MODERN PHYSICS BUT MOSTLY QUANTUM MECHANICSaoitani.net/Modern_Physics_2013.pdf · 2013. 8. 31. ·...

MODERN PHYSICSBUT MOSTLY

QUANTUM MECHANICS

Seeing the invisible, Proving the impossible, Thinking theunthinkable

Modern physics stretches your imagination beyondrecognition, pushes the intellectual envelope to the limit,

and takes you to the forefront of human knowledge.

Masayasu AOTANI(青谷正妥)

Kyoto UniversityKyoto, Japan

Anax parthenope （ギンヤンマ）

2013No insects, no life!

Modern Physics

Masayasu AOTANI（青谷正妥）

Spring 2013

Copyright c⃝1991–2013

Masayasu AOTANI

Permission is granted for personaluse only.

Contents

1 Classical Physics vs. Modern Physics 13

2 Mathematical Preliminaries 152.1 Linear Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . 152.2 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . 192.3 L2-Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.4 The Braket Notation . . . . . . . . . . . . . . . . . . . . . . . 272.5 Vector Subspace . . . . . . . . . . . . . . . . . . . . . . . . . 372.6 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . 372.7 Matrix Representation of Linear Operators . . . . . . . . . . . 39

2.7.1 Matrix Representations of Operator Products . . . . . 442.8 The Adjoint of an Operator . . . . . . . . . . . . . . . . . . . 452.9 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . 472.10 Special Types of Operators . . . . . . . . . . . . . . . . . . . 49

2.10.1 Hermitian Operators . . . . . . . . . . . . . . . . . . . 492.10.2 Simultaneous Diagonalization . . . . . . . . . . . . . . 59

2.11 Active and Passive Transformations . . . . . . . . . . . . . . . 60

3 The Postulates of Quantum Mechanics 653.1 The Fundamental Postulates . . . . . . . . . . . . . . . . . . . 663.2 Unitary Time Evolution . . . . . . . . . . . . . . . . . . . . . 703.3 Time-Independent Hamiltonian . . . . . . . . . . . . . . . . . 72

3.3.1 Propagator as a Power Series of H . . . . . . . . . . . 723.3.2 Eigenstate Expansion of the Propagator . . . . . . . . 75

3.4 The Uncertainty Principle . . . . . . . . . . . . . . . . . . . . 773.4.1 Uncertainty and Non-Commutation . . . . . . . . . . . 773.4.2 Position and Momentum . . . . . . . . . . . . . . . . . 793.4.3 Energy and Time . . . . . . . . . . . . . . . . . . . . . 79

3

3.4.3.1 Time rate of change of expectation values . . 803.4.3.2 Time-energy uncertainty . . . . . . . . . . . 80

3.5 Classical Mechanics and Quantum Mechanics . . . . . . . . . 813.5.1 Ehrenfest’s Theorem . . . . . . . . . . . . . . . . . . . 823.5.2 Exact Measurements and Expectation Values . . . . . 83

4 Spaces of Infinite Dimensionality 874.1 Two Types of Infinity . . . . . . . . . . . . . . . . . . . . . . . 884.2 Countably Infinite Dimensions . . . . . . . . . . . . . . . . . . 894.3 Uncountably Infinite Dimensions . . . . . . . . . . . . . . . . 904.4 Delta Function as a Limit of Gaussian Distribution . . . . . . 934.5 Delta Function and Fourier Transform . . . . . . . . . . . . . 934.6 f as a Vector with Uncountably Many Components . . . . . . 94

4.6.1 Finite Dimensions . . . . . . . . . . . . . . . . . . . . 944.6.2 Countably Infinite Dimensions . . . . . . . . . . . . . 944.6.3 Uncountably Infinite Dimensions . . . . . . . . . . . . 95

4.7 Hermiticity of the Momentum Operator . . . . . . . . . . . . 974.7.1 Checking [Pij] = [P ∗

ji] . . . . . . . . . . . . . . . . . . . 974.7.2 Hermiticity Condition in Uncountably Infinite Dimen-

sions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 984.7.3 The Eigenvalue Problem of the Momentum Operator P 99

4.8 Relations Between X and P . . . . . . . . . . . . . . . . . . . 1014.8.1 The Fourier Transform Connecting X and P . . . . . . 1014.8.2 X and P in the X Basis . . . . . . . . . . . . . . . . . 1024.8.3 Basis-Independent Descriptions in Reference to the X

Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1034.8.4 X and P in the P Basis . . . . . . . . . . . . . . . . . 1034.8.5 The Commutator [X,P ] . . . . . . . . . . . . . . . . . 105

5 Schrodinger’s Theory of Quantum Mechanics 1075.1 How was it derived? . . . . . . . . . . . . . . . . . . . . . . . 1075.2 Born’s Interpretation of Wavefunctions . . . . . . . . . . . . . 1125.3 Expectation Values . . . . . . . . . . . . . . . . . . . . . . . . 113

6 The Time-Independent Schrodinger Equation 121

7 Solutions of Time-Independent Schrodinger Equations in OneDimension 1277.1 The Zero Potential . . . . . . . . . . . . . . . . . . . . . . . . 1277.2 The Step Potential (E < V0) . . . . . . . . . . . . . . . . . . . 1297.3 The Step Potential (E > V0) . . . . . . . . . . . . . . . . . . . 1347.4 The Barrier Potential (E < V0) . . . . . . . . . . . . . . . . . 1377.5 The Infinite Square Well Potential . . . . . . . . . . . . . . . 1387.6 The Simple Harmonic Oscillator Potential . . . . . . . . . . . 141

7.6.1 Classic Solution . . . . . . . . . . . . . . . . . . . . . . 1417.6.2 Raising and Lowering Operators . . . . . . . . . . . . 150

7.6.2.1 Actions of a and a† . . . . . . . . . . . . . . 156

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

9 Angular Momentum 1759.1 Angular Momentum Operators . . . . . . . . . . . . . . . . . 1759.2 Quantum Mechanical Rotation Operator UR . . . . . . . . . . 1799.3 Rotationally Invariant Hamiltonian . . . . . . . . . . . . . . . 1829.4 Raising and Lowering Operators: L+ and L− . . . . . . . . . . 1839.5 Generalized Angular Momentum J . . . . . . . . . . . . . . . 189

10 The Hydrogen Atom 19310.1 2 Particles Instead of 1 . . . . . . . . . . . . . . . . . . . . . . 19410.2 Three-Dimensional System . . . . . . . . . . . . . . . . . . . . 19410.3 The Solutions for Φ . . . . . . . . . . . . . . . . . . . . . . . . 20510.4 The Solutions for Θ . . . . . . . . . . . . . . . . . . . . . . . . 20610.5 Associated Legendre Polynomials and Spherical Harmonics . . 21310.6 L2, Lz, and the Spherical Harmonics . . . . . . . . . . . . . . 21510.7 The Radial Function R . . . . . . . . . . . . . . . . . . . . . . 21510.8 Hydrogen-Like Atoms . . . . . . . . . . . . . . . . . . . . . . 22110.9 Simultaneous Diagonalization of H, L2, and Lz . . . . . . . . 22210.10Revisiting the Fundamental Postulates . . . . . . . . . . . . . 224

11 Electron Spin 23111.1 What is the Electron Spin? . . . . . . . . . . . . . . . . . . . 231

11.1.1 Stern-Gerlach Experiment . . . . . . . . . . . . . . . . 232

11.1.2 Fine Structure of the Hydrogen Spectrum: Spin-OrbitInteraction . . . . . . . . . . . . . . . . . . . . . . . . 233

11.2 Spin and Pauli Matrices . . . . . . . . . . . . . . . . . . . . . 23411.3 Sequential Measurements . . . . . . . . . . . . . . . . . . . . . 241

12 Molecular Rotation 24712.1 Rotational Kinetic Energy . . . . . . . . . . . . . . . . . . . . 24712.2 Moment of Inertia for a Diatomic Molecule . . . . . . . . . . . 24912.3 Two-Dimensional Rotation Confined to the x, y-Plane . . . . . 25012.4 Three-Dimensional Rotation . . . . . . . . . . . . . . . . . . . 251

Appendix A Matrix Algebra 257A.1 Invertible Matrices . . . . . . . . . . . . . . . . . . . . . . . . 257A.2 Rank of a Matrix . . . . . . . . . . . . . . . . . . . . . . . . . 258A.3 Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . 261A.4 Diagonalizability and Simultaneous Diagonalization . . . . . . 266

Appendix B Holomorphic Functional Calculus 277

Appendix C Wronskian and Linear Independence 279

Appendix D Newtonian, Lagrangian, and Hamiltonian Mechan-ics 283D.1 An Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . 283D.2 Newtonian Mechanics . . . . . . . . . . . . . . . . . . . . . . 284D.3 Lagrangian Mechanics . . . . . . . . . . . . . . . . . . . . . . 284D.4 Hamiltonian Mechanics . . . . . . . . . . . . . . . . . . . . . . 285

Appendix E Normalization Schemes for a Free Particle 287E.1 The Born Normalization . . . . . . . . . . . . . . . . . . . . . 287E.2 The Dirac Normalization . . . . . . . . . . . . . . . . . . . . . 287

E.2.1 Fourier Transform and the Delta Function . . . . . . . 288E.2.2 Wavefunctions as Momentum or Position Eigenfunctions288

E.3 The Unit-Flux Normalization . . . . . . . . . . . . . . . . . . 291

Appendix F Symmetries and Conserved Dynamical Variables 293F.1 Poisson Brackets and Constants of Motion . . . . . . . . . . . 293F.2 lz As a Generator . . . . . . . . . . . . . . . . . . . . . . . . . 296F.3 Rotation Around The Origin . . . . . . . . . . . . . . . . . . . 296

F.4 Infinitesimal Rotations Around the z-Axis . . . . . . . . . . . 297

Appendix G Commutators 299G.1 Commutator Identities . . . . . . . . . . . . . . . . . . . . . . 299G.2 Commutators involving X and P . . . . . . . . . . . . . . . . 300G.3 Commutators involving X, P , L, and r . . . . . . . . . . . . 300

Appendix H Probability Current 303

Appendix I Chain Rules in Partial Differentiation 309I.1 One Independent Variable . . . . . . . . . . . . . . . . . . . . 309I.2 Three Independent Variables . . . . . . . . . . . . . . . . . . . 309I.3 Reciprocal Relation is False . . . . . . . . . . . . . . . . . . . 310I.4 Inverse Function Theorems . . . . . . . . . . . . . . . . . . . . 311

Appendix J Laplacian Operator in Spherical Coordinates 313

Appendix K Legendre and Associated Legendre Polynomials 325K.1 Legendre Polynomials . . . . . . . . . . . . . . . . . . . . . . 325K.2 Associated Legendre Polynomials . . . . . . . . . . . . . . . . 326

Appendix L Laguerre and Associated Laguerre Polynomials 333

Bibliography 336

Answers to Exercises 337Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358

Subject Index 359

List of Tables

3.1 Physical observables and corresponding quantum operators . . 673.2 Quantum operators in three dimensions . . . . . . . . . . . . 67

10.1 Spherical Harmonics for l = 0, 1, and 2 . . . . . . . . . . . . . 216

9

List of Figures

10.1 Spherical Coordinate System (Source: Wikimedia Commons;Author: Andeggs) . . . . . . . . . . . . . . . . . . . . . . . . . 197

10.2 Spherical Coordinate System Used in Mathematics (Source:Wikimedia Commons; Author: Dmcq) . . . . . . . . . . . . . 198

10.3 Volume Element in Spherical Coordinates (Source: Victor J.Montemayor, Middle Tennessee State University) . . . . . . . 201

11.1 Stern-Gerlach Experiment (Diagram drawn by en wikipediaTheresa Knott.) . . . . . . . . . . . . . . . . . . . . . . . . . . 232

11.2 Results of Stern-Gerlach Experiment with and without a non-uniform external magnetic field (Source: Stern and Gerlach’soriginal paper [Gerlach and Stern, 1922, p.350]) . . . . . . . . 233

11

Chapter 1

Classical Physics vs. ModernPhysics

Classical Physics

• Newtonian Mechanics• Maxwell’s Electricity and Magnetism

Modern Physics

• Relativity• Quantum Mechanics

Newtonian mechanics no longer held under extreme conditions.

Two Types of Extreme Circumstances

1. Microscopic Phenomena (atomic and subatomic)Tunneling, Wave-particle duality

2. At High Speeds (near the speed of light denoted by c)Fast moving particles live longer. Gallilean transformation does notwork.

Historical Events Discoveries and Theories

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

13

14 CHAPTER 1. CLASSICAL PHYSICS VS. MODERN PHYSICS

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

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

4. Atomic spectra indicating discrete energy levels E1, E2, E3, . . . as op-posed to continuous distribution of energy levels E ∈ [0,∞)1

1We will later see that E = 0 is impossible in some quantum mechanical systems suchas a simple harmonic oscillator.

Chapter 2

Mathematical Preliminaries

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

2.1 Linear Vector Spaces

Consider the three dimensional Cartesian space R3 equipped with the x-,y-, and z-axes. We have three unit vectors, perpendicular/orthogonal toone another, denoted by (i, j,k), (ı, ȷ, k), or (x, y, z). In the componentnotation, we have i = (1, 0, 0), j = (0, 1, 0) and k = (0, 0, 1). Anypoint p in this space identified by three coordinates x, y, and z suchthat p = (x, y, z) can also be regarded as a vector xi + yj + zk or(x, y, z) in the component notation. Note that we use (x, y, z) bothfor the coordinates and components by abuse of notation. You mustbe familiar with such arithmetic operations as addition, subtraction,and scalar multiplication for vectors; which are basically component-wise operationsa. This is a canonical example of a linear vector space,and you should keep this example in mind as you study the followingdefinition of a linear vector space. If you are not so inclined, you cansafely skip this definition and still can understand the rest of this book

15

16 CHAPTER 2. MATHEMATICAL PRELIMINARIES

fully.aTo be more precise, these operations can be done component-wise though there

are other equivalent definitions/formulations for these arithmetic operations.

Definition 2.1 A linear vector space V is a set v1,v2,v3, . . . of objectscalled vectors for which “addition” and scalar multiplication are defined, suchthat

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

2. Addition and scalar multiplication obey the following axioms.

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

that0+ vi = vi + 0 = vi. (existence of the identity element)

(iv) For each vi, there exists a unique inverse (−vi) in V suchthatvi + (−vi) = 0. (existence of the inverse)1

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

(vi) (α + β)vi = αvi + βvi

(vii) α(βvi) = (αβ)vi

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

1. 0v = 0

2. α0 = 0

1The axioms (i) through (iv) means that a linear vector space form an abelian groupunder addition.

2.1. LINEAR VECTOR SPACES 17

3. (−1)v = (−v)

Definition 2.2 If the allowed values for the scalars α, β, γ, . . . comefrom some field F, we say the linear vector space V is defined over thefield F. In particular, if F is the field of real numbers R, V is a realvector space. Likewise, if F is the field of complex numbers C, V is acomplex vector space.

Clearly, R3 is not a complex vector space, but a real vector space. Com-plex vector spaces are the important vector spaces in quantum mechan-ics.

Definition 2.3 A set of vectors v1,v2, . . . ,vn is linearly indepen-dent ( LI) if

n∑i=1

αivi = 0 =⇒ α1 = α2 = . . . = αn = 0.

Definition 2.3 is equivalent to saying that no vector in v1,v2, . . . ,vncan be expressed as a linear combination of the other vectors in the set.

Vectors i, j, and k are linearly independent because

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

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

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

Definition 2.4 A vector space is n-dimensional if it has at most nvectors that are linearly independent.


Notation: An n-dimensional vector space over a field F is denoted byVn(F). The ones we encounter in this course are usually members ofVn(C), including the cases where n =∞.

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

Theorem 2.1 Suppose v1,v2, . . . ,vn are linearly independent vectors inan n-dimensional vector space Vn. Then, any vector v in Vn can be writtenas a linear combination of v1,v2, . . . ,vn.ProofAs Vn is n-dimensional, you can find a set of n+1 scalars α1, α2, . . . , αn, αn+1,not all zero, such that (

n∑i=1

αivi

)+ αn+1v = 0.

Else, v1,v2, . . . ,vn,v are linearly independent, and Vn is at least n + 1-dimensional, which is a clear contradiction. Furthermore, αn+1 , 0. If not,at least one of α1, α2, . . . , αn is not 0, and yet ∑n

i=1 αivi = 0, contradictingthe assumption that v1,v2, . . . ,vn are linearly independent. We now have

v =n∑i=1

−αiαn+1

vi. (2.1)

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

Theorem 2.2 The coefficients of Equation 2.1 are unique.ProofConsider any linear expression of v in terms of v1,v2, . . . ,vn

v =n∑i=1

βivi. (2.2)

Subtracting Equation 2.2 from Equation 2.1,

0 =n∑i=1

(−αiαn+1

− βi)

vi.

2.2. INNER PRODUCT SPACES 19

However, we know v1,v2, . . . ,vn are linearly independent. Hence,n∑i=1

(−αiαn+1

− βi)

vi = 0 =⇒ −αiαn+1

−βi = 0 or βi =−αiαn+1

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

In order to understand Definition 2.5 below, just imagine how any vectorin R3 can be expressed as a unique linear combination of i, j, and k.

Definition 2.5 A set of linearly independent vectors v1,v2, . . . ,vn,which can express any vector v in Vn as a linear combination (Equation2.2), is called a basis that spans Vn. The coefficients of the linear com-bination βi are called the components of v in the basis v1,v2, . . . ,vn.Turning the expression inside out, we also say that Vn is the vectorspace spanned by the basis v1,v2, . . . ,vna.

Once we pick a basis v1,v2, . . . ,vn, and express v as a linear combi-nation of the basis vectors v =

∑ni=1 βivi, we get a component notation

of v as follows.v =

n∑i=1

βivi = (β1, β2, . . . , βn).

With this notation, vector additions and scalar multiplications can bedone component by component. For vectors v = (β1, β2, . . . , βn), w =(γ1, γ2, . . . , γn) and a scalar α, we simply have

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

aThis means that we can define Vn as the collection of all objects of the form∑ni=1 αivi.

2.2 Inner Product Spaces

This is nothing but a generalization of the familiar scalar product forV3(R)a. Namely, it is an extension of the inner product defined on R3 tocomplex vectors. Readers should be familiar with the component-wisedefinition of the usual inner product on R3, also called dot product, suchthat a · b = (a1, a2, a3) · (b1, b2, b3) = a1b1 + a2b2 + a3b3. We can express


this as a matrix product of a row vector a and a column vector b.

[a1 a2 a3

] b1b2b3

=[a1b1 + a2b2 + a3b3

];

where the one-by-one matrix on the right-hand side is a scalar by defi-nition. This definition of inner product is extended to include complexcomponents by taking the complex conjugate of the components of a.So, the extended inner product, which we simply call an inner productin the rest of the book, is given by

[a1 a2 a3

]∗ b1b2b3

=[a∗1 a∗

2 a∗3

] b1b2b3

=[a∗1b1 + a∗

2b2 + a∗3b3

].

Definition 2.6 below is an abstract version of the familiar inner productdefined on R3. This definition contains the inner product defined on R3,but not all inner products are of that type. Nevertheless, you shouldkeep the inner product on R3 in mind as you read the following generaldefinition of inner product. In Definition 2.6, we are writing a∗ as ⟨a|and b as |b⟩ so that a∗ · b = ⟨a|b⟩b in anticipation of Dirac’s braketnotation to be introduced on p.27.

aThis is the notation suggested on p.18, which is the same as R3.bNaively, a∗ · b = ⟨a| |b⟩, but we use only one vertical line and write ⟨a|b⟩.

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

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

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

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

It is convenient to add the fourth property to this list, though it is not anaxiom but a consequence of Axioms (i) and (iii).


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

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

In order to see how (iv) follows from (ii) and (iii), see the derivation below.⟨αvi + βvj | vk⟩ = ⟨vk |αvi + βvj⟩∗ by (ii)

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

A moment’s

thought would tell us that (iii) and (iv) can be extended to the followingproperty. ⟨

n∑i=1

αivi

∣∣∣∣∣∣n∑j=1

βjvj

⟩=

n∑i,j=1

α∗iβj ⟨vi | vj⟩

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

Our canonical example R3 is naturally an inner product space.

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

Definition 2.9 (Metric Space) A metric space is a set where a notionof distance (called a ”metric”) between elements of the set is defined.2 Inparticular, a norm is a metric.

2There is a precise notion of a metric which is a real-valued function with two elementsof a set as the arguments. Let us use the canonical notation d(x, y). Then, a metric dshould satisfy the following conditions.

1. d(x, y) ≥ 0

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

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

4. d(x, z) ≤ d(x, y) + d(y, z)

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

2d(x, x) = 0.


Definition 2.10 (Complete Metric Space) A metric space M is calledcomplete if every Cauchy sequence of points in M converges in M ; that is, ifevery Cauchy sequence in M has a limit that is also in M.

Definition 2.11 (Hilbert Space) An inner product space which iscomplete with respect to the norm induced by the inner product is calleda Hilbert space.

R3 is also a Hilbert space.

Definition 2.12 (Orthogonality) Two vectors u and w are orthog-onal to each other if ⟨u |w⟩ = 0.

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

Definition 2.13 (Orthonormal Set) A set of vectors v1,v2, . . . ,vnform an orthonormal set if

⟨vi | vj⟩ = δij;

where the symbol δij is known as Kronecker’s delta with the followingstraightforward definition.

δij =

1 if i = j0 if i , j

Definition 2.14 (Completeness) a An orthonormal set of vectorsv1,v2, . . . ,vn is complete if

n∑i=1

|vi⟩ ⟨vi| = I. (2.3)

Alternatively, v1,v2, . . . ,vn is complete if any vector v in the Hilbertspace can be expressed as a linear combination of v1, v2, . . ., vn. Thatis, any v =

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

i=1 |vi⟩ ⟨vi| = I.


ProofSuppose ∑n

i=1 |vi⟩ ⟨vi| = I. Then,

v = Iv =

(n∑i=1

|vi⟩ ⟨vi|)|v⟩ =

n∑i=1

⟨vi|v⟩ |vi⟩ . (2.4)

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

(n∑i=1

|vi⟩ ⟨vi|)

v =

(n∑i=1

|vi⟩ ⟨vi|) n∑

j=1

aj |vj⟩

=n∑

i,j=1

|vi⟩ ⟨vi| aj |vj⟩

=n∑

i,j=1

aj |vi⟩ ⟨vi|vj⟩ =n∑

i,j=1

aj |vi⟩ δi,j =n∑j=1

aj |vj⟩ = v. (2.5)

Because (2.5) holds for an arbitrary vector v, we have ∑ni=1 |vi⟩ ⟨vi| = I

by definition.

Definition 2.15 (Orthonormal Basis) If a basis of a vector spaceVn forms an orthonormal set, it is called an orthonormal basis.b

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

aThis is also known as the closure condition.bNote that a set of vectors which is orthonormal and complete forms an orthonor-

mal basis.

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

⟨u |w⟩ =⟨

n∑i=1

uiei

∣∣∣∣∣∣n∑j=1

wjej

⟩=

n∑i,j=1

u∗iwj ⟨ei | ej⟩ =

n∑i,j=1

u∗iwjδij =

n∑i=1

u∗iwi


= u∗1w1+u

∗2w2+ . . .+u∗

n−1wn−1+u∗nwn.

3

Theorem 2.3 (Schwarz Inequality) Any inner product satisfies the fol-lowing inequality known as Schwarz Inequality.

|⟨vi | vj⟩|2 ≤ |vi|2|vj|2

ProofConsider the vector

v = vi − ⟨vj | vi⟩vj|vj|2

.

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

vj|vj|2

∣∣∣∣∣ vi − ⟨vj | vi⟩ vj|vj|2

⟩= ⟨vi | vi⟩−

⟨vj | vi⟩ ⟨vi | vj⟩|vj|2

−⟨vj | vi⟩∗ ⟨vj | vi⟩|vj|2

+⟨vj | vi⟩∗ ⟨vj | vi⟩ ⟨vj | vj⟩

|vj|4= |vi|2−


−⟨vj | vi⟩∗ ⟨vj | vi⟩|vj|2

+⟨vj | vi⟩∗ ⟨vj | vi⟩ |vj|2

|vj|4

= |vi|2−⟨vj | vi⟩ ⟨vi | vj⟩

|vj|2−⟨vj | vi⟩

∗ ⟨vj | vi⟩|vj|2

+⟨vj | vi⟩∗ ⟨vj | vi⟩

|vj|2= |vi|2−


≥ 0

=⇒ |vi|2|vj|2 ≥ |⟨vi | vj⟩|2 .We can also see that the equality holds only when

v = vi − ⟨vj | vi⟩vj|vj|2

= 0 or vi = ⟨vj | vi⟩vj|vj|2

.

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

⟨vj | vi⟩vj|vj|2

= ⟨vj |αvj⟩vj|vj|2

= ⟨vj | vj⟩αvj|vj|2

= |vj|2αvj|vj|2

= αvj = vi.

Therefore, the equality holds if and only if one vector is a scalar multiple ofthe other.4

3Don’t make a beginner’s mistake of applying this to vectors expressed in a basis thatis not orthonormal. For example, consider the space spanned by e1 and e2; i.e. thex, y-plane if you may. While e1, e2 form an orthonormal basis,

e1,

e1+e2√2

is a basis

composed of normalized vectors which are not mutually orthogonal. Let u = e1 andw = e1+e2√

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

1w1 + u∗2w2 = 1× 0 + 0× 1 = 0. But,

u ·w = e1 ·(

e1+e2√2

)= e1·e1+e1·e2√

2= 1√

2, 0.

4This is not surprising as ⟨vj | vi⟩ vj

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

2.3. L2-SPACE 25

2.3 L2-Space

It was mentioned on p.20 that there are other inner products than thefamiliar one on R3. And, there is an associated Hilbert space for eachof those inner products. One important class of Hilbert spaces for us isthe L2-space. It is a collection of functions with a common domain, forwhich an inner product is defined as an integral. Most of the Hilbertspaces we encounter in this course are L2-spaces. In the following,we will specialize to R as the domain of our functions, but the samedefinitions and descriptions apply to a general domaina. In particular,

if the domain is R2 or R3, we have a double integral,+∞!

−∞, or triple

integral,+∞#

−∞, respectively, rather than a single integral shown here.

Definition 2.16 (L2-Function) An L2-function f is a square inte-grable function; that is, f satisfies

∫+∞−∞ |f(x)|2 dx <∞. b

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

Definition 2.17 (L2-Inner Product) Given two square integrable func-tions f and g, we define an inner product by∫ +∞

−∞f(x)∗g(x) dxc;

where f(x)∗ signifies the complex conjugate.

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

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

2,+a

2) for a > 0. Using the notation given in Footnote b, we

will denote these L2-spaces by L2(R2), L2(R3), L2[0, 1], and L2(−a2,+a

2).


The central equation of quantum mechanics, called Schrodinger Equa-tion, is a differential equation, and its solutions are complex valued func-tions on a common domain. Therefore, the L2-space is our canonicalHilbert space.

aTo be precise, the domain should be a measure space so that an integral canexist. However, such purely mathematical details can be omitted as all the domainswe consider are measure spaces.

bWe often have an interval of finite length as the domain of f . The closed interval[0, 1] is a canonical example. In this case, we require

∫ 1

0|f(x)|2 dx < ∞. We use

notations such as L2(−∞,+∞) and L2[0, 1] to make this distinction.cWe can easily check that this integral satisfies Definition 2.6.

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

1. The trigonometric system ei2πnx+∞n=−∞ is an orthonormal basis for

L2[0, 1]. The expansion of a function in this basis is called the Fourierseries of that function.

2. The Legendre polynomials, which are obtained by taking the sequenceof monomials 1, x, x2, x3, . . . and applying the Gram-Schmidt orthog-onalization process to it, form an orthonormal basis for L2[−1, 1]. Thefirst few normalized polynomials are as follows.

p1(x) =√1/2

p2(x) = x√3/2

p3(x) =(3x2 − 1

) √5/8

p4(x) =(5x3 − 3x

) √7/2

2

p5(x) =(35x4 − 30x2 + 3

) √9/2

8

p6(x) =(63x5 − 70x3 + 15x

) √11/2

8

2.4. THE BRAKET NOTATION 27

2.4 The Braket Notation5

This is a notation first used by Dirac. In fact, we have already been usingit. You may have noticed that the inner product between v and w wasdenoted by ⟨v |w⟩, and not by the more familiar ⟨v, w⟩. We will see whyshortly. Without much ado, let us define a “bra” and a “ket”. On p.20, wehave already encountered ⟨a| = a∗ and |b⟩ = b such that the inner productbetween a and b is given by ⟨a|b⟩ = a∗

1b1+a∗2b2+a

∗3b3. Here, ⟨a| is called “bra

a”, and |b⟩ is called “ket b”. More generally, recall the component notationof a vector v = (v1, v2, . . . , vn). This usually means that we have picked aparticular orthonormal basis6 e1, e2, . . . , en and expressed v as a linearcombination of e1, e2, . . . , en such that v = v1e1 + v2e2 + . . . + vnen. Wecan arrange the components either horizontally or vertically, but it is moreconvenient for us to arrange it vertically as an n by 1 matrix, often called acolumn vector. In quantum mechanics, we denote this column vector by |v⟩called “ket v”.

|v⟩ =

v1...vn

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

⟨v| = [v∗1, . . . , v

∗n]

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

5Note that this is “bra + ket = braket”, and NOT “bracket”.6Actually, it does not have to be an orthonormal basis, but any basis would do. How-

ever, we will specialize to an orthonormal basis here in order to explain why we used thenotation ⟨u| |v⟩ for the inner product. Besides, orthonormal bases are by far the mostcommon and useful bases.


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

00...1...0

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

|v⟩ =n∑i=1

vi |i⟩

Similarly,

⟨v| =n∑i=1

v∗i ⟨i| .

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

|αv⟩ = α |v⟩ and ⟨αv| = α∗ ⟨v| = ⟨v|α∗.

As shown above, it is customary, for conserving symmetry, to write ⟨v|α∗

rather than α∗ ⟨v|. We say α |v⟩ and ⟨v|α∗ are adjoints of each other. Moregenerally, the adjoint of

n∑i=1

αi |vi⟩

isn∑i=1

⟨vi|α∗i ,

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


Let us now try some manipulations with bras and kets in order to getused to their properties.

First, consider two vectors u = u1e1 + u2e2 + . . . + unen and w =w1e1 + w2e2 + . . . + wnen in an n-dimensional space. We have

⟨u| = [u∗1, . . . , u

∗n] and |w⟩ =

w1...wn

.Hence,

⟨u| |w⟩ = [u∗1, . . . , u

∗n]

w1...wn

= u∗1w1 + u∗

2w2 + . . . + u∗nwn.

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

⟨u|w⟩ = [u∗1, . . . , u

∗n]

w1...wn

= u∗1w1 + u∗

2w2 + . . . + u∗nwn.

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

⟨ei | ej⟩ = ⟨i | j⟩ = δij.

Next, given an orthonormal basis |1⟩ , |2⟩ , . . . , |n⟩, any vector |v⟩ canbe expanded in this basis so that

|v⟩ =n∑i=1

vi |i⟩ , (2.6)

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

⟨j|v⟩ =⟨j∣∣∣ n∑i=1

vi∣∣∣i⟩ = n∑

i=1

vi ⟨j|i⟩ =n∑i=1

viδij = vj.


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

|v⟩ =n∑i=1

⟨i|v⟩ |i⟩ =n∑i=1

|i⟩ ⟨i|v⟩ . (2.7)

The second equality in Equation 2.7 may not be clear. So, we will show itbelow explicitly by writing out each vector. The term on the left-hand sideis as follows.

⟨i|v⟩ |i⟩ =

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

00...1...0

= vi

00...1...0

=

00...vi...0

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

|i⟩ ⟨i|v⟩ =

00...1...0

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

v1...vn

=

00...1...0

[vi] =

00...vi...0

. (2.8)

Hence, the second equality in Equation 2.7 indeed holds term by term. Inthe case of (2.8), we have a product, from the left, of an n-by-1 matrix,a 1-by-n matrix, and an n-by-1 matrix. By the associative law of matrix


multiplication, it is possible to compute |i⟩ ⟨i|7 first.

|i⟩ ⟨i|v⟩ =

00...1...0

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

v1...vn

=

1 i n

1 0 0 . . . 0 . . . 0

0. . .

......

.... . .

......

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

......

. . ....

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

v1...vn

= [δii]

v1...vn

=

00...vi...0

(2.9)

The symbol δij in (2.9) means an n by n matrix whose only nonzero entry isthe 1 at the intersection of the i-th row and the j-th column.

[δij] :=

1 j n

1 0 0 . . . 0 . . . 0

0. . .

......

.... . .

......

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

......

. . ....

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

It is easy to see that

7As (2.9) shows, |i⟩ ⟨i| extracts the i-th component of vector v. So, |i⟩ ⟨i| is called theprojection operator for the ket |i⟩.


|i >< j| = [δij].

And, it is also easy to see that

n∑i=1

|i >< i| =n∑i=1

[δii] = In; (2.10)

where In is the n-dimensional identity operator.8 With this identity, Equa-tion 2.7 reduces to triviality.

|v⟩ = I |v⟩ =(

n∑i=1

|i⟩ ⟨i|)|v⟩ =

(n∑i=1

|i⟩ ⟨i| |v⟩)=

n∑i=1

|i⟩ ⟨i|v⟩

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

i=1 |i⟩ ⟨i| anywhere during ourcomputation. This seemingly trivial observation combined with the associa-tivity of matrix multiplication will prove useful later in this book.

Example 2.1 (⟨i|, |i⟩, and |i⟩ ⟨i| in Two Dimensions) In two dimen-sions, we have

|1⟩ =[10

], |2⟩ =

[01

], ⟨1| =

[1 0

], and ⟨2| =

[0 1

](2.11)

Then,

2∑i=1

|i⟩ ⟨i| =[10

] [1 0

]+

[01

] [0 1

]=

[1 00 0

]+

[0 00 1

]=

[1 00 1

]. (2.12)

8When there is no possibility of confusion, we will simply use I instead of In.


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

|1⟩ ⟨1|v =

[10

] [1 0

] [ v1v2

]=

[1 00 0

] [v1v2

]=

[v10

](2.13)

|2⟩ ⟨2|v =

[01

] [0 1

] [ v1v2

]=

[0 00 1

] [v1v2

]=

[0v2

](2.14)

Let us consider the adjoint of (2.6).

⟨v| =n∑i=1

⟨i| v∗i (2.15)

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

⟨v| =n∑i=1

⟨i| (⟨i|v⟩)∗ =n∑i=1

⟨i| ⟨v|i⟩ =n∑i=1

⟨v|i⟩ ⟨i| .

Again, note the identity ∑ni=1 |i >< i| = ∑n

i=1[δii] = I. So,n∑i=1

⟨v|i⟩ ⟨i| = ⟨v|(

n∑i=1

|i⟩ ⟨i|)= ⟨v| I = ⟨v| .

It is convenient at this point to have a rule by which we can mechanicallyarrive at the adjoint of an expression.

Taking the Adjoint:

1. Reverse the order of all the factors in each term, including nu-merical coefficients, bras, and kets.

2. Take complex conjugate of the numerical coefficients.

3. Turn bras into kets and kets into bras.

The following example demonstrates what one should do.Consider

α |v⟩ = a |u⟩+ b |w⟩ ⟨s|t⟩ + . . . .

What is the adjoint?


After Step 1

|v⟩α = |u⟩ a + |t⟩ ⟨s|w⟩ b + . . .

After Step 2

|v⟩α∗ = |u⟩ a∗ + |t⟩ ⟨s|w⟩ b∗ + . . .

After Step 3

⟨v|α∗ = ⟨u| a∗ + ⟨t|s⟩ ⟨w| b∗ + . . .

Therefore, the adjoint of

α |v⟩ = a |u⟩+ b |w⟩ ⟨s|t⟩ + . . .

is⟨v|α∗ = ⟨u| a∗ + ⟨t|s⟩ ⟨w| b∗ + . . . .

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

⟨v1 | v1⟩.

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

|1′⟩ = |v1⟩ .

Next, let

|2′⟩ = |v2⟩ −|1′⟩ ⟨1′ | v2⟩⟨1′ | 1′⟩

, (2.16)

where|1′⟩ ⟨1′ | v2⟩⟨1′ | 1′⟩


is nothing but the projection of |v2⟩ on |1′⟩9. So, in (2.16), the component of|v2⟩ in the direction of |1′⟩ is subtracted from |v2⟩, making it orthogonal to|1′⟩. Indeed,

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

= 0.

The third ket is defined proceeding in the same manner.

|3′⟩ = |v3⟩ −|1′⟩ ⟨1′ | v3⟩⟨1′ | 1′⟩

− |2′⟩ ⟨2′ | v3⟩⟨2′ | 2′⟩

.

Then,

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

−⟨1′ | 2′⟩ ⟨2′ | v3⟩⟨2′ | 2′⟩

= ⟨1′ | v3⟩−⟨1′ | v3⟩−0 = 0

and

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

−⟨2′ | 2′⟩ ⟨2′ | v3⟩⟨2′ | 2′⟩

= ⟨2′ | v3⟩−0−⟨2′ | v3⟩−0 = 0.

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

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

− |2′⟩ ⟨2′ | vi+1⟩⟨2′ | 2′⟩

− . . . − |i′⟩ ⟨i′ | vi+1⟩⟨i′ | i′⟩

For any 1 ≤ j ≤ i,

⟨j′ | (i+ 1)′⟩ = ⟨j′ | vi+1⟩−i∑

k=1

⟨j′ | k′⟩ ⟨k′ | vi+1⟩⟨k′ | k′⟩

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

= 0.

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

|i⟩ = |i′⟩⟨i′ | i′⟩1/2

,

9In the familiar x, y-plane,

|1′⟩ ⟨1′ | v2⟩⟨1′ | 1′⟩

=

(∥1′∥ 1′

)(∥1′∥ ∥v2∥ cos θ)

∥1′∥2= ∥v2∥ cos θ 1′,

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


so that⟨i | i⟩ = ⟨i′ | i′⟩

⟨i′ | i′⟩1/2 ⟨i′ | i′⟩1/2= 1.

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

Question: Linear IndependenceIf you are a careful reader, you may have noticed that we did not, at leastexplicitly, use the fact that the vectors |v1⟩ , |v2⟩ , . . . , |vn⟩ are linearlyindependent. So, why is this condition necessary? In order to see this,suppose

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

− |2′⟩ ⟨2′ | vi+1⟩⟨2′ | 2′⟩

− . . . − |i′⟩ ⟨i′ | vi+1⟩⟨i′ | i′⟩

= 0

(2.17)for some i ≤ n−1. Because our construction of |i⟩ requires |vi⟩, this means wehave at most n−1 orthonormal vectors, and our construction fails. However,we know that |j′⟩ is a linear combination of |v1⟩ , |v2⟩ , . . . , and |vj⟩ for anyj by construction. Therefore,

−|1′⟩ ⟨1′ | vi+1⟩⟨1′ | 1′⟩

− |2′⟩ ⟨2′ | vi+1⟩⟨2′ | 2′⟩

− . . . − |i′⟩ ⟨i′ | vi+1⟩⟨i′ | i′⟩

in (2.17) is a linear combination of |v1⟩ , |v2⟩ , . . . , and |vi⟩. So, |(i+ 1)′⟩ is alinear combination of |v1⟩ , |v2⟩ , . . . , and |vi+1⟩ where the coefficient of |vi+1⟩is 1. But, this means that there is a linear combination of |v1⟩ , |v2⟩ , . . . ,and |vi+1⟩ which equals 0 though not all the coefficients are zero, violat-ing the linear independence assumption for |v1⟩ , |v2⟩ , . . . , |vn⟩. Turningthis inside out, we can conclude that |(i+ 1)′⟩ , 0 for all i ≤ n − 1 if|v1⟩ , |v2⟩ , . . . , |vn⟩ are linearly independent. This is why we needed thelinear independence of |v1⟩ , |v2⟩ , . . . , |vn⟩ in Theorem 2.4.

Theorem 2.5 The maximum number of mutually orthogonal vectors in aninner product space equals the maximum number of linearly independent vec-tors.ProofConsider an n-dimensional space Vn, which admits a maximum of n linearlyindependent vectors by definition. Suppose we have k mutually orthogonalvectors |v1⟩ , |v2⟩ , . . . , |vk⟩, and assume

k∑i=1

αi |vi⟩ = 0.

2.5. VECTOR SUBSPACE 37

Then,⟨vj

∣∣∣∣∣k∑i=1

αi |vi⟩⟩=

k∑i=1

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

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

2.5 Vector SubspaceDefinition 2.18 (Vector Subspaces) A subset of a vector space V thatform a vector space is called a subspace. Our notation for a subspace i ofdimensionality ni is Vni

i .

Definition 2.19 (Orthogonal Complement) The orthogonal complementof a subspace W of a vector space V equipped with a bilinear form B, of whichan inner product is the most frequently encountered example, is the set W⊥

of all vectors in V that are orthogonal to every vector in W. It is a subspaceof V.

Definition 2.20 (Direct Sum) The direct sum of two vector spaces V andW is the set V⊕W of pairs of vectors (v,w) in V andW, with the operations:(v,w) + (v′,w′) = (v + v′,w + w′)c(v,w) = (cv, cw)

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

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

Ω |u⟩ = |v⟩

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


⟨u|Ω = ⟨v| .

For an operator Ω to be a linear operator, it should have the following set ofproperties.

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

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

Recall that |u⟩ is a column vector and ⟨u| is a row vector. So, the best wayto understand a linear operator is to regard it as an n × n square matrix.Then, its action on |u⟩ and ⟨u| as well as linearity follows naturally. Thesignificance of linearity of an operator Ω is that its action on any vec-tor is completely specified by how it operates on the basis vectors|1⟩ , |2⟩ , . . . , |n⟩. Namely, if

|v⟩ =n∑i=1

αi |i⟩ ,

then,

Ω |v⟩ = Ω

(n∑i=1

αi |i⟩)=

n∑i=1

αiΩ |i⟩ .

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

(ΛΩ) |u⟩ = Λ (Ω |u⟩)

Ω acts on |u⟩ first followed by Λ. This is an obvious associative law if theoperators Λ and Ω are regarded as matrices as suggested already. With thisview, ΛΩ is nothing but a matrix multiplication, and as such, ΛΩ is not thesame as ΩΛ generally speaking.

2.7. MATRIX REPRESENTATION OF LINEAR OPERATORS 39

Definition 2.21 (Commutator) The commutator of operators Λ andΩ, denoted by [Λ,Ω] is defined as follows.

[Λ,Ω] := ΛΩ− ΩΛ

When the commutator [Λ,Ω] is zero, we say Λ and Ω commute, and theorder of operation does not affect the outcome.

We have the following commutator identities.

[Ω,ΛΓ] = Λ [Ω,Γ] + [Ω,Λ] Γ

[ΛΩ,Γ] = Λ [Ω,Γ] + [Λ,Γ]Ω

Definition 2.22 (Inverse) The inverse Ω−1 of an operator Ω satisfies

ΩΩ−1 = Ω−1Ω = I;

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

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

2.7 Matrix Representation of Linear Opera-tors

Recall that the action of a linear operator Ω is completely specified once itsaction on a basis is specified. Pick a basis |1⟩ , |2⟩ , . . . , |n⟩ of normalizedvectors which are not necessarily mutually orthogonal; that is, it is not nec-essarily an orthonormal basis. The action of Ω on any ket |v⟩ is completelyspecified if we know the coefficients Ωji for 1 ≤ i, j ≤ n such that


Ω |i⟩ =n∑j=1

Ωji |j⟩ . (2.18)

The coefficients in (2.18) can be computed as follows.

⟨k|Ω |i⟩ =⟨k

∣∣∣∣∣∣n∑j=1

Ωji

∣∣∣∣∣∣ j⟩=

n∑j=1

Ωji ⟨k| |j⟩ =n∑j=1

Ωjiδkj = Ωki

So, Ωji = ⟨j|Ω |i⟩ in (2.18). Note that ⟨k| |j⟩ was used instead of ⟨k | j⟩ tomake it clear that this matrix product

[ 1 k n

0 0 . . . 1 . . . 0]

00...1...0

1

j

n

is not necessarily an inner product unless our basis happens to be orthonor-mal.10

There is another way to find Ωji. From (2.10), we have ∑ni=1 |i⟩ ⟨i| = I, and

so,

Ω |i⟩ = IΩ |i⟩ =n∑j=1

|j⟩ ⟨j|Ω |i⟩ =n∑j=1

⟨j|Ω |i⟩ |j⟩ =⇒ Ωji = ⟨j|Ω |i⟩.

Now, let|v⟩ =

n∑i=1

vi |i⟩ and |v′⟩ = Ω |v⟩ =n∑j=1

v′j |j⟩ .

Then,

|v′⟩ = Ω |v⟩ = Ω

(n∑i=1

vi |i⟩)=

n∑i=1

viΩ |i⟩ =n∑i=1

vin∑j=1

Ωji |j⟩ =n∑j=1

(n∑i=1

Ωjivi |j⟩)

10See the footnote on p.24 for more detail.


=n∑j=1

(n∑i=1

Ωjivi

)|j⟩

impliesv′j =

n∑i=1

Ωjivi. (2.19)

Consider the matrix [Ωji] = [⟨j|Ω |i⟩], where Ωji is the j-th row and the i-thcolumn entry. Then, (2.19) implies

v′1......v′n

=

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

.... . .

...⟨n|Ω |1⟩ · · · ⟨n|Ω |n⟩

v1......vn

or

v′1......v′n

=

Ω11 · · · Ω1n

Ω21 · · · Ω2n...

. . ....

Ωn1 · · · Ωnn

v1......vn

.

Because

Ω11 · · · · · · Ω1k · · · Ω1n...

. . ....

......

. . ....

...

Ωk1 · · · · · · Ωkk · · · Ωkn...

.... . .

...Ωn1 · · · · · · Ωnk · · · Ωnn

0......

1...0

k

=

Ω1k......

Ωkk...

Ωnk

,

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

Example 2.2 (An Operator on C2) The column vectors[10

]and[

01

]form an orthonormal basis for C2. If Ω

[10

]=

[11

]and


Ω

[01

]=

[10

], we should have

Ω =

[1 11 0

]

because Ω

[10

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

and Ω

[01

]is the second column. Now consider an arbitrary vector v

given by

|v⟩ =[v1v2

]= v1

[10

]+ v2

[01

].

Using the matrix representation

Ω |v⟩ =[1 11 0

] [v1v2

]=

[v1 + v2v1

].

