some constant. Then, the radial part of the Schrodingerequation can be written as,

− ~2

2m

(d2

dr2− ` (` + 1)

r2

)χ(r) + Voχ(r) = Eχ(r), χ(r) = rR(r) (8.93)


which we rewrite as

− ~2

2m

d2

dr2χ(r) +

~2

2m

` (` + 1)

r2χ(r) = (E − Vo) χ(r) (8.94)

Suppose E > Vo, and define

k ≡√

2m (E − Vo)

~2. (8.95)

The radial equation takes the form

d2

dr2χ(r)− ` (` + 1)

r2+ k2χ(r) = 0 (8.96)

Let

ρ = kr,d

dr=

d

dρ

dρ

dr= k

d

dρ,

d2

dr2= k2 d2

dρ2, (8.97)

which produces

k2 d2

dρ2χ(ρ)− ` (` + 1)

k2

ρ2χ(ρ) + k2χ(ρ) = 0 (8.98)

ord2

dρ2χ(ρ)− ` (` + 1)

ρ2χ(ρ) + χ(ρ) = 0 (8.99)

Consider first the case where ` = 0. This is equivalent to the one–dimensional case: the possible solutions are χ0(ρ) = sin ρ or χ0(ρ) = cos ρ.

Here, we use the sine and cosine functions rather than the exponentialse±iρ because it is easier to enforce the requirement χ0(0) = 0; this means thatwe must keep only χ0(ρ) = sin ρ as a valid solution in a region that includesρ = 0. For ` 6= 0 and small ρ we have as usual χ(ρ) ∼ ρ`+1. (Note that,for large values of ρ, χ(ρ) ∼ A sin ρ + B cos ρ. Thus we set χ(ρ) = ρ`+1ξ(ρ).Now,

d

dρχ(ρ) = (` + 1) ρlξ(ρ) + ρ`+1ξ′(ρ) (8.100)

d2

dρ2χ(ρ) = ` (` + 1) ρ`−1ξ(ρ) + 2 (` + 1) ρ`ξ′(ρ) + ρ`+1ξ′′(ρ) (8.101)

Substituting in Eqn.(8.99), we find

ρ`+1ξ′′(ρ)+2 (` + 1) ρ`ξ′(ρ)+` (` + 1) ρ`−1ξ(ρ)−` (` + 1) ρ`−1ξ(ρ)+ρ`+1ξ(ρ) = 0(8.102)


or

ξ′′(ρ) +2 (` + 1)

ρξ′(ρ) + ξ(ρ) = 0 . (8.103)

Note that, contrary to the harmonic oscillator or to the hydrogen atom,we get no real mileage out of the condition ρ → ∞ because the solutionA sin ρ+B cos ρ. is oscillatory: this is because there is no “forbidden” regionwhere we expect the character of the solution to be a decreasing exponential.

Now consider the derivative of Eqn(8.103):

ξ′′′(ρ)− 2 (` + 1)

ρ2ξ′(ρ) +

2 (` + 1)

ρξ′′(ρ) + ξ′(ρ) (8.104)

Setting ξ′ = ρσ, we get

ξ′ = ρσ , (8.105)

ξ′′ = σ + ρσ′ , (8.106)

ξ′′′ = σ′ + σ′ + ρσ′′ = 2σ′ + ρσ′′ , (8.107)

Eqn.(8.104) becomes

ρσ′′ + 2σ′ − 2 (` + 1)

ρ2ρσ +

2 (` + 1)

ρ(σ + ρσ′) + ρσ = 0 (8.108)

which can be reorganized to

ρσ′′ + (2 + 2 (` + 1)) σ′ +(

2 (` + 1)

ρ− 2 (` + 1)

ρ+ ρ

)σ = 0 ,

⇒ σ′′ +2 (l + 2)

ρσ′ + σ = 0 . (8.109)

Thus, if ξ is a solution for `, then σ = 1ρξ′ is a solution for ` + 1. This is a

recursion relation for the solutions. We will write this recursion as

ξ`+1(ρ) =1

ρξ′` , (8.110)

using ` to differentiate the various solutions.To start the recursion, we note that we already know the form of the

solution for ` = 0: it is simply χ0(ρ) = sin ρ. Since our recursion is for ξ(ρ)


rather than χ(ρ), we have

ξ0(ρ) =1

ρsin ρ ,

ξ1(ρ) =1

ρξ′0(ρ) =

1

ρ

(1

ρcos ρ− 1

ρ2sin ρ

), (8.111)

and so forth.Now recall that χ(ρ) = ρ`+1ξ(ρ), and R(ρ) = χ(ρ)/ρ. So, using ` to label

the sequence of solutions that go to 0 as ρ → 0, we have

χ`(ρ) = ρ`+1ξ`(ρ) , R`(ρ) = ρ`ξ` , (8.112)

with

ξ`(ρ) =

(1

ρ

d

dρ

)ξ`−1(ρ) =

(1

ρ

d

dρ

)`

ξ0(ρ) =

(1

ρ

d

dρ

)`1

ρsin ρ . (8.113)

Now, we define the spherical Bessel function j`(ρ) by

j`(ρ) = (−1)`ρ`

(1

ρ

d

dρ

)`1

ρsin ρ , (8.114)

so that R`(ρ) = j`(ρ) (the overall sign is immaterial) is a solution that satisfiesthe boundary conditions at ρ → 0. j`(ρ) satisfies the recursion relations

j`+1(ρ) =

(2` + 1

ρ

)j`(ρ)− j`−1(ρ) (8.115)

d

dρj`(ρ) =

`

ρj`(ρ)− j`+1(ρ) . (8.116)

The first few j`(ρ) functions are:

` j`(ρ)

0 1ρsin ρ

1 −1ρcos ρ + 1

ρ2 sin ρ

2 −3 cos ρρ2 +

(3ρ3 − 1

ρ

)sin ρ

3(−15

ρ3 + 1ρ

)cos ρ +

(15ρ4 − 6

ρ2

)sin ρ

4(−105

ρ4 + 10ρ2

)cos ρ +

(105ρ5 − 45

ρ3 + 1ρ

)sin ρ

5(−945

ρ5 + 105ρ3 − 1

ρ

)cos ρ +

(945ρ6 − 420

ρ4 + 15ρ2

)sin ρ


5 10 15 20

-0.1

-0.05

0.05

0.1

0.15

0.2

0.25L=2

5 10 15 20

0.2

0.4

0.6

0.8

1

1.2

L=2

5 10 15 20

-0.1

0.1

0.2

0.3

L=2

5 10 15 20

0.2

0.4

0.6

0.8

1

1.2

L=2

5 10 15 20-0.1

0.1

0.2

0.3

0.4

L=1

5 10 15 20

0.2

0.4

0.6

0.8

1

L=1

5 10 15 20-0.2

0.2

0.4

0.6

0.8

1L=0

5 10 15 20

0.2

0.4

0.6

0.8

1L=0

Figure 12: The spherical Bessel functions of the first kind j`(ρ) (left) and thecorresponding probability densities (right) ρ2 j2

` (ρ). Notice how the probabilitydensity goes to 0 near ρ = 0, which makes these functions suitable to describeparticles that have access to the region of small ρ.


Of course, we have only considered solution that are 0 at ρ = 0. In someproblems, there are regions where ρ = 0 is excluded, for instance because abarrier or some other consideration preclude the particle from reaching theorigin. In such a problem, we must then also consider the solution χ0(ρ) =cos ρ as acceptable. Since the recursion relation simply starts with cos ρrather than sin ρ, we have another family of solutions, denoted by n`(ρ) anddefined by

n`(ρ) = (−1)`+1ρ`

(1

ρ

d

dρ

)`cos ρ

ρ(8.117)

For these we have:

` n`(ρ)

0 −1ρcos ρ

1 −1ρsin ρ− 1

ρ2 cos ρ

2 −3 sin ρρ

+(− 3

ρ3 + 1ρ

)cos ρ

3(−15

ρ3 + 1ρ

)sin ρ +

(−15ρ4 + 6

ρ2

)cos ρ

4(−105

ρ4 + 10ρ2

)sin ρ +

(−105

ρ5 + 45ρ3 − 1

ρ

)cos ρ

5(−945

ρ5 + 105ρ3 − 1

ρ

)sin ρ +

(−945

ρ6 + 420ρ4 − 15

ρ2

)cos ρ

When functions of an exponential type, such as eiρ, are required, one usespherical Hankel functions of the first kind, which are morally the equivalentof

eix = cos x + i sin x ,

and defined by

` h(1)` (ρ)

0 −ieiρ

ρ

1 −eiρ

ρ2 (i + ρ)

2 ieiρ

ρ3 (−3 + 3iρ + ρ2)

3 eiρ

ρ4 (−15i− 15ρ + 6iρ2 + ρ3)

4 eiρ

ρ5 (−105i− 105ρ + 445iρ2 + 10ρ3 − iρ4)


5 10 15 20

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1L=3

5 10 15 20

0.51

1.52

2.53

3.54

L=3

5 10 15 20

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1L=2

5 10 15 20

0.51

1.52

2.53

3.54

L=2

5 10 15 20

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1L=1

5 10 15 20

0.51

1.52

2.53

3.54

L=1

5 10 15 20

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1L=0

5 10 15 20

0.2

0.4

0.6

0.8

1L=0

Figure 13: The spherical Bessel functions of the second kind, or Neumann func-tions, n`(ρ) (left) and the corresponding probability densities (right) ρ2 n2

`(ρ). No-tice how the probability density does not go to 0 near ρ = 0: these functionscannot describe particles that have access to the region of small ρ.


There are also spherical Hankel functions of the second kind, which areconjugate to h

(1)` .

8.4.2 The Infinite Spherical Well

Consider the spherical well defined by

V (r) =

0 if r < a ,∞ if r > a .

Define, as before, k =√

2mE/~2 .By continuity of R(r), R(r) = 0 for all r > a. The only acceptable

solution to the radial equation is thus, R`(kr) = Aj`(kr), where A is somenormalization constant. The boundary condition R(r) = 0 at r = 0 precludesthe use of the n`(kr) function.

Since we must have R(r) = 0 at r = a, we see that the energy levels aredetermined via k through the condition j`(ka) = 0.

As j0(ka) = sin(ka)ka

, we have as usual, the condition

ka = nπ , n = 1, 2, 3, . . . or kn =nπ

a. (8.118)

These states have energies En` given by En0 = ~2kn2

2m= n2π2~2

2ma2 .For ` = 1 we need to solve j1(ka) = 0. Plotting this function we find the

approximate zeroes as:

ka = 4.4934, 7.725, 10.9041, 14.0662, 17.2208 , (` = 1)

so we have

E11 =~2

2m

(4.49)2

a2,

E21 =~2

2m

(7.725)2

a2,

E31 =~2

2m

(10.9041)2

a2, etc . (8.119)

Similarly for ` = 2, we need the zeroes at j2(ka), which are found to be

ka = 5.76, 9.10, 12.32, 15.51, 18.69 , (` = 2)


and so

E12 =~2

2m

(5.76)2

a2,

E22 =~2

2m

(9.10)2

a2, etc . (8.120)

It is therefore convenient to make a table of the zeroes of j`(ka)

n ` = 0 ` = 1 ` = 2 ` = 3 ` = 4 ` = 5 ` = 61 π 4.49 5.76 6.99 8.18 9.35 10.512 2π 7.73 9.10 10.41 11.70 12.97 14.203 3π 10.90 12.32 13.70 15.04 16.35 17.654 4π 14.07 15.51 16.92 18.30 19.65 20.985 5π 17.22 18.69 20.12 21.52 22.90 24.26

8.4.3 The Finite Spherical Well

Suppose now

V (r) =

−V0 if r < a ,0 if r > a ,

(8.121)

where V0 > 0. Let ε = −E so that, for bound states with E < 0, we haveε > 0. Define, as was done for the one–dimensional case in Eqn.(6.20),

k =

√2m (V0 − ε)

~2, κ =

√2m (ε)

~2. (8.122)

The wave function inside the well must be of the form R(r) = Aj`(kr).Outside the well, only the solution which decreases exponentially with r →∞is admissible. For ` = 0 this is clearly

R(r) ∼ e−κr

r= h

(1)0 (iκr) , (8.123)

so, for ` 6= 0 we find, by recursion, R(r) = Bh(1)` (iκr). (Note that, since the

problem has spherical symmetry, the angular momentum is conserved so thatthe same value of ` must be used on both sides of the discontinuity.


Now, in any region the wave function R`(r) is a linear combination ofspherical Bessel functions, which, in particular, satisfy (see Eqn.(8.115)):

d

drj`(kr) = k

d

dρj`(ρ) ,

= k

(`

krj`(kr)− j`+1(kr)

),

=`

rj`(kr)− k j`+1(kr) (8.124)

d

drh

(1)` (iκr) =

`

rh`(iκr)− iκ h`+1(iκr) (8.125)

and

j`+1(kr) =

(2` + 1

kr

)j`(kr)− j`−1(kr) (8.126)

h`+1(iκr) =

(2` + 1

iκr

)n`(iκr)− n`−1(iκr) (8.127)

for ` ≥ 1. We can use this to our advantage.From continuity at r = a, we have

Aj`(ka) = Bh(1)` (iκa) . (8.128)

From continuity of the derivatives at r = a, we have

Ad

drj`(kr)|r=a = B

d

drh`(kr)|r=a . (8.129)

Their ratio eliminates A and B to yield

j′`(ka)

j`(ka)=

h′`(iκa)

h`(iκa). (8.130)

Using Eqns.(8.124) and (8.125), we obtain

`

a− k

j`+1(ka)

j`(ka)=

`

a− iκ

h`+1(iκa)

h`(iκa). (8.131)

Using now Eqns.(8.126) and (8.127), this further simplifies to

2` + 1

a− k

j`−1(ka)

j`(ka)=

2` + 1

a− iκ

h`−1(iκa)

h`(iκa), (8.132)


or, finally,

kj`−1 (ka)

j` (ka)= iκ

h(1)`−1 (iκa)

h(1)` (iκa)

(8.133)

which is generally simpler since the spherical Bessel functions of lower degreehave less complicated denominators.

Let us apply this general analysis to ` = 0, just to see that we reallyrecover what we know from the analysis of odd functions (since R(r) = 0at the origin) of the one–dimensional case of section 6.3. We cannot applydirectly Eqn.(8.133) directly, because we have not defined j−1.

First, we need explicitly the following four functions

j0(ka) =sin(ka)

ka,

h(1)0 (iκa) = − i

(iκa)ei(iκa) = − 1

κae−κa ,

j1(ka) = −cos(ka)

ka+

sin(ka)

(ka)2,

h(1)1 = − 1

(iκa)2ei(iκa)(i + iκa) =

i

κ2a2e−κa(1 + κa) . (8.134)

The two parts of Eqn.(8.131) then read

kj1(ka)

j0(ka)=

−k cos(ka)

ka+

k sin(ka)

(ka)2

sin(ka)

ka

= −k cot(ka) +1

a, (8.135)

iκh

(1)1 (iκa)

h(1)0 (iκa)

=iκ

i

κ2a2e−κa(1 + κa)

− 1

κae−κa

=κ

κ2a2(κa)(1 + κa) ,

=1

a+ κ (8.136)

so that Eqn.(8.131) itself reads

−k cot(ka) +1

a=

1

a+ κ , ⇒ cot(ka) = −κ

k, (8.137)

as before.


Of course, the solutions are not necessarily restricted to spherical Besselfunctions of the first kind if the particle cannot reach the origin: in suchcases, there is no grounds to dismiss the second solution, of the type n`.

8.5 The Morse potential

Consider finally an example of a potential solvable exactly for l = 0, andwhich is useful for the analysis of many molecules.

V (r) = D(e−2α(r−r0)/r0 − 2e−α(r−r0)/r0

)(8.138)

For ` = 0, the radial part of the Schrodinger equation becomes;

d2

dξ2χ (ξ) +

(−β2 + 2γ2e−αξ − γ2e−2αξ)χ (ξ) = 0 (8.139)

where

ξ = (r − r0) /r0, γ2 =2mD

~2r20, β2 = −2mE

~2r20 > 0 . (8.140)

since E < 0 for bound state (as a consequence of V → 0 as r → ∞). Theparameters α and D depend on the molecule.

Let

y =2γ

αe−αξ . (8.141)

With this somewhat inspired change of variables,

d

dξ=

d

dy

dy

dξ= −2γe−αξ d

dy,

= −αyd

dy, (8.142)

d2

dξ2= −αy

d

dy

(−αy

d

dy

)= α2y

(d

dy+ y

d2

dy2

),

= αy2 d2

dy2+ α2y

d

dy(8.143)

The differential equation, written in terms of y, is then

α2y2χ′′(y) + α2yχ′(y) +

(−β2 + γy − α2

4y2

)χ(y) = 0 (8.144)


This is, in fact, closely related to the differential equations that we’ve seenbefore.

For “large” y (which corresponds physically to large e−α(r−r0)/r0 , i.e. tovalues of r near 0, we need the function to go to zero so that it can besquare-integrable.

In this regime we have

χ′′(y) + yχ′(y)− 1

4χ(y) ∼ 0 (8.145)

with solution χ(y) ∼ yse−y/2, where s is finite. This works since

χ′(y) = sys−1e−y/2 − ys 1

2e−y/2 (8.146)

χ′′(y) = s (s− 1) ys−2e−y/2 − sys−1e−y/2 + ys 1

4yse−y/2 (8.147)

so that

χ′′(y) +1

yχ′(y) +

1

4χ(y)

= e−y/2

(s (s− 1) ys−2 − sys−1 +

1

4ys + sys−2 − ys−1

2− 1

4ys

)

= e−y/2

(1

4+ s2y−2 − sy−1 − 1

2y−1 − 1

4

),

e−y/2

(s2

y2− s

y− 1

2y

)∼ 0 . (8.148)

On the other hand, for small values of y,

χ′′(y) +χ′(y)

y− β2

α2y2χ(y) ∼ 0 . (8.149)

Thus, if χ ∼ ys, we have

χ′′ = s (s− 1) ys−2, χ′ = sys−1 , (8.150)

which gives, for small y:

s(s− 1)ys−2 + sys−2 − β2

α2ys−2 = 0 ,

ys−2

(s2 − s + s− β2

α2

)= 0 , (8.151)


with solution s = ±βα. Since β > 0 and α is also > 0 on physical grounds,

we choose s = βα

so χ → 0 as y → 0. Thus we write:

χ(y) = yβ/αe−12yf(y)

where f(y) is a function which interpolates between y → 0 and y → ∞.With this ansatz, the differential equation

χ′′(y) +1

yχ′(y) +

(β2

α2

1

y2+

γ

α

1

y− 1

4

)χ(y) = 0

becomes, using

χ′(y) =1

2αe−y/2y−1+β/α [(2β − αy) f(y) + 2αyf ′(y)]

χ′′(y) =1

4α2e−y/2e−2+β/2

[(α2y2 − 4αβy + 4β (β − α)

)f(y)

+4αy (2β − αy) f ′(y) + 4α2y2f ′′(y)]

. (8.152)

After straightforward but tedious manipulations, we eventually obtain

e−y/2e−1+β/α

2α(2yαf ′′(y) + (2α− 2αy + 4β) f ′(y)− (α + 2β − 2γ) f(y)) = 0

(8.153)We now need to reorganize so that we can compare with the magic formula.

yf ′′(y) + (c− y) f ′(y)− af(y) = 0 ⇒ f(y) = 1F1 (a, c; y)

Dividing our differential equation by 2α :

yf ′′(y) +

(1− y +

2β

α

)f ′(y)−

(1

2+

β

α− γ

α

)f(y) = 0 , (8.154)

from which we extract

c = 1 +2β

α, a =

1

2+

β

α− γ

α. (8.155)

The hypergeometric function 1F1 (a, c; y) is a finite polynomial if a = −n,with n ∈ Z+. The series is required to terminate by asymptotic condition


yse−y/2 with s finite. Hence

1

2− β

α− γ

α= −n ⇒

(n +

1

2

)=

β

α+

γ

α⇒ α

(n +

1

2

)β + γ

⇒ β = α

(n +

1

2

)− γ

⇒ β2 = α2

(n +

1

2

)2

− 2αγ

(n +

1

2

)+ γ2 (8.156)

Going back to the original variables, with β2 = −2mEr20/~2, we find

E = − ~2

2mr20

(α2

(n +

1

2

)2

− 2αγ

(n +

1

2

))− ~2

2mr20

2mD

~2r20

= −D +2αγ~2

2mr20

(n +

1

2

)− ~2α2

2mr20

(n +

1

2

)2

(8.157)

Finally, since the classical frequency of small oscillation in this problem is

ω =√

2Dα2

mr20

, we note that

αγ~2

mr20

= ~2 α

m

√2mDr2

0

~2

1

r20

=~2

~

√2α2mDr2

0

m2r40

=~2

~

√2α2D

mr20

= ~ω (8.158)

and thus~2α2

2mr20

=~2α2

2mr20

(~ω)

αγ~2mr2

0 = ~ω(

α

2γ

)(8.159)

so that

E = −D + ~ω(

n +1

2

)−

(n +

1

2

)2

~ω(

α

2γ

)(8.160)

with

γ =

√2mDr2

0

~2. (8.161)

The angular equation 83

9 The angular equation

Recall that, in spherical coordinates, the Schrodinger equation for the wavefunction ψ(r, θ, ϕ) reads

− ~2

2m

[1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

∂2

∂ϕ2

]ψ(r, θ, ϕ)

+V (r)ψ(r, θ, ϕ) = Eψ(r, θ, ϕ) (9.1)

Separation of the angular and radial variables

ψ(r, θ, ϕ) = R(r)Y (θ, ϕ) . (9.2)

leads to

1

R(r)

d

dr

(r2 d

dr

)R(r) +

2mr2

~2(E − V (r))

= − 1

Y (θ, ϕ)

(1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

)Y (θ, ϕ) . (9.3)

Denoting the separation constant by −`(` + 1) for convenience, we arelead to the angular equation

(1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

)Y (θ, ϕ) = `(` + 1)Y (θ, ϕ) (9.4)

The solutions to Eqn.(9.4) are the topic of this section.

9.1 Separation of the azimuthal and polar variables.

Eqn.(9.4) is still a partial differential equation. To solve, we need furtherseparation between the polar and azimuthal angles. Let us write Y (θ, ϕ) =Θ(θ) Φ(ϕ). Inserting this in the angular equation, dividing through byΘ(θ)Φ(ϕ), multiplying by sin2 θ and rearranging gives

[1

Θ(θ)sin θ

∂

∂θ

(sin θ

∂

∂θ

)Θ(θ) + `(` + 1) sin2 θ

]= − 1

Φ(ϕ)

d2

dϕ2Φ(ϕ) (9.5)

The left hand side depends only on θ, and the right hand side only on ϕ.Thus, for the equality to hold always, both sides must be equal to a constant,denoted by −m2.


From this, we can solve immediately for Φ(ϕ):

d2

dϕ2Φ(ϕ) = −m2Φ(φ) =⇒ Φ(ϕ) =

eimϕ

√2π

(9.6)

Finally, if the function Φ(ϕ) is to be 2π–periodic, we must restrict mto integer values. In other words, since a rotation about z by 2π shouldphysically bring us back to where we started, we must impose the boundarycondition Φ(ϕ + 2π) = Φ(ϕ), which leads to the restriction on m.

9.2 The θ equation

Coming back to the θ-equation, we now get

1

sin θ

d

dθ

(sin θ

d

dθ

)Θ(θ) + `(` + 1)Θ(θ) =

m2

sin2 θΘ(θ) (9.7)

The same equation in the θ variable occurs for +m or −m. Let ξ = cos θ.Then

d

dξ=

d

dθ

dθ

dξ= − 1

sin θ

d

dθ(9.8)

d

dθ= −

√1− ξ2

d

dξ(9.9)

1

sin θ

d

dθ

(sin θ

d

dθ

)=

1

(1− ξ2)1/2

√1− ξ2

d

dξ

[(1− ξ2)

d

dξ

](9.10)

with the two minus signs from Eqn.(9.9) cancelling, so that, in terms of ξ ,we now have

d

dξ

[(1− ξ2)

d

dξ

]Θ(ξ) +

(`(` + 1)− m2

1− ξ2

)Θ(ξ)

= (1− ξ2)d2

dξ2Θ(ξ)− 2ξ

d

dξΘ(ξ) +

(`(` + 1)− m2

1− ξ2

)Θ(ξ) = 0 (9.11)

This is not quite the standard form. We first set

Θ(ξ) = (1− ξ2)|m|/2u(ξ) ⇔ Θ(θ) = (sin θ)|m| u(θ) (9.12)


and rewrite Eqn.(9.11) using:

d

dξ(1− ξ2)|m|/2 u(ξ) = −|m|ξ (1− ξ2)−1+|m|/2 u(ξ) + (1− ξ2)|m|/2u′(ξ) ,

d2

dξ2

((1− ξ2)|m|/2u(ξ)

)= (1− ξ2)−1+|m|/2

[(−1 + (|m|+ 1)ξ2) |m|u(ξ)

+2|m|ξ(1− ξ2) u′(ξ) + (1− ξ2)2u′′(ξ)]

. (9.13)

to obtain

(1−ξ2)u′′(ξ)−2 (|m|+ 1) ξ u′(ξ)+(`(` + 1)− |m|(|m|+ 1)) u(ξ) = 0 . (9.14)

To bring this to the standard form, we make a final change of variable:

ξ = 1− 2z , ⇔ z = 12(1− ξ) . (9.15)

Changing d/dξ to d/dz produces

z(1− z)u′′(z) + (|m|+ 1− 2(|m|+ 1)z) u′(z)

+ (`(` + 1)− |m|(|m|+ 1)) u(z) = 0 , (9.16)

which is of the 2F1 form with

c = |m|+ 1 , a = |m| − ` , b = |m|+ ` + 1 . (9.17)

We now need to look at the boundary conditions. In terms of ξ, oursolution is

u(ξ) = 2F1(|m| − ` , |m|+ ` + 1 , |m|+ 1; 12(1− ξ)) (9.18)

with ξ = cos θ. Hence, using Eqn.(9.12), we have Θ(θ) of the form

Θ`,m(θ) = (sin θ)|m| 2F1(|m| − ` , |m|+ ` + 1 , |m|+ 1; 12(1− cos θ)) (9.19)

where the indices ` and m have been introduces to uniquely identify eachsolution.

The variable cos θ can range between −1 and 1, so that ξ can cover thatrange as well, and z can also cover the same range. The difficulty with theboundary condition is that, at z = ±1, the 2F1 function diverges unless itis a polynomial containing finitely many terms. This will occur when a or b


is a negative integer, which in turn implies that |m| − ` must be a negativeinteger. Since we know that m is already an integer, we are left with no otheroption than imposing ` to be a positive integer greater than |m|.

We could, in principle, require that ` be a negative integer so that |m|+` + 1 is a negative integer. We will see, however, that this solution must berejected on physical grounds.

Thus we have almost completely justified the assumption that ` is a pos-itive integer, an assumption used to solve the radial equation.

9.3 Associated Legendre polynomials

Because of the numerous changes of variables involved in going from Eqn.(9.7)to Eqn.(9.19), it is more practical to study Θ`,m(ξ) directly rather than itssolution in terms of hypergeometric function.

The functions Θ`,m(ξ) are the so-called associated Legendre polynomials

P|m|` (ξ) . They are given as follows: for m = 0 , P 0

` (ξ) = P`(ξ) is the usual

Legendre polynomial. If |m| > ` , P|m|` (ξ) = 0 . For other values, one uses

look–up tables.

m ` P|m|` (ξ) P

|m|` (θ)

0 0 1 1

0 1 ξ cos θ

±1 1 −(1− ξ2)1/2 − sin θ

0 2 12(3ξ2 − 1) 1

2(3 cos2 θ − 1)

±1 2 3ξ(1− ξ)1/2 −3 sin θ cos θ

±2 2 3(1− ξ2) 3 sin2 θ

0 3 12ξ(5ξ2 − 3) 1

2cos θ(5 cos2 θ − 3)

±1 3 32(1− 5ξ2)(1− ξ2)1/2 −3

2(5 cos2 θ − 3)

±2 3 15ξ(1− ξ2) 15 cos θ sin2 θ

±2 3 −15(1− ξ2)3/2 −15 sin3 θ


They are normalized using

∫ 1

−1

dξPm` (ξ)Pm

`′ (ξ) =

0 if `′ 6= `

22`+1

(`+m)!(`−m)!

if `′= ` .

(9.20)

The associated Legendre polynomials are related to the Legendre poly-nomials P`(ξ) by

Pm` (ξ) = (1− ξ2)m/2 dm

dξmP`(ξ) . (9.21)

The Legendre polynomials are themselves obtained by numerous methods,one of which is the so–called Rodrigues’s formula:

P`(ξ) =1

2``!

d`

dξ`(ξ2 − 1)` . (9.22)

or by formal expansion of the generating function

G(ξ, h) = (1− 2ξh + h2)−1/2 = P0(ξ) + hP1(ξ) + h2P2(ξ) + . . .

=∞∑

`=0

h`P`(ξ) . (9.23)

9.4 Spherical harmonics

The full solutions to the angular equation are of the form Θ`,m(θ)Φm(ϕ). It isnormal usage to denote the properly normalized combination Θ`,m(θ)Φm(ϕ)by Y` m(θ, ϕ)

Y` m(θ, ϕ) = (−1)|m|√

(`−m)!(2` + 1)

4π (` + m)!P|m|` (cos θ)eimϕ (9.24)

These are known as the spherical harmonics. They satisfy the orthonormal-ization condition∫

dΩ (Y`1 m1(θ, ϕ))∗ Y`2 m2(θ, ϕ) =

δ`1`2δm1m2 = 1 if m1 = m2 and `1 = `2

0 otherwise.(9.25)

Note that dΩ = dϕ sin θdθ.The spherical harmonics are ubiquitous in three–dimensional problems

with spherical symmetry, because separation of variables always leads to


the same angular equations regardless of the problem. In other words, theangular part of this kind of problem is always the same, so that the solutionsneed to be found only once.

Because of their usefulness, spherical harmonics have been extensivelystudied and their properties have been tabulated by many sources. Beforewe study some of their properties, we establish some basic facts particularlypertinent to quantum mechanics.

First, Y` m(θ, ϕ) is complex unless m = 0, owing to the presence of a factoreimϕ. Next, the quantity

|Y` m(θo, ϕo)|2 dΩ =(`−m)!(2` + 1)

2 (` + m)!

(P|m|` (cos θo)

)2

dΩ (9.26)

where dΩ = sin θo dθdϕ, is the probability of finding a system with angularwavefunction given by Y` m(θ, ϕ)in a solid angle of opening dΩ centered a-round (θo, ϕo). This does not depend on the azimuthal angle ϕ. This doesnot mean that the probability density of a general system will not depend onϕ, just that, if the angular part of the wavefunction can be separated fromthe radial part and expressed by a single Y` m(θ, ϕ), the probability densitywill not depend on ϕ.

This situation is analogous to the separation of time and spatial variablesin the time–dependent wavefunction Ψ(r, θ, ϕ, t). If the wavefunction is sepa-rable in t and ~r, the probability density does not depend on t; the probabilitycan depend on t if the wavefunction is not separable.

The best way to plot the probability densities is to use polar plots. Insuch plots, the magnitude of |Y` m(θ, ϕ)|2 is given by the length of the lineextending from the origin to the point on the curve. The first few polargraphs for |Y` m(θ, ϕ)|2 are given in Figure 9.4.

9.5 Angular momentum operators

In this section, we would like to understand some the physics associated tothe spherical harmonics.

Consider the classical angular momentum vector ~L = ~r × ~p. In terms ofcomponents, and using

px 7→ px = −i~∂

∂x, etc. (9.27)


we find

Lx = ypz − zpy = −i~(

y∂

∂z− z

∂

∂y

)(9.28)

Ly = zpx − xpz = −i~(

z∂

∂x− x

∂

∂z

)(9.29)

Lz = xpy − ypx = −i~(

x∂

∂y− y

∂

∂x

)(9.30)

Converting to polar coordinates to bring out the θ, ϕ variables, we have

x = r sin θ cos ϕ y = r sin θ sin ϕ z = r cos ϕ (9.31)

and

Lz = −i~ r sin θ cos ϕ

(sin θ sin ϕ

∂

∂r+

cos θ

rsin ϕ

∂

∂θ+

cos ϕ

r sin θ

∂

∂ϕ

)

+i~ r sin θ sin ϕ

(sin θ cos ϕ

∂

∂ϕ+

cos θ

rcos ϕ

∂

∂θ− sin ϕ

r sin θ

∂

∂ϕ

)(9.32)

which, after some rapid simplification, can be written cleanly as

Lz = −i~∂

∂ϕ(9.33)

Similarly:

Lx = −i~(− sin ϕ

∂

∂θ− cot θ cos ϕ

∂

∂ϕ

)(9.34)

Ly = −i~(

cos ϕ∂

∂θ− cot θ sin ϕ

∂

∂ϕ

)(9.35)

Next, consider L2 ≡ L2x + L2

y + L2z. This can eventually be brought to the

form

L2 = −~2

[1

sin2 θ

∂2

∂ϕ2+

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)](9.36)

This form is important because, up to factors of ~2, this is just the angularequation. In other words, the angular equation can be rewritten

L2Y` m(θ, ϕ) = ~2`(` + 1)Y` m(θ, ϕ) (9.37)


The physics of Eqn.(9.37) is simple: spherical harmonics are eigenstates ofthe operator L2. Thus, we can associate with them a well–defined value ofthe classical quantity ~L·~L, in the sense that the fluctuation of L2 is zero whenthe angular wavefunction is of the form Y` m(θ, ϕ). In other words, sphericalharmonics describe states that have well–defined value for the magnitude ofthe total angular momentum. Since the square of the magnitude is ~2`(`+1),the magnitude itself is usually defined as ~

√`(` + 1).

Furthermore, note that Y` m(θ, ϕ) is also an eigenstate of Lz:

LzY` m(θ, ϕ) = −i~∂

∂ϕ(−1)|m|

√(`−m)!(2` + 1)

4π(` + m)!P|m|` (cos θ) eimϕ ,

= m~ (−1)|m|√

(`−m)!(2` + 1)

4π(` + m)!P|m|` (cos θ) eimϕ , (9.38)

= m~Y` m(θ, ϕ)

since the ϕ dependence of Y` m(θ, ϕ) is of the form eimϕ. Hence, when theangular portion of a wavefunction is of the form Y` m(θ, ϕ), the system de-scribed by this wavefunction has a well–defined value of the projection Lz ofthe angular momentum vector ~L, i.e. ∆Lz = 0 for such states.

We are now in a position to justify physically the choice of ` > 0 inEqn.(9.19). Clearly, the projection Lz of ~L on the z–axis cannot be greater

than the length√

~L · ~L = ~√

`(` + 1). In other words, |m| must always besmaller than `, which is the last piece required to show that ` must be apositive integer.

92

Part II

Bras and ketsIn this part, we will introduce a powerful notation: bras and kets. This allowsus to introduce the position and momentum representations, the notion ofstate vector, and to illustrate further properties of wavefunctions.

Bras and kets 93

10 Bras, kets and abstract state vectors

10.1 Introducing the ket vector

The bound state solutions ψi(ξ) of the time–independent Schrodinger equa-tion all have the following properties:

i) They are normalized, i.e.

∫dµξ ψ∗i (ξ)ψi(ξ) = 1 ,

where the integral covers the whole range of ξ and dµξ is the appropriateintegration element. (If you care to know, dµξ is called an invariantmeasure.)

ii) Two functions with different eigenvalues are orthogonal:

∫dµξ ψ∗i (ξ)ψj(ξ) = 0 , (i 6= j) .

This is neatly summarized in the single statement

∫dµξ ψ∗i (ξ)ψj(ξ) = δij , (10.1)

where the Kronecker δ symbol is defined as

δij =

1 if i = j0 if i 6= j .

(10.2)

Compare these properties with those of the basis vectors e1 = x, e2 =y, e3 = z. The former satisfy

ei · ej = δij , (10.3)

Thus we see that ∫dµξ ψ∗i (ξ)ψj(ξ) (10.4)

functions as some sort of inner product. In fact, the definition of Eqn.(10.4)satisfies all the properties of an inner product.

Bras and kets 94

What else do we know about the properties of basis vectors in ordinaryspace? Any vector in that space can be written as a linear combination ofthree linearly independent, preferably orthonormal vector, like e1, e2, e3.

Although we will not prove it just now, the solutions of the time–independentSchrodinger equation also form a basis of orthonormal vectors, in the sensethat any wavefunction can be written as a linear combination of these func-tions.

Now suppose you are given two arbitrary vectors in space, say ~r and ~s,and want to compute the inner product ~r · ~s. One way to do this is to firstwrite a vector in terms of its components:

~r = e1 r1 + e2 r2 + ez r3 . (10.5)

The component r1, r2 and r3 are obtained as

r1 = e1 · ~r , r2 = e2 · ~r , r3 = e3 · ~r .

so that Eqn.(10.5) can be written as

~r = e1 (e1 · ~r) + e2 (e2 · ~r) + ez (e3 · ~r) ,

= (e1 e1 ·+e2 e2 ·+e3 e3·)~r , (10.6)

=∑

i

ei (ei · ~r) .

In an expression like e1 e1·, the dot denotes the operation of taking the innerproduct, so e1· means take the inner product of e1 with whatever is on theright.

The same kind of expansion can be used for ~s, and the inner productobtained by the use of

~r · ~s = r1s1 + r2s2 + r3s3 =∑

i

risi . (10.7)

If the vectors are complex, then we need to use a slightly different form:

~r · ~s = r∗1s1 + r∗2s2 + r∗3s3 =∑

i

r∗i si . (10.8)

The complex conjugation is required if the inner product of a vector withitself is to be understood as the length of this vector, something that isnecessarily positive. Clearly, with this definition,

~r · ~s = (~s · ~r)∗ . (10.9)

Bras and kets 95

Let us compare all this, and Eqn.(10.7) in particular, with the expressionof the inner product of two arbitrary functions f(ξ) and g(ξ). For notationalpurposes, we use 〈f | g〉 to denote this inner product, rather than something

like ~f · ~g. Using Eqn.(10.4), we have

〈f | g〉 =

∫dµξ f ∗(ξ) g(ξ) , (10.10)

〈f | g〉∗ =

∫dµξ g∗(ξ)f(ξ) ,

= 〈g | f〉 (10.11)

If we remember that an integral is just a continuous sum, we can see that thesummation over the components in Eqn.(10.7) is replaced by integral over ξ.This suggests very strongly that, for some specified ξo, we may think of thenumber f(ξo) as some sort of component of the ”vector” f in the directionof the ”unit” vector ξo, something which we denote by f(ξo) = 〈ξo | f〉.

In other words, think of f as some abstract vector which we denote by|f〉. Just like ~r can be expressed as a column vector explicitly displaying itsthree components in the basis spanned by ei, i = 1, 2, 3,

~r =

rx

ry

rz

,

we can thing of |f〉 as a column vector explicitly displaying its ”components”in the basis spanned by the ”unit” vectors |ξi〉, i = −∞, . . . ,∞:

~r =

rx

ry

rz

=

e1 · ~re2 · ~re2 · ~r

⇔ |f〉 =

...

f(ξi−1)

f(ξi)

f(ξi+1)...

=

...

〈ξi−1 | f〉〈ξi | f〉〈ξi+1 | f〉

...

.

(10.12)Note that, for display purposes, one is forced to use discrete indices on thecoordinate ξ but it should really be continuous.

Thus, the ket vector |ψ〉 is the abstract version of the wavefunction ψ(ξ)in the basis spanned by |ξi〉, i = −∞, . . . ,∞ in the same sense that the

Bras and kets 96

vector ~r is the abstract version of the triple (r1, r2, r3) of components of ~rexpressed in the basis spanned by e1, e2, e3.

10.2 Wavefunctions in the braket notation

What are the advantages of using |ψ〉 rather than ψ(ξ)? The |ψ〉 does notdepend on the choice of basis, just like writing ~r denotes the vector quiteabstractly, without reference to any basis. This is important because, as youknow from experience, choosing the right basis often increases the insightinto a problem and can lead more rapidly to the correct result.

Please note carefully that the basis vectors ei have been left quite un-specified: they could be basis vectors in spherical, cartesian, or any othercoordinate convenient for the problem at hand.

Thus, for instance, there are some problems where an abstract ket |`m〉with specified ` and m is conveniently expressed in spherical coordinates, inwhich case its explicit expression becomes a Y` m(θ, ϕ):

Y` m(θ, ϕ) ≡ 〈θ, ϕ | `m〉 (10.13)

In other problems, the so–called polynomial basis simplifies the calculation,and we have

z`+m1 z`−m

2√(` + m)!(`−m)!

≡ 〈z1, z2 | `m〉 . (10.14)

Likewise, a wavefunction ψn`m(r, θ, ϕ) solution to the time–independentSchrodinger equation for the three–dimensional harmonic oscillator can bewritten as 〈rθϕ |n`m〉. In Cartesian coordinates, one would have

ψnxnynz(x, y, z) = 〈xyz |nxnynz〉 . (10.15)

The wavefunctions solutions to the one–dimensional harmonic oscillatorproblem are often denoted by the kets |n〉, so that, using x as the coordinate,we have

ψn(x) = 〈x |n〉 . (10.16)

10.3 The bra vector

Before continuing, it is good to explore the properties of the so–called bravector 〈f |. This is the continuous analog of the discrete version of symbol

Bras and kets 97

(~r·) for the vector ~r, in the sense that the expression

~r · ~s =∑

i

r∗i si (10.17)

for ordinary vectors becomes the expression

〈f | g〉 =

∫dµξf

∗(ξ)g(ξ) , (10.18)

if we use the continuous sum

〈f | ↔∫

dµξf∗(ξ) . (10.19)

Note that, just like Eqn.(10.17) has an implicit choice of basis in it (thebasis where the components of ~s and ~r are si and ri, respectively), Eqn.(10.19)is also basis dependent. This illustrates that, although the inner product ~r ·~sdoes not depend on the basis, one must choose a basis in order to actuallycompute the value of the inner product.

Formally, the bra vectors have all the properties of vectors, meaning theycan be added together, multiplied by scalars etc. They are called linearfunctionals in the mathematical literature.

10.4 Projection and closure

Consider the combinationPi = ei ei· (10.20)

It is quite clear that, if ~r is any vector

Pi ~r = ei ri , ri = ei · ~r . (10.21)

In other words, Pi projects from ~r the component of ~r that is along the unitvector ei. Pi is called a projection operator.

The sum ∑i

Pi = 1l , (10.22)

the unit operation since, when you ”feed” it an arbitrary vector ~r, you get ~rback: (∑

i

Pi

)~r =

∑i

ei ri = ~r , (10.23)

Bras and kets 98

i.e.nothing happens to ~r. Eqn.(10.22) is called the closure relation, and itexpresses the fact that the basis vectors ei span the space in which ~r lives.

When expressed in terms of bras and kets, Eqn.(10.22) looks like a con-tinuous sum:

1l =

∫dµξ|ξ〉〈ξ| (10.24)

Since 1l does nothing, we have

1l |ψ〉 = |ψ〉 ,〈φ |ψ〉 = 〈φ| 1l |ψ〉 ,

=

∫dµξ 〈φ | ξ〉〈ξ |ψ〉 , (10.25)

=

∫dµξ φ∗(ξ)ψ(ξ) , (10.26)

where 〈φ | ξ〉 = 〈ξ |φ〉∗, Eqn.(10.11), has been used.Hence, we recover from all these formal properties all the definitions that

we started with.

10.5 |x〉 and ψ(x)

We now specialize the discussion to one dimension only, with ξ = x, anddefine the ket vector |xo〉 as this vector that satisfies

x|xo〉 = xo|xo〉 , (10.27)

i.e. |xo〉 is an eigenvector of x with eigenvalue xo. We will use the set |x〉as basis vectors and assume Eqn.(10.24) holds and takes the form

1l =

∫ ∞

−∞dx |x〉〈x| . (10.28)

Then:

|xo〉 =

∫ ∞

−∞dx |x〉〈x |xo〉 , (10.29)

which implies〈x |xo〉 = δ(x− xo) . (10.30)

In Eqn.(10.30), δ(x − xo) is the Dirac δ function, a generalization to con-tinuous variable of the Kronecker delta of Eqn.(10.2). Properties of the δfunction are studied in Sec.10.6.

Bras and kets 99

For the moment, observe that, in accordance with Eqn.(10.12), we have

ψ(xo) = 〈xo |ψ〉 . (10.31)

In accordance with the statistical interpretation of Born, the quantityψ(xo)

∗ψ(xo) dx is the probability of finding the system in a bin of size dxcentered around xo. This can be expressed in terms of bras and kets as

ψ(xo)∗ψ(xo) dx = 〈ψ |xo〉〈xo |ψ〉 dx = |〈xo |ψ〉|2 dx , (10.32)

showing how inner products are related to probability densities. We willcome back to this later.

10.6 Properties of δ

10.6.1 δ(x) in one dimension

Basically, a δ-function is a generalization to continuous variables of the “dis-crete” δ:

δij =

1 if i = j0 if i 6= j

(10.33)

The Dirac δ-function is defined informally as

δ(x) =

∞ if x = 00 if x 6= 0

(10.34)

with ∫ ∞

−∞f(x)δ(x)dx = f(0) . (10.35)

As a mathematical object, δ is really a limit of functions. For instance,

limΓ→0

Γ2π

(x− b)2 + 14Γ2

= δ(x− b) (10.36)

is a δ-function because, if we integrate

∫ ∞

−∞limΓ→0

(Γ2π

(x− b)2 + 14Γ2

)f(x) dx

= limΓ→0

1

2π

∫ ∞

−∞

Γ

(x− b)2 + 14Γ2

f(x) dx , (10.37)

= f(b) . (10.38)

Bras and kets 100

Indeed, let(x− b) = 1

2Γ y , dx = 1

2Γdy . (10.39)

Then, the indefinite integral becomes

limΓ→0

1

2π

∫4Γ

Γ2(1 + y2)f(1

2Γ y + b) Γ

2dy

= limΓ→0

1

π

∫1

(1 + y2)

[f(b) +

Γy

2f ′(ξ)

]dy , b < ξ <

Γy

2+ b

=1

πf(b) arctan y + lim

Γ→0

Γ

2πf ′(ξ)

∫y

1 + y2dy , (10.40)

=1

πf(b) arctan y + 0 , (10.41)

and thus Eqn.(10.37) boils down to

limΓ→0

1

2π

∫ ∞

−∞

Γ

(x− b)2 + 14Γ2

f(x) dx =1

πf(b) arctan y

∣∣∣∞

−∞= f(b) . (10.42)

To show that

δ(x2 − a2) =1

|2a|(δ(x− a) + δ(x + a)) , (10.43)

letf(x) be any function with f(±∞) → 0. Then

∫ ∞

−∞dxδ(x2 − a2)f(x) =

∫ 0

−∞dxδ(x2 − a2)f(x)

+

∫ −∞

0

dxδ(x2 − a2)f(x) . (10.44)

Let y = x2 − a2, x2 = y + a2 x = ±√

y + a2 dx = ± 1

2|√

y+a2|dy. Making

Bras and kets 101

these substitutions, we find∫ ∞

−∞dxδ(x2 − a2)f(x)

=

∫ −a2

−∞

−dy

2√

y + a2δ(y) f(−

√y + a2) +

∫ ∞

−a2

dy

2√

y + a2δ(y) f(

√y + a2)

=

∫ ∞

−a2

+dy

2√

y + a2δ(y) f(−

√y + a2) +

∫ ∞

−a2

dy

2√

y + a2δ(y) f(

√y + a2)

=f(−

√a2)

2|a| +f(√

a)

2|a| =1

|a|(f(a) + f(−a))

=1

2|a|∫ ∞

−∞f(x)(δ(x− a) + δ(x + a))dx , (10.45)

which proves the result.Eqn.(10.36) is not the only function that ”behaves like δ” in some limit.