On the other hand,

Ω |v⟩ = Ω

(v1

[10

]+ v2

[01

])= v1Ω

[10

]+ v2Ω

[01

]

= v1

[1 11 0

] [10

]+ v2

[1 11 0

] [01

]

= v1

[11

]+ v2

[10

]=

[v1 + v2v1

].

This checks.

Recall that we denoted α |v⟩ by |αv⟩, and so, it is natural to denoteΩ |v⟩ by |Ωv⟩. Then, it is also natural to consider its adjoint ⟨Ωv|. For ascalar α, ⟨αv| = ⟨v|α∗ while |αv⟩ = α |v⟩. Therefore, it is not surprising if⟨Ωv| , ⟨v| [Ωij] though ⟨v| [Ωij] makes sense as a matrix multiplication. Wewill next compute the matrix elements of Ω when it comes out of the ⟨Ωv|and operate on ⟨v| from the right. Looking ahead, let us denote this by Ω†


such that ⟨Ωv| = ⟨v|Ω†. We have

|v⟩ =n∑i=1

vi |i⟩ and |v′⟩ = Ω |v⟩ = |Ωv⟩ =n∑j=1

v′j |j⟩ .

So,

⟨v| =n∑i=1

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

n∑j=1

⟨j| v′∗j ,

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

Ω |i⟩ =n∑j=1

Ωji |j⟩ ,

⟨v′| = adj (Ω |v⟩) = adj

(Ω

n∑i=1

vi |i⟩)= adj

(n∑i=1

viΩ |i⟩)= adj

n∑i=1

vin∑j=1

Ωji |j⟩

= adj

n∑j=1

n∑i=1

viΩji |j⟩

=n∑j=1

n∑i=1

⟨j| v∗iΩ

∗ji.

Hence,

v′∗j =

n∑i=1

v∗iΩ

∗ji for each 1 ≤ j ≤ n.

This translates to the following matrix relation.

[v′∗1 . . . v′∗

n

]=[v∗1 . . . v∗

n

]Ω∗

11 Ω∗21 · · · Ω∗

n1

Ω∗12 Ω∗

22 · · · Ω∗n2

......

. . ....

Ω∗1n Ω∗

2n · · · Ω∗nn

or

[v′∗1 . . . v′∗

n

]=[v∗1 . . . v∗

n

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

......

. . ....

⟨1|Ω |n⟩∗ ⟨2|Ω |n⟩∗ · · · ⟨n|Ω |n⟩∗


Because

[ k

0 · · · · · · 1 · · · 0]

Ω∗11 · · · · · · Ω∗

k1 · · · Ω∗n1

.... . .

......

.... . .

......

Ω∗1k · · · · · · Ω∗

kk · · · Ω∗nk

......

. . ....

Ω∗1n · · · · · · Ω∗

kn · · · Ω∗nn

=[Ω∗

1k · · · · · · Ω∗kk · · · Ω∗

nk

],

the k-th row of[Ω∗ji

]is nothing but ⟨kΩ|.

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

Ω = [Ωij] =

Ω11 · · · Ω1n

Ω21 · · · Ω2n...

. . ....

Ωn1 · · · Ωnn

,

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

Ω† =[Ω∗ji

]Ω∗

11 · · · Ω∗n1

Ω∗12 · · · Ω∗

n2...

. . ....

Ω∗1n · · · Ω∗

nn

.

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

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

k=1 |k⟩ ⟨k| = I. So,

(ΩΓ)ij = ⟨i|ΩΓ |j⟩ =n∑k=1

⟨i|Ω |k⟩ ⟨k|Γ |j⟩ =n∑k=1

ΩikΓkj.

Hence, the matrix representation of a product of two operators is the productof the matrix representations for the operators in the same order.Notation: We often denote [Ωij] by [Ω] for simplicity.

2.8. THE ADJOINT OF AN OPERATOR 45

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

Definition 2.23 (The Adjoint) Consider an operator Ω that operatesin a vector space V.

Ω |v⟩ = |Ωv⟩

If there exists an operator Ω† such that

⟨Ωv| = ⟨v|Ω†.

Then, Ω† is called the adjoint of Ω.

We can show that the adjoint indeed exists by explicitly computing the ma-trix elements in a basis |1⟩ , |2⟩ , . . . , |n⟩.

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

∗

As shown in Section 2.7, the matrix representing Ω† is the transposeconjugate of the matrix for Ω.

Example 2.3 (Adjoint of an Operator) Consider

Γ =

[i 10 1

].

Then,

Γ† =

[−i 01 1

].

Now, for v =

[v1v2

],

|Γv⟩ =[i 10 1

] [v1v2

]=

[iv1 + v2v2

]=⇒ ⟨Γv| =

[(iv1 + v2)

∗ v∗2

]


=[−iv∗

1 + v∗2 v∗

2

].

On the other hand,

⟨v|Γ† =[v∗1 v∗

2

] [ −i 01 1

]=[−iv∗

1 + v∗2 v∗

2

]This checks.

Fact 2.2 The adjoint of the product of two operators (ΩΓ)† is the prod-uct of their adjoints with the order switched; that is,

(ΩΓ)† = Γ†Ω†.

ProofFirst, regard ΩΓ as one operator. Then,

⟨ΩΓv| = ⟨(ΩΓ)v| = ⟨v| (ΩΓ)†.

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

⟨ΩΓv| = ⟨Ω(Γv)| = ⟨Γv|Ω†.

But,⟨Γv| = ⟨v|Γ†

means⟨ΩΓv| = ⟨v|Γ†Ω†.

Therefore,(ΩΓ)† = Γ†Ω†.

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

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

2. Take the complex conjugate of the scalars.

3. Switch bras to kets and kets to bras.

4. Take the adjoint of each operator.

2.9. EIGENVALUES AND EIGENVECTORS 47

For example, consider a linear expression

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

After Step 1

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

After Step 2

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

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

After Step 3

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

After Step 4

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

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

2.9 Eigenvalues and Eigenvectors

Definition 2.24 (Eigenvalues, Eigenvectors, and Eigenkets)Consider an operator Ω operating on a vector space V. When a nonzerovector v or a nonzero ket |v⟩ satisfies

Ωv = λv or Ω |v⟩ = λ |v⟩ ,

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


Example 2.4 (Eigenvalues and Eigenvectors) Consider

Ω =

[0.8 0.30.2 0.7

].

Then,

Ω

[0.60.4

]=

[0.8 0.30.2 0.7

] [0.60.4

]= 1 ·

[0.60.4

]and

Ω

[1−1

]=

[0.8 0.30.2 0.7

] [1−1

]=

[0.5−0.5

]=

1

2

[1−1

].

Hence, the eigenvalues of Ω are 1 and 12

with corresponding eigenvectorswhich are scalar multiples of the vectors given above.

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

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

Ω |v⟩ = λ |v⟩ .

This means that(Ω− λI) |v⟩ = |0⟩ ,

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

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

But, this is a contradiction because |v⟩ , |0⟩ if |v⟩ is an eigenvector. Nowconsider a matrix representation of the operator Ω in some basis. Then,from basic matrix theory, Ω− λI is invertible if and only if det(Ω− λI) , 0.If Ω is an n × n matrix, det(Ω − λI), called a characteristic polynomial, isan n-th degree polynomial with n roots11. We can find all the eigenvalues

11We do not always have n distinct roots. The total number of the roots is n if multi-plicity is taken into account. For example, if the polynomial has (x− 1)2 when factored,1 is a root with a multiplicity of 2, and it counts as two roots.

2.10. SPECIAL TYPES OF OPERATORS 49

by solving the characteristic equation det(Ω − λI) = 0. In order to com-pute det(Ω− λI) we need to pick a particular basis. However, the resultingeigenvalues are basis-independent because the characteristic polynomial isbasis-independent. For a proof, see Theorem A.4 in Appendix A.

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

2.10 Special Types of Operators2.10.1 Hermitian Operators12

Definition 2.25 (Hermitian Operators) A Hermitian operator isan operator which is the adjoint of itself; that is, an operator Ω isHermitian if and only if

Ω = Ω†.

In particular, this means that

⟨Ωv |w⟩ = ⟨v |Ωw⟩

as both sides are equal to⟨v|Ω |w⟩ .

In the linear algebra of real matrices, the matrix representation of aHermitian operator is nothing but a symmetric matrix. But, our ma-trices are complex, and the matrix representing a Hermitian operatorshould equal its transpose conjugate.

Definition 2.26 (Anti-Hermitian Operators) 13 An operator Ω is anti-12These are also called self-adjoint operators. You may hear that self-adjoint is the term

used by mathematicians and Hermitian is used by physicists. However, there is a subtledifference between ‘self-adjoint” and “Hermitian”. We will not want to get bogged downworrying too much about such details, but interested readers should check a book titled“Functional Analysis” by Michael Reed and Barry Simon [Reed and Simon, 1980, p.255].

13These are also known as anti-self-adjoint operators.


Hermitian ifΩ† = −Ω.

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

Ω =Ω+ Ω†

2+

Ω− Ω†

2,

then, the first term Ω+Ω†

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

2is anti-

Hermitian.

Hermitian operators play a central role in quantum mechanics due toits special characteristics.

Fact 2.3 (Eigenvalues and Eigenkets of a Hermitian Operator)

1. The eigenvalues of a Hermitian operator are real. Measurablevalues in physics such as position, momentum, and energy arereal numbers, and they are obtained as eigenvalues of Hermitianoperators in quantum mechanics.

2. The eigenkets of a Hermitian operator can always be chosen sothat they are mutually orthogonal. Because an eigenket can bemultiplied by any nonzero number to generate another eigenket ofany desired “length”, this means that we can always choose theeigenkets which form an orthonormal set.

3. The eigenkets of a Hermitian operator is a complete set in thesense that any vector can be expressed as a linear combination ofthe eigenkets. Therefore, if you know the action of a Hermitianoperator Ω on the eigenkets, you completely understand the actionof Ω on any vector in the space.Verification: Let v =

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

|λi⟩’s are the corresponding eigenkets. Then, Ωv = Ω(∑αi |λi⟩) =∑

αi (Ω |λi⟩)


Example 2.5 (A 2-by-2 Hermitian Operator/Matrix) Let

Ω =

[1 i−i 1

].

As the adjoint is the transpose conjugate, we have

Ω† =

[1 i−i 1

]= Ω,

and Ω is Hermitian. Observe that

Ω

[−i1

]=

[1 i−i 1

] [−i1

]=

[00

]= 0 ·

[−i1

]and

Ω

[i1

]=

[1 i−i 1

] [i1

]=

[2i2

]= 2 ·

[i1

]

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

|0⟩ =[−i1

]and |2⟩ =

[i1

].

Because

⟨0|2⟩ =[−i1

]† [i1

]=[i 1

] [ i1

]= 0,

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

|0⟩+ |2⟩2

=1

2

[02

]=

[01

]and

− |0⟩+ |2⟩2i

=1

2i

[2i0

]=

[10

].


Therefore, any ket vector v =

[v1v2

]can be expressed as a linear com-

bination of |0⟩ and |2⟩ as follows.

v =

[v1v2

]= v1

[10

]+ v2

[01

]= v1

(|0⟩+ |2⟩

2

)+ v2

(− |0⟩+ |2⟩

2i

)

=(v12− v2

2i

)|0⟩+

(v12

+v22i

)|2⟩

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

1√2

[−i1

]and 1√

2

[i1

]

form an orthonormal basis.

Let us prove 1.

Theorem 2.6 (Real Eigenvalues) The eigenvalues of a Hermitianoperator Ω are real.

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

λ∗⟨v∣∣∣ v⟩ = ⟨

λv∣∣∣ v⟩ = ⟨

Ωv∣∣∣ v⟩ = ⟨

v∣∣∣Ω†

∣∣∣ v⟩ = ⟨v∣∣∣Ω∣∣∣ v⟩ = ⟨

v∣∣∣λv⟩ = λ

⟨v∣∣∣ v⟩

=⇒ (λ∗−λ)⟨v∣∣∣ v⟩ = 0 =⇒ λ∗ = λ

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

Theorem 2.7 Eigenkets |v⟩ and |w⟩ of a Hermitian operator Ω corre-sponding to different eigenvalues λv and λw are orthogonal.

Proof

⟨ω |Ω|v⟩ = ⟨w |λv|v⟩ = λv ⟨w | v⟩


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

⟨w |Ω|v⟩ =⟨w∣∣∣Ω†|v

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

w ⟨v |w⟩∗ = λw ⟨w | v⟩ .

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

as λv , λw.

We will now prove 2 and 3 and some more.

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

the k-th column of [Ωij]B =

0......

λk...0

k

ProofFrom p.41, we know that the k-th column of [Ωij] is nothing but Ω |uk⟩. But,

Ω |uk⟩ = λkuk = λk

0......

1...0

k

=

0......

λk...0

k.


In matrix form,

Ω11 · · · · · · Ω1k · · · Ω1n...

. . ....

......

. . ....

...

Ωk1 · · · · · · Ωkk · · · Ωkn...

.... . .

...Ωn1 · · · · · · Ωnk · · · Ωnn

0......

1...0

k

=

Ω1k......

Ωkk...

Ωnk

= λk

0......

1...0

k

=

0......

λk...0

k

⇐⇒ Ωjk = λkδjk.

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

ProofΩjk = λkδjk follows from Lemma 2.1. Ωkj = λkδkj follows from the definitionof a Hermitian operator and Theorem 2.6. The matrix representation of Ω†,denoted here by [Ω†

ij]B, is the transpose conjugate of the matrix for Ω denotedby [Ωij]B. As the k-th column of [Ωij]B is

0......

λk...0

k,

the k-th row of [Ω†ij]B is

[ k

0 · · · · · · λ∗k · · · 0

].


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

0......

0 · · · · · · λk · · · 0...0

.

Lemma 2.2 (Block Structure of Hermitian Matrices) Consider an or-dered list of all the eigenvalues λ1, λ2, . . . , λn of a Hermitian operator Ω,where an eigenvalue λ is repeated mλ times if its multiplicity as a root ofthe characteristic polynomial is mλ. Without loss of generality, we can as-sume that the same eigenvalues appear next to each other without a differenteigenvalue in between.14 Then, the matrix representation of Ω in the basis|λ1⟩ , |λ2⟩ , . . . , |λn⟩

Including a root λ with multiplicity mλ mλ times,

Theorem 2.8 If an operator Ω on Vn is Hermitian, there is a basisof its orthonormal eigenvectors. In this basis, sometimes called aneigenbasis, the matrix representation of Ω is diagonal, and the diagonalentries are the eigenvalues. Note that this theorem only claims thatthere is at least one such basis. Theres is no claim about uniqueness asit is not true.

ProofConsider the characteristic polynomial which is of n-th degree. If we set itequal to zero, the equation, called the characteristic equation, has at least

14For simplicity, suppose we have a list of four eigenvalues consisting of two λ’s, one γ,and one ω. Then, the ordered list is λ, λ, ω, γ, λ, λ, γ, ω, ω, λ, λ, γ, γ, λ, λ, ω andnot λ, ω, λ.


one root λ1 and a corresponding normalized eigenket |λ1⟩. This can beextended to a full basis B according to Theorem A.5. Now, following theGram-Schmidt procedure, Theorem 2.4, we can find an orthonormal basisO = o1,o2,o3, . . . ,on such that o1 = |λ1⟩. In this basis, according toCorollary 2.1, Ω has the following matrix representation which we denote byM or [Ω] in keeping with the notation set on p.44.

M = [Ω] =

λ1 0 · · · 00...0

If we denote the shaded submatrix by N , it can be regarded as the ma-trix representation of an operator acting on the subspace O2 spanned byo2,o3, . . . ,on. The characteristic polynomial for M is given by

det(M − λIn) = (λ1 − λ)× det(N − λIn−1); (2.20)

where In is the n× n identity matrix, In−1 is the (n− 1)× (n− 1) identitymatrix, and det(N − λIn−1) is the characteristic polynomial of N . Becausedet(N − λIn−1) is P n−1(λ), an (n − 1)st degree polynomial in λ, it musthave at least one root λ2 and a normalized eigenket |λ2⟩ unless n = 1. Be-cause we are regarding N as operating on O2, |λ2⟩ is a linear combinationof o2,o3, . . . , and on, and is orthogonal to |λ1⟩. As before, we can find anorthonormal basis |λ2⟩ ,o′

3,o′4, . . . ,o

′n of O2. Note that λ2 is an eigenvalue

of M , |λ2⟩ is an eigenvector of M , and O1,2 = |λ1⟩ , |λ2⟩ ,o′3,o

′4, . . . ,o

′n is

an orthonormal basis of the original vector space Vn.

Let us take a detour here and see how the relations between M and Nwork explicitly. It may help you understand what is going on better.

First, λ2 is clearly an eigenvalue of M as it is a root of the character-istic polynomial for M because of the relation (2.20). Next, as |λ2⟩ is alinear combination of o2,o3, . . . , and on, it is also a linear combination of|λ1⟩ ,o2,o3, . . . , and on. We can write

|λ2⟩ = 0 · |λ1⟩+n∑i=2

eioi.


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

|λ2⟩ =

0e2...en

Hence,

M |λ2⟩ =

λ1 0 · · · 00...0

0e2...en

=

0

λ2e2...

λ2en

= λ2 |λ2⟩ ; (2.21)

where the second equality in (2.21) follows from0...0

0e2...en

=

[0

N |λ2⟩

]=

[0

λ2 |λ2⟩

]. (2.22)

Note that |λ2⟩ is regarded as a vector in Vn in (2.21) and as a vector inthe subspace spanned by o2,o3, . . . , and on in (2.22). There is a trade-offbetween avoiding abuse of notation and simplicity, and our choice here issimplicity.

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

M = [Ω] =

λ1 0 0 · · · 00 λ2 0 · · · 00 0...

...0 0

Repeating the same procedure, we will finally arrive at

M = [Ω] =

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

......

. . ....

0 0 0 · · · λn

.


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

Example 2.6 (Diagonalization of a Hermitian Matrix) Consider

Ω =

[1 11 1

].

(2.23)

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

Ω

[11

]=

[1 11 1

] [11

]=

[22

]= 2 ·

[11

]and

Ω

[1−1

]=

[1 11 1

] [1−1

]=

[00

]= 0 ·

[1−1

].

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

|e1⟩ =1√2

[11

]and |e2⟩ =

1√2

[1−1

](2.24)

Then,

Ω11 = ⟨e1|Ω|e1⟩ =1√2

[1 1

] [ 1 11 1

]1√2

[11

]

=1

2

[1 1

] [ 22

]=

1

2· 4 = 2. (2.25)

Similarly, we also get

Ω12 = ⟨e1|Ω|e2⟩ = 0

Ω21 = ⟨e2|Ω|e1⟩ = 0


andΩ22 = ⟨e2|Ω|e2⟩ = 0.

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

Fact 2.4 If Ω is an anti-Hermitian operator:

1. The eigenvalues of Ω are purely imaginary.

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

A |n⟩ = an |n⟩

A =∑n

|n⟩ an ⟨n|

the spectral decomposition of the Hermitian operator A

2.10.2 Simultaneous Diagonalization

Theorem 2.9 (Commuting Hermitian Operators) If Ω and Γ arecommuting Hermitian operators, there exists a basis consisting of com-mon eigenvectors, such that both Ω and Γ are diagonal in that basis.

This very useful theorem is actually a corollary of the following moregeneral theorems from matrix algebra. For details, check Appendix A.

Theorem 2.10 (Commuting Diagonalizable Matrices) Let A and Bbe diagonalizable n × n matrices. Then, AB = BA if and only if A andB are simultaneously diagonalizable.

Theorem 2.11 (A Family of Diagonalizable Matrices) Let F be a fam-ily of diagonalizable n × n matrices. Then, it is a commuting family if andonly if it is simultaneously diagonalizable.


2.11 Active and Passive TransformationsWe will mainly deal with active transformations in this book. So, you onlyneed to familiarize yourself with the general concept.

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

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

Rθ =

[cos θ − sin θsin θ cos θ

](2.26)

such that the new position (x′0, y

′0) is obtained as follows.[

x′0

y′0

]= Rθ

[x0y0

]=


] [x0y0

]=

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

](2.27)

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

• A passive transformation does not move the point (x0, y0) but onlychanges the coordinate system. To take a 2-dimensional rotation asan example, this corresponds to rotating the x-y coordinate systemwithout moving the point itself. Because the only change is in thevalue of the point’s x- and y-coordinates relative to the new rotatedaxes, and the position itself does not change in this case, the physicalenvironment of the point remains the same, and a dynamical variableω(q, p) will have the same value after the transformation.

Note that the new coordinates (x′0, y

′0) are numerically the same whether

you rotate the axes by θ or the point by −θ. Hence, the effect of passivetransformation by θ can be computed as below.[x′0

y′0

]= R−θ

[x0y0

]=

[cos(−θ) − sin(−θ)sin(−θ) cos(−θ)

] [x0y0

]=

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

](2.28)

2.11. ACTIVE AND PASSIVE TRANSFORMATIONS 61

Again, despite this relation, the two transformations should not be con-fused with each other because the active transformation is the only physi-cally meaningful transformation, in the sense that the values of dynamicalvariables may change.

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

⟨v′|Ω|v⟩ −→ ⟨Uv′|Ω|Uv⟩ = ⟨v′|U †ΩU |v⟩ (2.29)

In (2.29), the change is caused to the vector v in ⟨Uv′|Ω|Uv⟩, and to theoperator Ω in ⟨v′|U †ΩU |v⟩. Nevertheless, the matrix elements of Ω are thesame between the two formulations or interpretations.

• An active transformation corresponds to

⟨Uv′|Ω|Uv⟩ ; (2.30)

where the vector v is transformed as below.

v −→ Uv (2.31)

• A passive transformation picture is painted by

⟨v′|U †ΩU |v⟩ ; (2.32)

where the operator Ω is transformed as follows.

Ω −→ U †ΩU (2.33)

Because the essence of quantum mechanics is the matrix elements of opera-tors, and all of the physics of quantum mechanics lies in the matrix elements,the equality of (2.30) and (2.32) given in (2.29) indicates that these twoviews are completely equivalent. Hence, active and passive transformationsin quantum mechanics provide two equivalent ways to describe the samephysical transformation.

62

Exercises1. Consider

Ω =

[0 11 0

].

The eigenvalues are 1 and −1.

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

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

(d) Express an arbitrary vector v =

[v1v2

]as a linear combination of

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

2. Answer the following questions about the matrix representation of aHermitian operator.

(a) Given that the matrix

M =

[6 4α 0

]

is Hermitian, what is α?

(b) The eigenvalues are 8 and −2. Find NORMALIZED and REALeigenvectors |8⟩ and |−2⟩, remembering that |λ⟩ means an eigen-ket associated with the eigenvalue λ.

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

3. LetA =

[0 21 1

]and B =

[−1 21 0

].

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

63

(b) The eigenvalues for A are 2 and −1. Find two real normalizedeigenvectors, |2⟩A and |−1⟩A.

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

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

4. Consider two Hermitian matrices

M =

[1 −1−1 1

]and N =

[0 11 0

].

(a) Prove that M and N commute.(b) Find the eigenvalues and real normalized eigenvectors of M and

N .(c) Show that M and N are simultaneously diagonalizable.

Chapter 3

The Postulates of QuantumMechanics

This is where quantum mechanics markedly differs from mathematics asyou know ita. The postulates of quantum mechanics cannot be proved ordeduced in a purely mathematical fashion. The postulates are hypothe-ses, that are given to you from the beginning. If no disagreement withexperiments is found, the postulates are accepted as correct hypothe-ses. Sometimes, they are called axioms, which means the postulates areregarded true though not provable.

You should not feel excessively uncomfortable with this. Much ofwhat we know in physics cannot be proved. For example, consider thecelebrated Newton’s Second Law of Motion; F = ma. There is no purelymathematical derivation of this relationship. In physics, a theory isdeemed “correct” if there is a good agreement between observations andtheoretical predictions. While classical Newtonian mechanics suffices formacroscopic systems, a huge body of experimental evidence supportsthe view that the quantum postulates provide a consistent descriptionof reality on the atomic scale.

aI am saying ”mathematics as you know it” because mathematics also relies onmany postulates often known as axioms. However, most of us are used to encounter-ing definitions and proofs when doing mathematics and not necessarily postulates.

65

66 CHAPTER 3. THE POSTULATES OF QUANTUM MECHANICS

3.1 The Fundamental Postulates1. An isolated physical system is associated with a topologically sepa-

rable12 complex Hilbert space H with inner product ⟨ϕ|ψ⟩. Physicalstates can be identified with equivalence classes of vectors of length 1in H, where two vectors represent the same state if they differ only bya phase factor.

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

Each physical system has its associated Hilbert spaceH, so that there is one-to-one correspondence between aphysical state s of the system and a vector vs, or a ket|s⟩, of unit length in H.

2.

To every observable q in classical mechanics there corre-sponds a Hermitian operator Q in quantum mechanics.

Table 3.1 summarizes the correspondences.In Table 3.1, I am showing the correspondence between classical ob-servables and quantum mechanical operators for a single variable x forsimplicity. If we include y and z, parts of Table 3.1 should be modifiedas shown in Table 3.2.

1A topological space is separable if it contains a countable dense subset; that is, thereexists a sequence xi∞

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

2Separability means that the Hilbert space H has a basis consisting of countably manyvectors. Physically, this means that countably many observations suffice to uniquely de-termine the state.

3H or H is called Hamiltonian.

3.1. THE FUNDAMENTAL POSTULATES 67

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

(−i~ ∂

∂x

)Kinetic Energy T T or T − ~2

2m∂2

∂x2

Potential Energy V (x) V (x) or V (x) MV (x) : multiplicationTotal Energy E H or H3 − ~2

2m∂2

∂x2+ V (x)

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

∂y

)

Table 3.1: Physical observables and corresponding quantum operators

Observable q Q OperationMomentum p p or p −i~

(ı ∂∂x

+ ȷ ∂∂y

+ k ∂∂z

)Kinetic Energy T T or T − ~2

2m

(∂2

∂x2+ ∂2

∂y2+ ∂2

∂z2

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

2m

(∂2

∂x2+ ∂2

∂y2+ ∂2

∂z2

)+ V (r)

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

∂y

)ly Ly or Ly −i~

(z ∂∂x− x ∂

∂z

)lz Lz or Lz −i~

(x ∂∂y− y ∂

∂x

)

Table 3.2: Quantum operators in three dimensions

3.

When a measurement is made for an observable q asso-ciated with operator Q, the only possible outcomes arethe eigenvalues of Q denoted by λi.


4.

The set of eigenstates/eigenvectors of operator Q, de-noted by |λi⟩, forms a completea orthonormal set of theHilbert space H.

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

5.

Suppose a physical system is in state s. If the vectorvs ∈ H associated with the physical state s is given by

vs =n∑i=1

ai |λi⟩ ;

where n may go to ∞, the possible outcomes of a mea-surement of the physical quantity q are the eigenvaluesλ1, λ2, . . . , λk , . . ., and the probability that λk is observedis | ⟨λk |vs⟩ |2 = |ak|2.

6.

As soon as a measurement is conducted on a physicalstate s, yielding λk as the outcome, the state vector vscollapses to the corresponding eigenstate |λk⟩. Therefore,measurement affects the physical state of the system.a

aThis fact is often used in elaborate experimental verifications of quantummechanics.

7.

The average value of the observable q after a large num-ber of measurements is given by the expectation value

q or ⟨q⟩ = ⟨vs |Q|vs⟩ .

3.1. THE FUNDAMENTAL POSTULATES 69

8.

The state vector |s⟩ associated with a physical state stypically depends on time and the spatial coordinates;that is, |s⟩ = |s(r, t)⟩. The state evolves in time accordingto the time-dependent Schrodinger equation.

i~∂

∂t|s(r, t)⟩ = H |s(r, t)⟩ (3.1)

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

i~∂Ψ(r, t)∂t

= HΨ(r, t).

9. TO BE REWRITTEN: We can use the Schrödinger Equation toshow that the first derivative of the wave function should be continuous,unless the potential is infinite at the boundary. http://quantumme-chanics.ucsd.edu/ph130a/130 notes/node141.html

The wavefunction Ψ(r, t) and its spatial derivatives(∂/∂x)Ψ(r, t), (∂/∂y)Ψ(r, t), (∂/∂z)Ψ(r, t) are continuous inan isotropic mediuma.

aA good example of a non-isotropic medium is a medium where the poten-tial energy varies discontinuously with position. We will encounter a potentialof this kind when we discuss the infinite square well potential.

10. You can skip this for now.

The total wavefunction must be antisymmetric with re-spect to the interchange of all coordinates of one fermiona

with those of another. Electron spin must be included


in this set of coordinates. The Pauli exclusion principleis a direct consequence of this antisymmetry principle.Slater determinants provide a convenient means of en-forcing this property on electronic wavefunctions.

aParticles are classified into two categories, fermions and bosons. Fermionshave half-integral intrinsic spins such as 1/2, 3/2, and so forth, while bosonshave integral intrinsic spins such as 0, 1, and so forth. Quarks and leptons,as well as most composite particles, like protons and neutrons, are fermions.All the force carrier particles, such as photons, W and Z bosons, and gluonsare bosons, as are those composite particles with an even number of fermionparticles like mesons.

Of the 10 postulates, Postulates 8 is distinctly different from the othernine. The nine postulates describe the system at a given time t, while Pos-tulate 8 specifies how the system changes with time.

3.2 Unitary Time EvolutionTime evolution of a quantum system is always given by a unitary transfor-mation U , such that

|s(t)⟩ = U(t) |s(0)⟩ . (3.2)

The nature of U(t) depends on the system and the external forces it expe-riences. However, U(t) does not depend on the state |s⟩. Hence, U(α |s1⟩ +β |s2⟩) = αU |s1⟩ + βU |s2⟩, and the time evolution operator U is linear.Again, the key here is that the time evolution operator U is a function of thephysical system and not individual states. Sometimes U(t) is referred to asa propagator.

The unitarity of U follows from the time-dependent Schrodinger equation(3.1) as follows. We have

i~∂

∂t|s(t)⟩ = H(t) |s(t)⟩ or ∂

∂t|s(t)⟩ = −i

~H(t) |s(t)⟩ ;

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

|s(t)⟩ = U(t) |s(0)⟩

3.2. UNITARY TIME EVOLUTION 71

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

∂

∂t

(U(t) |s(0)⟩

)=−i~H(t)U(t) |s(0)⟩ =⇒

(∂

∂tU(t)

)|s(0)⟩ =

(−i~H(t)U(t)

)|s(0)⟩

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

∂

∂tU(t) and −i

~H(t)U(t)

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

∂

∂tU(t) =

−i~H(t)U(t)4. (3.3)

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

∂tU(t)

)†

=(−i~H(t)U(t)

)†=⇒ ∂

∂tU †(t) =

i

~U †(t)H†(t) =

i

~U †(t)H(t).

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

∂

∂t

(U †(t)U(t)

)=

(∂

∂tU †(t)

)U(t)+U †(t)

(∂

∂tU(t)

)=

(∂

∂tU †(t)

)U(t)+U †(t)

(∂

∂tU(t)

)

=i

~U †(t)H(t)U(t)+U †(t)

−i~H(t)U(t) =

i

~U †(t)

(H(t)− H(t)

)U(t) = 0.

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

∥ |s(t)⟩∥2 = ⟨s(t)|s(t)⟩ =⟨s(0)

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

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

4One way to understand what this equality means is to think in terms of their matrixrepresentations with respect to a basis.


3.3 Time-Independent HamiltonianSo far, we have considered a Hamiltonian with explicit time-dependence asis clear from the notation H(t). However, for many physical systems, theHamiltonian is not time-dependent. Fro example, consider a typical Hamil-tonian given by

H = − ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)+ V (r)

in Table 3.2. In the most general case, the potential energy term V canhave explicit time dependence so that V = V (r, t). However, if V does notdepend on t, and neither does H as shown above, there are simple relationsamong the time evolution operator U(t), the system Hamiltonian H, and theeigenstates of H.

3.3.1 Propagator as a Power Series of HWe start with the differential relation (3.3)

d

dtU(t) =

−i~H(t)U(t); (3.4)

where we switched to total time derivative as we will only focus on t here,holding all the other variables fixed. Let us now remove time-dependencefrom H in (3.4).

d

dtU(t) =

−i~HU(t); (3.5)

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

d

dtf(t) =

−i~Hf(t);

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

f(t) = f(0) exp[−iHt~

],

3.3. TIME-INDEPENDENT HAMILTONIAN 73

you may think the answer to the analogous operator differential equation(3.5) is something like

U(t) = U(0) exp[−iHt~

]= exp

[−iHt~

]; (3.6)

where the second equality holds as U(0) is the identity operator by necessity.As it turns out, this is the right answer with the interpretation of the right-hand side as a power series in operator H. This is explained in more detailin Appendix B. But, we will only check it formally here to convince ourselvesthat the solution really works. Maclaurin series of the exponential functiongives

exp[−iHt~

]=

∞∑n=0

[−iH~

]nn!

tn = I +∞∑n=1

[−iH~

]nn!

tn; (3.7)

where[

−iH~

]0= I by definition. Then,

d

dt

(exp

[−iHt~

])=

d

dt

I + ∞∑n=1

[−iH~

]nn!

tn

=d

dt

∞∑n=1

[−iH~

]nn!

tn

=∞∑n=1

[−iH~

]nn(n− 1)!

ntn−1

=−iH~

∞∑n=1

[−iH~

]n−1

(n− 1)!tn−1 =

−iH~

∞∑n=0

[−iH~

]nn!

tn =−i~H exp

[−iHt~

].

So,

U(t) = exp[−iHt~

]

satisfies the differential equation (3.5).

In Section 3.2, we showed that the time evolution operator U(t) is unitaryfor any Hamiltonian H. When H does not depend on time, we have an alter-native proof that U(t) is unitary. However, note that our discussion belowis somewhat heuristic and not completely rigorous mathematically speaking.


First, taking the term-by-term adjoint of the right-hand side of (3.7), weget

(exp

[−iHt~

])†=

∞∑n=0

[−iH~

]nn!

tn

†

=∞∑n=0

[(−iH~

)†]n

n!tn =

∞∑n=0

[+iH~

]nn!

tn = exp[iHt

~

].

It suffices to show

U(t)U †(t) = exp[−iHt~

]exp

[iHt

~

]= I.

Recall from Theorem 2.8 that a matrix representing a Hermitian operator isdiagonalizable with its eigenvalues on the diagonal. Because H is Hermitian,there exits an invertible matrix J such that

JHJ−1 = D =

λ1

. . .

λm

or H = J−1DJ = J−1

λ1

. . .

λm

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

Hn =(J−1DJ

)n= J−1DnJ = J−1

λ1

. . .

λm

n

J = J−1

λn1

. . .

λnm

J.Therefore,

U(t) = e− i~Ht =

∞∑n=0

(−i~

)nHntn

1

n!=

∞∑n=0

(−i~

)nJ−1DnJtn

1

n!=

∞∑n=0

J−1(−i~

)nDntn

1

n!J

= J−1

( ∞∑n=0

(−i~

)nDntn

1

n!

)J = J−1

∞∑n=0

(−i~

)n λ1

. . .

λm

n

tn1

n!

J

= J−1

∞∑n=0

(−i~

)n λn1

. . .

λnm

tn 1

n!

J

= J−1

∑∞n=0

(−i~

)nλn1 t

n 1n!

. . . ∑∞n=0

(−i~

)nλnmt

n 1n!

J

3.3. TIME-INDEPENDENT HAMILTONIAN 75

= J−1

∑∞n=0

(−i~λ1t

)n1n!

. . . ∑∞n=0

(−i~λmt

)n1n!

J

= J−1

exp

(−i~λ1t

). . .

exp(

−i~λmt

) J.

Similarly,

U †(t) = ei~Ht = J−1

exp

(i~λ1t

). . .

exp(i~λmt

) J.

We now have

U(t)U †(t) = J−1

exp

(−i~λ1t

). . .

exp(

−i~λmt

) JJ−1

exp

(i~λ1t

). . .

exp(i~λmt

) J

= J−1

exp

(−i~λ1t

). . .

exp(

−i~λmt

)

exp(i~λ1t

). . .

exp(i~λmt

) J

= J−1

1

. . .

1

J = I.

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

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

H |E⟩ = E |E⟩ ;


where E is an eigenvalue representing the total energy, and |E⟩ is an associ-ated normalized eigenket. From (2.10),∑

|E⟩ ⟨E| = I.

Hence,|s(t)⟩ =

∑|E⟩ ⟨E|s(t)⟩ ,

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

|s(t)⟩ =∑

cE(t) |E⟩ .

From Postulate 8,

i~Ψ(r, t)∂t

= HΨ(r, t) =⇒ i~∂ |s(t)⟩∂t

= H |s(t)⟩ =⇒ i~∂∑cE(t) |E⟩∂t

= H∑

cE(t) |E⟩

=⇒∑

i~∂cE(t)

∂t|E⟩ =

∑cE(t)E |E⟩ =⇒

∑(i~∂cE(t)

∂t− EcE(t)

)|E⟩ = 0.

As |E⟩’s are linearly independent,

i~∂cE(t)

∂t− EcE(t) = 0 for each value of E.

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

i~∂cE(t)

∂t− EcE(t) = 0 =⇒ ∂cE(t)

∂t=−iE~

cE(t) =⇒ cE(t) = cE(0)e−iEt/~.

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

|s(t)⟩ =∑⟨E|s(0)⟩ e−iEt/~ |E⟩ =

∑|E⟩ ⟨E|s(0)⟩ e−iEt/~.

Comparing this with (3.2), we get

U(t) =∑E

|E⟩ ⟨E| e−iEt/~.

3.4. THE UNCERTAINTY PRINCIPLE 77

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

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

U †(t) =

(∑E

|E⟩ ⟨E| e−iEt/~)†

=∑E

(|E⟩ ⟨E| e−iEt/~

)† Step 1−−−→∑E

e−iEt/~ ⟨E| |E⟩

Step 2−−−→∑E

eiEt/~ ⟨E| |E⟩ Step 3−−−→ U †(t) =∑E

eiEt/~ |E⟩ ⟨E| .

Hence,

U(t)U †(t) =

(∑E

|E⟩ ⟨E| e−iEt/~)(∑

E′eiE

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

=∑E=E′

|E⟩ ⟨E ′| e−i(E−E′)t/~ =∑E

|E⟩ ⟨E| = I.

This checks.

3.4 The Uncertainty PrincipleThe uncertainty principle is one of the celebrated consequences of quantummechanics, whose influence has gone beyond the realm of physics into hu-manities such as philosophy.

3.4.1 Uncertainty and Non-CommutationTheorem 3.1 (The Uncertainty Relation) Consider two physical observ-ables represented by Hermitian operators A and B. We denote the expectationvalue of an operator Ω by ⟨Ω⟩; that is, if the state of the system is |u⟩ afternormalization,

⟨Ω⟩ = ⟨u|Ω |u⟩ .

In addition, define A by


A =√⟨(A− ⟨A⟩I)2⟩ab.

aFrom the formula, you can see that A is the standard deviation of the observ-able A in the state |u⟩.

bThis definition means (A)2 = ⟨(A− ⟨A⟩)2⟩ = ⟨(A2−2A⟨A⟩+ ⟨A⟩2)⟩ = ⟨A2⟩−2⟨A⟩⟨A⟩+ ⟨A⟩2 = ⟨A2⟩ − ⟨A⟩2 or A =

√⟨A2⟩ − ⟨A⟩2.

Then, [A,B] = iα implies AB ≥ |α|2

; where α is a real scalar.

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

CA = A− ⟨A⟩I and CB = B − ⟨B⟩I

Then,

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

Next define two vectors by

|v⟩ = CA |u⟩ and |w⟩ = CB |u⟩ .

Then,

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

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

and

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

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

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

Now, by Schwarz Inequality (Theorem 2.3),

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

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

4| ⟨u|[CA, CB] + CA, CB|u⟩ |2;

3.4. THE UNCERTAINTY PRINCIPLE 79

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

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

= ⟨u|C†ACB|u⟩+⟨u|C

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

∗

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

(A)2 (B)2 ≥ 1

4|iα+ β|2 = 1

4(α2 + β2) ≥ 1

4α2.

Therefore,AB ≥ |α|

2. (3.8)

3.4.2 Position and MomentumFrom Table 3.1, p = −i~ d

dxand x =Mx. In order to compute the commutator

[x, p], we will let [x, p] act on a function f(x).

[x, p]f(x) = x(−i~ ddx

)f(x)−(−i~ ddxx)f(x) = −i~x d

dxf(x)+i~

d

dx(xf(x))

= −i~x ddxf(x)+i~f(x)+i~x

d

dxf(x) = i~f(x) =⇒ [x, p] = i~

Therefore,xp ≥ ~

2.

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

2has an energy-

time analog, Et ≥ ~2. However, this must be a different kind of relation-

ship because t is not a dynamical variable. This uncertainty cannot haveanything to do with lack of commutation.


3.4.3.1 Time rate of change of expectation values

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

d

dt⟨A⟩(t) = d

dt⟨u(t)|A|u(t)⟩ ;

where |u(t)⟩ is the state of the system at time t. From (3.1),

d

dt|u(t)⟩ = 1

i~H |u(t)⟩ and d

dt⟨u(t)| = − 1

i~⟨u(t)|H5.

So,

d

dt⟨A⟩(t) =

(d

dt⟨u(t)|

)A |u(t)⟩+⟨u(t)|A

(d

dt|u(t)⟩

)

= − 1

i~⟨u(t)|HA |u(t)⟩+⟨u(t)|A 1

i~H |u(t)⟩ = 1

i~⟨u(t)|−HA+AH |u(t)⟩ = 1

i~⟨[A,H]⟩(t).

We haved

dt⟨A⟩(t) = 1

i~⟨[A,H]⟩(t). (3.9)

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

3.4.3.2 Time-energy uncertainty

Our derivation of the general form of the uncertainty principle (3.8) concernstwo observables. However, as mentioned already, time t is not a dynamicalvariable. We need a way to make a connection between t and an observableA if we are to use the general inequality (3.8) to formulate time-energy un-certainty.

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

AE ≥ 1

2|⟨[A,H]⟩|

5One way to understand this is to regard |u(t)⟩ as an n×1 matrix (a column vector) and⟨u(t)| as its transpose conjugate (a row vector). With this interpretation, d

dt |U(t)⟩ becomesentry by entry differentiation of a matrix, and

(ddt |u(t)⟩

)†= d

dt ⟨u(t)| and(1i~H |u(t)⟩

)†=

− 1i~ ⟨u(t)|H naturally.

3.5. CLASSICAL MECHANICS AND QUANTUM MECHANICS 81

from (3.8). Now, from (3.9),

1

2|⟨[A,H]⟩| = 1

2

∣∣∣∣∣i~ ddt⟨A⟩∣∣∣∣∣ = ~2

∣∣∣∣∣ ddt⟨A⟩∣∣∣∣∣ .

Hence,

AE ≥ ~2

∣∣∣∣∣ ddt⟨A⟩∣∣∣∣∣ .

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

∣∣∣ ddt⟨A⟩

∣∣∣, is one candidate for t. Withthis definition, we have

t = A∣∣∣ ddt⟨A⟩

∣∣∣ .Then,

Et = EA∣∣∣ ddt⟨A⟩

∣∣∣ ≥~2

∣∣∣ ddt⟨A⟩

∣∣∣∣∣∣ ddt⟨A⟩

∣∣∣ =~

2=⇒Et ≥ ~

2.

Again this inequality is completely different from that for two observables. Itdepends on how you devise your way to measure time which is different from”the” usual time t. This may be the reason why time-energy uncertainty isso notoriously controversial.

3.5 Classical Mechanics and Quantum Me-chanics

According to the correspondence principle advocated by Niels Bohr, everynew physical theory must contain as a limiting case the old theory. So, quan-tum mechanics has to be a superset of classical mechanics. Indeed, classicalmechanics is a limiting case of quantum mechanics as h or ~ → 0. It wasalso shown by Ehrenfest that the expectation values of quantum mechani-cal observables satisfy the same equations as the corresponding variables inclassical mechanics.


3.5.1 Ehrenfest’s TheoremOne embodiment of the correspondence principle are the two relations provedby Ehrenfest that connect classical mechanics and quantum mechanics.Theorem 3.2 (Ehrenfest Theorem) If Ω is a quantum mechanical oper-ator and ⟨Ω⟩ is its expectation value, we have

d

dt⟨Ω⟩ = 1

i~⟨[Ω, H]⟩+

⟨∂Ω

∂t

⟩. (3.10)

ProofThis is a generalized version of (3.9); where the operator is possibly time-dependent. We will write out the integral for the expectation values explicitlyin order to see the reasons for the total derivative d

dtand the partial deriva-

tive ∂∂t

clearly. We will specialize to a one-dimensional case without lossof generality and let Ψ(x, t) be a normalized wavefunction representing thestate. Then,d

dt⟨Ω⟩ = d

dt

∫Ψ(x, t)∗Ω(x, t)Ψ(x, t) dx =

∫ ∂

∂t

(Ψ(x, t)∗Ω(x, t)Ψ(x, t)

)dx6

=∫ ∂

∂t

(Ψ(x, t)∗

)Ω(x, t)Ψ(x, t) dx

+∫

Ψ(x, t)∗ ∂

∂t

(Ω(x, t)

)Ψ(x, t) dx

+∫Ψ(x, t)∗Ω(x, t)

∂

∂t

(Ψ(x, t)

)dx. (3.11)

Now, from Postulate 8 on p.69, we have

i~∂

∂t|s(r, t)⟩ = H |s(r, t)⟩

=⇒ ∂

∂t|s(r, t)⟩ = 1

i~H |s(r, t)⟩ and ∂

∂t⟨s(r, t)| = 1

−i~⟨s(r, t)| H7. (3.12)

In our case (3.12) becomes∂

∂t

(Ψ(x, t)

)=

1

i~HΨ(x, t) and ∂

∂t

(Ψ(x, t)∗

)=−1i~

Ψ(x, t)∗H. (3.13)

6We need to switch from ddt to ∂

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

7Note that ∂∂t ⟨s(r, t)| =

1−i~ ⟨s(r, t)| H follows from

(∂∂t |s(r, t)⟩

)†=(

1i~H |s(r, t)⟩

)†.

3.5. CLASSICAL MECHANICS AND QUANTUM MECHANICS 83

Substituting these into (3.11),

d

dt⟨Ω⟩ =

∫ −1i~

Ψ(x, t)∗HΩ(x, t)Ψ(x, t) dx+

⟨∂Ω

∂t

⟩

+∫

Ψ(x, t)∗Ω(x, t)1

i~HΨ(x, t) dx

=1

i~

∫Ψ(x, t)∗(ΩH −HΩ)Ψ(x, t) dx+

⟨∂Ω

∂t

⟩

=1

i~⟨[Ω, H]⟩+

⟨∂Ω

∂t

⟩(3.14)

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

Corollary 3.1 (Time-Independent Hamiltonian) If the Hamiltonian Hdoes not have explicit time dependence, the time rate of change of the expecta-tion value of a variable Ω represented by operator Ω, < Ω >, is proportionalto the expectation value of the commutator between Ω and H, [Ω, H], with aproportionality constant −i

~. Therefore, < Ω > is conserved if Ω commutes

with a time-independent Hamiltonian H.