The following is a non–exhaustive list of equally valid choices:

δ(x− b) = limΓ→0

e−(x−b)2/Γ2

Γ√

π, (10.46)

= limΓ→0

e−|x−b|/Γ

2 Γ, (10.47)

= limΓ→0

sin2(

x−bΓ

)

πΓ(x−bΓ

)2. (10.48)

The δ function satisfies the formal identities:

i) δ(x) = δ(−x),

ii) xδ(x) = 0,

iii) δ(ax) = |a|−1 δ(x),

iv) δ(x2 − a2) = |2a|−1 [δ(x− a) + δ(x + a)],

v)∫

δ(a− x)δ(x− b) dx = δ(a− b),

vi) f(x)δ(x− a) = f(a)δ(x− a),

vii) δ[f(x)] = δ(x− xo)/|f ′(xo)|, where xo is such that f(xo) = 0.

To prove any of these statements, one must supply to an identity an arbitrarycontinuous function f(x) which vanishes at infinity, and integrate over x.

Bras and kets 102

10.6.2 δ(~r) in three dimensions

In three dimensions, the three–dimensional δ function is defined to haveproperties similar to its one–dimensional cousin. The basic property

∫d3~r δ(~r − ~ro)f(~r) = f(~ro) (10.49)

remains.If a cartesian system of coordinate is chosen, then Eqn.(10.49) can be

factored in the obvious way:

δ(~r − ~r0) = δ(x− xo)δ(y − yo)δ(z − zo) , (10.50)

with∫

d3~r = dx dy dz.In spherical coordinates, we have

δ(~r − ~r0) =1

r2 sin θδ(r − ro)δ(θ − θo)δ(ϕ− ϕo) , (10.51)

=1

r2δ(r − ro)δ(cos θ − cos θo)δ(ϕ− ϕo) . (10.52)

Please note that δ(~r − ~ro) has a vectorial argument, whereas δ(r − ro) has ascalar argument: the former is defined in three dimensions whereas the latteris a one–dimensional δ.

10.7 |p〉 and the Fourier transform

N.B.: This is covered in section 10.3 of Haberman. How to go from ournotation to his will be clarified below.

We now introduce the ket |p〉. This is an eigenstate of the operator p inthe sense that

p|po〉 = po|po〉 . (10.53)

Since |po〉 is an eigenstate of p, |po〉 has a well–defined value of momentum.Using the basis vectors |x〉, we can easily obtain the wavefunction de-

scribing a particle with a well–defined momentum: this is just a free particlewith wavefunction

〈x | p〉 = ψp(x) = Aeipx/~ , (10.54)

where ~k = p and where the constant A will be determined below. Hence,we have obtained (almost) an expression for 〈x | p〉.

Bras and kets 103

Consider now the general ket |ψ〉 and the expression

ψ(p) = 〈p |ψ〉 =

∫ ∞

−∞dx 〈p |x〉〈x |ψ〉 ,

=

∫ ∞

−∞dxA∗e−ipx/~ ψ(x) , (10.55)

where the complex conjugate of Eqn.(10.54) has been used. Thus, ψ(p) isthe Fourier transform of ψ(x).

Now, if Eqn.(10.32) gives the probability of finding the system in a binof size dx around xo, then we will say that

ψ(po)∗ψ(po) dp = |〈po |ψ〉|2 dp (10.56)

is the probability of finding the system with momentum p between po andpo + dp. The probability distribution in momentum space is the Fouriertransform of the probability distribution in position space!

We now pin down the multiplicative factor A in 〈x | p〉 = Aeipx/~. In thedefinition of Fourier transform pairs, there is an extra factor 1/2π that mustbe accounted for. The choice

〈x | p〉 =1√2π~

eipx/~ (10.57)

leads to the expressions

1l =

∫ ∞

−∞dx |x〉〈x| (10.58)

and

1l =

∫ ∞

−∞dp |p〉〈p| . (10.59)

for the closure relations in position and momentum space, and to the correctfactor of 1/2π in the expression of Fourier transform pairs. We will adoptthese definitions. (This is not quite the definition of Haberman. See below.).

We can use this to show a useful equality between the δ function and

Bras and kets 104

Fourier transform. Start from Eqn.(10.30) and insert Eqn.(10.59):

δ(x− xo) = 〈x |xo〉 , (10.60)

=

∫ ∞

−∞dp 〈x | p〉〈p |xo〉 , (10.61)

=1

2π~

∫ ∞

−∞dp eipx/~ e−ipxo/~ , (10.62)

=1

2π~

∫ ∞

−∞dp eip(x−xo)/~ , (10.63)

showing that δ(x) is the Fourier transform of constant momentum space func-tion ψ(p) = 1. In other words, a sharp value of x corresponds to a perfectlyundetermined momentum distribution, since the momentum distribution isconstant (every value of p is as probable as every other value of p).

Inversely, we have

δ(p− po) = 〈p | po〉 , (10.64)

=

∫ ∞

−∞dx 〈p |x〉〈x | po〉 , (10.65)

=1

2π~

∫ ∞

−∞dx e−ipx/~ eipox/~ , (10.66)

=1

2π~

∫ ∞

−∞dx e−i(p−po)x/~ , (10.67)

Note that Haberman, in the tradition of mathematicians, uses the con-vention different from the convention in physics.

Haberman ↔ us

x ↔ p/~

f(x) =

∫ ∞

−∞F (ω)e−iωx dω ↔ ψ(p) = 1√

2π~

∫ ∞

−∞ψ(x)e−ipx/~ dx

F (ω) = 12π

∫ ∞

−∞f(x)eiωx dx ↔ ψ(x) = 1√

2π~

∫ ∞

−∞ψ(p)eipx/~ dp

Bras and kets 105

10.8 The x and p representations

Let us now consider 〈x | p〉 = eipx/~. By definition p|p〉 = p|p〉 so we have

〈x|p|p〉 = p〈x | p〉 = peipx/~ . (10.68)

Hence we deduce

〈x|p|p〉 ↔ −i~∂

∂xeipx/~ , (10.69)

↔ −i~∂

∂x〈x | p〉 , (10.70)

Now, using closure on p, we have

〈x|p|ψ〉 =1

2π~

∫ ∞

−∞dp 〈x|p|p〉〈p |ψ〉 . (10.71)

Using Eqn.(10.70), this becomes

〈x|p|ψ〉 =1

2π~

∫ ∞

−∞dp

(−i~

∂

∂x〈x | p〉

)〈p |ψ〉 . (10.72)

But we are differentiating w/r to x and integrating w/r to p : we can pullthe derivative out of the integral sign:

〈x|p|ψ〉 = −i~∂

∂x

1

2π~

∫ ∞

−∞dp〈x | p〉〈p |ψ〉 , (10.73)

= −i~∂

∂x〈x|

(1

2π~

∫ ∞

−∞dp|p〉〈p|

)|ψ〉 , (10.74)

= −i~∂

∂x〈x| 1l |ψ〉 , (10.75)

= −i~∂

∂x〈x |ψ〉 , (10.76)

= −i~∂

∂xψ(x) . (10.77)

With the previous equation, we have just justified the use of rule

p 7→ −i~∂

∂x(10.78)

when p acts on any function of x.

Bras and kets 106

Likewise, since x|xo〉 by definition, we have

〈x|x|ψ〉 =

∫ ∞

−∞dx 〈x|x|x〉〈x |ψ〉 , (10.79)

=

∫ ∞

−∞dx x〈x | x〉x〈x |ψ〉 , (10.80)

=

∫ ∞

−∞dx x δ(x− x)xψ(x) , (10.81)

= xψ(x) , (10.82)

where Eqn.(10.30) has been used. This justifies the use of

x 7→ x (10.83)

when x acts on functions of x.Eqns.(10.78) and (10.82) are referred to as the x–representation of the

operators p and x. Likewise, ψ(x) is called the x–representation (or positionrepresentation) of the wavefunction or the state vector.

Besides functions of x, like ψ(x), we now know that it is possible toexpress the contents of a state vector |ψ〉 not as a function of the positionbut as a function of the momentum p. What are the expressions of p and xwhen they act on functions of p?

Start with

〈p|x|ψ〉 =

∫ ∞

−∞dx 〈p|x|x〉〈x |ψ〉 , (10.84)

=

∫ ∞

−∞dx x〈p | x〉〈x |ψ〉 , (10.85)

=

∫ ∞

−∞dx xe−ipx/~〈x |ψ〉 , (10.86)

=

∫ ∞

−∞dx

(i~

∂

∂pe−ipx/~

)〈x |ψ〉 , (10.87)

= i~∂

∂p

∫ ∞

−∞dx 〈p | x〉〈x |ψ〉 , (10.88)

= i~∂

∂p〈p|

(∫ ∞

−∞dx |x〉¯x

)|ψ〉, , (10.89)

= i~∂

∂p〈p |ψ〉 = i~

∂

∂pψ(p) . (10.90)

Bras and kets 107

Hence, when x acts on functions of p, we have the rule

x 7→ i~∂

∂p. (10.91)

Note the sign difference with Eqn.(10.78).Lastly, we have, using p|po〉 = po|po〉, we have

〈p|p|ψ〉 =

∫ ∞

−∞dp 〈p|p|p〉〈p |ψ〉 (10.92)

=

∫ ∞

−∞dp p 〈p | p〉〈p |ψ〉 (10.93)

=

∫ ∞

−∞dp p δ(p− p) 〈p |ψ〉 (10.94)

= p〈p |ψ〉 (10.95)

= pψ(p) , (10.96)

which shows that, acting on momentum wavefunctions, we have

p 7→ p . (10.97)

Using Eqns.(10.91) and (10.96), we can write the Schrodinger equationin momentum space:

1

2m〈p|p2|ψ〉+ 〈p|V (x)|ψ〉 = E〈p |ψ〉 , (10.98)

⇓ ⇓1

2mp2ψ(p) + V (−i~

∂

∂p)ψ(p) = Eψ(p) . (10.99)

This formulation is useful in some selected problem where V is constant,periodic (like in a crystal) or momentum dependent (as in some nuclearmodels).

10.9 On the use of |k〉In many problems, it is simpler to use the plane wave expression eikx, with~k = p. This has the advantage of not having any annoying factors of ~ in

Bras and kets 108

numerous formulae. The adjustments as simple to make: simply set ~ = 1in any formula involving p and replace every p by k. Thus:

〈x | k〉 =1√2π

eikx , (10.100)

δ(x− xo) =1

2π

∫ ∞

−∞dk eik(x−xo) , (10.101)

δ(k − ko) =1

2π

∫ ∞

−∞dx e−i(k−ko)x , (10.102)

ψ(k) =1√2π

∫ ∞

−∞ψ(x)e−ikx dk , (10.103)

ψ(x) =1√2π

∫ ∞

−∞ψ(k)eikx dx , (10.104)

10.10 Wave packets and Dirac notation

The objective of this section is to illustrate the versatility and power ofDirac notation by solving problems related to the propagation in time ofwave packets.

10.10.1 General formulation

Suppose that, at t = 0, we are given an arbitrary state vector |ψ(0)〉 for thesystem. What is |ψ(t)〉?

For the free particle, the p representation is particularly useful. We haveV = 0 and E = p2

2m. The solutions to the time–independent Schrodinger

equation form a continuous set |p〉. In the x–representation, we have

〈x | p〉 = eipx/~ , (10.105)

with E = p2

2m. Each |p〉 evolves in time according to

|p(t)〉 = |p〉eiEt/~ , (10.106)

where E = p2/2m. It is worth mentioning that, for every value of E, thereis +p and −p such that E = p2/2m.

At any time t, we have

1l =

∫ ∞

−∞dp |p(t)〉〈p(t)| =

∫ ∞

−∞dp |p〉〈p| . (10.107)

Bras and kets 109

As we know |p(t)〉, we can write

|ψ(0)〉 =

∫ ∞

−∞dp |p(0)〉〈p(0) |ψ(0)〉 ,

=

∫ ∞

−∞dp |p(0)〉〈p |ψ(0)〉 . (10.108)

Now, 〈p |ψ(0)〉 is an inner product, i.e. a number, depending on p, possiblycomplex, but still a number. In other words, it is a complex function of pbut not of t. Thus, it is time–independent, in the sense that once evaluatedat t = 0, this inner product remains the same. Let us call this number ψ(p).

It is worth mentioning that, for every value of E, there is +p and −p suchthat E = p2/2m. The time evolution of |ψ(t)〉 is then obtained by using theknown evolution of |p(0)〉 in time:

|ψ(t)〉 =

∫ ∞

−∞dp |p(t)〉ψ(p) , (10.109)

|ψ(t)〉 =

∫ ∞

−∞dp eip2t/2m~|p〉ψ(p) . (10.110)

As a wave function in position space, we have

ψ(x, t) = 〈x |ψ(t)〉 (10.111)

=

∫ ∞

−∞dp eip2t/2m~〈x | p〉ψ(p) , (10.112)

=1√2π~

∫ ∞

−∞dp eip2t/2m~eipx/~ψ(p) . (10.113)

Alternatively, using the |k〉 basis, one obtains in a similar way:

ψ(x, t) =1√2π

∫ ∞

−∞dk ei~k2t/2meikxψ(k) , (10.114)

whereψ(k) = 〈k |ψ(0)〉 . (10.115)

10.10.2 Example: the Gaussian wave packet

Suppose that

ψko(x, 0) =1

(2π)14

e−x2/4σ2

eikox , (10.116)

Bras and kets 110

where ko is real and measured in [m]−1 and σ is real positive, with unitsof [m]−1 also. The parameter σ is related to the width of the wave packet.Please note carefully the presence of the eikox factor. Although the spatialprobability density does not depend on ko:

|ψko(x, 0)|2 =1√2π

e−x2/2σ2

, (10.117)

the wavefunction ψko(x, 0) does and this will lead to different physical pre-dictions.

What is ψ(x, t)? The first step is to compute ψ(k, 0). This is just theFourier transform of ψ(x, 0). By definition,

ψ(k, 0) =1√2π

∫ ∞

−∞dx

1

(2π)14

eikxe−x2/4σ2

e−ikox (10.118)

There are several steps to evaluating this integral, the first of which is tocomplete the square:

ψ(k, 0) =1√2π

∫ ∞

−∞dx

e−( 14(x/σ)2+i(k−ko)x+(iσ(k−ko))2−(iσ(k−ko))2)

(2π)14

(10.119)

=1

(2π)34

∫ ∞

−∞dx e+(iσ(k−ko))2e−( x

2σ+i(k−ko))2

=1

(2π)34

e−σ2(k−ko)2∫ ∞

−∞dx e−( x

2σ+i(k−ko)2) (10.120)

To compute the integral, let y = x2σ

+iko so that dy = 12σ

dx, or dx = 2σdy.Then ∫ ∞

−∞dxe−( x

2σ+iko)2 = 2σ

∫ ∞

−∞dye−y2

= 2σ√

π (10.121)

(This integration is not strictly correct because the limits of integrationshould be adjusted to reflect the changes in variables, but the final resultis correct.) Therefore, we have found:

ψ(k) =

(2

π

)1/4

σ e−σ2(k−ko)2 . (10.122)

Notice how the factor of eikox has crept up in the momentum space wavefunction, which is now a Gaussian with center shifted to ko.

Bras and kets 111

Now, we need to plug this in

ψ(x, t) =1√2π

∫ ∞

−∞dk e−i~k2t/2meikxψ(k) (10.123)

=1√2π

(2

π

)1/4

σ

∫ ∞

−∞dk e−i~k2t/2meikxe−σ2(k−ko)2 (10.124)

It is interesting to verify that this is still a solution of the Schrodingerequation:

i~∂ψ(x, t)

∂t=

p2

2mψ(x, t) (remember that V=0)!

i~∂ψ(x, t)

∂t=

σ√2π

(2

π

)1/4 ∫ ∞

−∞dk

(i~

∂

∂t

)e−i~k2t/2meikxe−σ2(k−ko)2

=σ√2π

(2

π

)1/4 ∫ ∞

−∞dk i~

(−i~k2

2m

)e−i~k2t/2meikxe−σ2(k−ko)2

=σ√2π

(2

π

)1/4 ∫ ∞

−∞dk~2k2

2me−i~k2t/2meikxe−σ2(k−ko)2 .(10.125)

Note that

i~∂ψ(x, t)

∂t6= constant · ψ(x, t) , (10.126)

so this is certainly not a solution with well–defined energy.Next, we need to compute

p2

2mψ(x, t) . (10.127)

Since ψ(x, t) is expressed in the x-representation, we have

p = −i~∂

∂x, (10.128)

p2

2m= − ~

2

2m

∂2

∂x2, (10.129)

Bras and kets 112

so that

p2ψ(x, t) =1√2π

(2

π

)1/4

σ

∫ ∞

−∞dk e−i~k2t/2m

(− ~

2

2m

∂2

∂x2eikx

)e−σ2(k−ko)2

=σ√2π

(2

π

)1/4 ∫ ∞


(−~

2(−ik)2

2m

)eikxe−σ2(k−ko)2

=σ√2π

(2

π

)1/4 ∫ ∞


(~2k2

2m

)eikxe−σ2(k−ko)2

= i~∂

∂tψ(x, t)! (10.130)

It works!

10.10.3 Qualitative feature of the Gaussian wave packet

We now need to evaluate

σ√2π

(2

π

)1/4 ∫ ∞

−∞dk e−i~k2t/2m eikx e−σ2(k−ko)2 (10.131)

This particular integral can be obtained in closed form. However, to avoidinessential clutter, we will set m = 1 kg and σ = 1 meter.

Using tables, we find

∫ ∞

−∞dk e−i~k2t/2eikxe−(k−ko)2 =

√2π

e−(x2−2i~k2o−4ikox)/(2i~t+4σ2)

√2 + i~t

. (10.132)

So

ψ(x, t) =

(2

π

)1/4e(ix2−2~k2

ot+4kox)/(2~t−4i)

√2 + i~t

. (10.133)

Note that, for t = 0, we recover

ψ(x, 0) =

(2

π

)1/4e(ix2+4kox)/(−4i)

√2

, ,

=

(1

2π

)1/4

eikox−x2/4 , (10.134)

as required.

Bras and kets 113

The probability density ψ∗(x, t)ψ(x, t) is given by

Pψ(x, t) =

√2

π

e(−ix2−2~k2ot+4kox)/(2~t+4i)

√2− i~t

e(ix2−2~k2ot+4kox)/(2~t−4i)

√2 + i~t

,

=

√2

π

e−2(x−~kot)2/(4+~2t2)

√4 + ~2t2

(10.135)

This is a real function (ouf!) At t = 0, we have

Pψ(x, 0) =1

2

√2

πe−2x2/4 =

1√2π

e−x2/2 (10.136)

so that checks out.We can ”easily” calculate 〈x(t)〉. However, before we launch into the

mathematics, let us analyze the problem from a physical perspective. Forfixed t = to, the maximum of Pψ(x, to) is to be found at x = ~kot. If werecall that ~k = p, we see that, for successive values of time ti+1 > ti, themaximum of the probability density is displaced in the positive x at velocity~ko. Thus, the factor of eikox in the initial ψko(x, 0) makes a huge differencein the physics of the problem, this despite the fact that ψ∗ko

(x, 0)ψko(x, 0)does not depend on ko.

Next, the ”width” of the wave packet is proportional to the√~2t2 + 4,

which occurs in the argument of the exponential. For large negative t, thewidth is large, decreasing as t approaches zero, and increasing again forincreasing positive t. Broadly speaking, for t < 0, the wave packet ”collect”itself until it reaches its thinnest value at t = 0 and then broadens again fort > 0.

Let us now see this in more details. The average value of x(t) is given by

〈x(t)〉 =

∫ ∞

−∞dx x Pψ(x, t)

= ~kot . (10.137)

Thus, the average position of the wave packet, i.e. the ”center of gravity”,follows the maximum. This is not unexpected.

The average momentum is obtained by first computing the average k and

Bras and kets 114

multiplying by ~. Thus,

〈p(t)〉 = ~∫ ∞

−∞dk |ψ(k)|2k , (10.138)

= ~√

2

π

∫ ∞

−∞dk e−(k−ko)2k , (10.139)

= ~ko , (10.140)

again expected since the momentum distribution is symmetrical about k =ko. Alternatively, we have

〈p〉ψ =d

dt(〈x(t)〉) = ~ko , (recall m = 1.) (10.141)

Next, consider

〈x2(t)〉 =

∫ ∞

−∞dx x2 Pψ(x, t) (10.142)

=1

4

(4 + ~2(1 + 4k2

o )t2)

(10.143)

(If you’re worried about the units looking funny, it is because we have setm = 1 and σ = 1.) Thus

∆x(t) =√〈x2〉 − 〈x〉2 , (10.144)

=

√1 +

~2t2

4(10.145)

In particular, at t = 0, we have ∆x = 1.Now, 〈p2〉:

〈k2〉 =

∫ ∞

−∞dk k2|ψ(k)|2 ,

=1

4+ k2

o , (10.146)

and thus∆p = 1

2~ . (10.147)

The product

∆p∆x = 12~√

1 + ~2t2

4≥ ~

2(10.148)

in accordance with the Heisenberg uncertainty principle.

Scattering states 115

11 Scattering states

We now consider the problem of solving the time–independent Schrodingerequation for states with E > V (±∞). From our previous discussion of thequalitative features of the solutions of the time–independent Schrodingerequation, we have

− ~2

2m

d2

dη2ψ(η) = (E − V (η))ψ(η), (11.1)

with a positive right hand side: the solution will be of an oscillatory type,with ψ(±∞) not necessarily zero. This kind of solution, known as scatteringstates, describe ”free” (rather than bound) particles.

For convenience, we will limit our discussion to one–dimensional prob-lems, where η = x. Also for convenience, we will consider only cases whereV (x) is a rectangular barrier or rectangular well.

11.1 The potential step

Recall first the case where

V (x) =

0 if x ≤ 0,V0 if x > 0.

(11.2)

We assume E > V0.

V 0

0

x=0

E>V

Figure 15: The potential step.


In such problems, it is convenient to define

k1 =

√2mE

~2, k2 =

√2m(E − V0)

~2(11.3)

We assume a particle incoming from −∞, so that, in the region on the leftof the barrier, region 1, we have

ψ(x) = Aeik1x + AΓ e−ik1x, (11.4)

where A and R are complex. On the right of the barrier, region 2, we have

ψ(x) = Aτ eik2x, (11.5)

with T complex.As the discontinuity in the potential is finite, the wave function and its

derivative must be continuous at the point of discontinuity. This immedi-ately produces the two equations

A + AΓ = Aτ, (11.6)

ik1A− ik1AΓ = ik2τ. (11.7)

It is obvious that the amplitude A factors out of the problem. The solutionfor Γ and τ is easily found to be

Γ =k1 − k2

k1 + k2

, τ = 1 + Γ =2k1

k1 + k2

. (11.8)

This is the best we can do: we have exhausted all the conditions available tous. We cannot determine A. There is no restriction on k1 or k2, and thusno restriction on the possible energy E of the particle, other than the initialcondition that E > V0.

11.2 The potential barrier with E > V0

Consider now the problem of a particle, incident from x = −∞, hitting abarrier of heigth V0 and having width L :

V (x) =

0 if x ≤ 0,V0 if 0 < x < L,0 if L ≤ x.

(11.9)

It is now natural to divide the problem in three regions: region 1, on the leftof the barrier, region 2, inside the barrier, and region 3 on the right of thebarrier.


x=0 x=L

V0

0E>V

Figure 16: The potential barrier with E > V0.

We consider first the case where E > V0. The wavefunction for the systemis conveniently written as

ψ(x) =

Aeik1x + AΓ e−ik1x if x ≤ 0,ACeik2x + ADe−ik2x if 0 < x < L,Aτ eik1(x−L) if L ≤ x,

(11.10)

where

k1 = k3 =

√2mE

~2, k2 =

√2m (E − V0)

~2. (11.11)

Continuity of wave function and its derivative at x = 0 and x = L givesfour equations:

1 + Γ = C + D , (11.12)

k1(1− Γ) = k2(C −D) , (11.13)

Ceik2L + De−ik2L = τ , (11.14)

k2

(Ceik2L −De−ik2L

)= k1 τ . (11.15)

This is a system of four equations with four unknowns.The two coefficients C and D are given by

C = − 2k1 (k1 + k2)

e2ik2L (k1 − k2)2 − (k1 + k2)

2 , (11.16)

D =2eik2Lk1 (k1 − k2)

e2ik2L (k1 − k2)2 − (k1 + k2)

2 . (11.17)


We are primarily interested in Γ and τ , which are found to be

Γ =(k2

1 − k22) sin k2L

(k21 + k2

2) sin k2L + 2ik1k2 cos k2L, (11.18)

τ =2ik1k2

(k21 + k2

2) sin k2L + 2ik1k2 cos k2L, (11.19)

One notices an interesting feature of these quantities: if the so–called reso-nance condition is met:

k2L = nπ, (11.20)

then Γ is 0 and τ = 1 : the wave is completely transmitted through thebarrier. For this to happen, we must have

k22L

2 =2m (E − V0)

~2= n2π2 =⇒ En = V0 +

n2π2~2

2m. (11.21)

The label n has been attached to the n’th value of the energy for which thereis resonance. For other values of E, some of the wave is transmitted andsome is reflected. The probability amplitudes for transmission and reflectionsare just

ΓΓ∗ =(k2

1 − k22)

2sin2 k2L

(k21 + k2

2)2sin2 k2L + 4k2

1k22 cos2 k2L

, (11.22)

ττ ∗ =4k2

1k22

(k21 + k2

2)2sin2 k2L + 4k2

1k22 cos2 k2L

. (11.23)

The behaviour of these quantities is best illustrated graphically. For thispurpose, we introduce, as before, the dimensionless quantities

ξ =

√2mV0L2

~2, z =

E

V0

, k1L = ξ√

z , k2L = ξ√

z − 1. (11.24)

The range of z is, by construction, limited to 1 ≤ z. With this notation, ττ ∗

is written, for instance, as

ττ ∗ =4z (z − 1)

(2z − 1)2 sin2 ξ√

z − 1 + 4z (z − 1) cos2 ξ√

z − 1, (11.25)

=4z (z − 1)

4z(z − 1) + sin2 ξ√

z − 1(11.26)


The physical parameters of the system are all contained in ξ. For small valuesof ξ

√z − 1 (either ξ small or z ' 1) the term sin2 ξ

√1− z is small, so that

ττ ∗ is close to 1. For large values of z, the sine term saturates at 1: thez(z − 1) term dominates in the denominator and the ratio once again goesto one. For illustration purposes, it is therefore convenient to choose, say,ξ = 2 and ξ = 5.

2 4 6 8 10

0.2

0.4

0.6

0.8

1

2 4 6 8 10

0.2

0.4

0.6

0.8

1

Figure 17: The (relative) probability amplitude of the transmitted wave TT ∗ forξ = 2 (top) and ξ = 5 (bottom) for the case where E > V0 > 0.

11.3 The potential hole

The case of −V0 is treated much in the same way. If

V (x) =

0 if x ≤ 0,−V0 if 0 < x < L,

0 if L ≤ x,(11.27)


we now need

k1 = k3 =

√2mE

~2, k2 =

√2m (E + V0)

~2. (11.28)

Equations (11.12) , (11.13) , (11.14) and (11.15) remain identical, so that thesolutions for Γ and τ , when expressed in terms of k1 and k2, remain un-changed. The only difference is that

k2L = ξ√

z + 1 , z =E

V0

(11.29)

so that

ττ ∗ =4z (z + 1)

4z(z + 1) + sin2 ξ√

z + 1. (11.30)

The same general conclusions regarding the behavior of the transmitted waveremain. In particular, there are transmission resonances at

En = −V0 +n2π2~2

2m, (11.31)

and the range of z = E/V0 now extends from 0 to ∞.

11.4 The potential step with E < V0

The last case to consider if a barrier of height V0 but a particle incident fromx = −∞ having energy E < V0. Explicitly,

V (x) =

0 if x ≤ 0,V0 if 0 < x < L,0 if L ≤ x.

(11.32)

Again we have three regions, with wavefunction written as

ψ(x) =

Aeik1x + AΓ e−ik1x if x ≤ 0,ACe−κ2x + ADeκ2x if 0 < x < L,Aτ eik1(x−L) if L ≤ x,

(11.33)

with

k1 = k3 =

√2mE

~2, κ2 =

√2m (V0 − E)

~2. (11.34)


The solution can be obtained by comparaison with the E > V0 case, where

k2 =

√2m (E − V0)

~2, (11.35)

which implies

κ2 =

√2m (V0 − E)

~2=

√−2m (E − V0)

~2=√−1 k2 = ik2. (11.36)

In other words, we can simply take the solutions for E > V0 and replacek2 → iκ2 to obtain the solution for E < V0. To convert sin(k2L), for instance:

sin(k2L) =1

2i

(eik2L − e−ik2L

) → 1

2i

(e−κ2L − eκ2L

),

= i sinh (κ2L) . (11.37)

Likewise, cos(k2L) → 12

(eκ2L + e−κ2L

)= cosh (κ2L) so that

τE>V0 → τE<V0 =2ik1(iκ2)

(k21 − κ2

2)(i sinh(κ2L)) + 2ik1(iκ2) cosh(κ2L),

=−2ik1κ2

(κ22 − k2

1) sinh(κ2L)− 2ik1κ2 cosh(κ2L). (11.38)

It is once again possible to illustrate the behaviour of these coefficientsin terms of the dimensionless variables ξ and z. This is done on figure 11.4.

0.2 0.4 0.6 0.8 1

0.1

0.2

0.3

0.4

0.5

0.2 0.4 0.6 0.8 1

0.02

0.04

0.06

0.08

0.1

0.12

0.14

Figure 18: The (relative) probability amplitude of the transmitted wave ττ∗ forξ = 2 (left) and ξ = 5 (right) for the case where V0 > E > 0.


11.5 The S-matrix

Thus far, we have restricted ourselves to problems where particles are incidentfrom x = −∞. Our objective is to extend our results to the case whereparticles are incoming from both x = ∞ and x = −∞ : we can alwaysrecover the case of particle incident from only one direction by setting theappropriate amplitude equal to zero at the end of the problem.

In generalizing, we are able to setup a more realistic problem, which canbe pictured as below. Two particles, initially far apart, interact in a smallregion and emerge, usually with their energies and momenta changed, fromthe interaction region. This is illustrated in figure 11.5

?

Figure 19: The schematic representation of scattering: two particles enter aninteraction region and emerge with different energies and (in general) wave vectors.

In the one–dimensional case, this picture is somewhat naive but one cannevertheless easily imagine the situation. As always, it will be convenient todivide the problem in three regions: the interaction region is taken to extendbetween x = 0 and x = L. On the left of the interaction region, we have awavefunction of the type

Ψ(x, t) = ALe−i(ωt−k1x) + BLe−i(ωt+k1x), (11.39)

whereas, on the right of the interaction region, we have

Ψ(x, t) = BRei(ωt−k1x) + ARe−i(ωt+k1x). (11.40)

Notice how the amplitudes of the waves traveling towards the interaction re-gions need not have identical amplitudes. The wavefunctions have also beenwritten so that the reflected wave on the left and right have, respectively,


amplitude BL and BR rather than ALΓL and ARΓR. Thus, we can set, forinstance, ΓR = 0 to deal with particles incident from x = −∞ only. The rea-son for using Ψ(x, t) rather than the time–independent ψ(x) in to emphasizethe traveling wave nature of these solutions.

In general, the wavefunction inside the interaction region will have somecomplicated form, but it will be a superposition of a wave traveling to theright and a wave traveling to the left:

Ψ(x, t) = Cg(ωt− k2x) + Df(ωt + k2x). (11.41)

In general, the wave number k2 is a (possibly terribly complicated) functionof x through the energy E of the particle and the potential V (x) present inthe interaction region: k2 = k2(x).

It is usually assumed that these incident amplitudes are known, so thatthe problem is one of finding BL and BR in terms of the given amplitudesAR and AL. Clearly, the special case of a particle incident from x = −∞only can be recovered by setting AR = 0.

In general, we thus have six unknown amplitudes AL, BL, C, D, AR andBR but only four conditions: continuity of the wavefunction and its derivativeat x = 0, and the continuity of the wavefunction and its derivative at x = L.Thus we can solve for any four in terms of two. We choose to solve in termsof the known amplitudes AR and AL. In particular, the reflected waves ineach sector will be some linear combination AR and AL, something of thetype

BL = S11AL + S12AR, (11.42)

BR = S21AL + S22AR. (11.43)

This is most conveniently expressed in matrix form as

(BL

BR

)=

(S11 S12

S21 S22

) (AL

AR

). (11.44)

The matrix

S =

(S11 S12

S21 S22

)(11.45)

is known as the scattering matrix, or S matrix. Its coefficients are deter-mined by the continuity conditions on the wave functions and its derivative,and depend on the potential through these boundary conditions.


To recover what we already know from the case of a particle incident fromx = −∞ only, we need to set AR = 0, in which case the transmission andreflections coefficients become

Γ =|BL|2|AL|2

= |S11|2 , τ =|BR|2|AL|2

= |S21|2 . (11.46)

A similar interpretation of the coefficients |S22|2 and |S12|2 can be made interms of particles incident from x = +∞ only, when AR = 0.

We can squeeze more information from the S matrix. Suppose we havea bound state rather than a scattering state. Then, the energy E of theparticle is negative rather than positive. The expression for k1, which isk1 =

√2mE/~2 for scattering states (assuming V = 0 outside the interac-

tion region), becomes k1 →√−2m |E| /~2 = iκ1, with κ1 =

√2m |E| /~2.

Furthermore, the coefficients AR and AL must be zero, as these representwaves incident from the left and the right, respectively. Thus, for instance,if the particle is incident from the left, the coefficients Γ and τ in Eqn.(11.46)will become infinite. In other words, if you want to locate bound states, sub-stitute k1 → iκ1 in the S matrix and find the values of E (via k2) where thesecoefficients become infinite.

A good example of this procedure is afforded by the square hole. There,V = −V0 is constant, and we already have S11 and S12 through Eqns.(11.18)and (11.19): both have the same denominator

(k2

1 + k22

)sin k2L + 2ik1k2 cos k2L, (11.47)

which will make the expressions blow up when 0. Thus, rememberingthat, for the hole k2 =

√2m(− |E|+ V0)/~2 and substituting k1 → iκ1 =

i√

2m |E| /~2, we have

0 =(−κ2

1 + k22

)sin k2L +−2κ1k2 cos k2L (11.48)(

k22 − κ2

1

)sin k2L = 2κ1k2 cos k2L. (11.49)

This doesn’t look particularly familiar because we worked out the boundstates for a finite well of width 2L centered at x = 0 rather than a wellhaving the dimensions appropriate for this problem. This is not a seriousissue, though: we can easily work out the bound states of the hole of depthV0 and width L. The wave function must have the form

ψ(x) =

Aeκ1x , if x < 0,B sin k2x + C cos k2x , if 0 ≤ x ≤ L,De−κ1(x−L) , if x > L.

(11.50)


Matching the wave function and its derivative at x = 0 and x = L gives

A = C, (11.51)

κ1A = Bk2, (11.52)

D = B sin k2L + C cos k2L, (11.53)

−κ1D = k2 cos k2L− Ck2 sin k2L. (11.54)

Put together, these eventually produce

(−κ1

k2

+k2

κ1

)sin k2L = 2 cos k2L, (11.55)

with is just another form of Eqn.(11.48): everything is fine.

126

Part III

Formalism and calculation toolsIn this part, we introduce the tools and the formalism required to go beyondsimple quantization of energy.

The formalism of quantum mechanics borrows heavily from the theory oflinear vector spaces. The major difference between the quantum mechanicalvector spaces and ordinary, three–dimensional vector space that we are allfamiliar is that the quantum version is infinite–dimensional.

Many of the notions familiar to ordinary three–space, such as orthogonal-ity, linear dependence, completeness, carry over to the infinite–dimensionalcase, with the caveat that one must take care to make sure that the sumsor integral that arise are meaningful, i.e. they converge in the mathematicalsense.

Linear stuff 127

12 Linear stuff

12.1 Linear spaces

Schrodinger equation (the full time–dependent version) is a linear differentialequation: the sum of any two solutions is also a solution. Further, one canmultiply any solution by a complex scalar, and add it to another so multipliedsolution and still get a solution. These two properties are typical of vectorsin a linear space.

Linear spaces (or vector spaces) are populated by vectors, and the adjec-tive linear means that vectors which inhabit the space

i) can be added together to form other vectors, and

ii) can be multiplied by (real or complex) numbers to make new vectors.

The dimension of the vector space need not be finite.

Example 12.1 The following sets are all vector spaces:

1. The set of all n–tuples of real numbers. If x = (x1, x2, . . . , xn) andy = (y1, y2, . . . , yn) are vectors in this space, then addition is definedby x + y = (x1 + y1, x2 + y2, . . . , xn + yn). Multiplication of a vectorby a scalar χ is defined as χx = (χx1, χx2, . . . , χxn). This is an n–dimensional space;

2. The set of all infinite sequences of numbers (x1, x2, . . . , xk, . . .) such

that∞∑

k=1

|xk|2 < ∞. Addition of two vectors is defined as in 1.; this

infinite–dimensional space is called an `2–space;

3. The set of all those functions of a real variable x which can be dif-ferentiated infinitely many times. Addition of two vectors f and gdefined pointwise, i.e. (f + g)(x) = f(x) + g(x), and scalar multiplica-tion defined by (af)(x) = a(f(x)). This is an infinite–dimensionalspace; a basis for this space is given by the set of “basis vectors”1, x, x2, x3, . . .. Note that the dimension of the vector space (here,infinite–dimensional) is independent of the number of variables in theproblem (here, one). The component of the vector f(x) on the basisvector xn is just the coefficient cn of xn in the Taylor series expansion

Linear stuff 128

of xn. The set 1, x, x2, x3, . . . is not the only possible set of basisvectors: there are many (in fact, infinitely many) possible choices ofbasis functions.

4. The set of all functions ψ of a real variable x for which the integral∫|ψ(x)|2 dx makes sense and is < ∞. The range of integration can be

finite or infinite. Addition and multiplication by a scalar are definedas in 3.; this is an infinite–dimensional space known as an L2–space.

5. The set of all functions x of one real variable t which are solutions to theequation d2x(t)/dt2 = −ω2x(t), with addition and scalar multiplicationdefined as in 3.; this is either a two–dimensional real space or a one–dimensional complex space, depending on whether or not we allow thescalars to be real or complex. (Note again that the dimension of thevector space does not depend on the number of variables in the problem.In particular, we have a two–dimensional real vector space even if wehave a single independent variable, t.) A basis for the real space is given

by the two “basis vectors”

(10

)= cos(ωt) and

(01

)= sin(ωt). A

basis for the complex space is given by the basis vector eiωt. Clearly,this space is a subspace of the space described in 4.

From now on, we will denote a generic vecto space by V . An element inV , i.e. a vector, will be written as |ψ〉. The symbol ψ is just a symbol, with

no more significance than any other symbol; one could have written ~ψ, butit’s just more practical for future use to use |ψ〉. Thus, |a〉, |ξ〉, |α1〉 are allexample of abstract vectors. Technically speaking, |ψ〉 is called the “ket” ψ.

Two vectors |α1〉 and |α2〉 are said to be linearly independent is oneis not a multiple of the other, i.e. if there is no complex number β suchthat |α1〉 = β|α2〉. A vector |γ〉 is said to be linearly independent of a set|α1〉, |α2〉, . . . , |αn〉 if is it impossible to write |γ〉 as a linear combinationsof the vectors in the set, i.e. if the equation

|γ〉 =∑

i

βi|αi〉 , (12.1)

has no solution.

Linear stuff 129

The largest number of linearly independent vector in V is the dimensionof V .

It is also possible to choose a basis in V , which is a set of linearly inde-pendent vector having the same dimension as V . (One must be careful whenV is infinite–dimensional to make sure that we have “enough” vectors.) Ifthings are done right, any vector |α〉 in V can then be written as a uniquelinear combinations of the basis vectors |φi〉 in V :

|α〉 =∑

i

αi|φi〉 . (12.2)

The complex coefficients αi are called the components of |α〉 in the basis|φi〉.

12.2 Inner products and norms

We need now to generalize the notion of length (or norm) and angles betweenvectors in three dimensions to an arbitrary complex vector space.

An inner product is a rule that assigns to a pair of vectors a (generallycomplex) number. Oftentimes, this operator is denoted by a dot, ·, so that

the inner product of ~a with ~b is written ~a ·~b. We want to generalize this.Suppose that V is an n–dimensional vector space, and that vectors in V

can be represented by an ordered n-tuple:

|ψ〉 = (ψ1, . . . , ψn) ,

|φ〉 = (φ1, . . . , φn) . (12.3)

Their inner product will be denoted by 〈φ|ψ〉 and is defined as

〈φ|ψ〉 ≡n∑

i=1

φ∗i ψi , (12.4)

where φ∗i is the complex conjugate of the component φi.If V is an infinite–dimensional vector space, then the above definition of

inner products holds provided that the sum converges, i.e.

〈φ|ψ〉 ≡∞∑i=1

ψ∗i ψi < ∞ . (12.5)

Linear stuff 130

The complex conjugation in the definition of the inner product is im-portant. The length of a vector (also known as the norm of this vector), aquantity that should always be positive, is given by

‖ψ‖ =√〈ψ|ψ〉 , (12.6)

a definition possible only because 〈ψ|ψ〉 is necessarily non–negative underthe definition of the inner product. This definition of lenght implies that

‖ψ‖ = 0 ⇔ |ψ〉 = 0 . (12.7)

Two vectors are said to be orthogonal if their inner product is 0: 〈ψ|φ〉 =0.

Using this, it is possible to construct, starting from any basis in V , anorthonormal basis for V . An orthonormal basis is one where all basis vectorsare of unit length and any two distinct vectors are orthogonal to one another.In symbols, if |ei〉 denotes an normalized basis state

〈ei | ej〉 = δij , (12.8)

where the symbol

δij ≡

1 if i = j ,0 if i 6= j .

(12.9)

Example 12.2 Gram-Schmidt orthonormalization.Suppose we are given three vectors |1〉 , |2〉 and |3〉 . These vector are not

necessarily orthonormal, i.e. they don’t necessarily satisfy

〈i |j〉 = δij. (12.10)

How do we construct from these an orthonormal basis?.First, pick any one vector, say |1〉 . Which one is inessential. To obtain a

normalized vector from |1〉 , divide it by its norm. Thus, we define the firstnormalized vector to be

|e1〉 =|1〉√〈1 |1〉 . (12.11)

Select now a second vector, say |2〉 , and consider the combination

|2′〉 = |2〉+ β |e1〉 . (12.12)

Linear stuff 131

Orthogonality to |e1〉 implies

〈e1 |2′〉 = 0 = 〈e1 |2〉+ β 〈e1 |e1〉 (12.13)

= 〈e1 |2〉+ β, (12.14)

which implies

β = −〈e1 |2〉 = − 〈1 |2〉√〈1 |1〉 , (12.15)

|2′〉 = |2〉 − |e1〉〈e1 |2〉 = |2〉 − |1〉〈1 |2〉〈1 |1〉 , (12.16)

〈2′ |2′〉 = [〈2| − 〈2 |e1〉〈e1|] [|2〉 − 〈2 |e1〉 |e1〉] (12.17)

= 〈2 |2〉 − 2 〈e1 |2〉〈2 |e1〉+ 〈e1 |e1〉〈2 |e1〉〈2 |e1〉 (12.18)

= 〈2 |2〉 − |〈e1 |2〉|2 = 〈2 |2〉 − |〈1 |2〉|2〈1 |1〉 (12.19)

We can then normalize |2′〉 to obtain

|e2〉 =|2′〉√〈2′ |2′〉 , (12.20)

=|2〉 − |e1〉〈e1 |2〉√〈2 |2〉 − |〈e1 |2〉|2

. (12.21)

To find a third orthogonal vector, pick a vector other than the first two, say|3〉 , and write

|3′〉 = |3〉+ α |e1〉+ γ |e2〉 (12.22)

and impose

〈e1 |3′〉 = 0 = 〈e1 |3〉+ α + γ 〈e1 |e2〉 = 〈e1 |3〉+ α, (12.23)

〈e2 |3′〉 = 0 = 〈e2 |3〉+ α 〈e2 |e1〉+ γ = 〈e2 |3〉+ γ. (12.24)

This determines α, γ , from which |3′〉 can be normalized to obtain |e3〉 .As an example, consider the problem of finding an orthonormal basis two

dimensions given two vectors

|a〉 =

(2

1− i

), |b〉 =

( −1i

). (12.25)

Linear stuff 132

To find |e1〉 , start from |b〉 to get

|e1〉 =|b〉√〈b |b〉 =

1√2

( −1i

), 〈e1| = 1√

2(−1,−i) (12.26)

To obtain |e2〉 , we start from |a〉 and write

|a′〉 = |a〉 − |e1〉〈e1 |a〉=

(2

1− i

)− 1

2

( −1i

)(−1,−i)

(2

1− i

),

=

(2

1− i

)− 1

2

( −1i

)(−3− i)

=1

2

(4− 3− i

2− 2i + 3i− 1

)=

1

2

(1− i1 + i

). (12.27)

It happens, by luck, that |a′〉 is already normalized, so that

|e2〉 =1

2

(1− i1 + i

). (12.28)

One can verify that

〈e1 |e2〉 =1

2√

2(−1,−i)

(1− i1 + i

)=

1

2√

2(−1 + i− i + 1) = 0. (12.29)

Everything generalizes to the higher dimensions in an obvious way.

The general abstract definition of an inner product is obtained by requir-ing that the operation 〈ψ|φ〉 satisfy the following properties:

i. 〈ψ|φ〉 = 〈φ|ψ〉∗. Note that 〈ψ|φ〉 is a number, so that it makes sense totake its complex conjugate.

ii. 〈α1φ1 + α2φ2|φ〉 = α∗1〈φ1|φ〉 + α∗2〈φ2|φ〉 for any two arbitrary complexnumbers α1 and α2,

iii. 〈φ|φ〉 ≥ 0.

Linear stuff 133

For the space of square–integrable functions (this is an infinite–dimensionalspace), we obtain an inner product as the continuous version of Eq.(12.4):provided that ∫ b

a

dx |ψ(x)|2 < ∞ , (12.30)

then

〈ψ|φ〉 ≡∫ b

a

dxψ∗(x)φ(x) . (12.31)

Note that ‖ψ‖ = 0 implies that f(x) = 0 almost everywhere, i.e. except ona set of measure zero.

There also exists a useful inequality between the norms and inner productof two vectors, known as the Schwartz inequality:

|〈ψ|φ〉| ≤ ‖ψ‖ · ‖φ‖ . (12.32)

12.3 Linear operators

An operator (or an operation) is a rule that transforms any input (a vector)into an output (another vector) according to some prescribed instructions.For instance, an n× n matrix acting on an n–dimensional vector space is anoperator, as it transforms any vector to another vector according to the rulesof matrix multiplications.

An important family of operators are the linear operators: if you feeda sum of vectors ψ + φ to a linear operator O, the output will the sum ofthe outputs obtained by applying the operator on each member of the sumindividually:

O(ψ +φ) = (Oψ)+(Oφ) , and O(kψ) = k(Oψ) , k ∈ C . (12.33)

Matrices are examples of linear operators. Taking the square root is anexample of an operator that is not linear: acting on numbers (which one canthink of as vectors in a one–dimensional space), we have

√x + y 6= √

x+√

y.In the space of functions, the derivative operator d/dx is a linear operator:

d

dx(f(x) + g(x)) =

(d

dxf(x)

)+

(d

dxg(x)

). (12.34)

For the kinds of operators that we will be dealing with, and for the kindsof functions that we’ll be looking at, it is possible, in principle, to construct

Linear stuff 134

a matrix for the operator d/dx, although this matrix could turn out to beinfinite–dimensional (because the space of functions is infinite–dimensional).

Linear operators can usually be represented by matrices. Given a basisand an inner product, the matrix element of the operator O between twovectors |ψi〉 and |ψj〉 is defined by Oij = 〈ψi|O|ψj〉.Example 12.3 The operator x.

Let us compute the matrix representation of the operator x, with matrixelements computed between the first three solutions ψ0(x) , ψ1(x) and |ψ2〉(x)of the Schrodinger equation for the harmonic oscillator. For simplicity, wewill write these vectors as |0〉 , |1〉 and |2〉, respectively.

We now compute 〈n|x|r〉. To simplify the calculation, we will switch tothe variable ξ defined by √

~mω

ξ = x . (12.35)

Then

〈n|x|r〉 =1√

2nn!2rr!π

∫ ∞

−∞dξ e−ξ2

Hn(ξ)

√~

mωξ Hr(ξ)

=1√

2nn!2rr!π

√~

mω

∫ ∞

−∞dξe−ξ2

Hn(ξ)ξHr(ξ) (12.36)

Use the recursion relation for Hermite polynomials:

Hr+1(ξ) = 2ξHr(ξ)− 2rHr−1(ξ) , (12.37)

⇐⇒ ξHr(ξ) =1

2(Hr+1(ξ) + 2rHr−1(ξ)) , (12.38)

to get

〈n|x|r〉 =1√

2nn!2rr!π

√~

mω

1

2

×∫ ∞

−∞dξe−ξ2

Hn(ξ) (Hr+1(ξ) + 2rHr−1(ξ)) (12.39)

=

√~

mω

1

2

1√2n+rn!r!π

×

2r+1(r + 1)!√

π for n = r + 1 ,2r2r−1(r − 1)!

√π for n = r − 1 .

(12.40)

Linear stuff 135

Thus, we have two non–zero matrix elements:

〈r + 1|x|r〉 =

√~

mω

1

2

1

[22r+1(r + 1)!r!rπ]1/22r+1(r + 1)!

√π

=

√~

mω

1

2

1

2r√

2r!√

r + 12r+1(r + 1)r!

=

√~

2mω

√r + 1 (12.41)

〈r − 1|x|r〉 =

√~

mω

1

2

1

[22r+1(r − 1)!r!rπ]1/22r 2r−1(r − 1)!

√π

=

√~

mω

1

2

√2

2r

1

(r − 1)!

√r2r−1(r − 1)!2r

=

√~

mω

1√2

√r =

√~

2mω

√r (12.42)

Thus, we have the following matrix for the operator x√

~2mω

0 1 0

1 0√

2

0√

2 0

, (12.43)

where the basis states are ordered so that the first line, for instance, containsthe matrix elements x00, x01 and x02, in this order.

12.4 Eigenvectors are special

Suppose we have some linear operator T , and a normalized eigenvector |ψ〉such that

T |ψ〉 = λ|ψ〉 , (12.44)

Then, by definition,

〈T 〉ψ = 〈ψ|T |ψ〉 , (12.45)

= λ , (12.46)

〈T 2〉ψ = 〈ψ|(T )2|ψ〉 , (12.47)

= 〈ψ|T T |ψ〉 , (12.48)

= λ 〈ψ|T |ψ〉 , (12.49)

= λ2 . (12.50)

Linear stuff 136

so that ∆T =√〈T 2〉 − 〈T 〉2 = 0. Thus, we can associate to |ψ〉 a well–

defined value of T .The simplest example of this situation is when T is represented by a

diagonal matrix in some basis. Suppose |αi〉 , i = 1, ..., n is a basis for ourproblem, and that, in this basis, T has the form

T =

λ1 0 0 . . . 00 λ2 0 . . . 0... 0

. . . 00 . . . 0 λn

(12.51)

In this basis, we have, for instance,

|α1〉 7→

100...0

, |α2〉 7→

010...0

, etc . (12.52)

Then, clearly, using simple matrix multiplication,

T |αi〉 = λ|αi〉 . (12.53)

In other words, T is diagonal in the basis spanned by the eigenvectors of T .

12.5 The mechanics of matrix diagonalization

Recall again the motivation for our problem. Suppose we have an operatorT and a normalized state |ψ〉 such that

T |ψ〉 = λ|ψ〉. (12.54)

Then

〈ψ|T |ψ〉 = λ, (12.55)

〈ψ|T 2|ψ〉 = 〈ψ|T T |ψ〉 = λ2 (12.56)

and the uncertainty on the quantity T represented by the operator T is

∆T =

√〈ψ|T 2|ψ〉 − 〈ψ|T |ψ〉2 =

√λ2 − (λ)2 (12.57)

= 0. (12.58)

Linear stuff 137

Thus, if there is no uncertainty on the value of T (it is certainly λ ), we canuse the eigenvalue λ to distinguish various eigenvectors of T , i.e. we can useλ as a label so that

T |ψλ〉 = λ|ψλ〉. (12.59)

How do we find eigenvectors and eigenvalues of an operator? First, weneed a basis, i.e. a set |φi〉, i = 1, ..., n of orthonormal vectors which willbe used to calculate matrix elements of T via Tij = 〈φi|T |φj〉. There is no

reason to suspect that |φj〉 is an eigenvector of T . Once we have the matrix

representation of T , we find the eigenvalues of T , and from the eigenvalueswe recover the eigenvectors.

To illustrate the procedure, suppose we are given a basis in which someoperator T is represented by the 3× 3 matrix

T =

12

1 01 −1

2i

0 −i 12

, (12.60)

which has been computed using

|φ1〉 =

100

, |φ2〉 =

010

, |φ3〉 =

001

. (12.61)

The eigenvalues are those solutions to the equation

det(T − λ 1l

)= 0, (12.62)

where 1l is the (here, 3× 3) unit matrix

1l =

1 0 00 1 00 0 1

. (12.63)

Constructing T − λ 1l and obtaining the determinant yields

0 = det

12− λ 1 01 −1

2− λ i

0 −i 12− λ

(12.64)

= −λ3 + 12λ2 + 9

4λ− 9

8. (12.65)

Linear stuff 138

Unless you have a lot of experience with cubic, it is not clear that thiscan be rewritten as

0 =(λ− 3

2

) (λ + 3

2

) (λ− 1

2

). (12.66)

One practical way to find the solutions is to recall that the eigenvalues of ahermitian matrix must be real, plot

f(λ) = −λ3 +1

2λ2 +

9

4λ− 9

8(12.67)

and looks for values of λ where f(λ) crosses the axis. This is illustrated inFig. 20.

-2 -1 1 2

-2

-1

1

2

3

4

Figure 20: The polynomial f(λ) = −λ3 + 12λ2 + 9

4λ− 98 .

In any event, the eigenvalues are 32,−3

2and 1

2. Let us find the correspond-

ing eigenvectors. Assume

|ψ 32〉 = α 3

2|φ1〉+ β 3

2|φ2〉+ γ 3

2|φ3〉 =

α 32

β 32

γ 32

, (12.68)

Linear stuff 139

and insert this in

T |ψ 32〉 =

3

2|ψ 3

2〉 (12.69)

12

1 01 −1

2i

0 −i 12

α 32

β 32

γ 32

=

3

2

α 32

β 32

γ 32

, (12.70)

−1 1 01 −2 i0 −i −1

α 32

β 32

γ 32

= 0. (12.71)

This leads to the system

−α 32

+ β 32

= 0, (12.72)

α 32− 2β 3

2+ iγ 3

2= 0, (12.73)

−iβ 32− γ 3

2= 0. (12.74)

Notice how the sum of Eqn.(12.72) and Eqn.(12.73) add up to a multiple ofEqn.(12.74), indicating that we really have two equations for three unknowns.We will solve everything in terms of α 3

2, so

β 32

= α 32, (12.75)

γ 32

= −iβ 32

= −iα 32. (12.76)

Hence, the first eigenvector is

α 32(|φ1〉+ |φ2〉 − i|φ3〉) , (12.77)

which can be normalized to

|ψ 32〉 =

1√3|φ1〉+

1√3|φ2〉 − i√

3|φ3〉 =

1√3

11−i

. (12.78)

In a similar manner, one finds

|ψ− 32〉 =

1√6|φ1〉 − 2√

6|φ2〉 − i√

6|φ3〉, (12.79)

|ψ 12〉 =

1√2|φ1〉+

i√2|φ3〉. (12.80)

Linear stuff 140

One can verify explicitly, for instance, that

12

1 01 −1

2i

0 −i 12

1−2−i

=

12− 2

1 + 1 + 12i− i

2

=

−32

3−3i

2

, (12.81)

= −3

2

1−2−i

. (12.82)

Notice how the eigenvectors are orthogonal, and, because they are normal-

ized, in fact orthonormal. In the|ψ 3

2〉, |ψ− 3

2〉, |ψ 1

2〉

, the operator T is now

diagonal since, for instance,

〈ψ 32|T |ψ 3

2〉 =

3

2, (12.83)

〈ψ− 32|T |ψ 3

2〉 = 0, (12.84)

〈ψ 12|T |ψ 3

2〉 = 0. (12.85)

Indeed, in this basis, we have the matrix representation of T as

T =

32

0 00 −3

20

0 0 12

(12.86)

The change from the |φ1〉, |φ2〉, |φ3〉 basis to the|ψ 3

2〉, |ψ− 3

2〉, |ψ 1

2〉

basis can be written in matrix form by noting that

|φ1〉 = |ψ 32〉〈ψ 3

2|φ1〉+ |ψ− 3

2〉〈ψ− 3

2|φ1〉+ |ψ 1

2〉〈ψ 1

2|φ1〉, (12.87)

|φ2〉 = |ψ 32〉〈ψ 3

2|φ2〉+ |ψ− 3

2〉〈ψ− 3

2|φ2〉+ |ψ 1

2〉〈ψ 1

2|φ2〉, (12.88)

|φ3〉 = |ψ 32〉〈ψ 3

2|φ3〉+ |ψ− 3

2〉〈ψ− 3

2|φ3〉+ |ψ 1

2〉〈ψ 1

2|φ3〉, (12.89)

and constructing the matrix

Mab = 〈ψa|φb〉. (12.90)

In this way, quite clearly,

|φb〉 =3∑

a=1

|ψa〉Mab, (12.91)

〈φb | =3∑

a=1

M∗ba〈ψa | (12.92)

Linear stuff 141

Obviously, we have, for instance, from |ψ 32〉 = 1√

3|φ1〉+ 1√

3|φ2〉 − i√

3|φ3〉

〈ψ 32|φ1〉 =

1√3, (12.93)

〈ψ 32|φ2〉 =

1√3, (12.94)

〈ψ 32|φ3〉 = − i√

3, (12.95)

and so forth, which leads directly to the matrix

M =

1√3

1√6

1√2

1√3

− 2√6

0

− i√3− i√

6−i√

2

. (12.96)

Notice how M † = (M t)∗

= M−1, i.e. M is unitary and how each column ismade from the normalized eigenvectors of T Thus,

TM =

1√3

1√6

1√2

1√3

− 2√6

0

− i√3− i√

6−i√

2

32

0 00 −3

20

0 0 12

= M

32

0 00 −3

20

0 0 12

, (12.97)

M−1TM = M−1M

32

0 00 −3

20

0 0 12

, (12.98)

=

32

0 00 −3

20

0 0 12

. (12.99)

Thus, the change of basis obtained from M , which transform the |φi〉 tothe |ψa〉 basis as it expresses the |φi〉 as linear combination of the new basisvectors |ψa〉, makes the operator T diagonal, with the eigenvalues on thediagonal.

12.6 Application: neutrino mixing

The neutrino is a particle that is found to participate in reactions driven bythe weak interaction. It has long been thought that its mass is very small

Linear stuff 142

or zero and thus that it travels at the speed of light. In this section, wegive a practical application of diagonalization of a Hamiltonian by lookingat mixing 1 between two types of neutrinos, the so–called electron and muonneutrinos, νe and νµ, respectively, which occur in the reactions

νe + n → p + e−, (12.100)

νµ + n → p + µ−. (12.101)

The states |νe〉 and |νµ〉 are eigenstates of the weak interaction Hamiltonian

Hweak, but they are not eigenstates of the free–particle Hamiltonian Hfree

(basically, the mass of the particle), i.e.

Hfree |νe〉 6= mνe |νe〉 , (12.102)

Hfree |νµ〉 6= mνµ |νµ〉 . (12.103)

In other words, once the particles are no longer subject to Hweak, they willbecome free particles but not eigenstates Hfree. This gives rise to a phe-nomenon known as neutrino oscillation.

Let the eigenstates of Hfree be |ν1〉 and |ν2〉 , respectively, so that

Hfree |ν1〉 = m1c2 |ν1〉 , (12.104)

Hfree |ν2〉 = m2c2 |ν2〉 . (12.105)

Because these are eigenstates of the free Hfree, their time evolution, oncethey have escaped the interaction region, is just

|ν1〉 e−iE1t/~, (12.106)

|ν2〉 e−iE2t/~. (12.107)

We can write

|ν1〉 = cos θ |νe〉+ sin θ |νµ〉 , (12.108)

|ν2〉 = − sin θ |νe〉+ cos θ |νµ〉 (12.109)

for the normalized mass eigenstates, where θ is the so–called mixing angle.

1see: W.C. Haxton and B.R. Holstein, arXiv:hep-ph/9905257 for details

Linear stuff 144

The eigenvalues obtained from

det

(m1c

2 cos2 θ + m2c2 sin2 θ − λ (m1 −m2)c

2 cos θ sin θ(m1 −m2)c

2 cos θ sin θ m1c2 sin2 θ + m2c

2 cos2 θ − λ

)

= 0 (12.122)

=(m1c

2 cos2 θ + m2c2 sin2 θ − λ

) (m1c

2 sin2 θ + m2c2 cos2 θ − λ

)

−(m1 −m2)2c4 cos2 θ sin2 θ (12.123)

= λ2

−λ(m1c

2 cos2 θ + m2c2 sin2 θ + m1c

2 sin2 θ + m2c2 cos2 θ

)

+(m1c

2 cos2 θ + m2c2 sin2 θ

) (m1c

2 sin2 θ + m2c2 cos2 θ

)

− (m2

1 − 2m1m2 + m22

)c4 cos2 θ sin2 θ (12.124)

= λ2 − λ (m1 + m2) c2

+m1m2c4(cos4 θ + sin4 θ + 2 cos2 θ sin2 θ

)(12.125)

= λ2 − λ (m1 + m2) c2 + m1m2c4, (12.126)

which can be solved to yield

λ± =(m1 + m2) c2 ±

√(m1 + m2)

2 c4 − 4m1m2c4

2, (12.127)

=(m1 + m2) c2 ±

√(m1 + m2)

2 c4

2(12.128)

=(m1 + m2) c2 ± (m1 + m2) c2

2(12.129)

so that

λ1 = m1c2, (12.130)

λ2 = m2c2, (12.131)

as expected. One easily verifies that the eigenstates are correctly given byEqns. (12.108) and (12.109).

Suppose now some weak process yields, say, |νe〉 as a final product. Thetime evolution of this vector, once it has escaped the interaction region, is

|νe〉 → |νe(t)〉 = cos θ |ν1〉 e−iE1t/~ − sin θ |ν2〉 e−iE2t/~. (12.132)

Linear stuff 145

The probability that it will have ”oscillated” to |νµ〉 after a time t is givenas a postulate by

Pνe→νµ(t) = |〈νµ | νe(t)〉|2 (12.133)

=∣∣ 〈νµ |ν1〉 cos θe−iE1t/~ − 〈νµ |ν2〉 sin θe−iE2t/~∣∣2(12.134)

=∣∣ sin θ cos θe−iE1t/~ − cos θ sin θe−iE2t/~∣∣2 , (12.135)

=

∣∣∣∣ 2 sin θ cos θ sin

((E1 − E2) t

2~

)∣∣∣∣2

, (12.136)

=

∣∣∣∣ sin 2θ sin

((E1 − E2) t

2~

)∣∣∣∣2

. (12.137)

Thus, the oscillation frequency is half of the energy (or mass) difference be-tween the mass eigenstates. This phenomenon is a sensitive test of neutrinomasses, given m1 6= m2 and a non-zero mixing angle. If neutrinos (which areproduced at fantastic rates in the core of stars) were found to have non-zeromass, the cosmological consequences would be non-trivial.

In 1998, unambiguous evidence that some νµ were missing from atmo-spheric cosmic ray decay experiments. The rate of νµ coming directly fromabove agreed with predictions, but the rate of νµ passing straight throughthe earth was about half of what was predicted. It is suggested that thedifference be the result of the extra time-of-flight across the Earth. As aresult, one find that the mass difference between the neutrinos is in the range

10−1eV ≥ ∆E ≥ 10−2eV (12.138)

with a mixing angle consistent with θ = π/4.

12.7 The ammonia molecule revisited

Linear stuff 146

13 Hermitian matrices and their properties

13.1 Some definitions

Let M be an n× n square matrix. If M satisfies the basic property that

M † ≡ (MT )∗ = M , (13.1)

then M is called hermitian. Closely related is the definition of an antiher-mitian matrix: M † = −M . Note that, if M is antihermitian, then iM ishermitian.

A real hermitian matrix is symmetric (MT = M) and a real antihermitianmatrix is antisymmetric. Finally, a matrix M is unitary if, by definition,M−1 = M †.

These definitions are summarized as follows:

type definition remark

hermitian M † = M

antihermitian M † = −M

symmetric MT = M all real hermitian matrices

antisymmetric MT = −M all real antihermitian matrices

unitary M−1 = M † can be written as eiH ,

with H hermitian

orthogonal M−1 = MT can be written as eiL,

with L antisymmetric

A square matrix is invertible if its determinant is non–zero. The trace ofa square matrix is the sum of its diagonal elements. The determinant andtrace of products of matrices have the following properties. Let A and B betwo n× n matrices. Then,

Det(AB) = Det(BA) , Tr(AB) = Tr(BA) . (13.2)

Two matrices A and B are similar if there is a so–called intertwiningmatrix T so that

A = T B T−1 . (13.3)

Linear stuff 147

Similar matrices have identical traces and determinants, since

Det(A) = Det(T B T−1) = Det((T B) T−1

)

= Det(T−1 (T B)

)= Det

(T−1 T ) B)

),

= Det (B) , (13.4)

Tr(A) = Tr(T B T−1) = Tr((T B) T−1

)

= Tr(T−1 (T B)

)= Tr

((T−1 T ) B)

),

= Tr (B) . (13.5)

13.2 Hermitian operators

Let O be an operator acting on some vector space V , and let |ψi〉 be a basisfor this space. The matrix elements of O are defined, as usual, by

Oij = 〈ψi | Oψj〉 ≡ 〈ψi|O|ψj〉 . (13.6)

The hermitian adjoint of O is denoted by O†. Given any two vectors |φ〉and |ψ〉, O† is this operator with matrix elements defined by

〈φ | O†ψ〉 ≡ 〈Oφ |ψ〉 , (13.7)

= 〈ψ | Oφ〉∗ ,

= 〈Oφ |ψ〉 . (13.8)

Note that the adjoint of an operator is defined via matrix elements, whichimplies that some inner product is specified.

An operator O is hermitian if O† = O, i.e. if, for any two vectors,

〈Oφ |ψ〉 = 〈φ | Oψ〉 . (13.9)

Theorem 13.1 Every diagonal matrix element of a hermitian operator Tis real.

Proof: Consider 〈ψ | Tψ〉 = 〈T †ψ |ψ〉 = 〈Tψ |ψ〉 = 〈ψ | Tψ〉∗.Theorem 13.2 Every hermitian operator T is unitarily similar to a diagonaloperator, i.e. every hermitian operator can be written as T = U−1D U , whereU is unitary (U−1 = U †) and D is diagonal.

Proof: This is a proof by recursion. It starts by assuming that the theoremholds for 2× 2 matrices; one then assumes that it holds for n× n and showsthat the results holds for (n+1)× (n+1). The details will not be given here.

Linear stuff 148

Theorem 13.3 The eigenvalues of a hermitian operator are real.

Proof: Let |ψ〉 be an eigenvector corresponding to eigenvalue λ, i.e. T |ψ〉 =λ|ψ〉. Then:

〈ψ | Tψ〉 = λ〈ψ |ψ〉 , (13.10)

= 〈Tψ |ψ〉 = 〈ψ | Tψ〉∗ ,

= λ∗〈ψ |ψ〉 . (13.11)

Thus, we find λ∗ = λ, which proves the result.Theorem 13.4 Let T be hermitian, and let |ψ1〉 and |ψ2〉 be two eigenvectorscorresponding to respective eigenvalues λ1 and λ2. Then, 〈ψ1 |ψ2〉 = 0 unlessλ1 = λ2.

Proof: Start with

〈ψ1 | Tψ2〉 = λ2〈ψ1 |ψ2〉 , (13.12)

= 〈Tψ1 |ψ2〉 = λ∗1〈ψ1 |ψ2〉 ,= λ1〈ψ1 |ψ2〉 , (13.13)

where hermiticity of T and reality of λ1 has been used. Combining Eqn.(13.12)and Eqn.(13.13), we have

0 = (λ1 − λ2) 〈ψ1 |ψ2〉 , (13.14)

from which the claim follows. Thus, two eigenvectors of corresponding todistinct eigenvalues of a hermitian operator are necessarily orthogonal.Theorem 13.5 If T is hermitian, then one can find a set |ψi〉 of orthogonaleigenvectors of T that completely span the space V on which T acts. In otherwords, any vector in V can be written as a linear combination of eigenvectorsof T .

For practical applications, this theorem is usually written in the form ofa completeness relation, i.e.

1l =∑

i

|ψi〉〈ψi| . (13.15)

The combination |ψk〉〈ψk| (fixed k) is called a projector.

Finally, we observe that, if A is arbitrary, then A+ A† is hermitian. Thisis clear since 〈φ | (A + A†)ψ〉 = 〈φ | Aψ〉+ 〈φ | A†ψ〉 by linearity. Then,

〈φ | (A + A†)ψ〉 = 〈φ | Aψ〉+ 〈φ | A†ψ〉 = 〈A†φ |ψ〉+ 〈Aφ |ψ〉 ,= 〈(A + A†)φ |ψ〉 , (13.16)

which proves the claim.

Linear stuff 149

13.3 Properties of pairs of hermitian operators

We start by noting that, if A and B are arbitrary, then simple matrix ma-

nipulations show that(AB

)†= B† A†.

Define the commutator [A, B] of two arbitrary operators A and B by

[A, B] ≡ AB − BA . (13.17)

Theorem 13.6 If A and B are hermitian, and if [A, B] = 0, then AB ishermitian.

Proof: We will show that Eqn.(13.9) holds for the product AB under thecondition stated in the theorem. Start with

〈φ |(AB

)ψ〉 = 〈φ | ABψ〉 . (13.18)

Next, denote by η the state B|ψ〉 = |Bψ〉, so that we can rewrite the righthand side of Eqn.(13.18) as

|(AB)ψ〉 ≡ (AB)|ψ〉 = A|η〉 = |Aη〉 . (13.19)

Combining this with the fact that A is hermitian, we have

〈φ | ABψ〉 = 〈Aφ | Bψ〉 . (13.20)

Denote now by |ξ〉 the state A|φ〉, so that 〈Aφ| = 〈ξ|. Then, using now thefact that B is hermitian

〈φ | ABψ〉 = 〈ξ | Bψ〉 = 〈Bξ |ψ〉 = 〈BAφ |ψ〉 ,= 〈ABφ |ψ〉 , (13.21)

where the last step is allowed by the assumption that [A, B] = 0, so thatAB = BA. Eqn.(13.21) is identical to Eqn.(13.9), which proves the theorem.

13.4 Simultaneous eigenvectors of commuting hermi-tian operators

We now show the last but most important result, which deals with pairs ofcommuting hermitian operators.

Postulates 151

14 The postulates of Quantum Mechanics

The postulates of quantum mechanics provide a physical interpretation and aphysical background to the mathematical machinery of the previous sections.

Postulate I:

A physical system is completely defined by specifying a ket vector |Ψ(t)〉belonging to the space H of all possible states of the system. The dimensionof H is the number of possible distinct outcomes of all measurements thatcan be one on the system.

Note that specifying a ket dispenses with having to choose a particularrepresentation (x− or p−) and allows for the inclusion of states which haveno space or momentum wave functions.

Postulate II:

To every measurable physical quantity O (called an observable), therecorresponds a hermitian operation O acting on H.

Postulate III:

The possible results of a measurement of O are the eigenvalues of O .

Postulate IV:

Let Λ be a physical quantity with corresponding hermitian operator Λ.We denote by |ψi

λn〉 the kets such that

Λ|ψiλn〉 = λn|ψi

λn〉 . (14.1)

Suppose the eigenvalue λn occurs gn times. The label i is required to dif-ferentiate eigenvectors of Λ with the same eigenvalue λn; there are of courseprecisely gn such eigenvectors.

Suppose a physical system is described by the ket normalized |ψ〉. WhenΛ is measured on this system, the probability P (λn) of obtaining the valueλn for Λ is given by

Pλn =

gn∑i=1

|〈ψiλn|ψ〉|2 , (14.2)

with the sum running over all eigenstates having eigenvalue λn.

Postulate V:

Postulates 152

If the measurement of Λ in the state |ψ〉 gives λn, then immediately afterthe measurement, the state is given by

gn∑i=n

|ψiλn〉〈ψi

λn|ψ〉

(∑i |〈ψi

λn|ψ〉|2)1/2

(14.3)

This is known as the measurement postulate. It is quite controversial,and is the source of numerous paradoxes.

Consider the simplest case where λn occurs once. Then, if Λ is found tohave value λn . Then |ψ〉 ”instantly” becomes

|ψ〉 ⇒ |ψλn〉〈ψλn |ψ〉|〈ψλn |ψ〉|

(14.4)

If we measure Λ with result λn , it must be that, immediately after themeasurement, the value of Λ is certainly λn :|ψ〉 must be changed to |ψλn〉,which is the state for which Λ will certainly give λn as eigenvalue. Sinceλn occurs with probability 〈ψλn |ψ〉 , measuring Λ to be λn amounts to the(linear) projection

|ψλn〉〈ψλn| = πn (14.5)

Note thatπnπn = |ψλn〉〈ψλn | = πn (14.6)

Why is a measurement represented by |ψλn〉〈ψλn| ? Suppose we ask: isthe outcome of measuring Λ equal to λn ? The only two answers are either0 (i.e.no) or 1 (i.e.yes). This is the same as finding the two solutions to

σ(σ − 1) = 0 ⇒ σ2 − σ = 0 ⇒ σ2 = σ (14.7)

which is precisely the property πn · πn = πn .

Postulate VI:

The time evolution of a state |Ψ(t)〉 is governed by the Schrodinger equa-tion:

i~∂

∂t|Ψ(t)〉 = H|Ψ(t)〉 (14.8)

Note that, since the Schrodinger equation is a first order differential equa-tion, one needs only to specify |Ψ(0)〉 to completely determine |Ψ(t)〉 .

Postulates 153

We claim that

|Ψ(t)〉 =∑

n

e−iEnt/~|ψn〉〈ψn|Ψ(0)〉 , (14.9)

whereH|ψn〉 = En|ψn〉 . (14.10)

If the eigenvalue En occurs more than once, then the sum in Eq.(14.9) mustextend to every eigenvector with this eigenvalue.

To show this, observe that, at t = 0 , we have

|Ψ(0)〉 =∑

n

|ψn〉〈ψn|Ψ(0)〉 = |Ψ(0)〉 (14.11)

since ∑n

|ψn〉〈ψn| = 1l . (14.12)

Next,

i~∂

∂t|Ψ(t)〉 =

∑n

i~(−iEn

~

)e−iEnt/~|ψn〉〈ψn|Ψ(0)〉 (14.13)

=∑

n

Ene−iEnt/~|ψn〉〈ψn|Ψ(0)〉 (14.14)

=∑

n

He−iEnt/~|ψn〉〈ψn|Ψ(0)〉 (14.15)

= H|Ψ(t)〉 (14.16)

So |Ψ(t)〉 is solution to Schrodinger which satisfies the correct boundarycondition at t = 0 . By the uniqueness theorem on solutions to first orderdifferential equation, this is the only solution.

Finally, we mention Ehrenfest theorem:

d

dt〈Λ〉 = − i

~〈[Λ, H

]〉+ 〈∂Λ

∂t〉 . (14.17)

In particular, if d〈Λ〉dt

= 0 , i.e. Λ is a constant of motion, which implies[Λ, H

]= 0 (provided that Λ does not depend explicitly on t).

Matrix elements 154

15 Algebraic approach to some matrix ele-

ments

The calculation of matrix elements in some basis is essential if we want tofind eigenvectors and eigenvalues of some operators. This type of calculationis simplified in some problems using algebraic properties of operators. Thewhole setup does not require much in the way of integral and derivatives, butdisplaces the problem to one where heavy use of commutators is required.

15.1 Some properties of commutators

15.1.1 Linear combination of operators in a commutator

We start by noting that the commutator [A, B] of any two linear operatorssatisfies

[A, B] ≡ AB − BA , (15.1)

= −(BA− AB

), (15.2)

= −[B, A] , (15.3)

i.e. the commutator is antisymmetric in its arguments. In particular,

[A, A] = −[A, A] = 0 . (15.4)

Any operator commutes with itself.Now, consider

[A, B]† =(AB − BA

)†, (15.5)

=(AB

)†−

(BA

)†, (15.6)

= B†A† − A†B† , (15.7)

= −[A†, B†] . (15.8)

Next, for A, B and C linear, and α, β, γ arbitrary complex numbers,

[αA, βB + γC] = αA(βB + γC

)−

(βB + γC

)αA , (15.9)

= αβAB + αγAC − αβBA− αγCA , (15.10)

Matrix elements 155

since the operators are linear. We can rearrange the last line as

[αA, βB + γC] = αβ(AB − BA

)+ αγ

(AC − CA

), (15.11)

= αβ[A, B] + αγ[A, C] . (15.12)

Thus, a commutator containing a sum can be expanded as a sum of commu-tators, with multiplicative constants ”factored in front”.

15.1.2 [x, p]

We end by showing that [x, p] is just i~ 1l by considering an arbitrary ket |ψ〉using the x–representation. Thus

〈x| [x, p] |ψ〉 = x(−i~d

dx)ψ(x)− (−i~

d

dx)(xψ(x)) , (15.13)

= −ixdψ(x)

dx+ i~

(ψ(x) + x

dψ(x)

dx

), (15.14)

where the rule for the derivative of a product has been used. Since |ψ〉 isarbitrary, the right hand side simplifies to i~ψ(x) so that we have

[x, p] = i~ 1l , (15.15)

valid for any function.Alternatively, using the p–representation, we have

〈p| [x, p] |ψ〉 =

(+i~

d

dp

)(pψ(p))− p

(i~

dψ(p)

dp

), (15.16)

= i~(

ψ(p) + pdψ(p)

dp

)− p

(i~

dψ(p)

dp

), (15.17)

= i~ψ(p) , (15.18)

which again implies Eq.(15.15).

15.2 Harmonic oscillator and the operators a and a†

The Hamiltonian for the one–dimensional simple harmonic operator is

H =1

2mp2 + 1

2mω2x2 . (15.19)

Matrix elements 156

15.2.1 a and a†

Define two (non–hermitian) operators

a† =

√mω

2~x− i

1

2m~ωp ,

=

√mω

2~

(x− i

mωp

), (15.20)

a =

√mω

2~

(x +

i

mωp

). (15.21)

This choice is motivated by classical mechanics: the (complex) variables

α =mω

2

(x +

i

mωp

), α∗ =

mω

2

(x− i

mωp

)(15.22)

have simple time evolution.Consider now [a, a†]. We have

[a, a†

]=

mω

2~

[x +

i

mωp, x− i

mωp

], (15.23)

=mω

2~

([x, x] +

[x,− i

mωp

]+

[i

mωp, x

]+

[i

mωp,

i

mωp

]), (15.24)

=mω

2~

(− i

mω[x, p] +

i

mω[p, x]

), (15.25)

=mω

2~

(− i

mωi~ 1l +

i

mω(−i~ 1l)

), (15.26)

= 121l + 1

21l = 1l . (15.27)

Matrix elements 157

Furthermore,

a†a =mω

2~

(x− i

mωp

) (x +

i

mωp

), (15.28)

=mω

2~

(x2 − i

mωpx +

i

mωxp +

1

m2ω2p2

), (15.29)

=mω

2~

(x2 +

i

mω[x, p] +

1

m2ω2p2

), (15.30)

=1

~ω

(1

2mp2 + 1

2mω2x2

)− 1

21l , (15.31)

=1

~ωH − 1

21l , (15.32)

where H is the harmonic oscillator hamiltonian. Thus,

H = ~ω(a†a + 1

21l)

. (15.33)

Note that the combination a†a is hermitian. Let

N = a†a . (15.34)

Note that[N , a†

]=

[a†a, a†

], (15.35)

= a†aa† − a†a†a , (15.36)

= a†(aa† − a†a + a†a

)− a†a†a , (15.37)

= a†([

a, a†]+ a†a

)− a†a†a , (15.38)

= a† , (15.39)

where [a, a†] = 1l has been used. Likewise, one can show [N , a] = −a byexpanding everything, but it is easier to use Eq.(15.8) and see that

[N , a†

]†= (a†)† = a , (15.40)

= −[N †, (a†)†

], (15.41)

= −[N , a

], (15.42)

Matrix elements 158

where N † = N (hermiticity of N) and (a†)† = a (transpose of a transpose isunit, conjugate of conjugate is unit) have been used.

Since N is hermitian, it has a complete set of orthonormal eigenstates.Let |n〉 be such an eigenstate with eigenvalue n, i.e.

N |n〉 = n|n〉 . (15.43)

|n〉 is also assumed to be normalized. Since

H = ~ω(N + 12) , (15.44)

it is clear than |n〉 is also an eigenstate of H, with eigenvalue (n+ 12)~ω. From

Sec.(5.1), we know that no two eigenstates of H have the same eigenvalue,unless they are multiple one of the other. Thus, |n〉 is also the unique (up toa choice of phase in the normalization) eigenstate of N with eigenvalue n.

Observe first that (a|n〉)† = 〈n|a†. Let a|n〉 = |ξ〉. |ξ〉 may not benormalized, as we don’t know what is the result of operating on |n〉 with a.Consider now

0 ≤ 〈ξ | ξ〉 = 〈n|a†a|n〉 (15.45)

= 〈n|N |n〉 = n〈n |n〉 , (15.46)

= n . (15.47)

Thus, n ≥ 0.Next, let us look more closely at a|n〉. We can find more about this state

by using

N a|n〉 =(N a− aN + aN

)|n〉 , (15.48)

=([

N , a]

+ aN)|n〉 , (15.49)

=(−a + aN

)|n〉 , (15.50)

= −a|n〉+ n a|n〉 = (n− 1)a|n〉 . (15.51)

This show that a|n〉 is an eigenstate of N with eigenvalue n − 1, and musttherefore be proportional to the unique (up to a choice of phase in the nor-malization) eigenstate of N with eigenvalue n− 1, i.e.

a|n〉 = dn|n− 1〉 . (15.52)

Matrix elements 160

The choice cn =√

n + 1 is convenient, so we obtain

a†|n〉 =√

n + 1|n〉 . (15.67)

Summarizing, we have

a|n〉 =√

n|n− 1〉 , a†|n〉 =√

n + 1|n + 1〉 . (15.68)

15.2.2 Matrix elements of x, p and their powers

Let us illustrate the virtues of using a and a† in computing matrix elementsof x and p.

We will use quite a bit the results

aa† = aa† − a†a + a†a = [a, a†] + a†a = N + 1l , (15.69)

aN = aN − N a + N a = [a, N ] + N a = a + N a , (15.70)

a†N = a†N − N a† + N a† = [a†, N ] + N a† = −a† + N a† . (15.71)

Consider specifically the matrix elements of x3. Expressing x in terms ofa and a†, and taking powers, we obtain

x3 =

[√~

2mω

(a† + a

)]3

=

[~

2mω

]3/2 (a† + a

)3(15.72)

=

[~

2mω

]3/2 (a† + a

) (a† + a

) (a† + a

), (15.73)

=

[~

2mω

]3/2 (a† + a

) (a†a† + aa† + a†a + aa

). (15.74)

We now use Eqn.(15.69) to simplify to

x3 =

[~

2mω

]3/2 (a† + a

) (a†a† + 1l + 2N + aa

), (15.75)

and distribute further to obtain

x3 =

[~

2mω

]3/2 (a†a†a† + a† + 2a†N + a†aa

+aa†a† + a + 2aN + aaa)

. (15.76)

Matrix elements 161

Application of Eqns.(15.71) and (15.70) yields

x3 =

[~

2mω

]3/2 [(a†)3 + a† + 2a†N + N a

+(N + 1l

)a† + a + 2aN + a3

], (15.77)

=

[~

2mω

]3/2 [(a†)3 + a† + 2a†N + N a

+N a† + a† + a + 2aN + a3], (15.78)

=

[~

2mω

]3/2 [(a†)3 + 2a† + 2a†N + N a + Na† + a + 2aN + a3

].(15.79)

Using now Eqns.(15.69)–(15.71), we find

2a† + 2a†N + N a† = 3N a† , (15.80)

N a + a + 2aN = 3aN , (15.81)

so that

x3 =

[~

2mω

]3/2 [(a†)3 + 3Na† + 3aN + a3

]. (15.82)

Thus,

x3 |n〉 =

[~

2mω

]3/2 [√(n + 1) (n + 2) (n + 3) |n + 3〉

+3(n + 1)√

n + 1 |n + 1〉+3n

√n |n− 1〉+

√n (n− 1) (n− 2) |n− 3〉

].(15.83)

So, we have, in tabular form:

〈m| x3 |n〉 =

[~

2mω

]3/2

×

√(n + 1) (n + 2) (n + 3) if m = n + 3

3(n + 1)√

n + 1 if m = n + 1

3n√

n if m = n− 1√

n (n− 1) (n− 2) if m = n− 3

.

(15.84)

Matrix elements 162

15.2.3 Application: constructing the eigenfunctions of H using a†

We know that a|0〉 = 0. Using the x–representation, we have

0 = 〈x|a|0〉 , (15.85)

=

√mω

2~

(x− i

mωi~

d

dx

)〈x | 0〉 , (15.86)

= xψ0(x) +~

mω

dψ0(x)

dx. (15.87)

This first–order differential equation is easily solved:

dψ0(x)

dx= −mω

~x dx , (15.88)

ln ψ0(x)− ln N0 = −mω

2~x2 , (15.89)

ψ0(x) = N0e−mωx2/2~ , (15.90)

where the integration constant N0 must be determined from the normaliza-tion condition:

1 =

∫ ∞

−∞dxψ0(x)∗ψ0(x) = |N0|2

∫ ∞

−∞dx e−mωx2/~ , (15.91)

= |N0|2√

π~mω

, (15.92)

from which we find

N0 =(mω

~π

)1/4

, (15.93)

ψ0(x) =(mω

~π

)1/4

e−mωx2/2~ , (15.94)

in accordance with Eqns.(6.73) and (6.70).From

〈x|a†|0〉 =√

1 〈x | 1〉 = ψ1(x) , (15.95)

and the expression for a†, we find√

mω

2~〈x|

(x− i

mωp

)|0〉 =

√mω

2~

(x− ~

mω

d

dx

)〈x | 0〉 , (15.96)

=

√mω

2~

(xψ0(x)− ~

mω

dψ0(x)

dx

).(15.97)

Matrix elements 163

Using the explicit expression of Eq.(15.94), we have

d

dxe−mωx2/2~ = −mω

~xe−mωx2/2~ (15.98)

and so

ψ1(x) =(mω

~π

)1/4(√

mω

~πx

)e−mωx2/2~ . (15.99)

From〈x|a†|1〉 =

√2 〈x | 2〉 =

√2ψ2(x) , (15.100)

and Eq.(15.99), we find

√2 ψ2(x) =

√mω

2~

(x− ~

mω

d

dx

) (mω

~π

)1/4√

mω

~πxe−mωx2/2~ ,(15.101)

=(mω

~π

)1/4 1

2

(2mω

~x2 − 1

)e−mωx2/2~ , (15.102)

ψ2(x) =(mω

~π

)1/4 1

2√

2

(2mω

~x2 − 1

)e−mωx2/2~ , (15.103)

and so forth. The whole sequence can thus be constructed without solvingthe differential equation resulting from the use of the Schrodinger equationin the position representation.

15.2.4 Application: the harmonic oscillator coherent state

Let α ∈ C, and let |n〉 be the harmonic oscillator state with energy (n+ 12)~ω.

At t = 0, the “coherent state” |α(0)〉 is defined by

|α(0)〉 = e−|α|2/2

( ∞∑n=0

αn

√n!|n〉

)(15.104)

What is |α(t)〉, the coherent state at time t? We start with

|α(0)〉 = e−|α|2/2

( ∞∑n=0

αn

√n!|n〉

). (15.105)

Since |n〉 is an eigenstate of the harmonic oscillator hamiltonian H =(a†a + 1

2

)~ω with eigenvalue

(n + 1

2

)~ω, the time evolution of |n〉 is simply

Matrix elements 164

|n(t)〉 = e−i(n+ 12)ωt |n〉 and thus

|α(t)〉 = e−|α|2/2

( ∞∑n=0

αn

√n!

e−i(n+ 12)ωt |n〉

). (15.106)

Note that |α(t)〉 is normalized, as

〈α(t) |α(t)〉 = e−|α|2

( ∞∑m=0

(α∗)m

√m!

ei(m+ 12)ωt 〈m|

)( ∞∑n=0

αn

√n!

e−i(n+ 12)ωt |n〉

),(15.107)

= e−|α|2

∞∑m,n=0

(α∗)m

√m!

αn

√n!

ei(m−n)ωt 〈m| n〉 , (15.108)

= e−|α|2

∞∑m,n=0

(α∗)m

√m!

αn

√n!

ei(m−n)ωtδmn, (15.109)

= e−|α|2∞∑

n=0

|α|2n

n!= e−|α|

2

e|α|2

= 1. (15.110)

Let us show that a|α(t)〉 = α ei~ωt|α(t)〉. Recall that a |n〉 =√

n |n− 1〉 .Then, since a is linear,

a |α(t)〉 = e−|α|2/2

( ∞∑n=0

αn

√n!

e−i(n+ 12)ωta |n〉

), (15.111)

= e−|α|2/2

( ∞∑n=0

αn

√n!

e−i(n+ 12)ωt√

n |n− 1〉)

, (15.112)

= e−|α|2/2

( ∞∑n=0

αn

√(n− 1)!

e−i(n+ 12)ωt |n− 1〉

), (15.113)

= αe−iωte−|α|2/2

( ∞∑n=0

αn−1

√(n− 1)!

e−i(n−1+ 12)ωt |n− 1〉

).(15.114)

The sum properly starts at n = 1 since the n = 0 term does not exist. Thus,setting m = n − 1, we can rewrite this sum in terms of m, with m startingat m = 0. Hence, we have

a |α(t)〉 = αe−iωt

[e−|α|

2/2

( ∞∑m=0

αm

√m!

e−i(m+ 12)ωt |m〉

)](15.115)

= αe−iωt |α(t)〉 . (15.116)

Matrix elements 166

Finally, let us compute ∆x∆p for |α(t)〉 and verify the Heisenberg uncer-tainty principle. We need x2 and p2. From x and p, we find

x2 =~

2mω

(a† + a

)2=

~2mω

((a†

)2+ a†a + aa† + (a)2

), (15.128)

=~

2mω

((a†

)2+ 2a†a + 1 + (a)2

), (15.129)

p2 = −mω~2

(a− a†

)2= −mω~

2

((a†

)2 − a†a− aa† + (a)2)

,(15.130)

= −mω~2

((a†

)2 − 2a†a− 1 + (a)2)

, (15.131)

where

aa† = aa† − a†a + a†a =[a, a†

]+ a†a = 1 + a†a (15.132)

has been used. Thus,

⟨x2(t)

⟩=

~2mω

[〈α(t)| (a†)2 |α(t)〉+ 2 〈α(t)| a†a |α(t)〉

+1 + 〈α(t)| a2 |α(t)〉] ,(15.133)

=~

2mω

[(α∗eiωt

)2+ 2α∗α + 1 +

(αe−iωt

)2]

, (15.134)

=~

2mω

[(α∗eiωt + αe−iωt

)2+ 1

], (15.135)

=~

2mω

[4 |α|2 cos2 (ωt− θ) + 1

]. (15.136)

⟨p2(t)

⟩= −mω~

2

[〈α(t)| (a†)2 |α(t)〉 − 2 〈α(t)| a†a |α(t)〉

−1 + 〈α(t)| a2 |α(t)〉] ,(15.137)

= −mω~2

[(α∗eiωt

)2 − 2α∗α− 1 +(αe−iωt

)2]

, (15.138)

= −mω~2

[(α∗eiωt − αe−iωt

)2 − 1], (15.139)

= −mω~2

[−4 |α|2 sin2 (ωt− θ)− 1]

, (15.140)

=mω~

2

[4 |α|2 sin2 (ωt− θ) + 1

]. (15.141)

Matrix elements 167

We now need

(∆x(t))2 =⟨x2(t)

⟩− 〈x(t)〉2 , (15.142)

=~

2mω

[4 |α|2 cos2 (ωt− θ) + 1

]

− 2~mω

|α|2 cos2 (ωt− θ) (15.143)

=~

2mω. (15.144)

(∆p(t))2 =⟨p2(t)

⟩− 〈p(t)〉2 , (15.145)

=mω~

2

[4 |α|2 sin2 (ωt− θ) + 1

]

−2mω~ |α|2 sin2 (ωt− θ) (15.146)

=mω~

2. (15.147)

Hence,

∆x(t)∆p(t) =

√~

2mω× mω~

2=~2. (15.148)

The coherent state ”saturates” the uncertainty relation for all times.