3.5.2 Exact Measurements and Expectation ValuesWe have seen in Tables 3.1 and 3.2 that each classical observable has itsassociated quantum mechanical operator. This is an exact one-to-one corre-spondence.

However, in terms of actual measurements, a quantum mechanical opera-tor does not generate one exact and fixed value under repeated measurementsas in the classical measurement. Instead, each quantum mechanical operatorspecifies a probability distribution of possible outcomes of the measurementby way of a sate function or wavefunction. This necessitates a probabilisticinterpretation of classical laws of physics.


As we saw in Ehrenfest Theorem, the general strategy is to use expec-tation values rather than exact measured values. In the quantum me-chanical formulation of classical laws of physics, the expectationvalues play the role of the classical variables.

85

Exercises1. Let Q be a quantum mechanical operator representing a physical ob-

servable q. An example of Q is the momentum operator P = −i~ ∂∂x

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

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

(b) Now consider the set |λi⟩ of all normalized eigenkets8 of Q.i. What is ⟨λi|λj⟩? This property makes λi an orthonormal

set.ii. The set λi is complete. What does this mean?

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

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

(b) With what probabilities will λi and λj be observed if a measure-ment is made?

(c) A measurement has been made for q and the value obtained wasλi. What is the state the system is in after the measurement? Youcan only answer this up to an arbitrary phase eiθ.

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

8Recall we use eigenket, eigenvector, and eigenfunction interchangeably.

Chapter 4

Spaces of InfiniteDimensionality

So far, we have focused mostly on finite-dimensional spaces. This was be-cause finite dimensions make it easy to explain mathematics and physics forthe author and also easy to understand mathematics and physics for thereadership. However, the attentive reader may have wondered why finitedimensions will suffice.

The simplest example is the position x and its quantum mechanical rep-resentation (operator) X or Mx. The position observable is continuously dis-tributed over the interval (−∞,+∞), and hence, there are infinitely manydistinct eigenvalues for the operator X, which in turn implies that thereare infinitely many eigenvectors which are mutually orthogonal according toTheorem 2.7. Needless to say, the dimensionality of the Hilbert space inwhich the operator X operates has to be infinite.

Another example is the discrete energy distribution as indicated in 4on p.14. This states that the total energy E can only take discrete valuesE1, E2, E3, . . . and not any value E ∈ [0,∞) as in classical mechanics whereE is continuously distributed. However, it does not place an upper limit onE, and infinitely many discrete values of E are possible.

87

88 CHAPTER 4. SPACES OF INFINITE DIMENSIONALITY

4.1 Two Types of InfinityTwo kinds of infinite dimensional vector spaces are encountered in quantummechanics. One infinity is known as countable or denumerable infinity, andthe other infinity is called uncountable/non-denumerable infinity. Discreteenergy level distribution is an example of countable infinity, and the positionx is an example of uncountable infinity. A canonical example of countableinfinity is the number of natural numbers N = 1, 2, 3, . . ., and a goodexample of uncountable infinity is the number of points in the interval (0, 1).In fact, countable infinity is far “smaller” than uncountable infinity. Namely,countably infinite number of points, when put next to each other, fit in a linesegment of length 0, while uncountably many points form a line of strictlypositive length. In order to understand the difference, let us see how we canshow N is “of length 0”.

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

1 ∈(1− ε(1− ε)

2, 1 +

ε(1− ε)2

), 2 ∈

(2− ε2(1− ε)

2, 2 +

ε2(1− ε)2

),

. . . , n ∈(n− εn(1− ε)

2, n+

εn(1− ε)2

), . . . .

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

∞∑n=1

εn(1− ε).

Noting that a geometric series ∑∞n=1 ar

n for 0 < r < 1 converges to a1−r and

we have (a, r) = (ε(1− ε), ε),∞∑n=1

εn(1− ε) = ε(1− ε)1− ε

= ε.

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

4.2. COUNTABLY INFINITE DIMENSIONS 89

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

n=1 εn(1 − ε)

because this is nothing but counting, and you cannot count the uncountable.Here is an additional qualitative description of what is happening. Eachpoint is, by definition, of length 0. If there are only countably many points,the total length is 0 even if there are infinitely many points. However, ifthere are uncountably many points, the total length is strictly positive asuncountable infinity overwhelms the length of 0 of each point.

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

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

x→∞f(x) = 0 and lim

x→∞g(x) =

∞, but limx→∞

f(x)g(x) = limx→∞

1/√x = 0.

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

f(x) = 0 and limx→∞

g(x) =∞,but lim

x→∞f(x)g(x) = lim

x→∞x =∞.

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

It is now clear that countable infinity and uncountable infinity should betreated separately.

4.2 Countably Infinite DimensionsIf the Hilbert space is of countably infinite dimensions, we have a completeorthonormal basis

|e1⟩ , |e2⟩ , . . . , |en⟩ , . . . or |ek⟩ ; k = 1, 2, . . . , ∞, (4.1)

satisfying

⟨ei|ej⟩ = δij (orthonormality) (4.2)

and∞∑i=1

|ei⟩ ⟨ei| = I (completeness). (4.3)


In this case, aside from the infinitely many components for vectors and in-finitely many elements matrices carry, everything is the same as for finitedimensional Hilbert spaces.

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

⟨x|x′⟩ = δ(x− x′) (orthonormality) (4.4)

and ∫ +∞

−∞|x⟩ ⟨x| dx = I (completeness); (4.5)

where the right-hand side of (4.4) is a real-valued function called the Diracdelta function with the following definition and properties. Delta function isnecessary whenever the basis kets are labeled by a continuous index includ-ing the position variable x.

Heuristic DescriptionQualitatively, the delta function can be described by

δ(x) =

∞ x = 00 x , 0

(4.6)

and ∫ +∞

−∞δ(x) dx = 1. (4.7)

Needless to say, this is an “extended” function, and no usual function hasthese properties. Rather than regarding the delta function as a function aswe know them, we should think of it as a device to make quantum mechanicsas well as other theories of physics work in a consistent fashion.

Properties of the Dirac Delta FunctionLet us look at the properties of δ(x− x′) found in (4.4).

4.3. UNCOUNTABLY INFINITE DIMENSIONS 91

1. As before, we haveδ(x− x′) = 0 if x , x′

and (4.8)∫ +∞

−∞δ(x− x′) dx′ = 1.

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

−∞δ(x− x′)f(x′) dx′ = f(x). (4.9)

3. The delta function is even as shown below.δ(x− x′) = ⟨x|x′⟩ = ⟨x′|x⟩∗ = δ(x′ − x)∗ = δ(x′ − x) (4.10)

The last equality holds because the delta function is a real-valued func-tion as stated on p.90.

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

dxf(x− x′) = lim

h→0

f(x− x′)− f((x− h)− x′)

h(4.11)

andd

dx′f(x− x′) = lim

h→0

f(x− (x′ + h))− f(x− x′)

h

= limh→0

f(x− h− x′)− f(x− x′)

h

= − limh→0

f(x− x′)− f((x− h)− x′)

h(4.12)

= − d

dxf(x− x′) (4.13)

imply the following relation for the “formal” derivatives of the delta function.1

δ′(x− x′) =d

dxδ(x− x′) = − d

dx′ δ(x− x′) (4.14)

1As the delta function is even, its “derivative” is an odd function. Generally speaking,let f(x) be a differentiable even function and y = −x, then, for an arbitrary value x0,d

dxf(x)

∣∣∣∣x=x0

=d

dxf(−x)

∣∣∣∣x0

= − d

d(−x)f(−x)

∣∣∣∣x0

= − d

dyf(y)

∣∣∣∣y=−x0

= − d

dxf(x)

∣∣∣∣x=−x0

.

So, we have f ′(x0) = −f ′(−x0) for any x0.


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

dnδ(x− x′)

dxn= δ(x− x′)

dn

dx′n . (4.15)

Therefore,

∫δ′(x− x′)f(x′) dx′ =

∫δ(x− x′)

d

dx′f(x′) dx′

=∫δ(x− x′)f ′(x′) dx′ = f ′(x), (4.16)

and more generally,

∫δn(x− x′)f(x′) dx′ =

∫δ(x− x′)

dn

dxnf(x′) dx′

=∫δ(x− x′)fn(x′) dx′ = fn(x). (4.17)

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

= a =⇒ dx = dya

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

∫ +∞

−∞f(x)δ(ax) dx =

(∫ +∞

−∞f(y

a

)δ(y) dy

)1

a

=1

af(0

a

)=

1

af(0) if a > 0 (4.18)

and∫ +∞

−∞f(x)δ(ax) dx =

(∫ −∞

+∞f(y

a

)δ(y) dy

)1

a=(∫ +∞

−∞f(y

a

)δ(y) dy

)1

−a

=1

−af(0

a

)=

1

−af(0) if a < 0. (4.19)

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

δ(ax) =δ(x)

|a|for any a , 0 (4.20)

4.4. DELTA FUNCTION AS A LIMIT OF GAUSSIAN DISTRIBUTION 93

4.4 Delta Function as a Limit of GaussianDistribution

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

gσ(x− x′) =1

(πσ2)12

exp[−(x− x′)2

σ2.

](4.21)

The Gaussian function gσ is symmetric about x = x′, has width2 σ, and itspeak height is (πσ)− 1

2 at x = x′. Furthermore, the area under the curveis unity irrespective of the value of the standard deviation σ or variance σ2.Hence, it is at least qualitatively obvious that gσ provides a better and betterapproximation of the delta function as σ approaches zero.

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

f(k) =1√2π

∫ +∞

−∞e−ikxf(x) dx, (4.22)

while the inverse Fourier transform is

f(x′) =1√2π

∫ +∞

−∞eikx

′f(k) dk. (4.23)

Substituting (4.22) into (4.23), we obtain the following relation.

f(x′) =∫ +∞

−∞

(1

2π

∫ +∞

−∞eik(x

′−x) dk)f(x) dx (4.24)

Now, compare (4.24) with (4.9). Then, we can see1

2π

∫ +∞

−∞eik(x

′−x) dk = δ(x′ − x) (4.25)

or1

2π

∫ +∞

−∞eik(x−x′) dk = δ(x− x′). (4.26)

This is another characterization of the delta function.2The “width” is defined as the distance between the two points of inflection of gσ.


4.6 f as a Vector with Uncountably ManyComponents

In order to understand a function f and its value at x, normally denotedby f(x), we will consider vectors in a finite dimensional space, a space ofcountably infinite dimension, and a space spanned by uncountably manybasis vectors, in this order. Without loss of generality, we will considerorthonormal bases.

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

|v⟩ =

31

32...3n

B

(4.27)

We have

|v⟩ =n∑i=1

3i |ei⟩ (4.28)

and3i = ⟨ei|v⟩ . (4.29)

4.6.2 Countably Infinite DimensionsIf the vector space has a countably infinite dimension, and an orthonormalbasis B = |e1⟩ , |e2⟩ , . . . , |en⟩ , . . ., we can express |v⟩ as an infinitely longcolumn vector with respect to B.

|v⟩ =

31

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

B

(4.30)

4.6. F AS A VECTOR WITH UNCOUNTABLY MANY COMPONENTS 95

We have

|v⟩ =∞∑i=1

3i |ei⟩ (4.31)

and3i = ⟨ei|v⟩ . (4.32)

4.6.3 Uncountably Infinite DimensionsWhen the vector space has an uncountably infinite dimension and an or-thonormal basis B = |ex⟩x∈R or x∈(−∞,+∞); where the kets in the basis Bare indexed by a continuous variable x, each vector v has uncountably manycomponents 3xx∈R with a continuous index x ∈ R , as opposed to a discreteindex i ∈ N. We can still regard |v⟩ as a column vector.

|v⟩ =

3−∞↑...3x...↓3+∞

B

(4.33)

By abuse of notation, we can write|v⟩ =

∑x∈R3x |ex⟩ (4.34)

and, as before,3x = ⟨ex|v⟩ . (4.35)

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

|f⟩ =

f−∞↑...fx...↓

f+∞

B

, (4.36)


|f⟩ =∑x∈R

fx |x⟩ , (4.37)

andfx = ⟨x|f⟩ . (4.38)

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

|f⟩ =∑x∈R

f(x) |x⟩ , (4.39)

andf(x) = ⟨x|f⟩ . (4.40)

Notationally more appropriate expression for |f⟩ than (4.39) is obtained usingthe completeness relation (4.5).

|f⟩ = I |f⟩ =∫ +∞

−∞|x⟩ ⟨x|f⟩ dx =

∫ +∞

−∞f(x) |x⟩ dx (4.41)

In this view, a function |f⟩ is a vector which is a linear combination oforthonormal basis vectors |x⟩x∈R, and its coefficients, components of thecolumn vector with respect to the basis |x⟩x∈R, are our familiar f(x). Wecan rewrite (4.36) as follows.

|f⟩ =

f(−∞)↑...

f(x)...↓

f(+∞)

|x⟩x∈R

(4.42)

Finally, you can see that the inner product is the one defined for L2-functionsin Section 2.3. All we need to do is to use the completeness property (4.5).

⟨f |g⟩ = ⟨f |I|g⟩ =⟨f

∣∣∣∣∫ +∞

−∞|x⟩ ⟨x| dx

∣∣∣∣ g⟩ =∫ +∞

−∞⟨f |x⟩ ⟨x|g⟩ dx

=∫ +∞

−∞⟨x|f⟩∗ ⟨x|g⟩ dx =

∫ +∞

−∞f(x)∗g(x) dx (4.43)

4.7. HERMITICITY OF THE MOMENTUM OPERATOR 97

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

P = −i~ ∂∂x

= −i~ ddx. (4.44)

This operator acts on differentiable L2 functions f(x) and has to be Her-mitian as the momentum p is a physical observable. Let us verify this. Wehave the following in our notation.

P |f⟩ =∣∣∣∣∣−i~ dfdx

⟩(4.45)

⟨x|P |f⟩ =⟨x

∣∣∣∣∣− i~ ddx∣∣∣∣∣f⟩=

⟨x

∣∣∣∣∣ −i~ dfdx⟩= −i~ df

dx(x) = −i~df(x)

dx(4.46)

4.7.1 Checking [Pij] = [P ∗ji]

In this section, we will check if P is represented by a matrix which is equalto its transpose conjugate. Because our basis consists of uncountably manyorthonormal vectors |x⟩x∈R, it is actually more appropriate to write [Pxx′ ] =[P ∗x′x]; where Pxx′ = ⟨x|P |x′⟩. Using the completeness property (4.5), we can

rewrite (4.46).

⟨x|P |f⟩ = ⟨x|PI|f⟩ =⟨x

∣∣∣∣∣P∫ b

a|x′⟩ ⟨x′| dx′

∣∣∣∣∣ f⟩

=∫ b

a⟨x|P |x′⟩ ⟨x′|f⟩ dx′ =

∫ b

a⟨x|P |x′⟩ f(x′) dx′ = −i~df(x)

dx(4.47)

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

Now compare (4.16) and (4.47) to obtain

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

dx′ . (4.48)

As shown in the footnote on p.91, the first derivative of the real-valued δ-function δ′(y) is an odd function. Hence,

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

= (−i~)∗(δ′(x′ − x))∗


= i~δ′(x′ − x) = i~(−δ′(x− x′)) = −i~δ′(x− x′) = Pxx′ . (4.49)

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

, denoted by [Pxx′ ], is equalto its transpose conjugate. This was a necessary and sufficient condition foran operator to be Hermitian if we have a finite dimensional Hilbert space(See Definition 2.25). However, this condition alone is not sufficient whenthe dimension is uncountably infinite.

4.7.2 Hermiticity Condition in Uncountably InfiniteDimensions

We will go back to the definition of Hermiticity. Definition 2.25 states thatan operator Ω is Hermitian if and only if

Ω = Ω†, (4.50)

which is equivalent to

⟨Ωv |w⟩ = ⟨v |Ωw⟩ (4.51)

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

, we would like to see if⟨−i~ df

dx

∣∣∣∣∣ g⟩=

⟨f

∣∣∣∣∣− i~dgdx⟩⇐⇒ i~

⟨df

dx

∣∣∣∣∣ g⟩= −i~

⟨f

∣∣∣∣∣ dgdx⟩

⇐⇒⟨df

dx

∣∣∣∣∣ g⟩= −

⟨f

∣∣∣∣∣ dgdx⟩. (4.52)

Let us first work on the left-hand side. Completeness condition and integra-tion by parts give us the following.⟨

df

dx

∣∣∣∣∣ g⟩=

⟨df

dx

∣∣∣∣∣ I∣∣∣∣∣ g⟩=

⟨df

dx

∣∣∣∣∣∫ b

a|x⟩ ⟨x| dx

∣∣∣∣∣ g⟩

=∫ b

a

⟨df

dx

∣∣∣∣∣x⟩⟨x| g⟩ dx =

∫ b

a

(df(x)

dx

)∗

g(x) dx =∫ b

a

df ∗(x)

dxg(x) dx

= f ∗(x)g(x)∣∣∣+∞

−∞−∫ b

af ∗(x)g′(x) dx. (4.53)

On the other hand, the right-hand side is

−⟨f

∣∣∣∣∣ dgdx⟩= −

⟨f

∣∣∣∣∣∫ b

a|x⟩ ⟨x| dx

∣∣∣∣∣ dgdx⟩= −

∫ b

a⟨f |x⟩

⟨x

∣∣∣∣∣ dgdx⟩dx

4.7. HERMITICITY OF THE MOMENTUM OPERATOR 99

= −∫ b

af ∗(x)g′(x) dx. (4.54)

The expressions (4.53) and (4.54) are equal if and only if

f ∗(x)g(x)∣∣∣ba= 0. (4.55)

Needless to say, this is not satisfied by all functions f and g. However,there are a couple of notable cases where (4.55) holds.

Single-Valuedness : When the spherical coordinates (r, θ, ϕ) areused as in Chapter 10, (r, θ, 0) and (r, θ, 2π) are the same point inthe three-dimensional space. Therefore, we require Ψ(r, θ, 0, t) =Ψ(r, θ, 2π, t) due to the single-valuedness condition, Condition 2, onp.124. In this case, a = 0 and b = 2π, and we indeed have

f ∗(x)g(x)∣∣∣ba= f ∗(ϕ)g(ϕ)

∣∣∣2π0

= Ψ∗(r, θ, ϕ, t)Ψ(r, θ, ϕ, t)∣∣∣2π0

= 0

with respect to the ϕ-integral.3 One class of functions that satisfy(4.55) consists of periodic functions. For example, if a = 0 and b =2π, trigonometric functions such as sin x and cos x and their complexcounterparts including eix satisfy this condition.

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

f ∗(x)g(x)∣∣∣+∞

−∞= 0. (4.56)

In particular, this equality holds for L2-functions, described on p.25of Section 2.3, as f ∈ L2(−∞,+∞) implies lim

|x|→∞f(x) = 0.

4.7.3 The Eigenvalue Problem of the Momentum Op-erator P

The eigenvalue problem for P in an arbitrary basis is

P |λ⟩ = λ |λ⟩ . (4.57)3We will see in Chapter 10 that we can separate the four variables (r, θ, ϕ, t) and write

Ψ as a product of functions of r, θ, ϕ, and t; namely Ψ(r, θ, ϕ, t) = R(r)Θ(θ)Φ(ϕ)T (t).


Taking the projection on ⟨x|,

⟨x|P |λ⟩ = λ ⟨x|λ⟩ . (4.58)

Using completeness,

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

)|λ⟩ =

∫⟨x|P |x′⟩ ⟨x′|λ⟩ dx′

= λ ⟨x|λ⟩ . (4.59)

Now, from (4.48), we have

⟨x|P |x′⟩ = −i~δ(x− x′)d

dx′ . (4.60)

Hence,

⟨x|P |λ⟩ =∫⟨x|P |x′⟩ ⟨x′|λ⟩ dx′ =

∫−i~δ(x− x′)

d

dx′ ⟨x′|λ⟩ dx′

= −i~ ddx⟨x|λ⟩ (4.61)

If we denote ⟨x|λ⟩ by fλ(x), we have

−i~ ddxfλ(x) = λfλ(x). (4.62)

Had we started with the X-basis made up of the eigenvectors of the positionoperator X, we would have obtained (4.62) immediately as P = −i~ d

dxin

this basis.

Solving the ordinary differential equation (4.62), we get

fλ(x) = Ceiλx/~; (4.63)

where C is an arbitrary constant. It is now clear that P takes any realnumber λ as its eigenvalue with the corresponding eigenfunction Ceiλx in theX-basis. In order to normalize the eigenfunctions, we let C = 1√

2π~. Then,

⟨x|λ⟩ = fλ(x) =1√2π~

eiλx/~, (4.64)

4.8. RELATIONS BETWEEN X AND P 101

and, due to (4.26) and (4.20),

⟨λ|λ′⟩ =∫ +∞

−∞⟨λ|x⟩ ⟨x|λ′⟩ dx =

1√2π~

1√2π~

∫ +∞

−∞e−iλx/~eiλ

′x/~ dx

=1

2π~

∫ +∞

−∞ei

1~(λ′−λ)x dx =

1

~

(1

2π

∫ +∞

−∞ei

1~(λ′−λ)x dx

)=

1

~δ(1

~(λ− λ′)

)=

1

~

δ(λ− λ′)

|1/~|= δ(λ− λ′). (4.65)

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

4.8 Relations Between X and P

4.8.1 The Fourier Transform Connecting X and P

Given a physical system S , there is an associated Hilbert space H accordingto Postulate 1 on p.66. This Hilbert space is spanned either by the eigenketsof P (the P basis) or the eigenkets of X (the X basis) as both X and P areHermitian operators. Given a ket |f⟩, it can be expanded either in the Xbasis |x⟩ or in the P basis |p⟩.

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

Recall that we have

|f⟩ =∑x∈R⟨x|f⟩ |x⟩ and |f⟩ =

∑p∈R⟨p|f⟩ |p⟩ . (4.66)

So, f(x) = ⟨x|f⟩ is the “x-th” coefficient/component of |f⟩, and f(p) = ⟨p|f⟩is the “p-th” coefficient/component of |f⟩. They are coefficients in view ofthe expansion (4.66), but they are also components of the column vector |f⟩.Let us compare f(p) and f(x) using (4.64).

f(p) = ⟨p|f⟩ =∫ +∞

−∞⟨p|x⟩ ⟨x|f⟩ dx =

∫ +∞

−∞⟨x|p⟩∗ ⟨x|f⟩ dx

=⇒ f(p) =1√2π~

∫ +∞

−∞e−ipx/~f(x) dx (4.67)

f(x) = ⟨x|f⟩ =∫ +∞

−∞⟨x|p⟩ ⟨p|f⟩ dp


=⇒ f(x) =1√2π~

∫ +∞

−∞eipx/~f(p) dp (4.68)

These are nothing but the Fourier transform and the Fourier inverse trans-form. Therefore, the expressions of a function |f⟩ in this Hilbert space interms of the complete X basis and complete P basis are related through thefamiliar Fourier transform.

You may note that the more familiar form of the Fourier transform canbe obtained if we set ~ = 1.4 Indeed, if we consider the operator K = P/~and its orthonormal eigenbasis |k⟩k∈R, which satisfies ⟨x|k⟩ = 1√

2πeikx as

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

f(k) = ⟨k|f⟩ =∫ +∞

−∞⟨k|x⟩ ⟨x|f⟩ dx =

∫ +∞

−∞⟨x|k⟩∗ ⟨x|f⟩ dx

=1√2π

∫ +∞

−∞e−ikxf(x) dx (4.69)

f(x) = ⟨x|f⟩ =∫ +∞

−∞⟨x|k⟩ ⟨k|f⟩ dk =

1√2π

∫ +∞

−∞eikxf(k) dk (4.70)

As advertised, these are the Fourier transform and its inverse in their morefamiliar forms.

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

dx. We can check the action of X in the framework

of this chapter as follows.

X |x⟩ = x |x⟩ =⇒ ⟨x′|X|x⟩ = ⟨x′|x|x⟩ = x ⟨x′|x⟩ = xδ(x′ − x) (4.71)

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

)|f⟩ =

∫⟨x|X|x′⟩ ⟨x′|f⟩ dx′

=∫x′δ(x− x′)f(x′) dx′ = xf(x) (4.72)

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

4This is the reason why we left 2π~ as it was even though 2π~ = 2π h2π = h.


4.8.3 Basis-Independent Descriptions in Reference tothe X Basis

Recall that |f⟩ is a basis-independent expression for a vector (function) inour Hilbert space. One way to specify f is by giving ⟨x|f⟩ = f(x) for theorthonormal basis |x⟩x∈R. Differently put, we cannot really specify thenature of f unless we go to some basis such as the position basis xx∈R.With this specification, it is now possible to express f as a basis-independentket in reference to the position orthonormal basis |x⟩x∈R. This potentiallyconfusing strategy reduces to writing the basis-independent ket |f⟩ as |f(x)⟩.What does this mean? It means the following.

1. |f⟩ itself is basis-independent.

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

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

For example, from (4.72), the coefficients of X |f⟩ with respect to |x⟩x∈R,namely ⟨x|X|f⟩x∈R, are xf(x)x∈R. Therefore, the action of X can beexpressed in a basis-independent manner, but in reference to the orthonormalposition basis |x⟩x∈R, as below.

X |f(x)⟩ = |xf(x)⟩ (4.73)

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

P |f(x)⟩ =∣∣∣∣∣−i~df(x)dx

⟩. (4.74)

4.8.4 X and P in the P BasisFrom (4.57), we have

P |p⟩ = p |p⟩ (4.75)

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

⟨p|P |p′⟩ = ⟨p|p′|p′⟩ = p′ ⟨p|p′⟩ = p′δ(p− p′). (4.76)


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

First, note that

i~d

dp

(e−ipx/~

)= i~(−ix/~)e−ipx/~ = xe−ipx/~. (4.77)

Using (4.64) and (4.77), we get

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

)|p′⟩ =

∫⟨p|X|x⟩ ⟨x|p′⟩ dx

=∫⟨p|x|x⟩ ⟨x|p′⟩ dx =

∫x ⟨p|x⟩ ⟨x|p′⟩ dx =

∫x ⟨x|p⟩∗ ⟨x|p′⟩ dx

=∫x

(1√2π~

eipx/~)∗ (

1√2π~

eip′x/~

)dx =

1

2π~

∫xe−ipx/~eip

′x/~ dx

=1

2π~

∫i~d

dp

(e−ipx/~eip

′x/~)dx = i

d

dp

(1

2π

∫ei(p

′−p)x/~ dx)

= i~d

dpδ(p− p′) = i~δ′(p− p′). (4.78)

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

1

2π

∫ +∞

−∞ei(p

′−p)x/~ dx =1

2π

∫ +∞

−∞ei(p

′−p)y ~dy = ~(

1

2π

∫ +∞

−∞ei(p

′−p)y dy)

= ~δ(p′ − p) = ~δ(p− p′). (4.79)

You may realize that this can be modified to generate an alternative proofof Property 5 or (4.20) on p.92.

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

Xf(k) = i~df(k)

dk. (4.80)

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

X |f(k)⟩ =∣∣∣∣∣i~ ddkf(k)

⟩. (4.81)


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

• In the X basis:

X ←→ x or Mx (4.82)

P ←→ −i~ ddx

(4.83)

• In the P basis:

X ←→ i~d

dp(4.84)

P ←→ p or Mp (4.85)

4.8.5 The Commutator [X,P ]

Let [X,P ]X and [X,P ]P be commutators of X and P in the X and P bases,respectively. Then,

[X,P ]Xf(x) =

[Mx

(−i~ d

dx

)−(−i~ d

dxMx

)]f(x)

= −i~xdf(x)dx

+ i~d

dx(xf(x)) = −i~xdf(x)

dx+ i~

(f(x) + x

df(x)

dx

)= i~f(x) =⇒ [X,P ]X = i~I (4.86)

and

[X,P ]Pf(p) =

[(i~d

dp

)Mp −Mp

(i~d

dp

)]f(p)

= i~d

dp(pf(p))− i~pdf(p)

dp= i~

(f(p) + p

df(p)

dp

)− i~pdf(p)

dp

= i~f(p) =⇒ [X,P ]p = i~I. (4.87)

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

[X,P ] = i~I (4.88)

in either basis.

Chapter 5

Schrodinger’s Theory ofQuantum Mechanics

In 1925 an Austrian physicist Erwin Rudolf Josef Alexander Schrodingerproposed the Schrodinger Equation which is a differential equation whosesolutions are wavefunctions. He chose a differential equation to describe newphysics since a differential equation was the most common type of equationknown to the physicists with a function for a solution. In Chapter 3, thepostulates of quantum mechanics were presented. However, these postulateswere put together in a post hoc fashion. In this chapter, we will learn howquantum mechanical framework was initially constructed, drawing on theanalogy to classical mechanics.

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

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

Ψ(x, t) = sin 2π(x

λ− νt

)(5.1)

2. The de Broglie-Einstein postulates

λ =h

p, ν =

E

h(5.2)

107

108 CHAPTER 5. SCHRODINGER’S THEORY OF QUANTUM MECHANICS

3. The Newtonian energy equation

E =P 2

2m+ V (5.3)

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

Ψ1(x, t), Ψ2(x, t) (5.4)

are two solutions of the Schrodinger Equation, then any linear combi-nation of the two

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

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

Here is what we want.

• A differential equation, later named the Schrodinger Equation, whichlooks like the energy equation.

• Wave functions, which are the solutions of the Schrodinger Equationand behave like classical waves, including interference and superposi-tion.

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

Let k = 2πλ

1 and ω = 2πν. Then, we have

Ψ(x, t) = sin(kx− ωt). (5.6)

On the other hand, the energy equation is

E =P 2

2m+ V.

1This k is called the wavenumber or angular/circular wavenumber to make a clear dis-tinction from another wavenumber 1/λ, which is also called the spectroscopic wavenumber.By definition, k is the number of wavelengths per 2π units of distance.

5.1. HOW WAS IT DERIVED? 109

Plug in P = hλ

and E = hν to obtain

h2

2mλ2+ V (x, t) = hν. (5.7)

Recall that k = 2πλ

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

(1

λ

)2

=

(k

2π

)2

ν =ω

2π(5.8)

We now substitute these back into the energy equation.

E =P 2

2m+ V =⇒ h2

2mλ2+ V (x, t) = hν =⇒ h2k2

(2π)2· 1

2m+ V (x, t) = h

(ω

2π

)

=⇒(h

2π

)2

k2 · 1

2m+ V (x, t) =

(h

2π

)ω (5.9)

Now we introduce a new notation.

~ =h

2π(5.10)

When you read it aloud, it is pronouced either h bar or h cross, with h barseeming more dominant these days. With this notation, we finally arrive at

~2k2

2m+ V (x, t) = ~ω. (5.11)

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

Ψ(x, t) = sin(kx− ωt).

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

∂Ψ

∂x= k cos(kx− ωt) (5.12)

∂2Ψ

∂x2= −k2 sin(kx− ωt) (5.13)

∂Ψ

∂t= −ω cos(kx− ωt) (5.14)


From these, we arrive at the implications below.

=⇒

∂2

∂x2gets us k2.

∂∂t

gets us ω.

(5.15)

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

By abuse of notation, let Ψ(x, t) be the desired wave function which actslike the traveling wave.

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

~2k2

2m⇐⇒ ∂2

∂x2Ψ(x, t) (5.16)

~ω ⇐⇒ ∂

∂tΨ(x, t) (5.17)

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

α∂2

∂x2Ψ(x, t) + V (x, t) = β

∂Ψ(x, t)

∂t(5.18)

Why α and β? Recall de Broglie. We were off there by a constant.

How about the fourth assumption; the linearity assumption?

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

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

=⇒ α(Ψ1,xx +Ψ2,xx) + 2V = β(Ψ1,t +Ψ2,t). (5.19)

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

And one way around this problem is to modify the equation in the fol-lowing way.

α∂2

∂x2Ψ(x, t) + V (x, t)Ψ(x, t) = β

∂Ψ(x, t)

∂t(5.20)

5.1. HOW WAS IT DERIVED? 111

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

Ψ(x, t) = ei(kx−ωt). (5.21)

Substituting V = 0 and (5.21) into (5.20),

α∂2

∂x2Ψ(x, t) + V (x, t)Ψ(x, t) = β

∂Ψ(x, t)

∂t

−→ α∂2

∂x2ei(kx−ωt) = β

∂

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

=⇒ −αk2 = −iωβ (5.22)

Comparing (5.22) with (5.11) when V (x, t) = 0, we get

−αk2 = −iωβ ←→ ~2k2

2m= ~ω. (5.23)

Therefore,

α =−~2

2mand β =

~

−i= i~. (5.24)

And our end product is

− ~2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂t. (5.25)

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

− ~2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t(5.26)

The solution set Ψ(x, t) represents the wave functions which are to beassociated with the motion of a particle of mass m under the influnece of theforce F described by V (x, t); namely

F (x, t) = −∂V (x, t)

∂x. (5.27)

Agreement with experiment? How do you check it? We first need a physicalinterpretation of Ψ to do so.


5.2 Born’s Interpretation of WavefunctionsWe now have

− ~2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂tor[

− ~2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t.

(5.28)

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

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

Max Born realized that Ψ∗Ψ = |Ψ|2 is real while Ψ itself is not and pos-tulated that P (x, t) = Ψ(x, t)∗Ψ(x, t) can be regarded as a probability. Hepostulated that the probability P (x, t)dx of finding the particle at a coordi-nate between x and x+ dx is equal to Ψ(x, t)∗Ψ(x, t)dx.

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

− ~2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂t.

This means that the probability is not unique, which of course is not accept-able. Furthermore, the total probability is not necessarily one if we used anarbitrary solution. Our necessary condition is∫ +∞

−∞P (x, t)dx = 1. (5.29)

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

−∞P (x, t)dx =

∫ +∞

−∞Ψ∗(x, t)Ψ(x, t)dx = 1. (5.30)

This is the physically meaningful Ψ called a normalized wavefunction. Thisprocess is called normalization.

5.3. EXPECTATION VALUES 113

5.3 Expectation ValuesRecall that for a die∑

value× probability = 1× 1

6+ 2× 1

6+ 3× 1

6+ 4× 1

6+ 5× 1

6+ 6× 1

6

= (1 + 2 + 3 + 4 + 5 + 6) · 16

gives the expectation value.

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

Answer : By computing the expectation value of the position x de-noted by ⟨x⟩ or x.

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

⟨x⟩ =∑

x× probability. (5.31)

The continuous version of the above is

⟨x⟩ = x =∫ +∞

−∞xΨ∗(x, t)Ψ(x, t)dx =

∫ +∞

−∞Ψ∗(x, t)xΨ(x, t)dx. (5.32)

How about P and E? We are tempted to write

P =∫ +∞

−∞Ψ∗(x, t)PΨ(x, t)dx (5.33)

and

E =∫ +∞

−∞Ψ∗(x, t)EΨ(x, t)dx. (5.34)

It turns out . . . . . .

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

− ~2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t


=⇒ − ~2

2m

∂2

∂x2Ψ(x, t) = i~

∂Ψ(x, t)

∂t(5.35)

We will solve this in detail in the next chapter, but the answer is the classicaltraveling wave.

Previously, we used

Ψ(x, t) = sin(kx− ωt).

Now we will use its complex version.

Ψ(x, t) = cos(kx− ωt) + i sin(kx− ωt) = ei(kx−ωt) (5.36)

Taking the first partial derivative with respect to x,

∂Ψ

∂x= ikΨ(x, t) = i

P

~Ψ(x, t); (5.37)

where the last equality follows from

k =2π

λ=

2π

h/p=

P

h/2π=P

~.

Therefore,

P [Ψ(x, t)] = −i~ ∂∂x

[Ψ(x, t)] . (5.38)

Similarly,

∂Ψ(x, t)

∂t= −iωΨ(x, t) = −iE

~Ψ(x, t); (5.39)

where the last equality follows from

ω = 2πν = 2πE

h=

E

h/2π=E

~.

Therefore,

E [Ψ(x, t)] = i~∂

∂t[Ψ(x, t)] . (5.40)


On the other hand, let us compare

P 2

2m+ V (x, t) = E

with [− ~

2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t

term by term.The first terms give us

−~2

2m

∂2

∂x2=P 2

2m.

Working formally,

√−~2

√∂2

∂x2=√P 2 =⇒ ±i~ ∂

∂x= P

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

E = i~∂

∂t.

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

We have P [Ψ(x, t)] = −i~ ∂

∂x[Ψ(x, t)]

E [Ψ(x, t)] = i~ ∂∂t[Ψ(x, t)]

. (5.41)

So,

⟨P ⟩ =∫ +∞

−∞Ψ∗(x, t)PΨ(x, t)dx =

∫ +∞

−∞Ψ∗(x, t)

(−i~ ∂

∂x

)Ψ(x, t)dx

= −i~∫ +∞

−∞Ψ∗ ∂

∂xΨdx (5.42)

and

⟨E⟩ =∫ +∞

−∞Ψ∗(x, t)EΨ(x, t)dx = i~

∫ +∞

−∞Ψ∗ ∂

∂tΨdx. (5.43)


Incidentally, this is the same as

⟨E⟩ =∫ +∞

−∞Ψ∗(x, t)

(− ~

2

2m

∂2

∂x2+ V (x, t)

)Ψ(x, t)dx. (5.44)

An Infinite Square Well Potential

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

V (x, t) =

0 −a

2< x < a

2

(a > 0)∞ |x| ≥ a

2

(5.45)

One normalized solution to the Schrodinger Equation is

Ψ(x, t) =

A cos πx

ae−iEt/~ −a

2< x < a

2

(a > 0)0 |x| ≥ a

2

; (5.46)

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

∂Ψ

∂t=−iE~

e−iEt/~A cos πxa

=−iE~

Ψ(x, t), (5.47)

we get

⟨E⟩ = i~∫ +∞

−∞Ψ∗(x, t)

∂Ψ(x, t)

∂tdx = i~

∫ +∞

−∞Ψ∗(x, t)

−iE~

Ψ(x, t)

= E∫ +∞

−∞Ψ∗Ψdx. (5.48)

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

−∞Ψ∗Ψdx = 1

and⟨E⟩ = E.


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

−∞Ψ∗Ψdx =

∫ +a2

− a2

(A cos πx

ae−iEt/~

)∗A cos πx

ae−iEt/~dx

=∫ +a

2

− a2

A∗ cos πxae+iEt/~A cos πx

ae−iEt/~dx =

∫ +a2

− a2

A∗A cos2 πxae+iEt/~e−iEt/~dx

= |A|2∫ +a

2

− a2

cos2 πxadx = 2|A|2

∫ +a2

0cos2 πx

adx = 1 (5.49)

If we let y = πxa

, we get dx = aπdy and y : 0→ π

2as x : 0→ a

2. So, integration

by this subsitution gives us∫ +a

2

0cos2 πx

adx =

∫ +π2

0

(a

π

)cos2 ydy =

(a

π

) [y

2+

sin 2y

4

]π/20

=a

π· π4=a

4. (5.50)

Therefore,

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

2= 1 =⇒ |A| =

√2

a. (5.51)

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

|A| =√2

a=⇒ A =

√2

aeiθ; (5.52)

where θ is any real number. This fact that the normalization constant A canonly be specified up to the argument is another reason why the wavefunctionitself does not represent physical reality.

118

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

below; where λ is the wavelength and ν is the frequency so that νλ = 3(the speed of its propagation).

Ψ(x, t) = A sin[2π

λ(x− 3t)

]Show that

Ψ(x, t) = Ψ(x+ 3t0, t+ t0),

and briefly explain what this means.2

2. Write down, but do not derive, the time-dependent Schrodinger equa-tion for V (x, t). This is a one-dimensional case, and the only variablesare the spatial variable x and the temporal variable t.

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

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

(a) eiθeiϕ = ei(θ+ϕ)

(b)(eiθ)n

= einθ

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

2,+a

2], but is strictly prohibited from

going outside this region. This corresponds to what is called an infinitesquare well potential V (x) given by V (x) = 0 for −a

2< x < +a

2and

2Note thatνλ = 3 =⇒ ω = 2πν =

2π

λ3.

Therefore, we have

Ψ(x, t) = A sin[2π

λx− ωt

].

119

V (x) = ∞ elsewhere. If we solve the Schrodinger equation for thisV (x), one of the solutions is

Ψ(x, t) =

A cos

(πxa

)e−iEt/~ −a

2< x < a

2

(a > 0)0 |x| ≥ a

2

.

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

tion reduces to

− ~2

2m

∂2

∂x2Ψ(x, t) = i~

∂Ψ(x, t)

∂t.

Plug Ψ(x, t) = A cos(πxa

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

π2~2

2ma2.

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

integral table for this. (Of course, you can always try contourintegration or some such. But, that is beyond this course.)

(e) Evaluate ⟨P 2⟩ for the same wavefunction. You will not need anintegral table if you use the fact that the function is normalized.Of course, a brute force computation will yield the same result aswell.

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

(a) What is the implication of this fact as to the uniqueness of wave-function?

(b) What is the implication of this fact as to the probability densityΨ∗(x, t)Ψ(x, t)? How about ⟨P ⟩, ⟨P 2⟩, and ⟨x2⟩?

Chapter 6

The Time-IndependentSchrodinger Equation

In the Schrodinger Equation[− ~

2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t(6.1)

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

V (x, t) =

0 −a

2< x < a

2

(a > 0)∞ |x| ≥ a

2

(6.2)

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

Ψ(x, t) = ψ(x)ϕ(t) (6.3)

This technique is called “separation of variables”. Let us substitute the aboveΨ into the Schrodinger Equation.[

− ~2

2m

∂2

∂x2+ V (x)

]ψ(x)ϕ(t) = i~

∂ψ(x)ϕ(t)

∂t(6.4)

=⇒ − ~2

2m

∂2

∂x2ψ(x)ϕ(t) + V (x)ψ(x)ϕ(t) = i~

∂ψ(x)ϕ(t)

∂t

121

122 CHAPTER 6. THE TIME-INDEPENDENT SCHRODINGER EQUATION

=⇒ − ~2

2mϕ(t)

∂2

∂x2ψ(x) + V (x)ψ(x)ϕ(t) = ψ(x)i~

∂ϕ(t)

∂t

=⇒ ϕ(t)

[− ~

2

2m

∂2

∂x2ψ(x) + V (x)ψ(x)

]= ψ(x)i~

∂ϕ(t)

∂t(6.5)

Dividing through by ψ(x)ϕ(t), we get

1

ψ(x)

[− ~

2

2m

d2

dx2ψ(x) + V (x)ψ(x)

]︸︷︷︸

H(x)

= i~1

ϕ(t)

dϕ(t)

dt︸︷︷︸K(t)

; (6.6)

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

Generally speaking,

H(x) = K(t) for all pairs (x, t) (6.7)

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

K(t) = G⇐⇒ i~1

ϕ(t)

dϕ(t)

dt= G⇐⇒ dϕ(t)

dt=

1

i~Gϕ(t) =

−i~Gϕ(t). (6.8)

Therefore,

ϕ(t) = Ae−iGt/~. (6.9)

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

ei(kx−ωt) = eikxe−iωt. (6.10)

We realize that G~= 2πG

h“should be” ω.

2πG

h= ω =⇒ G =

ωh

2π=

2πνh

2π= hν = E. (6.11)

We now have G = E, and

ϕ(t) = e−iEt/~. (6.12)

123

This also implies

H(x) =1

ψ(x)

[− ~

2

2m

d2

dx2ψ(x) + V (x)ψ(x)

]= E

=⇒ − ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x); (6.13)

where E is the total energy.

− ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x) (6.14)

or[− ~

2

2m

d2

dx2+ V (x)

]ψ(x) = Eψ(x) (6.15)

is the Time-Independent Schrodinger Equation.

Note here that the full wavefunction is given by

Ψ(x, t) = ψ(x)e−iEt/~. (6.16)

In the rest of this course, we will solve the Time-Independent SchrodingerEquation for various systems, culminating in a solution for the hydrogenatom.

Physical Sense Revisited: Previously, I mentioned that only normalizedsolutions are the physically meaningful solutions in order for the Born’s inter-pretation to make sense. However, this restriction alone would not eliminateall the mathematically correct solutions which are not acceptable on physi-cal grounds. Hence, we need to put further restrictions on the nature of thesolutions. We will place three types of restrictions on both ψ(x) and dψ(x)

dx.

To be precise, the wavefunction ψ will most likly depend on x, y, and z foractual physical systems, and we will have ψ(x, y, z) instead of ψ(x). Theseare ad-hoc and post-hoc conditions as usual albeit physically very reasonable.

1. Finiteness and Integrability: The solution and its first derivative haveto take a finite value for all x. Furthermore, the wavefunction has tobe normalizable; i.e.

∫+∞−∞ ψ∗ψdx <∞.

124 CHAPTER 6. THE TIME-INDEPENDENT SCHRODINGER EQUATION

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

3. Continuity: Both the solution and its first derivative have to be con-tinuous everywhere.

We will find solutions of this type. Incidentally, Condition 3 on thederivative can not be imposed if the potential V (x) itself has a discontinuity.We will see such an example in Chapter 7.

125

Exercises1. Write down, but do not derive, the time-independent Schrodinger equa-

tion for V (x), that is, V is now a function of x only. This is a one-dimensional case, and the only spatial variable is x.

2. The time-dependent Schrodinger Equation is

− ~2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂t.

Consider a case where the potential does not depend on time; i.e.V (x, t) = V (x). Assume Ψ(x, t) = ψ(x)e−iEt/~ and derive the time-independent Schrodinger Equation.

3. Discuss in concrete terms and sufficient detail why Φ(ϕ) = sinϕ is anacceptable wavefunction while Φ(ϕ) = ϕ is not. We are using sphericalpolar coordinates here.

Chapter 7

Solutions of Time-IndependentSchrodinger Equations in OneDimension

Though you may, perhaps naively, think our world is three-dimensional, thereare many physical systems which are of lower dimensions. For example, a thinfilm forms a two-dimensional system, and a particle moving freely withoutany external force is the simplest one-dimensional system. Furthermore, theSchrodinger equations for one-dimensional systems are easier to solve. Hence,it makes good mathematical and physical sense to start with one dimensionalcases.