15.3 Angular momentum theory and the operators L+

and L−We looked already in section 9 at the radial differential equation

1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)Y (θ, ϕ) +

1

sin2 θ

∂2

∂ϕ2Y (θ, ϕ) = `(` + 1)Y (θ, ϕ) , (15.149)

and showed that the solutions where the spherical harmonics Y` m(θ, ϕ). Inthis section, we present an alternative method of analyzing this equation,which emphasizes the operator aspect of angular momentum operators.

Matrix elements 168

15.3.1 Commutation relations for angular momentum operators

Recall the three angular momentum operators


y∂

∂z− z

∂

∂y

)(15.150)

= −i~(

cos ϕ∂

∂θ− cotan θ sin ϕ

∂

∂ϕ

), (15.151)

Ly = zpx − xpz = −i~(

z∂

∂x− x

∂

∂z

), (15.152)

= −i~(− sin ϕ

∂

∂θ− cotan θ cos ϕ

∂

∂ϕ

), (15.153)

Lz = xpy − ypx = −i~(

x∂

∂y− y

∂

∂x

), (15.154)

= −i~∂

∂ϕ. (15.155)

(15.156)

They are all hermitian operators since they are constructed from commutinghermitian operators like x and py.

The operator~L · ~L ≡ L2

x + L2y + L2

z , (15.157)

is likewise hermitian.We now systematically consider commutators of angular momentum op-

erators. We recall first that

[x, px] = i~ 1l , [x, py] = 0 , (15.158)

etc., which is most easily shown in the position representation, where

x 7→ x , px 7→ −i~∂

∂x, (15.159)

and similarly for y, py, z and pz. Thus, if f(x, y, z) is an arbitrary function,

[x, px]ψ = xpx f(x, y, z)− pxx f(x, y, z) (15.160)

= −i~x∂

∂xf(x, y, z) + i~

(1 + x

∂

∂x

)f(x, y, z) , (15.161)

= i~ 1l f(x, y, z) , (15.162)

Matrix elements 169

using the product rule for derivative in Eq.(15.161). Eq.(15.159) follows fromthe observation that f(x, y, z) is arbitrary.

Next, one can show, most easily in Cartesian coordinates, that

[Lx, Ly] = i~Lz , [Ly, Lz] = i~Lx , [Lz, Ly] = i~Ly . (15.163)

Notice how the indices are cyclically permuted between x, y and z.. Let ussummarize this as follows by defining εijk, where i, j, k can take the valuesx, y, z, as be a symbol defined as follows:

εijk = 0 if i = j, j = k or i = k , (15.164)

εxyz = εyzx = εzxy = +1 (15.165)

εyxz = εxzy = εzyx = −1 (15.166)

The commutators are then easily summarized as

[Lj, Lk] = i~εjklLl . (15.167)

It is also easy to show that[L2, Li

]= 0 for i = x, y, z. To show this,

consider, for instance, i = 3.[L2, Lz

]=

[L2

x + L2y + Lz, Lz

](15.168)

=[L2

x, Lz

]+

[L2

y, Lz

]+

[L2

z, Lz

], (15.169)

= −[Lz, L

2y

]−

[Lz, L

2y

](15.170)

= −[Lz, Ly

]Lx − Lx

[Lz, Lx

]

−[Lz, Ly

]Ly − Ly

[Lz, Ly

](15.171)

= −(i~Ly)Lx − Ly(i~Ly)− (i~Lx)Ly − Ly(i~Lx)(15.172)

= −i~LyLx + i~LyLx − i~LyLy + i~LyLy , (15.173)

= 0 . (15.174)

Thus, L2 and one component of angular momentum can be chosen to besimultaneously diagonal, i.e it is possible to assign to find angular states withsimultaneous definite values of L2 and one of Li. If we choose i = z, thenthese angular states, expressed in spherical coordinates, are nothing but theY` m(θ, ϕ) that we found by solving the angular differential equation.

Matrix elements 170

Let us establish some facts. The operations Li are all hermitian. Let |ϕ〉be any vector (not necessarily normalized), and define

|χ〉 ≡(Li|ϕ〉

). (15.175)

The corresponding 〈χ| is then

〈χ| =(Li|ϕ〉

)†, (15.176)

= 〈ϕ|(Li

)†= 〈ϕ|Li , (15.177)

where the hermiticity of Li has been used.By the definition of the inner product, we have

0 ≤ 〈χ |χ〉 = 〈ϕ|LiLi|ϕ〉 , (15.178)

= 〈ϕ|(Li

)2

|ϕ〉 . (15.179)

Hence, for any vector |ψ〉,〈ψ|L2|ψ〉 ≥ 0 , (15.180)

sinceL2 =

∑i=x,y,z

L2i . (15.181)

15.3.2 L+ and L−

We now introduce two operators:

L+ = Lx + iLy , (15.182)

L− = Lx − iLy . (15.183)

Note that L+ and L− are not hermitian. In fact,(L+

)†= L− . (15.184)

We then have[Lz, L+

]=

[Lz, Lx + iLy

]=

[Lz, Lx

]+ i~

[Lz, Ly

](15.185)

= i~Ly − i(i~Lx) = ~(Lx + iLy) (15.186)

= ~L+ (15.187)

Matrix elements 171

Similarly [Lz, L−

]= −~L− (15.188)

Furthermore,

[L+, L−

]=

[Lx + iLy, Lx − iLy

](15.189)

=[(Lx,−iLy

]+

[iLy, Lx

](15.190)

= −2i[Lx, Ly

]= −2i~Lz = 2~Lz (15.191)

Also:

L+L− = (Lx + iLy)(Lx − iLy) (15.192)

= L2x + iLyLx − iLxLy + L2

y (15.193)

= L2x + L2

y + i[Ly, Lx

]= L2

x + L2y + i(−i~)Lz (15.194)

= L2x + L2

y + L2z − L2

z + ~Lz = L2 − L2z + ~Lz (15.195)

From this

〈`m|L+L−|`m〉 = 〈`m|L†−L−|`m〉 (15.196)

= ‖L−|`m〉‖2 ≥ 0 (15.197)

= 〈`m|(L2 − L2z + ~Lz)|`m〉 (15.198)

= ~2(`(` + 1)−m2 + m)〈`m|`m〉 (15.199)

`(` + 1)−m(m− 1) ≥ 0 → `(` + 1) ≥ m(m− 1) (15.200)

or (` +

1

2

)2

≥(

m− 1

2

)2

. (15.201)

Similarly, from 〈`m|L−L+|`m〉 ≥ 0, we find

~2(`(` + 1)−m(m + 1)) ≥ 0 (15.202)

`(` + 1) ≥ m(m + 1) (15.203)(` +

1

2

)2

≥(

m +1

2

)2

(15.204)

Matrix elements 172

From this, we conclude, depending on the relative signs of the squareroots, that

` +1

2≥ m− 1

2⇒ m ≤ ` + 1 , (15.205)

` +1

2≥ −m +

1

2⇒ m ≥ −` , (15.206)

` +1

2≥ m +

1

2⇒ m ≤ ` , (15.207)

` +1

2≥ −m− 1

2⇒ m ≥ −`− 1 . (15.208)

Putting these together, we have

−` ≤ m ≤ ` . (15.209)

To obtain relations between the various |`m〉 , consider the state L−|`m〉.Then:

LzL−|`m〉 =(LzL− − L−Lz + L−Lz

)|`m〉 (15.210)

=([

Lz, L−]

+ L−Lz

)|`m〉 (15.211)

=(−~L− + L−Lz

)|`m〉 (15.212)

= ~L−|`m〉+ m~L−|`m〉 = (m− 1)~L−|`m〉 (15.213)

Thus, L−|`m〉 is an eigenstate of Lz with eigenvalue m − 1. Hence,L−|`m〉 is proportional to |l, m− 1〉 since Lz|`,m− 1〉 = ~(m− 1)|`,m− 1〉.Let

L−|`m〉 = C−|`,m− 1〉 , (15.214)

where C− is the proportionality factor to be found. Observe that, for L−|`m〉to be normalized

〈`m|L†−L−|`m〉 = 1 = 〈`m|L+L−|`m〉 , (15.215)

= |C−|2〈l, m− 1|l,m− 1〉 = |C−|2 (15.216)

= ~2 [`(` + 1)−m(m− 1)] (15.217)

Thus, we can choose C− = ~√

(` + m)(`−m + 1) . So that,

L−|`m〉 = ~√

(` + m)(`−m + 1)|`,m− 1〉 (15.218)

Matrix elements 173

Similarly, consider L+|`m〉 .

LzL+|`m〉 = (LzL+ − L+Lz + L+Lz)|`m〉 (15.219)

= ([Lz, L+

]+ L+Lz)|`m〉 (15.220)

= (~L+ + L+Lz)|`m〉 = ~L+|`m〉+ L+~m|`m〉(15.221)

= ~(m + 1)L+|`m〉 (15.222)

Thus, L+|`m〉 = C+|`,m + 1〉. To find C+, start with

〈`m|L†+L+|`m〉 = 1 = |C+|2 (15.223)

= 〈`m|L−L+|`m〉 (15.224)

= 〈`m|(L2 − L2z − ~Lz)`m〉 (15.225)

= ~2[`(` + 1)−m2 −m

] 〈`m|`m〉 , (15.226)

from which we obtain C+ = ~√

(`−m)(` + m + 1). and

L+|`m〉 = ~√

(`−m)(` + m + 1)|`, m + 1〉 (15.227)

L±|`m〉 = ~√

(`∓m)(`±m + 1)|`,m± 1〉 (15.228)

15.3.3 Matrix elements of Lx and Ly

SinceL+ = Lx + iLy , L− = Lx − iLy ,

we obtain the matrix elements of Lx and Ly through

Lx = 12

(L+ + L−

), Ly = 1

2i

(L+ − L−

). (15.229)

Using Eq.(15.228), we find

〈`′m′|Lx|`m〉 = 12~

√(`−m)(` + m + 1)δ`′` δm′,m+1

+12~

√(` + m)(`−m + 1)δ`′` δm′,m−1 , (15.230)

〈`′m′|Ly|`m〉 =1

2i~

√(`−m)(` + m + 1)δ`′` δm′,m+1

− 1

2i~

√(` + m)(`−m + 1)δ`′` δm′,m−1 . (15.231)

Matrix elements 174

Thus, for a state with ` = 1, we have the possible kets |`m〉 = |11〉, |10〉and |1,−1〉. The matrix representation of Lx in this basis is obtained bynoting that the non–zero matrix elements are

〈10|Lx|11〉 = 12〈10|L−|11〉 = 1

2~

√(1 + 1)(1− 1 + 1) , (15.232)

=

√2

2~ , (15.233)

〈11|Lx|10〉 = 〈10|L†x|11〉∗ = 〈10|Lx|11〉∗ =

√2

2~ , (15.234)

〈1,−1|Lx|10〉 = 12〈1,−1|L−|10〉 = 1

2

√(1 + 0)(1− 0 + 1)~ ,(15.235)

=

√2

2~ , (15.236)

〈10|Lx|1,−1〉 = 〈1,−1|L†x|10〉∗ = 〈1,−1|Lx|10〉∗ =

√2

2~ . (15.237)

Thus, in the three–dimensional space spanned by |11〉, |10〉 and |1,−1〉, wehave

Lx 7→√

2

2~

0 1 01 0 10 1 0

. (15.238)

This is a hermitian matrix, as expected since Lx is a hermitian operator.Likewise, one readily finds that, when ` = 1,

Ly 7→√

2

2i~

0 1 0−1 0 10 −1 0

. (15.239)

Finally, we have

Lz 7→ ~

1 0 00 0 00 0 −1

. (15.240)

Note that, as matrices, we have, of course, [Lx, Ly] = i~Lz etc.

15.3.4 Constructing the simultaneous eigenfunctions of L2 and Lz.

Let 〈θ, ϕ | `m〉 = Y` m(θ, ϕ). From the condition −` ≤ m ≤ `, it follows thatthere must be a lowest m, i.e there must exists mmin such that L−|mmin〉 = 0.

Matrix elements 175

If not, then successive application of L− on |`m〉 would eventually producea state where m < −`. Obviously, the smallest value of m is −`.

Let us consider, using spherical coordinates,

L− = Lx − iLy (15.241)

= i~(

sin ϕ∂

∂θ+ cos ϕcotan θ

∂

∂ϕ

)

−~(

cos ϕ∂

∂θ− sin ϕcotan θ

∂

∂ϕ

)(15.242)

= ~e−iϕ

(∂

∂θ+ cotan θ(i2 sin ϕ− i cos ϕ)

d

dϕ

)(15.243)

= ~e−iϕ

(− ∂

∂θ+ i cotan θ

d

dϕ

), (15.244)

= ~e−iϕ

(− ∂

∂θ+ i cotan θ

∂

∂ϕ

)(15.245)

We also have, in spherical coordinates,

Lz = −i~∂

∂ϕ. (15.246)

Using this on the ket |`,−`〉 in the 〈θ, ϕ| basis, we find

LzY`,−`(θ, ϕ) = −i~∂

∂ϕY`,−`(θ, ϕ) = −` ~Y`,−`(θ, ϕ) , (15.247)

⇒ Y`,−`(θ, ϕ) = g(θ)e−i`ϕ (15.248)

Since m = −` is the smallest possible value of m, we have

L−Y`,−`(θ, ϕ) = 0 . (15.249)

This translates into the differential equation for g(θ):

(− d

dθ+ i(−i`)cotan θ

)g(θ) = 0 (15.250)

with solutiong(θ) = C` (sin θ)` , (15.251)

Matrix elements 176

so thatY`,−`(θ, ϕ) = C` (sin θ)`e−i`ϕ (15.252)

The integration constant C` is determined from the normalization condi-tion.

∫dΩ|Y`,−`(θ, ϕ)|2 = 1 = |C`|2

∫ 2π

0

dϕ

∫ π

0

dθ sin θ(sin θ)2` (15.253)

= 2π|C`|2∫ π

0

dθ sin θ(sin θ)2` (15.254)

To integrate, let ξ = cos θ and dξ = − sin θ dθ. The limits of integrationmust be adjusted. When θ = 0, ξ = 1 and when θ = π, ξ = −1 so

∫ π

0

dθ sin θ[(sin θ)2

]`=

∫ 1

−1

dξ(1− ξ2)` (15.255)

=

∫ 1

−1

dξ∑n=0

(−1)n `!

n!(n− `)!ξ2n (15.256)

=∑n=0

(−1)n `!

n!(n− `)!

ξ2n+1

2n + 1

∣∣∣1

−1, (15.257)

= 2∑n=0

(−1)n

(l

n

)1

2n + 1, (15.258)

=22`+1

(2` + 1)!(`!)2 (15.259)

To see this, let ` = 1.

∫ 1

−1

dξ(1− ξ2) =

[ξ − ξ3

ξ

]1

−1

= 1− 1

3−

(−1 +

1

3

)=

4

3=

23

3!=

8

6. (15.260)

Similarly, for ` = 2:

∫ 1

−1

dξ(1− ξ2)2 = 2

∫ 1

0

dξ(1− 2ξ2 + ξ4) = 2

[ξ − 2ξ3

3+

ξ3

5

]1

0

,(15.261)

= 2

[1− 2

3+

1

5

]= 2 · (15− 10 + 4)

15=

16

15, (15.262)

=25 · 2 · 2

5!=

128

2 · 3 · 4 · 5 =16

15(15.263)

Matrix elements 177

and so forth. Thus,

Y`,−`(θ, ϕ) =

√(2` + 1)!

4π

1

`!2`(sin θ)`e−i`ϕ (15.264)

To obtain the remaining spherical harmonics, consider 〈θ, ϕ|(L+)`+m|`,−`〉and recall that

L+|`m〉 =√

(`−m)(` + m + 1) |`,m + 1〉 (15.265)

Then:

(L+)`+m|`,−`〉= (L+)`+m−1|`,−` + 1〉

√2` · 1~ (15.266)

= (L+)`+m−2|`,−` + 2〉√

(2`) ·√

(2`− 1) · 2 ~2 (15.267)

= (L+)`+m−3|`,−` + 3〉√

(2`)(2`− 1)(2`− 2) · 1 · 2 · 3 ~3(15.268)

and generally:

〈θ, ϕ|(L+)`+m|`,−`〉 =

√(2`)!(` + m)!

(`−m)!〈θ, ϕ|`,m〉 ~`+m (15.269)

Hence,

Y`,m(θ, ϕ) = 〈θ, ϕ|`m〉 =

√(`−m)!

(2`)!(` + m)!〈θ, ϕ|(L+)`+m|`,−`〉 1

~`+m

(15.270)Using

L+ = −i~eiϕ

[i

∂

∂θ− cotan θ

∂

∂ϕ

](15.271)

= ~eiϕ

[∂

∂θ+ i cotan θ

∂

∂ϕ

](15.272)

Y`,−`(θ, ϕ) =

√(2` + 1)!

4π

1

`!2`(sin θ)`e−i`ϕ (15.273)

and

〈θ, ϕ|L`+m+ |`,−`〉 = ~`+mei(`+m)ϕ

(∂

∂θ+ i cotan

∂

∂ϕ

)`+m

Y`,−`(θ, ϕ) ,

(15.274)

Matrix elements 178

we finally arrive at

Y`,m(θ, ϕ) =

√(`−m)!

(2`)!(` + m)!

√(2` + 1)!

4π

1

`!2`

×[eiϕ

(∂

∂θ+ i cotan θ

∂

∂ϕ

)]`+m

(sin θ)`e−i`ϕ .(15.275)

For example, consider the case where ` = 1 . Then,

Y1,−1(θ, ϕ) =

√(2 · 1 + 1)!

4π

1

1!

1

2!sin θe−iϕ =

√6

16πsin θe−iϕ ,(15.276)

=

√3

8πsin θe−iϕ (15.277)

Y1,0(θ, ϕ) =

√(1− 0)!3

(1 + 0!4π

1

1!

1

21ei(1+0)ϕ

×(

∂

∂θ− i cos θ

∂

∂ϕ

)sin θe−iϕ (15.278)

=

√3

4π

1

2eiϕ

[(cos θ + i

cos θ

sin θ(−i) sin θ

)e−iϕ

], (15.279)

=

√3

4πcos θ , (15.280)

Y1,1(θ, ϕ) =

√(1− 1)!

(1 + 1)!· 3

4π

1

1!

1

2!

×[eiϕ

(∂

∂θ+ i cotan θ

∂

∂ϕ

)]2

sin θe−iϕ (15.281)

=

√3

2 · 4π1

2e−iϕ

(∂

∂θ+ i cotan θ

∂

∂ϕ

)

×eiϕ

(∂

∂θ+ icotan θ

∂

∂ϕ

)sin θe−iϕ (15.282)

=

√3

8π

1

2eiϕ

(∂

∂θ+ i cotan θ

∂

∂ϕ

)2 cos θ (15.283)

=

√3

8πeiϕ(− sin θ) = −

√3

8πeiϕ sin θ (15.284)

Matrix elements 179

An alternative is to work directly in Cartesian coordinates.


y∂

∂z− z

∂

∂y

), (15.285)

Lz = xpy − ypx (15.286)

Ly = −xpz + zpx = −i~(−x

∂

∂z+ z

∂

∂x

). (15.287)

The expressions for L+ and L− are then

L+ = Lx + iLy = ~(

z

(∂

∂x+ i

∂

∂y

)− (x + iy)

∂

∂z

)(15.288)

L− = Lx − iLy = ~(

z

(∂

∂x− i

∂

∂y

)− (x− iy)

∂

∂z

)(15.289)

Clearly,

L−(x− iy) = ~z(1− i(−i)) = 0 (15.290)

Lz(x− iy) = i~(−ix− y · 1) = ~(−x + iy) = −~(x− iy)(15.291)

Thus, (x− iy) is proportional to Y1,−1(θ, ϕ), or

〈x, y, z|1,−1〉 = α1Y−11 (15.292)

Quite clearly, then:

L−(x− iy)` = 0 , Lz(x− iy)` = −`(x− iy)l` (15.293)

so(x− iy)` = α`Y`,−`(θ, ϕ) (15.294)

Using spherical coordinates:

x− iy = r sin θ cos ϕ− ir sin θ sin ϕ = re−iϕ(sin θ)` (15.295)

(x− iy)` = r`e−i`ϕ(sin θ)` (15.296)

= r` · 2``!

√4π

(2` + 1)!Y`,−`(θ, ϕ) (15.297)

Let

〈x, y, z|`,−`〉 = (x− iy)` = r` 2``!

√4π

(2` + 1)!〈θ, ϕ|`,−`〉 . (15.298)

Matrix elements 180

Then

〈x, y, z|(L+)`+m|`,−`〉

=

√(2`)!(` + m)!

(`−m)!〈x, y, z|`m〉(~)`+m , (15.299)

= r`2``!

√4π

(2` + 1)!〈θ, ϕ|(L+)`+m|`,−`〉 , (15.300)

= r`2``!

√4π

(2` + 1)!

√(2`)!(` + m)!

(`−m)!Y`,m(θ, ϕ) (15.301)

On the other hand,

〈x, y, z|(L+)`+m|`,−`〉 (15.302)

= (L+)`+m(x− iy)` (15.303)

= (~)`+m

(z

(∂

∂x+ i

∂

∂y

)− (x + iy)

∂

∂z

)`+m

(x− iy)` (15.304)

= (~)`+mr`2``!

√4π

(2l + 1)

(` + m)!

(`−m)!Y`,m(θ, ϕ) (15.305)

or

Y`,m(θ, ϕ) =1

r`

1

2``!

√(`−m)!(2` + 1)

(` + m)!4π

×(

z∂

∂x+

∂

∂y− (x + iy)

∂

∂z

)l+m

(x− iy)` .(15.306)

Matrix elements 181

For instance:

Y1,−1(θ, ϕ) =1

r

1

21 · 1!

√2! · 30!4π

(x− iy) =1

r

√3

8π(x− iy) (15.307)

Y1,0(θ, ϕ) =1

r

1

2 · 1

√1 · 31 · 4π

×(

z

(∂

∂x+ i

∂

∂y

)− (x− iy)

∂

∂z

)(x− iy)(15.308)

=1

r

1

2

√3

4π(2z − 0) =

√3

4π

z

r(15.309)

Y1,1(θ, ϕ) =1

r

1

2·√

0! · 32!4π

×(

z

(∂

∂x+ i

∂

∂y

)− (x− iy)

∂

∂z

)2

(x− iy)(15.310)

=1

r

√3

8π

(z

(∂

∂y+ i

∂

y

)− (x + iy)

∂

∂z·)

z (15.311)

= −√

3

8π

(x + iy)

r(15.312)

Note that Y ∗`,m(θ, ϕ) = (−1)mY`,−m(θ, ϕ) .

Uncertainty relations 182

16 Uncertainty relations

The purpose of this section is to present a rigorous discussion of the uncer-tainty relation.

16.1 〈∆A〉〈∆B〉 ≥ 12 |〈[A, B]〉|

Let |ψ〉 be any normalized state. If 〈ψ|X|ψ〉 = 〈X〉, then, by definition, wehave, for any observable X:

∆X ≡ X − 〈X〉 , (16.1)

〈(∆X)2〉 ≡ 〈ψ|(X − 〈X〉

)2

|ψ〉= 〈ψ|X2 − 2X〈X〉+ 〈X〉2|ψ〉= 〈ψ|X2|ψ〉 − 2〈ψ|X|ψ〉〈X〉+ 〈X〉2= 〈ψ|X2|ψ〉 − 〈X〉2 ,

= 〈X2〉 − 〈X〉2 ≥ 0 . (16.2)

We now show the uncertainty relation:

〈(∆A)2〉〈(∆B)2〉 ≥ 1

4|〈[A, B]〉|2 . (16.3)

Proof: Let

|ψA〉 ≡ ∆A|ψ〉 = A|ψ〉 − |ψ〉〈ψ|A|ψ〉 ,|ψB〉 ≡ ∆B|ψ〉 = B|ψ〉 − |ψ〉〈ψ|B|ψ〉 (16.4)

where |ψA〉 is any normalized state. (Be mindful of this condition: the proofcan break down if |ψA〉 is non normalizable.) Note, for future reference, thatthe strict equality holds only if |ψB〉 is proportional to |ψA〉.

Then, by Schwarz inequality,

〈ψA|ψA〉〈ψB|ψB〉 ≥ |〈ψA|ψB〉|2 (16.5)

Assuming that A and B are hermitian, we have

〈ψA|ψA〉 = 〈ψ|∆A∆A|ψ〉 = 〈(∆A)2〉 , (16.6)


and similarly for 〈ψB|ψB〉. Still under the assumption that the operatorsare hermitian, the inner product on the right-hand side of Eq.(16.5) can bedevelopped into

〈ψA|ψB〉 = 〈ψ|∆A∆B|ψ〉 (16.7)

= 〈ψ|12(∆A∆B −∆B∆A) + 1

2(∆A∆B + ∆B∆A)|ψ〉 .(16.8)

Consider now the first term of this last sum:

〈ψ|12(∆A∆B −∆B∆A)|ψ〉 = 〈ψ|1

2[∆A, ∆B]|ψ〉 = 1

2〈ψ|[A, B]|ψ〉 . (16.9)

The operator [A, B] = AB − BA is clearly anti-hermitian, since

[A, B]† = (AB − BA)† ,

= (AB)† − (BA)† ,

= B†A† − A†B†

= BA− AB = −[A, B] , (16.10)

using the fact that A and B are hermitian. Now, the expectation value ofan anti-hermitian operator C (i.e. such that C† = −C) is purely imaginary,since it changes sign under complex conjugation:

〈ψ|C|ψ〉∗ = 〈ψ|C†|ψ〉 = −〈ψ|C|ψ〉 . (16.11)

Let us therefore denote 〈ψ|[A, B]|ψ〉 ≡ ic−, with c− ∈ R.On the other hand, the operator ∆A∆B + ∆B∆A is clearly hermitian;

its expectation value is purely real and denoted by c+.Hence

〈ψA|ψB〉 = 12(ic− + c+) ,

|〈ψA|ψB〉|2 =1

4(c2− + c2

+) ≥ 1

4c2−

≥ 1

4|〈[A, B]〉|2 . (16.12)

Putting it all together, we get

〈(∆A)2〉〈(∆B)2〉 ≥ |〈ψA|ψB〉|2 ≥ 1

4|〈[A, B]〉|2 , (16.13)

which proves the claim.In particular, if A = x , B = p so that [x, p] = i~ 1l, we then obtain

〈∆x〉〈∆p〉 ≥ 12~ . (16.14)


16.2 Understanding the assumptions

Suppose that |x0〉 is an eigenstate of x, i.e. x|x0〉 = x0|x0〉. Then, clearly,∆x = 0 for this state, and

∆x ∆p ≥ 12|〈x0|[x, p]|x0〉| ,

0 ·∆p = 12|〈x0|i~ 1l |x0〉| ,

= 12~〈x0|x0〉 . (16.15)

If we go with the assumption that |x0〉 is normalizable to 1 and that p ishermitian, then we are in trouble: the r.h. side is 1

2~ while the l.h. side is

0 if ∆p is finite. If ∆p → ∞, the l.h. side must be considered much morecarefully: one must use Fourier analysis arguments to show, using a sequenceof normalizable states approaching |x0〉, that ∆p → ∞ when ∆x → 0. Thestate |x0〉 is itself non–normalizable.

Alternatively, if we insist on 〈x0|x0〉 = 1, then it follows that p cannot behermitian.


Figure 21: As soon as they are under observation, the atoms behave differently.

Complex numbers 186

A Complex numbers

The entire field of classical mechanics can studied using real numbers only. Itis possible to use complex numbers (or quaternions, or octonions), but theirintroduction is incidental and represents merely a (sometimes good) trick tohelp with the calculations.

By contrast, complex numbers are essential to quantum mechanics.

A.1 A bit of history

If you are given the linear equation

mz + b = 0 , m, b ∈ R , (A.1)

then the solution z to this equation is also a real number. However, if youare asked to solve the quadratic equation

az2 + bz + c = 0 , (A.2)

where, once again, the coefficients a, b, c ∈ R, then it may be that the solu-tions

z± =−b±√b2 − 4ac

2a(A.3)

do not exists over the reals if the discriminant b2−4ac < 0. In order to solveall quadratic equations, we need to introduce a new kind of number calledthe complex number.

The simplest complex number is i, and its basic property is that

i2 = −1 . (A.4)

Suppose b2 − 4ac < 0. Then, write

b2 − 4ac = −(4ac− b2) = i2(4ac− b2) (A.5)

where 4ac− b2 > 0. Since we know (in principle) how to compute the squareroot of a positive number like 4ac − b2, the square root of a square like i2,and the square root of a product, we find that

z± =−b±

√i2(4ac− b2)

2a=−b± i

√4ac− b2

2a. (A.6)

Complex numbers 187

Example A.1 The roots of the quadratic equation 3z2 − 2z + 1 = 0 arez+ = 1

3(1 + i

√2) and z− = 1

3(1− i

√2).

When b2 − 4ac < 0, the solutions z± to Eq.(A.2) are made of two parts:the real part, −b/2a, and another part, called the imaginary part,

√4ac− b2,

which comes multiplied by the imaginary unit i. Please note that both thereal and imaginary parts are real numbers, as a, b and c are real and, byassumption, 4ac− b2 > 0.

One writes a general arbitrary complex number z = x+ iy, where x and yare both real. x is the real part of z, often denoted x = Re(z), and y = Im(z)is the imaginary part of z. Two complex numbers z1 and z2 are equal if theirreal and imaginary parts are respectively equal, i.e. if z1 = x1 + iy1 andz2 = x2 + iy2, then

z1 = z2 ⇔ x1 = x2 and y1 = y2. (A.7)

Example A.2 Consider the complex number z = 3 + iπ. Its real part is 3and its imaginary part is π.

A number is real if its imaginary part is 0, otherwise it is complex. If thereal part of a complex number is 0, then one often says that this number ispurely imaginary.

If you are now given a cubic equation:

az3 + bz2 + cz + d = 0 , a, b, c, d ∈ R , (A.8)

then you may think one would require the introduction of yet another typeof number, the simplest of which (call it putatively j), would be such thatj3 = −1 (or some similar property related to the third power of j). It turnsout, however, that this is not so. The great triumph of algebra, as containedin the Fundamental Theorem of Algebra, is the observation that, for anyequation of type

anzn + an−1zn−1 + . . . + a0 =

n∑p=0

apzp = 0 , (A.9)

with ap ∈ R, one can find all the solutions using at most complex numbers.You never need to use anything more complicated than complex numbers tosolve an algebraic equation of any degree if the coefficients are real.

Complex numbers 188

A.2 Adding complex numbers

It is convenient to associate with a complex z = x + iy a point (x, y) onthe R2 plane. Usually, the horizontal axis of the plane is chosen as the realaxis, while the vertical axis is the imaginary axis. Hence, all real numbers lieon the horizontal axis, and purely imaginary numbers on the vertical axis.This representation is known as the Cartesian representation of the complexnumbers. It is illustrated in fig.22.

z=(x,y)

x

y

Re(z)

Im(z)

Figure 22: The Cartesian representation of a complex number z = x + iy.

The Cartesian representation is useful to picture the operation of additionof two (or more) complex numbers. If z1 = x, +iy, and z2 = x2+iy2, then thesum z3 = z1 + z2 is this complex number x3 + iy3 such that x3 = x1 + x2 andy3 = y1 +y2. Adding complex numbers therefore corresponds, graphically, toadding vectors in R2 having, respectively, (x1, y1) and (x2, y2) as components.

The general properties of addition of complex numbers are obvious:

1. z1 +z2 = z2 +z1, i.e. the order in which they are added doesn’t matter,

2. z1 + (z2 + z3) = (z1 + z2) + z3,

3. 0 = 0 + 0i = 0i is the neutral element: z + 0 = 0 + z = z,

4. if z1 = x, +iy,, then its inverse −z1 = −x1 − iy1 is unique and suchthat z1 + (−z1) = 0,

5. if k ∈ R, then k(z1 + z2) = kz1 + kz2 = k(x1 + x2) + ik(y1 + y2) =kx1 + kx2 + iky1 + iky2,

Complex numbers 189

6. if k, m ∈ R, then (k + m)z = kz + mz,

7. if k, m ∈ R, then k(mz) = (km)z,

8. 1z = z.

Example A.3 Let z1 = 4 − 3i, z2 = e2 + i ln(2) be two complex numbers.Then

z1 + z2 = (4 + e2) + i(−3 + ln(2)) , 7z1 = 7(4− 3i) = 28− 21i , etc.

A.3 The complex conjugate

There is a very important complex number associated to z: it is called thecomplex conjugate of z and is usually denoted by z∗ (or z). z∗ is the complexnumber z∗ = x− iy, obtained by changing i → −i everywhere in z.

One sees that the real part x of z can be obtained from x = 12(z + z∗)

while the imaginary part is obtained from

z − z∗ = 2iy ⇒ −i(z − z∗) = 2y ⇒ y = − i

2(z − z∗) . (A.10)

Example A.4 Let z1 = 4 − 3i, z2 = e2 + i ln(2) be two complex numbers.Their respective complex conjugates are

z∗1 = 4 + 3i , z∗2 = e2 − i ln(2) .

Next, note that

z · z∗ = (x + iy) · (x− iy) = x2 + y2 ≥ 0 . (A.11)

Thus, z · z∗ is necessarily real and non-negative. In fact, it is zero only ifz = 0; thus z · z∗ measures the square of the “length” of the complex numberz in the R2 plane. It is customary to denote the “length” of z, called themodulus of z, by

ρ ≡ |z| ≡ +√

z · z∗ . (A.12)

Complex numbers 190

Example A.5 Let z = 3 + 4i. Then

z∗ = 3− 4i, z · z∗ = 25 |z| = 5 .

A.4 Multiplication and division

A.4.1 Multiplying and dividing in the Cartesian representation

Let z1 = x, +iy, and z2 = x2 + iy2 be two complex numbers. Their productz3 = z1 · z2 is defined by simply considering i like any other number andapplying the usual distributivity of multiplication:

z1 · z2 = (x1 + iy1) · (x2 + iy2) ,

= x1x2 + iy1x2 + ix1y2 + i2y1y2 ,

= (x1x2 − y1y2) + i(x1y2 + y1x2) , (A.13)

where we have used i2 = −1.

Example A.6 Let z1 = 4− 3i, z2 = e2 + i ln(2). Then

z1 · z2 = (4e2 + 3 ln(2)) + i(−3e2 + 4 ln(2)).

Division by a complex number requires the use of the complex conjugate.We define the quotient

z3 =z1

z2

≡ z1

z2

× z∗2z∗2

=z1 · z∗2z2 · z∗2

. (A.14)

The denominator z2 · z∗2 is a real number, so that

z3 =

(x1

z2 · z∗2+ i

y1

z2 · z∗2

)(x2

z2 · z∗2− i

y2

z2 · z∗2

)=

(x1x2 + y1y2

x22 + y2

2

)+i

(x2y1 − x1y2

x22 + y2

2

).

(A.15)This kind of manipulation can become quite cumbersome, especially if thenumerator and denominators contain multiple products of complex numbers.

Complex numbers 191

Example A.7 Let z1 = 4− 3i, z2 = e2 + i ln(2). Then

z1

z2

=4e2 − 3 ln(2)

e4 + ln(2)2+ i

(−3e2 − 4 ln(2)

e4 + ln(2)2

).

A.4.2 The polar representation

z=( cos( ) , sin( ))

sin( )

Re(z)

Im(z)

θ

θ

θθ

θρ

cos( )ρ

ρρ

ρ

Figure 23: The polar representation of a complex number z = ρ(cos θ + i sin θ).

To multiply complex numbers, it is convenient to use the so–called polarrepresentation. Since z · z∗ is the square of the “length” of z = x + iy, weconsider the vector z in the plane R2 and note that, by simple trigonometry,we can define an angle θ such that

x = ρ cos θ , y = ρ sin θ , ρ = |z| =√

zz∗ . (A.16)

Thus, when θ = 0, z is real, whereas, when θ = π/2, z is purely imaginary.It is then a simple matter to write

z = x + iy = ρ (cos θ + i sin θ) , (A.17)

withρ = |z| =

√z · z∗ , tan θ =

y

x. (A.18)

Complex numbers 192

The combinaison cos θ + i sin θ will soon become supremely important.For the moment, let us write it in the shorthand notation

cos θ + i sin θ ≡ cis(θ) . (A.19)

In particular , if z = ρ cis(θ), then the complex conjuage z∗ = ρ cis(−θ).

Example A.8 Let z1 = −1 + i√

3, z2 = 3 + 4i. Their polar forms are,respectively,

z1 = 2 cis(2π3

) , z2 = 5 cis(arctan(4/3)) .

If z1 = x1 + iy1 , z2 = x2 + iy2, then their product is simply expressed, inthe polar representation, as

z1 · z2 = ρ1ρ2 (cos θ1 + i sin θ1) (cos θ2 + i sin θ2) ,

= ρ1ρ2 (cos θ1 cos θ2 − sin θ1 sin θ2)

+iρ1ρ2 (cos θ1 sin θ2 + sin θ1 cos θ2) ,

= ρ1ρ2 cos(θ1 + θ2) + iρ1ρ2 sin(θ1 + θ2) ,

= ρ1ρ2 cis(θ1 + θ2) . (A.20)

One thus observes the fundamental property that

cis(θ1) cis(θ2) = cis(θ1 + θ2) . (A.21)

Division is also quite easy:

z1

z2

=z1 · z∗2z2 · z∗2

=ρ1ρ2

ρ22

cis(θ1 − θ2) =ρ1

ρ2

cis(θ1 − θ2) . (A.22)

Example A.9 Let z1 = 3i , z2 = −1 + i√

3. The polar representation ofz1 is z1 = 3 cis(π

2), and the polar representation of z2 was obtained in the

previous example. We have, on the one hand,

z1 · z2 = 3i(−1 + i√

3) = −3√

3− 3i = 6 cis(7π6

)

while, on the other hand,

z1 · z2 = 3 cis(π2) · 2 cis(2π

3) = 6 cis(7π

6) ,

Complex numbers 193

as expected.

Example A.10 Let z1 = 3i , z2 = −1 + i√

3. Then

z1

z2

=3

2cis(−π

6)

as can be confirmed using the Cartesian representation.

A.4.3 Euler’s theorem

Eq.(A.21) is quite remarkable. If one thinks of cis as a function, then it is wiseto ask what are the known functions that possess this property cis(θ) cis(ϕ) =cis(θ+ϕ) . The answer is that (besides the trivial constant function) there isonly one: the exponential function. Indeed, if f(x) = ex, then f(θ) · f(ϕ) =f(θ + ϕ). What we have observed is that there is (apparently, at least) aconnection between cis and the exponential function. There are several waysof finding the link: let us assume that

cos θ + i sin θ = Aekθ , A, k ∈ C . (A.23)

Recall now the Taylor series for cos θ, sin θ and ekx:

cos θ = 1− 1

2!θ2 +

1

4!θ4 + . . . ,

sin θ = θ − 1

3!θ3 +

1

5!θ5 + . . . ,

ekθ = 1 + kθ +1

2!k2θ2 +

1

3!k3θ3 + . . . (A.24)

Since Eq.(A.23) is a functional relation, it must hold for every θ, so that wecan compare the l.h.s and r.h.s of Eq.(A.23) term by term. Writing the firstfew terms of this expansion:

1 + iθ − 1

2!θ2 + . . . = A + Akθ + A

1

2!k2θ2 + . . . . (A.25)

From the terms in θ0 (i.e. the constant terms), we have A = 1. From theterms in θ, we have i = Ak. Using A = 1, we immediately find that k = i.

Complex numbers 194

We can verify that this choice holds for all powers in θ. Hence, we find afamous formula, called Euler’s theorem:

cos θ + i sin θ = eiθ . (A.26)

(Euler proved hundreds of theorems in just about every field of mathematicsthat existed when Euler was around; these theorems are called Euler’s theo-rem on this or Euler’s theorem on that, depending on the particular theoremat hand. Eq.(A.26) is so important that it is does not have a special name:it’s simply called Euler’s theorem.)

A first consequence of Eq.(A.26) is that

cos θ = 12

(eiθ + e−iθ

), sin θ =

1

2i

(eiθ − e−iθ

). (A.27)

Eq.(A.27) is extremely useful in proving trigonometric identities.

Example A.11 Using Eq.(A.27), one can prove that

cos(θ + φ) = cos(θ) cos(φ)− sin(θ) sin(φ) .

From the expression for cos(θ) and sin(θ), we have

cos(θ) cos(φ) = 12

(eiθ + e−iθ

)12

(eiφ + e−iφ

),

= 14

(ei(θ+φ) + e−i(θ−φ) + ei(θ−φ) + e−i(θ+φ)

),

= 12cos(θ + φ) + 1

2cos(θ − φ) ,

sin(θ) sin(φ) = 12i

(eiθ − e−iθ

)12i

(eiφ − e−iφ

),

= −14

(ei(θ+φ) − e−i(θ−φ) − ei(θ−φ) + e−i(θ+φ)

),

= −12cos(θ + φ) + 1

2cos(θ − φ) .

We can add these to results to obtain the desired trigonometric identity.

Example A.12 Let m,n be integers. Show that

∫ 2π

0

cos(mθ) cos(nθ) dθ =

0, if m 6= n ,π, if m = n .

To show this, we use, from the previous example,

cos(mθ) cos(nθ) =1

4

(ei(m+n) + e−i(m−n)θ + ei(m−n)θ + e−i(m+n)

),

Complex numbers 195

so that, by integrating the exponential functions, and assuming that m 6= n,

∫ 2π

0

cos(mθ) cos(nθ) dθ

=1

4

[1

i(m + n)ei(n+m)θ +

1

−i(m− n)e−i(m−n)θ

+1

i(m− n)ei(m−n)θ +

1

−i(m + n)e−i(m−n)θ

]2π

0

=1

4i

[1

(m + n)

(ei2π(n+m) − 1

)− 1

(m− n)

(e−i2π(m−n) − 1

)

+1

(m− n)

(ei2π(m−n) − 1

)− 1

(m + n)

(e−i2π(m−n) − 1

)]

=1

4i[1− 1 + 1− 1 + 1− 1 + 1− 1] ,

= 0 .

If m = n, then

cos(mθ) cos(mθ) =1

4

(e2imθ + 2 + e−2imθ

)= 1

2(cos(2mθ) + 1) ,

so that ∫ 2π

0

cos(mθ) cos(mθ) dθ =1

2

[sin(2mθ)

2m+ θ

]2π

0

= π ,

which completes the proof of the result.

Another consequence of Euler’s theoreom is as follows: set θ = π. Then

eiπ + 1 = 0 . (A.28)

Euler couldn’t imagine that these seemingly completely unrelated numbers,e, i, the angle π, and the numbers 1 and 0, could come together in this thesimplest of all mathematical relations. Indeed, Euler considered this resulta proof of the existence of God.

Putting all of this together, we finally have the polar form of a complexnumber z as

z = ρ(cos(θ) + i sin(θ)) = ρ eiθ = ρ cis(θ) . (A.29)

Complex numbers 196

A.4.4 Taking powers and roots

The use of Eq.(A.29) makes it particularly easy to take powers of complexnumbers. Thus, if p is an integer and z = ρ eiθ, then

zp = ρpeipθ = ρp cispθ. (A.30)

Example A.13 Let z = −1 + i√

3. To compute z3, we first use (a + b)3 =a3 + 3a2b + 3ab2 + b3 and expand explicitly (−1 + i

√3)3 with a = −1 and

b = i√

3 to findz3 = 8 .

Using the polar form with z = 2 cis(2π3

), we immediately find

z3 = 23 cis(2π) = 8 .

Try computing z17 with the Cartesian representation!

One would think that the extension of Eq.(A.30) to the case where p isfractional (say, 1

/2) would be easy:

z1/2 =

√ρeiθ/2 . (A.31)

However, there is a subtelty involved in taking roots. Even with real numbers,we know that √

25 = ±5 , (A.32)

that is, there are two solutions. Thus, we expect that

z1/2 = ±√ρeiθ/2 ,

=√

ρe±iπ/2eiθ/2 . (A.33)

How can we generalize this to taking the n’th root? One would expect tofind n solutions. For instance, the four (complex) numbers z1, z2, z3 and z4

such that z4k = 1 are simply ±1,±i since (±i)4 = 1.

Let us go back to the logic of finding the two numbers z1 and z2 such thatz21 = z2

2 = 25. We start by finding one number, say η, such that η2 = 25. Letη = 5. Next, consider the all the numbers whose square is 1: there are twobecause we’re calculating a square root, and they are ξ1 = 1 and ξ2 = −1.

Complex numbers 197

One then easily sees that the two solutions to z2 = 25 are z1 = ξ1η = +5and z2 = ξ2η = −5. This works because

zk = ξkη ,

z2k = (ξkη)2 ,

= (ξk)2 η2 ,

= 1× 25 , (A.34)

since, by construction, ξ2k = 1 and η2 = 25.

The procedure to obtain all the n’th root of the complex number z is basedon this line of thought, along with the observation that ξk = (e2ikπ/p)p =e2ikπ = 1.

To find the n’th root of z, start with one complex number η such thatηn = z. Finding η is actually quite simple: just express z = ρeiθ and thenη = ρ1/neiθ/n. Then, multiply η by all the numbers ξk such that ξn

k = 1.The complete list of complex numbers ζk such that ζn

k = z is then simplyobtained by multiplying in turn η by every ξk = e2ikπ/n.

Example A.14 Let z = −1/2 + i√

3/2. To find all the numbers z1/4, let usfirst find one number z1 such that z4

1 = z. We use the first polar form of z,from which we obtain

z1 = z1/4 = ( cis(2π3

))1/4 = cis(2π12

) =√

32

+ 12i.