7.1 The Zero PotentialThis is the case where V (x) = 0 for all x.

Our Time-Independent Schrodinger Equation is

−~2

2m

d2ψ(x)

dx2= Eψ(x). (7.1)

The most general solution for this second order linear ordinary differentialequation is of the form

ψ(x) = A1 sin√2mE

~x+ A2 cos

√2mE

~x = C1e

ikx + C2e−ikx; (7.2)

127

128CHAPTER 7. SOLUTIONS OF TIME-INDEPENDENT SCHRODINGER

EQUATIONS IN ONE DIMENSION

where A1, A2, C1, and C2 are arbitrary constants, and k =√2mE~

.

On the other hand, we always have

ϕ(t) = e−iEt/~ = e−ihνt/(h/2π) = e−i2πνt = e−iωt (7.3)

for time dependence.

Therefore, the full time-dependent wavefunction is

Ψ(x, t) = ψ(x)ϕ(t) =(C1e

ikx + C2e−ikx

)e−iωt

= C1ei(kx−ωt) + C2e

−i(kx+ωt). (7.4)

Let us take

Ψ(x, t) = C1ei(kx−ωt)

(= C1e

ikxe−iEt/~).

Then,

P = ⟨P ⟩ =∫ +∞

−∞Ψ∗PΨdx =

∫ +∞

−∞C∗

1e−ikxeiEt/~(−i~) ∂

∂xC1e

ikxe−iEt/~dx

=∫ +∞

−∞C∗

1e−ikxeiEt/~(−i~)(ik)C1e

ikxe−iEt/~dx =(∫ +∞

−∞Ψ∗Ψdx

)= ~k

∫ +∞

−∞ψ∗ψdx = ~k =

h

2π· 2mE~

=√2mE. (7.5)

This makes sense as

P 2

2m= E =⇒ P =

√2mE

classically, and (7.5) serves as yet another consistency checking device.

As simple as the concept and the wavefunctions of a free particle are, thereis one serious complication regarding normalization. To see this, consider aneigenfunction ψ(x) = Aeikx. In order for this function to be normalizable, ithas to be integrable to begin with. However,∫ +∞

−∞

(Aeikx

)∗ (Aeikx

)dx = |A|2

∫ +∞

−∞e−ikxeikx dx = |A|2

∫ +∞

−∞1 dx =∞,

7.2. THE STEP POTENTIAL (E < V0) 129

and ψ(x) is not normalizable in the usual way.

In order to circumvent this problem, three normalization schemes havebeen proposed; namely, the Born normalization, Dirac normalization, andunit-flux normalization. As these normalization procedures are not crucialfor the rest of this book, their discussions are relegated to Appendix ??.

7.2 The Step Potential (E < V0)We now consider a step potential given by

V (x) =

V0 x > 00 x ≤ 0

(7.6)

for E < V0.

In Newtonian (classical) mechanics, the particle can not enter the region(0,∞). We have

E = K.E.+ P.E. = K.E.+ V0 in (0,∞) =⇒ K.E. = E − V0 < 0.

This does not make any sense of any kind classically. Is this also the case inquantum mechanics? Let us just plunge in and solve it!

The time-independent Schrodinger Equation is

− ~2

2m

d2ψ(x)

dx2+ V (x)ψ(x) = Eψ(x).

In our case, we have the following.

−~2

2m

d2ψ(x)

dx2= Eψ(x) x ≤ 0 (7.7)

−~2

2m

d2ψ(x)

dx2+ V0ψ(x) = Eψ(x) x > 0 (7.8)

Equation 7.7 is that for a free particle. Therefore, the general solution is

ψ(x) = Aeik1x +Be−ik1x; (7.9)



where k1 =√2mE~

, and A and B are arbitrary constants.

On the other hand, (7.8) can be rewritten as follows.

−~2

2m

d2ψ(x)

dx2= (E − V0)ψ(x)⇐⇒

~2

2m

d2ψ(x)

dx2= (V0 − E)ψ(x)

⇐⇒ d2ψ(x)

dx2=

2m(V0 − E)~2

ψ(x) (7.10)

The general solution for x > 0 is

ψ(x) = Cek2x +De−k2x; (7.11)

where k2 =

√2m(V0−E)

~, and C and D are arbitrary constants. Note that

V0 − E is positive in this region.

As stated above, these are indeed the solutions, or the (most) generalsolutions, since we have linear second order ordinary differential equations.

We have ψ(x) = Aeik1x +Be−ik1x (k1 =√2mE~

) x ≤ 0

ψ(x) = Cek2x +De−k2x (k2 =

√2m(V0−E)

~) x > 0

. (7.12)

Recall the conditions on (ψ, dψdx).

1. Finite

2. Single-valued

3. Continuous

We will determine A, B, C, and D using these conditions.

Let us begin with the region x > 0. In this region,

ψ(x) = Cek2x +De−k2x; where k2 =

√2m(V0 − E)~

. (7.13)

Therefore,

ψ∗(x)ψ(x) =(C∗ek2x +D∗e−k2x

) (Cek2x +De−k2x

)


= |C|2e2k2x +2ReC∗D︷︸︸︷

D∗C + C∗D+|D|2e−2k2x. (7.14)

Since the wavefunction has to be normalizable, it has to satisfy the finitenesscondition ∫ +∞

−∞ψ∗(x)ψ(x)dx =

∫ +∞

−∞|ψ(x)|2dx <∞. (7.15)

Noting that this ψ is the wavefuntion only in the region x > 0, we actuallyneed ∫ +∞

0|ψ(x)|2dx <∞. (7.16)

Hence, we have to haveC = 0.

At this point, we have ψ(x) = Aeik1x +Be−ik1x k1 =√2mE~

x ≤ 0

ψ(x) = De−k2x (k2 =

√2m(V0−E)

~) x > 0

. (7.17)

Next, we will use the continuity condition on ψ(x) and dψ(x)dx

.Continuity of ψ(x) at x = 0 gives

De−k2·0 = Aeik1·0 +Be−ik1·0 =⇒ D = A+B. (7.18)

Since the first derivative isdψ(x)dx

= A(ik1)eik1x +B(−ik1)e−ik1x k1 =

√2mE~

x ≤ 0dψ(x)dx

= D(−k2)e−k2x (k2 =

√2m(V0−E)

~) x > 0

, (7.19)

continuity of dψ(x)dx

at x = 0 gives

−k2De−k2·0 = ik1Aeik1·0 − ik1Be−ik1·0 =⇒ −k2D = ik1(A−B)

=⇒ −k2ik1

D = A−B. (7.20)



We now have a simultaneous equationik2k1D = A−BD = A+B

(7.21)

for A, B, and D with the solution

A = 12

(1 + ik2

k1

)D

B = 12

(1− ik2

k1

)D. (7.22)

This gives

ψ(x) =

D2(1 + ik2/k1)e

ik1x + D2(1− ik2/k1)e−ik1x x ≤ 0

De−k2x x > 0. (7.23)

So, the full solution is

Ψ(x, t) = ψ(x)ϕ(t) = ψ(x)e−iEt/~

=

D2(1 + ik2/k1)e

i(k1x−Et/~) + D2(1− ik2/k1)e−i(k1x+Et/~) x ≤ 0

De−k2xe−iEt/~ x > 0. (7.24)

Recall thate−iEt/~ = e−iωt.

This is because

Et/~ = hνt/~ = hνt/(h

2π) = 2πνt = ωt.

Therefore, in the region x ≤ 0,

Ψ(x, t) = Aei(k1x−ωt) +Bei(−k1x−ωt). (7.25)

Note thatAei(k1x−ωt)

is traveling to the right while

Bei(−k1x−ωt)

is traveling to the left. In other words,

Aei(k1x−ωt)


is the incident wave, incident on the step form the left, while

Bei(−k1x−ωt)

is the wave reflected by the step.

What is the reflection coefficient? Since the reflection coefficient is theratio of the amplitude of the reflected wave and the amplitude of the incidentwave, we have

the reflection coefficient = the amplitude of the reflected wavethe amplitude of the incident wave

=|Bei(−k1x−ωt)|2

|Aei(k1x−ωt)|2=|B|2|ei(−k1x−ωt)|2

|A|2|ei(k1x−ωt)|2=|B|2

|A|2=B∗B

A∗A

=

D∗

2

(1− ik2

k1

)∗D2

(1− ik2

k1

)D∗

2

(1 + ik2

k1

)∗D2

(1 + ik2

k1

) =

(1 + ik2

k1

) (1− ik2

k1

)(1− ik2

k1

) (1 + ik2

k1

) = 1 (7.26)

And, we have a total reflection.

Indeed, we have a standing wave in the region x ≤ 0. To see this, plugeik1x = cos k1x+ i sin k1x into

ψ(x) =

D2

[(1 + ik2/k1)e

ik1x + (1− ik2/k1)e−ik1x]x ≤ 0

De−k2x x > 0(7.27)

to obtain

ψ(x) =

D cos k1x−D k2

k1sin k1x x ≤ 0

De−k2x x > 0. (7.28)

So, for x ≤ 0,

Ψ(x, t) = D

(cos k1x−

k2k1

sin k1x)e−iEt/~. (7.29)

Remember the following from high school math,

α cos kx− β sin kx =

(α√

α2 + β2cos kx− β√

α2 + β2sin kx

) √α2 + β2



= cos(kx+ θ); (7.30)

where θ is such that

cos θ = α√α2 + β2

and sin θ = β√α2 + β2

. (7.31)

Therefore, the nodes of Ψ(x, t)∗Ψ(x, t) do not move, and we indeed have astanding wave.

Now, in the region x > 0,

Ψ∗(x, t)Ψ(x, t) = D∗De−2k2x.

Whilelimx→∞

Ψ∗Ψ = 0,

Ψ∗Ψ , 0

for ∀x > 0. This means that the probability of finding the particle to theright of the step is not zero even if the total energy E is smaller than thestep height V0. This phenomenon is definitely non-classical and is known as“barrier penetration”.

7.3 The Step Potential (E > V0)We will now consider the case where the total energy E is greater than thestep height.

Classically, we have the following.

1. Not a total reflection

2. Has to be a total penetration into (0,∞)

But, quantum mechanically, the Schrodinger Equation predicts the following.

1. Not a total reflection (agreement with the classical theory)

2. There is partial reflection at the boundary x = 0 (disagreement)

7.3. THE STEP POTENTIAL (E > V0) 135

Let us see how it works.

As usual, Ψ(x, t) = ψ(x)e−iEt/~, and the Time-Independent SchrodingerEquation is

−~22m

d2ψ(x)dx2

= Eψ(x) x < 0−~22m

d2ψ(x)dx2

= (E − V0)ψ(x) x > 0(7.32)

As E− V0 > 0 this time, the two equations are basically the same. They areboth free particles, and the solutions are

ψ(x) =

Aeik1x +Be−ik1x x < 0Ceik2x +De−ik2x x > 0

; (7.33)

where k1 =√2mE~

and k2 =

√2m(E−V0)~

.

Suppose the particle is in the region x < 0 or (−∞, 0) at t = 0. In otherwords, the particle is incident on the step from the left. Then, since the waveis travelling to the right in the region (0,∞), we should set D = 0. This issimply because e−ik2xe−iEt/~ is a wave travelling to the left.

Imposing the continuity condition on ψ and dψdx

at x = 0, we can expressthe constants B and C in terms of A as follows.

ψ(x)|x=0− = ψ(x)|x=0+dψ(x)dx

∣∣∣x=0−

= dψ(x)dx

∣∣∣x=0+

=⇒

A+B = Ck1(A−B) = k2C

=⇒B = Ak1−k2

k1+k2

C = A 2k1k1+k2

(7.34)

Therefoere,

ψ(x) =

Aeik1x + Ak1−k2

k1+k2e−ik1x

A 2k1k1+k2

eik2x(7.35)

The reflection coefficient R is given by

R =B∗B

A∗A=

(k1 − k2k1 + k2

)2

. (7.36)



Since the transmission coefficient T is related to R via R + T = 1,

T = 1−R = 1−(k1 − k2k1 + k2

)2

=(k1 + k2 + k1 − k2)(k1 + k2 − k1 + k2)

(k1 + k2)2

=4k1k2

(k1 + k2)2. (7.37)

As advertised, we do not have a total reflection, and we do not have a totalpenetration into (0,∞), either.

We now solve the same problem for a wave incident on the potentialboundary at x = 0 from the right. In other words, the particle is in theregion (0,∞) at t = 0 and traveling to the left.

Our general solution is

ψ(x) =

Aeik1x +Be−ik1x x < 0Ceik2x +De−ik2x x > 0

; (7.38)

where k1 =√2mE~

and k2 =

√2m(E−V0)~

.

Since there is no boundary to reflect the wave in the region (−∞, 0), thewave should be traveling to the left in this region. We set A = 0.

ψ(x) =

Be−ik1x x < 0Ceik2x +De−ik2x x > 0

(7.39)

The first derivative with respect to x is

ψ′(x) =

−ik1Be−ik1x x < 0ik2Ce

ik2x + (−ik2)De−ik2x x > 0. (7.40)

Imposing the continuity condition on ψ(x) and dψ(x)dx

at x = 0, we getB = C +D

−ik1B = ik2(C −D)=⇒

C +D = BC −D = −k1

k2B

=⇒C = k2−k1

2k2B

D = k2+k12k2

B. (7.41)

7.4. THE BARRIER POTENTIAL (E < V0) 137

What are the reflection coefficient and the transmission coefficient in thiscase?

R =C∗C

D∗D=

(k2−k12k2

)2B∗B(

k2+k12k2

)2B∗B

=

(k1 − k2k1 + k2

)2

(7.42)

and

T = 1−R =4k1k2

(k1 + k2)2(7.43)

Note that these are the same as before when we computed R and T for theparticle moving in from the left.

Incidentally, B , 0 in (7.33) can be proved on purely mathematicalgorunds as follows.

If B = 0,

ψ(x) =

Aeik1x x < 0Ceik2x x > 0

=⇒

A = Ck1A = k2C

.

But, it is obvious that this can not be solved for A and C unless we acceptA = C = 0 which is physically meaningless. Hence, B , 0 on this ground.

7.4 The Barrier Potential (E < V0)The potential V (x) is V0 in the region (0, a) and 0 elsewhere.

V (x) =

V0 0 < x < a0 x < 0, a < x

(7.44)

Our Time-Independent Schrodinger Equation is

ψ(x) =

Aeik1x + Be−ik1x x < 0Fe−k2x + Gek2x 0 < x < aCeik1x + De−ik1x x > a

; (7.45)



where k1 =√2mE~

and k2 =

√2m(V0−E)

~.

Assuming x < 0 at t = 0, D = 0.On the other hand, our boundary conditions are

ψ(x)|x=0− = ψ(x)|x=0+

ψ(x)|x=a− = ψ(x)|x=a+dψ(x)dx

∣∣∣x=0−

= dψ(x)dx

∣∣∣x=0+

dψ(x)dx

∣∣∣x=a−

= dψ(x)dx

∣∣∣x=a+

. (7.46)

We have the above four equations and the five unknowns A, B, C, F ,and G. But, we also have a normalization condition. So, we can solve forthe five unknowns.

As it turned out, C , 0, and the probability density Ψ∗(x, t)Ψ(x, t) =ψ∗(x)ψ(x) has a (nonzero) tail in the region x > a. Of course, it is nonzeroalso inside the barrier.Therefore, we have both barrier penetration and tunneling.

7.5 The Infinite Square Well PotentialThe potential is given by

V (x) =

∞ |x| > a

2

0 |x| < a2

, (7.47)

and we get, as we have already seen,

ψ(x) =

0 |x| > a

2

A sin kx+B cos kx |x| < a2

; (7.48)

where k =√2mE~

. As usual, the full wavefunction is

Ψ(x, t) = ψ(x)e−iEt/~. (7.49)

It turned out the continuity condition of the first derivative is too restrictivefor this system as the potential blows up to ∞ unlike any known physical

7.5. THE INFINITE SQUARE WELL POTENTIAL 139

system.The continuity condition on ψ(x) at |x| = a

2gives

A sin ka2

+ B cos ka2

= 0−A sin ka

2+ B cos ka

2= 0

=⇒

2B cos ka2

= 02A sin ka

2= 0

=⇒A = 0 and cos ka

2= 0

B = 0 and sin ka2= 0

. (7.50)

Therefore, ψ(x) = B cos kx where cos ka

2= 0

orψ(x) = A sin kx where sin ka

2= 0

. (7.51)

We have two families of solutions.ψn(x) = Bn cos knx where kn = nπ

an = 1, 3, 5, . . .

ψn(x) = An sin knx where kn = nπa

n = 2, 4, 6, . . .. (7.52)

Note that n is a positive integer in both cases because kn > 0. We can alsonote that negative n’s give redundant solutions and n = 0 gives a physicallymeaningless trivial solution. Indeed,

n = 0 =⇒ ψ0(x) = A0 sin 0 = 0.

Let us compute the normalization constants An and Bn. We will obtain realand positive constants.∫ +a

2

− a2

(Bn cos knx)∗ (Bn cos knx) dx = |Bn|2∫ +a

2

− a2

cos2 knx dx

= |Bn|2∫ +a

2

− a2

1 + cos 2knx2

dx =|Bn|2

2

[x+

1

2knsin 2knx

]+a2

− a2

=|Bn|2

2

[x+

a

2nπsin 2nπ

ax]+a

2

− a2

=|Bn|2

2

[(a

2+

a

2nπsinnπ

)−(−a2− a

2nπsinnπ

)]=|Bn|2

2· a = 1

=⇒ Bn =

√2

a


EQUATIONS IN ONE DIMENSION∫ +a2

− a2

(An sin knx)∗ (An sin knx) dx = |An|2∫ +a

2

− a2

sin2 knx dx

= |An|2∫ +a

2

− a2

1− cos 2knx2

dx =|An|2

2

[x− 1

2knsin 2knx

]+a2

− a2

=|An|2

2

[x− a

2nπsin 2nπ

ax]+a

2

− a2

=|An|2

2

[(a

2− a

2nπsinnπ

)−(−a2+

a

2nπsinnπ

)]=|An|2

2· a = 1

=⇒ An =

√2

a

Hence, (7.52) becomes ψn(x) =√

2a

cos knx where kn = nπa

n = 1, 3, 5, . . .

ψn(x) =√

2a

sin knx where kn = nπa

n = 2, 4, 6, . . .. (7.53)

We can also write (7.53) as follows.

ψn(x) =

√2

asin

nπ(x+ a

2

)a

=

√2

asin

[nπ

(x

a+

1

2

)](7.54)

Furthermore, if we shift the x-axis so that V (x) = 0 for 0 < x < a, we willget

ψn(x) =

√2

asin

(nπx

a

). (7.55)

Because kn =√2mEn/~, the total energy En can only take discrete values.

kn =

√2mEn~

=⇒ 2mEn = ~2k2n

=⇒ En =~2k2n2m

=π2~2n2

2ma2n = 1, 2, 3, 4, 5, . . . (7.56)

Note that En ∝ n2 and the spacing between neighboring energy levels

En+1 − En =π2~2(n+ 1)2

2ma2− π2~2n2

2ma2=π2~2(2n+ 1)

2ma2(7.57)

7.6. THE SIMPLE HARMONIC OSCILLATOR POTENTIAL 141

is proportional to 2n + 1. This is the reason for the widening of the gap asn→∞.

More significant implication of this is that the particle can not have zerototal energy. Indeed, the lowest energy level

E1 =π2~2

2ka2> 0. (7.58)

This is one reason why absolute zero degree temperature can not be achieved.While the infinite square well potential is a mathematical idealization, it ap-proximates the potential well experienced by atoms or nuclei in a crystal forexample.

Another type of bound system frequently encountered in nature is a di-atomic molecule.

7.6 The Simple Harmonic Oscillator Poten-tial

7.6.1 Classic Solution

Consider a typical spring encountered in high school physics with aspring constant k such that the restoring force F is given by F = −kxfor the displacement x from the equilibrium position. For no other rea-son than convention, we will use c instead of k for the spring constant.

Because

V (x) =1

2cx2, (7.59)

the Time-Independent Schrodinger Equation is

−~2

2m

d2ψ(x)

dx2+c

2x2ψ(x) = Eψ(x). (7.60)

We will need some fancy footwork to solve this equation. As it turns out,it is convenient to introduce a new variable ν analogously to the classicaltheory. Hence,



ν =1

2πω =

1

2π

√c

ma.

After substituting c = (2π)2ν2m into the equation and dividing throughby −~2

2m, we get

d2ψ

dx2+

[2mE

~2−(2πmν

~

)2

x2]ψ = 0. (7.61)

Now, let α = 2πmν/~ and β = 2mE/~2 to simplify the equation to

d2ψ

dx2+ (β − α2x2)ψ = 0. (7.62)

aTo see this, we start with the classical equation F = mα. For us,

−cx = md2x

dt2=⇒ d2x

dt2= − c

mx =⇒ x(t) = A sin

(√c

mt

)+B cos

(√c

mt

)

=√A2 +B2 sin

(√c

mt+ θ

);

where cos θ = A√A2+B2

and sin θ = B√A2+B2

. Therefore, we have

ω =

√c

m.

Here are further manipulations.

u =√αx =

[2πm

~2π(c

m)1/2

]1/2x =

(cm)1/4

~1/2x (7.63)

dψ

dx=dψ

du

du

dx=√αdψ

du(7.64)


d2ψ

dx2=

d

dx

(dψ

dx

)=d(dψ

dx)

dx=d(√αdψdu)

dx=du

dx

d

du

(√αdψ

du

)

=√α√αd

du

(dψ

du

)= α

d2ψ

du2(7.65)

You may be more comfortable with the following.

du =√αdx =⇒ dx =

1√αdu (7.66)

d2ψ

dx2=

d1√αdu

dψ1√αdu

= αd2ψ

du2(7.67)

Either way is fine, and we now have

d2ψ

dx2+ (β − α2x2)ψ = α

d2ψ

du2+ (β − αu2)ψ = 0

=⇒ d2ψ

du2+

(β

α− u2

)ψ = 0. (7.68)

Here comes a very inexact argument.

As |u| → ∞, the differential equation behaves like

d2ψ

du2− u2ψ = 0 (7.69)

or

d2ψ

du2= u2ψ (7.70)

as β/α is a constant.The (mathematical) general solution of this differential equation is ap-



proximatelya

ψ = Ae−u2/2 +Beu2/2 as |u| → ∞. (7.71)

But, we have to setB = 0

to satisfy the finiteness condition. So,

ψ(u) = Ae−u2/2 as |u| → ∞. (7.72)

We will look for a solution of the form

ψ(u) = Ae−u2/2H(u); (7.73)

where H(u) must be slowly varying in comparison to e−u2/2.Compute d2ψ(u)

du2, then plug it back into the equation

d2ψ(u)

du2+

(β

α− u2

)ψ(u) = 0

to obtain

d2H

du2− 2u

dH

du+

(β

α− 1

)H = 0 (7.74)

as follows.a

d2

du2

(e−u2/2

)=

d

du

(−ue−u2/2

)= −e−u2/2 − u

(−ue−u2/2

)= (u2 − 1)e−u2/2

−−−−−→|u|−→∞

u2e−u2/2

d2

du2

(eu2/2

)=

d

du

(ueu2/2

)= eu2/2 + u

(ueu2/2

)= (u2 + 1)eu2/2

−−−−−→|u|−→∞

u2eu2/2

Hence, (7.71) holds.


Without loss of generality, set A = 1, and consider

ψ(u) = e−u2/2H(u). (7.75)

Then, we have

dψ

du= −ue−u2/2H(U) + e−u2/2H ′(u)

andd2ψ

du2= −e−u2/2H(u)− u

(−ue−u2/2

)H(u)− ue−u2/2H ′(u)

−ue−u2/2H ′(u) + e−u2/2H ′′(u)

= −e−u2/2H(u) + u2e−u2/2H(u)− 2ue−u2/2H ′(u) + e−u2/2H ′′(u).

Therefore,

d2ψ(u)

du2+

(β

α− u2

)ψ(u) = −e−u2/2H(u) + u2e−u2/2H(u)− 2ue−u2/2H ′(u)

+e−u2/2H ′′(u) +

(β

α− u2

)e−u2/2H(u)

= −e−u2/2H(u)− 2ue−u2/2H ′(u) + e−u2/2H ′′(u) +β

αe−u2/2H(u) = 0. (7.76)

Dividing through by e−u2/2, we get

−H(u)− 2uH ′(u) +H ′′(u) +β

αH(u)

= H ′′(u)− 2uH ′(u) +

(β

α− 1

)H(u) = 0,

which is (7.74).

We will try a power series solution for H(u).Let

H(u) =∞∑l=0

al · ul = a0 + a1u+ a2u2 + a3u

3 + · · · . (7.77)



Then,

udH(u)

du= (a1 + 2a2u+ 3a3u

2 + 4a4u3 + · · · )u =

( ∞∑l=0

(l + 1)al+1ul

)· u

=∞∑l=0

(l + 1)al+1ul+1 =

∞∑l=1

lalul (7.78)

and

d2H(u)

du2= 2a2 + 2 · 3a3u+ 3 · 4a4u2 + · · · =

∞∑l=0

(l + 1)(l + 2)al+2ul.

(7.79)

Plugging the above into

d2H

du2− 2u

dH

du+

(β

α− 1

)H = 0

gives us

∞∑l=0

(l + 1)(l + 2)al+2ul − 2

∞∑l=0

lalul +

(β

α− 1

) ∞∑l=0

alul = 0 (7.80)

or∞∑l=0

[(l + 1)(l + 2)al+2 − 2lal +

(β

α− 1

)al

]ul = 0. (7.81)

For a power series to be zero, each term has to be zero; that is to sayall the coefficients have to be zero.We now have the following recursion relation.

al+2 = −(β/α− 1− 2l)

(l + 1)(l + 2)al (7.82)

The general solution is a sum of even and odd series; namely,

H(u) = a0

(1 +

a2a0u2 +

a4a0u4 +

a6a0u6 + · · ·

)


+a1

(u+

a3a1u3 +

a5a1u5 +

a7a1u7 + · · ·

)(7.83)

= a0(1 + b1u2 + b2u

4 + b3u6 + . . .) + a1u(1 + c1u

2 + c2u4 + c3u

6 + . . .)

= a0∑l even

b l2ul + a1u

∑l even

c l2ul; (7.84)

where

b l2=ala0

and c l2=al+1

a1for l = 2, 4, 6, . . . , (7.85)

and the reason for the awkward notations b l2

and c l2

for the coefficients willbecome clear when compared with (7.88). What (7.83) means is that one canget all even coefficients using (7.82) if a0 is given1, and all odd coefficientscan be computed if a1 is chosen. However, there is no relation between theeven and odd coefficients, and the even series and odd series of (7.83) or(7.84) are completely independent.

As l→∞,

al+2

al= −(β/α− 1− 2l)

(l + 1)(l + 2)→ 2l

l2→ 2

l. (7.86)

So, we can see immediately that

b l2+1

b l2

=al+2

al

l→∞−−−−−→ 2

land

c l2+1

c l2

=al+3

al+1

l→∞−−−−−→ 2

l. (7.87)

Now compare this with the Taylor series expansion (about 0) of eu2 wherewe change the running index from i to l = 2i and denote 1

( l2)!

by d l2.

eu2

=∞∑i=0

((u2)i)

i!=

∞∑i=0

u2i

i!=

even∑l=0

ul(l2

)!=

even∑l=0

d l2ul (7.88)

Observe that

d l2+1

d l2

=

1

( l2+1)!1

( l2)!

=

(l2

)!(

l2+ 1

)!=

(l2

)!(

l2+ 1

) (l2

)!

1For example, a6

a0= a6

a4

a4

a2

a2

a0.



=1

l2+ 1

l→∞−−−−−→ 1

l/2=

2

l. (7.89)

Therefore, both the even series and the odd series of (7.84) behave likethe Taylor expansion of eu2 for higher order terms, or as l→∞. Guidedby this, we assume H(u) behaves like

H(u) = a0Keu2 + a1K

′ueu2

. (7.90)

So, e−u2/2H(u) of (7.75) and (7.73) behaves like

a0Keu2/2 + a1K

′ueu2/2 as |u| → ∞a. (7.91)

In order to satisfy the integrability condition, Condition 1 on p.123,H(u) has to terminate after some n to prevent the exponential termsin (7.91) from blowing up. This means we have to set β/α = 2n+1 forn = 0, 1, 2, 3, 4, 5, . . . due to (7.82).

When β/α = 2n+1 we get a Hermite polynomial denoted by Hn(u).This gives us a series of eigenfunctions

ψn(u) = Ane−u2/2Hn(u).

The first six functions look like this.

n = 0 ψ0 = A0e−u2/2

1 ψ1 = A1ue−u2/2

2 ψ2 = A2(1− 2u2)e−u2/2

3 ψ3 = A3(3u− 2u3)e−u2/2

4 ψ4 = A4(3− 12u2 + 4u4)e−u2/2

5 ψ5 = A5(15u− 20u3 + 4u5)e−u2/2

(7.92)

aAs |u| → ∞, higher order terms with large values of l become dominant, andthis is where the series in (7.91) begin to behave like the Taylor expansion (7.88).

Now recall that u =√αx, α = 2πmν/~, and β = 2mE/~2.


β/α =2mE/~2

2πmν/~=

E

~πν=

Eh2π· πν

=2E

hν

Setting this equal to 2n+ 1,

β/α = 2n+ 1 =⇒ 2Enhν

= 2n+ 1 =⇒ En =(n+

1

2

)hν

n = 0, 1, 2, 3, 4, 5, . . . . (7.93)

Recall that

ν =1

2π

√c

m.

Therefore, if the force constant c and the mass m are known, we know theenergy levels of this harmonic oscillator.

These are the two important facts about the simple harmonic oscillator.

1. The energy levels are equally spaced. This is markedly differentfrom the infinite square well. This means many higher energylevles can be achieved more easily.

2. The lowest energy posssible is not zero but E0 = 12hν for n = 0.

Since vibrations of diatomic molecules can be closely approxi-mated by the simple harmonic oscillator, this is another reasonwhy the absolute zero degree temperature can not be achieved.

When a molecule drops from one energy level to a lower level, aphoton carrying that much energy is released. On the other hand, amolecule moves to a higher energy level if it absorbs a photon carryingthe energy corresponding to the difference between the two levels. Thesephenomena are called emission and absorption respectively.



7.6.2 Raising and Lowering OperatorsOur primary goal in this section is to simplify the computation of the en-ergy levels for a simple harmonic oscillator. The beauty of this alternativeapproach is that we will be able to obtain the energy levels without explicitlycomputing the wavefunctions. The concept of raising and lowering operatorshas applications to angular momentum discussed in Chapter 9 and electronspin, a special type of angular momentum, described in Chapter 11.

From (7.60), we have

−~2

2m

d2ψ(x)

dx2+c

2x2ψ(x) = Eψ(x). (7.94)

This equation is often expressed in terms of ω = 2πν.

ν =1

2π

√c

m=⇒ ω = 2πν =

√c

m=⇒ c = mω2

Substituting mω2 for c, we get[−~2

2m

d2

dx2+

1

2mω2x2

]ψ(x) = Eψ(x). (7.95)

Our first step is to modify and rescale the variables. Let

X =

√mω

~x (7.96)

and

P =1√m~ω

p; (7.97)

where x is the usual multiplication operator Mx and p = −i~ ∂∂x

as given inTable 3.1. We will not use capital letters X and P in this section in order to


make a clear distinction between the derived (X, P ) and the original (X,P ).X and P are dimensionless variables satisfying a simpler commutation rela-tion. Indeed,

XP − P X =

√mω

~x

1√m~ω

p− 1√m~ω

p

√mω

~x =

1

~(xp− px), (7.98)

and we have

[X, P ] = i. (7.99)

We also introduce a rescaled Hamiltonian H defined by

H =1

~ωH. (7.100)

Now note that

X2 + P 2 =mω

~x2 +

1

m~ω

(−i~ d

dx

)(−i~ d

dx

)=mω2

~ωx2 − ~2

m~ω

d2

dx2

=2

~ω

(−~2

2m

d2

dx2+

1

2mω2x2

)= 2

1

~ωH. (7.101)

Therefore,

H =1

2

(X2 + P 2

). (7.102)

We will now define a new operator a and its adjoint a†, which will turn outto be ”raising” and ”lowering” operators in the title of this section.

a =1√2

(X + iP

)(7.103)



a† =1√2

(X − iP

)(7.104)

Going backwards, we get the following if the above relations are solved forX and P .

X =1√2

(a† + a

)(7.105)

P =1√2

(a† − a

)(7.106)

Next, we will compute the commutator of a and a†.[1√2

(X + iP

),

1√2

(X − iP

)]=

1

2

[X + iP , X − iP

]=i

2

[P , X

]−[X, P

]=i

2(−i− i) = 12 (7.107)

Therefore,

[a, a†] = 1. (7.108)

Now,

a†a =1

2

(X − iP

) (X + iP

)=

1

2

(X2 + P 2 + iXP − iP X

)=

1

2

(X2 + P 2 − 1

)= H − 1

2. (7.109)

So,

H = a†a+1

2. (7.110)

On the other hand,

aa† = [a, a†] + a†a = 1 + H − 1

2= H +

1

2. (7.111)

And,


H = aa† − 1

2. (7.112)

At this point, let us introduce another self-adjoint operator N defined by

N = a†a. (7.113)

Note that(a†a

)†= a†

(a†)†

= a†a, which proves that N is self-adjoint. Theoperator N satisfies the following commutation relations.

[N, a] = [a†a, a] = a†aa+(a†aa− a†aa)− aa†a = [a†, a]a = −a (7.114)[N, a†] = [a†a, a†] = a†aa† + (a†a†a− a†a†a)− a†a†a = a†[a, a†] = a†

(7.115)

We have now replaced the initial eigenvalue problem for H, which is a func-tion of x and p, with that of N , which is a product of a and a†. Because wehave

H = ~ωH, H = N +1

2=⇒ H = ~ω

(N +

1

2

), (7.116)

N |n⟩ = n |n⟩ =⇒ H |n⟩ = ~ω(n+

1

2

)|n⟩ ; (7.117)

where we are looking ahead and already using suggestive notations n and|n⟩, as opposed to the usual λ and |λ⟩. However, it is only on p.155 that weactually show that the eigenvalues of N are nonnegative integers. Relations(7.116) and (7.117) show that the eigenvalue problem for N is equivalent tothe eigenvalue problem for H. When the eigenvalue problem for N is solved,we will get En =

(n+ 1

2

)~ω as the total energy for the simple harmonic

oscillator.



We will need the following three facts before solving the eigenvalue prob-lem for N .

Fact 7.1 Each eigenvalue n of the operator N is nonnegative.

ProofSuppose N |n⟩ = n |n⟩. Then,

⟨n|N |n⟩ = n ⟨n|n⟩ . (7.118)

On the other hand,

⟨n|N |n⟩ = ⟨n|a†a|n⟩ = ∥a |n⟩ ∥2. (7.119)

Therefore,

n ⟨n|n⟩ = ∥a |n⟩ ∥2 =⇒ n =∥a |n⟩ ∥2

⟨n|n⟩≥ 0. (7.120)

Fact 7.2 If the eigenvalue n = 0 for N , a |0⟩ = 0. Conversely, ifa |v⟩ = 0, N |v⟩ = 0 |v⟩. For other values of n, i.e. n > 0 from Fact7.1, a |n⟩ is an eigenvector of N with the eigenvalue n− 1.

ProofSuppose N |0⟩ = 0 |0⟩. Then,

∥a |0⟩ ∥2 = ⟨0|a†a|0⟩ = ⟨0|N |0⟩ = ⟨0|0|0⟩ = 0 =⇒ a |0⟩ = 0. (7.121)

Now suppose a |v⟩ = 0, then,

a |v⟩ = 0 =⇒ a†a |v⟩ = N |v⟩ = 0 |v⟩ . (7.122)

So, any nonzero vector |v⟩ with a |v⟩ = 0 is an eigenvector of N with theeigenvalue n = 0.


Next, consider a strictly positive eigenvalue n of the operator N . Then, from(7.114),

[N, a] |n⟩ = Na |n⟩ − aN |n⟩ = Na |n⟩ − an |n⟩ = Na |n⟩ − na |n⟩ = −a |n⟩=⇒ N (a |n⟩) = na |n⟩ − a |n⟩ = (n− 1) (a |n⟩) . (7.123)

Therefore, a |n⟩ is an eigenvector of N with the eigenvalue n− 1.

Fact 7.3 If |n⟩ is an eigenvector of N with the eigenvalue n. Then,a† |n⟩ is an eigenvector of N with the eigenvalue n+ 1.

ProofWe will first show a† |n⟩ is nonzero. From (7.108),

∥a† |n⟩ ∥2 = ⟨n|aa†|n⟩ = ⟨n|[a, a†] + a†a|n⟩ = ⟨n|1 +N |n⟩= ⟨n|n⟩+ n ⟨n|n⟩ = (n+ 1)∥ |n⟩ ∥2. (7.124)

As |n⟩ is an eigenvector of N , it is nonzero, and a† |n⟩ is nonzero.Now from (7.108),

[N, a†] |n⟩ = Na† |n⟩ − a†N |n⟩ = Na† |n⟩ − a†n |n⟩ = Na† |n⟩ − na† |n⟩ = a† |n⟩=⇒ N

(a† |n⟩

)= (n+ 1)

(a† |n⟩

). (7.125)

Therefore, a† |n⟩ is an eigenvector of N with the eigenvalue n+ 1.

We are now ready to show that the eigenvalues of N are non-negativeintegers. Suppose an eigenvalue λ of N is not an integer, and let |λ⟩ denotean associated eigenvector. From Fact 7.1, we already know that λ is strictlypositive. From Fact 7.2, we know that a |λ⟩ is an eigenvector of N with theeigenvalue λ − 1. After repeated applications of a, on |λ⟩, k times, we getak |λ⟩ which is an eigenvector of N with the associated eigenvalue λ− k. Ask can be as large as we wish it to be, this implies that N has strictly negativeeigenvalues, contradicting Fact 7.1. Therefore, the eigenvalues of N are allnonnegative integers.



Consider an eigenvalue n, which is a positive integer, and an associatedeigenvector denoted by |n⟩. Then, due to Fact 7.2, ak |n⟩ > is an eigenvectorfor k = 1, 2, . . . , n. When k = n, the eigenvalue associated with an |n⟩ is0. Then, from Fact 7.2, a (an |n⟩) = 0, terminating the chain of eigenvectors|n⟩ , a |n⟩ , a2 |n⟩ , . . . at k = n. So, an eigenvalue n of N can only be a non-negative integer.

On the other hand, given an eigenvector |0⟩ of N , we can apply a† to thevector any number of times, with each application generating another eigen-vector whose associated eigenvalue is greater than the previous eigenvalue by1. This is guaranteed by Fact 7.3. Therefore, the eigenvalues of N , H, andH are all nonnegative integers.

We have

En =(n+

1

2

)~ω for n = 0, 1, 2, . . . (7.126)

as before.

7.6.2.1 Actions of a and a†

Let |0⟩ , |1⟩ , |2⟩ , . . . be an orthonormal basis3 for the Hilbert space for H,H, or N . Recall that the eigenvectors are the same for these three operators.We would like to find exactly what a and a† do to these basis vectors.

We already know

a |n⟩ = cn |n− 1⟩ (7.127)

for some constant cn. Taking the adjoint,

⟨n| a† = ⟨n− 1| c∗n. (7.128)

So,

⟨n|a†a|n⟩ = ⟨n− 1|n− 1⟩ c∗ncn = c∗

ncn =⇒ ⟨n|N |n⟩ = c∗ncn

3Because these are eigenvectors of a Hermitian operator H, they form a basis automat-ically. The only additional condition imposed here is normalization of each ket.


=⇒ ⟨n|n|n⟩ = c∗ncn =⇒ n = |cn|2. (7.129)

We have

cn =√neiϕ. (7.130)

By convention, we normally set ϕ = 0. Then,

cn =√n and a |n⟩ =

√n |n− 1⟩ . (7.131)

Similarly, we also know

a† |n⟩ = dn |n+ 1⟩ (7.132)

for some constant dn. Taking the adjoint,

⟨n| a = ⟨n+ 1| d∗n. (7.133)

This gives us

⟨n|aa†|n⟩ = ⟨n+ 1|d∗ndn|n+ 1⟩ . (7.134)

The left-hand side can be transformed as below using the relation (7.108).

⟨n|aa†|n⟩ = ⟨n|[a, a†] + a†a|n⟩ = ⟨n|1 +N |n⟩ = 1 + n (7.135)

Therefore,

|dn|2 = n+ 1. (7.136)

Setting the phase equal to 0 as before, we get

dn =√n+ 1 and a† |n⟩ =

√n+ 1 |n+ 1⟩ . (7.137)

We can represent the actions of a and a† as matrices with infinitely manyentries.

a ↔

0√1 0 0 . . .

0 0√2 0 . . .

0 0 0√3 . . .

0 0 0 0 . . ....

......

.... . .

(7.138)



a† ↔

0 0 0 0 . . .√1 0 0 0 . . .

0√2 0 0 . . .

0 0√3 0 . . .

......

....... . .

(7.139)

I hope it is now clear to you why a is the lowering operator and a† is theraising operator4.

4Note that a and a† are also called destruction and creation operators, respectively.

159

Exercises1. This problem is the same as Chapter 5 Problem 5.

Consider a particle of mass m and total energy E which can move freelyalong the x-axis in the interval [−a

2,+a

2], but is strictly prohibited from

going outside this region. This corresponds to what is called an infinitesquare well potential V (x) given by V (x) = 0 for −a

2< x < +a

2and

V (x) = ∞ elsewhere. If we solve the Schrodinger equation for thisV (x), one of the solutions is

Ψ(x, t) =

A cos

(πxa

)e−iEt/~ −a

2< x < a

2

(a > 0)0 |x| ≥ a

2

.

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

tion reduces to

− ~2

2m

∂2

∂x2Ψ(x, t) = i~

∂Ψ(x, t)

∂t.

Plug Ψ(x, t) = A cos(πxa

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

π2~2

2ma2.

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

integral table for this. (Of course, you can always try contourintegration or some such. But, that is beyond this course.)

(e) Evaluate ⟨P 2⟩ for the same wavefunction. You will not need anintegral table if you use the fact that the function is normalized.Of course, a brute force computation will yield the same result aswell.

2. This problem is the same as Chapter 5 Problem 6.If you are a careful student who pays attention to details, you mayhave realized that A in Problem 1 can only be determined up to theargument, or up to the sign if you assume A is a real number.

(a) What is the implication of this fact as to the uniqueness of wave-function?

160

(b) What is the implication of this fact as to the probability densityΨ∗(x, t)Ψ(x, t)? How about ⟨P ⟩, ⟨P 2⟩, and ⟨x2⟩?

3. Solve the time-independent Schrodinger equation for a free particle.(Note that a free particle is a particle with no force acting on it. Interms of the potential V (x), this means that it is a constant. Forsimplicity, assume that V (x) = 0 in this problem.)

4. Consider a particle of mass m which can move freely along the x-axis anywhere from x = −a/2 to x = +a/2 for some a > 0, but isstrictly confined to the region [−a/2,+a/2]. This corresponds to whatis called an infinite square well potential V (x) given by V (x) = 0 for−a/2 < x < +a/2 and V (x) = ∞ elsewhere. Where the potential isinfinite, the wavefunction is zero. Use the continuity condition on ψ(x)at |x| = a/2, but not on ψ′(x), to find at least two of the solutions.

5. Give a rigorous mathematical proof that “H(x) = K(t) for all pairs ofreal numbers (x, t)” implies that “H(x) and K(t) are a constant”.

6. Given a potential V (x) with the following profile

V (x) =

0 x < 0V0 x > 0

solve the Time-Independent Schrodinger Equation for a particle withthe total energy E < V0 traveling from x = −∞ to x = +∞. In fact,we have already solved this problem in this chapter. I just want you toreview the procedure.

7. Show that we have total reflection in 6 above by direct computationusing probability current described in Appendix H.

8. Consider the potential V (x) given by V (x) = 0 for x < 0 and V (x) =V0 > E for x > 0, and show that we have a standing wave in the regionx < 0 for a particle incident on the discontinuity at x = 0 from theleft, that is, from the region characterized by negative values of x.

9. Consider the potential V (x) given by V (x) = 0 for x < 0 and V (x) =V0 < E.

161

(a) Study the complete solution of the time-independent Schrodingerequation we solved in this chapter.

(b) Sow that the transmission coefficient is given by 4k1k2(k1+k2)2

.

10. Consider a step potential given by

V (x) =

0 x < 0V0 x > 0

.

(a) Write down, but do not derive or solve, the time-independentSchrodinger equation for a particle of mass m and total energyE > V0 for each region.

(b) For a particle moving toward the step from the left, the solutionsare

ψ(x) = Aeik1x +Be−ik1x (x < 0)

andψ(x) = Ceik2x (x > 0),

where k1 =√2mE~

and k2 =

√2m(E−V0)~

. Use the two boundaryconditions at x = 0 to express C in terms of A, k1, and k2.

(c) As it turns out, we also get B = k1−k2k1+k2

A. Compute the reflectioncoefficient R; that is, express it in terms of k1 and k2.

11. In order to solve the Time-Independent Schrodinger Equation for aparticle incident on the potential barrier given by

V (x) =

V0 > E 0 < x < a0 elsewhere

from the left, that is, x < 0 at t = 0, you need to consider the wave-function given by

ψ(x) =

Aeik1x + Be−ik1x x < 0Fe−k2x + Gek2x 0 < x < aCeik1x + De−ik1x x > a

; (7.140)

with the following set of boundary conditions.

ψ(x)|x=0− = ψ(x)|x=0+

ψ(x)|x=a− = ψ(x)|x=a+dψ(x)dx

∣∣∣x=0−

= dψ(x)dx

∣∣∣x=0+

dψ(x)dx

∣∣∣x=a−

= dψ(x)dx

∣∣∣x=a+

(7.141)

162

(a) What are k1 and k2 in (7.140)?(b) Why should D in (7.140) be 0?(c) Write down the boundary conditions (7.141) after setting D = 0.

12. Consider a particle of energy E > V0 moving from x = +∞ to the leftin the step potential:

V (x) =

V0 x > 0

0 x < 0.

(a) Write the time-independent Schrodinger equations in the regionsx < 0 and x > 0.

(b) The general solutions are of the formA× (wave traveling to the right) + B × (wave traveling to the left) x < 0

C × (wave traveling to the right) +D × (wave traveling to the left) x > 0.

Complete the general solutions; i.e. substitute actual formulas for(wave traveling to the right) and (wave traveling to the left). Forsimplicity of notation, let k1 =

√2mE~

and k2 =

√2m(E−V0)~

.(c) Determine the value of one of the constants based on the fact that

there is no wave coming back from x = −∞ and traveling to theright.

(d) Use the boundary conditions at x = 0 to get the relations amongthe constants.

(e) Compute the reflection coefficient R.

13. Consider a barrier potential given by

V (x) =

0 (x < −a)V0 (−a < x < 0)0 (x > 0)

.

A stream of particles are originating at x = −∞ and traveling to theright. The total energy E is less than the barrier height V0.

(a) Write down, but do not derive or solve, the time-independentSchrodinger equation for a particle of mass m and total energyE < V0 for each region.

163

(b) For a particle moving toward the step from the left, the solutionsare

ψ(x) =

Aeik1x +Be−ik1x (x < −a)Cek2x +De−k2x (−a < x < 0)Geik1x (x > 0)

where k1 =√2mE~

and k2 =

√2m(V0−E)

~.

i. Explain in words why we have ψ(x) = Geik1x and not ψ(x) =Geik1x +He−ik1x in the region x > 0.

ii. Give the two continuity coonditions at x = 0.iii. On physical grounds, we also need continuity of ψ∗ψ as well

as its first derivative at the boundaries. Use this continuitycondition on dψ∗ψ

dxat x = 0 to show neither C nor D is 0.

14. Show the following for the infinite square well potential described inthe class.

V (x) =

0 −a

2< x < a

2

∞ elsewhere

(a) Two general solutions ψ(x) = A sin(kx) + B cos(kx) and ψ(x) =Ceikx+De−ikx are equivalent. Here, k =

√2mE/~, and A, B, C,

and D are arbitrary constants.(b) The usual boundary conditions at x = 0 and a allow only dis-

crete values of k, which in turn implies that the total energy isdiscretized.

15. The potential for the infinite square well is given by

V (x) =

∞ |x| > a

2

0 |x| < a2

,

and the solutions for the time-independent Schrodinger equation givenin the class were

ψn(x) = Bn cos knx where kn = nπa

n = 1, 3, 5, . . .ψn(x) = An sin knx where kn = nπ

an = 2, 4, 6, . . .

.

(a) Find a normalized solution for n = 2 for which the normalizationconstant A2 is purely imaginary.

164

(b) What is the expectation value of the position x for this solution?Use the full wavefunction to solve this problem.

(c) Plug the above solution into the lefthand side of the time-independentSchrodinger equation to determine E2, the total energy associatedwith this state.

16. The potential for the infinite square well is given by

V (x) =

∞ |x| > a

2

0 |x| < a2

.

(a) Write down, but do not derive or solve, the time-independentSchrodinger equation for a particle of mass m and total energy Ein the region −a

2< x < a

2.

(b) Find a solution of the form ψ(x) = A sin(Bx) for −a2

< x <a2, where A and B are constants, and compute the total energy

following the steps below.i. Express B as a function of E so that ψ(x) = A sin(Bx) is

a solution of the Shcrodinger equation. (Choose the positivesquare root.)

ii. Find the smallest value of B that satisfies the boundary con-ditions.

iii. Compute the total energy corresponding to the B above.iv. Determine A so that the wavefunction is properly normalized.

17. Consider an infinite square well whose potential is given by

V (x) =

∞ x < 00 0 < x < L∞ L < x

.

(a) Write down, but do not derive or solve, the time-independentSchrodinger equation for a particle of mass m and total energy Ein the region 0 < x < L.

(b) The most general solution of the time-independent Schrodingerequation in the region 0 < x < L is of the form

ψ(x) = A sin kx+B cos kx;

where A and B are arbitrary constants and k =√2mE~

.

165

i. Using one of the boundary conditions on ψ(x), determine thevalue of B.

ii. Find all possible positive values k can take by applying one ofthe boundary conditions on ψ(x). Express k in terms of π, L,and an arbitrary natural number n. We will refer to each ofthese values as kn. With this, we can write our wavefunctionas

ψn(x) = An sin knx+Bn cos knxin order to make (potential) dependence on n more explicit.

iii. Impose the normalization condition to find the positive realvalue of An given the above kn.

(c) Now, use the relation

kn =

√2mEn~

to determine all possible values the total energy can take. In otherwords, express En in terms of m, h, L, and n.

18. Try a series solutionf(x) =

∞∑j=0

ajxj

for the differential equation

df(x)

dx+ f(x) = 3x2 + 8x+ 3,

and show that

aj + (j + 1)aj+1 = 0 for j ≥ 3.

19. For a simple harmonic oscillator, show that the wavefunction corre-sponding to n = 1 is given by

ψ1 = A1ue−u2/2,

and the wavefunction for n = 2 takes the form

ψ2 = A2(1− 2u2)e−u2/2.

166

20. A time-independent ground state wavefunction of the simple harmonicoscillator is given by Φ0 = A0exp[−u2

2], where u = [(Cm)

14/~

12 ]x. Find

the expectation value of x as a function of the normalization constantA0. Use the full wavefunction with the total energy E0.

21. This question concerns the simple harmonic oscillator.

(a) The full ground state wavefunction Ψ0 can be expressed as

ψ0e−iEt/~

with

ψ0 = A0e−u2/2;

where A0 is a normalization constant, and u =[(Cm)

14~− 1

2

]x.

Compute the expectation value < x > of the position for thisstate.

(b) Show by explicit computation that the first two wavefunctions ofa simple harmonic oscillator are orthogonal. Use the functionsgiven below, where A0 and A1 are normalization constants.

ψ0 = A0e−u2/2

ψ1 = A1ue−u2/2

(c) Why do you know the third and the fifth eigenfunctions, denotedby ψ2 and ψ4 as the first eigenfunction is ψ0, are orthogonal toeach other without conducting explicit integration?

ψ2 = A2(1− 2u2)e−u2/2

ψ4 = A4(3− 12u2 + 4u4)e−u2/2

(d) Explain in fewer than 50 words why a solid made of diatomicmolecules can not be cooled to absolute zero. Here, I am onlyasking you to reproduce the crude argument I gave you in theclass as incomplete as it was.

22. Suppose |n⟩ is an eigenket of a†a with the eigenvalue n. Show thata† |n⟩ is an eigenket of a†a with the eigenvalue n+ 1.

167

23. Describe, in a sentence or two, one of the differences between classicalmechanics and quantum mechanics. Your answer should be about anactual physical phenomenon or its interpretation. Hence, for exam-ple, the difference in form between the Schrodinger equation and theNewton’s equation is not an acceptable answer.

Chapter 8

Higher Spatial Dimensions

So far, we have only dealt with one dimensional cases partly for simplicity.However, most physical systems have multiple spatial dimensions, and thereare a few new phenomena which you can observe only if you have more thanone dimension. We will discuss two of such new features; degeneracy andangular momentum.

8.1 DegeneracyIn a typical dictionary, degeneracy is defined as a state of being degener-ate, and the primary or ordinary meanings of the adjective degenerate are“Having declined, as in function or nature, from a former or original state”and “Having fallen to an inferior or undesirable state, especially in mentalor moral qualities.” But, degeneracy in quantum mechanics does not havemuch to do with declining or becoming inferior. According to Wikipedia[Wikipedia contributors, nd], degeneracy in quantum mechanics has the fol-lowing definition.

In quantum mechanics, a branch of physics, two or more differentstates of a system are said to be degenerate if they are all at the sameenergy level. It is represented mathematically by the Hamiltonian forthe system having more than one linearly independent eigenstate withthe same eigenvalue. Conversely, an energy level is said to be degener-ate if it contains two or more different states. The number of differentstates at a particular energy level is called the level’s degeneracy, andthis phenomenon is generally known as a quantum degeneracy.

169

170 CHAPTER 8. HIGHER SPATIAL DIMENSIONS

From the perspective of quantum statistical mechanics, several degen-erate states at the same level are all equally probable of being filled.

One example of this is the hydrogen atom discussed in Chapter 10. In thischapter, we will give a general proof that there is no degeneracy for “bound”states in one dimension, leaving discussions of degeneracy for specific casesto other chapters.

Definition 8.1 (Bound States) Bound states occur whenever the particlecannot move to infinity. That is, the particle is confined or bound at allenergies to move within a finite and limited region of space which is delimitedby two classical turning points.

As the particle cannot move to infinity, we need to have

ψ(x −→ ±∞) −→ 0 (8.1)

for any bound state.

Theorem 8.1 There is no degeneracy for one-dimensional bound states.

ProofSuppose there is degeneracy for total energy E. This means there are twounit vectors |u⟩ and |w⟩, representing distinct physical states, such that

H |u⟩ = E |u⟩ and H |w⟩ = E |w⟩ .

Let x be the only position variable in this one-dimensional space. Then,writing ψu(x) and ψw(x) for |u⟩ and |w⟩ respectively, we get

− ~2

2m

d2

dx2ψu(x) + V ψu(x) = Eψu(x) (2)

and

− ~2

2m

d2

dx2ψw(x) + V ψw(x) = Eψw(x). (3)

Multiplying (2) by ψw(x) and (3) by ψu(x) gives us(− ~

2

2m

d2

dx2ψu(x)

)ψw(x) + V ψu(x)ψw(x) = Eψu(x)ψw(x) (4)

8.1. DEGENERACY 171

and(− ~

2

2m

d2

dx2ψw(x)

)ψu(x) + V ψw(x)ψu(x) = Eψw(x)ψu(x). (5)

Now, subtracting (5) from (4) leads to

− ~2

2m

(d2

dx2ψu(x)

)ψw(x)−

(− ~

2

2m

)(d2

dx2ψw(x)

)ψu(x) = 0

=⇒(d2

dx2ψu(x)

)ψw(x)−

(d2

dx2ψw(x)

)ψu(x) = 0. (6)

Noting that (d2

dx2ψu(x)

)ψw(x)−

(d2

dx2ψw(x)

)ψu(x)

=

(d2

dx2ψu(x)

)ψw(x) +

d

dxψu(x)

d

dxψw(x)

− d

dxψu(x)

d

dxψw(x)−

(d2

dx2ψw(x)

)ψu(x)

=

(d2

dx2ψu(x)

)ψw(x) +

d

dxψu(x)

d

dxψw(x)

−(d

dxψu(x)

d

dxψw(x) +

(d2

dx2ψw(x)

)ψu(x)

)

=d

dx

[(d

dxψu(x)

)ψw(x)

]− d

dx

[(d

dxψw(x)

)ψu(x)

]

=d

dx

[(d

dxψu(x)

)ψw(x)−

(d

dxψw(x)

)ψu(x)

], (7)

(6) implies (d

dxψu(x)

)ψw(x)−

(d

dxψw(x)

)ψu(x) = c (8)

where c is some constant. Because ψu and ψw are bound states,

ψu(x)x→±∞−−−−−→ 0 and ψw(x)

x→±∞−−−−−→ 0. (9)


Therefore, the constant c in (8) is 0.1 Naively, i.e. assuming ψu and ψw arepositive for now, we have

1

ψu(x)

d

dxψu(x) =

1

ψw(x)

d

dxψw(x) =⇒ d lnψu(x)

dx=d lnψw(x)

dx

=⇒∫ d lnψu(x)

dxdx =

∫ d lnψw(x)dx

dx =⇒ lnψu(x) = lnψw(x) + k

=⇒ lnψu(x) = ln ekψw(x) =⇒ ψu(x) = ekψw(x). (10)

So, in this case, ψu(x) is a scalar multiple of ψw(x), and they represent thesame physical state. This in turn implies that there is no degeneracy. If youthink there is too much hand-waving in this “proof”, you are extremely right!I gave this naive argument first in order to make this proof more accessiblefor, say, freshmen.

Having said that, a more rigorous proof can be given resorting to theWronskian argument described in Appendix C. According to Appendix C,two solutions, ψu and ψw, of a linear second-order ordinary differential equa-tion of the form

ψ′′(x) + p(x)ψ′(x) + q(x)ψ(x) = 0 (11)

where p(x) and q(x) are continuous, are linearly dependent if and only if theWronskian, W (ψu, ψw), given by

W (ψu, ψw)(x) = ψu(x)ψ′w(x)− ψ′

u(x)ψw(x) (12)

is 0 for all x in the domain. As this is the case for us, we have ψu(x) = Kψw(x)for some constant K, and this proves the theorem.

We have now shown that degeneracy, which requires two linearly indepen-dent eigenfunctions with the same energy eigenvalue E, is a phenomenonspecific to more than one dimension.

In passing, let us make a note of the following theorem found on p.217of “Quantum Mechanics: Concepts and Applications (second edition)” by

1A more rigorous proof also requires an examination of the behavior of ddxψu(x) and

ddxψw(x) as x tends to ±∞. For now, just think of exponential functions e∓kx with k > 0and their first derivatives as x→ ±∞.

8.2. ANGULAR MOMENTUM 173

Nouredine Zettili [Zettili, 2009, p.217], which includes discreteness of energylevels for bound states.

Theorem 8.2 (Discrete and Nondegenerate) In a one-dimensional prob-lem the energy levels of a bound state system are discrete and not degenerate.

The condition (8.1) together with Condition 3 (continuity conditions) onp.124 can be satisfied only for certain special values of energy, making allowedtotal energy values discrete. We saw how this works repeatedly in Chapter7. Zettili also provides the following theorem.

Theorem 8.3 The wave function ψn(x) of a one-dimensional bound statesystem has n nodes (i.e., ψn(x) vanishes n times) if n = 0 corresponds to theground state and (n− 1) nodes if n = 1 corresponds to the ground state.

Remark 8.1 (Unbound One-Dimensional States) Note that we dohave degeneracy if we have unbound states. A good example is a freeparticle traveling wave where we have

ei(kx−ωt) and e−i(kx+ωt) (8.13)

according to (7.4).

These are linearly independent states corresponding to particles/waves trav-eling to the right ei(kx−ωt) and to the left e−i(kx+ωt). Because k =

√2mE~

forboth waves, they have the same energy, and we have two-fold degeneracy.

8.2 Angular MomentumThe only momentum we encounter in a one-dimensional system is the linearmomentum p = mv as we only have a linear motion in one-dimension. How-ever, in higher dimensions, rotational motions are also possible, and we haveanother variety of momentum called the angular momentum. The classicaldefinition of the angular momentum L of a particle of mass m with respectto a chosen origin is given by


L = r × p; (8.14)

where r is the position vector connecting the origin and the particle, p = mvis the linear momentum, and × is the usual vector cross product. If we denotethe angle formed by r and p by θ, we have

∥L∥ = ∥p∥∥r∥ sin θ = m∥v∥∥r∥ sin θ or L = mvr sin θ.a (8.15)

aWe can choose either angle for this purpose as sin θ = sin (π − θ).

The angular momentum is subject to the fundamental constraints of theconservation of angular momentum principle if there is no external torque,rotation-causing force, on the object. Using the usual substitutions

xi ←→Mxiand pi ←→ −i~

∂

∂xi; (8.16)

where x1 = x, x2 = y, x3 = z, and likewise for pi’s, we can “derive” quan-tum mechanical angular momentum operators Lx, Ly, and Lz drawing onthe classical relation (8.14).

Because the study of the angular momentum is a large and thriving indus-try, we will devote the entire Chapter 9 to a detailed discussion of quantummechanical angular momentum.

Chapter 9

Angular Momentum

As we move from one-dimensional systems to higher dimensionality, moreand more interesting physics emerges. In this chapter, we will discuss howangular momentum is dealt with in quantum mechanics and the consequencesof quantum mechanical computations of angular momenta. The angular mo-mentum relations are very important in the study of atoms and nuclei. Likethe total energy E, the magnitude of angular momentum, usually denotedby L, as well as its z-component, Lz are conserved. Furthermore, angularmomentum is quantized like energy levels.

9.1 Angular Momentum OperatorsQuantum mechanical operators representing the x-, y-, and z-component ofangular momentum were already presented in Table 3.2 without derivation.Now, we will see how we can “derive” quantum mechanical counterpart draw-ing on the classical theory of angular momentum.

The angular momentum of a particle about the origin is given by

L = r×p =⇒ lxı+ly ȷ+lzk =

∣∣∣∣∣∣∣ı ȷ kx y zpx py pz

∣∣∣∣∣∣∣ =⇒

lxlylz

=

ypz − zpyzpx − xpzxpy − ypx

;(9.1)

175

176 CHAPTER 9. ANGULAR MOMENTUM

where the expression L = r×p is coordinate-independent, but the component-by-component expressions on the right are in Cartesian coordinates. Thequantum mechanical operator L = Lxı+ Ly ȷ+ Lzk is obtained as follows.

First, noting that px ↔ −i~ ∂∂x , py ↔ −i~ ∂∂y , and pz ↔ −i~ ∂∂z , we get:

Lx = y

(−i~ ∂

∂z

)− z

(−i~ ∂

∂y

)= −i~

(y∂

∂z− z ∂

∂y

)

Ly = z

(−i~ ∂

∂x

)− x

(−i~ ∂

∂z

)= −i~

(z∂

∂x− x ∂

∂z

)(9.2)

Lz = x

(−i~ ∂

∂y

)− y

(−i~ ∂

∂x

)= −i~

(x∂

∂y− y ∂

∂x

)

This book culminates in a full solution of the Schrodinger equation forthe hydrogen atom; where we only have a central force, and the sphericalcoordinate system is employed to simplify the computation. In sphericalcoordinates, we get:

Lx = i~

(sinϕ ∂

∂θ+ cot θ cosϕ ∂

∂ϕ

)

Ly = i~

(− cosϕ ∂

∂θ+ cot θ sinϕ ∂

∂ϕ

)(9.3)

Lz = −i~∂

∂ϕ.

Note that Lz is particularly simple. As for the magnitude of angularmomentum, we will work with the squared magnitude, denoted by L2.

L2 = L2x + L2

y + L2z = −~2

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

](9.4)

We have the following commutation relations.

9.1. ANGULAR MOMENTUM OPERATORS 177

[Lx, Ly] = i~Lz (9.5)[Ly, Lz] = i~Lx (9.6)[Lz, Lx] = i~Ly (9.7)

[L2, Lx] = [L2, Ly] = [L2, Lz] = 0 (9.8)

The relations (9.5), (9.6), and (9.7) can be summarized as

L×L = i~L. (9.9)

Let us verify (9.5) only, as (9.6) and (9.7) can be verified in the same manner.We will verify (9.5) in three different ways for the sake of comparison. Thefirst verification uses (9.2). For simplicity of notation, we will denote ∂

∂x, ∂∂y

,and ∂

∂zby ∂x, ∂y, ∂z, respectively. You might as well familiarize yourself with

this notational scheme at this point, unless you are already comfortable withit, because it is widely used in physics and engineering.

[Lx, Ly] = (−i~)2 [(y∂z − z∂y) (z∂x − x∂z)− (z∂x − x∂z) (y∂z − z∂y)]= (−i~)2 [y∂zz∂x − y∂zx∂z − z∂yz∂x + z∂yx∂z

− (z∂xy∂z − z∂xz∂y − x∂zy∂z + x∂zz∂y)]

= (−i~)2(y∂x + yz∂z∂x − yx∂2z − z2∂y∂x + zx∂y∂z − zy∂x∂z

+z2∂x∂y + xy∂2z − x∂y − xz∂z∂y)

= i~(−i~) (x∂y − y∂x) = i~(−i~)(x∂

∂y− y ∂

∂x

)= i~Lz (9.10)

Next, we will verify (9.5) in spherical coordinates.

[Lx, Ly] = LxLy − LyLx= i~ (sinϕ∂θ + cot θ cosϕ∂ϕ) i~ (− cosϕ∂θ + cot θ sinϕ∂ϕ)− i~ (− cosϕ∂θ + cot θ sinϕ∂ϕ) i~ (sinϕ∂θ + cot θ cosϕ∂ϕ)

= (i~)2 [sinϕ∂θ (− cosϕ∂θ) + sinϕ∂θ (cot θ sinϕ∂ϕ)+ cot θ cosϕ∂ϕ (− cosϕ∂θ) + cot θ cosϕ∂ϕ (cot θ sinϕ∂ϕ)]− (i~)2 [− cosϕ∂θ(sinϕ∂θ)− cosϕ∂θ(cot θ cosϕ∂ϕ)+ cot θ sinϕ∂ϕ(sinϕ∂θ) + cot θ sinϕ∂ϕ(cot θ cosϕ∂ϕ)]


= (i~)2(− sinϕ cosϕ∂2θ + sin2 ϕ

−1sin2 θ

∂ϕ + sin2 ϕ cot θ∂θ∂ϕ

+ cot θ cosϕ sinϕ∂θ − cot θ cos2 ϕ∂ϕ∂θ + cot2 θ cos2 ϕ∂ϕ+ cot2 θ sinϕ cosϕ∂2ϕ

)− (i~)2

(− sinϕ cosϕ∂2θ − cos2 ϕ −1sin2 θ

∂ϕ − cos2 ϕ cot θ∂θ∂ϕ

+ cot θ sinϕ cosϕ∂θ + sin2 ϕ cot θ∂ϕ∂θ − cot2 θ sin2 ϕ∂ϕ

+ cot2 θ sinϕ cosϕ∂2ϕ)

= (i~)2(

sin2 ϕ−1

sin2 θ∂ϕ + cot2 θ cos2 ϕ∂ϕ + cos2 ϕ −1sin2 θ

∂ϕ

+ cot2 θ sin2 ϕ∂ϕ)

= (i~)2( −1

sin2 θ∂ϕ + cot2 θ∂ϕ

)= i~(−i~)

(1

sin2 θ− cot2 θ

)∂ϕ

= i~(−i~)1− cos2 θsin2 θ

∂ϕ = i~(−i~) ∂∂ϕ

= i~Lz (9.11)

Finally, we will use the commutator identity (G.5).

[Lx, Ly] = [ypz − zpy, zpx − xpz]= [ypz, zpx]− [ypz, xpz]− [zpy, zpx] + [zpy, xpz]

= y[pz, z]px + z[y, px]pz + zy[pz, px] + [y, z]pzpx

− (y[pz, x]pz + x[y, pz]pz + xy[pz, pz] + [y, x]pzpz)

− (z[py, z]px + z[z, px]py + zz[py, px] + [z, z]pypx)

+ z[py, x]pz + x[z, pz]py + xz[py, pz] + [z, x]pypz

= y(−i~I)px + z · 0 · pz + zy · 0 + 0 · pzpx− (y · 0 · pz + x · 0 · pz + xy · 0 + 0 · pzpz)− (z · 0 · px + z · 0 · py + zz · 0 + 0 · pypx)+ z · 0 · pz + x(i~I)py + xz · 0 + 0 · pypz

= i~(xpy − ypx) = i~Lz (9.12)

Now, we will verify (9.8).

[L2, Lx] = [L2x + L2

y + L2z, Lx] = [L2

x, Lx] + [L2y, Lx] + [L2

z, Lx]

= 0 + [Ly, Lx]Ly + Ly[Ly, Lx] + [Lz, Lx]Lz + Lz[Lz, Lx]

= −i~LzLy − i~LyLz + i~LyLz + i~LzLy = 0 (9.13)

9.2. QUANTUM MECHANICAL ROTATION OPERATOR UR 179

Likewise for Ly and Lz.

9.2 Quantum Mechanical Rotation OperatorUR

Let R(ϕk) be a physical rotation, a counterclockwise rotation around thez-axis by ϕ, and UR be the quantum mechanical operator associated withthe physical rotation R. Then, the action of UR is characterized by

UR |x, y, z⟩ = |x cosϕ− y sinϕ, x sinϕ+ y cosϕ, z⟩ (9.14)

In order to derive UR, we will first consider an infinitesimal physical rota-tion R(εk) and its quantum mechanical counterpart URε given to first orderin ε by

URε = I + εΩ. (9.15)

As it turns out, the correct form is

URε = I − iε

~Lz; (9.16)

where Lz is the quantum mechanical angular momentum operator aroundthe z-axis (9.2). We can convince ourselves that (9.16) is indeed the desiredquantum mechanical operator as follows. Note that this is not a very rigor-ous proof, but a rough sketch of how it should be shown.

By definition, we have

URε |x, y, z⟩ = |x− yε, y + xε, z⟩ . (9.17)

It follows from (9.17) that

URε |ψ⟩ = URε

+∞$−∞

|x, y, z⟩ ⟨x, y, z|ψ⟩ dxdydz

=

+∞$−∞

URε |x, y, z⟩ ⟨x, y, z|ψ⟩ dxdydz


=

+∞$−∞

|x− yε, y + xε, z⟩ ⟨x, y, z|ψ⟩ dxdydz. (9.18)

Now, let x′ = x− yε, y′ = y + xε, and z′ = z. Then, the Jacobian is∣∣∣∣∣∣∣∣∂x′

∂x∂y′

∂x∂z′

∂x∂x′

∂y∂y′

∂y∂z′

∂y∂x′

∂z∂y′

∂z∂z′

∂z

∣∣∣∣∣∣∣∣−1

=

∣∣∣∣∣∣∣1 ε 0−ε 1 00 0 1

∣∣∣∣∣∣∣−1

=1

1 + ε2≈ 1− ε2 = 1

to first order in ε. So,

URε |ψ⟩ =+∞$

−∞

|x′, y′, z′⟩ ⟨x, y, z′|ψ⟩ 1

1 + ε2dx′dy′dz′

=

+∞$−∞

|x′, y′, z′⟩ ⟨x, y, z′|ψ⟩ dx′dy′dz′

=

+∞$−∞

|x′, y′, z′⟩ ⟨x′ + yε, y′ − xε, z′|ψ⟩ dx′dy′dz′ (9.19)

to first order in ε. Hence,

⟨x, y, z|URε |ψ⟩ = ⟨x, y, z|+∞$

−∞

|x′, y′, z′⟩ ⟨x′ + yε, y′ − xε, z′|ψ⟩ dx′dy′dz′

=

+∞$−∞

⟨x, y, z|x′, y′, z′⟩ ⟨x′ + yε, y′ − xε, z′|ψ⟩ dx′dy′dz′

=

+∞$−∞

δ(x− x′)δ(y − y′)δ(z − z′) ⟨x′ + yε, y′ − xε, z′|ψ⟩ dx′dy′dz′

= ⟨x+ yε, y − xε, z|ψ⟩ .

So, we have

=⇒ ⟨x, y, z|URε |ψ⟩ = (URε |ψ⟩) (x, y, z) = ψ(x+ yε, y − xϵ, z). (9.20)

9.2. QUANTUM MECHANICAL ROTATION OPERATOR UR 181

Expanding ψ(x+ yε, y − xε, z) in a Taylor series to order ε, we obtain

ψ(x+ yε, y − xε, z) = ψ(x, y, z) +∂ψ

∂x(yε) +

∂ψ

∂y(−xε). (9.21)

On the other hand,

⟨x, y, z|URε|ψ⟩ = ⟨x, y, z|I + εΩ|ψ⟩ = ⟨x, y, z|ψ⟩+ ε ⟨x, y, z|Ω|ψ⟩= ψ(x, y, z) + ε ⟨x, y, z|Ω|ψ⟩ . (9.22)

Therefore,

⟨x, y, z|URε |ψ⟩ = ψ(x+ yε, y − xϵ, z)

=⇒ ψ(x, y, z) + ε ⟨x, y, z|Ω|ψ⟩ = ψ(x, y, z) +∂ψ

∂x(yε) +

∂ψ

∂y(−xε)

=⇒ ε ⟨x, y, z|Ω|ψ⟩ = ε

[y∂ψ

∂x+ (−x)∂ψ

∂y

]

=⇒ (Ω |ψ⟩) (x, y, z) =[(y∂

∂x− x ∂

∂y

)ψ

](x, y, z)

=⇒ Ω = y∂

∂x− x ∂

∂y=⇒ i~Ω = i~

(y∂

∂x− x ∂

∂y

)

=⇒ −i~(x∂

∂y− y ∂

∂x

)= Lz =⇒ Ω =

Lzi~

=−i~Lz (9.23)

This verifies (9.16).

URε = I − iε

~Lz

The quantum mechanical operator UR for a finite rotation R(ϕ0k) can beconstructed from (9.16) as follows.

UR = limN→∞

(I − i

~

ϕ0

NLz

)N= exp (−iϕ0Lz/~) (9.24)

The exponential here is to be regarded as a convergent power series.

exp (−iϕ0Lz/~) =∞∑n=0

(−iϕ0/~)n

n!Lnz (9.25)


Now, from (9.3),Lz = −i~

∂

∂ϕ.

So, in spherical coordinates,

UR = exp (−iϕ0Lz/~) = exp(−iϕ0

(−i~ ∂

∂ϕ

)/~

)= exp

(−ϕ0

∂

∂ϕ

). (9.26)

9.3 Rotationally Invariant HamiltonianSuppose the only force in the system is a central force F (r), which derivesfrom a central potential V (r) where r is the distance from the origin, so that

F = −∇V (r) = −dV (r)

dr.

Then, the HamiltonianH =

p2

2m+ V (r)

is invariant under arbitrary rotations, and it commutes with Lx, Ly, Lz, andL2. From (G.19) and (G.20), we have the following.

[H,Lx] = [H,Ly] = [H,Lz] = 0 and [H,L2] = 0

Therefore, L2, Lx, Ly, and Lz are constants of motion and conserved. Weknow from Theorem 2.9 that commuting Hermitian operators Ω and Γ havecommon eigenvectors which diagonalize Ω and Γ simultaneously.

Because Lx, Ly, and Lz do not commute with each other, we cannot di-agonalize H, L2, Lx, Ly, and Lz simultaneously. The best we can do is todiagonalize H, L2, and only one of the components of L. It is customary tochoose Lz. So, we will diagonalize H, L2, and Lz simultaneously 1, whichamounts to finding all eigenfunctions ψ(r, θ, ϕ) which are, by definition,

1 We can draw a contradiction if we assume there is a common eigenstate |l⟩ for Lz

and Lx such thatLx |l⟩ = lx |l⟩ and Lz |l⟩ = lz |l⟩ .

This implies

[Lz, Lx] |l⟩ = i~Ly |l⟩ = LzLx |l⟩ − LxLz |l⟩ = lzlx |l⟩ − lxlz |l⟩ = 0.

9.4. RAISING AND LOWERING OPERATORS: L+ AND L− 183

stationary under H, L2, and Lz.

The explicit forms of the eigenfunctions will be found in Chapter 10. But,there is a way to solve the eigenvalue problem of L2 and Lz using raising andlowering operators as for the simple harmonic oscillator discussed in Section7.6.2.

9.4 Raising and Lowering Operators: L+ andL−

In this section, we will find the eigenvalues of L2 and Lz without findingthe eigenfunctions explicitly. Looking ahead, we will denote the eigenvaluesof L2 by ~2l(l + 1) and those of Lz by ml~. We will write |l,ml⟩ for theorthonormal eigenstates with associated eigenvalues ~2l(l + 1) and ml~.

L2 |l,ml⟩ = ~2l(l + 1) |l,ml⟩Lz |l,ml⟩ = ml~ |l,ml⟩ (9.27)⟨l,ml|l′m′

l⟩ = δll′δml,m′l

We will show that l is a nonnegative integer andml = −l,−l+1, . . . , 0, . . . , l−1, l. The number l is referred to as an orbital quantum number, and ml iscalled either a magnetic2 quantum number or an azimuthal3 quantum num-ber. We first define the raising and lowering operators as follows. The reasonfor the naming will become clear shortly.

This in turn implies

[Lx, Ly] |l⟩ = i~Lz |l⟩ = LxLy |l⟩ − LyLx |l⟩ = 0− Ly(lx |l⟩) = 0− 0 = 0.

We now haveLy |l⟩ = Lz |l⟩ = 0 =⇒ Lx |l⟩ = 0.

Therefore, only possible common states are those for zero angular momentum. Needlessto say, we cannot find a set of orthonormal eigenstates that simultaneously diagonalizesLx, Ly, and Lz.

2As we will see later, the energy levels among the states with different ml’s are thesame in the absence of an external magnetic field.

3It is so called as the angle θ in spherical coordinate system is called an azimuth.


L+ = Lx + iLy (9.28)L− = Lx − iLy (9.29)

Because Lx and Ly are Hermitian, it follows that

L†+ = (Lx + iLy)

† = L†x − iL†

y = Lx − iLy = L− (9.30)and

L†− = (Lx − iLy)† = L†

x + iL†y = Lx + iLy = L+. (9.31)

Incidentally, using (9.3), we can express L+ and L− in spherical coordinates.

L± = ~e±iϕ(± ∂

∂θ+ i cot θ ∂

∂ϕ

)(9.32)

As we already know L2 commutes with Lx and Ly, we have

[L2, L+] = [L2, L−] = 0. (9.33)

Let us also compute other commutators.

[L+, L−] = [Lx + iLy, Lx − iLy] = [Lx, Lx] + [Lx,−iLy]+ [iLy, Lx] + [iLy,−iLy]

= 0 + (−i)i~Lz + i(−i~Lz) + 0 = 2~Lz (9.34)

[L+, Lz] = [Lx + iLy, Lz] = [Lx, Lz] + i[Ly, Lz] = −i~Ly + i(i~Lx)

= −~Lx − ~Lx = −~(Lx + iLy) = −~L+ (9.35)

[L−, Lz] = [Lx − iLy, Lz] = [Lx, Lz]− i[Ly, Lz] = −i~Ly − i(i~Lx)= ~Lx − i~Ly = ~(Lx − iLy) = ~L− (9.36)

We also have the following.

L+L− = (Lx + iLy)(Lx − iLy) = L2x − iLxLy + iLyLx + L2

y


= L2 − L2z − i[Lx, Ly] = L2 − L2

z − i(i~Lz) = L2 − L2z + ~Lz (9.37)

L−L+ = (Lx − iLy)(Lx + iLy) = L2x + iLxLy − iLyLx + L2

y

= L2 − L2z + i[Lx, Ly] = L2 − L2

z + i(i~Lz) = L2 − L2z − ~Lz (9.38)

Because the operators Lx, Ly, and Lz are all Hermitian,

⟨l,ml|L2|l,ml⟩ = ⟨l,m|L2x + L2

y + L2z|l,ml⟩

= ⟨l,m|L2x|l,ml⟩+ ⟨l,m|L2

y|l,ml⟩+ ⟨l,m|L2z|l,ml⟩

= ⟨l,m|L†xLx|l,ml⟩+ ⟨l,m|L†

yLy|l,ml⟩+ ⟨l,m|L†zLz|l,ml⟩

= ⟨Lx(l,ml)|Lx(l,ml)⟩+ ⟨Ly(l,ml)|Ly(l,ml)⟩ (9.39)+ ⟨Lz(l,ml)|Lz(l,ml)⟩

= ∥ |Lx(l,ml)⟩ ∥2 + ∥ |Ly(l,ml)⟩ ∥2 + ∥ |Lz(l,ml)⟩ ∥2 ≥ 0

On the other hand,

⟨l,ml|L2|l,ml⟩ = ~2l(l + 1) ⟨l,ml|l,ml⟩ = ~2l(l + 1). (9.40)

Therefore,

~2l(l + 1) ≥ 0 =⇒ l ≥ 0 or l ≤ −1. (9.41)

Because the parabola y = x(x+ 1) is symmetric about the axis x = −12, we

can choose l ≥ 0 and still get all possible nonnegative values l(l + 1) can take.

Now, let

|+⟩ := L+ |l,ml⟩ ; (9.42)

where |l,ml⟩ is a normalized eigenstate of L2 with the eigenvalue ~2l(l + 1)and also a normalized eigenstate of Lz with the eigenvalue ml from (9.27).Then,

L2 |+⟩ = L2L+ |l,ml⟩ = L+L2 |l,ml⟩ = L+

(~2l(l + 1) |l,ml⟩

)= ~2l(l + 1)L+ |l,ml⟩ = ~2l(l + 1) |+⟩ . (9.43)


Hence, |+⟩ is still an eigenstate of L2 with the same eigenvalue ~2l(l + 1).On the other hand, due to (9.27) and (9.35),

Lz |+⟩ = LzL+ |l,ml⟩ = ([Lz, L+] + L+Lz) |l,ml⟩= (~L+ +ml~L+) |l,ml⟩ = (ml + 1)L+ |l,ml⟩ = (ml + 1)~ |+⟩ . (9.44)

Hence, |+⟩ is still an eigenstate of Lz but with an eigenvalue which is ~greater than the original ml~. Likewise, if we let

|−⟩ := L− |l,ml⟩ , (9.45)

we get

L2 |−⟩ = L2L− |l,ml⟩ = L−L2 |l,ml⟩ = L−

(~2l(l + 1) |l,ml⟩

)= ~2l(l + 1)L− |l,ml⟩ = ~2l(l + 1) |−⟩ (9.46)

and

Lz |−⟩ = LzL− |l,ml⟩ = ([Lz, L−] + L−Lz) |l,ml⟩= (−~L− +ml~L−) |l,ml⟩ = (ml − 1)L− |l,ml⟩

= (ml − 1)~ |−⟩ . (9.47)

This shows that |−⟩ is an eigenstate of L2 with an eigenvalue ~2l(l + 1) andis an eigenstate of Lz with an eigenvalue that is ~ smaller than the originalml~. In sum, we have shown that L+/L− raises/lowers the magnetic quan-tum number ml by one, but preserves the total angular momentum l. Inother words, L+ |l,ml⟩ is proportional to |l,ml + 1⟩, and L− |l,ml⟩ is propor-tional to |l,ml − 1⟩.

Let us now compute the square of the norm for |+⟩ and |−⟩, respectively.

∥ |+⟩ ∥2 = ⟨+|+⟩ = ⟨L+(l,ml)|L+(l,ml)⟩


= ⟨l,ml|L−L+|l,ml⟩ = ⟨l,ml|L2 − L2z − ~Lz|l,ml⟩

=(~2l(l + 1)− ~2m2

l − ~2ml

)⟨l,ml|l,ml⟩ = ~2(l(l + 1)−ml(ml + 1))

= ~2(l2 + l −m2l −ml) = ~

2((l +ml)(l −ml) + (l −ml))

= ~2(l −ml)(l +ml + 1) (9.48)

∥ |−⟩ ∥2 = ⟨−|−⟩ = ⟨L−(l,ml)|L−(l,ml)⟩= ⟨l,ml|L+L−|l,ml⟩ = ⟨l,ml|L2 − L2

z + ~Lz|l,ml⟩=(~2l(l + 1)− ~2m2

l + ~2ml

)⟨l,ml|l,ml⟩ = ~2(l(l + 1)−ml(ml − 1))

= ~2(l2 + l −m2l +ml) = ~

2((l +ml)(l −ml) + (l +ml))

= ~2(l +ml)(l −ml + 1) (9.49)

Because we have to have ∥ |+⟩ ∥2 ≥ 0 and ∥ |−⟩ ∥2 ≥ 0, from (9.48) and(9.49), we can see that

~2(l −ml)(l +ml + 1) ≥ 0 (9.50)and

~2(l +ml)(l −ml + 1) ≥ 0. (9.51)

The inequality (9.50) requires either

I. l ≥ ml and l ≥ −ml − 1

or

II. l ≤ ml and l ≤ −ml − 1;

while the other inequality (9.51)implies either

III. l ≥ −ml and l ≥ ml − 1

or

IV. l ≤ −ml and l ≤ ml − 1.

Therefore, we should have either condition I or condition II holding true, andat the same time, one of conditions III and IV to be true.

First, suppose ml ≥ 0. As specified on p.185, we chose l ≥ 0. If ml

is nonnegative, l ≤ −ml − 1 is impossible. Hence, condition I should be


satisfied. In particular, we need l ≥ ml as l ≥ −ml − 1 is automaticallysatisfied in this case. Consider conditions III and IV next. Condition IIIreduces to l ≥ ml − 1, and condition IV is impossible because l ≤ −ml andl ≤ ml − 1 implies 2l ≤ −1 which contradicts l ≥ 0. So, we have to satisfyl ≥ ml and l ≥ ml − 1 simultaneously. Therefore,

ml ≤ l if ml ≥ 0. (9.52)

Next, suppose ml < 0. Condition I reduces to l ≥ −ml−1, and conditionII is impossible as l > 0 > ml. Now, condition III reduces to l ≥ −ml asl ≥ ml − 1 is automatically satisfied, and condition IV is impossible becausel ≥ 0 > ml > ml−1. Hence, we require l ≥ −ml−1 and l ≥ −ml. Therefore,

ml ≥ −l if ml < 0. (9.53)

The inequalities (9.52) and (9.53) together with the fact that L+/L−raises/lowers the magnetic quantum number ml by one imply

ml = −l,−l + 1, . . . , l − 1,+l ; where l ≥ 0. (9.54)

As the spacing for each pair of neighboring ml values is 1, it is necessarythat l − (−l) = 2l is a nonnegative integer. But, this is possible only for anintegral or half-integral l. It turns out l is an integer for the orbital angularmomentum4.

Before closing this section, we will briefly discuss a simple implication of(9.43), (9.44), and (9.48), which will be useful later. A similar result followsfrom (9.46), (9.47), and (9.49). From (9.43), we can see that L+ |l,ml⟩ is aneigenvector of L2 with the eigenvalue ~2l(l+1). Furthermore, (9.44) indicatesthat L+ |l,ml⟩ is an eigenvector of Lz with the eigenvalue (ml + 1)~. Hence,we have

L+ |l,ml⟩ = C |l,ml + 1⟩ (9.55)

for some scalar C. However, we already know from (9.48) that

4Half-integral l’s correspond to an internal or intrinsic angular momentum called thespin of a particle. We will discuss electron spin in Chapter 11.

9.5. GENERALIZED ANGULAR MOMENTUM J 189

∥L+ |l,ml⟩ ∥2 = ⟨l,ml + 1|C∗C|l,ml + 1⟩ = |C|2

= ~2(l −ml)(l +ml + 1), (9.56)

and so,

C = eiθ~√(l −ml)(l +ml + 1). (9.57)

According to standard convention, we choose θ = 0 or eiθ = 1 [Shankar, 1980,p.336] to get

C = ~√(l −ml)(l +ml + 1) =⇒

L+ |l,ml⟩ = ~√(l −ml)(l +ml + 1) |l,ml + 1⟩ . (9.58)

Similarly, we can also show

L− |l,ml⟩ = ~√(l +ml)(l −ml + 1) |l,ml − 1⟩ . (9.59)

Finally, noting that |l,−l⟩, |l,−l + 1⟩, . . . , |l, l − 1⟩, and |l, l⟩ are the onlyeigenvectors of Lz from (9.54), we have to have

L+ |l, l⟩ = 0 (9.60)and

L− |l,−l⟩ = 0 (9.61)

as in Section 7.6.2.

9.5 Generalized Angular Momentum J

As shown on p.177, the three operators associated with the components ofan arbitrary classical angular momentum, Lx, Ly, and Lz, satisfy the com-mutation relations (9.5), (9.6), and (9.7). These relations originate from


the geometric properties of rotations in three-dimensional space. However,in quantum mechanics it is sometimes necessary to consider more abstract“angular momentum” with no counterpart in classical mechanics. What wedo is to go backwards, from the commutation relations to “rotations”, anddefine quantum mechanical angular momentum J as any set of three observ-ables/Hermitian operators Jx, Jy, and Jz which satisfies the same commuta-tion relations as Lx, Ly, and Lz. Namely,

[Jx, Jy] = i~Jz, (9.62)[Jy, Jz] = i~Jx, (9.63)[Jz, Jx] = i~Jy, (9.64)

and[J2, Jx] = [J2, Jy] = [J2, Jz] = 0; (9.65)

where (9.65) follows from (9.62), (9.63), and (9.64) as in (9.13) on p.178.Note that J2 = J2

x + J2y + J2

z (a scalar) by definition, and (9.65) implies[J2, J ] = 0. With this theoretical framework, quantum mechanical angularmomentum is based entirely on the commutation relations (9.62), (9.63), and(9.64).

If you examine the computations we conducted for L, L2, Lx, Ly, andLz, you can see that we only used the commutation relations (9.5), (9.6),and (9.7), which are equivalent to (9.62), (9.63), and (9.64). Therefore,all the results we obtained for L apply to J without modification.In particular, we can define raising and lowering operators by

J+ = Jx + iJy (9.66)and

J− = Jx − iJy, (9.67)

which satisfy

J+ |j,mj⟩ = ~√(j −mj)(j +mj + 1) |j,mj + 1⟩ (9.68)

9.5. GENERALIZED ANGULAR MOMENTUM J 191

and

J− |j,mj⟩ = ~√(j +mj)(j −mj + 1) |j,mj − 1⟩ (9.69)

as well as

J+ |j, j⟩ = 0 (9.70)and

J− |j,−j⟩ = 0. (9.71)

So, we have

−j ≤ mj ≤ +j for j ≥ 0, (9.72)J2 |j,mj⟩ = ~2j(j + 1) |j,mj⟩ , (9.73)

andJz |j,mj⟩ = mj~ |j,mj⟩ . (9.74)

Finally, recall that one immediate implication of this theory for experi-mental physics is that simultaneous measurements of two components of J isimpossible while J2 and one component of J , typically chosen to be Jz, canbe determined at the same time. We will revisit the concept of generalizedangular momentum in Chapter 11, where we discuss the electron spin.

192

Exercises1. Show that

[Lz, Lx] = i~Ly.

Chapter 10

The Hydrogen Atom

So far

1. TDSE (the Time-Dependent Schrodinger Equation)[− ~

2

2m

∂2

∂x2+ V (x, t)

]Ψ(x, t) = i~

∂Ψ(x, t)

∂t(10.1)

2. TISE (the Time-Independent Schrodinger Equation)[− ~

2

2m

d2

dx2+ V (x)

]ψ(x) = Eψ(x) (10.2)

3. The Relation between Ψ and ψ

Ψ(x, t) = ψ(x)e−iEt/~ (10.3)

Terminology

In 2 above,ψ(x) is an eigenfunction, andE is an eigenvalue.

Here is a more familiar 2-by-2 matrix example.[0 20 2

] [11

]=

[22

]= 2

[11

]

193

194 CHAPTER 10. THE HYDROGEN ATOM

[11

]is an eigenvector for

[0 20 2

]with an eigenvalue 2.

We have not really tried any real system yet. But, we need to carry outa real computation for a real system for an experimental confirmation of thecorrectness of the Schrodinger Equation

One of the simplest systems is a hydrogen atom.

However, this poses a couple of difficulties since (1) two particles areinvolved now and also since (2) it is no longer one-dimensional, but three-dimensional.

10.1 2 Particles Instead of 1Let M be the mass of the nucleus and m the mass of the electron. Thenucleus and the electron are moving about their fixed center of mass. But,introduction of the reduced mass µ given by

µ =Mm

M +m(10.4)

solves this problem. In particular, for a hydrogen atom,

µ =(

M

m+M

)m ≈ m (10.5)

and we can simply use m instead of µ.

10.2 Three-Dimensional SystemThe classical energy equation is

p2

2µ+ V = E (10.6)

10.2. THREE-DIMENSIONAL SYSTEM 195

or

1

2µ

(p2x + p2y + p2z

)+ V (x, y, z) = E. (10.7)

We want to transform this into a three-dimensional Schrodinger Equation.Our solution, the wavefunction, now depends on four variables x, y, z, andt.