Next, note that ξ1 = e2iπ/4 = i , ξ2 = e2·2iπ/4 = −1 , ξ3 = e2·3iπ/4 = −i , ξ4 =e2·4iπ/4 = 1 , are the four numbers such that ξ4

k = 1. Thus, the four roots are

ξ1 · z1 = i(√

32

+ 12i)

= −12

+ i√

32

= eiπ/6eiπ/2 = e2iπ/3

ξ2 · z1 = −1(√

32

+ 12i)

= eiπ/6eiπ = e7iπ/6

ξ3 · z1 = −i(√

32

+ 12i)

= e−iπ/3

ξ4 · z1 = z1 =(√

32

+ 12i)

. (A.35)

Example A.15 Let ω be a complex number such that ωn = 1 (n is apositive integer). Show that, if ω 6= 1, then

ω + ω2 + ω3 + . . . + ωn−1 + ωn =n∑

p=1

ωp = 0 .

Complex numbers 198

Proof: Suppose that∑n

p=1 = A , A ∈ C . Then,

ω

(n∑

p=1

ωp

)= ωA =

n∑p=1

ωp+1 =n∑

p=1

ωp , = A , (A.36)

since ωn+1 = ωnω = ω. Thus, A = ωA. There are two solutions: ω = 1,which contrary to the original assumption that ω 6= 1 and so must be rejected,or A = 0, which is the only solution compatible with ω 6= 1. QED.

Simpson’s rule 199

B Numerical integration and Simpson’s rule

It is always possible to find the derivative of a function, but it is not alwayspossible to reverse the process and find an expression for

∫dx g(x) . (B.1)

Two examples will suffice to illustrate this: one cannot “ solve” the indefiniteintegrals ∫

dx e−x2

, or

∫dx

sin x

x. (B.2)

Quite generally, it may therefore not be possible to evaluate the definiteintegral ∫ b

a

dx g(x) (B.3)

by simple means. One must then proceed numerically, keeping in mind thatEq.(B.3) represents the area under the curve.

B.1 Preliminaries

There are many ways of understanding the so–called “Simpson ’s rule” , butthe following is particularly intuitive. Take the function g(x) in Eq.(B.3) andapproximate it over the interval [a, b] by a polynomial:

g(x) ≈ c0 + c1x + c2x2 + ...cnx

n, (B.4)

with the coefficients c0, ..., cn chosen in some way to be specified. Then,∫ b

a

dx g(x) ≈∫ b

a

dx(c0 + c1x + c2x

2 + ...cnxn), (B.5)

which can be integrated easily.Consider, as a very simple example, the problem of numerically integrat-

ing ∫ 2

0

e−xdx. (B.6)

Of course, this is a made up problem: we can easily get a primitive for thecorresponding indefinite integral and find the exact result

∫ 2

0

e−xdx = −e−x∣∣∣2

0= −e−2 + 1 ≈ 0.8647. (B.7)


We will use this simply as a guide to see how close various numerical schemescome to the “true” answer.

One could approximate this integral by replacing the e−x by a straightline:

e−x ≈ c0 + c1x. (B.8)

To determine the coefficients, use x = 0 and x = 2 to get the system

e−0 = c0, e−2 = c0 + 2c1 (B.9)

which immediately yields

c0 = 1 , c1 =e−2 − 1

2, (B.10)

∫ 2

0

e−xdx ≈∫ 2

0

(c0 + c1x) dx =(c0x + 1

2c1x

2) ∣∣∣

2

0

= 2c0 + 412c1 = 2(c0 + c1) = 1.1359 . (B.11)

We can get a feel for what is going on by considering Fig.24. We seethat the linear approximation is always above the function e−x, and the areaunder the straight line is visible larger than the area under e−x.

0.5 1 1.5 2

0.2

0.4

0.6

0.8

1

x

e-x

c0+c1x

Figure 24: The function e−x and its linear approximation by c0 + c1x over therange [0, 2].

Alternatively, we could consider replacing e−x by the quadratic

e−x ≈ c0 + c1x + c2x2, (B.12)


using the three points x = 0, x = 1 and x = 2 to get:

e−0 = c0,

e−1 = c0 + c1 + c2, (B.13)

e−2 = c0 + 2c1 + 4c2.

Solving for the coefficients yields

c0 = 1 , c1 = −1− 4e + 3e2

2e2, c2 =

1− 2e + e2

2e2(B.14)

so that∫ 2

0

e−xdx ≈∫ 2

0

(c0 + c1x + c2x

2)dx (B.15)

= 2c0 + 2c1 + 83c2 = 0.8689. (B.16)

The situation is now illustrated on Fig.25. The quadratic curve does amuch better job of approximating e−x than the straight line, with the resultthat the approximate value of the integral is much closer to the “exact” valuethan corresponding result for the straight line.

0.5 1 1.5 2

0.2

0.4

0.6

0.8

1

c0+c1x+c2x2

e-x

x

Figure 25: The function e−x and its linear approximation by c0 + c1x + c2x2 over

the range [0, 2].

Intuitively, it is clear that, as we add more and more terms to the poly-nomial on the right hand side of Eq.(B.4), we are increasing the accuracy ofthe calculation. However, the improvement in accuracy is done at the cost


of increasing the difficulty in specifying the coefficients ci. If we keep veryfew terms, the coefficients will be easy to evaluate but the calculation willnot be so precise.

A nice way to balance this is to start dividing the interval of integration,i.e. use the result

∫ b

a

dx f(x) =

∫ β

a

dx f(x) +

∫ b

β

dx f(x). (B.17)

In our example, we can choose β = 1 so that

∫ 2

0

e−x dx =

∫ 1

0

e−x dx +

∫ 2

1

e−x dx. (B.18)

We can then approximate the exponential over each subinterval. Thus, inthe interval [0, 1], we can use

e−x ≈ c10 + c1

1x + c12x

2, (B.19)

and the points x = 0, x = 1/2 and x = 1 to get

e−0 = c10, (B.20)

e−1/2 = c10 + 1

2c11 + 1

4c12, (B.21)

e−1 = c10 + c1

1 + c12, (B.22)

so that

c10 = 1 , c1

1 = −1− 4e1/2 + 3e

2e, c1

2 = 2

(1− 2e1/2 + e

e

), (B.23)

∫ 1

0

e−xdx ≈ (c10 x + 1

2c11 x2 + 1

3c12 x3)

∣∣∣1

0

= c10 + 1

2c11 + 1

3c12 = 0.6223. (B.24)

In the interval [1, 2], we choose x = 1, x = 3/2 and x = 2 so that

e−1 = c20 + c2

1 + c22,

e−3/2 = c20 + 3

2c21 + 9

4c22, (B.25)

e−2 = c20 + 2 c2

1 + 4 c22,


and

c20 =

3− 8e1/2 + 6e

e2, c2

1 = −5− 12e1/2 + 7e

2e2,

c22 = 2

(1− 2e1/2 + e

e2

), (B.26)

∫ 2

1

e−x dx ≈ (c20 x + 1

2c21 x2 + 1

3c22 x3)

∣∣∣2

1

= c20 + 3

2c21 + 7

3c22 = 0.2326 (B.27)

and thus ∫ 2

0

e−x dx ≈ 0.8649 . (B.28)

This is illustrated in Fig.26. We see that there is practically no visibledifference between the piecewise approximation by a quadratic and the exactcurve. The final area is close to the “true” area under the curve e−x.

0.5 1 1.5 2

0.2

0.4

0.6

0.8

1

x

Figure 26: The function e−x and its linear approximation by c0 + c1x + c2x2 over

the range [0, 2].

Before we discuss the generalization of this scheme, let us point out someof the features of the previous calculation. First, note that we used the pointsx = 0, 1

2, 1, 3

2and 2. The point x = 1 was used twice. We will label these

points asx0 = 0 , x1 = 1

2, x2 = 1 , x3 = 3

2, x4 = 2 . (B.29)


If we define a segment as a segment of the curve between xn and xn+1, we seethat we used four segments to compute the integral, each segment having thesame length xn+1 − xn = 1

2. You can verify for yourself that, had we written

∫ 2

0

e−x dx =

∫ 2/3

0

e−x dx +

∫ 4/3

2/3

e−x dx +

∫ 2

4/3

e−x dx , (B.30)

we would have used six segments. In fact, it is quite clear that, if we approx-imate g(x) on each segment by a quadratic, the total number of segmentswill always be even.

B.2 Simpson’s rule

A systematic method to proceed with the numerical integration is as follows.To evaluate Eqn.(B.1), one first divides the interval [a, b] into an even numbern = 2m of equal segments of length h = (b− a)/2m. The boundaries of thei’th segment will be denoted by xi and xi+1, i.e. the i’th segment runsbetween [xi, xi+1], with x0 = a, x1 = a + h, xn = b and xi+1 − xi = h.

We start by evaluating the function g(x) at every one of the 2m points,thus obtaining a table of values containing xi and g(xi). Next, we considerthree consecutive values g(xi), g(xi+1), g(xi+2) and fit through those pointsthe parabola Ax2 + Bx + C, with A,B,C unknowns to be adjusted so thatthe parabola goes exactly through g(xi), g(xi+1), g(xi+2). These are the ci

j’sof the previous section.

In order to find A,B, C, we therefore need to solve the three equations

Ax2i + Bxi + C = g(xi)

Ax2i+1 + Bxi+1 + C = g(xi+1) (B.31)

Ax2i+2 + Bxi+2 + C = g(xi+2)

Suppose now we have found A,B,C. We then have, on the consecutivesegments [xi, xi+1] and [xi+1, xi+2] as approximation of g(x) as Ax2 +Bx+C.We can then integrate this:

∫ xi+2

xi

dx g(x) ≈∫ xi+2

xi

dx(Ax2 + Bx + C

),

=

(A

3x3 +

B

2x2 + C x

) ∣∣∣xi+2

xi

. (B.32)


If we remember that xi+1 = xi + h, xi+2 = xi + 2h, we eventually find,using the solutions for A,B,C, that

∫ xi+2

xi

dx g(x) ≈ h

3(g(xi) + 4g(xi+1) + g(xi+2)) . (B.33)

Eqn.(B.33) is the basis for Simpson’s rule. Clearly, we have now

∫ xi+4

xi

dx g(x) =

∫ xi+2

xi

dx g(x) +

∫ xi+4

xi+2

dx g(x) ,

≈ h

3(g(xi) + 4g(xi+1) + g(xi+2)

+g(xi+2) + 4g(xi+3) + g(xi+4)) (B.34)

and so forth, until we finally obtain

∫ b

a

dx g(x) ≈ (b− a)

6m[g(x0) + g(x2m)

+2 (g(x2) + g(x4) + . . . + g(x2m−2))

+4 (g(x1) + g(x3) + g(x5) + . . . + g(x2m−1)](B.35)

Note that the accuracy of this method increases with decreasing h, i.e. withincreasing number of intervals n = 2m.

B.3 Application: quantum particle outside a harmonicwell

As an example of application, we consider the probability of finding a particledescribed by the first excited state of a harmonic oscillator to be “outside”the classically allowed region.

First, we need to find the classical turning points, i.e. those values of xfor which the potential energy is equal to the total energy of a particle; inprinciple, a particle has no kinetic energy at that point and must therefore“turn around” its motion.

To find the turning points, we need the energy of a particle described bythe wavefunction ψ1(x) given in Eq.(6.76):

ψ1(x) =(mω

~π

)3/4√

2

π1/4x e−mωx2/2~ .


The potential for this system is V (x) = 12mω2x2, so ψ1(x) is solution of

− ~2

2m

d2

dx2ψ1(x) + 1

2mω2x2 ψ1(x) = E1ψ1(x) , (B.36)

from which we can infer the energy E1 of the particle.Using

d

dxψ1(x) =

(mω

~π

)3/4√

2

π1/4

(e−mωx2/2~ − mω

~x2 e−mωx2/2~

),

=(mω

~π

)3/4√

2

π1/4e−mωx2/2~

(1− mω

~x2

), (B.37)

d2

dx2ψ1(x) =

(mω

~π

)3/4√

2

π1/4e−mωx2/2~

(−mω

~x − 2

mω

~x +

m2ω2

~2x3

),

=(mω

~π

)3/4√

2

π1/4x e−mωx2/2~

(−3mω

~+

m2ω2

~2x2

), (B.38)

=

(−3mω

~+

m2ω2

~2x2

)ψ1(x) , (B.39)

we rapidly find

− ~2

2m

d2

dx2ψ1(x) + 1

2mω2 x2 ψ1(x)

= − ~2

2m

(−3mω

~+

m2ω2

~2x2

)ψ1(x) + 1

2mω2 x2 ψ1(x) , (B.40)

=

(3~ω2

− mω2 x2

2+

mω2 x2

2

)ψ1(x) , (B.41)

= 32~ωψ1(x) . (B.42)

Comparing this with

− ~2

2m

d2

dx2ψ1(x) + 1

2mω2 x2 ψ1(x) = E1ψ1(x) (B.43)

givesE1 = 3

2~ω . (B.44)

The turning points are thus located at

12mω2 x2 = 3

2~ω ⇒ x± = ±

√3~mω

. (B.45)


The probability of find the particle in the classically forbidden region isthus ∫ x−

−∞ψ∗1(x) ψ1(x) dx +

∫ ∞

x+

ψ∗1(x) ψ1(x) dx

=

∫ x−

−∞

(mω

~

)3/2 2√π

x2 e−mωx2/~ dx

+

∫ ∞

x+

(mω

~

)3/2 2√π

x2 e−mωx2/~ dx (B.46)

= 2×∫ ∞√

3~mω

(mω

~

)3/2 2√π

x2 e−mωx2/~ dx . (B.47)

Alternatively, the probability of finding the particle in the forbidden regionis just 1−(Probability of finding the particle in the classical region), i.e

1− 2×∫ √

3~mω

0

(mω

~

)3/2 2√π

x2 e−mωx2/~ dx . (B.48)

Using

ξ =

√mω

~x , dξ =

√mω

~dx , (B.49)

brings Eq.(B.48) and Eq.(B.47) to the form

1− 4√π

∫ √3

0

dξ ξ2 e−ξ2

, and4√π

∫ ∞

√3

dξ ξ2 e−ξ2

. (B.50)

Let us work on

∫ √3

0

dξ ξ2 e−ξ2

. Choose n = 10 segments so that m =

5, (b− a) =√

3, h =√

3/10 and g(ξ) = ξ2 e−ξ2. We have the following:

n ξn g(ξn) n ξn g(ξn)

0 0 0

1√

310

0.02911 6 3√

35

0.36676

2√

35

0.10643 7 7√

310

0.33799

3 3√

310

0.20611 8 4√

35

0.28149

4 2√

35

0.29702 9 9√

310

0.21393

5√

32

0.35428 10√

3 0.149361

(B.51)


Using Eq.(B.35) gives

∫ √3

0

dξ ξ2 e−ξ2

≈√

3

30(0 + 0.149361

+2(0.10643 + 0.29702 + 0.36676 + 0.28149)

+4(0.02911 + 0.20611 + 0.35428 + 0.33799 + 0.21393)) , (B.52)

= 0.3937 . (B.53)

Thus, the probability is approximately

1− 4√π× 0.3937 = 0.112 . (B.54)

Finding zeroes 209

C Finding zeroes of an equation

There are many situation where one cannot find an algebraic solution to theproblem at hand. Two examples are the solutions to Eqs.(6.43) and (6.50).Here, we briefly review two methods. The first one is simple bracketing, whilethe second is known as Newton’s method for numerically solving equations.

C.1 Bracketing

Bracketing is easy. Suppose you want to find the solution to the equationx2 = 2. We rewrite this as

y(x) = x2 − 2 (C.1)

and notice that y(1) = −1 but y(2) = 2. Thus (because y(x) is continuous),the correct value lies between x = 1 and x = 2. Split the interval and evaluatefor x = 3/2. Then: y(3/2) = 1/4 so the correct value lies between x = 1 andx = 3/2. Split again and evaluate for x = 5/4 to get y(5/4) = −7/16, so thecorrect value lies between 5/4 and 3/2. Next, try x = 11/8 to get −7/64 andso forth, until the interval is sufficiently narrow to properly estimate

√2.

C.2 Newton’s method

Near a point x, a function can always be approximated by a straight line

f(x + ∆x) = f(x) + ∆x f ′(x) , (C.2)

where

f ′(x) =df

dx

∣∣∣x. (C.3)

Suppose that we have xn so that f(xn) is approximately (but not exactly) 0.Then, we replace f(x) near xn by a straight line and find the value xn+1 forwhich this straight line crosses 0, i.e. set ∆x = xn+1 − xn so that

0 = f(xn) + (xn+1 − xn)f ′(xn) , (C.4)

which immediately gives

xn+1 = xn − f(xn)

f ′(xn). (C.5)

If this is done right, then f(xn+1) should be closer to 0 than f(xn).

Finding zeroes 210

We can illustrate the procedure, and the geometrical interpretation of theprocedure, by looking for a numerical approximation to

√2, i.e. looking for

a value of x so thatf(x) = x2 − 2 = 0 . (C.6)

We start withdf

dx= 2x , (C.7)

so that we have the iteration formula

xn+1 = xn − x2n − 2

2xn

= 12xn + 1

xn. (C.8)

We use as initial guess x0 = 1. This yields x1 = 3/2. Using now x1 = 3/2,we obtain x2 = 17/12 ≈ 1.41667 and, in sequence, x3 = 577/408 ≈ 1.41422and x4 = 665857/470832 ≈ 1.41421.

0.5 1 1.5 2

-2

-1

1

2

x

x2-2

0.5 1 1.5 2

-3

-2

-1

1

2

x

x2-2

2x-3

Figure 27: Left: The curve x2 − 2 near x0 = 1. Right: the same curve with itsapproximation by the straight line 2x− 3 near x0 = 1. Note that the straight linecrosses the axis at x1 = 3/2; this becomes the new value for the next iteration.

Different values of the initial guess x0 affect the rate of convergence. Forinstance, using x0 = 7/5 (which is justified because x2

0 = 49/25 is obvi-ously very close to 2) produces the sequence x0 = 7/5 = 1.2, x1 = 99/70 ≈1.41429, x2 = 19601/13860 ≈ 1.41421.

Hypergeometric and confluent solutions 211

D Remarks on some second order differential

equations

This section is heavily influenced by J.B. Seaborn, Hypergeometric Functionsand their applications 1991 Springer-Verlag, NY. Regrettably, our librarydoes not have a copy of this excellent text.

D.1 The Gauss hypergeometric 2F1(a, b, c; z)

Any second order differential equations can be written in the form

u′′(z) + P (z)u′(z) + Q(z)u(z) = 0 . (D.1)

The problem consists in finding u as an explicit function of z.When the functions P (z) and Q(z) are rational, it is always possible, usu-

ally after many changes of variables, to write Eqn.(D.1) in the standardizedform

z(1− z)u′′ + [c− (a + b + 1)z]u′ − abu = 0 (D.2)

where a, b and c are constant. Eqn.(D.2) is known as the Gauss or thehypergeometric differential equation.

To solve, we expand u(z) in series:

u(z) =∞∑

n=0

an zn+s , (D.3)

where an are expansion coefficients (unrelated to the a of Eqn.(D.2)). Thelowest power of z in the expansion is taken to be s: there is no a priori reasonto suppose s = 0.

Inserting Eqn.(D.3) into Eqn.(D.2) we obtain

0 = z(1− z)∞∑

n=0

an (n + s)(n + s− 1) zn+s−2

+[c− (a + b + 1)z]∞∑

n=0

an (n + s) zn+s−1 − a b

∞∑n=0

an zn+s . (D.4)


Collecting like powers of z yields

0 =∞∑

n=0

an (n + s)(n + s + c− 1) zn+s−1

−∞∑

n=0

an [(n + s)(n + s + a + b) + ab] zn+s . (D.5)

One can isolate the term zs−1 and relabel the first sum so as to obtain

0 = a0 s (s + c− 1) zs−1

+∞∑

n=0

[(n + s + 1)(n + s + c) an+1

− [(n + s + a + b)(n + s) + ab] an] zn+s . (D.6)

Since this must hold for every z, every coefficient of individual powers of zmust vanish independently. This produces

s (s + c− 1) a0 = 0 , (D.7)

an+1 =(n + s)(n + s + a + b) + a b

(n + s + 1)(n + s + c)an , (D.8)

which is a recursion relation between consecutive coefficient in the series.The series is obviously interesting only if a0 6= 0, which implies s = 1− c

or s = 0. If we set s = 0, we obtain

an+1 =(n + a)(n + b)

(n + 1)(n + c)an , (D.9)

so that

a1 =a b

ca0 ,

a2 =a (a + 1) b (b + 1)

1× 2 c(c + 1)a0 , (D.10)

a2 =a (a + 1) (a + 2) b (b + 1) (b + 2)

1× 2× 3 c(c + 1)(c + 2)a0 ,

an =a (a + 1) . . . (a + n− 1) b (b + 1) . . . (b + n− 1)

n! c (b + 1) . . . (c + n− 1)a0 . (D.11)


Let

(a)n = a(a + 1)(a + 2)...(a + n− 1) =(a + n− 1)!

(a− 1)!, a0 = 1 (D.12)

denote the rising factorial. The solution with s = 0 can then be written as

u(z) =∞∑

n=0

(a)n(b)n

(c)n

zn

n!, (D.13)

≡ 2F1(a, b, c; z) (D.14)

The form 2F1 indicates that the numerator contains two rising factorials andthe denominator only one. If a or b is a negative integer, then 2F1 is apolynomial.

One can show that, if s = 1− c, then

u(z) = z1−c2F1(a + 1− c, b + 1− c, 2− c; z), (c 6= 2, 3, 4 . . .) (D.15)

is also solution.Many functions are expressible as hypergeometric functions. For instance,

1

xln(1 + x) = 2F1(1, 1, 2;−x)

= 1 +(1)(1)

2(−x) +

(1)(2)(1)(2)

(2)(3)

(−x)2

2!

+(2)(3)(2)(3)

(2)(3)(4)

(−x3)

3!. . .

= 1− x

2+

x2

3− x3

4. . . (D.16)

∫ π/2

0

1√1− k2 sin2 θ

dθ =π

22F1(

1

2,1

2, 1; k2) (D.17)

The hypergeometric function 2F1 is convenient to denote solutions of dif-ferential equations. For instance, the solution to Legendre equation

(1− x2)u′′ − 2xu′ + n(n + 1)u = 0 (D.18)

can be transformed into the standard form of Eqn.(D.2) using z = (1−x)/2,or x = 1− 2z :

(1− x2) = (1− x)(1 + x) = 2z(2− 2z) = 4z(1− z) (D.19)

d

dx=

d

dz

dz

dx= −1

2

d

dz(D.20)


so that Eqn.(D.18), written in terms of z, becomes

4z(1− z)1

4

d2

dz2u(z)− 2(1− 2z)(−1

2)

d

dzu(z) + n(n + 1)u(z) = 0 ,

z(1− z)d2

dz2u + (1− 2z)

d

dzu + n(n + 1)u = 0 .(D.21)

The solution is then instantaneous. Comparing with the standard form ofEqn.(D.2), we have immediately a = −n, b = (n + 1) and

c− (a + b + 1)z = c− (−n + n + 1 + 1)z = c− 2z = 1− 2z ⇒ c = 1 .

Thus, the solution to Eqn.(D.21) is simply

z(1− z)d2

dz2u + (1− 2z)

d

dzu + n(n + 1)u = 0 ⇒ u(z) = 2F1(−n, n + 1, 1; z) .

(D.22)In terms of the original variable x:

u(x) = 2F1(−n, n + 1, 1;1− x

2). (D.23)

D.2 The confluent hypergeometric function 1F1(a, c; z)

Another family of differential equations is of the type

zu′′(z) + (c− z)u′(z)− au(z) = 0 (D.24)

with solution u(z) = 1F1(a, c; z) where

1F1(a, c; z) = 1 +a

c

z

1!+

a(a + 1)

c(c + 1)

z2

2!+

a(a + 1)(a + 2)

c(c + 1)(c + 2)

z3

3!+ . . . (D.25)

The second solution is u(z) = z1−c1F1(a+1− c, 2− c; z). (Note that one can

show 1F1(a, c; z) = ez1F1(c− a, c;−z))

An alternate form of Eqn.(G.44) is obtained if we make the change z = ρ2.Then

d

dz=

d

dρ

dρ

dz=

1

2√

z

d

dρ=

1

2ρ

d

dρ, (D.26)

d2

dz2=

1

2ρ

d

dρ

(1

2ρ

d

dρ

)=

1

4ρ

(− 1

ρ2

d

dρ+

1

ρ

d2

dρ2

). (D.27)


so that Eqn.(G.44) becomes

ρ2

[1

4ρ

d2

dρ2u(ρ2)− 1

4ρ2

d

dρu(ρ3)

]

+(c− ρ2)1

2ρ

d

dρu(ρ2)− 4au(ρ2) = 0

d2

dρ2u(ρ2) +

(2c− 1

ρ− 2ρ

)d

dρu(ρ2)− 4au(ρ2) = 0 . (D.28)

As an example, consider Hermite’s differential equation:

H ′′n − 2xH ′

n + 2ny = 0 .

It has the solution

H2n(x) = (−1)n (2n)!

n!1F1(−n,

1

2; x2) (D.29)

H2n+1(x) = (−1)n (2n + 1)!

n!2x 1F1(−n,

3

2; x2) (D.30)

Another useful example is Bessel’s differential equation:

x2J ′′ν (x) + xJ ′ν(x) + (x2 − ν2)Jν = 0 , (D.31)

with solution

Jν(x) =e−ix

ν!

(x

2

)ν

1F1(ν +1

2, 2ν + 1; 2ix) . (D.32)

Two other examples are worth mentioning. The solutions to Laguerre’sdifferential equation

xy′′(x) + (1− x)y′(x) + ny(x) = 0 , (D.33)

can be expressed asLn(x) = 1F1(−n, 1; x) . (D.34)

Closely related are associated Laguerre functions

Lmn (x) =

(n + m)!

n!m!1F1(−n, m + 1; x) (D.35)

which are solutions to

xy′′(x) + (m + 1− x)y(x) + ny(x) = 0 . (D.36)

Hermite polynomials 216

E Hermite polynomials

E.1 Series solution

We need to solve

H ′′(ξ)− 2ξH ′(ξ) + 2nH(ξ) = 0 (E.1)

Write H(ξ) in the form of a power series

H(ξ) = a0 + a1ξ + a2ξ2 + a3ξ

3 + . . . =∞∑

r=0

arξr (E.2)

and place Eqn.(E.2) into Eqn.(E.1) to get:

H ′ = a1 + 2a2ξ + 3a3ξ2 + 4a4ξ

3 + . . . (E.3)

2ξ H ′ = 2(a1ξ + 2a2ξ

2 + 3a3ξ3 + . . .

)= 2

∞∑r=0

r arξr , (E.4)

H ′′(ξ) = 2a2 + 2× 3a3ξ + 3× 4a4ξ2 + 4× 5ξ3 + . . . , (E.5)

=∞∑

r=0

r(r − 1) arξr−2 , (E.6)

The first non–zero term of this last sum is for r = 2, so we can shift thesummation index and rewrite it as

H ′′(ξ)) =∞∑

r=0

(r + 1)(r + 2)ar+2ξr . (E.7)

Collecting terms, Eq.(E.1) now becomes

0 =∞∑

r=0

(r + 2)(r + 1)ar+2ξr − 2

∞∑r=0

r arξr + 2n

∞∑r=0

arξr , (E.8)

=∞∑

r=0

[(r + 1)(r + 2)ar+2 + 2(n− r)ar] ξr . (E.9)

Since this must hold for every value of ξ , the coefficient of each power ofξ must be 0 : i.e.

(r + 1)(r + 2)ar+2 + 2ar(n− r) = 0 (E.10)

ar+2 =2ar(r − n)

(r + 1)(r + 2)(E.11)


Let us now look at the behavior of the recursion relation of Eq.(E.11).For very large r, we have

ar+2 ∼ 2ar r

r2=

2ar

r. (E.12)

At this point we recall that

e2ξ2

= 1 + (2ξ2) + 12(2ξ2)2 + 1

3!(2ξ2)3 + . . . , (E.13)

=∞∑

n=0

2n

n!ξ2n =

∞∑n=0

c2n ξ2n , (E.14)

where

c2n =2n

n!, (E.15)

and note that we have

c2n+2 =2n+1

(n + 1)!

n!

2nc2n =

2

nc2n . (E.16)

Comparing Eqs.(E.16) and (E.12), we see that the series solution of Eq.(E.2)will eventually grow like

y(ξ) ∼ e2ξ2

, (E.17)

unless the series is truncated.To produce a polynomial and thus avoid the behavior of Eq.(E.17), we

must select some integer r = n so that all values an+2, an+4, . . . obtained viaEq.(E.11) are 0.

Thus, choosing for instance n = 0 yields the sequence

r = 0 : a2 =2a0(0− 0)

1× 2= 0 , (E.18)

r = 2 : a4 =2a2(2− 0)

3× 4=

2(2− 0)

3× 4× 0 , (E.19)

since a2 = 0. It is clear that all subsequent choices of even values of r willyield ar = 0. We do not have any conditions on the odd values of r becauseof the form of the recursion relation. However, one trivial solution is to seta1 = 0 so that all ar’s are 0. Thus, for this choice of n and with a1 = 0, a0 isthe only non–zero coefficient. Thus, our first polynomial solution to Eq.(E.1)is

H0(ξ) = a0 . (E.20)


Next, choose n = 1. Here, we see that we will not have any condition ofthe even r’s, so we will choose the trivial solution a0 = 0 so that a2n = 0.Thus, we have

r = 1 : a3 =2a1(1− 1)

2× 3= 0 , (E.21)

r = 3 : a5 =2a3(3− 1)

4× 5= 0 (E.22)

since a3 = 0. Hence, the series stops at n = 1 and yields

H1(ξ) = a1ξ . (E.23)

Choosing now n = 2 yields the sequence

r = 0 : a2 =2a0(0− 2)

1× 2= −2a0 , (E.24)

r = 2 : a4 =2a2(2− 2)

3× 4= 0 , (E.25)

r = 4 : a6 =2a4(4− 2)

5× 6= 0 , (E.26)

since now a4 = 0. For this choice of n and with a1 = 0, this polynomial is

H2(ξ) = a0 − 2a0ξ2 = a0(1− 2ξ2) . (E.27)

The process is easily summarized:

r = n ai Hn(ξ)

0 a2 = 2a0(0−0)1·2 = 0

a4 = 2a2·(2−0)3·4 = 0 . . . H0(ξ) = a0

1 a3 = 2a1(1−1)2·3 = 0

a5 = 2a3(3−1)4·5 = 0 . . . H1(ξ) = a1ξ

2 a2 = 2·a0(0−2)1·2 = −2a0

a4 = 2a0(2−2)3·4 = 0 . . . H2(ξ) = a0(1− 2ξ2)

3 a3 = 2a1(1−3)2·3 = −2

3a1

a5 = 2a1(3−3)4·5 = 0 . . . H3(ξ) = a1

(ξ − 2

3ξ3

)

4 a2 = 2a0(0−4)1·2 = −4a0

a4 = 2a2(2−4)3·4 = −1

3a2 = 4

3a0 H4(ξ) = a0

(1− 4ξ2 + 4

3ξ4

)


The values of a0, a1 for each polynomial Hn(ξ) is determined historicallyby choosing the coefficient of ξn in Hn(ξ) to be +2n. This in turn leads to

∫ ∞

−∞dξe−ξ2

H2n(ξ) = 2nn!

√π (E.28)

E.2 Generating function

Alternatively, the set of all Hermite polynomials can be obtained from thegenerating function

Φ(x, h) = e2xh−h2

=∞∑

n=0

Hn(x)hn

n!. (E.29)

Thus, expanding the generating function in powers of h:

Φ(x, h) = 1 + (2xh− h2) +1

2(2xh− h2)2 +

1

3!(2xh− h2)3 + ...

= 1 + 2xh− h2 +1

2(4x2h2 − 4xh3 + h4)

+1

3!(8x3h3 − 3 · 4x2h2h2 + 3 · 2xhh4−h) + ...

= 1 + h2x + h2(2x− 1) +h3

3!(8x3 + 12x) + ...

= H0(x) + hH1(x) +1

2h2H2(x) +

1

3!h3H3(x)... (E.30)

and so, for instance,

H0(x) = 1 , H1(x) = 2x , H2(x) = 4x2− 2 , H3(x) = 8x3− 12x (E.31)

E.3 Rodrigues formula and orthogonality

The Hermite polynomials can also be obtained by the so–called Rodriguesformula:

Hn(ξ) = (−1)ne+ξ2 dn

dξne−ξ2

. (E.32)

Thus, we have

n = 0 : H0(ξ) = 1 , (E.33)

n = 1 : H1(ξ) = (−1)eξ2

(d

dξ

)e−ξ2

= −eξ2

(−2ξ)e−ξ2

= 2ξ , (E.34)


etc.Consider now, for any function y(ξ),

eξ2 d

dξ

(e−ξ2

y′(ξ))

= eξ2(−2ξe−ξ2

y′ + e−ξ2

y′′)

, (E.35)

= y′′ − 2ξ y′ , (E.36)

where y′(ξ) ≡ dy(ξ)/dξ. Substituting Hn(ξ) for y(ξ), we can write Eq.(E.1)in the form

eξ2 d

dξ

(e−ξ2

H ′n(ξ)

)+ 2nHn(ξ) = 0 , (E.37)

eξ2 d

dξ

(e−ξ2

H ′m(ξ)

)+ 2m Hm(ξ) = 0 . (E.38)

Multiplying the first by Hm(ξ) and the second by Hn(ξ) yields

Hm(ξ) eξ2 d

dξ

(e−ξ2

H ′n(ξ)

)+ 2nHm(ξ) Hn(ξ) = 0 , (E.39)

Hn(ξ) eξ2 d

dξ

(e−ξ2

H ′m(ξ)

)+ 2m Hn(ξ) Hm(ξ) = 0 . (E.40)

Subtracting produces

eξ2

(Hm(ξ)

d

dξ

(e−ξ2

H ′n(ξ)

)−Hn(ξ)

d

dξ

(e−ξ2

H ′m(ξ)

))

+2(n−m)Hn(ξ) Hm(ξ) = 0 , (E.41)

which can be rearranged, after premultipliation by e−ξ2, into

d

dξ

(e−ξ2

(Hm(ξ) H ′n(ξ)−H ′

m(ξ)Hn(ξ)))

+2(n−m)Hn(ξ) Hm(ξ) e−ξ2

= 0 . (E.42)

Integrating from −∞ to +∞ yields∫ ∞

−∞dx

d

dξ

(e−ξ2

(Hm(ξ) H ′n(ξ)−H ′

m(ξ)Hn(ξ)))

+2(n−m)

∫ ∞

−∞dxHn(ξ) Hm(ξ) e−ξ2

= 0 , (E.43)

e−ξ2

(Hm(ξ) H ′n(ξ)−H ′

m(ξ)Hn(ξ))∣∣∣∞

−∞

+2(n−m)

∫ ∞


= 0 . (E.44)


Using

e−ξ2

(Hm(ξ) H ′n(ξ)−H ′

m(ξ)Hn(ξ))∣∣∣∞

−∞= 0 , (E.45)

we finally obtain

2(n−m)

∫ ∞


= 0 . (E.46)

Eq.(E.46) shows that two different Hermite polynomials are orthogonalw/r to the weight function e−ξ2

. In particular, it follows from Eq.(6.73)that any two solutions of the time–independent Schrodinger equation for theharmonic oscillator are orthogonal.

Eigenvectors 222

F Eigenvectors: On the importance of being

proper

In quantum mechanics, things live in a Hilbert space. This is an abstractspace, populated by vectors which, in examples that we’ve seen so far, arefunctions solutions of the Schrodinger equation.

From the Schrodinger equation, we have not only concluded that functionspace is relevant, but also that a rather important object is a linear operatorH, which, in one dimension, reads

H = − ~2

2m

∂2

∂x2+ V (x). (F.1)

There are more general linear operators: we have studied their abstract prop-erties in the previous two sections, focusing in particular on hermitian oper-ators. We want to capitalize on these properties.

Up to now, our strategy to solve the (time–dependent) Schrodinger equa-tion has been to attack the equation Schrodinger, i.e. solve (using kets tomake this as general as possible)

i~∂

∂t|Ψ(t)〉 = H |Ψ(t)〉 (F.2)

by solving the differential equation.A special subset of solutions are particularly useful: the separable solu-

tions: if |ψE〉 satisfiesH |ψE〉 = E |ψE〉 , (F.3)

then|ΨE(t)〉 = e−iEt/~ |ψE〉 , (F.4)

In some specific cases, it is possible to solve directly this differential equa-tion (using either the x– or p–representation.) However, in most cases, it isnot possible. Indeed, for more general systems, in more than one dimensionor for more than one particle, solving Eqn.(F.3) directly is practically impos-sible. In conclusion, we need a change in the mechanics of solving Eqn.(F.3)so that this equation can be solved quite generally.

We have seen that, under broad assumptions, p and x are hermitianoperators, and so is H. Thus, instead of solving Eqn.(F.3) as a differential

Eigenvectors 223

equation, we will try to solve it as a matrix equation. The difference betweenthe two approaches is this: in the differential approach, we try to solve for|ψE〉 directly in a single step. In the matrix approach, we start by choosingsome basis states to obtain the matrix realization of H. If we are lucky inthe choice of our basis states, then H will be diagonal; the basis states arethen the solutions |ψE〉 themselves.

In general, the choice of basis states is based on experience and educatedguesses; H will not be a diagonal matrix. The solutions to Eqn.(F.3) arethen expressed as a linear combination of the basis states, and these specificlinear combination will have an ”easy” time–evolution. The matrix approachis more versatile because we have complete freedom in choosing the basisstates. Of course, a ”bad” choice of basis means that we will have to do a lotof work to solve Eqn.(F.3), but we can still obtain a solution. This is betterthat directly solving the differential equation for |ψE〉 which, as mentionedbefore, sometimes can’t be done.

It is important to realize that, in both cases, we are aiming to find |ψE〉 ,because, if we know |ψE〉 , we instantly know the time evolution of |ΨE(t)〉through Eqn.(F.4). Besides their mathematical practicality in computing thetime–evolution of the state, |ψE〉 is also physically important because that

the fluctuation in energy for this state, ∆E =

√〈ψE| H2 − 〈H〉2 |ψE〉 = 0,

i.e. |ψE〉 has a well–defined energy. This is true quite generally, if |ψ〉 satisfiesT |ψ〉 = λ |ψ〉 , then ∆T = 0, so states with the property T |ψ〉 = λ |ψ〉 areobviously of great physical importance.

To solve T |ψ〉 = λ |ψ〉 in general, and to solve H |ψE〉 = E |ψE〉 in par-ticular, one must solve the associated eigenvalue problem. This is done byan algorithm presented in the next section.

F.1 General eigenvalue problem

Suppose that you are given some n×n matrix M . Then, its eigenvalues are,by definition, the roots of the characteristic equation

det (M − λ 1ln) = 0 , (F.5)

where 1ln is the n × n unit matrix. This is a definition. There’s nothing tounderstand about it; it’s just something that one needs to remember.

Eigenvectors 224

Example F.1 The eigenvalues of the matrix

K =

( −2 11 −2

)(F.6)

are obtained by solving the characteristic equation

det

(( −2 11 −2

)− λ

(1 00 1

))= det

( −2− λ 11 −2− λ

)= λ2+4λ+3 = 0 .

The solutions are λ1 = −3 and λ2 = −1. These are the eigenvalues of thematrix.

Example F.2 The eigenvalues of the matrix

1 −27 −151 −2 −3−2 −3 2

are obtained by solving the characteristic equation

det

1 −27 −151 −2 −3−2 −3 2

− λ

1 0 00 1 00 0 1

= det

1− λ −27 −151 −2− λ −3−2 −3 2− λ

= −λ3 + λ2 + 16λ− 16 = 0 .

It is possible, in principle, to obtain the roots algebraically. A moreexpedient but approximate method is to plot the polynomial and locate thezeroes graphically. If more accuracy is required, then one can simply use theintersection method, using our approximate location as initial guesses. (Thedrawback is that this makes it difficult to locate multiple occurences of a givenroot.) In any event, the solutions for the case at hand are λ1 = 4, λ2 = 1 andλ3 = −4.

The eigenvector |ψ〉 associated with an eigenvalue λ is just this vector |ψ〉that satisfies the equation

M |ψ〉 = λ|ψ〉 (F.7)

To find |ψ〉, one must know λ. Plugging λ into Eq.(F.7), one obtains alinear system of n equations with n unknowns. In this system, only n − 1

Eigenvectors 225

equations will be linearly independent, which means that it is possible to findn − 1 components of |ψ〉 in terms of the last one. To completely determinethe problem, we will always require that 〈ψ|ψ〉 = 1. The unknown phase ofthe last entry will always be taken as 1.

Example F.3 The eigenvectors of the matrix

( −2 11 −2

)(F.8)

are found as follows. Choose the first eigenvalue λ1 = −1 and let |ξ〉 denotethe corresponding eigenvector. Assume a general form for |ξ〉:

|ξ〉 =

(ξ1

ξ2

)(F.9)

and consider ( −2 11 −2

)(ξ1

ξ2

)= −1

(ξ1

ξ2

), (F.10)

from which we obtain the system of equations

−2ξ1 + ξ2 = −ξ1 ⇒ −ξ1 + ξ2 = 0 , (F.11)

ξ1 − 2ξ2 = −ξ2 ⇒ ξ1 − ξ2 = 0 . (F.12)

Obviously, these equations are not linearly independent. There is a singlelinearly independent equation for two variables, so we choose to solve ξ2 interms of ξ1: the immediately gives ξ2 = ξ1 and thence

|ξ〉 = ξ1

(11

)(F.13)

After normalization: 〈ξ | ξ〉 = 1, we finally obtain the normalized eigenvector

|ξ〉 =1√2

(11

). (F.14)

The (normalized) eigenvector η, corresponding to the eigenvalue λ = −3, iseasily shown to be, following the same procedure,

|η〉 =1√2

(1−1

). (F.15)

Eigenvectors 226

Example F.4 The eigenvectors of the matrix

1 −27 −151 −2 −3−2 −3 2

.

are obtained as follows. The first eigenvector, (let’s call it |ψ1〉) is associatedwith the eigenvalue λ1 = 4 and is solution to

1 −27 −151 −2 −3−2 −3 2

·

ψ11

ψ12

ψ13

= 4

ψ11

ψ12

ψ13

. (F.16)

This gives the linear system

−3ψ11 −27ψ12 −15ψ13 = 0ψ11 −6ψ12 −3ψ13 = 0

−2ψ11 −3ψ12 −2ψ13 = 0(F.17)

with solution ψ11 = −ψ13/5 , ψ12 = −8ψ13/15. From the condition 〈ψ1|ψ1〉 =1, we obtain

ψ∗13ψ13

(1

25+

64

225+ 1

)= ψ∗13ψ13

(298

225

)= 1 ⇒ ψ13 =

15√298

(F.18)

from which it follows that the normalized |ψ1〉 is simply

|ψ1〉 =15√298

−1/5−8/15

1

. (F.19)

For the second eigenvector, |ψ2〉, associated to λ2 = 1, we have the system

−27ψ22 −15ψ23 = 0ψ21 −3ψ22 −3ψ23 = 0

−2ψ21 −3ψ22 +ψ23 = 0, (F.20)

with solution ψ21 = 4ψ23/3 , ψ22 = −5ψ23/9. From the normalization condi-tion 〈ψ2|ψ2〉 = 1, we obtain

16ψ23ψ∗23

9+

25ψ23ψ∗23

81+ ψ23ψ

∗23 , (F.21)

Eigenvectors 227

from which it follows that250

81ψ23ψ

∗23 = 1, or ψ23 =

9

5√

10. Thus,

|ψ2〉 =9

5√

10

4/3−5/9

1

. (F.22)

Finally, the last eigenvector, |ψ3〉 is associated with the eigenvalue λ3 =−4. |ψ3〉 is given by

|ψ3〉 =1√10

301

. (F.23)

The universal recipe for finding the eigenvalues and eigenvectors of amatrix M is thus:

1. Find the solution to the characteristic equation Det(M−λ 1l) = 0. Thiswill give you all the eigenvalues.

2. For each eigenvalue, find the corresponding eigenvector by solving M |ψi〉 =λi|ψi〉.

3. Do not forget to normalize your eigenvectors such that 〈ψi|ψi〉 = 1.

F.2 Diagonalizable matrices

An n× n matrix M is called diagonalizable if and only if there is a set of nlinearly independent vectors, each of which is non–null and is an eigenvectorof M . If M is diagonalizable, then it can be written in the form

M = U ·D · U−1 , (F.24)

where D is a diagonal matrix containing as entries the eigenvalues of M .Thus, if λ1, . . . , λn are the eigenvalues of M , then, if M is diagonalizable,we can write

M = U ·

λ1

. . .

λn

· U−1 (F.25)

Eigenvectors 228

Furthermore, the first column of the matrix U−1 is the eigenvector associatedwith λ1, the second column is the eigenvector associated with λ2, etc. Pleasenote that, technically, a multiple of the i’th eigenvector can appear in columni, i.e. it is not strictly necessary for the eigenvectors to be normalized toobtain U−1. Once U−1 is found, then its inverse U can be obtained in theusual way.

Example F.5 The matrix K of Eqn.(F.6) is diagonalizable. The matrixU−1 is obtained from the eigenvectors of K, and it is easy to find that

U−1 =1√2

(1 11 −1

)⇒ U =

1√2

(1 11 −1

). (F.26)

(The matrices U and U−1 turn out to be the same in this case; this is nottrue in general.) It is then easy to verify that

( −2 11 −2

)=

1√2

(1 11 −1

)·( −1 0

0 −3

)· 1√

2

(1 11 −1

). (F.27)

Example F.6 The matrix σ+ =

(0 10 0

)is not diagonalizable. Both of its

eigenvalues are 0. The first (unnormalized) eigenvector is (1, 0)t, while thesecond is the null vector (0, 0)t

F.3 Tricks of the trade

F.3.1 Powers of a matrix

What can one do with diagonalizable matrices. Well, first observe that, ifM = U ·D · U−1, then

M2 = U ·D · U−1 · U ·D · U−1 = U ·D2 · U−1

Mp+1 = Mp ·M ,= U ·D · U−1 · U ·D · U−1

= U ·Dp+1 · U−1 ,

(F.28)

It is easy to compute powers of D since it is just a diagonal matrix:

Dp+1 =

λp+11

. . .

λp+1n

. (F.29)

Eigenvectors 229

This is a very numerically efficient way of computing powers of M sinceone must calculate the matrices D, U and U−1 only once. It is not only fasterbut also more accurate than multiplying the matrix M many times by itself.

Suppose now that M is a diagonalizable matrix such that, for some integerp > 0, we have Mp = 1l. Since we can write

M = U ·D · U−1 , (F.30)

with D the diagonal matrix containing the eigenvalues, we find

Mp = 1l = U ·Dp · U−1 . (F.31)

Multiplying this equality from the left by U−1 and from the right by U , wehave

U · 1l · U−1 = 1l = Dp =

λp1

. . .