Ψ(x, t) =⇒ Ψ(x, y, z, t) (10.8)

Here are the operator correspondences we are going to use.

px ⇐⇒ −i~∂

∂xpy ⇐⇒ −i~

∂

∂ypz ⇐⇒ −i~

∂

∂zE ⇐⇒ i~

∂

∂t(10.9)

Of course, this means we have the following.

p2x = −~2∂2

∂x2p2y = −~2

∂2

∂y2p2z = −~2

∂2

∂z2(10.10)

Therefore,

p2

2µ=p2x + p2y + p2z

2µ=

1

2µ

[−~2 ∂

2

∂x2+

(−~2 ∂

2

∂y2

)+

(−~2 ∂

2

∂z2

)]

=−~2

2µ

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

). (10.11)

On the other hand, the potential energy is now a function of x, y, and z asbelow.


V = V (x, y, z) =−e2

4πε0√x2 + y2 + z2

(10.12)

We now have the following Time-Dependent Schrodinger Equation inthree dimensions.

−~2

2µ

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ(x, y, z, t) + V (x, y, z)Ψ(x, y, z, t)

= i~∂Ψ(x, y, z, t)

∂t(10.13)

Let us introduce a Laplacian operator or “del squared” ∇2 defined by

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2. (10.14)

This simplifies the equation to the following form.

−~2µ∇2Ψ+ VΨ = i~

∂Ψ

∂t; (10.15)

where

V = V (x, y, z) and Ψ = Ψ(x, y, z, t). (10.16)

Now let

Ψ(x, y, z, t) = ψ(x, y, z)e−iEt/~. (10.17)

Then, we get the Time-Independent Schrodinger Equation.


−~2

2µ∇2ψ(x, y, z) + V (x, y, z)ψ(x, y, z) = Eψ(x, y, z) (10.18)

As it turns out, it is easier to solve this equation in spherical (polar) co-ordinates, where (x, y, z) are replaced by (r, θ, ϕ).1 Since a hydrogen atompossesses spherical symmetry, this is a better coordinate system. Figure 10.1shows the spherical coordinates used in physics, while Figure 10.2 is the con-vention used in mathematics. It is unfortunate that the roles of θ and ϕ areswitched between the two systems, but there is not much chance of confusiononce you become used to the spherical coordinate system. In this class, wewill use the physicists’ convention.

x

y

z

(r, θ, φ)

φ

θ

r

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

Right away, V (x, y, z) simplifies as follows.

1Note that in pure math books θ and ϕ are usually switched.


x

y

z

(r, θ, φ)

φ

θ

r

Figure 10.2: Spherical Coordinate System Used in Mathematics (Source:Wikimedia Commons; Author: Dmcq)

V (x, y, z) =−e2

4πε0√x2 + y2 + z2

=−e2

4πε0r= V (r, θ, φ) (10.19)

In fact, the potential energy only depends on the distance from the origin.

V (r, θ, φ) =−e2

4πε0r= V (r) (10.20)

Of course,

ψ(x, y, z) =⇒ ψ(r, θ, φ). (10.21)

This means that we have to convert∇2 = ∂2

∂x2+ ∂2

∂y2+ ∂2

∂z2to another equivalent

operator expressed as derivatives with respect to r, θ, and φ.Then, we will have


−~2

2µ∇2ψ(r, θ, φ) + V (r)ψ(r, θ, φ) = Eψ(r, θ, φ). (10.22)

The answer is

∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2, (10.23)

and we can see that ∇2 appears far more complicated in the spherical coor-dinates.

Here is how it is done. We begin with the three relations between (x, y, z)and (r, θ, φ).

x = r sin θ cosφy = r sin θ sinφz = r cos θ

(10.24)

We can see how ∂2

∂x2etc. can be converted. What we should do is to keep

using chain rules and relations like ∂x∂r

= sin θ cosφ. But, the actual operationis quite tedious.

Let us prove a part of this conversion.

Consider a one-variable function ψ(r) for r =√x2 + y2 + z2.

As

∂ψ

∂x=∂r

∂x

∂ψ

∂r=

x√x2 + y2 + z2

∂ψ

∂r=x

r

∂ψ

∂r(10.25)

and

∂r

∂x=

1

2

(x2 + y2 + z2

)−1/2· 2 =

x

r, (10.26)


we have

∂2ψ

∂x2=

∂

∂x

(x

r

∂ψ

∂r

)=

∂

∂x

(x · 1

r

∂ψ

∂r

)=∂x

∂x

(1

r

∂ψ

∂r

)+ x

∂

∂x

(1

r

∂ψ

∂r

)

=1

r

∂ψ

∂r+ x

∂r

∂x

∂

∂r

(1

r

∂ψ

∂r

)=

1

r

∂ψ

∂r+ x · x

r

∂

∂r

(1

r

∂ψ

∂r

)

=1

r

∂ψ

∂r+x2

r

∂

∂r

(1

r

∂ψ

∂r

). (10.27)

Similarly,

∂2ψ

∂y2=

1

r

∂ψ

∂r+y2

r

∂

∂r

(1

r

∂ψ

∂r

)(10.28)

and

∂2ψ

∂z2=

1

r

∂ψ

∂r+z2

r

∂

∂r

(1

r

∂ψ

∂r

). (10.29)

Therefore,

∇2ψ =3

r

∂ψ

∂r+x2 + y2 + z2

r

∂

∂r

(1

r

∂ψ

∂r

)=

3

r

∂ψ

∂r+ r

∂

∂r

(1

r

∂ψ

∂r

)

=3

r

∂ψ

∂r+ r

(− 1

r2∂ψ

∂r+

1

r

∂2ψ

∂r2

)=

2

r

∂ψ

∂r+∂2ψ

∂r2=

1

r2∂

∂r

(r2∂ψ

∂r

). (10.30)

In order to get other parts, try ψ = ψ(φ) and ψ = ψ(θ).

In preparation for the rest of the course as well as your career as a physicistor just as someone who needs to understand and use physics, please do getused to the spherical coordinates. For example, the volume element

dxdydz

becomesr2dr sin θdθdφ or − r2drd(cos θ)dφ


Figure 10.3: Volume Element in Spherical Coordinates (Source: Victor J.Montemayor, Middle Tennessee State University)

in spherical coordinates as shown in Figure 10.3.

Let us pause for a minute here, take a deep breath, and summarize whatwe have achieved so far to clearly understand where we are now, as well as,“hopefully”, where we are heading.

1. Due to the spherical symmetry inherent in the hydrogen atom, wemade a decision to use the spherical coordinates as opposed to thefamiliar Cartesian coordinates. In particular, this simplifies theexpression for the potential energy greatly. It is now a functiononly of the radial distance between the electron and the protonnucleus.a

V (x, y, z) = V (r) =−e2

4πε0r(10.31)

2. However, a rather heavy trade-off was the conversion of ∇2 to the


equivalent expression in the spherical coordinates. The conversionprocess is cumbersome, and the resulting expression is far lesspalatable.

∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2

(10.32)aIf the charge on the nucleus is Ze rather than just e, we have V (x, y, z) = V (r) =

−Ze2

4πε0r instead. This makes extending the results obtained for the hydrogen atom tohydrogen-like atoms quite straightforward as we will see in Section 10.8.

We are now ready to solve the Time-Independent Schrodinger Equation inspherical coordinates given by

−~2

2µ∇2ψ(r, θ, φ) + V (r)ψ(r, θ, φ) = Eψ(r, θ, φ). (10.33)

We will resort to the separation of variables technique yet one more time.So, consider ψ as a product of three single variable functions of r, θ, and φrespectively.

ψ(r, θ, φ) = R(r)Θ(θ)Φ(φ) (10.34)

Then, the Time-Independent Schrodinger Equation is

[−~2

2µ∇2

(r,θ,φ) + V (r)

]R(r)Θ(θ)Φ(φ) = ER(r)Θ(θ)Φ(φ). (10.35)

Let us write it out in its full glory and start computing!

−~2

2µ

[1

r2∂

∂r

(r2∂(RΘΦ)

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ∂(RΘΦ)

∂θ

)+

1

r2 sin2 θ

∂2(RΘΦ)

∂φ2

]

+V (r)R(r)Θ(θ)Φ(φ) = ER(r)Θ(θ)Φ(φ)


=⇒ −~2

2µ

[1

r2∂

∂r

(r2∂R

∂rΘΦ

)+

1

r2 sin θ∂

∂θ

(sin θ∂Θ

∂θRΦ

)+

1

r2 sin2 θ

∂2Φ

∂φ2RΘ

]

+V (r)RΘΦ

=−~2

2µ

[ΘΦ

1

r2∂

∂r

(r2∂R

∂r

)+

1

r2 sin θRΦ∂

∂θ

(sin θ∂Θ

∂θ

)+

1

r2 sin2 θRΘ

∂2Φ

∂φ2

]

+V (r)RΘΦ

=−~2

2µ

[ΘΦ

1

r2d

dr

(r2dR

dr

)+

1

r2 sin θRΦd

dθ

(sin θdΘ

dθ

)+

1

r2 sin2 θRΘ

d2Φ

dφ2

]

+V (r)RΘΦ = ER(r)Θ(θ)Φ(φ) (10.36)

Now, multiply through by−2µ~2· r2 sin2 θ

1

RΘΦ.

−~2

2µ· −2µ~2

r2 sin2 θ1

RΘΦ

[ΘΦ

1

r2d

dr

(r2dR

dr

)+

1

r2 sin θRΦd

dθ

(sin θdΘ

dθ

)

+1

r2 sin2 θRΘ

d2Φ

dφ2

]+ V (r)RΘΦ

−2µ~2

r2 sin2 θ1

RΘΦ

= ERΘΦ · −2µ~2

r2 sin2 θ1

RΘΦ(10.37)

=⇒ sin2 θ

R

d

dr

(r2dR

dr

)+

sin θΘ

d

dθ

(sin θdΘ

dθ

)+

1

Φ

d2Φ

dφ2+−2µ~2

r2 sin2 θV (r)

=−2µ~2

r2 sin2 θE

=⇒ 1

Φ

d2Φ

dφ2= −sin2 θ

R

d

dr

(r2dR

dr

)− sin θ

Θ

d

dθ

(sin θdΘ

dθ

)− 2µ

~2r2 sin2 θ [E − V (r)]

(10.38)


The LHS (lefthand side) is a function of φ, and the RHS (righthand side)is a function of r and θ. And the equality holds for all values of r, θ, andφ. Therefore, both sides have to equal a constant. For later convenience, wedenote this by −m2

l . So, we have

1

Φ

d2Φ

dφ2= −m2

l (10.39)

or

d2Φ

dφ2= −m2

lΦ. (10.40)

According to (10.38), this also means

−sin2 θ

R

d

dr

(r2dR

dr

)− sin θ

Θ

d

dθ

(sin θdΘ

dθ

)− 2µ

~2r2 sin2 θ [E − V (r)] = −m2

l .

(10.41)

Dividing through by sin2 θ,

− 1

R

d

dr

(r2dR

dr

)− 1

Θ sin θd

dθ

(sin θdΘ

dθ

)− 2µ

~2r2 [E − V (r)] = − m2

l

sin2 θ

=⇒ 1

R

d

dr

(r2dR

dr

)+

2µ

~2r2 [E − V (r)] =

m2l

sin2 θ− 1

Θ sin θd

dθ

(sin θdΘ

dθ

).

(10.42)

The LHS of (10.42) depends on r, and the RHS is a function of θ. As theequality holds for any values of r and θ, both sides have to be a constant.Denote that constant by l(l+1)2, for the reason that will become clear shortly,to get

2We are placing any restriction on the values l can take. Because l can be a complexnumber, any scalar can be represented as l(l+1). However, we will later show that l is infact a nonnegative integer.

10.3. THE SOLUTIONS FOR Φ 205

−1sin θ

d

dθ

(sin θdΘ

dθ

)+m2lΘ

sin2 θ= l(l + 1)Θ (10.43)

and

1

r2d

dr

(r2dR

dr

)+

2µ

~2[E − V (r)]R = l(l + 1)

R

r2. (10.44)

We have now reduced the problem to that of solving the following threeequations, each in one variable.

d2Φ

dφ2= −m2

lΦ (10.45)

−1sin θ

d

dθ

(sin θdΘ

dθ

)+m2lΘ

sin2 θ= l(l + 1)Θ (10.46)

1

r2d

dr

(r2dR

dr

)+

2µ

~2[E − V (r)]R = l(l + 1)

R

r2(10.47)

10.3 The Solutions for Φ

The easiest to solve is equation (10.45).

d2Φ

dφ2= −m2

lΦ

So,

Φ(φ) = eimlφ. (10.48)

To be precise, we have

Aeimlφ (10.49)


for an arbitrary scalar A. However, we can adjust the constant when wenormalize the total wavefunction Ψ(r, φ, θ, t). So, we have set A = 1 for now.

Consider single-valuedness.

Φ(0) = Φ(2π) =⇒ ei·ml·0 = ei·ml·2π

=⇒ 1 = eiml·2π = cos(ml2π) + i sin(ml2π) (10.50)

Now, let ml = a+ bi for (a, b ∈ R).

|1| =∣∣∣eiml2π

∣∣∣ = ∣∣∣ei(a+bi)2π∣∣∣ = ∣∣∣eia2π∣∣∣ ∣∣∣e−b2π∣∣∣ (10.51)∣∣∣e−b2π

∣∣∣ = e−b2π = 1 (10.52)

2πb = 0 =⇒ b = 0 (10.53)

It is now obvious that ml is a real number and

|ml| = 0, 1, 2, 3, 4, . . . . . . . (10.54)

For each value of ml, there is a corresponding solution

Φml(φ) = eimlφ. (10.55)

The numbers ml are known as magnetic quantum numbers.

Now we need to solve equations (10.46) and (10.47).

10.4 The Solutions for Θ

Equation (10.46) is known as the angular equation.

− 1

sin θd

dθ

(sin θdΘ

dθ

)+m2lΘ

sin2 θ= l(l + 1)Θ

Note that

d cos θ = − sin θdθ.

10.4. THE SOLUTIONS FOR Θ 207

So,

−1sin θ

d

dθ=

d

d cos θ ,

and

sin θ ddθ

= sin θ dd cos θ− sin θ

= − sin2 θd

d cos θ = −(1− cos2 θ) d

d cos θ .

This suggests that (10.46) can be simplified if we let z = cos θ. We have

d

d cos θ

(−(1− cos2 θ) dΘ

d cos θ

)+

m2lΘ

1− cos2 θ = l(l + 1)Θ,

which leads to

d

dz

(−(1− z2)dΘ

dz

)+

m2lΘ

1− z2= l(l + 1)Θ =⇒

d

dz

((1− z2)dΘ

dz

)+

(l(l + 1)− m2

l

1− z2

)Θ = 0 (10.56)

At the time Schrodinger was solving his equation for the hydrogen atom,this was already a well-known differential equation, and the solutions arecalled the associated Legendre functions denoted by Θlml

(z). This is relatedto a better known set of functions called the Legendre polynomials denotedby Pl(z). The relation between Θlml

(z) and Pl(z) is as follows.

Θlml(z) = (1− z2)|ml|/2d

|ml|Pl(z)

dz|ml|(10.57)

Our first job is to prove this.

Now Pl(z) is a solution to


(1− z2)d2Pldz2− 2z

dPldz

+ l(l + 1)Pl = 0. (10.58)

Differentiate both sides with respect to z to obtain

−2z d2Pldz2

+ (1− z2)d3Pldz3− 2

dPldz− 2z

d2Pldz2

+ l(l + 1)dPldz

= (1− z2)d3Pldz3− 4z

d2Pldz2

+

((l(l + 1)− 2)

dPldz

)= 0. (10.59)

Try ddz

again.

−2d2Pldz2− 2z

d3Pldz3− 2z

d3Pldz3

+ (1− z2)d4Pldz4− 2

d2Pldz2− 2

d2Pldz2

−2z d3Pldz3

+ l(l + 1)d2Pldz2

= (1− z2)d4Pldz4− 6z

d3Pldz3

+ (l(l + 1)− 6)d2Pldz2

= 0 (10.60)

Suppose we get

(1− z2)dk+2Pldzk+2

− 2(k + 1)zdk+1Pldzk+1

+ (l(l + 1)− k(k + 1))dk

dzkPl = 0

(10.61)

after differentiating k times. Relation (10.61) holds for k = 0, 1, and 2 ac-cording to (10.58), (10.59), and (10.60).Now apply d

dzone more time.

−2z dk+2Pldzk+2

+ (1− z2)dk+3Pldzk+3

− 2(k + 1)dk+1Pldzk+1

− 2(k + 1)zdk+2Pldzk+2

+(l(l + 1)− k(k + 1))dk+1Pldzk+1

= (1− z2)d(k+1)+2Pldz(k+1)+2

− 2((k + 1) + 1)zd(k+2)+1Pldz(k+2)+1

+(l(l + 1)− (k + 1)((k + 1) + 1)dk+1Pldzk+1

= 0 (10.62)


So, by mathematical induction,

(1− z2) dk+2

dzk+2Pl − 2(k + 1)z

dk+1Pldzk+1

+ (l(l + 1)− k(k + 1))dkPldzk

= 0

(10.63)

when we apply dk

dzk to both sides of (10.58).

On the other hand, consider

Θlml= (1− z2)|ml|/2Γ(z) (10.64)

and substitute this into

d

dz

((1− z2)dΘ

dz

)+

(l(l + 1)− m2

l

1− z2

)Θ = 0. (10.65)

We have the following lengthy computation.

d

dz

((1− z2) d

dz(1− z2)k/2Γ

)+

(l(l + 1)− k2

1− z2

)(1− z2)k/2Γ

= −2z ddz

((1− z2)k/2Γ) + (1− z2) d2

dz2((1− z2)k/2Γ)

+

(l(l + 1)− k2

1− z2

)(1− z2)k/2Γ

= −2z(k

2(1− z2)(k/2−1)(−2z)Γ + (1− z2)k/2dΓ

dz

)

+(1− z2) ddz

(k

2(1− z2)(k/2−1)(−2z)Γ + (1− z2)k/2dΓ

dz

)

+

(l(l + 1)− k2

1− z2

)(1− z2)k/2Γ

= −2z(kz(1− z2)(k/2−1)Γ + (1− z2)k/2dΓ

dz

)


+(1− z2)[k

2(k/2− 1)(1− z2)(k/2−2)(−2z)(−2z)Γ +

k

2(1− z2)(k/2−1)(−2)Γ

+k

2(1− z2)(k/2−1)(−2z)dΓ

dz+k

2(1− z2)(k/2−1)(−2z)dΓ

dz+ (1− z2)k/2d

2Γ

dz2

]

+

(l(l + 1)− k2

1− z2

)(1− z2)k/2Γ

= 2kz2(1− z2)(k/2−1)Γ− 2z(1− z2)k/2dΓdz

+(1− z2)[2kz2

(k

2− 1

)(1− z2)(k/2−2)Γ− k(1− z2)(k/2−1)Γ

−kz(1− z2)(k/2−1)dΓ

dz− kz(1− z2)(k/2−1)dΓ

dz+ (1− z2)k/2d

2Γ

dz2

]

+

(l(l + 1)− k2

1− z2

)(1− z2)k/2Γ = 0

Multiplying through by (1− z2)−k/2,

2kz2

1− z2Γ− 2z

dΓ

dz+ (1− z2)

[2kz2

(k

2− 1

)(1− z2)−2Γ− k(1− z2)−1Γ

−kz(1− z2)−1dΓ

dz− kz(1− z2)−1dΓ

dz+d2Γ

dz2

]+

(l(l + 1)− k2

1− z2

)Γ

=2kz2

1− z2Γ− 2z

dΓ

dz+ 2kz2

(k

2− 1

)(1− z2)−1Γ− kΓ− kzdΓ

dz− kzdΓ

dz

+(1− z2)d2Γ

dz2+

(l(l + 1)− k2

1− z2

)Γ

= (1− z2)d2Γ

dz2+ (−2kz − 2z)

dΓ

dz+

l(l + 1) +2kz2 + 2kz2

(k2− 1

)− k2

1− z2− k

Γ

= (1− z2)d2Γ

dz2− 2(k + 1)z

dΓ

dz+

[l(l + 1) +

k2z2 − k2

1− z2− k

]Γ


= (1− z2)d2Γ

dz2− 2(k + 1)z

dΓ

dz+[l(l + 1)− k2 − k

]Γ = 0. (10.66)

So, we have shown

(1− z2)d2Γ

dz2− 2(k + 1)z

dΓ

dz+ [l(l + 1)− k(k + 1)] Γ = 0 (10.67)

or

(1− z2)d2Γ

dz2− 2(|ml|+ 1)z

dΓ

dz+ [l(l + 1)− |ml| (|ml|+ 1)] Γ = 0. (10.68)

Comparing this with (10.61), which is given below again for reference,

(1− z2) dk+2

dzk+2Pl − 2(k + 1)z

dk+1Pldzk+1

+ (l(l + 1)− k(k + 1))dkPldzk

= 0,

we conclude

Θlml= (1− z2)|ml|/2d

|ml|Pldz|ml|

. (10.69)

It remains to solve (10.58)

(1− z2)d2Pldz2− 2z

dPldz

+ l(l + 1)Pl = 0

for Pl. Try

Pl(z) =∞∑k=0

akzk. (10.70)

We get a recursion relation as we did before on p.146 in Section 7.6.1 for thesimple harmonic oscillator.


aj+2 =j(j + 1)− l(l + 1)

(j + 2)(j + 1)aj (10.71)

Also as before, we do not want an infinite series; that is, we want the series toterminate after a finite number of terms, so that the function is integrable/normalizable. We need

j(j + 1)− l(l + 1) = j2 + j − l2 − l = (j + l)(j − l) + j − l= (j − l)(j + l + 1) = 0 =⇒ l = j or − j − 1. (10.72)

As j is a nonnegative integer, l can be any integer, positive, 0, or negative.However, the only meaningful quantity is l(l + 1). Noting that the parabola

y(l) = l(l + 1) =(l +

1

2

)2

−(1

2

)2

is symmetric about l = −12, it is sufficient to consider l = 0, 1, 2, 3, . . . . So,

l = 0, 1, 2, 3, . . . give all acceptable solutions. In fact, y(−l − 1) = (−l −1)(−l−1+1) = (−l−1)(−l) = l(l+1) proves that the pairs (0,−1), (1,−2),(2,−3), (3,−4) . . . give the same value for l(l+1), and negative l values areclearly redundant.We get

P0 = 1, P1 = z, P2 = 1− 3z2, P3 = 3z − 5z3, . . . .

So, the corresponding Θlml’s are

Θ00 = 1, Θ10 = z, Θ1±1 = (1− z2)1/2, Θ20 = 1− 3z2,

Θ2±1 = (1− z2)1/2z, Θ2±2 = 1− z2, . . .

Note that Pl(z) is an lth degree polynomial. Therefore, for each l, we have

ml = −l, − l + 1, . . . , 0, . . . , l − 1, + l. (10.73)

10.5. ASSOCIATED LEGENDRE POLYNOMIALS AND SPHERICAL HARMONICS

10.5 Associated Legendre Polynomials and Spher-ical Harmonics

Recall that Pl(z)’s = Pl(cos θ)’s are called the Legendre polynomials, andΘlml

’s are called associated Legendre functions/polynomials as explained onp.207. The associated Legendre polynomials are also denoted by Pml

l (cos θ)or simply by Pm

l (cos θ). So,

Θlml(cos θ) = Pml

l (cos θ). (10.74)

We now introduce a new set of functions known as the spherical harmonics,which are the products of Φ and Θ up to the normalization constant. It iscustomary to denote each spherical harmonic by Y ml

l or simply by Y ml . So,

we have

Y mll (θ, ϕ) ∝ Φml

(ϕ)Θlml(cos θ) = Φml

(ϕ)Pmll (cos θ) = eimlϕPml

l (cos θ).(10.75)

With the appropriate normalization constant3,

Y mll (θ, ϕ) = (−1)ml

√√√√2l + 1

4π

(l −ml)!

(l +ml)!Pmll (cos θ)eimlϕ; (10.76)

where the normalization is chosen such that∫ 2π

0

∫ π

0(Y ml

l (θ, ϕ))∗ Y mll (θ, ϕ) sin θdθdϕ

=∫ 2π

0

∫ +1

−1(Y ml

l (θ, ϕ))∗ Y mll (θ, ϕ)d(cos θ)dϕ = 1. (10.77)

In fact, Y mll ’s are orthonormal such that∫ 2π

0

∫ π

0(Y ml

l (θ, ϕ))∗ Ym′

ll′ (θ, ϕ) sin θdθdϕ

3Different definitions for spherical harmonics are used in different fields such as geodesy,magnetics, quantum mechanics, and seismology. The definition given here is commonlyadopted in the quantum mechanics community.


=∫ 2π

0

∫ +1

−1(Y ml

l (θ, ϕ))∗ Ym′

ll′ (θ, ϕ)d(cos θ)dϕ = δll′δmlm

′l. (10.78)

We will see how this comes about in Section 10.6. In passing let us makea note of the fact that the normalization constant from the ϕ-dependentportion of the integral is

√12π

, and the normalization constant from the θ-

dependent portion of the integral is√

(2l+1)(l−m)!2(l+ml)!

.

Spherical harmonics for ml < 0 has the following alternative formula

Y mll (θ, ϕ) =

√√√√2l + 1

4π

(l − |ml|)!(l + |ml|)!

P|ml|l (cos θ)eimlϕ. (10.79)

This is a direct consequence of the following identity satisfied by the associ-ated Legendre polynomials.4 We will write m instead of ml for simplicity.

P−ml (x) = (−1)m (l −m)!

(l +m)!Pml (x) (10.80)

Indeed, for m = −|m| < 0, we have

Y ml (θ, ϕ) = Y

−|m|l (θ, ϕ) = (−1)−|m|

√√√√2l + 1

4π

(l + |m|)!(l − |m|)!

P−|m|l (cos θ)e−i|m|ϕ

= (−1)−|m|

√√√√2l + 1

4π

(l + |m|)!(l − |m|)!

[(−1)|m| (l − |m|)!

(l + |m|)!P

|m|l (cos θ)

]ei(−|m|)ϕ

=

√√√√2l + 1

4π

(l − |m|)!(l + |m|)!

P|m|l (cos θ)eimϕ. (10.81)

We can combine (10.76) and (10.81) in the following manner.

Y mll (θ, ϕ) = (−1)(ml+|ml|)/2

√√√√2l + 1

4π

(l − |ml|)!(l + |ml|)!

P|ml|l (cos θ)eimlϕ (10.82)

Note that the phase factor (−1)ml or (−1)(ml+|ml|)/2 does not change thephysical observables, and it is more or less a mathematical book keepingdevice.

4See Fact K.1 of Appendix K on p.326 for a proof.

10.6. L2, LZ , AND THE SPHERICAL HARMONICS

10.6 L2, Lz, and the Spherical HarmonicsThe orthonormality relation (10.78) is a consequence of the following fact.

Fact 10.1 The spherical harmonics are simultaneous eigenstates of L2

and Lz corresponding to the eigenvalues l(l + 1)~2 and ml~.

L2Y mll (θ, ϕ) = l(l + 1)~2Y ml

l (θ, ϕ) (10.83)and

LzYmll (θ, ϕ) = ml~Y

mll (θ, ϕ) (10.84)

Because L2 and Lz are Hermitian operators, and also because Y mll ’s are

normalized, the set

Y mll

−l≤ml≤+ll=0,1,2,...

forms an orthonormal basis. Hence, Y mll ’s are not only orthonormal, but also

complete, and any function of θ and ϕ can be represented as a superposition(linear combination) of spherical harmonics.

Recall that for each of l = 0, 1, 2, 3, . . . we have ml = −l,−l + 1,−l +2, . . . 0, . . . , l− 2, l− 1, l. All the spherical harmonics for l = 0, 1, and 2 arelisted in Table 10.1.

Next on our agenda is the radial function R.

10.7 The Radial Function R

The general radial equation for a one-electron atom is

1

r2d

dr

(r2dR

dr

)+

2µ

~2

[E +

Ze2

4πε0r

]R = l(l + 1)

R

r2; (10.85)

where Ze is the charge on the nucleus. We will only solve this for Z = 1 orfor the hydrogen atom.


Spherical Harmonicsl = 0 ml = 0 Y 0

0 (θ, ϕ) =12

√1π

l = 1ml = −1 Y −1

1 (θ, ϕ) = 12

√32πe−iϕ sin θ = 1

2

√32π

x−iyr

ml = 0 Y 01 (θ, ϕ) =

12

√3π

cos θ = 12

√3πzr

ml = +1 Y 11 (θ, ϕ) =

−12

√32πeiϕ sin θ = −1

2

√32π

x+iyr

l = 2

ml = −2 Y −22 (θ, ϕ) = 1

4

√152πe−i2ϕ sin2 θ = 1

4

√152π

(x−iy)2r2

ml = −1 Y −12 (θ, ϕ) = 1

2

√152πe−iϕ sin θ cos θ = 1

2

√152π

(x−iy)zr2

ml = 0 Y 02 (θ, ϕ) =

14

√5π(3 cos2 θ − 1) = 1

4

√5π2z2−x2−y2

r2

ml = +1 Y 12 (θ, ϕ) =

−12

√152πeiϕ sin θ cos θ = −1

2

√152π

(x+iy)zr2

ml = +2 Y 22 (θ, ϕ) =

14

√152πei2ϕ sin2 θ = 1

4

√152π

(x+iy)2

r2

Table 10.1: Spherical Harmonics for l = 0, 1, and 2

Let us first invoke some substitutions to put the radial equation in a moremanageable form.

β2 =2µE

~(β > 0) (10.86)

ρ = 2βr (10.87)

γ =µe2

4πε0~2β(10.88)

With these substitutions, we get

1

ρ2d

dρ

(ρ2dR

dρ

)+

[−1

4− l(l + 1)

ρ2+γ

ρ

]R = 0. (10.89)

As ρ→∞

R(ρ) ≈ e−ρ/2. (10.90)

10.7. THE RADIAL FUNCTION R 217

So, let us write

R(ρ) = e−ρ/2F (ρ), (10.91)

and substitute this into the radial equation to get

d2F

dρ2+

(2

ρ− 1

)dF

dρ+

[γ − 1

ρ− l(l + 1)

ρ2

]F = 0. (10.92)

Once again, we will try a series solution with

F (ρ) = ρs∞∑k=0

akρk (a0 , 0, s ≥ 0); (10.93)

where the condition on s is in place to prevent F from blowing up at theorigin, while a0 , 0 assures that our solution is not trivial as we will see in(10.96).

After substituting this trial series solution into the differential equation(10.92), we get

[s(s+ 1)− l(l + 1)]a0ρs−2 +

∞∑j=0

[(s+ j + 1)(s+ j + 2)− l(l + 1)]aj+1

−(s+ j + 1− γ)ajρs+j−1 = 0. (10.94)

For the left-hand side to be zero for any value of ρ, the following two relationsshould hold.

s(s+ 1)− l(l + 1) = 0

aj+1 =s+ j + 1− γ

(s+ j + 1)(s+ j + 2)− l(l + 1)aj

(10.95)

(10.96)

The first equality (10.95) gives s = l and s = −(l + 1). But, s = −(l + 1)should be rejected as s ≥ 0. The second relation (10.96)


aj+1 =j + l + 1− γ

(j + l + 1)(j + l + 2)− l(l + 1)aj

indicates that

γ = j + l + 1 = n (Call this n.) (10.97)

assures termination of the series after a finite number of terms; namely, afterthe jth term. Hence,

n = l + 1, l + 2, l + 3, · · ·

as j goes from 0 to ∞ with l = 0, 1, 2, 3, . . ..

Now recall from (10.86) on p.216

β2 =2µE

~2

and from (10.88) on p.216

γ

=

n

=µe2

4πε0~2β

So,

En = − µe4

(4πε0)22~2n2= −

(µe4

32π2ε20~2

)1

n2= −

(~2

2µa20

)1

n2

for n = 1, 2, 3, · · · . (10.98)

Here,

10.7. THE RADIAL FUNCTION R 219

a0 =4πε0~

2

µe2(10.99)

is analogous to the Bohr radius aB defined by

aB =4πε0~

2

mee2,

where me is the rest mass of an electron.

Putting it all together, we get

ψnlml(r, θ, φ) = Rnl(r)Y

mll (θ, φ) = Rnl(r)Θlml

(θ)Φml(φ); (10.100)

where

Φml(φ) = eimlφ |ml| = 0, 1, 2, 3, · · · , (10.101)

Θlml(θ) = Θlml

(cos θ) = Pmll (cos θ) = sin|ml| θFl|ml|(cos θ), (10.102)

and

Rnl(r) = e−r/na0(r

a0

)2

Gnl

(r

a0

). (10.103)

The functions F and G are both polynomials;

Fl|ml|(cos θ)

is a polynomial in cos θ, and


Gnl

(r

a0

)

is a polynomial in r/a0.

More specifically, the normalized radial wavefunctions are5

Rnl(r) =

√√√√( 2

na0

)3 (n− l − 1)!

2n[(n+ l)!]3e−r/na0

(2r

na0

)lL2l+1n−l−1

(2r

na0

)(10.104)

and the normalized full wavefunctions are

ψnlml= Rnl(r)Y

mll (θ, ϕ)

=

√√√√( 2

na0

)3 (n− l − 1)!

2n[(n+ l)!]3e−r/na0

(2r

na0

)lL2l+1n−l−1

(2r

na0

)×

√√√√2l + 1

4π

(l −ml)!

(l +ml)!Pmll (cos θ)eimlϕ

. (10.105)

Let us now summarize the relations among n, l, andml. We have obtained

|ml| = 0, 1, 2, 3, · · · , (10.106)l = |ml| , |ml|+ 1, · · · , |ml|+ 3, · · · , (10.107)

andn = l + 1, l + 2, l + 3, · · · (10.108)

in this order, but this can be reorganized as follows.

n = 1, 2, 3, 4, · · · (10.109)l = 0, 1, 2, · · · , n− 1 (10.110)

ml = −l,−l + 1, · · · , 0, · · · ,+l (10.111)

5L in (10.104) is an associated Laguerre polynomial described in Appendix L.

10.8. HYDROGEN-LIKE ATOMS 221

The first number n is associated with R(r) and is called the principal quan-tum number. The second number l is associated with Θ(θ) and is calledthe orbital quantum number or the angular momentum quantum number.Finally, the number ml is called the magnetic quantum number which isassociated with Φ(φ). In addition to these, we also have a spin quantumnumber ms, which we will discuss in Chapter 11.

10.8 Hydrogen-Like AtomsA hydrogen-like atom, or hydrogen-like ion, is an atomic nucleus with oneelectron. Some examples other than the hydrogen atom itself are He+, Li2+,Be3+, and B4+. If we denote the atomic number by Z, the charge carriedby hydrogen-like ions are e(Z − 1). With this notation, we can derive allthe wavefunctions for the hydrogen-like ions from the wavefunctions for thehydrogen atom by replacing the charge e on proton by Ze. You can see thisfrom the time-independent Hamiltonian.

−~2

2µ

[1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)

+1

r2 sin2 θ

∂2

∂φ2

]− Ze2

4πε0r

ψ(r, θ, φ) = Eψ(r, θ, φ)

(10.112)

The only difference is Z appearing once in the potential term. The resultingwavefunctions are

ψnlml,Z = Rnl,Z(r)Ymll (θ, ϕ)

=

√√√√( 2Z

na0

)3 (n− l − 1)!

2n[(n+ l)!]3e−Zr/na0

(2Zr

na0

)lL2l+1n−l−1

(2Zr

na0

)×

√√√√2l + 1

4π

(l −ml)!

(l +ml)!Pmll (cos θ)eimlϕ

. (10.113)

And, the total energy EZ is given by


EZ,n = − Z2µe4

(4πε0)22~2n2= −

(Z2µe4

32π2ε20~2

)1

n2= −

(Z2~2

2µa20

)1

n2

for n = 1, 2, 3, · · · . (10.114)

10.9 Simultaneous Diagonalization of H, L2,and Lz

Let us examine the big picture now. In Section 10.6, we saw that the sphericalharmonics

Y mll

−l≤ml≤+ll=0,1,2,...

form an orthonormal basis that simultaneously diagonalizes L2 and Lz. Like-wise, the set of normalized eigenfunctions or eigenkets

ψnlml or |n, l,ml⟩ (10.115)

for n = 1, 2, 3, . . ., l = 0, 1, . . . , n−1, and ml = −l,−l+1, . . . 0 . . . , l−1,+lform an orthonormal basis that simultaneously diagonalizes the HamiltonianH, the magnitude of the orbital angular momentum squared L2, and thez-component of the orbital angular momentum Lz. We have the followingset of simultaneous eigenvalue problems with a common solution |n, l,ml⟩.

H |ψ⟩ = En |ψ⟩L2 |ψ⟩ = l(l + 1)~2 |ψ⟩ (10.116)Lz |ψ⟩ = ml~ |ψ⟩

The reason why such common eigenkets can be found is that the commutators[H,L2] and [H,Lz] are both zero in addition to [L2, Lz] = 0 which we knowfrom (9.8). As you can see in Appendix J, there are alternative forms of ∇2.The form given in (10.23) is

∇2 =1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2,

10.9. SIMULTANEOUS DIAGONALIZATION OF H, L2, AND LZ 223

and one alternative form is

∇2 =∂2

∂r2+

2

r

∂

∂r+

1

r2∂2

∂θ2+

cos θr2 sin θ

∂

∂θ+

1

r2 sin2 θ

∂2

∂ϕ2(10.117)

or

∇2 =∂2

∂r2+

2

r

∂

∂r+

1

r2

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

]. (10.118)

From (9.4) and (10.118), we have

∇2 =∂2

∂r2+

2

r

∂

∂r+

1

r2

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

]

=∂2

∂r2+

2

r

∂

∂r− 1

~2r2(−~2)

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

]

=∂2

∂r2+

2

r

∂

∂r− L2

~2r2. (10.119)

We will use this form of ∇2 in order to show [H,L2] = 0 and [H,Lz] = 0.6

Fact 10.2 ([H, L2] = 0) The full Hamiltonian

H =−~2

2µ∇2 + V (r) =

−~2

2µ

[∂2

∂r2+

2

r

∂

∂r− L2

~2r2

]+ V (r) (10.120)

commutes with the squared magnitude of the angular momentum

L2 = −~2[

1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

]. (10.121)

ProofThe important point here is that the expression of H contains only the radialvariable r and L2, while the expression of L2 contains the azimuthal andpolar variables ϕ and θ. With this in mind, the commutator [H,L2] can becomputed as follows.

[H,L2

]=

[−~2

2µ

[∂2

∂r2+

2

r

∂

∂r− L2

~2r2

]+ V (r), L2

]6Both [H,L2] = 0 and [H,Lz] = 0 are proved in a different manner in Appendix G.


=

[−~2

2µ

[∂2

∂r2+

2

r

∂

∂r− L2

~2r2

], L2

]+[V (r), L2

]=

[−~2

2µ

[∂2

∂r2+

2

r

∂

∂r

], L2

]+

[L2

2µr2, L2

]+[V (r), L2

]. (10.122)

It is now clear that each term of (10.122) is zero, and we have verified[H,L2

]= 0. (10.123)

Fact 10.3 ([H, Lz] = 0) The full Hamiltonian

H =−~2

2µ∇2 + V (r) =

−~2

2µ

[∂2

∂r2+

2

r

∂

∂r− L2

~2r2

]+ V (r) (10.124)

commutes with the z-component of the angular momentum

Lz = −i~∂

∂ϕ. (10.125)

Proof

[H,Lz] =

[−~2

2µ

[∂2

∂r2+

2

r

∂

∂r− L2

~2r2

]+ V (r), Lz

]

=

[−~2

2µ

[∂2

∂r2+

2

r

∂

∂r

], Lz

]+

[L2

2µr2, Lz

]+ [V (r), Lz] = 0.

10.10 Revisiting the Fundamental PostulatesLet us recall the Fundamental Postulates presented in Section 3.1. Thefollowing is the set of guiding principles that connects the physical realitywith mathematical abstraction.

10.10. REVISITING THE FUNDAMENTAL POSTULATES 225

• Each physical system S has an associated Hilbert space H.

• Each physical state s of the physical system S has an associated nor-malized ket |s⟩, called a state ket, in H.

• Every physical observable q is represented by a Hermitian Operator Qwhose domain is H.

• Measurement of q in the physical system S is mirrored in the abstractquantum mechanical system by the action of the corresponding opera-tor Q on the state ket |s⟩.

• The only possible outcomes of quantum mechanical measurements arethe eigenvalues of the corresponding operator Q.

• A state ket is initially a linear combination of normalized eigenkets ofQ. However, after the measurement, the state ket becomes the eigenket associated with the measured eigenvalue.

You can note here that the quantum mechanical system is composed of thethreesome (H, |s⟩, Q), which are purely mathematical and abstract con-structs. In particular, no reference is made to any coordinate system or afunction defined relative to such coordinates.

Therefore, the generic simultaneous eigenvalue problems (10.116) and theirsolutions (10.115) are actually coordinate free. In other words, the eigenkets|n, l,ml⟩ “live” in some abstract unspecified Hilbert spaceH. From Section4.6, we can expand |n, l,ml⟩ in the coordinate basis |r, θ, ϕ⟩ to get

|n, l,ml⟩ = I |n, l,ml⟩ =

∑0≤r<+∞

∑0≤θ≤π

∑0≤ϕ<2π

|r, θ, ϕ⟩ ⟨r, θ, ϕ|

|n, l,ml⟩

=∑

0≤r<+∞

∑0≤θ≤π

∑0≤ϕ<2π

⟨r, θ, ϕ|n, l,ml⟩ |r, θ, ϕ⟩ (10.126)

=∑

0≤r<+∞

∑0≤θ≤π

∑0≤ϕ<2π

ψnlml(r, θ, ϕ) |r, θ, ϕ⟩ (10.127)

or more appropriately

|n, l,ml⟩ =

∞∫0

π∫0

2π∫0

|r, θ, ϕ⟩ ⟨r, θ, ϕ| r2dr sin θdθdϕ |n, l,ml⟩


=

∞∫0

π∫0

2π∫0

⟨r, θ, ϕ|n, l,ml⟩ |r, θ, ϕ⟩ r2dr sin θdθdϕ (10.128)

=

∞∫0

π∫0

2π∫0

ψnlml(r, θ, ϕ) |r, θ, ϕ⟩ r2dr sin θdθdϕ; (10.129)

where we used the identities∑

0≤r<+∞

∑0≤θ≤π

∑0≤ϕ<2π

|r, θ, ϕ⟩ ⟨r, θ, ϕ| = I

and∞∫0

π∫0

2π∫0

|r, θ, ϕ⟩ ⟨r, θ, ϕ| r2dr sin θdθdϕ = I.

When we write

ψnlml(r, θ, ϕ) = ⟨r, θ, ϕ|n, l,ml⟩ (10.130)

for the coefficients in (10.126) and (10.128), we are taking the projectionof |n, l,ml⟩ onto the infinite-dimensional coordinate system whose axes aredefined by |r, θ, ϕ⟩, to use a geometric language. In a more common lan-guage, this amounts to solving the set of eigenvalue problems (10.116) usingthe spherical coordinate system (r, θ, ϕ) in order to obtain the eigenvectors asfunctions of r, θ, and ϕ. The operators H, L2, and Lz are also abstract andcoordinate-independent generic objects whose representations in the spheri-cal coordinate system are

H =−~2

2µ

[1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2

]+ V (r),

(10.131)

L2 = −~2[

1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂ϕ2

], (10.132)

and

Lz = −i~∂

∂ϕ. (10.133)

10.10. REVISITING THE FUNDAMENTAL POSTULATES 227

Similarly, we can project |n, l,ml⟩ onto the Cartesian system defined by thefamiliar |x, y, z⟩.7

ψnlml(x, y, z) = ⟨x, y, z|n, l,ml⟩ (10.134)

In this case, we already know

H =−~2

2µ

[∂2

∂x2+

∂2

∂y2+

∂2

∂z2

]+ V (x, y, z) (10.135)

and

Lz = −i~(x∂

∂y− y ∂

∂x

). 8 (10.136)

7This is not to be confused with the usual projection onto the x-, y-, and z-axes ofa vector v, which allows us to write v = (vx, vy, vz). Our coordinate system here is notthe usual three-dimensional system, but it is of infinite dimensions, each axis of which isdefined by one of |x, y, z⟩; where −∞ < x, y, z < +∞.

8As we know Lx = −i~(y ∂

∂z − z∂

∂y

)and Ly = −i~

(z ∂

∂x − x∂

∂z

)as well as L2 =

L2x + L2

y + L2z, we can also express L2 in terms of x, y, z and ∂

∂x ,∂

∂y ,∂

∂z . However, theresulting messy formula is not of much instructional value or other use to us. That is thereason why L2 is not included in this list.

228

Exercises1. Consider a one-variable function ψ(r) for r =

√x2 + y2 + z2, and de-

rive the relation below.

∂2ψ

∂x2=

1

r

∂ψ

∂r+x2

r

∂

∂r

(1

r

∂ψ

∂r

)

2. Without showing detailed computations, simply sketch the proof that

Θlml= (1− z2)|ml|/2d

|ml|Pldz|ml|

.

3. Hydrogen, deuterium, and singly ionized helium are all examples ofone-electron atoms. The deuterium nucleus has the same charge as thehydrogen nucleus, and almost exactly twice the mass. The helium nu-cleus has twice the charge of the hydrogen nucleus, and almost exactlyfour times the mass. Make an approximate prediction of the ratios ofthe ground state energies of these atoms.

4. Consider n = 3 state of a hydrogen atom.

(a) What l values are possible?(b) For each value of l, what ml values are possible?(c) Finally, how many degenerate states do we have for n = 3?

5. What is the energy of a photon emitted when the electron drops fromthe 3rd highest energy level (n = 3) to the ground state (n = 1)? Leaveµ, e, π, ε0, and ~ as they are.

6. Answer the following questions about the hydrogen atom.

(a) The differential equation of the ϕ-dependent function Φ(ϕ) is givenby d2Φ

dϕ2= −k2Φ with a solution Φ(ϕ) = eikϕ. Find all values

of k = a + bi (a, b real) consistent with the single-valuednesscondition. Show all your work.

(b) For each principal quantum number n, there are n2 degenerateenergy levels. Prove this.

7. Answer the following questions about the hydrogen atom.

229

(a) Consider n = 4.i. What is the largest allowed value of l?ii. What is the magnitude of the corresponding angular momen-

tum? Leave ~ as it is.iii. How many different z-components may this angular momen-

tum vector have?iv. What is the magnitude of the largest z component? Leave ~

as it is.(b) Consider an electron in the ground state of the hydrogen atom

characterized by n = 0. The normalized ground state wavefunc-tion is given by

ψ000 =1√π

(1

a0

)3/2

e−r/a0 .