λpn

(F.32)

Thus, λpi = 1 ∀i, i.e. the p’th power of every eigenvalue is 1. In other words,

all the eigenvalues are to be found in the set of p’th root of unity.

Example F.7 Consider the matrix

Q =

0 i 0−i 0 00 0 −1

. (F.33)

It is easy to verify the Q2 = 1l, and that the eigenvalues of Q are 1,−1,−1.

F.3.2 The determinant and trace conditions

We can also make the following statements, valid for any n× n matrix.The trace of a matrix is the sum of its diagonal elements. Thus, the trace

of a matrix S is given by tr(S) = s11 + s22 + . . . + snn. One can show thatthe trace of S is equal to the sum of the eigenvalues. For instance, the traceof the matrix Q of the previous example is tr(Q) = 0 + 0− 1 = −1 is indeedequal to the sum of the three eigenvalues.

The determinant of a matrix S is computed in the usual way, and is equalto the product of the eigenvalues: det(S) = λ1 · λ2 · . . . · λn. It is also easyto verify this property with the matrix Q of previous example or, for thatmatter, for all the matrices of all the examples presented thus far.

Problems 230

G Collected Problems 2001-2006

1. Let z =√

6(1 + i)−√2(1− i).

a) What is the real part of z?

b) What is the imaginary part of z?

c) What is z∗?

d) Express z in polar form.

e) Compute z3 using the Cartesian and the polar representations of z.

f) Find all the complex numbers solutions to z1/5.

g) Find all the complex numbers solutions to z3/5 by

- using z3 and finding these numbers solutions to (z3)1/5,

- using the solutions to z1/5 and raising them to the third power.

- Are both calculations identical?

2. Let

M =18

3 9√

3 −14√

39√

3 −15 −14−14

√3 −14 12

(G.1)

a) What is the determinant of M?

b) What is the characteristic polynomial of this matrix?

c) Find the eigenvalues of M

- by making a plot of the characteristic polynomial,

- identifying one or all of the roots of the characteristic equation. If youidentify one root, you may factor it out of the cubic equation and solvethe resulting quadratic.

d) Verify the trace condition on the eigenvalues,

e) Verify the determinant condition on the eigenvalues.

Problems 231

3. (15 points) Using our definition of the fluctuation of a physical quantity, showthat the fluctuation ∆O of the physical quantity O associated with the operatorO can be written as

∆O =√〈(O − 〈O〉)2〉 . (G.2)

4. (10 points) Consider the wave function

ψ(x, t) =[Aeipx/~ + Be−ipx/~

]e−ip2t/2m~ . (G.3)

Find the probability current corresponding to this wave function.

5. (25 points) (Easy) Let ψm(ϕ) = 〈ϕ |ψm〉 = eimϕ. Let

O = 12eiϕ

(1 + i

d

dϕ

)+ 1

2e−iϕ

(1− i

d

dϕ

), (G.4)

and define the inner product of two functions φ(ϕ) and ξ(ϕ) by

〈φ | ξ〉 =12π

∫ 2π

0dϕ φ(ϕ)∗ ξ(ϕ) . (G.5)

1. Given the above definition of the inner product, show that the states |ψ−1〉, |ψ0〉and |ψ1〉 are normalized and pairwise orthogonal,

2. Construct a 3×3 matrix O with matrix elements Oij given by 〈ψi|O|ψj〉 fori, j = −1, 0, 1. Is the matrix O hermitian?

3. What are the eigenvalues of O? Is it true that the sum of eigenvalues ofO equals the trace of O? Is it true that the determinant of O equals theproduct of the eigenvalues of O?

4. What are the eigenvectors of O? Are they orthogonal?

6. (25 points) (Moderately difficult) Getting familiar with commutatorsThe commutator of two matrices A and B is defined by [A,B ] ≡ AB − BA.

The commutator of two operators A and B is defined as follows. Let |ψ〉 be anyvector. Then

[ A, B ]|ψ〉 ≡(AB − BA

)|ψ〉 . (G.6)

i) Show that, for three operators A, B and C, [ A, B C ] = [ A, B ]C + B[ A, C ] .(This is know as the “derivative” property.)

Problems 232

ii) What is [ x, p ]?

iii) Show that, if |ψ〉 is a solution to Schrodinger’s equation, thend

dtO =

− i

~[ H, O ]. A consequence of this is that, if an operator O commutes with

the Hamiltonian, the corresponding physical quantity is conserved in time(i.e. it is a constant of motion).

iv) Show that, if X and O are operators, then

eXOe−X = O+ [ X, O ] +12!

[ X, [ X, O ] ] +13!

[ X, [ X, [ X, O ] ] ] + . . . (G.7)

(Hint: use the definition eX = 1l + X +12!

(X)2 + . . .)

7. (30 points) (Easy) Application of the square well.Let us suppose that the potential felt by a neutron in the nucleus can be

approximated an infinite square well of the type

V (x) =

0 if 0 ≤ x < L ,∞ if x ≥ L .

(G.8)

with L = 10−12 meters.

1. What is the minimal kinetic energy of a neutron in this well?

2. Assuming that the potential of Eqn(G.8) is also applicable to an electron ina nucleus:

i) What is the minimal kinetic energy of an electron in this well?

ii) What is the potential energy of an electron located at a distance L =10−12m (i.e. just outside the well) if the nucleus has an electric chargeof 50e?

iii) Is the potential energy calculated in ii) enough to bind an electronhaving the kinetic energy calculated in i)?

3. Verify that Heisenberg’s uncertainly relation ∆x∆p ≥ 12~ holds for the

stationary states of this well.

8. (30 points) (Fairly difficult) Bound states of Schrodinger’s equation

Problems 233

Let us define a potential V (x) by

V (x) =

+∞ if x < 0 ,0 if 0 ≤ x < L ,V0 if x ≥ L .

(G.9)

Find the bound states of the Schrodinger equation using “brute force” (somethingwhich, admittedly, one should only do as a last result):

1. The wave function that is solution to Schrodinger’s equation for this poten-tial must have amplitude 0∀ x < 0. Why?

2. The solution must have the form A sin(kx + ϕ) in the region 0 ≤ x < L.Why? What must be the value of ϕ if the wave function is to have noamplitude at x = 0?

3. The wave function must have the form Ce−κ(x−L) for x ≥ L. Why?

4. Show that, at x = L, the coefficients k and κ are related by

γ√

z = nπ − arctan√

z

1− z, (G.10)

where n is an integer (positive or negative), z = E/V and γ =√

2mV0L/~.

5.) How many bound states will this potential accomodate if γ = 9? What arethe energies of these bound states?

6.) What must be (approximately) the value of V0 for this potential to bind aneutron having a lowest energy of 8 MeV if L = 10−12m? (Note: 1 Mev=1.602× 10−13 Joule)

9. (15 points) Consider the following potential

V (x) =

∞ if x < −L/2 ,Uδ(x) if − L/2 < x < L/2 ,∞ if x > L/2 .

(G.11)

This potential is illustrated in figure 29Find the bound states for this potential.

1) For x < 0, write Ψ(x) = A cos(kx) + B sin(kx), while, for x > 0, writeΨ(x) = C cos(kx) + D sin(kx).

Problems 234

x=0 x=Lx=−L /2 /2

V

Figure 28: The potential of Eqn.(G.21).

2) Show that the boundary conditions imply:

A = C , B = −D =mUA

~2. (G.12)

3) Show that the energies are given by the solutions to

tankL

2=

mU

~2. (G.13)

10. (20 points) (Looks harder than it is). Double δ well.Consider a (very crude) model of a singly ionized hydrogen molecule H+

2 ,where the sole electron sees the sum of the attractive Coulomb potentials of thetwo separated protons. Our model for the potential is a sum of two δ functions:

V (x) = −U δ(x + 12L)− U δ(x− 1

2L) , (G.14)

such that meU2/2~2 = 13.6eV, i.e. U = 0.53 × 27.2 A eV so that the binding

energy of this electron is the same as if there was a single (more complicated)potential well rather than two discrete wells.

Show that this potential accomodates two bound states (one even and one odd)with energies given by

e−κL = ±(1− 2κ

α) , (G.15)

with E = −~2κ2/2m and α = 2meU/~2, and find these energies for the electronwith the values of U given above and L = ~2/meU .

11. (20 points) (Straightforward but somewhat tedious) Separation of variables

Problems 235

1. Consider the Hamiltonian for a one-dimensional system of two particles ofmasses m1 and m2, each subjected to their own potential V1(x1) and V2(x2)without interacting with one another:

H = H1 + H2 ,

=p21

2m1+ V1(x1) +

p22

2m2+ V2(x2) . (G.16)

Assuming that the wave function ψ(t, x1, x2) for the system can be factor-ized:

ψ(t, x1, x2) = φ(t)Ψ(x1)ξ(x2) , (G.17)

and that ψ(t, x1, x2) satisfies the Schrodinger equation: Show that Ψ(x1)and ξ(x2) respectively satisfy

H1Ψ(x1) = E1Ψ(x1) , H2ξ(x2) = E2ξ(x2) (G.18)

and that φ(t) = e−iEt/~ with E = E1 + E2.

2. Suppose now that the particles are subjected to a potential which only de-pends on their relative position x1 − x2:

H =p21

2m1+

p22

2m2+ V (x1 − x2) . (G.19)

i) Write the Schrodinger equation using the new variables x and X, where

x = x1 − x2 relative distance

X =m1x1 + m2x2

m1 + m2center of mass . (G.20)

ii) Using separation of variables, show that the time dependence is givenby e−iEt/~.

iii) Using separation of variables, find the equation of motions (i.e. theSchrodinger equation) for the relative and center of mass coordinates.

iv) Show that wave function describing the center of mass coordinates isthat of a free particle with mass m1 + m2 with energy Ecm,

v) Show that the wave function describing the relative motion of the parti-

cles is that of a particle of reduced massm1 + m2

m1m2with energy E−Ecm.

Problems 236

12. (Requires a good deal of thoughts) A qualitative picture of what is requiredfor bound states.

Consider the following potential

V (x) =

0 if 0 < |x| < L ,V0 if L < |x| < L + ∆ ,0 if |x| > L + ∆.

(G.21)

This potential is illustrated in figure 29

x=0 x=Lx=−L /2 /2

V

Figure 29: The potential of Eqn.(G.21).

Using your knowledge of the solution for the potential barrier of finite width,show that this potential cannot have stationary states:

1) (5 points) The wave functions Ψ(x) for this potential must satisfy eitherΨ(0) = 0 or else dΨ(0)/dx = 0. Why?

2) Show that there are no solutions to the above two equations, which meansthat this potential cannot accomodate stationary states. Explain why this isso by comparing the potential with the case where ∆ →∞ or with the po-tential of figure 30. Why could this second potential accomodate stationarystates?

This potential is illustrated in figure 29

13. A particle of mass m is trapped in a harmonic potential such that ω =√k/m. At t = 0, the wave function describing this particle is given by ψ(x, 0) =

12s

∑n

Ψn(x) , where the sum n goes from N − s to N + s with N À s À 1.

Problems 237

x=0 x=Lx=−L

V V o o

∆∆

Figure 30: The alternate to the potential of Eqn.(G.21).

1. (5 points) What is the form of the wave function ψ(x, t) at any time t?

2. (15 points) Show that < x > (t) varies (co)sinusoidally in time with ampli-tude

√2~N/mω, and compare the frequency of oscillation with the classical

result.

14. (30 points) Compute the following matrix elements for the one–dimensionalharmonic oscillator: 〈n|x2|m〉 , 〈n|p2|m〉 , and 〈n|x3|m〉 . Hint: there are exactly 10non–zero matrix elements.

15. The simplest molecules of noble gasses, such as Ne, arise from interactionsbetween the two atoms that can be modeled by the Lennard–Jones potential:

V (x1 − x2) = 4Vo

[(σ

x1 − x2

)12

−(

σ

x1 − x2

)6]

, (G.22)

where σ and Vo are constants determined for each noble gas (σ = 2.74−10m,Vo =0.0031eV for Ne).

1. (5 points) Use the relative and center of mass coordinate to separate theproblem into center of mass and relative coordinates and write down theSchrodinger equation for the relative coordinate.

2. (5 points) Find the minimum xo of the potential, and evaluate what is thisminimum value known as the equilibrium point.

Problems 238

3. (10 points) Expand the potential V around the equilibrium point xo to showthat V can be approximately written

V (x) = Vmin + 12k(x− xo)2 + χ(x− xo)3 + O[(x− xo)4] . (G.23)

Estimate numerically k and χ using the values of Vo and σ for Ne. χ shouldbe significantly smaller than k.

4. (5 points) Ignoring the cubic term on the ground that it is small, estimatewhat is the ground state energy of the molecule, assuming that the center–of–mass energy is known.

5. (10 points) Using Ψ0(x−xo) and Ψ1(x−xo) as basis states and the harmonicoscillator matrix elements from the previous problem, construct the 2 × 2matrix for the operator H describing the relative motion of the molecules,using as potential V (x) = Vmin + 1

2k(x− xo)2 + χ(x− xo)3.

6. (10 points) Diagonalize this 2 × 2 matrix and find the (improved) groundstate energy of the molecule (again assuming that the center–of–mass energyis known).

16. (A problem that will require numerical integration).The classical turning points for the motion of a particle in one dimension are

those values x± of x between which a particle is constrained to move by the lawof classical mechanics.

For a particle in a one–dimensional harmonic well,

1. (5 points) Find the classical turning points as a function of the energy En

of the n’th eigenstates of the harmonic oscillator Hamiltonian.

2. (15 points) Compute the probability of finding the particle in the classicallyforbidden region, i.e. in the region where x > x+ and where x < x−, for then = 0, 1 and 2 states.

3. (10 points) From the previous result, and using common sense, guess theprobability of finding the particle in the classically forbidden region if n →∞. Explain briefly your answer.

17. Let us define, in the usual way, the operators a, a†, b and b† so that

[a, a†] = 1l , [a, b] = [a†, b] = [a, b†] = [a†, b†] = 0 , [b, b†] = 1l (G.24)

Let |na, nb〉 be the common eigenstates of a†a and b†b.

Problems 239

1. (10 points) Show that the three operators a†b, ab†, a†a− b†b satisfy the samecommutation relations as the operators L+, L−, Lz.

2. (10 points) Compute the operator ~L · ~L in terms of a, a†, b and b†.

3. (5 points) Show that |na, nb〉 is an eigenstate of ~L · ~L (expressed in terms ofa†, a, b† and b). What is the corresponding eigenvalue?

4. (5 points) What is the correspondence between |na, nb〉 and |lm〉? (i.e.express j and m in terms of na and nb.)

5. (5 points) Using the previous result, show that 〈lm′|L+|lm〉 = 〈na + 1, nb −1|a†b|na, nb〉, and that 〈lm′|L−|lm〉 = 〈na − 1, nb + 1|ab†|na, nb〉.

18. (20 points) Compute the matrix elements 〈l′m′|Lx|lm〉 and 〈l′m′|Ly|lm〉

1. By expressing Lx and Ly in terms of L+ and L−,

2. By expressing Lx and Ly in spherical coordinates and using properties ofY m

l (θ, ϕ).

19. (25 points) Show explicitly that the l = 2 spherical harmonics are given by

Y ±22 (θ, ϕ) =

√1532π

1r2

(x± iy)2 =

√1532π

sin2 θe±2iϕ ,

Y ±12 (θ, ϕ) = ∓

√158π

1r2

z(x± iy) = ∓√

158π

cos θ sin θe±iϕ , (G.25)

Y 02 (θ, ϕ) =

√5

16π

1r2

(2z2 − x2 − y2) =

√5

16π(3 cos2 θ − 1) .

20. (20 points) Let

ψnlm(ρ, θ, ϕ) = Rn(ρ)Y ml (θ, ϕ) = 〈ρ, θ, ϕ |ψnlm〉 ,

with Rnl(ρ) some function of the radial coordinates.

1. Write the matrix element 〈ψn′,l=2,m=0|x2|ψn,l=0,m=0〉 as an integral over theradial coordinate multiplied by an integral over the angular coordinates.

2. Evaluate the radial part of this matrix element (i.e. the integral that con-tains Y m

l (θ, ϕ)’s.)

Problems 240

21. Consider a particle in a central potential. Given that |lm〉 is an eigenstate of~L · ~L and Lz:

1. (20 points) Compute the sum (∆Lx)2 + (∆Ly)2,

2. (5 points) For which values of l and m does the above sum vanish?

22. (30 points) Using the normalized eigenfuctions of the hydrogen atom in then = 1 states,

1. For all the allowed values of l and m, obtain the probability densities Pnl(r) =4πr2|Rnl(r)|2, where Rnl(r) is the radial part of the wave function.

2. Compute 〈r〉nl in all cases, and express your answer in terms of the Bohrradius,

3. Find the maximum of Pnl(r) for each allowed value of l and m. Expressyour answer in terms of the Bohr radius.

23. (15 points) Let Q0 = (2z2− y2− x2). Express this operator in terms of a radialpart multiplied by a spherical harmonics. Compute the following matrix elementsusing the states |nlm〉 of the spherical three–dimensional harmonic oscillator:

1. 〈220|Q0|000〉, 2. 〈420|Q0|220〉, 3. 〈440|Q0|220〉.

24. The Kratzer molecular potential (total: 65 points). This is a nice examproblem: it combines many

To investigate the rotation–vibration spectrum of a diatomic molecule, Kratzerhas used the potential

V (r) = −2D

(a

r− a2

2r2

).

The solution to the Schrodinger equation can be factorized into

Ψ(r, θ, ϕ) =1rχ(r)Y m

l (θ, ϕ) (G.26)

1. (5 points) Using the dimensionless abbreviations

ξ =r

a, β2 = −2

ma2

~2E , γ2 = 2

ma2

~2D , (G.27)

with γ > 0 and real β > 0, show that the radial equation for χ(r) takes theform

d2χ(ξ)dξ2

+[−β2 + 2

γ2

ξ− γ2 + l(l + 1)

ξ2

]χ(ξ) . (G.28)

Problems 241

2. (5 points) Justify the choice of the form

χ(ξ) = ξλ e−βξ f(ξ) . (G.29)

3. (10 points) Show that f(ξ) satisfies

ξf′′(ξ) + (2λ− 2βξ)f

′(ξ) + (−2λβ + 2γ2)f(ξ) = 0 . (G.30)

4. (5 points) Show that a solution to the above differential equation is

f(ξ) = 1F 1(−λ +γ2

β, 2λ; 2βξ) , (G.31)

where λ is solution to λ(λ− 1) = γ2 + l(l + 1) .

5. (5 points) Show that the energies are quantized, such that

En = − ~2

2ma2

γ4

(n + λ)2. (G.32)

6. (5 points) For most molecules, γ >> 1. Express λ in terms of γ and expandyour expression for En to order 1/γ2.

We now consider a perturbative approach to this problem

7. (5 points) Find the minimum r0 of the potential and expand V (r) aroundthat minimum, keeping all terms of degree 3 or lower in (r − r0).

8. (5 points) Write the total Hamiltonian as a sum of H0 + Hpert. Identifyclearly the perturbative term.

9. (15 points) Using the states |nlm〉 of the harmonic oscillator, find the firstand second order correction to the ground state energy.

10. (5 points) Check your answer by comparing with the result of [6.]

25. (20 points) Consider the matrix

M =

38

1348

1748

13

518

718

1972

1348

67144

. (G.33)

Problems 242

a) Show that λ = 1 is an eigenvalue of M .

b) Find the eigenvectors of M .

c) Find the matrix M1/2.

d) Find the matrix M∞.

26. (100 points) Consider the problem of a particle trapped in a infinite potentialwell of width L. Inside the well, there is an extra potential “bump” V (x) of widthw given by

V (x) =

0 , if x < 12(L− w)

V0 , if 12(L− w) < x < 1

2(L + w)

0 , if x > 12(L + w)

. (G.34)

a) Find the conditions that the wave function must satisfy at x = 12(L − w)

and x = 12(L + w).

Suppose that V0 is 12 of the energy of the lowest eigenstate for a square well of

width L, with w = L/10.

b) (20 points) Find the lowest three energies for this well, and compare themwith the lowest three energies of the square well .

c) (20 points) Find an expression for the lowest energy eigenstate valid every-where inside the well. Make sure that your wave function is normalized.

d) (5 points) Make a plot of the probability density of this wave function.

e) (5 points) What is the overlap between your solution and the lowest wave-function for the square well problem?

f-g-h-i) Repeat the previous calculations assuming now that V0 is 32 times the energy

of the lowest eigenstate for a square well of width L with w = L/10.

27. (30 points) Let α ∈ C, and let |n〉 be the harmonic oscillator state with energy(n + 1

2)~ω.At t = 0, the “coherent state” |α(0)〉 is defined by

|α(0)〉 = e−|α|2/2

( ∞∑

n=0

αn

√n!|n〉

)(G.35)

Problems 243

a) (5 points) What is |α(t)〉, the coherent state at time t?

b) (5 points) Show that a|α(t)〉 = α e−i~ωt|α(t)〉.c) (5 points) Find 〈p(t)〉 and 〈x(t)〉 for |α(t)〉.d) (10 points) Find 〈p2(t)〉 and 〈x2(t)〉 for |α(t)〉.e) (5 points) Compute ∆x∆p for |α(t)〉.

28. (15 points) A particle of mass m is trapped in a harmonic potential suchthat ω =

√k/m. At t = 0, the wave function describing this particle is given

by |Ψ(0)〉 =12s

∑n

|n〉 , where the sum over n goes from N − s to N + s with

N À s À 1.

a) (5 points) What is the form of the wave function |Ψ(t)〉 at any time t?

b) (10 points) Show that 〈x(t)〉 varies (co)sinusoidally in time with amplitude√2~N/mω, and compare the frequency of oscillation with the classical re-

sult.

29. (20 points) Compute the matrix elements 〈n|x3|m〉 and 〈n|x4|m〉 for the one–dimensional harmonic oscillator.

30. (50 points) The Morse potential. To investivate the vibrational spectrum of adiatomic molecule, Morse has introduced the potential

V (x) = D(e−2αx − 2e−αx

). (G.36)

a) (5 points) Find the point x0 where this potential is minimum. This is theequilibrium point.

b) (5 points) By expanding V in series about x0, write an approximate expres-sion for V up to and including terms in (x− x0)4:

V (x) ≈ V0 + 12mω2(x− x0)2 + χ(x− x0)3 + η(x− x0)4 . (G.37)

c) (5 points) Converting to ξ = (x − x0)/x0, write the time–independentSchrodinger equation for this problem using the approximate expression forV found in Eq.(G.55). Identify explicitly the Hamiltonian operator H.

Problems 244

d) (10 points) Using the reduced mass formula: 1/m = 1/mCl + 1/mH andthe parameters given below, evaluate ω, χ and η for an electron in potentialspecified by the parameters describing the HCl molecule if: D = 37244 cm−1,~2/2mx2

0 = 10.5930 cm−1, α = 2.380. What is the equilibrium separationdistance x0 between the two atoms in the molecule?

e) (5 points) Using the previous values for the paramaters of your potential,construct a 3 × 3 matrix with matrix elements Hmn = 〈m|H|n〉 using thestates |0〉 , |1〉 and |2〉. (This should be a matrix of number.)

f) (10 points) Determine the characteristic polynomial of this matrix, and findthe roots by plotting the polynomial and identifying the zeroes. (Aside: Yourenergy eigenvalues will have units of cm−1: to convert to eV, use E(eV ) =E(cm−1) × 1.2398 × 10−4. The eigenvalues should be near ~ω , 3

2~ω , 52~ω.

Why?)

g) (5 points) Compare the lowest eigenvalue with the energy 12~ω of the ground

state of the oscillator. (The difference between the eigenvalue and 12~ω is

due to the cubic and quartic terms in Eq.(G.55).)

h) (5 points) What is the wavelength λ of the photon emitted when a state isde–excited from the first excited state to the ground state? Is this in thevisible, infra–red or ultra–violet?

31. (40 points) Consider the three angular momentum states |1, 1〉z , |1, 0〉z and|1,−1〉z, with Lz|1,m〉z = ~m|1, m〉z .

a) Construct the 3 × 3 matrix for the operator Lx. Is this matrix hermitian?Show that the eigenvalues of this matrix are ±1, 0, i.e. they are identical tothe eigenvalues of the operator Lz.

b) Repeat this calculation for the operator Ly.

c) Find the linear combinations of |1,m〉z state that are eigenstates of Ly.

d) State the Heisenberg uncertainty relation for ∆Lx ∆Ly, and verify that therelation holds for each of the states |1,m〉z.

32. (30 points) Suppose that an electron is in a state defined by the wave function

ψ(r, θ, ϕ) =1√4π

(eiϕ sin θ + cos θ

)g(r) , (G.38)

Problems 245

where ∫ ∞

0|g(r)|2r2 dr = 1 . (G.39)

a) What are the possible results of a measurement of the z-component of theangular momentum of the electron in this state?

b) What is the probability of obtaining each of the possible results found in a)?

c) What is the expectation value of Lz?

33. (20 points) The quadrupole moment operator is defined as Q0 = 2z2−x2−y2.

a) Using spherical coordinates, show that this operator is proportional to r2Y 02 (θ, ϕ).

b) Let 〈r, θ, ϕ |ψ〉 = R(r)Y m` (θ, ϕ). Assuming ` = 1, which magnetic substate

(or substates) has the largest average quadrupole moment 〈Q0〉.

34. (40 points) Consider a particle represented by the wave function ψ = K(x +y + 2z)e−αr, where r =

√x2 + y2 + z2, with K,α ∈ R constants.

a) What is the total angular momentum of this particle?

b) What is the expectation value of the z-component of angular momentum?

c) If the z-component of angular momentum, Lz, were measured, what wouldbe the probability that the result would be +~?

d) What is the probability of finding the particle at θ, ϕ and in the solid angledΩ? (θ and ϕ are the usual angles in spherical coordinates.)

35. Design a hydrogenoid state |ψ〉 which satisfies the following requirements:i)Repeated measurements of the energy of a particle described by |ψ〉 produceonly the values E = −13.6 eV or E = −(13.6/4) eV, ii) 〈E〉 = 5

8 × (−13.6) eV, iii)〈Lz〉 = 1

8 ~, iv) 〈Lx〉 = 0 ~, v) 〈Ly〉 = 14 ~.

36. Find the lowest possible energy of a nucleon having angular momentum ` = 2~,trapped in the spherical well of depth 8MeV in a nucleus containing 65 nucleons,given that the nuclear radius is r = 1.5× 10−15 A1/3, with A the nucleon number,and compare with the energy of such a nucleon in an infinite spherical well havingthe same radius.

37. Let A and B be any operators.

Problems 246

a) Let

eA = 1l +11!

A +12!

A2 +13!

A3 + . . . (G.40)

Show that

eA B e−A = B + [A, B] + 12 [A, [A, B]] + 1

3! [A, [A, [A, B]]] + . . . (G.41)

b) Use Eq.(G.41) to show that

eiθLy Lze−iθLy = Lz cos θ − Lx sin θ .

38. (A review problem on angular momentum) Let ζ = − tan(θ)eiϕ. Define theangular momentum coherent state at t = 0 by

|ζ〉 =1

(1 + |ζ|2)jeζL+ |`,−`〉 , (G.42)

where the exponential of an operator is defined in Eq.(G.40), and where |`, m〉 isthe state such that

L2|`,m〉 = ~2`(` + 1)|`,m〉 , Lz|`,m〉 = m~ |`,m〉 .a) Write an explicit expression in terms of θ and ϕ for |ζ〉 if ` = 1.

b) For the special case of ` = 1, show that |ζ〉 is properly normalized.

c) Let H = 12AL2 +ωLz, where ω is the rotational frequency. Show, on dimen-

sional grounds, that 1/A is proportional to the moment of internia of thesystem having this Hamiltonian.

d) Find |ζ(t)〉 when ` = 1.

e) Compute 〈E〉 and ∆E.

f) Compute 〈Lz〉 ,∆Lz , 〈Ly〉 and ∆Ly when ` = 1, and verify that the uncer-tainty relation is satisfied.

39. (A review problem on harmonic oscillators and radial wave functions) Considerthe two–dimensional isotropic harmonic oscillator, with potential V = 1

2mω2(x2 +y2). Let

ax =

√~

2mω

(x +

i

mωpx

)a†x =

√~

2mω

(x− i

mωpx

)

ay =

√~

2mω

(y +

i

2ωpy

)a†y =

√~

2mω

(y − i

mωpy

),

where [ai, a†j ] = δij , i, j = x, y.

Problems 247

a) Let |ψ〉 = N a†xa†y|0〉. Find a constant N that will normalize correctly |ψ〉.b) Find |Ψ(t)〉, where |Ψ(0)〉 = |ψ〉.

The coordinate approach.

c) Using the known wave functions for the 1-d harmonic oscillator and theexpression of x and y in polar coordinates, express |ψ〉 in polar coordinates〈r, ϕ |ψ〉 = ψ(r, ϕ)

d) Let Lz = i~∂/∂ϕ. Compute 〈Lz〉 and ∆Lz for the state |ψ〉.e) What is the probability of measuring Lz = 0~ for a particle prepared in the

state |ψ〉?f) Given the Laplacian in 2-d:

∇2 =1r

∂

∂r

(r

∂

∂r

)+

1r2

∂2

∂ϕ2,

use separation of variables to show that eigenstates of H are of the formf(r)eimϕ. Find the differential equation that must be satisfied by the func-tion f(r).

g) Using the boundary conditions at r = ∞ and r = 0, show that f(r) is of theform

f(r) = r|m|e−λr2/2F (r) ,

where F (r) satisfies

d2

dr2F (r) +

(2|m|+ 1

r− 2λr

)d

drF (r)− (

2λ(|m|+ 1)− k2)F (r) = 0 ,

(G.43)where λ = mω/~ , E = ~2k2/2m.

h) Show that the substitution t = λr2 transforms Eq.(G.43) into

td2

dt2F (t) + (|m|+ 1− t)

d

dtF (t)− 1

2

(|m|+ 1− k2

4λ

)F (t) = 0 , (G.44)

i) Compare Eq.(G.44) with the standard generalized Laguerre functions, andfind the condition for which F (t) will become a polynomial of finite degree.Deduce the possible values of E from that condition.

Problems 248

j) Write all the possible wave functions having energy E1 = 2~ω, and normalizethem properly. (Hint: these are functions of r and ϕ.)

The Hilbert space approach

k) Express H, the Hamiltonian for this system, in terms of ax, a†x, ay and a†y.

l) Is |ψ〉 an eigenstate of H? If yes, what is the energy associated with thisstate? If no, what is ∆E for this state?

m) Show that the operators

i~a†xay , 12 i~

(a†xax − a†yay

), i~axa†y

have commutation relations identical to L+, L−, Lz, and that H commuteswith 1

2 i~(a†xax − a†yay

).

n) Using 12 i~


), compute 〈Lz〉 and ∆Lz.

o) Using the polar coordinates, find an explicit expression for the operator12 i~


).

40. (20 points) (see Griffiths p.13 prob. 1.8) Consider the wave function

Ψ(x, t) = Ae−λ|x| e−iωt, (G.45)

where A, λ, ω are real constants.

a) Normalize Ψ,

b) Determine the expectation values of x and x2 ,

c) Find the standard deviation of x. Sketch the graph of |Ψ(x, t)|2 as a functionof x, and mark the points (〈x〉+ σ) and (〈x〉 − σ) to illustrate the sense inwhich σ represents the spread in x. What is the probability that the particlewould be found outside this range?

41. (20 points) (Griffiths p.13 prob. 1.9) Let Pab(t) be the probability of findingthe particle in the range a < x < b at time t.

Problems 249

a) Show thatdPab(t)

dt= J(a, t)− J(b, t) , (G.46)

where

J(x, t) ≡ i~2m

(Ψ(x, t)

∂Ψ∗(x, t)∂x

−Ψ∗(x, t)∂Ψ(x, t)

∂x

).

What are the units of J(x, t)?

b) Find the probability current for the wave function of Eq.(G.45).

42. (20 points) (Griffiths p. 14 prob. 1.10) Suppose you wanted to describe anunstable particle that spontaneously disintegrates with a ”lifetime” τ . In thatcase, the total probability of finding the particle somewhere should not be constant,but should decrease with time at (say) some exponential rate:

P (t) ≡∫ ∞

−∞|Ψ(x, t)|2 dx = e−t/τ .

A crude way of achieving this is to assign to the potential of the Schrodingerequation an imaginary part:

V = V0 − iΓ , (G.47)

where V0 is the true potential energy and Γ is a real positive constant.

a) Show thatd

dt

∫ ∞

−∞|Ψ(x, t)|2 dx = −2Γ

~P . (G.48)

b) Solve for P (t) and find the lifetime of the particle in terms of Γ.

43. (40 points) (Griffiths p.19 prob. 1.14) A particle of mass m is described bythe wave function

Ψ(x, t) = Aea[(mx2/~)+it] , (G.49)

where A and a are positive real constants.

a) Find A.

b) For what potential energy function V (x) does Ψ(x, t) satisfy the Schrodingerequation?

c) Calculate the expectation values of x, x2, p and p2.

Problems 250

d) Find σx and σp. Is their product consistent with the uncertainty principle?

44. 30 (see Griffiths p.29 prob. 2.6&2.7) A particle in the infinite square well hasas its initial wave function an even mixture of the first two stationary states:

Ψ(x, 0) = A (ψ1(x) + ψ2(x)) . (G.50)

a) (5 points) Normalize Ψ(x, 0).

b) (5 points) Find Ψ(x, t) and |Ψ(x, t)|2, and express the result in terms of sineand cosine functions of time, using ω = π2~/2ma2.

c) (5 points) Compute 〈x〉. What is the frequency of oscillation of 〈x〉? Whatis the amplitude of oscillation?

d) (5 points) Compute 〈p〉e) (5 points) Find the expectation value of H. How does it compare with E1

and E2?

Although the overall phase constant of the wave function is of no physicalsignificance, the relative phase of the coefficients in Eqn. (G.50) does matter. Forexample, suppose we change the relative phase of ψ1(x) and ψ2(x) to

Ψ(x, 0) = A(ψ1(x) + eiφψ2(x)

), (G.51)

where φ is a constant.

f) (5 points) Find Ψ(x, t), |Ψ(x, t)|2 and 〈x〉, and compare your results withthe results obtained using Eqn.(G.50), where φ was set to 0.

45. 35 The “radial” part of Schrodinger’s equation in three dimensions is given by(see. Eqn.(4.16))

1R(r)

d

dr

(r2 dR

dr

)− 2mr2

~2(V (r)− E) = ` (` + 1), (G.52)

where R(r) is the radial wave function and r ≥ 0 is the radius.

a) (5 points) Show that the change of variable χ(r) = rR(r) will tranform theradial equation into an equivalent one–dimensional problem subject to thecondition r ≥ 0. What is the equivalent one–dimensional potential for thisproblem?

Problems 251

The nuclear potential can be modeled as a well having a finite depth of ≈57MeV and radius r0 ≈ 1.25A1/3 with A the total number of nucleons.

b) (15 points) How many states with angular momentum ` = 0 can you findfor 56

26Fe? What are the energies of these states?

c) (10 points) Write down explicitly the normalized wave function for the lowestenergy state of 56

26Fe. Find 〈r〉 for this state if the radial probability densityis given by

P (r) = r2|R(r)|2 . (G.53)

d) (5 points) What is the probability of finding a nucleon “outside” the nucleusif it is described by an ` = 0 wavefunction having the lowest possible energyin the above well?

46. 35 The Morse potential. To investigate the vibrational spectrum of a diatomicmolecule, Morse has introduced the potential

V (x) = D(e−2αx − 2e−αx

), x =

r − r0

r0. (G.54)

a) (5 points) Sketch this potential, and show that it has a minimum at r = r0.This is the equilibrium point.

b) (5 points) By expanding V in series about r0, write an approximate expres-sion for V up to and including terms in (r − r0)2:

V (x) ≈ V0 + 12mω2(r − r0)2 . (G.55)

c) (10 points) Using the reduced mass formula: 1/m = 1/mCl + 1/mH andthe parameters given below, evaluate ω for an electron in potential spec-ified by the parameters describing the HCl molecule if: D = 37244 cm−1,~2/2mr2

0 = 10.5930 cm−1, α = 2.380. To convert from cm−1 to eV, useE(eV ) = E(cm−1) × 1.2398 × 10−4). What is the equilibrium separationdistance r0 between the two atoms in the molecule?

d) (5 points) By comparing with the harmonic oscillator, obtain the (approxi-mate) values for the lowest two energy levels of HCl.

e) (10 points) By comparing with the wave functions of the harmonic oscillator,write the (approximate) radial wave function R(r) for the state of lowestenergy. Compute 〈r〉 using Eqn.(G.53)

Problems 252

47. 40 points (see Flugge prob. 42). Consider the two–dimensional isotropicharmonic oscillator, with potential V = 1

2mω2(x2 + y2). Given

H = − ~2

2m∇2 + 1

2mω2 r2 , ∇2 =1r

∂

∂r

(r

∂

∂r

)+

1r2

∂2

∂ϕ2,

in polar coordinates,

a) Using separation of variables, show that the eigenstates of H are of the formf(r)eimϕ.

b) Show that f(r) satisfies the differential equation(

f ′′(r) +1rf ′(r)− m2

r2f(r)

)+(k2−λ2r2)f(r) = 0 , k2 =

2mE

~2, λ =

mω

~.

(G.56)

c) Show that, for large values of r, the solution must be of the form f(r) ∼rs e−λr2/2 with finite s, and that, for small values of r, the solution must beof the form f(r) ∼ r|m|.

d) Use the ansatz f(r) = r|m| e−λr2/2 F (r) to show that the interpolating func-tion F (r) satisfies

F ′′ +(

2|m|+ 1r

− 2λr

)F ′ − [

2λ(|m|+ 1)− k2]F = 0 . (G.57)

e) Make the change of variable t = λr2 and show that F (t) now satisfies

td2F

dt2+ [(|m|+ 1)− t]

dF

dt− 1

2

[(|m|+ 1)− k2

4λ

]F = 0 . (G.58)

f) Show that the polynomial solutions to the previous differential equation areF (t) =1 F1(a, |m|+ 1; t), where

a = 12

[(|m|+ 1)− k2

4λ

]≡ −nr , (G.59)

with nr a positive integer.

g) Show that this last condition leads to Enr,m = ~ω(|m|+ 1 + 2nr)

Problems 253

48. (15 points) The nuclear potential can be modeled as a well having a finitedepth of ≈ 57MeV and radius r0 ≈ 1.25A1/3 with A the total number of nucleons.How many nuclear states with angular momentum ` = 1 can you find for 56

26Fe?What are the energies of these states?

49. (20 points) A particle of mass m is constrained to move between two concentricimpermeables spheres of radii r = a and r = b. There is no other potential.

a) (10 points) Show that the lowest ` = 0 state has energy

E =~2π2

2m(b− a)2. (G.60)

b) (10 points) Find the radial wave function R(r) for this lowest state, andobtain the normalized wave function ψ(~r) if the normalization conditionreads ∫

dΩ r2|ψ(~r)|2 dr = 1 . (G.61)

50. (25 points) For a particle of mass m = 1/2, a Schrodinger equation in onedimension, in units where ~ = 1, reads

(− d2

dx2− 2sech2 x

)ψ(x) = εψ(x) . (G.62)

a) (5 points) Determine the constant η in eikx(tanhx + η) that will make thisfunction a solution of this Schrodinger equation.

b) (10 points) Calculate the S–matrix for this problem.

c) (10 points) The wave function ψ(x) = sechx happens to satisfy the Schrodingerequation. Calculate the energy of the corresponding bound state and givea simple argument that it must be the lowest energy state of this potential.(Hint: number of nodes).

51. (30 points) Conserved quantities.

a) (10 points) prob. 3.54 page 118 from Griffiths.

Problems 254

(Particle in a magnetic field) The Hamiltonian for a spinless charged particlein a magnetic field ~B = ∇× ~A is

H =1

2m

(~p− e

c~A)

, (G.63)

where e is the charge of the particle, ~p = (px, py, pz) is the momentum and ~A

is the potential vector. Let ~A = −B0yx, so that the magnetic field ~B = B0zis constant along z.

b) Prove that px and pz are constants of motions by showing that they commutewith the Hamiltonian operator H of the system.

c) Show thatψ(x, y, z) = ei(xpx+zpz)φ(y) (G.64)

is an eigenstate of px and pz with respective eigenvalues px and pz.

d) Write the time–independent Schrodinger equation for the y coordinates interms of px and pz, and

52. (30 points) (based on Griffiths page 120 prob. 3.58) Imagine a system in whichthere are just two linearly independent states:

|1〉 =(

10

), |2〉 =

(01

). (G.65)

The most general state is a normalized linear combination

|Ψ〉 = a|1〉+ b|2〉 =(

ab

), (G.66)

with aa∗ + bb∗− = 1.Suppose the matrix representing the Hamiltonian is

H =(

ε0 gg∗ ε0

), (G.67)

where ε0 ∈ R and g ∈ C. The time–dependent Schrodinger equation is

i~∂

∂t|Ψ〉 = H|Ψ〉 . (G.68)

a) Find the eigenvalues and normalized eigenvectors of H.

Problems 255

b) Suppose the system starts out (at t = 0) in state |1〉. What is the state ofthe system at time t?

The ammonia molecule, NH3, has a pyramidal shape, with the nitrogen atomat vertex and three hydrogen atoms at the base. Classically, the nitrogen atommay be at one of two symmetric equilibrium positions: to the “left” or “right” theplane of the hydrogen atoms. These states are denoted, respectively, by |NH3〉and |H3N〉.

c) Given the energy separation ∆E = 9.84 × 10−5eV , can you reconstruct allthe matrix elements (magnitude and phase) of the Hamiltonian matrix?

d) Which of the two eigenstates has the lowest energy? (Hint: even/oddnessof corresponding wave functions and number of nodes.)

e) Find the frequency of oscillation between the |NH3〉 and |H3N〉 states.

53. (30 points) Let α ∈ C, and let |n〉 be the harmonic oscillator state with energy(n + 1

2)~ω.At t = 0, the “coherent state” |α(0)〉 is defined by

|α(0)〉 = e−|α|2/2

( ∞∑

n=0

αn

√n!|n〉

)(G.69)

a) (5 points) What is |α(t)〉, the coherent state at time t?

b) (5 points) Show that a|α(t)〉 = α e−i~ωt|α(t)〉.c) (5 points) Find 〈p(t)〉 and 〈x(t)〉 for |α(t)〉.d) (10 points) Find 〈p2(t)〉 and 〈x2(t)〉 for |α(t)〉.e) (5 points) Compute ∆x∆p for |α(t)〉.

54. (40 points) Morse revisited.

a) (5 points) Compute the matrix elements 〈n|x3|m〉 for the one–dimensionalharmonic oscillator.

b) (5 points) Expand the Morse potential:

V (x) = D(e−2αx − 2e−αx

), x =

r − r0

r0. (G.70)

near its equilibrium point, including terms in (r − r0)3.

Problems 256

c) (5 points) Using this expansion for V (x), write the (approximate) Hamilto-nian operator H in terms of the operators a and a†.

d) (10 points)Using |0〉, |1〉 and |2〉, obtain the 3× 3 matrix representing H inthis subspace.

e) (5 points)Using the reduced mass formula: 1/m = 1/mCl + 1/mH and theparameters given below for the HCl molecule: D = 37244 cm−1, ~2/2mr2

0 =10.5930 cm−1, α = 2.380, write the numerical form of the matrix for H. (Toconvert from cm−1 to eV, use E(eV ) = E(cm−1)× 1.2398× 10−4).

f) (5 points)Obtain the characteristic equation Det(H − λ 1l) = 0 and find thelowest eigenvalue. This will have to be done numerically by plotting thepolynomial in λ and looking for the points where it goes through zero.

g) (5 points)Compare this lowest eigenvalue with the one obtained in ProblemSet 2. Is this new eigenvalue an improvement when compared with theexam energy (which you can find from the section in the notes on the Morsepotential)?

55. (20 points) Let 〈θϕ | `m〉 = Y`m(θ, ϕ).

a) Obtain the matrix representation Mx of the operators Lx between states ofangular momentum ` = 1.

b) Find the eigenvalues and normalized eigenvectors |m〉x of Lx.

c) Construct the matrix U †, having as its first column the first eigenvector ofLx, as its second column the second eigenvector of Lx etc. Show the U † isunitary by verifying that U · U † = 1l.

d) Verify that U ·Mx · U † is diagonal.

56. (20 points) Let µ ∈ C, and consider

|µ〉 =1

1 + |µ|22∑

k=0

µk

k!

(L−

)k|1, 1〉 , (G.71)

where |`, m〉 is the state

~L · ~L|`,m〉 = ~2 `(` + 1)ket`,m , Lz|`,m〉 = m ~|`,m〉 . (G.72)

Problems 257

a) (5 points) Construct explicitly |µ〉 as a linear combination of the |1,m〉’s.b) (5 points) Show that |µ〉 is normalized.

c) (10 points) Compute ∆Lx∆Ly for this state, and verify that this productsatisfies the uncertainty relation.

57. (20 points) Let ψn`m(r, θ, ϕ) = 〈r, θ, ϕ |n, `,m〉 be a wave function for a three–dimensional harmonic oscillator.

a) (5 points) Write explicitly the two wave functions having total energy 72~ω

and magnetic quantum number m = 0.

b) (5 points) The quadrupole moment operator Q0 is defined as 3z2 − x2 − y2.Find an expression for the operator in terms of ` = 2 spherical harmonics.

c) (10 points) Compute the matrix element 〈000|Q0|220〉 using the explicitexpression of the appropriate wave functions.

58. (25 points) Let |n`m〉 denote the state of a hydrogen atom with energy−13.6/n2eV, angular momentum ` and magnetic quantum number m. If, at timet = 0, the state of a hydrogen atome is

|Ψ(0)〉 =1√10

(|100〉+ |210〉+

√2|211〉+

√3|21,−1〉

), (G.73)

a) Find the average energy of this state,

b) What is |Ψ(t)〉?c) What is the probability of finding the system with m = 0 as a function of

time?

d) At time t = 0, what is the probability of finding the electron within 10−12mof the proton? (It is valid to use expansions for small r if need be.)

e) Suppose a measurement is made which shows that ` = 1 and m = 1. De-scribe the (normalized) state immediately after this measurement.

These problems are designed to review some of the material of Modern PhysicsI & II.

59. (10 points) Basic properties of the Schrodinger equation.

Problems 258

a) The time–dependent Schrodinger equation in one dimension reads

i~∂

∂tΨ(x, t) = − ~

2

2m

∂2

∂x2Ψ(x, t) + V (x)Ψ(x, t) . (G.74)

Show that, if Ψ1(x, t) and Ψ2(x, t) are separately solutions of Eqn.(G.127),then the arbitrary linear combination Ψ(x, t) = α1Ψ1(x, t) + α2Ψ2(x, t) isalso solution for arbitrary complex constants α1 and α2.

b) The time–independent Schrodinger equation in one dimension reads

− ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x) , (G.75)

where E is the eigenvalue of associated to ψ(x). Show that, if ψ1(x) andψ2(x) are separately solutions of Eqn.(G.128) with eigenvalues E1 and E2 re-spectively, then the arbitrary linear combination ψ(x) = α1ψ1(x)+α2ψ2(x) ,with α1 and α2 arbitrary complex constants, is not a solution of Eqn.(G.128)unless E1 = E2.