You may use∫x2ebx dx = ebx

(x2

b− 2x

b2+

2

b3

)+ C

and∫x3ebx dx = ebx

(x3

b− 3x2

b2+

6x

b3− 6

b4

)+ C.

i. What is the probability P (0 ≤ r < a0) that the electron liesinside a sphere of radius a0 centered at the origin? Leave e asit is.

ii. Find the average distance < r > of the electron from thenucleus.

(c) For a hydrogen-like atom with Z = 3, what is the energy of a pho-ton emitted when the electron drops from the 3rd highest energylevel (n = 3) to the ground state (n = 1)? Leave µ, e, π, ε0, and~ as they are. Do not use a0 in your answer.

Chapter 11

Electron Spin

There are quantum mechanical phenomena that do not fit in the frameworkbased on the postulates given in Section 3.1 in a straightforward manner.While the postulates draw heavily on the analogy based on the correspondingclassical mechanical system, these phenomena have no classical analogues.They require an additional variable called intrinsic spin and a special Hilbertspace on which the spin operators act. The intrinsic spin S is a very im-portant signature for many particles. We will devote this chapter to thediscussion of electron spin as it is the most representative example. Intrinsicspin can be characterized and understood best if we regard it as another kindof angular momentum that requires a separate treatment different from thatfor the familiar orbital angular momentum represented by Lx, Ly, and Lz.

11.1 What is the Electron Spin?The electron was discovered in 1897 by Joseph John Thompson. It wasthe first subatomic and fundamental particle discovered. Physicists knewfrom the beginning that an electron had charge and mass. But, the conceptof electron spin was proposed by Samuel Abraham Goudsmit and GeorgeUhlenbeck much later in 1925. Because the electron was usually regardedas a point particle with no internal structure and no physical dimension, itwas actually impossible for the electron to spin like a top. Therefore, it wasnecessary to assume that the electron was born with intrinsic angular mo-mentum, not associated with its orbital motion, called electron spin. Thefirst sign of electron spin came from an experiment conducted by Otto Stern

231

232 CHAPTER 11. ELECTRON SPIN

and Walther Gerlach in 1921.

11.1.1 Stern-Gerlach ExperimentIn this experiment, electrically neutral silver atoms passed through a nonuni-form magnetic field B = Bz, oriented along the z-axis, which deflected theatoms. The atoms, after passing through the magnetic field, hit a photo-graphic plate generating small visible dots. The deflection is along the z-axisand the magnitude of the force on the atoms is proportional to dB

dz. In the

classical picture, the orbiting electron produces magnetic dipole momentswhose magnitudes are continuously distributed. If this is the case, the ex-tent of vertical deflection should also be continuously distributed forming a“blob” on the plate. On the other hand, in the quantum mechanical picture,there are 2l + 1 z-components, ml’s, for l = 1, 2, 3, . . .. In particular, notethat ml = 0 is always possible. When ml = 0, there is no force on the atomand the atoms with ml = 0 will not be deflected. Hence, if l = 0, no atomwill be deflected.

Figure 11.1: Stern-Gerlach Experiment (Diagram drawn by en wikipediaTheresa Knott.)

The results of Stern-Gerlach experiment presented in their original paper[Gerlach and Stern, 1922, p.350] are reproduced in Figure 11.2. The pictureon the left shows only one line as there was no external magnetic field. Whenan external non-uniform magnetic field was turned on, there were two lines

11.1. WHAT IS THE ELECTRON SPIN? 233

shown on the right due to the force proportional to dBdz

. However, this wasinconsistent with their expectations in two ways.

1. There was no line at the center corresponding to ml = 0.

2. There should be an odd number of lines; namely 2l+1. But, theyobserved two lines.

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

Later in 1927, T. E. Phipps and J. B. Taylor reproduced this effect usinghydrogen atoms in the ground state (l = 0 and ml = 0). This was a con-vincing piece of evidence that there was “something” other than the orbitalangular momentum because l = 0 implies the orbital angular momentum is0.

11.1.2 Fine Structure of the Hydrogen Spectrum: Spin-Orbit Interaction

Generally speaking, the term fine structure refers to the splitting of spectrallines due to first order relativistic effects. These relativistic effects add three


fine structure terms to the Hamiltonian known as kinetic, spin-orbit interac-tion, and Darwin terms. And it is the spin-orbit splitting that provided oneof the first pieces of evidence for electron spin.

11.2 Spin and Pauli MatricesOne way to account for these apparent discrepancies is to assume the exis-tence of a fourth quantum number known as a spin quantum number denotedby s.1 We can easily see that if we assume the magnitude of the spin S isgiven by S = ~

√s(s+ 1) and s = 1/2, the number of values its z-component

Sz = ms~ can take is 2s+ 1 = 2, and we can successfully explain the two-lines observed. In a nutshell, this is what we are going to do. While we willspecialize to the electron shortly, spin itself is an intrinsic property possessedby any fundamental particle. Every fundamental particle has its own uniquesignature value of s that characterizes the particle. While ms varies between−s and +s, s itself does not.

As mentioned on p.231, spin is best understood as intrinsic angular mo-mentum. According to Section 9.5, this means that the three componentsof the spin satisfy the same commutation relations as Lx, Ly, and Lz; ortheir generalized counterparts Jx, Jy, and Jz. Because spin is a generalizedangular momentum, we can understand spin if we understand the quantummechanical properties of angular momentum. Here are the basic propertiesof spin as a quantum mechanical angular momentum.

S = Sxx + Syy + Szz (11.1)S2 = S2

x + S2y + S2

z (11.2)[Sx, Sy] = i~Sz (11.3)[Sy, Sz] = i~Sx (11.4)

1In the framework of our quantum theory based on the Schrodinger equation, spin is ana posteriori consideration. However, spin appears naturally when the relativistic analogof the Schrodinger equation, known as the Dirac equation, is used. Spin is a consequenceof demanding that the Dirac equation be first order in time and space.

11.2. SPIN AND PAULI MATRICES 235

[Sz, Sx] = i~Sy (11.5)[S2, Sx] = [S2, Sy] = [S2, Sz] = [S2, S] = 0 (11.6)

From these properties and the results of actual measurements, we can con-struct the following framework to describe and understand the electron spin.We will find an orthonormal basis that diagonalizes S2 and Sz simultaneouslyand find (non-diagonal) matrix representations for Sx and Sy in that basis.

First, based on the outcomes of various experiments, we know s = 1/2 andms = ±1/2. This means S2 = s(s+ 1)~2I = 1

2

(12+ 1

)~2I = 3

4~2I, where I

is the identity operator, and the eigenvalues of Sz are ± ~2. Now, due to

Properties 2 and 3 on p.50 as well as Theorem 2.7 on p.52, two normalizedeigenvectors of Sz form an orthonormal basis for the corresponding Hilbertspace. We will denote a normalized eigenvector for the eigenvalue + ~

2by |↑⟩

and that for the eigenvalue − ~2

by |↓⟩. Using the notation of Section 9.5 ongeneralized angular momentum, |j,mj⟩, we have

|↑⟩ = |s = 1/2,ms = 1/2⟩ = |1/2, 1/2⟩ (11.7)and

|↓⟩ = |1/2,−1/2⟩ . (11.8)

Because our Hilbert space is two-dimensional, all operators can be repre-sented by 2-by-2 matrices. Let us compute the matrix elements for S2, Sz,Sx, and Sy.

Let us arrange |↑⟩ and |↓⟩ in this order, so that

|↑⟩ =[10

](11.9)

and

|↓⟩ =[01

], (11.10)

and express the i-th row and j-th column component of the (2-by-2) matrix


representation of an operator Ω by Ωij. For S2,

S211 = ⟨↑ |S2| ↑⟩ =

⟨↑∣∣∣∣34~2I

∣∣∣∣ ↑⟩ =3

4~2 ⟨↑ | ↑⟩ = 3

4~2

S212 = ⟨↑ |S2| ↓⟩ =

⟨↑∣∣∣∣34~2I

∣∣∣∣ ↓⟩ =3

4~2 ⟨↑ | ↓⟩ = 0

S221 = ⟨↓ |S2| ↑⟩ =

⟨↓∣∣∣∣34~2I

∣∣∣∣ ↑⟩ =3

4~2 ⟨↓ | ↑⟩ = 0

S222 = ⟨↓ |S2| ↓⟩ =

⟨↓∣∣∣∣34~2I

∣∣∣∣ ↓⟩ =3

4~2 ⟨↓ | ↓⟩ = 3

4~2

⇓

S2 =

[34~2 00 3

4~2

]=

3

4~2[1 00 1

]. (11.11)

For Sz,

Sz11 = ⟨↑ |Sz| ↑⟩ =⟨↑∣∣∣∣+~2

∣∣∣∣ ↑⟩ = +~

2⟨↑ | ↑⟩ = +

~

2

Sz12 = ⟨↑ |Sz| ↓⟩ =⟨↑∣∣∣∣−~2

∣∣∣∣ ↓⟩ = −~2⟨↑ | ↓⟩ = 0

Sz21 = ⟨↓ |Sz| ↑⟩ =⟨↓∣∣∣∣+~2

∣∣∣∣ ↑⟩ = +~

2⟨↓ | ↑⟩ = 0

Sz22 = ⟨↓ |Sz| ↓⟩ =⟨↓∣∣∣∣−~2

∣∣∣∣ ↓⟩ = −~2⟨↓ | ↓⟩ = −~

2

⇓

Sz =

[+ ~

20

0 − ~2

]=~

2

[1 00 −1

]. (11.12)

In order to compute the matrix representations of Sx and Sy, it is conve-nient to use raising and lowering operators S+ = Sx + iSy and S− = Sx − iSy.Solving for Sx and Sy, we get


Sx =1

2(S+ + S−) (11.13)

and

Sy =1

2i(S+ − S−). (11.14)

From Section 9.5, we have

S+ |s,ms⟩ = ~√(s−ms)(s+ms + 1) |s,ms + 1⟩ (11.15)

and

S− |s,ms⟩ = ~√(s+ms)(s−ms + 1) |s,ms − 1⟩ .a (11.16)

aRecall that these relations hold for any generalized momentum, including, ofcourse, electron spin. All we require are the commutation relations

[Jx, Jy] = i~Jz,

[Jy, Jz] = i~Jx,

and[Jz, Jx] = i~Jy;

where J is S for us.

These relations give the following.

S+ |↑⟩ = S+ |s = 1/2,ms = 1/2⟩ = 0 (11.17)

S+ |↓⟩ = S+ |1/2,−1/2⟩ = ~√(

1

2−(−1

2

))(1

2+(−1

2

)+ 1

)|1/2, 1/2⟩

= ~ |1/2, 1/2⟩ = ~ |↑⟩ (11.18)

S− |↑⟩ = S− |1/2, 1/2⟩ = ~√(

1

2+

1

2

)(1

2− 1

2+ 1

)|1/2,−1/2⟩


= ~ |1/2,−1/2⟩ = ~ |↓⟩ (11.19)S− |↓⟩ = S− |1/2,−1/2⟩ = 0 (11.20)

For Sx = 12(S+ + S−),

Sx11 = ⟨↑ |Sx| ↑⟩ =⟨↑∣∣∣∣12 (S+ + S−)

∣∣∣∣ ↑⟩ =1

2(⟨↑ |S+| ↑⟩+ ⟨↑ |S−| ↑⟩)

=1

2(0 + ⟨↑ |~| ↓⟩) = 1

2(0 + 0) = 0

Sx12 = ⟨↑ |Sx| ↓⟩ =⟨↑∣∣∣∣12 (S+ + S−)

∣∣∣∣ ↓⟩ =1

2(⟨↑ |S+| ↓⟩+ ⟨↑ |S−| ↓⟩)

=1

2(⟨↑ |~| ↑⟩+ 0) =

~

2

Sx21 = ⟨↓ |Sx| ↑⟩ =⟨↓∣∣∣∣12 (S+ + S−)

∣∣∣∣ ↑⟩ =1

2(⟨↓ |S+| ↑⟩+ ⟨↓ |S−| ↑⟩)

=1

2(0 + ⟨↓ |~| ↓⟩) = ~

2

Sx22 = ⟨↓ |Sx| ↓⟩ =⟨↓∣∣∣∣12 (S+ + S−)

∣∣∣∣ ↓⟩ =1

2(⟨↓ |S+| ↓⟩+ ⟨↓ |S−| ↓⟩)

=1

2(⟨↓ |~| ↑⟩+ 0) =

1

2(0 + 0) = 0

⇓

Sx =

[0 ~

2~2

0

]= ~

2

[0 11 0

]. (11.21)

For Sy = 12i(S+ − S−),

Sy11 = ⟨↑ |Sy| ↑⟩ =⟨↑∣∣∣∣ 12i (S+ − S−)

∣∣∣∣ ↑⟩ =1

2i(⟨↑ |S+| ↑⟩ − ⟨↑ |S−| ↑⟩)

=1

2i(0− ⟨↑ |~| ↓⟩) = 1

2i(0− 0) = 0

Sy12 = ⟨↑ |Sy| ↓⟩ =⟨↑∣∣∣∣ 12i (S+ − S−)

∣∣∣∣ ↓⟩ =1

2i(⟨↑ |S+| ↓⟩ − ⟨↑ |S−| ↓⟩)

=1

2i(⟨↑ |~| ↑⟩+ 0) =

~

2i= −i~

2


Sy21 = ⟨↓ |Sy| ↑⟩ =⟨↓∣∣∣∣ 12i (S+ − S−)

∣∣∣∣ ↑⟩ =1

2i(⟨↓ |S+| ↑⟩ − ⟨↓ |S−| ↑⟩)

=1

2i(0− ⟨↓ |~| ↓⟩) = − ~

2i= i~

2

Sy22 = ⟨↓ |Sy| ↓⟩ =⟨↓∣∣∣∣ 12i (S+ − S−)

∣∣∣∣ ↓⟩ =1

2i(⟨↓ |S+| ↓⟩ − ⟨↓ |S−| ↓⟩)

=1

2i(⟨↓ |~| ↑⟩+ 0) =

1

2i(0 + 0) = 0

⇓

Sy =

[0 −i ~

2

i ~2

0

]= ~

2

[0 −ii 0

]. (11.22)

Definition 11.1 (Pauli Matrices) The Pauli matrices σ is defined by

S =~

2σ; (11.23)

where S = (Sx, Sy, Sz) and σ = (σx, σy, σz). Component by component, wehave the following matrix representations of σi’s in the orthonormal basisconsisting of normalized eigenvectors of Sz.

σx =

[0 11 0

], σy =

[0 −ii 0

], and σz =

[1 00 −1

]. (11.24)

In addition, we often define σ0 as follows; namely, σ0 is the identity matrix.

σ0 =

[1 00 1

](11.25)

Noting that

iσy =

[0 1−1 0

],


we can see

1

2(σx + iσy) =

1

2

([0 11 0

]+

[0 1−1 0

])=

[0 10 0

],

Similarly,

1

2(σx − iσy) =

1

2

([0 11 0

]−[

0 1−1 0

])=

1

2

[0 02 0

]=

[0 01 0

].

1

2(σ0 + σz) =

1

2

([1 00 1

]+

[1 00 −1

])=

[1 00 0

],

and1

2(σ0 − σz) =

1

2

([1 00 1

]−[1 00 −1

])=

[0 00 1

].

Therefore, σ0, σx, σy, σz forms a basis for the vector space spanned by

[1 00 0

],

[0 10 0

],

[0 01 0

], and

[0 00 1

].

Properties of Pauli Matrices

1. |↑⟩ and |↓⟩ are normalized eigenvectors of σz, and also Sz. Similarly,|↑⟩x and |↓⟩x are normalized eigenvectors of σx, and |↑⟩y and |↓⟩y arenormalized eigenvectors of σy. Here, |↑⟩x and |↓⟩x are the states corre-sponding to spin up and spin down along the x-axis; and likewise for|↑⟩y and |↓⟩y.2

2. Pauli spin matrices are Hermitian. For example,

σ†y =

[0 −ii 0

]†

=

[0 −ii 0

]= σy, (11.26)

and similarly for σx and σz.2The eigenvalues of Sx, Sy, and Sz are ± ~2 ; while the eigenvalues are ±1 for the rescaled

σx, σy, and σz.

11.3. SEQUENTIAL MEASUREMENTS 241

3. Pauli matrices are unitary. For example,

σ†yσy =

[0 −ii 0

]† [0 −ii 0

]=

[0 −ii 0

] [0 −ii 0

]

=

[(−i)i 0

0 i(−i)

]=

[1 00 1

]. (11.27)

Likewise for σx and σz.

4. Pauli matrices anti-commute; i.e. the anti-commutator

[σi, σj]+ = σiσj + σjσi = 0 if i , j and i, j , 0.3 (11.28)

5. We have the following cyclic relations.

σxσy = iσz, σyσz = iσx, and σzσx = iσy. (11.29)

6. Pauli matrices are traceless.

Tr σx = Tr σy = Tr σz = 0 (11.30)

7. Pauli matrices are idempotent, which is a direct consequence of Prop-erties 2 and 3.

σ2i = I. (11.31)

11.3 Sequential MeasurementsWe will now consider consecutive Stern-Gerlach type experiments. This willdemonstrate how Postulates 3, 5, and 6 from Chapter 3 work.

We need to do some preliminary work first. In the Sz basis |↑⟩ , |↓⟩, wehave

σx|↑⟩+ |↓⟩√

2=

1√2

[0 11 0

]([10

]+

[01

])=

1√2

[0 11 0

] [11

]3Some authors use for the anti-commutator. However, is used for the Poisson

bracket in classical mechanics. So, we will use [ ]+ instead.


=1√2

[11

]= 1 · 1√

2

([10

]+

[01

])= 1 · |↑⟩+ |↓⟩√

2(11.32)

and

σx|↑⟩ − |↓⟩√

2=

1√2

[0 11 0

]([10

]−[01

])=

1√2

[0 11 0

] [1−1

]

=1√2

[−11

]= −1 · 1√

2

([10

]−[01

])= −1 · |↑⟩ − |↓⟩√

2. (11.33)

Therefore,

|↑⟩x =1√2(|↑⟩+ |↓⟩) (11.34)

and

|↓⟩x =1√2(|↑⟩ − |↓⟩). (11.35)

Similarly,

σy|↑⟩+ i |↓⟩√

2=

1√2

[0 −ii 0

]([10

]+

[0i

])=

1√2

[0 −ii 0

] [1i

]

=1√2

[1i

]= 1 · 1√

2

([10

]+

[0i

])= 1 · |↑⟩+ i |↓⟩√

2(11.36)

and

σy|↑⟩ − i |↓⟩√

2=

1√2

[0 −ii 0

]([10

]−[0i

])=

1√2

[0 −ii 0

] [1−i

]

=1√2

[−1i

]= −1 · 1√

2

([10

]−[0i

])= −1 · |↑⟩ − i |↓⟩√

2. (11.37)

Therefore,

|↑⟩y =1√2(|↑⟩+ i |↓⟩) (11.38)

and


|↓⟩y =1√2(|↑⟩ − i |↓⟩). (11.39)

Note that (11.34), (11.35), (11.38), and (11.39) also mean

|↑⟩ = 1√2(|↑⟩x + |↓⟩x) =

1√2(|↑⟩y + |↓⟩y) (11.40)

and

|↓⟩ = 1√2(|↑⟩x − |↓⟩x) =

1√2i(|↑⟩y − |↓⟩y). (11.41)

Now we are ready to embark on a study of sequential Stern-Gerlach ex-periments/measurements. Stern-Gerlach device does not only allow us todetect two components, spin up and spin down, but also enables us to isolateone variety of electrons, be it spin up or spin down. This is the feature wewill need.

We start with a state

α |↑⟩+ β |↓⟩ with |α|2 + |β|2 = 1. (11.42)

The condition on α and β assures that the linear combination α |↑⟩ + β |↓⟩is of unit length; i.e. it is indeed a state. The fact that we can only observeSz = + ~

2or Sz = ~

2is a consequence of Postulate 3 and the matrix repre-

sentation of Sz given by (11.12), which implies that the eigenvalues of Szare indeed + ~

2and − ~

2. After the electrons in this state go through the first

Stern-Gerlach device, you will observe spin up |↑⟩ and spin down |↓⟩ stateswith probabilities |α|2 and |β|2, respectively according to Postulate 5.

If an electron with the spin wavefunction (11.42) is found to have an up-spin, Postulate 6 forces its wavefunction to be |↑⟩ after the measurement. So,we have


α |↑⟩+ β |↓⟩ first measurement−−−−−−−−−−−−−−−−−−→ |↑⟩ . (11.43)

Having isolated |↑⟩, we now have a pure eigenstate with an up-spin, and thesecond Stern-Gerlach device find |↑⟩ with a probability 1.0. This obviousconsequence is described in (11.44).

α |↑⟩+ β |↓⟩ first measurement−−−−−−−−−−−−−−−−−−→ |↑⟩ second measurement−−−−−−−−−−−−−−−−−−−−→ |↑⟩(11.44)

Next, let us see what will happen if the second Stern-Gerlach device is rotatedby 90 so that it is now in the x-direction. With what probabilities will weobserve |↑⟩x and |↓⟩x? The answer lies in (11.40), which gives

|↑⟩ = 1√2(|↑⟩x + |↓⟩x). (11.45)

According to Postulate 5, the probabilities of observing each eigenstate isgiven by

∣∣∣ 1√2

∣∣∣2. Therefore, we will observe the up-spin 50% of the time andthe down-spin 50% of the time.

α |↑⟩+ β |↓⟩ first measurement−−−−−−−−−−−−−−−−−−−−−−−−−−→to choose the spin-up state

|↑⟩

second measurement−−−−−−−−−−−−−−−−−−−−→in the x-direction

|↑⟩x (50%) and |↓⟩x (50%) (11.46)

Similarly, we have

α |↑⟩+ β |↓⟩ first measurement−−−−−−−−−−−−−−−−−−−−−−−−−−−−→to choose the spin-down state

|↓⟩


second measurement−−−−−−−−−−−−−−−−−−−−→in the x-direction

|↑⟩x (50%) and |↓⟩x (50%) (11.47)

as

|↓⟩ = 1√2(|↑⟩x − |↓⟩x) (11.48)

from (11.41).

What if |↑⟩x goes through a Stern-Gerlach device oriented in the y-direction? By symmetry, we should observe the up-spin and down-spin stateswith a probability of 0.5, respectively. Let us double check this by an explicitcomputation. By (11.34), (11.40), and (11.41),

|↑⟩x =1√2(|↑⟩+ |↓⟩) = 1√

2

(1√2(|↑⟩y + |↓⟩y) +

1√2i(|↑⟩y − |↓⟩y)

)

=1

2(|↑⟩y + |↓⟩y) +

1

2i(|↑⟩y − |↓⟩y) =

1

2(|↑⟩y + |↓⟩y)−

1

2i(|↑⟩y − |↓⟩y)

=1− i2|↑⟩y +

1 + i

2|↓⟩y . (11.49)

Because

∣∣∣∣1− i2

∣∣∣∣2 =( √

2

2

)2

=1

2and

∣∣∣∣1 + i

2

∣∣∣∣2 =( √

2

2

)2

=1

2, (11.50)

our intuition is supported mathematically according to Postulate 5.

246

Exercises1. Consider an electron in the initial spin state

1√2|↑⟩+ 1√

2|↓⟩ .

(a) If two Stern-Gerlach devices are sequentially arranged so that thefirst device measures the spin in the z-direction and the second inthe y-direction, what is the probability that the first device mea-sures an up-spin in the z-direction and the second device measuresa down-spin in the y-direction?

(b) In the above, what is the final spin state of the electron?

Chapter 12

Molecular Rotation

There is only one three-dimensional motion for a point mass such as an elec-tron; namely translation or simply the motion of its center of mass, i.e. itsposition, in the three-dimensional space. However, for entities with morecomplex structures such as molecules, there are three types of motion; trans-lation, vibration, and rotation. Translation is the change of the position ofits center of mass and vibration is as in a simple harmonic oscillator. Frorotation, think of a top spinning while remaining at the same spot. The pointis that translation is the only motion that involves a change in the center ofgravity, and three dimensional motions of molecules can be decomposed intotranslation, vibration, and rotation. As we have already discussed transla-tional motion of particles and the simple harmonic oscillator as an exampleof vibration of diatomic molecules, rotation is our final frontier. We will onlyconsider a rigid body in this chapter.

12.1 Rotational Kinetic Energy1

The classical formula for the rotational kinetic energy E of a freely rotatingmolecule is

1There is no such thing as rotational potential energy, and we only concern ourselveswith rotational kinetic energy.

247

248 CHAPTER 12. MOLECULAR ROTATION

E =1

2Ixω

2x +

1

2Iyω

2y +

1

2Izω

2z , (12.1)

where x, y, and z are principal axes of rotation, and Ix, Iy, and Iz are themoments of inertia about the x-, y-, and z-axes. As the angular momentaabout the principal axes are given by

Lx = Ixωx, Ly = Iyωy, and Lz = Izωz, (12.2)

we can also express the rotational kinetic energy as follows in terms of theangular momenta about the principal axes.

E =L2x

2Ix+L2y

2Iy+L2z

2Iz(12.3)

The quantum mechanical Hamiltonian is obtained by the usual substitutionof classical observables by the corresponding quantum mechanical operators.Namely,

H =L2x

2Ix+L2y

2Iy+L2z

2Iz; (12.4)

where

Lx = −i~(y∂

∂z− z ∂

∂y

)= i~

(sinϕ ∂

∂θ+ cot θ cosϕ ∂

∂ϕ

)

Ly = −i~(z∂

∂x− x ∂

∂z

)= i~

(− cosϕ ∂


∂ϕ

)(12.5)

Lz = −i~(x∂

∂y− y ∂

∂x

)= −i~ ∂

∂ϕ.

from (9.2) and (9.3).

12.2. MOMENT OF INERTIA FOR A DIATOMIC MOLECULE 249

12.2 Moment of Inertia for a Diatomic MoleculeMoment of inertia is defined relative to a chosen axis of rotation. For a pointmass, the moment of inertia is defined as the product of the mass and thesquare of perpendicular distance to the axis of rotation. Therefore,

I = mr2, (12.6)

where m is the mass and r is the distance between the point mass and the axisof rotation. To a first approximation, a diatomic molecule can be regraded astwo pint masses M and m connected by a rigid massless rod of length r. Nowintroduce a canonical coordinate axes, call them the x- and y-axes, so thatthe massless rod lies on the x-axis with the center of mass of the molecule atthe origin. Suppose the coordinate of the point mass M is (r1, 0), and that ofm is (−r2, 0), where the positive numbers r1 and r2 satisfy r1 + r2 = r. Ouraxis of rotation is the y-axis that goes through the center of mass. Then, themoment of inertia I for this diatomic molecule is the sum of the moment forthe two point masses to a first approximation.

I =Mr21 +mr22 (12.7)

Now, because the center of mass is at the origin, we have

Mr1 −mr2M +m

= 0 =⇒Mr1 = mr2. (12.8)

We can use this relation to express r1 and r2 in terms of r.

Mr1 = mr2 =⇒ r1 =m

Mr2 =

m

M(r − r1) =⇒

(1 +

m

M

)r1 =

m

Mr

M +m

Mr1 =

m

Mr =⇒ r1 =

m

M +mr =⇒ r2 =

M

mr1 =

M

M +mr (12.9)

Plugging these into (12.7),

I =Mr21 +mr22 =Mm2

(M +m)2r2 +m

M2

(M +m)2r2

=Mm(m+M)

(M +m)2r2 =

Mm

M +mr2. (12.10)

Recall that µ = MmM+m

is the reduced mass. Therefore, we get


I = µr2. (12.11)

12.3 Two-Dimensional Rotation Confined tothe x, y-Plane

The angular momentum vector is always parallel to the z-axis; i.e. J = Jzk2.

But, the difference from the classical counterpart is the quantization of Jz.Without loss of generality, we will specialize to rotations about the z-axis.As Lx and Ly are 0 in (12.4) and (12.5), we get

H =L2z

2I=

(−i~ ∂

∂ϕ

) (−i~ ∂

∂ϕ

)2I

= − ~2

2I

∂2

∂ϕ2; (12.12)

where we dropped the subscript z from Iz because Iz is the only nonzeromoment of inertia in our situation. Our Schrodinger equation is

Hψ = Eψ =⇒ − ~2

2I

∂2ψ

∂ϕ2= Eψ. (12.13)

Because ψ here is a function of ϕ alone, and also because it is basically thesame as Φ encountered in Chapter 10 “The Hydrogen Atom”, it is customaryto denote this wavefunction by Φ. With this convention, we now have

− ~2

2I

∂2Φ(ϕ)

∂ϕ2= EΦ(ϕ). (12.14)

The solutions are

2In the literature about molecular rotation, the angular momentum vector is typicallydenoted by J rather than L.

12.4. THREE-DIMENSIONAL ROTATION 251

Φm(ϕ) =

√1

2πeimϕ for m = 0,±1,±2, . . . . (12.15)

As in the case of the hydrogen atom, possible values for m are determined bythe single-valuedness condition; namely, Φ(ϕ) = Φ(ϕ + 2nπ) for any integern. For the total energy, we have

m = ±√2IE

~=⇒ E =

m2~2

2Ifor m = 0,±1,±2, . . . . (12.16)

The term m2 in the expression for the energy implies that we have two-folddegeneracy for each value of m except for m = 0, which means there is norotation. In the classical picture, we can regard this as deriving from thesense of rotation, clockwise or counter-clockwise.

12.4 Three-Dimensional RotationIn this section, we still consider a rotational motion confined to a plane.However, unlike in Section 12.3 where the rotation was confined to the x, y-plane, we now allow the plane of rotation to be tilted relative to the z-axis.In other words, it is necessary to give the magnitude of angular momentum∥J∥ and its tilt from the z-axis, measured by Jz as before, to completelyspecify the rotation.

From (10.32) the full Hamiltonian in spherical coordinates is

−~2

2µ

[1

r2∂

∂r

(r2∂

∂r

)+

1

r2 sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2

](12.17)

=−~2

2µr2

[∂

∂r

(r2∂

∂r

)+

1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

]. (12.18)

Because we are dealing with a rigid molecule, r is fixed, and there is no


kinetic energy of radial motion. So, the term

1

r2∂

∂r

(r2∂

∂r

)(12.19)

vanishes. Also, we know from (12.11) that I = µr2. Therefore, the Hamilto-nian for three-dimensional rotation of a rigid body is given by

−~2

2I

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

]. (12.20)

This is basically the same as the θ- and φ-dependent parts of the Hamil-tonian for the hydrogen atom given in Chapter 10. The time-independentSchrodinger equation is

−~2

2I

[1

sin θ∂

∂θ

(sin θ ∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

]ψ(θ, φ) = Eψ(θ, φ). (12.21)

Separation of variables work here as well. So, let

ψ(θ, φ) = Θ(θ)Φ(φ). (12.22)

Then, we will get

d2Φ

dφ2= −m2

JΦ (12.23)

and~2

2I

[1

sin θd

dθ

(sin θdΘ

dθ

)− m2

JΘ

sin2 θ

]+ EΘ = 0. (12.24)

Due to single-valuedness and normalizability conditions on the wavefunc-tions, we get

12.4. THREE-DIMENSIONAL ROTATION 253

E =~2

2IJ(J + 1) for J = 1, 2, 3, . . . (12.25)

andmJ = −J,−J + 1, . . . , 0, . . . , J − 1, J. (12.26)

The wavefunctions are

ΦmJ(φ) =

1

2πeimJφ (12.27)

and

ΘmJJ(θ) =

[(2J + 1)(J − |mJ |)!

2(J + |mJ |)!

] 12

P|mJ |J (cos θ); (12.28)

where P |mJ |J (cos θ)’s denote the associated Legendre polynomials first encoun-

tered in (10.74) on p.213. We usually use the symbol B for ~22I

, and so, wehave

E = BJ(J + 1). (12.29)

As for mJ , it is proportional to the z-component Jz of the rotational angularmomentum; namely,

Jz = mJ~. (12.30)

This is another example of degeneracy in higher dimensions discussed in Sec-tion 8.1. The total rotational energy depends only on J , but there are 2J+1distinct states with mJ = −J,−J + 1, . . . , 0, . . . , J − 1, J which share thesame J , and hence, have the same total energy E.

In the most general case, we have Ix, Iy, and Iz which are different fromeach other. For a rigid body like this, we need to introduce three constantsA = ~2

2Ix, B = ~2

2Iy, and C = ~2

2Izrather than just B = ~2

2I. Interested readers


should consult “Microwave Molecular Spectra (3rd edition)” by Gordy andCook [Gordy and Cook, 1984].

Appendices

255

Appendix A

Matrix Algebra

A.1 Invertible MatricesFact A.1 (Invertible Matrix) Let A be a square n× n matrix. Then thefollowing statements are equivalent. Hence, they are either all true or allfalse.

1. A is an invertible matrix.

2. A is bijective; that is, A is one-to-one and onto.

3. A is row equivalent to the n × n identity matrix.

4. A has n pivot positions.

5. The equation Ax = 0 has only the trivial solution 0.

6. The null space of A is 0.

7. The dimension of the null space of A is 0.

8. The columns of A form a linearly independent set.

9. The linear transformation x 7→ Ax is one-to-one.

10. The equation Ax = b has at least one solution for each b ∈ Rn.

11. The columns of A span Rn.

12. The linear transformation x 7→ Ax maps Rn onto Rn.

257

258 APPENDIX A. MATRIX ALGEBRA

13. There is an n× n matrix L such that LA = I.

14. There is an n× n matrix R such that AR = I.

15. At is an invertible matrix.

16. The columns of A form a basis of Rn.

17. The column space C(A) = Rn.

18. The dimension of C(A) = n.

19. rank A = n

20. The number 0 is not an eigenvalue of A.

21. det A , 0

A.2 Rank of a MatrixDefinition A.1 (Rank of a Matrix) The column rank of a matrix A isthe maximum number of linearly independent column vectors of A. The rowrank of A is the maximum number of linearly independent row vectors of A.Equivalently, the column rank of A is the dimension of the column space of A,while the row rank of A is the dimension of the row space of A. The columnrank and the row rank are always equal, and we call this number simply therank of A, denoted by rank A.1

Theorem A.1 (Column Rank = Row Rank) The column rank and therow rank of a matrix A are equal.

The following proofs are basically from Wikipedia.

ProofConsider an m × n matrix A whose i-th row and j-th column entry is Aijand column rank is r. Pick an arbitrary basis v1,v2, . . . ,vr of the column

1Alternatively, the row rank of A is the number of non-zero rows in the row reducedfrom of A.

A.2. RANK OF A MATRIX 259

space of A. Each basis vector is an m × 1 column vector, and we can forman m× r matrix

C = [v1 v2 . . . vr] = [vkl]1≤l≤r1≤k≤m. (A.1)

Note that we are writing the k-th component of the m× 1 column vector vlas vkl with 1 ≤ k ≤ m such that

vl =

v1lv2l...vml

(A.2)

for bookkeeping purposes. The j-th column [Aj] of A can be expressed as alinear combination of v1,v2, . . . ,vr, or vi for short, with coefficients bijsuch that

[Aj] =r∑p=1

bpjvp for 1 ≤ j ≤ n. (A.3)

In its full glory, (A.3) can be written as follows.A1j

A2j...

Amj

=

∑rp=1 v1pbpj∑rp=1 v2pbpj

...∑rp=1 vmpbpj

(A.4)

Hence,

Aij =r∑p=1

vipbpj =⇒ A = CB; (A.5)

where A = [Aij]1≤j≤n1≤i≤m is the original m × n matrix, C = [vkl]

1≤l≤r1≤k≤m is the

m × r matrix whose columns are the basis vectors v1,v2, . . . ,vr, andB = [bpj]

1≤j≤n1≤p≤r is the r×n coefficient matrix defined by (A.3). Now consider

the i-th row of A denoted here by [Ai].

[Ai] = [Ai1 Ai2 . . . Ain] =

r∑p=1

vipbp1r∑p=1

vipbp2 . . .r∑p=1

vipbpn


=r∑p=1

vip [bp1 bp2 . . . bpn] (A.6)

Because [bp1 bp2 . . . bpn] is the p-th row of B, the rows Ai of the m × nmatrix A are linear combinations of the rows of B, we have row rank A ≤row rank B ≤ r = column rank A. We have now shown that row rank A ≤column rank A. Furthermore, if we subject At to the same procedure, we willarrive at row rank At ≤ column rank At =⇒ column rank A ≤ row rank A.This proves the theorem.

An Alternative Proof for Real MatricesLet A be an m×n real matrix whose row rank is r; i.e. the dimension of therow space of A is r. Choose a basis v1,v2, . . . ,vr of the row space of A.Note that vi’s are 1 × n row vectors, (vi)t’s are n × 1 column vectors, andA(vi)

t’s are m× 1 column vectors. Now, consider a linear combinationr∑i=1

ciA(vi)t = A

[r∑i=1

ci(vi)t

]= Awt = 0; (A.7)

where ci’s are some scalars, and w :=r∑i=1

civi is some vector in the row space

of A. Because Aw = 0 and the i-th component of the m× 1 column vectorAwt is given by ∑n

j=1Aijwj, which is the inner product between the i-th rowof A and w, (A.7) implies that w is orthogonal to each row vector of A.Hence, w is orthogonal to all the vectors in the row space of A. But, weknow w belongs to the row space of A. Therefore, w =

r∑i=1

civi = 0, which

in turn implies that all ci’s are 0 because vi’s are linearly independent. Itnow follows that A(v1)

t, A(v2)t, . . . , A(vr)

t are linearly independent. Notingthat [A(vi)

t]k =∑nl=1Akl(vi)

tl , we have

A(vi)t =

∑nl=1A1l(vi)

tl∑n

l=1A2l(vi)tl

...∑nl=1Aml(vi)

tl

=

A11(vi)

t1 + A12(vi)

t2 + . . .+ A1n(vi)

tn

A21(vi)t1 + A22(vi)

t2 + . . .+ A2n(vi)

tn

...Am1(vi)

t1 + Am2(vi)

t2 + . . .+ Amn(vi)

tn

= (vi)t1

A11

A21...

Am1

+ (vi)t2

A12

A22...

Am2

+ . . .+ (vi)tn

A1n

A2n...

Amn

, (A.8)

A.3. CHANGE OF BASIS 261

which is a linear combination of the column vectors of A. Hence, we have rlinearly independent vectors A(v1)

t, A(v2)t, . . . , A(vr)

t in the column spaceof A. We conclude that the dimension of the column space is at least r, andwe have shown that column rank A ≥ row rank A. We next consider At andgo through the same procedure to show that column rank At ≥ row rank At.But, column rank At = row rank A and row rank At = column rank Agive us row rank A ≥ column rank A. This concludes the proof.

A.3 Change of BasisTheorem A.2 (Change of Basis: Transition Matrix) [Anton, 2010, p.218] Consider two bases B = u1,u2, . . . ,un and B′ = u′

1,u′2, . . . ,u

′n of

an n-dimensional vector space Vn; where we regard ui’s and u′i’s as n × 1

column vectors. If vB is the n× 1 column matrix representation of a vectorv ∈ Vn in B, and vB′ the same in B′, then, the relation between vB and vB′

can be described byvB = PB′→BvB′ ;

where PB′→B, the transition matrix from B′ to B, is an n×n matrix whosecolumns are u′

1B,u′2B, . . . ,u

′nB. Note that with our notation u′

jB is the j-thbasis vector of B′ expressed as a column vector in the basis B. Similarly, dueto symmetry, with another transition matrix PB→B′ whose columns are ujB′,we have

vB′ = PB→B′vB.

In fact, we can show

PB′→BPB→B′ = PB→B′PB′→B = I

and PB′→B is the inverse of PB→B′.

ProofAs B and B′ are bases, each vector in B can be expressed as a unique linearcombination of the vectors in B′, and each vector in B′ can be expressed as aunique linear combination of the vectors in B. Hence, we can find the uniquecoefficients Sij and Tij, for 1 ≤ j, i ≤ n, or differently put, unique matricesS = [Sij] and T = [Tij] such that

ui =n∑j=1

Sjiu′j and u′

i =n∑j=1

Tjiuj (A.9)


or in matrix form

[ u1 u2 · · · un ] = [ u′1 u′

2 · · · u′n ]

S11 · · · S1n...

. . ....

Sn1 · · · Snn

and

[ u′1 u′

2 · · · u′n ] = [ u1 u2 · · · un ]

T11 · · · T1n...

. . ....

Tn1 · · · Tnn

.If we combine the two relations of (A.9), we get

ui =n∑j=1

Sjiu′j =

n∑j=1

Sjin∑k=1

Tkjuk =n∑k=1

n∑j=1

TkjSji

uk

and

u′i =

n∑j=1

Tjiuj =n∑j=1

Tjin∑k=1

Skju′k =

n∑k=1

n∑j=1

SkjTji

u′k.

However, because B = u1,u2, . . . ,un and B′ = u′1,u

′2, . . . ,u

′n are

bases, ui’s and u′i’s are linearly independent. Hence,

n∑j=1

TkjSji =n∑j=1

SkjTji = δki.

Recall thatδki =

1 i = k0 i , k

.

If we form the matrices S = [Sij] and T = [Tij],∑nj=1 SkjTji is the k-th row

and the i-th column entry of the matrix product ST , and ∑nj=1 TkjSji is the

k-th row and the i-th column entry of the matrix product TS. Therefore,

ST = TS = I,

and T is the inverse of S. Getting back to (A.9),

ui =n∑j=1

Sjiu′j = S1iu

′1 + S2iu

′2 + . . . + Sniu

′n =

S1i

S2i...Sni

B′


and

u′i =

n∑j=1

Tjiuj = T1iu1 + T2iu2 + . . . + Tniun =

T1iT2i...Tni

B

;

where []B and []B′ mean they are with respect to B and B′, respectively. Thisproves that the j-th column of [Sij] is ujB′, and that the j-th column of [Tij]is u′

jB. It remains to show that

[Tij] = PB′→B and [Sij] = PB→B′ (A.10)

as described in the theorem. Take any vector v ∈ Vn and express it as linearcombinations of ui’s and u′

i’s, respectively.

v =

v1v2...vn

B

=n∑i=1

viui and v =

v′1

v′2...v′n

B′

=n∑i=1

v′iu

′i

We get

v =n∑i=1

v′iu

′i =

n∑i=1

v′i

n∑j=1

Tjiuj =n∑j=1

(n∑i=1

Tjiv′i

)uj

=

(n∑i=1

T1iv′i

)u1+

(n∑i=1

T2iv′i

)u2+ . . .+

(n∑i=1

Tniv′i

)un =

∑ni=1 T1iv

′i∑n

i=1 T2iv′i

...∑ni=1 Tniv

′i

B

=⇒ [v]B = [Tij][v]B′

and

v =n∑i=1

viui =n∑i=1

vin∑j=1

Sjiu′j =

n∑j=1

(n∑i=1

Sjivi

)u′j

=

(n∑i=1

S1ivi

)u′

1+

(n∑i=1

S2ivi

)u′

2+ . . .+

(n∑i=1

Snivi

)u′n =

∑ni=1 S1ivi∑ni=1 S2ivi...∑n

i=1 Snivi

B′


=⇒ [v]B′ = [Sij][v]B

This proves (A.10), and we have now proved the theorem by an explicitconstruction of the transition matrices PB′→B and PB→B′.

Theorem A.3 (Change of Basis and Linear Transformations) Let Ωbe a linear operator operating on a vector space Vn. In other words, Ω is alinear map from Vn into Vn. Consider two bases B = u1,u2, . . . ,un andB′ = u′

1,u′2, . . . ,u

′n of Vn and the corresponding matrix representations

of Ω in those bases, denoted by [Ωij] and [Ω′ij] respectively. Then,

Ω′ij =

n∑k,l=1

T−1ik ΩklTlj or [Ω′

ij] = [T−1ij ][Ωij][Tij] or Ω′ = T−1ΩT ;

where T = PB′→B.

ProofFrom (2.18) on p.40,

Ωui =n∑j=1

Ωjiuj (A.11)

andΩu′

i =n∑j=1

Ω′jiu

′j. (A.12)

Next, from (A.9), we have

u′i =

n∑j=1

Tjiuj. (A.13)

Substituting (A.13) into the right-hand side of (A.12),

n∑j=1

Ω′jiu

′j =

n∑j=1

Ω′ji

(n∑k=1

Tkjuk

)=

n∑k=1

n∑j=1

TkjΩ′ji

uk. (A.14)

Now, substituting (A.13) and (A.11) successively into the left-hand side of(A.12),

Ωu′i = Ω

n∑j=1

Tjiuj

=n∑j=1

TjiΩuj =n∑j=1

Tji

(n∑k=1

Ωkjuk

)=

n∑k=1

n∑j=1

ΩkjTji

uk.

(A.15)


Because uk’s are linearly independent, (A.14) and (A.15) imply that

n∑j=1

TkjΩ′ji =

n∑j=1

ΩkjTji

for all 1 ≤ i, k ≤ n. This means that the (k, i)-entry of the matrix productTΩ′ equals the (k, i)-entry of the matrix product ΩT for all 1 ≤ i, k ≤ n.Therefore,

TΩ′ = ΩT =⇒ Ω′ = T−1ΩT ;

where we know T = PB′→B from Theorem A.2.

Theorem A.4 (Basis Independence of Characteristic Polynomials)The characteristic polynomial of a linear transformation Ω is well defined.That is, it is basis-independent and does not depend on the choice of a basis.

ProofConsider two bases B = ui and B′ = v′

i, and denote the matrix repre-sentations of a linear transformation OMEGA with respect to these basesby Ω and Ω′. If T := PB′→B so that Tv′

k = uk, then, from Theorem A.3 wehave Ω′ = T−1ΩT . This gives us

det(Ω′ − λI) = det(T−1ΩT − λI) = det(T−1ΩT − T−1(λI)T )

= det(T−1(Ω− λI)T ) = det(T−1)det(Ω− λI))detT =1

detTdet(Ω− λI)detT

= det(Ω− λI), (A.16)

which is the desired result.

Theorem A.5 If there is a linearly independent set S, it can be extended toa basis of the vector space V; that is, if u1,u2, . . . ,um (m < n) are linearlyindependent, we can find um+1 . . . ,un such that u1,u2, . . . ,um+1 . . . unis a basis for V.

Proof by Zorn’s Lemma


A.4 Diagonalizability and Simultaneous Di-agonalization

In the following, we will denote the set of all n× n matrices by Mn.

Definition A.2 (Similarity) A matrix B ∈ Mn is similar to another ma-trix A ∈Mn if there exists an invertible matrix S ∈Mn such that

B = S−1AS. (A.17)

Definition A.3 (Diagonalizability) A matrix A ∈ Mn is diagonalizableif it is similar to a diagonal matrix.