60. (50 points) Playing with wave functions. Let

ψ0(x) =(

a√π

) 12

e−a2x2/2 , ψ1(x) =(

a

2√

π

) 12

2ax e−a2x2/2 , (G.76)

where ω =√

k/m , a =(

mω~

) 12 .

a) (5 points) Show ψ0(x) and ψ1(x) are separately solutions of Eqn.(G.128)with V (x) = 1

2kx2, and find the corresponding eigenvalues E0 and E1,

b) (15 points) Show ψ0(x) and ψ1(x) are normalized and orthogonal w/r to thestandard `2 inner product:

〈ψi |ψj〉 =∫ ∞

−∞dxψ∗i (x)ψj(x) = δij ≡

0 if i 6= j ,1 if i = j .

(Suggestion: Find∫ ∞

−∞dξ ξn e−ξ2

as a function of n knowing∫ ∞

−∞dξ e−ξ2

=√

π.)

c) (10 points) Write an expression for Ψ0(x, t) and Ψ1(x, t), and use the previ-ous results to show that Ψ0(x, t) and Ψ1(x, t) also satisfy 〈Ψi(t) |Ψj(t)〉 = δij .

Problems 259

d) (5 points) Let Ψ(x, t) = (cos η)Ψ0(x, t) + eiϕ(sin η)Ψ1(x, t), where η and ϕare real parameters. Show that Ψ(x, t) is normalized.

e) (5 points) Using Ψ(x, t) above, find the average value of x as a function oftime, i.e. compute

〈x(t)〉 =∫ ∞

−∞dxΨ∗(x, t) xΨ(x, t) . (G.77)

61. (15 points) A problem on quantum weirdness. A classical turning point X is avalue of x for which the potential energy is equal to the total energy of a particle;in principle, a particle has no kinetic energy at that point and must therefore “turnaround” its motion.

a) Use the wave function ψ0(x) of Eqn.(G.76) and its energy found in 2.1 tofind the turning points of a classical particle in a harmonic oscillator well.(Hint: there are two turning points...)

b) The region between the classical turning points will be referred to as theclassically allowed region. The regions not classically allowed are calledclassically forbidden. Find a definite integral expressing the probability offinding the particle in the classically forbidden region.

c) Evaluate the previous integral. You will need to do some numerical integra-tion to obtain the final result

62. (30 points) Consider the motion of a quantum particle trapped in an infinitewell “with a step”. The total width of potential is a, the height of the step is V0,and the width of the step is a/2. The potential is illustrated in figure G.

Letψ(x) = A sin(πx/a) . (G.78)

a) (2 points) Is the wavefunction of Eqn.(G.78) an eigenstate of the Hamiltonianof this system?

b) (3 points) Discuss why the wavefunction of Eqn.(G.78) is a valid wavefunc-tion for this system.

c) (2 points) Normalize ψ(x, t).

Problems 260

Figure 31: The potential for this problem.

d) (3 points) Find 〈x〉 for a particle described by ψ(x).

e) (5 points) Find 〈p〉 for a particle described by ψ(x), and explain why theresults is expected on physical grounds.

f) (5 points) Find 〈E〉 for a particle described by ψ(x).

g) (10 points) What is the maximum height of the potential step V0 that willresult in no bound state with energy E < V0?

63. (30 points) Consider a particle trapped in an infinitely deep well with wave-function (at t = 0)

Ψ(x, 0) = A (ψ1(x)− 3iψ3(x))

where ψn(x) is the n solution to the time–independent Schrodinger for this poten-tial.

a) (5 points) Find Ψ(x, t).

b) (5 points) Find a value for the constant A that will normalize Ψ(x, t).

c) (5 points) Compute the probability current J(x, t) for this particle.

d) (10 points) Using Ψ(x, 0), find 〈x〉,∆x, 〈p〉,∆p and verify the uncertaintyrelation.

e) (5 points) Using Ψ(x, 0), find 〈E〉 and ∆E.

64. (20 points) The deuteron is a composite particle made from one proton and oneneutron. The relative motion of the constituents can be rewritten as the motionof a single particle, having reduced mass µ given by

1µ

=1

mn+

1mp

,

Problems 261

and evolving in a spherical well of finite depth −V0 (V0 > 0). The deuteron hasonly one bound state with angular momentum ` = 0. In such circumstances, thesolutions of the Schrodinger equation in three dimension reduces to the solutionsof the one–dimensional problem provided that the motion is restricted to positivecoordinates.

Knowing the binding energy −E0 = −2.22 MeV , obtain an estimate of V0 ifthe width r0 of the spherical potential is r0 ≈ (~c/140)MeV −1×10−15m , mn c2 ≈mp c2 ≈ 940MeV , ~c = 197MeV × 10−15m.

65. (30 points) (This is an augmentation of a midterm problem from last year.)The ammonia molecule, NH3, is in the form of pyramid, with the nitrogen atomat vertex and three hydrogen atoms at the base. The nitrogen atom may be atone of two symmetric equilibrium positions: “above” or “below” the plane of thehydrogen atoms.

The potential energy for the motion of nitrogen is illustrated in figure 1 as the“W”–shape curve. The figure also illustrates the approximation of this potential bya infinite one–dimensional square well with a central bump of height V0 = 0.245eV.The infinite walls are located at ±b = ±0.5548×10−10m; the central bump extendsfrom −a to +a, with a = 0.2298× 10−10m.

V

a

b

o

b−a

Figure 32: The potential for the nitrogen atom of the ammonia molecule. Theexact potential (wavy line) is approximated by an infinite square well with a bumpin the middle (heavy line).

a) (2 points) Let x denote the position of the nitrogen atom. Explain why thispotential is even under transformation of x → −x.

Problems 262

b) (5 points) Show the wave function

ψ(x) =

B sin(k(b + x)) if − b ≤ x ≤ a ,

A (eκx + e−κx) if − a < x < a ,


(G.79)

is even under the transformation x → −x, and find a general expression fora wavefunction odd under the transformation x → −x.

c) (3 points) Justify the use of these wave function in searching for solutions ofa problem with even potential. In particular, justify the assumptions behindthe use of the exponentials to describe the wavefunction in the central region.

d) (10 points) Let

κ =

√2m(V0 − E)

~2, k =

√2mE

~2. (G.80)

Show the energies of the even wave functions can be found from the consis-tency equation:

1κ

(eκa + e−κa

eκa − e−κa

)= −1

ktan(k(b− a)) , (G.81)

and find the corresponding transcendental equation valid for the odd wave-function.

e) (5 points) Using the graphical method described in class, find the energy ofall the even wavefunctions with E < V0 knowing that mNH3 = 2300MeV/c2

and ~c = 197.5MeV× 10−15m.

f) (5 points) Determine if there are odd wavefunctions with energy smallerthan the height of the central barrier. If there are such a functions, findtheir energy; if there is not such function, find the smallest value of V0 thatwill allow for one odd state.

Part I: Harmonic vibrations of molecules and solidsYou may want to read prob. 5.11 of Schaum and Beiser before starting these

two problems.

66. The interaction between ions in a simple molecular solid can be modeled bythe Lennard-Jones potential:

V (r) = 4V0

[(r0

r

)12−

(r0

r

)6]

. (G.82)

(This parametrization differs from that of the Schaum.)

Problems 263

a) Show that this potential has a minimum at rmin = 21/6r0.

b) This potential represents the interaction of a single molecule with the restof the crystal. For a crystal of N2, the parameters of the potential areV0 = 0.00791eV and r0 = 3.92× 10−10m. Make a plot of V (r) applicable toN2. Choose a range of r that clearly illustrates the depth of the potentialand the position of the minimum.

c) (5 points) Expand V (r) in a power series around r = rmin, and show thepotential has the approximate form

V (r) ≈ −V0 + 12 mω2 (r − rmin)2 . (G.83)

The expression −V0 + 12 mω2 (r − rmin)2 is the osculating parabola for this

potential and transforms V (r) into an approximately harmonic oscillatorpotential for r − rmin, shifted in energy by −V0.

d) (10 points) Use the series expansion to estimate the energies of the lowestthree vibration levels of the molecule in the crystal. Express your energiesin eV . Be careful about your choice of mass parameter!

e) (7 points) What is the approximate energy of a photon emitted in a transi-tion from the n = 1 to the n = 0 vibrational state? Is this in the infrared,visible or ultraviolet light?

67. (30 points) The potential between the two atoms of a diatomic molecule canbe modeled by the Morse potential:

V (r) = D(e−2a(r−r0) − 2e−a(r−r0)) , (G.84)

where D is the dissociation energy, r0 is the equilibrium position and a is a pa-rameter. The variable r represents the relative distance between the atoms. Itis possible to solve the Schrodinger equation exactly for the Morse potential, butmany important features can be understood by looking at the harmonic approxi-mation to this potential.

Start by recovering the original article by Morse. The complete reference ofthe paper is: P.M. Morse, Physical Review vol. 34 page 57 (1929). You will needto do this from a LU computer or from home using first the ”Connect from Home”feature available from the Library webpage. Once you have ”logged in”, find thewebsite for Physical Review (any section) and go to PROLA to access older papers.In the small dialog box where you choose the journal, select Phys. Rev (no section;next to last choice) and enter the correct volume and page numbers. Note thatMorse denotes his potential by E(r) rather than V (r).

Problems 264

a) (3 points) From the data provided on page 62 of the paper, make a plotof V (r) for 12Σ configuration of the N2 molecule. Note that r0 is given in10−10m and that a is measured in 1/(10−10m). The units of D are cm−1,which can be converted to Joules by multiplying by hc, where h is thePlanck constant and c is the velocity of light (in cm/sec). Choose a range0 ≤ r ≤ rmax that clearly shows the position of the minimum and the depthof the potential.

b) (5 points) Show that the minimum of V (r) is located at r = r0. This is theclassical equilibrium position.

c) (5 points) Expand the Morse potential in a power series in r − r0 aroundr = r0, and show the potential has the approximate form

V (r) ≈ −D + Da2(r − r0)2 . (G.85)

d) (10 points) Use the series expansion to estimate the energies of the lowestthree vibration levels of the molecule. Express your energies in eV . (Here,what vibrates are the atoms one relative to the other. The effect of thisvibration on the electrons is ignored.)

e) (7 points) What is the approximate energy of a photon emitted in a transi-tion from the n = 1 to the n = 0 vibrational state? Is this in the infrared,visible or ultraviolet light?

f) (5 points) If we use the exact potential rather than its approximate ex-pression, will the exact energies be higher or lower than the correspondingapproximate ones? You need to think about qualitative features of the wave-function, such as curvature, in order to get a sense of that.

Note that the analysis of Prob. 1 is valid for solids, whereas Prob. 2 isapplicable to gases...

Part II: The Schrodinger equation

68. (10 points) (see Schaum prob. 3.29 page 49) Consider a particle of mass mheld in the potential

V (x) = −V0

(δ(x− 1

2r0) + δ(x + 12r0)

). (G.86)

(Note that this potential is symmetric, whereas Schaum’s isn’t.) Show there aretwo bound states and find a transcendental equation for their energy.

Problems 265

69. (55 points) Find the bound state solutions E < 0 to the Schrodinger equationfor the potential

V (x) = − V0

cosh2(xa )

. (G.87)

a) (5 points) Make a plot of this potential for ξ = x/a and V0 = 1 from−5 ≤ ξ ≤ 5.

b) (10 points) Make the substitutions

ψ(x) =(cosh

x

a

)−2λu(x) , λ =

14

(√8mV0a2~−2 + 1− 1

), (G.88)

to convert Schrodinger equation to

d2u

dx2−

(4λ

atanh(

x

a))

du

dx+

4a2

(λ2−κ2)u = 0 , κ =

√mεa2

2~2, ε = −E > 0 .

(G.89)

c) (10 points) Introduce the new independent variable z = − sinh2(x

a

)and

show this leads the hypergeometric equation

z(1− z)d2u

dz2+

[12 − (1− 2λ)z

] du

dz+ (λ2 − κ2)u = 0 . (G.90)

d) (5 points) Show that the two solutions that give even and odd wavefunctionsare, respectively,

u1 = 2F1(−λ + κ,−λ− κ, 12 ; z) ,

u2 =√

z 2F1(−λ + κ + 12 ,−λ− κ + 1

2 , 32 ; z) . (G.91)

To show the statement, use the series solution expression for 2F1(a, b, c; z).

e) (5 points) Show that the solutions of Eqn.(G.91) remain finite at z = 0.

f) (5 points) Show that the choice k = λ − κ transforms the hypergometricfunction u1 into a polynomial, thereby guaranteeing that the wavefunctionψ(x) = (cosh x

a )−2λ u1(x) goes to 0 at ∞. Find the energies of the evenfunctions as a function of k = 0, 1, 2, .... This deals with the even solutions.

g) (5 points) Show that the choice k′ = λ−κ− 12 transforms the hypergeometric

function u2 into a polynomial so that the odd ψ(x)’s go to 0 at ∞. Find theenergies of the odd wavefunctions as a function of k′ = 0, 1, ....

Problems 266

h) (5 points) Combine these results to show

En = − ~2

2ma2

(12

√8mV0a

2

~2+ 1− (n + 1

2)

)2

, n = 0, 1, ... . (G.92)

i) (5 points) Show the number of discrete levels is equal to the largest integerN satisfying the inequality

N < 12

√8mV0a

2

~2+ 1− 1

2 (G.93)

Independent multiparticle systems

70. Consider a system with two non–interacting identical particles subject to thesame potential, so that the total Hamiltonian H of the system can be written as

H = − ~2

2m

(∂2

∂ξ21

+∂2

∂ξ22

)+ (V (ξ1) + V (ξ2))

= H1 + H2 , (G.94)

where ξ1, ξ2 denote the coordinates of particles 1 and 2, respectively.Let ψn(ξ1) be the n’th eigenstate of H1, and ψm(ξ2) the m’th eigenstate of H2,

i.e. assume

H1ψn(ξ1) = Enψn(ξ1) , H2ψm(ξ2) = Emψm(ξ2) . (G.95)

a) (5 points) Show explicitly that ψ(ξ1, ξ2)n,m = ψn(ξ1)ψm(ξ2) is an eigenstateof H and show that the eigenvalue is E = En + Em.

b) (10 points) Show that, if ψ(ξj) satisfies∫

dξj ψ∗n(ξj)ψs(ξj) = δns , (G.96)

then ψn(ξ1)ψm(ξ2) satisfies∫

dξ1 dξ2 (ψn(ξ1)ψm(ξ2))∗ ψr(ξ1)ψs(ξ2) = δnrδms . (G.97)

Use this to show that ψ(ξ1, ξ2)n,m is properly normalized.

Problems 267

c) (5 points) Show that the combination

A (ψ(ξ1, ξ2)n,m + ψ(ξ1, ξ2)m,n) , (G.98)

where A is an arbitrary constant, is an eigenstate of H with eigenvalueE = En + Em.

d) (10 points) Find the constant A that will normalize the eigenfunction ofEqn.(G.98). Explain why you need to treat the cases m = n and m 6= nseparately.

e) (5 points) Suppose m = n. Compare your result with Eqn.(9.23) of Beiserand discuss how this author doesn’t know what he’s doing (once more).

In the Fermi-gas model of the nucleus, the nuclear constituents move indepen-dently one from the other in an average potential, much in the way that electronsmove independently from each other in a crystal. The nuclear potential has a formgiven by

V (r) = −V0

[1 + exp

((r −R0)

a

)]−1

. (G.99)

(This is called the Woods–Saxon potential, or the inverted Fermi potential.) Theparameters are:

R0 = r0 A1/3 ; V0 ≈ 50 MeV ; a ≈ 0.5×10−15 m ; r0 ≈ 1.2×10−15 m.

Here, A is the total number of nucleons (protons plus neutrons).

Problems 268

The potential actually has a finite but negligible slope at r = 0. Thus, forcalculation purposes, one often uses one of three possible approximations:

a) The 3–dimensional harmonic oscillator:

V (r) = −V0

[1− r

R0

]= 1

2mω20(r

2 −R20) ,

b) The infinite spherical well:

V (r) = −V0 for r ≤ R0 ,

∞ for r > R0 .

c) The finite spherical well:

V (r) = −V0 for r ≤ R0 ,

0 for r > R0 .

Consider the case of 5626Fe, containing 26 protons and 30 neutrons (A = 56).

For the parameter m, use the mass of a nucleon ≈ mp ≈ mn (protons and neutronsare assumed to have identical mass).

71. (35 points) Let’s approximate the Woods–Saxon potential by a harmonic os-cillator, as indicated above.

a) (5 points) Make a plot of the Woods–Saxon potential appropriate for 5626Fe.

b) (10 points) List the first 20 energy levels (accounting for multiplicities)in increasing order of energy by giving the cartesian quantum numbers(nx, ny, nz) of these levels. In this notation, the lowest energy level is (0, 0, 0).

c) (10 points) List the first 20 energy levels (accounting for multiplicities) inincreasing order of energy by giving the quantum numbers (n, `) and thecorresponding energies. (Hint: In this notation, the lowest energy level is(0, 0).) You may assume that each (n, `) state contains 2` + 1 substates.

d) (10 points) In its simplest form, the Pauli exclusion principle states that nomore than two fermion can have the same set of quantum numbers. Whatis the lowest possible energy of 56

26Fe, containing (Neutrons and protonsare fermions, but they are not identical.) You may assume that, for eachseparate value of `, there are 2` + 1 states.

72. Repeat items iii) and iv) previous problem

Problems 269

a) (10 points) Using the infinite spherical well approximation. (No need toreplot the potential.)

b) (20 points) Using the finite spherical well approximation.

73. Suppose some system is described by the Hamiltonian

H =1

2mp2

ξ +12mω2ξ2 + εξ , (G.100)

where ε is a constant.

a) By completing the square, and making change of variable ξ → η, show thatit is possible to write H in the form

H =1

2mp2

η +12mω2η2 + κ , (G.101)

where κ is a constant.

b) Using the solutions to Eqn.(G.101), find the energy levels of the system interms of the initial variable ξ.

c) Write an explicit expression for the lowest energy wavefunction if ξ is thevariable x, which can range from −∞ to +∞.

d) Write an explicit expression for the lowest energy wavefunction if ξ is thevariable r, which can range from 0 to ∞.

74. Most probable 6= average. The objective of this problem set is to make surethat you become familiar with some important properties of the wavefunctionsthat we’ve obtained thus far. (See Schaum, in particular Eqn.(8.42) and prob.8.12)

Using the definition of the confluent hypergeometric function 1F1 and the re-lation between χn`(r) and 1F1:

a) Show that the most probable value of r in any hydrogen radial state Rn,n−1

with ` = n− 1 is exactly identical to the prediction Bohr’s theory.

b) Find a closed form expression for the average value of r when the system isin such a state.

Problems 270

You need to recall that the probability density for the radial coordinate isrelated to the radial wavefunction R(r) by r2R∗(r)R(r).

75. Muonic atoms. (See Schaum prob. 8.8) One way to investigate wavefunctionsof hydrogen–like systems is by way of the muonic atom, which is formed whena negative muon µ− is captured by a proton. Because the mass of µ− is much207me, the reduced mass of the proton–muon system is much smaller than thereduced mass of the proton–electron system. This change affects energy levels andwavefunctions in a very profitable way.

a) Start by going back to the hydrogen solution. Calculate the binding energyE of the hydrogen atom using the reduced mass of the proton–electronsystem. Find the difference (expressed in percentage) between the exactcalculation using the reduced mass and the approximate calculation (donein class) using the only the mass of the electron.

b) Repeat the previous calculation for the muonic atom, and find the ”Bohrradius” of this new system.

c) The radius of a proton is about 10−15m. Using the hydrogen wavefunc-tions adjusted for the muonic system, calculate the probability of find theµ− inside the proton for the following states: n = 3, ` = 0;n = 3, ` =1;n = 3, ` = 2. To simplify calculations, you might want to justify using anapproximation of the form e−ξ ≈ 1− ξ if ξ is sufficiently small.

d) Discuss the trend with ` of your previous calculation in the context of thecentrifugal term `(` + 1)/r2 appearing in the effective potential that occursin the Schrdinger equation.

76. (10 points) Let |n`m〉 be any ket. Show r|n`m〉 must satisfy the selection rule

〈n′`′m′|r|n`m〉 = 〈n′`|r|n`〉δ``′δmm′ , (G.102)

where 〈n′`|r|n`〉 is the radial integral∫ ∞

0drr3R∗

n′`(r)Rn`(r) . (G.103)

(Note: you cannot evaluate this integral unless you are given supplementary infor-mation about the wavefunction.) You may assume r 7→ r in spherical coordinates.(Hint: use orthonormality of Y`m(θ, ϕ), i.e. 〈`′m′ | `m〉 = δ``′δmm′ .

77. (20 points) Converting from cartesian to spherical.

Problems 271

a) (2 points) Using Cartesian coordinates, write down explicitly the wave func-tion ψnxnynz(x, y, z) for the three–dimensional harmonic oscillator state withnx = 0, ny = 0, nz = 2.

b) (2 points) What is the energy E of this state.

c) (6 points) Convert z2 to spherical coordinates, and express the result as asum of Y`,m(θ, ϕ), i.e. write

z2 = r2

∑

`,m

c`mY`m(θ, ϕ)

. (G.104)

(Hint: only ` = 2 and ` = 0 should appear.)

d) (5 points) Using this, convert your Cartesian wavefunction to a linear com-bination of spherical wavefunctions ψn`m(r, θ, ϕ). Verify that the result isproperly normalized.

e) (5 points) Show that all the spherical wavefunctions ψn`m(r, θ, ϕ) enteringin the expression of ψnxnynz(x, y, z) have energy E identical to the energy ofthe system described by ψnxnynz(x, y, z).

78. (20 points) At time t = 0, a hydrogen system is described by the ket

|ψ(0)〉 = A(2|100〉+ |210〉+

√2|211〉+

√3|2, 1,−1〉

)(G.105)

where |n`m〉 is an eigenket of the hydrogen–atom Hamiltonian.

a) (3 points) Use orthonormality of ket vectors 〈n′`′m′ |n`m〉 = δn′nδ`′`δm′m tofind a constant A that will normalize |ψ(0)〉.

b) (3 points) Find 〈E〉.c) (10 points) What is the probability of finding the electron within 10−12

meters of the proton (at t = 0)? You may simplify calculations by justifyingthe use, in the appropriate context, of the approximation e−x ≈ 1−x. (Hint:the probability works out to ≈ 3.6× 10−6.)

d) (4 points) Find |ψ(t)〉.

79. (10 points) The p–representation.

Problems 272

a) (3 points) Write the Schrodinger equation for the one–dimensional harmonicoscillator using the p–representation of operators.

b) (7 points) By comparing with the solutions in the x–representation, find anexplicit expression for the two lowest energy wavefunctions ψ0(p) and ψ1(p)in p–space.

80. (10 points) [Griffiths # 2.35 .] A particle of mass m and kinetic energy E > 0approaches an abrupt potential drop V0. The potential for this problem is thus:

V (x) =

0 if x < 0−V0 if x ≥ 0.

(G.106)

a) What is the probability that it will reflect back if E = V0/3?

b) Do not do part b)

c) When a free neutron enters a nucleus, it experiences a sudden drop in po-tential from V = 0 to V = −12 MeV inside. Suppose a neutron, emittedwith kinetic energy 4MeV by a fission event, strikes such a nucleus. What isthe probability that it will be absorbed, thereby initiating another fission?(Hint: you calculated the probability of reflection in part i). Use T = 1−Rto get the probability of transmission through the surface.)

For this problem, Griffiths’ B is what we call Γ whereas his F is our τ . You areencouraged to read the comment in Prob. 2.34c), which, for us, translates as thefollowing: for a potential step of height V0, the probability of transmission is notsimply ττ∗ but rather

√(E − V0)/V0 ττ∗.

81. (15 points) [Griffiths # 2.38 .] A particle of mass m is in the ground state of theinfinite square well. Suddenly, the well expands to twice its original size, the rightwall moving from a to 2a– leaving the wavefunction momentarily undisturbed.The energy of the particle is now measured.

a) What is the most probable result for E? What is the probability of gettingthis result?

b) What is the next most probable result, and what is its probability?

c) What is the expectation value 〈E〉 of the energy? (Hint: If you find yourselfconfronted with an infinite series, try another method.)

82. (10 points) [Griffiths # 3.2 .]

Problems 273

a) For what range of ν is the function f(x) in the Hilbert space of square–integrable function on the interval (0, 1), i.e. for what range of ν is

∫ 1

0dx |f(x)|2 < ∞ ? (G.107)

b) For the specific case ν = 12 , is f(x) in the Hilbert space? What about xf(x)?

What about (d/dx)f(x)?

83. (15 points) [Griffiths # 3.22 .] Consider a three–dimensional vector spacespanned by an orthonormal basis |1〉, |2〉, |3〉. Kets |α〉 and |β〉 are given by

|α〉 = i|1〉 − 2|2〉 − i|3〉 , |β〉 = i|1〉+ 2|3〉 . (G.108)

a) Find 〈α| and 〈β| in terms of 〈1|, 〈2|, 〈3|.b) Find 〈α |β〉 and 〈β |α〉 confirm that one is the complex conjugate of the

other.

c) Find all nine matrix elements of the operator A = |α〉〈β| in the basis spannedby |1〉, |2〉, |3〉, i.e. find the nine quantities 〈i|A|j〉, where |i〉, |j〉 are elementsof the basis. Construct the matrix A representing the operator A. Is Ahermitian?

84. (20 points) [Griffiths # 3.37 .] The Hamiltonian operator H for a certainthree–level system is represented by the matrix

H 7→ H =

a 0 b0 c 0b 0 a

, (G.109)

where a, b and c are real.

a) If the system is initially in the state

|ψ(0)〉 =

010

, (G.110)

what is |ψ(t)〉? (Hint: only the eigenvectors of H have a simple time evolu-tion, and these eigenvectors form a complete set.)

Problems 274

b) If the system is initially in the state

|ϕ(0)〉 =

001

, (G.111)

what is |ϕ(t)〉?

85. (15 points) Creation and destruction operator. A particle of mass m is trappedin a harmonic potential such that ω =

√k/m. At t = 0, the wave function

describing this particle is given by

|Ψ(0)〉 =1√2s

∑n

|n〉 , (G.112)

where the sum over n goes from N − s to N + s with N À s À 1.



sult.

86. (20 points) Commutators of product operators.

a) (5 points) (The derivative rule). Show that, if A, B and C are any threeoperators, then

[A, BC] = [A, B]C + B[A, C] . (G.113)

This is known as the derivative rule since it resembles the rule for derivationof a product of functions.

Consider two sets of harmonic oscillator creation and destruction operators,a, a† and b, b† such that

[a, a†] = [b, b†] = 1l , [a, b] = [a, b†] = [a†, b] = [a†, b†] = 0 . (G.114)

b) (5 points) Let

X = 12~

(a†b + b†a

), Y = 1

2i~(a†b− b†a

), Z = 1

2~(a†a + b†b

).

Verify that

[X, Y ] = i~Z , [Y , Z] = i~X , [Z, X] = i~Y , (G.115)

i.e. these operators have commutation relations identical to the angularmomentum operators Lx, Ly, Lz. (Hint: use the derivative rule.)

Problems 275

c) (5 points) Find the operators J+ and J− that play the role of L+ and L− inthe theory of angular momentum.

d) (5 points) Consider the two–dimensional normalized oscillator state |na, nb〉,such that a|na, nb〉 =

√na|na − 1, nb〉 etc.. Show that, if na = 1

2(j + m) andnb = 1

2(j −m), then

J+|na, nb〉 =√

(j + m + 1)(j −m)|na + 1, nb − 1〉 , (G.116)

like the angular momentum result L+|jm〉.

87. (15 points) The virial theorem. When a force can be derived from a potential,the virial theorem of classical mechanics states that 〈T 〉 = 1

2〈~r · ∇V 〉 . For thespecial case of a potential of the form V = k rn, we then obtain

〈T 〉 = 12n 〈V 〉 . (G.117)

This problem examines the quantum version of this result.

a) Let A be any well–defined hermitian operator, and let |ψ〉 be any normalizedeigenstate of H:

H|ψ〉 = E|ψ〉 . (G.118)

Find 〈[A, H]〉 = 〈ψ|[A, H]|ψ〉.b) Let H = 1

2m(p)2 + V (r) be the Hamiltonian for a three–dimensional particlein a spherically–symmetric potential. Show that

[H, r] = − i~m

p , [H, p] = i~ (∇V (r)) . (G.119)

c) Use the result of part i) with A = r · p to show that

2〈ψ|T |ψ〉 = 〈ψ|(r

∂V

∂r

)|ψ〉 , (G.120)

where T is the kinetic energy.

d) If V (r) is of the form rn, show that the energy of an eigenstate with energyE is given by

E =(1 +

n

2

)〈ψ|V |ψ〉 . (G.121)

Problems 276

88. (25 points) The Pauli matrices [see Griffiths sec. 4.4]. Consider the three 2×2matrices

σx =(

0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

). (G.122)

Let Sx = 12~σx etc.

a) (5 points) Using matrix multiplication, show explicitly that the matricesSx, Sy, Sz satisfy the commutation relation

[Sx, Sy

]= i~Sz ,

[Sy, Sz

]= i~Sx ,

[Sz, Sx

]= i~Sy . (G.123)

(Note that these are identical to the commutation relations for Lx, Ly andLz.)

b) (5 points) What are the possible outcomes of measuring Sz, Sy and Sx,respectively?

c) (5 points) Suppose the system is a state with well–defined value of Sz. Whatis the probability of obtaining 1

2~ when measuring Sy?

d) (10 points) The states |ψ〉 that satisfy ∆Sy∆Sz = 12~|〈ψ|Sz|ψ〉|2 are called

intelligent. Show that, if a state |ψ〉 is an eigenstate of Sx − iαSy, where αis real, then |ψ〉 is intelligent.

89. (10 points) Let A be any hermitian operator. Let |ψ〉 be any state, and let|ψ⊥〉 be such that

A|ψ〉 = α|ψ〉+ β|ψ⊥〉 , (G.124)〈ψ⊥ |ψ〉 = 0 , (G.125)

with β real and non–negative. Show

α = 〈A〉 , β = ∆A =√〈A2〉+ 〈A〉2 . (G.126)

90. (10 points) Basic properties of the Schrodinger equation. (see Beiser Exercise5.8)

Problems 277

a) The time–dependent Schrodinger equation in one dimension reads

i~∂

∂tΨ(x, t) = − ~

2

2m

∂2

∂x2Ψ(x, t) + V (x)Ψ(x, t) . (G.127)

Show that, if Ψ1(x, t) and Ψ2(x, t) are separately solutions of Eqn.(G.127),then the linear combination Ψ(x, t) = α1Ψ1(x, t)+α2Ψ2(x, t) is also solutionfor arbitrary complex constants α1 and α2.

b) The time–independent Schrodinger equation in one dimension reads

− ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x) , (G.128)

where E is the eigenvalue of associated to ψ(x). Show that, if ψ1(x) andψ2(x) are separately solutions of Eqn.(G.128) with eigenvalues E1 and E2 re-spectively, then the arbitrary linear combination ψ(x) = α1ψ1(x)+α2ψ2(x) ,with α1 and α2 arbitrary complex constants, is not a solution of Eqn.(G.128)unless E1 = E2.

91. (40 points) Playing with wave functions. (You will need material from Griffithssections 1.3 to 1.6, and you may want to review Beiser sections 5.5 to 5.8 andGriffiths section 2.2, in particular Eqn.[2.28].)

A particle of mass m is confined to a one–dimensional region 0 ≤ x ≤ a. Thisis modeled by the potential

V (x) =

0 , if 0 ≤ x ≤ a ,∞ , otherwise.

(G.129)

At t = 0, the wavefunction is given by

ψ(x) = N(1 + cos

(πx

a

))sin

(πx

a

). (G.130)

a) (5 points) Find the constant N that will normalize the wavefunction.

b) (5 points) Show that the normalized wavefunction ψ(x) can be written asa sum of two normalized solutions to the time–independent Schrodingerequation for this problem. Give the energy of these two solutions.

c) (5 points) Find Ψ(x, t). (Hint: Griffiths Eqn.[2.17], Example 2.1)

d) (20 points) Compute ∆x and ∆p as a function of time using Ψ(x, t).

Problems 278

e) (5 points) Verify that Heisenberg’s uncertainty relation is satisfied for alltimes.

92. (10 points) Consider the one–dimensional wavefunction

ψ(x) = N(

x

x0

)n

e−x/x0 , (G.131)

where N , n and x0 are constants.Using Schrodinger’s equation, find the potential V (x) and energy E for which

this wavefunction is an eigenfunction. You may assume V (x) → 0 as x → ∞.(Hint: obtain an expression for E − V (x) independent of any solution ψ(x) anduse the condition at ∞ to deduce E, then V (x) for any x.)

93. (20 points) A problem on quantum weirdness. (see caption of Fig.5.12 ofBeiser) A classical turning point A is a value of x for which the potential energy isequal to the total energy of a particle; in principle, a particle has no kinetic energyat that point and must therefore “turn around” its motion.

a) Start with the harmonic oscillator wave function ψ1(x) of Eqn.(6.67) of thecourse notes, or Eqn.[2.85] of Griffiths (together with Eqn.[2.71]). Find itsenergy by inserting this wavefunction into the time–independent Schrodingerequation.

Note: the (miserable) discussion of Beiser around Eqn.(5.34) of his book isincomplete; the condition that a certain dynamical variable G be restrictedto discrete values Gn is that the wavefunction satisfy Eqn.(5.34) and benormalizable.)

b) Find the turning points of a classical particle in a harmonic oscillator well.

c) The region between the classical turning points will be referred to as theclassically allowed region. The regions not classically allowed are calledclassically forbidden. Find a definite integral expressing the probability offinding the particle in the classically forbidden region.

d) Evaluate the previous integral. You will need to do some numerical integra-tion to obtain the final result

In the Fermi-gas model of the nucleus, the nuclear constituents move indepen-dently one from the other in an average potential, much in the way that electrons

Problems 279

move independently from each other in a crystal. The nuclear potential has a formgiven by

V (r) = −V0

[1 + exp

((r −R0)

a

)]−1

. (G.132)

(This is called the Woods–Saxon potential, or the inverted Fermi potential.) Theparameters are:

R0 = r0 A1/3 ; V0 ≈ 50 MeV ; a ≈ 0.5×10−15 m ; r0 ≈ 1.2×10−15 m.

Here, A is the total number of nucleons (protons plus neutrons), and R0 is thenuclear radius.

This is the net potential felt by individual nucleons (protons and neutrons).We assume both nucleons to have identical masses:

m = mn = mp = 938.919MeV/c2 . (G.133)

You will also need the conversion factor ~c = 197.326MeV× 10−15m.

94. (20 points) ` = 2 states of 13250 Sn

The Woods–Saxon potential of Eqn.(G.132) is often approximated by a finitespherical well:

V (r) = −V0 for r ≤ R0 ,

0 for r > R0 .

V=0

r=R o

V=−Vo MeV

Figure 33: An approximation to the nuclear well.

a) Find the energy of all the ` = 2 states of 13250 Sn using this potential.

b) What is the largest possible energy of a bound state in this potential, i.e.what is the largest E so that E < 0? What is the angular quantum number` of this state?

Problems 280

95. The Woods–Saxon potential of Eqn.(G.132) is often approximated by thethree–dimensional harmonic potential

V (r) ≈ 12 mω2

0 r2 − 12 mω2

0 R20. (G.134)

a) (5 points) What is the largest possible energy of a bound state in this po-tential, i.e. what is the largest E so that E < 0?

b) (10 points) What is the probability of finding a nucleon outside the nucleusif the nucleon is in the lowest possible energy state?

c) (5 points) What is the average 〈r〉 for a nucleon in the lowest possible statewith ` = 1?

96. (20 points) Hydrogen atom.

a) Show that the most probable value of r for an electron in the lowest energystate of a hydrogen atom is a0.

b) Show that the most probable value of r for an electron in the nr = 2, ` = 1state of the hydrogen atom is 4a0.

c) Find 〈r〉 for an electron in the nr = 3, ` = 0 state of a hydrogen atom.

d) Find 〈r〉 for an electron in the nr = 3, ` = 1 state of a hydrogen atom.Compare this with the previous result. Is this compatible with the classicalpicture of increasing centripetal force with increasing value of `?

97. (10 points) Muonic atoms. Hydrogen–like atoms can be formed in which amuon is bound to a nucleus of charge Ze; such states are called muonic atoms.Muons are ”heavy electrons” in that they have the same electric charge (−e) aselectrons but are much heavier: mµc2 = 105.6MeV whereas mec

2 = 0.511MeV.Use the appropriately–modified hydrogenoid wavefunctions to calculate the

probability of finding a muon inside the 20983 Bi nucleus assuming the muon is in the

lowest energy state of this system. Use the parameters given at the start of theassignment to obtain the radius of 209

83 Bi. (Hint: reduced mass).

98. (10 points) Orthogonality of the angular functions. The angular part of thewavefunction is of the form Y (θ, ϕ) = Θ(θ)eimϕ, where Θ(θ) satisfies

1sin θ

d

dθ

(sin θ

dΘdθ

)+

(` (` + 1)− m2

sin2 θ

)Θ(θ) = 0 . (G.135)

Problems 281

a) (1 points) Make the change of variable x = cos θ to obtain the so–calledassociated Legendre equation:

(1− x2)Θ(x)− 2x Θ(x) +(

` (` + 1)− m2

1− x2

)Θ(x) = 0 . (G.136)

Let Θm` (x) denote a solution to Eqn.(G.136).

b) (2 point) Show this can be written as

d

dx

[(1− x2) (Θm

` (x))′]+

(` (` + 1)− m2

1− x2

)Θm

` (x) = 0 , (G.137)

where ′ denote differentiation w/r to the argument.

c) (5 points) Multiply Eqn.(G.137) by ΘmL (x). Write Eqn.(G.137) for L and

multiply it by Θm` (x), then subtract the two to recover

d

dx

[(1− x2)

(Θm

` (x)(ΘmL (x))′ − (Θm

` (x))′ΘmL (x)

)]+(L(L + 1)− ` (` + 1))Θm

` (x)ΘmL (x) = 0 .

(G.138)

d) (2 points) Integrate from −1 to 1 to show orthogonality over that interval.

99. (15 points) Orthonormality of wavefunctions. Let |nr `m〉 denote a hydrogenket, so that

〈r, θ, ϕ |nr `m〉 = ψnr`m(r, θ, ϕ) = Rnr`(r)Y m` (θ, ϕ) , (G.139)

is a hydrogen wave function. Consider now the state vector at t = 0

|Ψ(0)〉 =1√10

(|100〉+ |210〉+

√2 |211〉+

√3 |21,−1〉

). (G.140)

a) (2 points) Show that, at least, 〈n′r `′m′ |nr `m〉 = δ`,`′δm,m′ . (Hint: use the〈rθϕ| basis and orthonormality of Y m

` (θ, ϕ).)

b) (2 points) Show that |Ψ(0)〉 is properly normalized, i.e. 〈ψ |ψ〉 = 1.

c) (2 points) Find |Ψ(t)〉 as a linear combination of kets.

For a system described by the state vector |Ψ(0)〉,d) (3 points) What is the average value of energy ?

Problems 282

e) (3 points) What is the average value of ~L · ~L?

f) (3 points) What is the average value of Lz?

100. (15 points) p-representation of the harmonic oscillator.

a) (5 points) Write the wavefunction ψ0(p) for the lowest energy state of a one–dimensional harmonic oscillator in the p-representation. (Hint: you mayeither write the Schrodinger equation in p-space and repeat the solution ofthe differential equation, or compute the Fourier transform of ψ0(x).)

b) (10 points) Find the probability that p2/(2m) > E0.

101. (20 points) The Wigner distribution in one dimension. Consider the function

W (x, p) =1~π

∫ ∞

−∞dy ψ∗(x + y) ψ(x− y) e2ipy/~ , (G.141)

known in the literature as a Wigner function or Wigner distribution. Using prop-erties of the δ-function, show

a) the probability density in position space

|ψ(x)|2 =∫ ∞

−∞dpW (x, p) , (G.142)

b) the probability density in momentum space

|ψ(p)|2 =∫ ∞

−∞dxW (x, p) . (G.143)

102. (Bonus: 10 points) The radial wavefunction R(r) near r = 0. In solv-ing the radial equation, we have introduced the auxiliary function χ(r) = r R(r)and showed that the solution to the radial equation for χ(r) is similar to a one–dimensional problem where a particle is restricted to r ≥ 0 in a potential Veff .The condition that the wavefunction be integrable is then

∫ ∞

0|R(r) |2 r2 dr =

∫ ∞

0|χ(r) |2 dr ≤ ∞ . (G.144)

In this problem, we want to justify the condition

limr→0

χ(r) → 0 . (G.145)

For small values of r and V not worse than 1/r at the origin, the wavefunctionχ(r) must behave like ∼ r`+1 or ∼ r−` since the `(`+1)/r2 term dominates. If weselect χ(r) ∼ r`+1, Eqn.(G.145) is automatically satisfied as ` ≥ 0.

Problems 283

a) (2 points) Suppose we choose instead χ(r) ∼ r−`, with ` 6= 0. Show that itis impossible to satisfy Eqn.(G.144) by showing that

∫ ∞

0|χ(r) |2 dr =

∫ r0

0|χ(r) |2 dr +

∫ ∞

r0

|χ(r) |2 dr (G.146)

is not normalizable even if r0 is chosen to be so small that χ(r) ∼ r−` holdsfor the first integral.

There remains χ(r) ∼ r−` with ` = 0, i.e. χ(r) ∼ c, where c is some constant.This implies R(r) ∼ c/r for small r, and thus ψ(r, θ, ϕ) ∼ c/r.

b) (5 points) Show that∇2(1/r) = −4πδ3(~r), where∇2 = ~∇·~∇ is the Laplacianin spherical coordinate. (Hint: First show that ∇2(1/r) = 0 if r 6= 0. To seewhat happens at r = 0, consider a small sphere centered at the origin and useGauss’ theorem (the divergence theorem) and the identity ∇2ψ = ~∇·

(~∇ψ

).

c) (3 points) Thus, argue that unless Veff contains a δ function at the origin(which we assume it does not), the choice c 6= 0 is untenable.

103. (30 points) Imagine a system in which there are just two linearly independentstates:

|1〉 =(

10

), |2〉 =

(01

). (G.147)

The most general state is a normalized linear combination

|Ψ〉 = a|1〉+ b|2〉 =(

ab

), (G.148)

with aa∗ + bb∗− = 1.Suppose the matrix representing the Hamiltonian is

H =(

ε0 gg∗ ε0

), (G.149)

where ε0 ∈ R and g ∈ C. The time–dependent Schrodinger equation is given inEqn.(??).

a) (5 points) Find the eigenvalues and normalized eigenvectors of H.

b) (10 points) Suppose the system starts out (at t = 0) in state |1〉. What isthe state of the system at time t?

Problems 284

The ammonia molecule, NH3, has a pyramidal shape, with the nitrogen atomat vertex and three hydrogen atoms at the base. Classically, the nitrogen atommay be at one of two symmetric equilibrium positions: to the “left” or “right” theplane of the hydrogen atoms. These states are denoted, respectively, by |NH3〉and |H3N〉.

c) (5 points) Given the energy separation ∆E = 9.84×10−5eV , can you recon-struct all the matrix elements (magnitude and phase) of the Hamiltonianmatrix?

d) (5 points) Which of the two eigenstates has the lowest energy? (Hint:even/oddness of corresponding wave functions and number of nodes.)

e) (5 points) Find the frequency of oscillation between the |NH3〉 and |H3N〉states.

104. (20 points) Given the matrix

X =

0 ω 00 0 ω2

1 0 0

, (G.150)

where ω = e2πi/3,

a) Find the determinant of this matrix,

b) Given that X can be written as X = U−1 ·D · U , where D is diagonal, findthe eigenvalues of X and the corresponding eigenvectors.

c) Verify that the sum of the eigenvalues is the trace of the matrix.

d) Verify that the product of the eigenvalues is the determinant of the matrix

105. (15 points) A particle of mass m is trapped in a harmonic potential suchthat ω =

√k/m. At t = 0, the wave function describing this particle is given

by |Ψ(0)〉 =1√2s

∑n

|n〉 , where the sum over n goes from N − s to N + s with

N À s À 1.


Problems 285


sult.

106. (60 points) (A review problem on harmonic oscillators and radial wave func-tions) Consider the two–dimensional isotropic harmonic oscillator, with potentialV = 1

2mω2(x2 + y2). Let

ax =

√~

2mω

(x +

i

mωpx

)a†x =

√~

2mω

(x− i

mωpx

)

ay =

√~

2mω

(y +

i

2ωpy

)a†y =

√~

2mω

(y − i

mωpy

),

where [ai, a†j ] = δij , i, j = x, y.

a) Let |ψ〉 = N a†xa†y|0〉, with ax|0〉 = ay|0〉 = 0. Find a constant N that willnormalize correctly |ψ〉.

b) Express H, the Hamiltonian for this system, in terms of ax, a†x, ay and a†y.

c) Is |ψ〉 an eigenstate of H? If yes, what is the energy associated with thisstate? If no, what is ∆E for this state?

d) Find |Ψ(t)〉, where |Ψ(0)〉 = |ψ〉.e) Using the known wave functions for the 1-d harmonic oscillator and the

expression of x and y in polar coordinates, express |ψ〉 in polar coordinates〈r, ϕ |ψ〉 = ψ(r, ϕ)

f) Let Lz = i~∂/∂ϕ. Show that Lz commutes with H.

g) What are the possible outcomes of measuring Lz for a system prepared inthe state |ψ〉?

h) What is the probability of measuring Lz = 0~ for a particle prepared in thestate |ψ〉?

i) Compute 〈Lz〉 and ∆Lz for the state |ψ〉.j) Show that the operators

i~a†xay , 12 i~


), i~axa†y

have commutation relations identical to L+, L−, Lz, and that H commuteswith 1

2 i~(a†xax − a†yay

).

Problems 286

k) Using 12 i~


), compute 〈Lz〉 and ∆Lz.

l) Using the polar coordinates, find an explicit expression for the operator12 i~


).

107. (30 points) Let |n`m〉 be denote the usual hydrogen atom ket and considerthe state

|Ψ(0)〉 = A(|311〉 − µ2|21,−1〉) , (G.151)

at t = 0, where µ is real.

a) Find a constant A that will normalize this state.

b) Find |Ψ(t)〉.c) Find 〈~L · ~L〉.d) Find ∆Lx,∆Ly and show the uncertainty relation is verified.

e) What is the probability of measuring the value Ly = 0~ for a system de-scribed by |Ψ(t)〉?

f) What is the most probable value of Ly at t = 0?

108. (25 points) Radial momentum operator. We know that px 7→ −i~ ∂∂x , x 7→ x ,

with[x, p ] = i~ . (G.152)

Let

pr 7→ −i~(

∂

∂r+

1r

), r 7→ r , (G.153)

and f(r) be any function of r.

a) Compute pr r f(r)− r pr f(r) and compare with Eqn.(G.152).

b) We have seen the radial part of the Schrodinger equation can be written as

− ~2

2mr2

d

dr

(r2 d

dr

)R(r) + (V (r)−E)R(r)− ~2

2m

`(` + 1)r2

R(r) = 0 .

Show that the term

− ~2

2mr2

d

dr

(r2 d

dr

)R(r)

is nothing but (pr)2 R(r).

Problems 287

c) To establish the conditions under which pr is hermitian, show that the condi-tion 0 = 〈ψ | prψ〉−〈ψ | prψ〉∗, where ψ(r, θ, ϕ) is a square integrable function,leads to the restriction

limr→0

r ψ(r, θϕ) = 0 , (G.154)

showing that χ(r) = r R(r) must go to zero at the origin, as shown ourstudy of the radial solutions.

d) From the previous calculations, one defines the radial momentum operatorto be the operator of Eqn.(G.153). Find 〈pr〉 and 〈p2

r〉 for a hydrogen atomsystem with n = 1 and ` = 0. Compute ∆pr (Remember to properlyintroduce an extra r2 factor in your calculation of the average value.)

e) Use the results for 〈r〉 and 〈r2〉 found at Eqn.(8.42) and (8.43) of the Schaumto compute ∆r, and compute the uncertainty product ∆pr∆r. Is this com-patible with the uncertainty principle?

Nota: although pr is hermitian, no observable is associated with this operator.To show this, note that, for any ω, the solution to pr f(r) = ω f(r) is, to within aconstant,

f(r) ∝ eiωr/~

r, (G.155)

which never satisfies the condition of Eqn.(G.154). The eigenvalue problem for pr

has no physically valid solution.

PHY3113: Quantum mechanics I October 27, 2002

Mid-Term examination

If need be, you may refer to the following results: for m,n ∈ Z:

1

π

∫ 2π

0

dx sin(mx) sin(nx) =1

π

∫ 2π

0

dx cos(mx) cos(nx) = δmn

∫ 2π

0

dx sin(mx) cos(nx) = 0 .

1. (total: 40 points) Let

R(θ) =

(cos θ − sin θsin θ cos θ

). (1)

a) (10 points) Knowing that λ1 = eiθ is an eigenvalue of R(θ), Find all the other eigenvalues ofR(θ).

b) (10 points) Is R(θ) a hermitian matrix? Justify your answer.

c) (20 points) Find the eigenvectors of R(θ) as a function of the angle θ.

2. (total: 50 points) Consider a particle with E > 0 trapped in the potential

V (x) =

∞ for x < 0,0 for 0 < x < L/2,−V0 for L/2 < x < L,∞ for x > L

(2)

a) (5 points) What is the time–independent Schrodinger equation for this system?

b) (5 points) In the regions x < 0 and x > L, ψ(x) must be 0. Explain briefly why.

c) (10 points) In the region 0 < x < L/2, ψ(x) can be written as ψ(x) = A sin(kx). Explain brieflywhy.

d) (10 points) In the region L/2 < x < L, ψ(x) can be written as ψ(x) = F sin(k′(x−L)). Explainbriefly why?

e) (10 points) Find an equation that will determine the possible energies of a particle in this well.

f) (10 points) Using the graphs that are supplied, find the number of states having energy E > 0but less than 25V0 assuming V0 = 9~2/2mL2. Estimate the energies of the lowest three statessatisfying the above conditions.

3. (total: 25 points) Given the eigenstates for the infinite square-well potential between 0 and L:

〈x |ψn 〉 = ψn(x) =

√2

Lsin(nπx/L) , (3)

a) (10 points) Show that 〈 p 〉n ≡ 〈ψn| p |ψn〉 = 0 .

b) (5 points) Explain briefly how you would prove the more general result 〈 p 〉ψ = 0 for any wavefunction ψ(x) describing a particle in an infinite well (ψ(x) is not necessarily an eigenstate of

H).

b) (10 points) Explain briefly how 〈 p 〉ψ = 0 can be expected if |ψ〉 describes any bound state (i.e.any state for which ψ(±∞) = 0.

1

1

PHY3113: Quantum Mechanics I March 2, 2004

MidTerm examination 2004

This is an open book exam.There are three questions on two pages. The last page contains graphs relevant to one question.You are allowed 90 minutes.The maximum mark of this exam is 40. If you accumulate more than 40 point, you final mark will be 40/40.

1. (20 points) Under some circumstances, the motion of an electron in a solid is restricted to two dimensions, withthe electron evolving in a harmonic potential of the form

V0(x, y) = 12mω2x2 + 1

2mω2y2 (1)

When subjected to a uniform electric field ~E = yE directed along the y axis, the potential is modified to

V (x, y) = 12mω2x2 + 1

2mω2y2 − eEy . (2)

1. (5 points) Write the time–independent Schrodinger equation for the particle in the potential V0, and give thetwo lowest energies of the electron in the potential V0 (Just state what these energies are.)

2. (5 points) Show the lowest two energies of the electron in the potential V (x, y) are ~ω +D and 2~ω +D, whereD is a constant. (Hint: 1

2mω2y2 − eEy = 12mω2(y − eE

ω2m )2 − 12

e2E2

ω2m .

3. (5 points) Give an explict expression in terms of x, y and t for the time–dependent wave function Ψ0(x, y, t),solution of the Schrodinger equation with potential V (x, y) and having the lowest energy.

4. (5 points) Estimate the average values 〈x〉 and 〈y〉 for an electron described by Ψ(x, y, t).

2. (10 points) Consider the wave function at t = 0

Ψ(x, 0) = Ax(x− L) , A ∈ C . (3)

1. Why can this wave function describe a particle trapped in a one–dimensional well having infinitely high wallsat x = 0 and x = L? Explain your answer carefully.

2. Is this an eigenstate of the Hamiltonian operator for this system?

3. Why can this wave function describe a particle trapped in this well? Explain your answer carefully.

4. Is it possible to normalize the wave function? If yes, do so. If not, explain why it is not possible.

5. For this wave function, is it true that Ψ(x, t) = Ax(x− L)e−iEt/~? If yes, find E; if no, explain why this is nottrue.

Please turn over

2

3. (20 points) The ammonia molecule. The ammonia molecule, NH3, is in the form of pyramid, with the nitrogenatom at vertex and three hydrogen atoms at the base. The nitrogen atom may be at one of two symmetric equilibriumpositions: “above” or “below” the plane of the hydrogen atoms.

The potential energy for the motion of nitrogen is illustrated in figure 1 as the “W”–shape curve. The figure alsoillustrates the approximation of this potential by a infinite one–dimensional square well with a central bump of heightV0 = 0.245eV. The infinite walls are located at ±b = ±0.5548 × 10−10m; the central bump extends from −a to +a,with a = 0.2298× 10−10m.

V

a

b

o

b−a

FIG. 1: The potential for the nitrogen atom of the ammonia molecule. The exact potential (wavy line) is approximated by aninfinite square well with a bump in the middle (heavy line).

1. (2 points) Let x denote the position of the nitrogen atom. Show the wave function

ψ(x) =

B sin(k(b + x)) if − b ≤ x ≤ a ,

A (eκx + e−κx) if − a < x < a ,


(4)

is even under the transformation x → −x.

2. (6 points) Justify the use of this wave function in searching for even solutions with E < V0: i) discuss why oneneed a combination of exponentials in the central region and an oscillatory outside the bump, and ii) identifythe boundary conditions at x = ±b and show ψ(x) satisfies these conditions.

3. (6 points) Let

κ =

√2m(V0 − E)

~2, k =

√2mE

~2. (5)

Show the energies of the even wave functions can be found from the consistency equation

1κ

(eκa + e−κa

eκa − e−κa

)= −1

ktan(k(b− a)) . (6)

4. (6 points) Let

ξ =

√2mV0 a2

~2, z =

E

V0. (7)

Using the correct graph from the ones supplied on the next page, identify the number of states with E < V0

and estimate their respective energy. You will need the following data:

mNH3 : 2300 MeV/c2 ~c : 197.3 MeV× 10−15mb : 0.5548× 10−10m a : 0.2298× 10−10m

2mV0 a2

~2: 15.3182

√2mV0 a2

~2: 3.9138

−1 + b/a : 1.4143

(8)

In case of emergency, use graph 1.

3

–4

–3

–2

–1

00.2 0.4 0.6 0.8 1z

Graph 1: Solutions to − (e3.91√

1−z + e−3.91√

1−z)(e3.91

√1−z − e−3.91

√1−z)

√1− z

=tan(5.54

√z)√

z.

–4

–3

–2

–1

00.2 0.4 0.6 0.8 1z


1−z + e−3.91√

1−z)(e3.91

√1−z − e−3.91

√1−z)

√1− z

=tan(3.91

√z)√

z.

–4

–3

–2

–1

00.2 0.4 0.6 0.8 1z


z + e−3.91√

z)(e3.91

√z − e−3.91

√z)√

z=

tan(5.54√

1− z)√1− z

PHY3113: Intermediate electricity and Magnetism March 7th, 2006

MidTerm examinationTime allocated: 80 minutes

This exam contains 3 questions.Only pen or pencil, eraser and calculator allowed. Everything else is supplied with the exam.Remember to obtain a cheat sheet. If you use any equation from the cheat sheet, please indicate so at theappropriate place on your copy.

This maximum score for this exam is 25.If you accumulate more than 25 marks, your score will be 25.

1. (10 points) As a first approximation, the force between a neutron and a proton can be described by an attractivesquare-well potential, given in Eqn.(1) and illustrated in Fig. 1.

V (r) = −V0 for r ≤ R0 ,

0 for r > R0 .(1)

Here, r is the relative radius between the two nucleons.

V=0

r=R o

V=−Vo MeV

FIG. 1: An approximation to the nuclear well.

The range R0 of the nuclear force is known to be of the order of the Compton wavelength of the pion:

R0 =~

mπc, (2)

where mπ = 139MeV/c2 is the mass of the pionKnowing that the proton–neutron has a only one bound state with energy −2.22MeV, and knowing this state has

angular momentum ` = 0, find the depth of the potential V0 as follows:

a) Show that V0 can be found from the solution to the consistency equation

tan(kR0) = −k

κ, (3)

where k =

√2µ(V0 − ε)

~2, κ =

√2µε

~2and µ the reduced mass of the proton–neutron system. Here ε = 2.22 is

the negative of the bound state energy.

b) Let z =V0

ε. Reduce the transcendental equation to an equation in terms of z and use one of the graphs supplied

at the end of the test to estimate the value of V0. The mass of the proton can be taken to be the mass of theneutron, with both equal to 940MeV/c2. You may also need ~c = 197MeV×10−15m.

1

2. Let

ψn(ξ) =

√2L

sin(

nπξ

L

), (4)

denote a solution to the time–independent Schrodinger equation for a infinitely deep well. Consider the time–dependent wavefunction

Ψ(x, y, 0) = iψ1(x− x0)ψ2(y) , (5)

at t = 0, where x0 is a real constant.

a) (2 points) Find 〈E 〉 at t = 0.

b) (6 points) Find Ψ(x, y, t) and the corresponding probability density.

c) (2 points) Given the relations∫

dξ ξ sin2 (nπ ξ) =ξ2

4− cos (2nπξ)

8n2π2− ξ sin (2nπξ)

4nπ, (6)

∫dξ sin2 (nπ ξ) =

ξ

2− sin (2nπξ)

4nπ, (7)

find 〈x 〉. (In case of doubt, use the wavefunction at t = 0.)

3. (10 points) A positronium atom is formed when a positron (e+) and an electron (e−) ”orbit” around one another.Given that the reduced mass of the system if 1

2me, where me is the electron mass, find

a) the most probable distance between the positron and the electron if the system is in its lowest energy state,

b) the average distance between e+ and e− if the system is in its lowest energy state.

You might need to know that the Bohr radius for the hydrogen problem is

a0 =4πε0~2

me e2. (8)

and the following integral:∫

dξ ξn e−αξ = −ξn

αe−ξ +

n

α

∫dξ ξn−1 e−αξ . (9)

Please turn over for graphs

Examinations are formidable even to the best prepared, for the greatest fool may ask more than the wisest can answer.- Charles Caleb Colton

2

Some possibly useful graphs

10 20 30 40

-10

-8

-6

-4

-2

10 20 30 40

-10

-8

-6

-4

-2

FIG. 2:Left: The graphical intersection of the curves tan(0.4648

√z − 1) and −√z − 1.

Right: The graphical intersection of the curves tan(0.3289√

z − 1) and −√z − 1.

10 20 30 40 50 60

-2

-1.75

-1.5

-1.25

-1

-0.75

-0.5

-0.25

10 20 30 40

-2

-1.75

-1.5

-1.25

-1

-0.75

-0.5

-0.25

FIG. 3:Left: The graphical intersection of the curves tan(0.3289

√z − 1) and −

p(z − 1)/z.

Right: The graphical intersection of the curves tan(0.4648√

z − 1) and −p

(z − 1)/z.

3

PHY3113: Quantum Mechanics I Feb. 25th, 2007

MidTerm examination

Time allocated: 80 minutes. This exam contains 3 questions.Only pen or pencil, eraser and calculator allowed. Everything else is supplied with the exam.Remember to obtain a cheat sheet. If you use any equation from the cheat sheet, please indicate so at theappropriate place on your copy.


You may (or you may not) need the following:

d

dxsechx = − sechx tanh x ,

d2

dx2sechx = − sech3x + sechx

(1− sech2x

),

d3

dx3sechx = 5 sech3x tanh x− sechx tanh x

(1− sech2x

). (1)

1. (17 points) The wavefunction at t = 0 for a particle of mass m evolving in a one–dimensional harmonic potential is

Ψ(x, 0) = A(ψ0(x) +

√5 i ψ2(x)

), A ∈ C , (2)

where ψn(x) is the n’th excited state of the system.

a) (3 points) What is Ψ(x, t)?

b) (3 points) Is this an eigenstate of the Hamiltonian for the system?

c) (2 points) Find a value for the constant A that will normalize Ψ(x, 0). Assume∫∞−∞ dxψ∗n(x)ψm(x) = δmn.

d) (5 points) What is 〈E〉? What is ∆E?

e) (2 points) Is the wavefunction even, odd or neither even nor odd under the transformation x → −x?

f) (2 points) Find 〈x〉 at t = 0.

2. (6 points) A particle of mass m = 1/2 evolves in the potential V (x) = −2 sech2x (~ = 1), where sech x = 1/ cosh x.

a) (4 points) Show that the wavefunction ψ(x) = A sechx with A a constant satisfies the time–independentSchrodinger equation and find the energy corresponding to this solution.

b) (2 points) Given the graph of Fig. 1, explain if this wavefunction is or is not the solution with lowest energy.

-4 -2 2 4

0.2

0.4

0.6

0.8

1

FIG. 1: The graph for the function sech x.

1

3. (7 points) A particle evolves in a plane under the influence of the potential V (x, y) = 12mω2(x2 + y2).

a) (3 points) Write the wavefunction for the lowest energy solution for this problem. Express you answer in termsof ρ, where ρ2 = x2 + y2. What is the energy of this state?

b) (2 points) Knowing∫∞0

e−ξ2dξ =

√π/2, find 〈ρ〉 for this state or at least obtain an integral expression for 〈ρ〉.

(Hint: ρ dρ dϕ)

c) (2 points) Find the most probable value of ρ for this state.

2

1

PHY3113: Quantum mechanics I Dec.13th, 2001

Final examination

Nota: If you accumulate more than 100 points, you total mark will be 100%.

Section 1. Basics. Maximum: 25 points

1.1. (5 points) True or false: a complex number necessarily has a non–zero complex part. Explain your answer.

1.2. (15 points) Let M =(

3 22 0

). Find M

12 .

1.3. (10 points) Using Euler’s theorem, prove the equality

sin α + sin β = 2 sin 12 (α + β) cos 1

2 (α− β) . (1)

1.4. (10 points) Compute [x, p].1.5. (5 points) Evaluate all the solutions to i1/3.

Section 2. One–dimensional problems. Maximum: 30 points

2.1. (5 points) True or false: all solutions of the (one–dimensional) time–dependent Schrodinger equation are of theform ψ(x, t) = e−iEtΨ(x). Explain your answer.2.2. (20 points) Let α ∈ C, and let |n〉 be the harmonic oscillator state with energy (n + 1

2 )~ω.At t = 0, the “coherent state” |α(0)〉 is defined by

|α(0)〉 = e−|α|2/2

( ∞∑n=0

αn

√n!|n〉

)(2)

a) What is |α(t)〉, the coherent state at time t?

b) Show that a|α(t)〉 = α e−i~ωt|α(t)〉.c) Find 〈p〉(t) and 〈x〉(t) for |α(t)〉.d) Find 〈p2〉(t) and 〈x2〉(t) for |α(t)〉.e) Compute ∆x∆p for |α(t)〉.

2.3. (15 points) Let us define a potential V (x) by

V (x) =

+∞ if x < 0 ,0 if 0 ≤ x < L ,V0 if x ≥ L .

(3)

1. The wave function that is solution to Schrodinger’s equation for this potential must have amplitude 0∀ x < 0.Why?

2. The solution must have the form A sin(kx + ϕ) in the region 0 ≤ x < L. Why? What must be the value of ϕ ifthe wave function is to have no amplitude at x = 0?

3. The wave function must have the form Ce−κ(x−L) for x ≥ L. Why?

4. Show that, at x = L, the coefficients k and κ are related by

γ√

z = nπ − arctan√

z

1− z, (4)

where n is an integer (positive or negative), z = E/V0 and γ =√

2mV0L/~.

2

2.4. (15 points) Let

Ψ0(x) =(

1a√

π

) 12

e−x2/2a2, Ψ1(x) =

(1

2a√

π

)12

2(x

a

)e−x2/2a2

, (5)

where ω =√

k/m , a = (~/√

mk)12 , be two wave functions describing a particle of mass m in a harmonic potential

V = 12kx2 at t = 0.

a) What are the energies of Ψ0(x) and Ψ1(x)?

b) Let φ(x, t) = (iΨ0(x, t) − Ψ1(x, t))/√

2. What is the probability of finding the particle described by φ(x, t) inthe state Ψ0(x, t)? Does this probability depend on t?

Section 3. 3–dimensional problems. Maximum: 45 points

3.1. (5 points) True or false: In angular momentum theory, the choice of diagonalizing Lz rather than, say, Ly or Lx,is arbitrary. Explain your answer.3.2. (20 points) Consider a particle in a central potential. Given that |lm〉 is an eigenstate of ~L · ~L and Lz:

1. (15 points) Compute the sum (∆Lx)2 + (∆Ly)2,

2. (5 points) For which values of l and m does the above sum vanish?

3.3. (20 points) In some problems, it is more convenient to diagonalize Lx rather than Lz.

a) If |l, mz〉z is an eigenstate of ~L · ~L and Lz such that Lz|l, mz〉z = mz~|l,mz〉z, show that

|1〉x =1√6

(|1, 1〉z + 2|1, 0〉z + |1, 1〉z) (6)

is an eigenstate of Lx.

b) What are the quantum numbers l and mx specifying |1〉x?

c) Find two operators L+ and L− such that [Lx, L±] = ±L±.

d) Find all the other states having the same value of l but different values of mx.

3.4. (20 points) The (unnormalized) wave function

Ψ(r, θ, ϕ) =(

6− r

a0

)r

a0e−r/3a0 sin θ sin ϕ (7)

is an eigenstate of the Hamiltonian for the hydrogen atom.

a) What is the energy of this state?

b) Is this an eigenstate of ~L · ~L?

c) Construct explicitly another state, having the same energy as Ψ(r, θ, ϕ), but orthogonal to Ψ(r, θ, ϕ).

3.5. (20 points) To investivate the rotation–vibration spectrum of a diatomic molecule, Kratzer proposed the potential

V (r) = −2D

(a

r− a2

2r2

).

For this potential, the solution to the Schrodinger equation can be factorized into a product Ψ(r, θ, ϕ) =1rχ(r)Y m

l (θ, ϕ). Using the dimensionless abbreviations

ξ =r

a, β2 = −2

ma2

~2E , γ2 = 2

ma2

~2D , (8)

with γ > 0 and real β > 0, the radial equation for χ(r) takes the form

d2χ(ξ)dξ2

+[−β2 + 2

γ2

ξ− γ2 + l(l + 1)

ξ2

]χ(ξ) . (9)

3

a) (5 points) To solve, one sets

χ(ξ) = ξλ e−βξ f(ξ) . (10)

Explain why this form is justified.

b) (5 points) If we substitute Eqn.(10) in the radial equation, one finds that f(ξ) must satisfy

ξf′′(ξ) + (2λ− 2βξ)f

′(ξ) + (−2λβ + 2γ2)f(ξ) = 0 . (11)

with solution

f(ξ) = 1F 1(λ−γ2

β, 2λ; 2βξ) , (12)

where λ is solution to λ(λ− 1) = γ2 + l(l + 1) .

Explain how this leads to the quantization of energy, i.e. discrete energy levels En, and write En in terms ofthe mass of the particle and the parameters of the potential.

c) (5 points) Are there restrictions on the possible values of l and m for a given n?

d) (5 points) Write the complete spatial eigenfunction Ψ(r, θ, ϕ) for the lowest energy level of a particle in thispotential.

Section 4. Perturbation. Maximum: 10 points

4.1. (5 points) Is it possible to use perturbation theory to diagonalize the matrix(

2 44 1

)? Explain why?

4.2. (10 points) Suppose that a one–dimensional harmonic oscillator is perturbed by a potential of the type Vpert =−Fx. Find the first non–vanishing correction to the energy of the ground state, and the first non–vanishing correctionto the wave function of the ground state.

PHY3113: Quantum mechanics I December 2002Final Examination

Part 1

You must complete the following three problems.

1. (15 points) Let

V (x) = 12mω2x2 − 3

16

√mω

2~~ωx . (1)

1) If a particle trapped in the potential V (x) of Eq.(1) is restricted to the harmonic oscillator states |0〉 and |1〉,find the 2× 2 matrix representation of the operator H = p2/2m + V (x)

2) Find the eigenvalues and eigenvectors of H, expressed as linear combinations of |0〉 and |1〉.3) State the Heisenberg uncertainty principle for x and p, and verify that it holds for a particle described by the

state |ψ−〉, which is the eigenstate of the H with the lowest energy.

2. (15 points) A particle of mass m is subject to the three–dimensional potential

V (r) =

0 if r < r0

V0 if r > r0. (2)

1) Show that the minimum value of V0 needed to achieve two bound states (E ≤ V0) with zero angular momentumis V0 = 9π2~2/8mr2

0. You may refer to the graphs below for guidance.

2) With a potential of this depth, find the energy of the other bound state with ` = 0. You may refer to the graphsbelow.

0.2 0.4 0.6 0.8 1

-2

-1.5

-1

-0.5

0.5

1

1.5

2

0.2 0.4 0.6 0.8 1

-2

-1.5

-1

-0.5

0.5

1

1.5

2

0.2 0.4 0.6 0.8 1

-3

-2

-1

1

2

3

a) b) c)

The graphs for a) cot(π√

z/2) and −√

(1− z)/z, b) cot(3π√

z/2) and −√

(1− z)/z, c) cot(5π√

z/2) and−

√(1− z)/z, as a function of z.

3. (10 points) Let |`,m〉 denote angular momentum states with ~L · ~L|`,m〉 = ~2`(`+1)|`,m〉 and Lz|`,m〉 = m~|`,m〉.Compute the product ∆Lx∆Ly for the state |ψ〉 =

√23 |2, 1〉 − i

√13 |2, 0〉, and verify the Heisenberg uncertainty

relation.

1

Part 2

Complete any two of the following three problems. If you attempt more three problems, only the result of the best twowill be kept.

4. (30 points) Consider the three–dimensional harmonic oscillator state

|ψ〉 = N (1l + a†x)2|0〉 , (3)

where |0〉 is the ground state with ax|0〉 = ay|0〉 = az|0〉 = 0.

1) What are the possible outcomes of a measurement of the energy E of this state?

2) What is 〈E〉?3) If, at t = 0, a system is described by |Ψ(t = 0)〉 = |ψ〉, what is |Ψ(t)〉?4) What are the possible outcomes of a measurement of the angular momentum of a particle described by the state

|ψ〉?5) What is 〈Lx〉 for this state?

6) What is the probability of measuring Lz = 0~ for this state?

5. (30 points) Let ζ = − tan(θ)eiϕ. Define the angular momentum coherent state at t = 0 by

|ζ〉 =1

(1 + |ζ|2)`eζL+/~|`,−`〉 , (4)

where the exponential of an operator is defined by

eA = 1l + A + 12 A2 + 1

3! A3 + 1

4! A4 + . . . ,

and where |`,m〉 is the state such that

L2|`,m〉 = ~2`(` + 1)|`,m〉 , Lz|`,m〉 = m~ |`, m〉 .

1) (10 points) Let H = 12AL2 + ωLz, where ω is the rotational frequency. Find |ζ(t)〉 when ` = 1 and use this to

compute 〈Lz(t)〉.2) (10 points) Compute 〈E〉 and ∆E.

3) (10 points) At t = 0, compute 〈Lz〉 , ∆Lz , 〈Ly〉 and ∆Ly when ` = 1, and verify that the uncertainty relation issatisfied.

6. (30 points) At time t = 0, the wave function for the hydrogen atom is

|Ψ(0)〉 =1√10

(2i|1, 0, 0〉 − i|2, 1, 0〉+

√2|2, 1, 1〉+

√3|2, 1,−1〉

), (5)

where n, `, m are quantum numbers for hydrogen states.

1) (10 points) Calculate 〈E〉 for this state. What is the most probable value of E that one would measure in anexperiment?

2) (5 points) What is the probability of finding the particle described by this wave function with ` = 1 at t = 0?

3) (10 points) What is the probability of finding the electron withing 10−10 meter of the origin (at t = 0). Youmay use e−αr/a0 ≈ 1− αr/a0 and give your final answer in terms of the Bohr radius a0.

4) (5 points) Suppose that a measurement is made which shows that ` = 1 and Lz = +1~. Describe the stateimmediately after this measurement.

2

1

PHY3113: Quantum mechanics I April 20th, 2004Final Examination

This exam contains four questions on two pages.You are allowed 3 hours to complete all questions.An equation sheet is supplied. You can use your calculator.The maximum for this exam is 100. If you accumulate more than 100 points, your mark will be 100.

1. (30 points) Let |n〉 denote an eigenket of the operator

H =1

2m(p)2 + 1

2 mω2 (x)2 , (1)

and consider the state vector at t = 0

|Ψ(0)〉 = A (4|0〉 − 3i|1〉) . (2)

1.1) (5 points) Find a constant A that will normalize |Ψ(0)〉.

1.2) (5 points) Find |Ψ(t)〉.

1.3) (5 points) Find 〈x(t)〉, and show it is real.

1.4) (5 points) Find 〈p(t)〉, and show it is real.

1.5) (10 points) For |Ψ(0)〉, obtain ∆x∆p and show this product satisfies the Heisenberg uncertainty relation.

2. (30 points) Let |n`m〉 denote the eigenket describing an electron trapped in the potential of a hydrogen nucleus.The quantum numbers n, ` and m are the principal, angular momentum and azimutal quantum numbers, respectively.Let

|ψ〉 = 12

(√3|210〉 − |211〉

). (3)

2.1) (10 points) Find ∆E.

2.2) (5 points) What is the probability of measuring the value E2 = −13.6/4 eV for the energy of this system?

2.3) (5 points) What is the probability of measuring Lz = +1~ for the projection of ~L along z?

2.4) (10 points) Find ∆Lx ∆Ly and verify that this product satisfies the Heisenberg inequality.

3. (25 points) Let |nx, ny〉 be a two–dimensional harmonic oscillator ket in Cartesian coordinates, such that

ax|nx, ny〉 =√nx|nx − 1, ny〉 , a†x|nx, ny〉 =

√nx + 1|nx + 1, ny〉 , (4)

and similarly for ay and a†y.Let the Hamiltonian operator be H = ~ω

(a†xax + a†yay + 1 + a†xay + axa

†y

).

3.1) (5 points) Show the matrix representing H in the space spanned by the two states |1x, 0y〉 and |0x, 1y〉 is

H = ~ω(

2 11 2

). (5)

3.2) (5 points) What are the possible outcomes of measuring the energy of this system?

3.3) (5 points) Find the eigenvectors of H.

3.4) (10 points) If the system is in the state |Ψ(0)〉 = |1x, 0y〉 at t = 0, what is the probability of finding the systemin this state at later times t?

2

4. (20 points) Let us model the potential felt by single nucleon inside a 234U nucleus as an infinite spherical well ofradius r0 ≈ 1.25A1/3 × 10−15m, where A is the total number of nucleons.

4.1) (10 points) If |12m〉 is the ket describing the lowest energy state with angular momentum ` = 2~ havingprojection m~, and |100〉 the ket describing the lowest energy state with angular momentum ` = 0, identifywhich |12m〉 ket will have a non–zero matrix element 〈100|Q|12m〉 if

Q =

√16π5

r2 Y 02 (θ, ϕ) , (6)

(Hint: use orthogonality of the Y m` (θ, ϕ).)

4.2) (5 points) Write an explicit expression for the (unnormalized) wave function 〈 r, θ, ϕ | 12m 〉 for which Eqn.(6) isnon–zero.

4.3) (5 points) What is the energy of the transition |12m〉 → |100〉?

PHY3113: Quantum Mechanics I April 11th, 2006

Final examinationTime allocated: 180 minutes

This exam contains 4 compulsory questions. There is a choice of one between two bonus questions at the end.Only pen or pencil, eraser, calculator and cheat sheet are allowed. Remember to get a cheat sheet.The maximum for this exam is 55. If you accumulate more than 55 points, your final mark will be 55.

Problem 1. (10 points) Consider a particle of mass m trapped in a one-dimensional harmonic well. Let |n〉 denotea time-independent 1-dimensional harmonic oscillator ket.

a) (1 point) Write the Hamiltonian for this system in terms of the operators p and x.

Consider the system described by

|Ψ(0)〉 = α|0〉+ β|1〉 , (1)

where α and β are real, and such that α2 + β2 = 1.

b) (2 points) What is |Ψ(t)〉?c) (4 points) Find the values of t for which 〈x(t)〉 is 0.

d) (2 points) For α = β = 1/√

2, find ∆E as a function of time.

e) (1 point) Show that the maximum possible value of 〈x(t)〉 occurs when α = β = 1/√

2.

Problem 2. (14 points) A particle trapped in an infinitely deep one–dimensional square well with walls at x = 0 andx = L is described at t = 0 by the ket |φ〉, such that, inside the well,

φ(x) ≡ 〈x |φ 〉 =

√30L5

x (x− L) . (2)

a) (1 point) Write H in the x–representation inside the well.

b) (1 point) What is 〈x |φ 〉 outside the well? Justify briefly your answer.

c) (2 points) Is φ(x) an eigenstate of the Hamiltonian?

d) (2 points) Find 〈 p 〉 for a system described by φ(x).

Let 〈x |ψn 〉 be an eigenstate of H for this problem. Given

∫ L

0

dxx(x− L)

√2L

sin(nπx

L

)= −4

√2 L5/2

π3×

1n3

n odd,

0 n even ,(3)

e) (3 points) What are the possible outcomes of measuring the energy of this system?

f) (2 points) What is the probability of measuring E = E1 = π2~22mL2 ?

g) (3 points) If a δ-barrier is added at x = L/2, i.e. in the middle of the well, so that the potential inside the wellis now

V (x) =

∞ x < 0 ,Ω δ(x− L/2) 0 ≤ x ≤ L ,∞ , x > L ,

(4)

with Ω > 0, show that the new energies of the system are determined by solving the transcendental equation

tan√

2mEL

2~= −

√2E

m

~Ω

. (5)

1

Problem 3. (18 points) Let |nr`mz〉 be a familiar hydrogen atom eigenket. A particular hydrogen atom is describedat t = 0 by one of the following kets:

i) |Ψ(0)〉 =1√2

(|211〉 − |21,−1〉) , ii) |Ψ(0)〉 =1√2

(|333〉 − i|11,−1〉) , iii) |Ψ(0)〉 =i√2

(|200〉+ |102〉) .

a) (2 points) Identify the only ket that can represent the state of a hydrogen atom. Explain carefully your answer.

b) (2 points) What is |Ψ(t)〉 for your ket?

c) (5 points) What is 〈r(t)〉 for a system described by your ket?

d) (4 points) Show that the correct |Ψ(0)〉 is an eigenstate of Lx.

e) (2 points) What is the possible outcome, or what are the possible outcomes, of measuring the observable Lx fora system described by the correct |Ψ(0)〉?

f) (3 points) Find ∆Lx ∆Ly and show the uncertainty relation is satisfied.

Problem 4. (13 points) An idealized atom contains only two levels, denoted by |ψ1〉 and |ψ2〉. |ψ1〉 and |ψ2〉 areorthonormal. The Hamiltonian describing this idealized atom is plunged in an electric field of frequency ωf is given,in matrix form, by

H →

∆2

g eiωf t

g e−iωf t −∆2

, (6)

where ∆ = E1 − E2 − ~ωf and g is real.

a) (3 points) What are the eigenvalues of H?

b) (3 points) Suppose ∆ = 0. What are the eigenstates of H?

c) (2 points) Show the eigenstates are orthogonal.

d) (5 points) If ∆ = 0 and the system starts at t = 0 in the state |ψ1〉, what is the probability that the system willbe found in the state |ψ2〉 as a function of time?

Problem 5. Bonus You may attempt one (1) of the following two questions for bonus.

1) (5 points) A particle of mass m is constrained to move in 3d between two concentric spherical wall of radii r = aand r = b. In other words, the potential for this problem is

V (r) =

∞ , 0 < r ≤ a ,0 , a < r < b ,∞ , b ≤ r .

(7)

Find an expression for the energy of the lowest energy wavefunction with ` = 0.

2) (5 points) Let Lx, Ly and Lz be the usual angular momentum hermitian operators Show that, if |ψx〉 is aneigenvector of Lx, then 〈ψx|Ly|ψx〉 = 〈ψx|Lz|ψx〉 = 0, irrespective of the value of ` or any other quantumnumber.

2


Final examination

Time allocated: 180 minutes. This exam contains 4 compulsory questions plus one choice of bonus question.Only pen or pencil, eraser and calculator allowed. Everything else is supplied with the exam.Remember to obtain a cheat sheet. If you use any equation from the cheat sheet, please indicate so at theappropriate place on your copy.


You may (or you may not) need the following:∫

dx sin2(x) =x

2− 1

4sin(2x) ,

∫dxx sin2(x) =

x2

4− 1

8cos(2x)− 1

4x sin(2x) .

1. (10 points) A particle of mass m evolves in the potential

V (x) =

∞ if x < 00 if 0 ≤ x < L/2 ,

V0 if L/2 ≤ x < L ,

∞ if x > L ,

(1)

where V0 = 4~2 π2/2mL2. This potential is illustrated on the left of Fig.1.

L/2

Vo

L/2

0.2 0.4 0.6 0.8 1

-2

-1.5

-1

-0.5

0.5

1

1.5

2√1 − z tan π √z

− √z tanh π√ 1 − z

FIG. 1: Left: The potential of Eqn.(1). Right: The intersection of the curves√

1− z tan(π√

z) and −√z tanh(π√

1− z), wheretanh(x) = (ex − e−x)/(ex + e−x).

a) (2 points) Is the wavefunction

ψ(x) =

0 if x < 0√

2/L sin(3πx/L) if 0 ≤ x ≤ L ,

0 if x > L,

(2)

an eigenstate of the Hamiltonian for this system?

b) (2 points) What is the average energy 〈E〉 for a particle described by ψ(x) of Eqn.(2)?

c) (6 points) How many eigenstates of H have energy less than V0? (Hint: eκ(x−L) − e−κ(x−L)) If there is one ormore bound state, what is (or what are) the energy of these states? You may or may not need the graph on theright of Fig.1 to answer this. Justify your answer carefully.

1

2. (15 points) Suppose a particle of mass m trapped in a harmonic well is found in the state |n〉, where H|n〉 = En|n〉.You may assume

H =1

2mp2 + 1

2 mω2x2 = ~ω(a†a + 1

2 1l)

. (3)

a) (5 points) Show that [ x, H ] = i~p/m and [ p, H ] = −i~mω2x.

b) (5 points) Show that

〈n|[H, x p ]|n〉 = 0 . (4)

c) (5 points) Show that 〈n|p2|n〉/2m = 12mω2〈n|x2|n〉. One way (but not necessarily the only way) to do this is to

use the result of part a) to evaluate the commutator in Eqn.(4).

3. (15 points) Consider the linear combination of hydrogen–atom kets

|ψ〉 = − 12 |211〉 − i

√2

2 |210〉+ 12 |21,−1〉 . (5)

a) (5 points) Show this is an eigenstate of Ly. What is the outcome of measuring Ly for a system described by|ψ〉?

b) (2 points) What are the possible outcomes of measuring Lz for a state described by |ψ〉?c) (3 points) What is the most probable value of Lz?

d) (5 points) What the most probable value of r for a system described by |ψ〉?

4. (15 points) An idealized two–level system is described by the Hamiltonian matrix

H = ~ωo

(1

√3√

3 −1

). (6)

a) (2 points) What are the possible value of the energy of this system?

b) (3 points) What are the eigenvectors of the system?

c) (10 points) If the system starts initially in a state |Ψ(0)〉 =(

01

), what is the probability that it will be found

in the state(

10

)as a function of time?

(Hint: the eigenvectors have a relatively simple form.)

Please turn over for bonus questions

2

Bonus questions

You may attempt one of the following two questions for 5 additional marks.

5. (10 points) Let A be any hermitian operator. Let |ψ〉 be any normalized state, and let |ψ⊥〉 be another normalizedstate such that

A|ψ〉 = α|ψ〉+ β|ψ⊥〉 , (7)〈ψ⊥ |ψ 〉 = 0 , (8)


α = 〈A〉 , β = ∆A =√〈A2〉+ 〈A〉2 . (9)

6. (10 points) Show that Heisenberg’s uncertainty relation holds for x and p when a quantum state is described by aharmonic oscillator ket |n〉.

3

PHY3113: Quantum Mechanics I Feb. 27th, 2009

MidTerm examination

Time allocated: 55 minutes. This exam contains 3 questions plus one bonus. This exam is out of 20, with amaximum mark of 20.Only pen or pencil, eraser and calculator allowed. Everything else is supplied with the exam.Remember to obtain a cheat sheet. If you use any equation from the cheat sheet, please indicate so at theappropriate place on your copy.

1. (9 points) A particle of mass m is trapped in a harmonic potential V = 12mω2x2. At t = 0, the wavefunction

describing this system is

Ψ(x, 0) =(

1 + i

2

)ψ0(x) +

1√2

ψ1(x) . (1)

a) (1 point) What is Ψ(x, t)?

b) (1 point) Is Ψ(x, 0) correctly normalized? If no, normalize it. You may find it convenient to assume orthogonalityof the harmonic oscillator wavefunction: 〈ψi |ψj 〉 = δij .

c) (3 points) What is the average value of energy for a system described by Ψ(x, 0)?

d) (4 points) What is 〈x〉 for a system described by Ψ(x, 0)? You may find it useful to know that∫ ∞

−∞dx e−λx2

=√

π√λ

,

∫ ∞

−∞dxx2n+1e−λx2

= 0 ,

∫ ∞

−∞dxx2e−λx2

=√

π

2λ3/2(2)

2. (9 points) Consider the problem of a particle trapped of mass m in a finite well, extending from −a to +a, havingdepth −V0, where V0 > 0. The particle is bound so E = −ε < 0, i.e. ε > 0.

a) (5 points) Given that the lowest energy solution is even, show that, if

a =~π

4√

mV0

(3)

the lowest energy state has energy E = −V0/2. (Hint: Assume ε = V0/2, recall tan(π/4) = 1 and solve for a.)

b) (4 points) Using the value of a given in Eq.(3) and one of the graphs provided on the next page, determine ifthere is an odd bound state solution. Indicate which one of the graphs is appropriate, and explain your answercarefully.

3. (4 points) A particle of mass m is trapped in a potential of the form

V (x) = V0e+λx2/2 (4)

a) (2 points) Determine approximately the energies of the lowest two states of the particle.

b) (2 points) Do you expect your estimates to be greater or smaller than the first two true energies of the system?

(Hint: expand near the minimum, or alternatively ez ≈ 1 + z for small z.)

4. (5 points) Bonus. A particle of mass m is trapped in a δ potential given by V (x) = −Ω δ(x). The particle is bound,so that E = −ε < 0, so that ε > 0.

Given that the wavefunction describing this system can be written in the form

ψ(x) =

Aeκx for x < 0,Ae−κx for x > 0. , (5)

a) (1 point) Show that the choice A = i√

κ normalizes ψ(x).

b) (4 points) Show that, if

x0 =~2

2mΩln(2) (6)

then the probability of finding the particle in the range −x0 to +x0 is 12 . (Hint: to complete the calculation,

you will need to relate κ to the parameter Ω of the potential by find the energy of the bound state.)

1

0.2 0.4 0.6 0.8 1.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

(a)tan(√

24

π√

1− z) andp

z/(1− z)

0.2 0.4 0.6 0.8 1.0

0.5

1.0

1.5

2.0

2.5

3.0

(b)tan( π4√

2

√1− z) and

pz/(1− z)

0.2 0.4 0.6 0.8 1.0

-6

-4

-2

2

4

(c)tan(√

2 π√

1− z) and −p

(1− z)/z

0.2 0.4 0.6 0.8 1.0

-5

-4

-3

-2

-1

1

2

(d)tan(√

24

π√

1− z) and −p

(1− z)/z

FIG. 1: Intersections of some functions.

2


Final examination

Time allocated: 180 minutes. This exam contains 6 questions plus one bonus. This exam is out of 55, with amaximum mark of 55.Only pen or pencil, eraser and calculator allowed. Everything else is supplied with the exam.Remember to obtain a cheat sheet. If you use any equation from the cheat sheet, please indicate so at theappropriate place on your copy.

You may (or you may not) need the following:

d

dxsechx = − sechx tanhx ,

d2

dx2sechx = − sech3x+ sechx

(1− sech2x

),

d3

dx3sechx = 5 sech3x tanh x− sechx tanh x

(1− sech2x

). (1)

1. (6 points) A particle of mass m = 12 evolves in the potential V (x) = −2 sech2x (~ = 1), where sechx = 1/ coshx.

a) (2 points) Write the time–independent Schrodinger equation for this problem (take ~ = 1 and remember m = 12 ).

b) (4 points) Show that the wavefunction ψ(x) = A sechx with A a constant satisfies the time–independentSchrodinger equation and find the energy corresponding to this solution.

c) (2 points) Given the graph of Fig. 1, explain if this wavefunction is or is not the solution with lowest energy.

FIG. 1: The graph for the function sech x.

2. Let |n〉 denote a solution to the Schrodinger equation for the 1–dimensional infinite potential. A particle of massm is described by

|ψ〉 = N(−|1〉+ i

√2|3〉

)(2)

a) Find a value of N that will normalize this ket and write 〈x |ψ 〉.

b) Is |ψ〉 an eigenstate of the Hamiltonian? If yes, what is E? If no, what is ∆E?

c) What is the most probable value of E, and what is the probability of obtaining this value?

d) What is 〈x〉? (Hint:∫ 1

0dxx sin2(nπx) = 1

4 for integer n.)

e) What is |Ψ(t)〉 if |Ψ(0)〉 = |ψ〉?

3. Show that, for a system described by the harmonic oscillator ket |n〉,

∆p∆x = (n+ 12 )~ (3)

1

4. The Pauli matrices have the form

σx = 12

(0 11 0

), σy = 1

2

(0 −ii 0

), σz = 1

2

(1 00 −1

)(4)

Consider the matrix

Sα = σx − iασy (5)

where α is a real number.

a) Is Sα hermitian?

b) What are the eigenvalues of Sα?

c) Suppose α = 2. Show that |ψ1〉 = 12

(−i√

3

)is an eigenvector of S2.

d) What is the other eigenvector? Is it orthogonal to |ψ1〉?

e) Find (∆σx) (∆σy) for |ψ1〉 and show that this product is in agreement with Heisenberg’s uncertainty relation.

5. Let |nr`mz〉 be a familiar hydrogen atom eigenket. A particular hydrogen atom is described by

|φ〉 = 12 |311〉 − 1

2 |31,−1〉+ i2 |211〉 − i

2 |21,−1〉 (6)

a) Show that |φ〉 is an eigenstate of Lx.

b) What are 〈Lz〉, 〈Ly〉 and 〈Lx〉 for the valid ket?

c) What are the possible outcomes of measuring Lz, and what are the probabilities of these outcomes?

d) What is the probability of measuring the value Lx = +~ for a system described by |φ〉?

e) Given

x =

√2π3r(Y −1

1 (θ, ϕ)− Y −11 (θ, ϕ)

)(7)

find 〈ψ|x|ψ〉 using |φ〉. (Hints:∫dϕ cos3(ϕ) = 3 sinϕ

4 + sin 3ϕ12 ,

∫dθ sin4(θ) = 3θ

4 −sin 2θ

4 + sin 4θ32 .)

6. For a hydrogen atom described by ψ200(r, θ, ϕ):

a) Show that a minimum in the radial probability density occurs at r = 2a0.

b) What is the probability of finding an electron within a distance 2a0 of the nucleus? (Hints:∫dxx2e−x =

−e−x(x2 + 2x+ 2),∫dxx3e−x = −e−x(x3 + 3x2 + 6x+ 6),

∫dxx4e−x = −e−x(x4 + 4x3 + 12x2 + 24x+ 24).)

7. Bonus You may attempt one (1) of the following two questions for bonus.

1) (5 points) A particle of mass m is constrained to move in 3d between two concentric spherical walls of radiir = a and r = b. In other words, the potential for this problem is

V (r) =

∞ , 0 < r ≤ a ,0 , a < r < b ,∞ , b ≤ r .

(8)

Find an expression for the energy of the lowest energy wavefunction with ` = 0.

2) (5 points) Let A be any hermitian operator. Let |ψ〉 be any normalized state, and let |ψ⊥〉 be another normalizedstate such that

A|ψ〉 = α|ψ〉+ β|ψ⊥〉 , (9)〈ψ⊥ |ψ 〉 = 0 , (10)


α = 〈A〉 , β = ∆A =√〈A2〉+ 〈A〉2 . (11)

2

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