Theorem A.6 A matrix A ∈Mn is diagonalizable if and only if there existsa set of n linearly independent eigenvectors of A.

Proof(=⇒)Let S be an invertible matrix such that S−1AS = D, where D is a diagonalmatrix. We have S−1AS = D =⇒ S (S−1AS) = SD =⇒ AS = SD. Now,denote the j-th column of S by sj, which can also be regarded as an n × 1column vector such that

sj = [sij] with 1 ≤ i ≤ n, (A.18)

where sij is the i-th row and j-th column entry of S. With this notation,

Asj =

∑i

A1isij

...∑i

Akisij

...∑i

Anisij

.

(A.19)

So, Asj is the j-th column of AS, and we have

AS = [As1 As2 . . . Asn]. (A.20)

A.4. DIAGONALIZABILITY AND SIMULTANEOUS DIAGONALIZATION 267

Next, let D = [dii], where dii is the i-th diagonal entry of D. Then, the i-throw and the j-th column entry of SD denoted by (SD)ij is given by

(SD)ij =∑k

sikdkj = sijdjj = djjsij. (A.21)

Hence, the j-th column of SD is given bydjjs1j...

djjskjdjjsnj

= djjsj, (A.22)

which in turn indicates

SD = [d11s1 d22s2 . . . dnnsn]. (A.23)

Putting these all together,

AS = SD =⇒ [As1 As2 . . . Asn] = [d11s1 d22s2 . . . dnnsn] =⇒ Asj = djjsj

for 1 ≤ j ≤ n. (A.24)

Therefore, each sj is an eigenvector of A with corresponding eigenvalue djj.Because S is invertible, rank S = n, and sj’s are linearly independent.(=⇒)Conversely, suppose there is a linearly independent set s1 s2 . . . sn ofeigenvectors of A with associated eigenvalues s1 s2 . . . sn, and form S =[s1 s2 . . . sn]. As before, AS = [As1 As2 . . . Asn] = [λ1s1 λ2s2 . . . λnsn].If we let D be the diagonal matrix whose k-th diagonal entry is λk. Then,SD = [λ1s1 λ2s2 . . . λnsn] = AS. Because sj is a linearly independentset, S is of full rank, and S−1 exists. We now have AS = SD =⇒ S−1AS =S−1SD = D, and A is diagonalizable.

Definition A.4 (Simultaneous Diagonalization) Two matrices A,B ∈Mn are said to be simultaneously diagonalizable if there exists an invertiblematrix S such that both S−1AS and S−1BS are diagonal matrices.

Theorem A.7 (Simultaneous Diagonalization and Commutativity)Let A,B ∈ Mn be diagonalizable. Then AB = BA if and only if A and Bare simultaneously diagonalizable.


In order to prove Theorem A.7, we need the following two lemmas.

Lemma A.1 Two matrices A ∈ Mn and B ∈ Mm are diagonalizable if andonly if the composite block matrix

C =

[A 00 B

]∈Mn+m (A.25)

is diagonalizable.

Proof(=⇒)Suppose A ∈ Mn and B ∈ Mm are diagonalizable. Then, we can find twoinvertible matrices SA ∈ Mn and SB ∈ Mm such that S−1

A ASA and S−1B BSB

are diagonal matrices. Define S ∈Mn+m by

S :=

[SA 00 SB

]. (A.26)

Then,

S−1 =

[S−1A 00 S−1

B

], (A.27)

and

S−1CS = S−1

[A 00 B

]S =

[S−1A 00 S−1

B

] [A 00 B

] [SA 00 SB

]

=

[S−1A ASA 00 S−1

B BSB

](A.28)

is a diagonal matrix, and C is diagonalizable.(⇐=)Conversely, assume that C is diagonalizable. Then, there exists an invertiblematrix S ∈ Mn+m such that D := S−1CS is diagonal. Now, denote thej-th column of S by sj ∈ Cn+m. We can write S = [s1 s2 . . . sn+m], andbecause CS = SD, with the usual representation D = [di], where di is thei-th diagonal entry, we have

[Cs1 Cs2 . . . Csn+m] = [d1s1 d2s2 . . . dn+msn+m] (A.29)


as in (A.24). This shows that Csk = dksk, and sk’s are eigenvectors of Cwith the associated eigenvalues dk. Now, we write each sk in composite form

sk =

[xkyk

], (A.30)

where xk ∈ Cn and yk ∈ Cm. Due to the block structure of S, Csk = dkskrequires Axk = dkxk and Byk = dkyk. Denoting [x1 x2 . . . xn+m] by X and[y1 y2 . . . yn+m] by Y , we have

S =

[XY

]. (A.31)

Because X is an n× (n+m) matrix and Y is an m× (n+m) matrix,

rank X ≤ n and rank Y ≤ m. (A.32)

On the other hand,

rank S = n+m (A.33)

as S in invertible. We have the following implications.

n+m = rank S ≤ rank X + rank Y ≤ n+m

=⇒ rank X + rank Y = n+m =⇒ rank X = n, rank Y = m (A.34)

Therefore, X contains n linearly independent columns and Y contains mlinearly independent columns. Recalling that A ∈ Mn operates on vectorsin Cn and the columns of X are eigenvectors of A, we can now see thatA is diagonalizable by Theorem A.6. Likewise, B is also diagonalizable asB ∈Mm operates on vectors in Cm and the columns of Y are eigenvectors ofB.

Lemma A.2 For a diagonal matrix

D =

d1 0 . . . 0 00 d2 . . . 0 0...

.... . .

......

0 0 . . . dn−1 00 0 . . . 0 dn

∈Mn, (A.35)

there exists an invertible matrix P ∈ Mn such that P−1DP switches thepositions of di and dj for any 1 ≤ i, j ≤ n. In fact, such P can be obtainedby switching the i-th row and the j-th row of the identity matrix I ∈Mn.


Before embarking on a general proof of this lemma. Let us see how P workswith a 4× 4 matrix

D =

d1 0 0 00 d2 0 00 0 d3 00 0 0 d4

. (A.36)

We will try to switch d1 and d3 to obtain

Dd1↔d3 =

d3 0 0 00 d2 0 00 0 d1 00 0 0 d4

. (A.37)

According to our program, P is obtained by switching the first and the thirdrows of I ∈M4. Hence,

P =

0 0 1 00 1 0 01 0 0 00 0 0 1

and P−1 =

0 0 1 00 1 0 01 0 0 00 0 0 1

. (A.38)

With this P , we have

P−1DP =

0 0 1 00 1 0 01 0 0 00 0 0 1

d1 0 0 00 d2 0 00 0 d3 00 0 0 d4

0 0 1 00 1 0 01 0 0 00 0 0 1

=

0 0 1 00 1 0 01 0 0 00 0 0 1

0 0 d1 00 d2 0 0d3 0 0 00 0 0 d4

=

d3 0 0 00 d2 0 00 0 d1 00 0 0 d4

= Dd1↔d3

(A.39)

Proof of Lemma A.2Our proof is by way of an explicit computation. Let us see how the matrix Pgiven by (A.40) switches di and dj. In (A.40), “. . .”, “...”, and a blank space


signify 0, while “. . .” stands for 1.

i j

1 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0

0 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0...

.... . .

......

0 0 . . . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 0

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

0 0 . . . . . . 0 1 . . . . . . . . . 0 . . . . . . . . . 0 0...

... . . . . . .... . . .

. . . . . . . . .... . . . . . . . . .

......

...... . . . . . .

... . . . . . .. . . . . .

... . . . . . . . . ....

...

0 0 . . . . . . 0 . . . . . . . . . 1 0 . . . . . . . . . 0 0

j 0 0 . . . . . . 1 . . . . . . . . . . . . 0 . . . . . . . . . 0 0

0 0 . . . . . . . . . . . . . . . . . . . . . . . . 1 . . . . . . 0 0...

.... . .

......

......

. . ....

...

0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 0

0 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 1

(A.40)

Note first that P has the following block diagonal structure; where Ii−1 isthe (i − 1) × (i − 1) identity matrix, In−j is the (n − j) × (n − j) identity


matrix, “. . .” stands for 0, and “. . .” denotes 1.

P =

i j

Ii−1 . . . . . . . . . . . . . . . . . . . . .

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

. . . 0 1 . . . . . . . . . 0 . . .

. . .... . . .

. . . . . . . . .... . . .

. . .... . . . . . .

. . . . . .... . . .

. . . 0 . . . . . . . . . 1 0 . . .

j . . . 1 . . . . . . . . . . . . 0 . . .

. . . . . . . . . . . . . . . . . . . . . In−j

=

Ii−1 . . . . . .. . . Q . . .. . . . . . In−j

;

(A.41)

where Q ∈Mj−i+1 is defined by

Q =

0 . . . . . . . . . . . . 10 1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . 1 01 . . . . . . . . . . . . 0

. (A.42)

It is easy to see Q2 = I or Q−1 = Q. Due to the block diagonal structure ofP , it suffices to consider the action of Q on the following submatrix D of D.

D :=

di . . . . . . . . . . . . 00 di+1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . dj−1 00 . . . . . . . . . . . . dj

(A.43)


We have

Q−1DQ =

0 . . . . . . . . . . . . 10 1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . 1 01 . . . . . . . . . . . . 0

di . . . . . . . . . . . . 00 di+1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . dj−1 00 . . . . . . . . . . . . dj

0 . . . . . . . . . . . . 10 1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . 1 01 . . . . . . . . . . . . 0

=

0 . . . . . . . . . . . . 10 1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . 1 01 . . . . . . . . . . . . 0

0 . . . . . . . . . . . . di0 di+1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . dj−1 0dj . . . . . . . . . . . . 0

=

dj . . . . . . . . . . . . 00 di+1 . . . . . . . . . 0... . . .

. . . . . . . . ....

... . . . . . .. . . . . .

...0 . . . . . . . . . dj−1 00 . . . . . . . . . . . . di

. (A.44)

This proves Lemma A.2.

There is an obvious corollary to Lemma A.2.

Corollary A.1 Given a diagonal matrix D ∈Mn and D′permutation ∈Mn,

where D′permutation is another diagonal matrix whose diagonal entries are a

permuted version of the original D, there exists an invertible matrix Q ∈Mn


such that D′permutation = Q−1DQ. In other words, one can permute the

diagonal entries by some similarity transformation Q−1DQ.

Proof of Theorem A.7(simultaneous diagonalizability =⇒ commutativity)As S−1AS and S−1BS are both diagonal, and diagonal matrices commutewith each other, we get

AB = S(S−1AS

) (S−1BS

)S−1 = S

(S−1BS

) (S−1AS

)S−1 = BA.

(A.45)

(commutativity =⇒ simultaneous diagonalizability)As A is diagonalizable, there is an invertible matrix S ∈ Mn such thatD = S−1AS is diagonal. By Corollary A.1, we can permute the diagonalentries of D by multiplying S by an appropriate invertible matrix. Hence,we may assume that

D =

λ1Im1 0 . . . 00 λ2Im2 0...

. . ....

0 . . . 0 λrImr

; (A.46)

where λi’s are distinct, mj is the multiplicity of λj, and Imjis the mj ×mj

identity matrix. As AB = BA,(S−1AS

) (S−1BS

)= S−1ABS = S−1BAS =

(S−1BS

) (S−1AS

). (A.47)

Denoting (S−1BS) by C, we have DC = CD. Now, let cij, dij, (cd)ij, and(dc)ij signify the i-th row and the j-th column entries of C, D, CD, and DCrespectively. As D is diagonal, we can write dij = δijdjj. So,

(cd)ij =∑k

cikdkj =∑k

cik(δkjdjj) = cijdjj (A.48)

and(dc)ij =

∑k

dikckj =∑k

(δikdii)ckj = diicij. (A.49)

Because CD = DC means (cd)ij = (dc)ij for all 1 ≤ i, j ≤ n, we have

(cd)ij = (dc)ij =⇒ cijdjj = diicij =⇒ (dii − djj)cij = 0 for 1 ≤ i, j ≤ n.(A.50)


Therefore, cij = 0 unless dii = djj. But, dii = djj only if dii and djj belongto the same diagonal block of (A.46). This implies that cij = 0 unless diiand djj belong to the same block of (A.46). Hence, the matrix C is blockdiagonal whose i-th block is an mi ×mi submatrix. In other words, C hasthe same block structure as D, namely

C =

C1 0 . . . 00 C2 0...

. . ....

0 . . . 0 Cr

; where Cl ∈Mmlfor each l. (A.51)

As B is also diagonalizable, there is an invertible matrix R such that R−1BRis diagonal. So,

R−1SCS−1R = R−1SS−1BSS−1R = R−1BR, (A.52)

and C is diagonalizable. From Lemma A.1, all the blocks of C are diagonal-izable; that is, for each j, there exists an invertible matrix Tj ∈ Mmj

suchthat T−1

j CjTj is diagonal. Next, define a block diagonal matrix T ∈Mn by

T =

T1 0 . . . 00 T2 0...

. . ....

0 . . . 0 Tr

. (A.53)

Then, in a straightforward manner,

T−1 =

T−11 0 . . . 00 T−1

2 0...

. . ....

0 . . . 0 T−1r

. (A.54)

We now have

(ST )−1B(ST ) = T−1(S−1BS)T = T−1CT

=

T−11 C1T1 0 . . . 0

0 T−12 C2T2 0

.... . .

...0 . . . 0 T−1

r CrTr

(A.55)


and(ST )−1A(ST ) = T−1(S−1AS)T = T−1DT

=

T−11 λ1Im1T1 0 . . . 0

0 T−12 λ2Im2T2 0

.... . .

...0 . . . 0 T−1

r λrImrTr

=

λ1Im1 0 . . . 00 λ2Im2 0...

. . ....

0 . . . 0 λrImr

= D. (A.56)

As both (A.55) and (A.56) are diagonal, A and B are simultaneously diago-nalizable by ST ∈Mn.

Every vector space has a basis

SEPTEMBER 24, 2011 6 COMMENTSIf you haven’t seen a Zorn’s lemma argument before, this will probably be bewildering. It is here for the sake of completeness.Theorem:Every vector space has a basis.Proof:Consider the collection of linearly independent sets ordered by inclusion. So if and are linearly independent sets, and , We would say . Of course, it may be that we have sets neither of which is a subset of the other. This is totally fine, since we are in a partially ordered set, or poset.Zorn’s lemma says that if every chain has an upper bound, then there is a maximal element. That is, if we take a bunch of linearly independent sets that are all comparable (one contains the other), and they have an upper bound, then there is a set which is maximal in the sense that nothing contains it. Such a maximal set must have , for if it didn’t, we could add another element to it from , so it wouldn’t be maximal. By virtue of being in the partially ordered set to begin with, must be linearly independent. As is linearly independent and spanning, it is by definition a basis.So we only need to show that every chain has an upper bound. If I have a bunch of linearly independent sets (possibly infinitely many, and a really big infinity too) and they form a chain of containment, we want to find a really big linearly independent set that contains all of the . It turns out that

works. Why? We just need to show that it is linearly independent. Well, the trick is that linear combinations have to be finite. So if I had some linear dependence,There must be some for which the . But then, since is linearly independent, each of the must be zero as desired.

Zorn’s lemma is always weird if you haven’t done it before. Luckily, you can try your own. Another theorem is that if we have a linearly independent set , we can extend it to a basis for our vector space. Try to recreate Zorn’s lemma argument. Instead of starting with all linearly independent sets, start with all linearly independent sets which contain .

Appendix B

Holomorphic FunctionalCalculus

Consider a function f : C → C and an operator Ω. If f can be expanded ina Taylor series about z = 01 such that

f(z) =∞∑n=0

f (n)(0)

n!zn,

we can define a map f : Ω 7→ f(Ω) by

f(Ω) =∞∑n=0

f (n)(0)

n!Ωn.

One way to understand this formalism is to consider a matrix representationof Ω. Then, the sum is also a square matrix and convergence of the series isnothing but an entry-by-entry convergence.

Fact B.1

Fact B.2 (Basis Independence of the Characteristic Polynomial) Thecharacteristic polynomial is basis independent. Hence, the characteristic

1The general form of Taylor series about point z0 is

f(z) =

∞∑n=0

f (n)(z0)

n!(z − z0)n.

When the Taylor expansion is about 0, it is also called a Maclaurin series.

277

278 APPENDIX B. HOLOMORPHIC FUNCTIONAL CALCULUS

equation, the eigenvalues, and their multiplicities are basis-independent.ProofConsider two bases B = b1, b2, . . . bn and D = d1,d2, . . . dn, and letM = [Mkl] be the matrix of basis change, such that

bi =n∑j=1

Mijdj

Appendix C

Wronskian and LinearIndependence

The following theorems were taken from lecture notes prepared by KiamHeong Kwa [Kwa, nd] except for the last one.

Definition C.1 (The Wronskian) Let y1 and y2 be two differentiable func-tions of x. Then, the Wronskian W (y1, y2), associated to y1 and y2, is thefunction

W (y1, y2)(x) =

∣∣∣∣∣ y1(x) y2(x)y′1(x) y′

2(x)

∣∣∣∣∣ = y1(x)y′2(x)− y′

1(x)y2(x). (C.1)

Theorem C.1 [Bocher, 1901] If f(x) and g(x) are differentiable functionson an open interval I such that W (f, g)(x0) , 0 for some point x0 in I, thenf(x) and g(x) are linearly independent on I. Equivalently, if f(x) and g(x)are linearly dependent on I, then W (f, g)(x) = 0 for all x ∈ I.

Remark C.1 The converse of Theorem C.1 does not hold without additionalconditions on the pair of functions involved. That is to say, it is not truein general that two differentiable functions f(x) and g(x) with a vanish-ing Wronskian on an open interval must be linearly dependent on the sameinterval.

Theorem C.2 If f(x) and g(x) are differentiable functions on an open in-terval I such that f(x) , 0 and W (f, g)(x) = 0 for all x ∈ I, then f(x) andg(x) are linearly dependent. In particular, g(x) = kf(x) for a constant k.

279

280 APPENDIX C. WRONSKIAN AND LINEAR INDEPENDENCE

Theorem C.3 (Abel’s Theorem) Consider a linear second-order ordinarydifferential equation

y′′ + p(x)y′ + q(x)y = 0; (C.2)

where the coefficients p(x) and q(x) are continuous on an open interval I.The Wronskian W (y1, y2) is given by the following integral on I for someconstant that depends on y1(x) and y2(x).

W (y1, y2)(x) = c exp[−∫p(x) dx

](C.3)

To be more explicit, if x0 is any given point in I,

W (y1, y2)(x) = W (y1, y2)(x0) exp[−∫ x

x0p(s) ds

](C.4)

for all x ∈ I.

Remark C.2 By (C.3) and (C.4), the Wronskian of any two solutions of(C.2) is either always negative, always zero, or always positive on the intervalI.

Theorem C.4 [Bostan and Dumas, 2010] Let y1(x) and y2(x) be solutionsof (C.2). Then y1(x) and y2(x) are linearly independent if and only ifW (y1, y2)(x) , 0 for all x ∈ I. Equivalently, y1(x) and y2(x) are linearlydependent if and only if W (y1, y2)(x) = 0 for all x ∈ I.

Definition C.2 (Analytic Functions) A function f(z) is said to be ana-lytic about a point z0 if it has a convergent power series representation in aneighborhood of z0; that is,

f(z) =∞∑k=0

fk(z − z0)k (C.5)

is convergent in an open set, or interval if z is real, containing z0. In addition,the coefficients fk are uniquely determined by

fk =f (k)(z0)

k!(C.6)

for each k = 0, 1, 2, . . . .Therefore, a function is (globally) analytic if and only if its Taylor seriesabout z0 converges to the function in some neighborhood for every z0 in itsdomain.

281

Theorem C.5 Let f(x) and g(x) be analytic functions on an open intervalI. If W (f, g)(x) = 0 on I, then f(x) and g(x) are linearly dependent.

Fact C.1 (Principle of Superposition) Let y1 and y2 be solutions of ahomogeneous linear differential equation of the general form

A0(x)dny(x)

dxn+ A1(x)

dn−1y(x)

dxn−1+ . . .

+An−1(x)dy(x)

dx+ An(x)y(x) = 0 (C.7)

orn∑i=0

Ai(x)dn−iy(x)

dxn−i =

[n∑i=0

Ai(x)dn−i

dxn−i

]y(x) = 0; (C.8)

where d0

dx0y(x) is to be regarded as y(x) by definition. Then,

c1y1(x) + c2y2(x) (C.9)

is a solution of the same differential equation for any constants c1 and c2.

Theorem C.6 (Fundamental Set) If y1(x) and y2(x) are two linearly in-dependent solutions of the equation

y′′(x) + p(x)y′(x) + q(x)y(x) = 0, (C.10)

then, any solution y(x) is given by

y(x) = c1y1(x) + c2y2(x) (C.11)

for some constants c1 and c2. In this case, the set y1(x), y2(x) is called afundamental set of solutions.

Corollary C.1 Two polynomials p(x) and q(x) are linearly dependent if andonly if W (p, q)(x) = 0 everywhere.

282 APPENDIX C. WRONSKIAN AND LINEAR INDEPENDENCE

Appendix D

Newtonian, Lagrangian, andHamiltonian Mechanics

Newtonian mechanics, Lagrangian mechanics, and Hamiltonian mechanicsare three equivalent formulations of classical mechanics. Therefore, any ofthese formulations can be used to solve a problem in classical mechanics.However, depending on the nature of the problem, one of these formulationsleads to equations which are much easier to solve. One major difference be-tween Newtonian mechanics and the other two is that Newtonian mechanicsdescribes the system in terms of force, while Lagrangian and Hamiltonianmechanics do so in terms of energy. The following is a very brief review anddescription of the three formulations.

D.1 An OverviewHere is a summary of the qualitative differences among Newtonian, La-grangian, and Hamiltonian formulations.

Newtonian formulation considers all the forces acting in/on the system.If you know the forces and the mass, you can get a differential equation forthe position of each particle.

Lagrangian formulation is based on an action principle. In a conservativefield, action should be minimized for the actual trajectory (path of motion)of a particle. The action in this case is what is known as the Lagrangian

283

APPENDIX D. NEWTONIAN, LAGRANGIAN, AND HAMILTONIAN MECHANICS

L = T − V , total energy minus potential energy, and its minimization leadsto a description of the behavior of each particle by way of Lagrange’s (dif-ferential) equations. The variables involved are generalized positions (oftendenoted by q), generalized velocities (often denoted by q), and time t; i.e.L = L(q, q, t).

Hamiltonian formulation is usually derived from the Lagrangian. Whenthe total energy is conserved, which is the case in this book, the Hamilto-nian H is nothing but the total energy of the system, H = T + V . Thevariables are generalized positions q, generalized momenta p, and time t; i.e.H = H(p, q, t). This is the formulation employed almost exclusively in quan-tum mechanics.

More quantitative/mathematical descriptions will follow.

D.2 Newtonian MechanicsNewton’s second law is

Fj = maj or Fj = md2xjdt2

for j = 1, . . . , n.

The position xj(t) is obtained by solving this second-order ordinary differen-tial equation with some boundary condition such as the initial position.

The Lagrangian and Hamiltonian are formulated in an n-dimensionalconfiguration space.

D.3 Lagrangian MechanicsThe Lagrangian, L, is defined as the difference between the kinetic energy Tand the potential energy V .

L = T − V

The motion is described by Lagrange’s equations of motion

d

dt

(∂L

∂qj

)=∂L

∂qjfor j = 1, . . . , n;

D.4. HAMILTONIAN MECHANICS 285

where qj’s are generalized coordinates (positions) and qj’s are generalizedvelocities1.

D.4 Hamiltonian MechanicsThe HamiltonianH is defined as the sum of the kinetic and potential energies.

H = T + V

Hamilton’s equations of motion are as follows.

∂H

∂qj= −pj and ∂H

∂pj= qj for j = 1, . . . , n; (D.1)

where qj’s are generalized coordinates (positions) and pj’s are generalizedmomenta. We also have the following time derivatives connecting H and L.

∂H

∂t= −∂L

∂t

One apparent difference between the Lagrangian formulation and Hamil-tonian formulation is that Lagrange’s equations describe the system by a setof n second-order differential equations, while Hamilton’s equations are n setsof coupled first-order differential equations. Hence, Hamiltonian mechanicsrequires 2n first-order differential equations while Lagrangian mechanics re-quires n second-order differential equations. As mentioned in the beginningon p.283, one can also note that “force” enters the equation of motion ex-plicitly only for Newtonian mechanics.

1Generalized coordinates refer to the parameters that describe the configuration of thesystem relative to some reference configuration. Their time derivatives are the generalizedvelocities.

APPENDIX D. NEWTONIAN, LAGRANGIAN, AND HAMILTONIAN MECHANICS

Appendix E

Normalization Schemes for aFree Particle

We will only consider ψ(x) = Aeikx for k =√2mE~

. Computations for ψ(x) =Ae−ikx are completely analogous.

E.1 The Born NormalizationThis highly controversial but widely used procedure involves restricting therange of integration to a finite length L and letting L to infinity afterwards.If we only consider real and positive normalization constants,∫ +L/2

−L/2

(Aeikx

)∗ (Aeikx

)dx = |A|2

∫ +L/2

−L/2e−ikxeikx dx = |A|2L = 1

=⇒ A =1√L. (E.1)

Hence, the Born-normalized wavefunction is

ψ(x) =1√Leikx. (E.2)

E.2 The Dirac NormalizationDirac normalization is based on the Dirac delta function δ(x−x′) described inSection 4.3. In particular, we will use the relation between Fourier transformand the delta function.

287

APPENDIX E. NORMALIZATION SCHEMES FOR A FREE PARTICLE

E.2.1 Fourier Transform and the Delta FunctionThe derivations found in Section 4.5 is reproduced below with slightly dif-ferent notations more fitting for this argument.1 For a function f(x), wehave

F (k) =1√2π

∫ +∞

−∞f(x)e−ikx dx the Fourier transform (E.3)

and

f(x′) =1

2π

∫ +∞

−∞F (k)eikx

′dk the inverse Fourier transform (E.4)

Substituting (E.3) for F (k) in (E.4),

f(x′) =∫ +∞

−∞f(x)

[1

2π

∫ +∞

−∞eik(x−x′) dk

]dx. (E.5)

This leads to one representation of Dirac delta function

δ(x′ − x) = 1

2π

∫ +∞

−∞eik(x−x′) dk =

1

2π~

∫ +∞

−∞eip(x

′−x)/~ dp; (E.6)

where we used the relation k =√2mE~

=

√2m p2

2m

~= p/~. Similarly,

1

2π

∫ +∞

−∞e−ik′xeikx dx =

1

2π

∫ +∞

−∞eix(k−k′) dx = δ(k − k′). (E.7)

E.2.2 Wavefunctions as Momentum or Position Eigen-functions

For a free particle, we have

H =P 2

2m(E.8)

and the commutator

[H,P ] = 0. (E.9)1In the author’s modest and personal opinion, seeing/understanding the same deriva-

tion and/or relation expressed with different notations is one good way to solidify yourgrasp of a mathematical and physical concept.

E.2. THE DIRAC NORMALIZATION 289

Therefore, the stationary solutions, or the eigenfunctions, of the HamiltonianH are also eigenfunctions of the momentum operator P . Indeed,

PAeikx = −i~ ddxAei

p~x = −i~

(ip

~

)Aei

p~x = pAeikx. (E.10)

In the case of a free particle, the distribution of the momentum eigenvaluesp is continuous ranging from −∞ to +∞ just like the position x. This isthe situation described in Section 4.8. We will label each normalized eigenketof P by its associated momentum p to obtain |p⟩ or Apei

p~x; where Ap’s

are normalization constants. Now, from (4.84), the representation of X inthe P basis is given by

X = i~d

dp. (E.11)

Noting that

XAe−ikx = i~d

dpAe−i p

~x = i~

(−ix~

)Ae−i p

~x = xAe−ikx, (E.12)

we will label each normalized eigenket of X by its associated position eigen-value x to obtain |x⟩ or Axe−ix

~p; where Ax’s are normalization constants.

It is important to understand the reason behind writing |p⟩ as Apeip~x and

|x⟩ as Axe−ix~p. When we interpret p as a fixed value and x as the index in

the X space, we write ei p~x, while we write e−ix

~p when x is fixed and p is the

index in the P space. It is all a matter of what is interpreted as fixed andwhat is interpreted as a running index.

In the braket notation, we get

⟨x|p⟩ = Apei p~x (E.13)

and⟨p|x⟩ = Axe

−ix~p. (E.14)

Needless to say, Ax = A∗p is a necessary condition because ⟨p|x⟩ = ⟨x|p⟩∗

gives

Axe−ix~p = A∗

pe−i p~x =⇒ Ax = A∗

p. (E.15)

Based on (4.4), the appropriate normalization scheme suggests itself at thispoint.


Definition E.1 (The Dirac Normalization) The free particle eigenfunc-tions |p⟩ are normalized a la Dirac if

⟨p′|p⟩ = ⟨p|p′⟩ =∫ +∞

−∞

(Ape

i p′~x)∗ (

Apei p~x)dx = δ(p− p′). (E.16)

We have ∫ +∞

−∞

(Ape

ik′x)∗ (

Apeikx)dx = |Ap|2

∫ +∞

−∞e−ik′xeikx dx

= |Ap|2∫ +∞

−∞e−i p′

~xei

p~x dx = |Ap|2

∫ +∞

−∞e−ix(p′−p)/~ dx

= 2π~|Ap|2[1

2π

∫ +∞

−∞e−iy(p′−p) dy

]= 2π~|Ap|2δ(p− p′); (E.17)

where we let y = x/~. The expression (E.17) becomes δ(p − p′) if we letAp =

1√2π~

. This gives us |p⟩ = ψp with

⟨x|p⟩ = ψp(x) =1√2π~

eikx or 1√2π~

eip~x. (E.18)

with the proposed normalization (E.16)

⟨p|p′⟩ =∫ +∞

−∞ψp(x)

∗ψp′(x) dx =1

2π

∫ +∞

−∞e−iy(p−p′) dy = δ(p− p′).

We also have∫ +∞

−∞

(Axe

−ik′x)∗ (

Axe−ikx

)dp = |Ax|2

∫ +∞

−∞eik

′xe−ikx dp

= |Ax|2∫ +∞

−∞ei

x′~pe−ix

~p dp = |Ax|2

∫ +∞

−∞eip(x

′−x)/~ dp

= 2π~|Ax|2[1

2π

∫ +∞

−∞eiy(x

′−x) dy]= 2π~|Ax|2δ(x− x′); (E.19)

where we let y = p/~. The expression (E.19) becomes δ(x − x′) if we letAx =

1√2π~

. This gives us |x⟩ = ψx with

⟨p|x⟩ = ψx(p) =1√2π~

eikx or 1√2π~

eix~p (E.20)

with the normalization

⟨x|x′⟩ =∫ +∞

−∞ψx(p)

∗ψx′(p) dp =1

2π

∫ +∞

−∞eiy(x

′−x) dy = δ(x− x′).

E.3. THE UNIT-FLUX NORMALIZATION 291

We have made sure Ap = A∗x as required by (E.15).

Finally, note again that there is symmetry between the Dirac normaliza-tion condition ⟨p|p′⟩ = δ(p− p′) for the eigenkets |p⟩ and the completeness(or closure) condition ⟨x|x′⟩ = δ(x− x′) for the position kets |x⟩.

E.3 The Unit-Flux NormalizationThis is based on the concept of particle current density discussed in AppendixH. A wavefunction is normalized if it has a unit current density. Namely, thefollowing functions are normalized in the unit flux sense.

ψ±(x) =1√~k/m

e±ikx ; where k =

√2mE

~(E.21)

Let us verify that we indeed have a unit current density.

j =i~

2m

(ψ∂

∂xψ∗ − ψ∗ ∂

∂xψ

)

=i~

2m

1√~k/m

e±ikx ∂

∂x

1√~k/m

e∓ikx − 1√~k/m

e∓ikx ∂

∂x

1√~k/m

e±ikx

=

i~

2m

1

(~k/m)

(e±ikx(∓ik)e∓ikx − e∓ikx (±ik) e±ikx

)=

i

2k(∓2ik) = ±1 (E.22)

The ± sign signifies the direction of the flow relative to the way the x-axisis set up. Now note that

~k/m =(~P

~

)/m =

P

m=mv

m= v. (E.23)

Hence, the probability density is given by

ψ±(x)∗ψ±(x) =

1√~k/m

e±ikx

∗1√~k/m

e±ikx =1

~k/m=

1

v, (E.24)

which provides a classical explanation for the unit flux because the “amount”of probability that flows through any point x = x0 per unit time, say onesecond, is 1

v× v = 1.

Appendix F

Symmetries and ConservedDynamical Variables

The description and consequences of symmetry are important and beautifulcomponents of both classical mechanics and quantum mechanics. In the con-text of physics, symmetry means that a certain observable remains constantunder some transformation. In other words, symmetry is always in referenceto a given transformation.

F.1 Poisson Brackets and Constants of Mo-tion

Definition F.1 (Poisson Bracket) Let p and q denote sets of n gen-eralized momenta and generalized coordinates; p = p1, p2, . . . , pn andq = q1, q2, . . . , qn. The Poisson bracket between ω(p, q) and λ(p, q), de-noted by ω, λ, is defined as below.

ω, λ =n∑i=1

(∂ω

∂qi

∂λ

∂pi− ∂ω

∂pi

∂λ

∂qi

)(F.1)

Poisson bracket arises naturally as the total time derivative of a functionω(p, q), which does not depend on time t explicitly, as follows.

dω

dt=

n∑i=1

(∂ω

∂qiqi +

∂ω

∂pipi

)

293

294 APPENDIX F. SYMMETRIES AND CONSERVED DYNAMICAL VARIABLES

=n∑i=1

(∂ω

∂qi

∂H

∂pi+∂ω

∂pi

∂H

∂qi

)= ω,H (F.2)

Fact F.1 From (F.2), we can see that a variable γ whose Poisson bracketwith H vanishes, γ,H = 0, is a constant of motion (COM), or preserved.

Theorem F.1 (Invariant H and Conservation of a Dynamical Variable)Consider a dynamical variable g(p, q) which generates1 the following infinites-imal transformation

qi → qi = qi + ε ∂g∂pi

= qi + δqi

pi → pi = pi − ε ∂g∂qi= pi + δpi

. (F.3)

If H is invariant under the infinitesimal transformation (F.3), then g isconserved; i.e. g is a constant of motion (COM).

ProofThe Taylor series Tf (x1, x2, . . . , xn) of a multivariate function f(x1, x2, . . . , xn)about a point (a1, a2, . . . , an) is given by

Tf (x1, x2, . . . , xn) =∞∑i1=0

∞∑i2=0

· · ·∞∑in=0

(x1 − a1)i1(x2 − a2)i2 · · · (xn − an)ini1!i2! · · · in!(

∂i1+i2+···+inf

∂xi11 ∂xi22 · · · ∂xinn

)(a1, a2, . . . , an).

Therefore, the Taylor series expansion TH (p1, p2, . . . , pn, q1, q2, . . . , qn) ofour Hamiltonian

H (p, q) = H (p1, p2, . . . , pn, q1, q2, . . . , qn)

about the point (a1, a2, . . . , a2n) is given by

TH (p1, p2, . . . , pn, q1, q2, . . . , qn) =∞∑i1=0

∞∑i2=0

· · ·∞∑

i2n=0

(p1 − a1)i1 · · · (qn − a2n)i2n

i1!i2! · · · (i2n)!(∂i1+i2+···+i2nH

∂pi11 ∂pi22 · · · ∂qi2n

n

)(a1, a2, . . . , a2n).

(F.4)1We say g is the generator of the infinitesimal transformation (F.3).

F.1. POISSON BRACKETS AND CONSTANTS OF MOTION 295

Now we let pi = pi, qi = qi, ai = pi for 1 ≤ i ≤ n, and aj = qj forn+ 1 ≤ j ≤ 2n in (F.4).

TH (p1, p2, . . . , pn, q1, q2, . . . , qn) =∞∑i1=0

∞∑i2=0

· · ·∞∑

i2n=0

(p1 − p1)i1 · · · (qn − qn)i2n

i1!i2! · · · (i2n)!(∂i1+i2+···+i2nH

∂pi11 ∂pi22 · · · ∂qi2n

n

)(p1, p2, . . . , qn).

(F.5)

Note that pi − pi = −ε ∂g∂qiand qi − qi = ε ∂g

∂pifrom (F.3). Substituting these

into (F.5), we get

TH (p1, p2, . . . , pn, q1, q2, . . . , qn) =∞∑i1=0

∞∑i2=0

· · ·∞∑

i2n=0

(−ε ∂g

∂qi

)i1 · · · (ε ∂g∂pi

)i2n

i1!i2! · · · (i2n)!

×(∂i1+i2+···+i2nH

∂pi11 ∂pi22 · · · ∂qi2n

n

)(p1, p2, . . . , qn)

= H (p1, p2, . . . , qn) +n∑i=1

[∂H

∂pi

(−ε ∂g

∂qi

)+∂H

∂qi

(ε∂g

∂pi

)]+O(ε2).

(F.6)

As H is invariant under (F.3), we have

H (p1, p2, . . . , qn)−H (p1, p2, . . . , qn)

=n∑i=1

[∂H

∂pi

(−ε ∂g

∂qi

)+∂H

∂qi

(ε∂g

∂pi

)]+O(ε2)

= εn∑i=1

(−∂H∂pi

∂g

∂qi+∂H

∂qi

∂g

∂pi

)+O(ε2) = εH , g+O(ε2)

= ε [H , g+O(ε)] = 0 (F.7)

Suppose H , g(p, q), which is H , g evaluated at (p, q), is γ > 0 orδ < 0. As O(ε) ε→0−−→ 0, |H , g+O(ε)| > 0 if ε is sufficiently small. But,this contradicts (F.7). Therefore, H , g = 0, and this in turn implies thatthe dynamical variable g is conserved due to (F.2).


F.2 lz As a GeneratorLet g in (F.3) be the z-component of angular momentum; i.e. g = lz =xpy − ypx from (9.1). Then, we have the following.

x→ x = x+ ε∂g

∂px= x− yε (F.8)

y → y = y + ε∂g

∂py= xε+ y (F.9)

z → z = z + ε∂g

∂pz= z (F.10)

Comparing this with (F.16), we can see that this is an infinitesimal rotationaround the z-axis. Hence, the angular momentum around the z-axis is thegenerator of rotations around the z-axis. Therefore, lz is conserved if His invariant under rotations of a physical system around the z-axis. Froexample, if we only have a central force F (r), and hence a central potentialV (r), there is no force in the tangential direction as we move around a circle∥r∥ = r0, θ = θ0, whose center is on the z-axis. Because we have no torquein this case, it follows that lz is a constant of motion (COM).

F.3 Rotation Around The OriginOne simple example of such an observable-transformation pair is the distancefrom the origin, f(x, y, z) =

√x2 + y2 + z2, of a particle under a rotation

about the origin denoted by R. In this case, if the rotation is by ϕ about thex-axis, followed by a rotation by θ about the y-axis, followed by a rotation byψ about the z-axis, all in the counter-clockwise direction, the transformationR is given by the following matrix.

R =

[cos θ cosϕ cosϕ sinψ + sinϕ sin θ cosψ sinϕ sinψ − cosϕ sinψ − cosϕ sin θ cosψ

− cos θ sinψ cosϕ cosψ − sinϕ sin θ sinψ sinϕ cosψ + cosϕ sin θ sinψsin θ − sinϕ cos θ cosϕ cos θ

](F.11)

As it is very cumbersome to carry on a computation with the full three-dimensional rotation matrix, we will only consider a counter-clockwise rota-tion Rz about the z-axis given by the matrix below.

Rz =

cos θ − sin θ 0sin θ cos θ 00 0 1

(F.12)

F.4. INFINITESIMAL ROTATIONS AROUND THE Z-AXIS 297

Under the transformation, (x, y, z) becomes (x, y, z). xyz

= Rz

xyz

=

cos θ − sin θ 0sin θ cos θ 00 0 1

xyz

=

x cos θ − y sin θx sin θ + y cos θ

z

(F.13)

So,

f(Rz[x, y, z]) = f(x, y, z) = f(x cos θ−y sin θ, x sin θ+y cos θ, z)

=√(x cos θ − y sin θ)2 + (x sin θ + y cos θ)2 + z2

=√x2 cos2 θ − 2xy sin θ cos θ + y2 sin2 θ + x2 sin2 θ + 2xy sin θ cos θ + y2 cos2 θ + z2

=√x2(sin2 θ + cos2 θ) + y2(sin2 θ + cos2 θ)− 2xy sin θ cos θ + 2xy sin θ cos θ + z2

=√x2 + y2 + z2 = f(x, y, z),

and the distance from the origin f is invariant under a rotation about thez-axis. Similarly, we can also show

f(Rx[x, y, z]) = f(x, y, z) and f(Ry[x, y, z]) = f(x, y, z);

where Rx and Ry are rotations about the x- and y-axes, respectively. As anyrotation is a combination of Rx, Ry, and Rz, this proves that the distanceform the origin is invariant under any rotation about the origin as it shouldbe.

F.4 Infinitesimal Rotations Around the z-AxisWe will first specialize to rotations about the origin in two dimensions forcomputational simplicity. The results can be extended to rotations aboutthe z-axis in three dimensions in a straightforward manner.

In two dimensions, we only have Rz. The effect of Rz by θ is as follows.[xy

]= Rz

[xy

]=


] [xy

]=

[x cos θ − y sin θx sin θ + y cos θ

](F.14)


Next, consider an infinitesimal transformation which differs from the identitytransformation I by angle ε. In the limit as ε approaches 0, cos ε approaches1 and sin ε tends to ε. Hence,[

xy

]=

[x cos ε− y sin εx sin ε+ y cos ε

]ε → 0−−−→

[x− yεxε+ y

]. (F.15)

This extends to three dimensions as below. xyz

=

x cos ε− y sin εx sin ε+ y cos ε

z

ε → 0−−−→

x− yεxε+ yz

. (F.16)

Appendix G

Commutators

G.1 Commutator IdentitiesHere are some basic and useful commutator identities.

[A,B + C] = [A,B] + [A,C] (G.1)[A+B,C] = [A,C] + [B,C] (G.2)[A,BC] = [A,B]C +B[A,C] (G.3)[AB,C] = [A,C]B + A[B,C] (G.4)

[AB,CD] = A[B,C]D + C[A,D]B + CA[B,D] + [A,C]BD (G.5)[A, [B,C]] + [C, [A,B]] + [B, [C,A]] = 0 (G.6)

Let us verify (G.3), (G.4), and (G.5). It is straightforward for (G.3) and (G.4).

[A,BC] = ABC −BCA = ABC −BAC +BAC −BCA = (AB −BA)C +B(AC − CA)= [A,B]C +B[A,C]

Likewise for (G.4). Verification of (G.5) is similar.

[AB,CD] = [(AB), CD] = [AB,C]D + C[AB,D]

= ([A,C]B + A[B,C])D + C([A,D]B + A[B,D])

= A[B,C]D + C[A,D]B + CA[B,D] + [A,C]BD (G.7)

299

300 APPENDIX G. COMMUTATORS

G.2 Commutators involving X and P

From Tables 3.1 and 3.2, we have Mxi(multiplication by xi) and xi

(−i~ ∂

∂xi

)for (x1, x2, x3) = (x, y, z) and (x1, x2, x3) =

(i, j, k

). Of 18 possible com-

binations for the commutator, there are only three that are nonzero. Wehave:

[Xi, Xj] = [Mxi,Mxj

] = 0 for all (i, j) (G.8)

[Pi, Pj] =

[−i~ ∂

∂xi,−i~ ∂

∂xj

]= 0 for all (i, j) (G.9)

[Pi, Xj] = −i~δij for all (i, j). (G.10)

These relations are very useful in computing other commutators such as thoseamong Lx, Ly, Lz, L2, and H.

G.3 Commutators involving X, P , L, and r

[Lz, x] = i~y (G.11)[Lz, y] = −i~x (G.12)[Lz, z] = 0 (G.13)

[Lz, px] = i~py (G.14)[Lz, py] = −i~px (G.15)

[Lz, pz] = 0 (G.16)

[Lz, r2] = 0 (G.17)

[Lz, p2] = 0 (G.18)

[H,Lx] = [H,Ly] = [H,Lz] = 01 when V = V (r) (G.19)[H,L2] = 0 (G.20)

Verification

[Lz, x] = [xpy − ypx, x] = [xpy, x]− [ypx, x] = [x, x]py + x[py, x]− ([y, x]px + y[px, x])

1One can write these as [H,L] = 0.

G.3. COMMUTATORS INVOLVING X, P , L, AND R 301

= 0 + 0− 0− y[px, x] = −y(−i~) = i~y (G.11)

[Lz, px] = [xpy − ypx, px] = [xpy, px]− [ypx, px] = [x, px]py + x[py, px]

−([y, px]px + y[px, px]) = i~py + 0− 0− 0 = i~py (G.14)

Recall from (9.3) thatLz = −i~

∂

∂ϕ.

Because Lz is a partial derivative with respect to ϕ, it commutes with mul-tiplication by any function of r, including r2. This proves (G.17).

[Lz, p2] = [Lz, p

2x + p2y + p2z] = [Lz, p

2x] + [Lz, p

2y] + [Lz, p

2z]

= [Lz, px]px + px[Lz, px] + [Lz, py]py + py[Lz, py] + [Lz, pz]pz + pz[Lz, pz]

= i~pypx + i~pxpy − i~pxpy − i~pypx + 0 + 0 = 0 (G.18)

[H,Lz] =

[p2

2m+ V (r), Lz

]=

1

2m[p2, Lz] + [V (r), Lz] = 0 + 0 = 0 (G.19)

[H,L2] = [H,L2x + L2

y + L2z] = [H,L2

x] + [H,L2y] + [H,L2

z]

= [H,Lx]Lx + Lx[H,Lx] + [H,Ly]Ly + Ly[H,Ly] + [H,Lz]Lz + Lz[H,Lz]

= 0 + 0 + 0 + 0 + 0 + 0 (G.20)

302 APPENDIX G. COMMUTATORS

Appendix H

Probability Current

The following is based on a lecture note of Professor Steven Carlip at UCDavis [Carlip, 2013].

Let R be a region in R3 with a boundary ∂R, and consider a particlewhose wavefunction is Ψ(x, t). Then, the probability P (R) that the particleis found in R is given by

P (R) =∫RΨ∗(x, t)Ψ(x, t) dxdydz. (H.1)

If the particle is moving in the classical sense, or the magnitude of thewavefunction is changing with time in the quantum mechanical sense, P (R)changes with time, and the first derivative is given by

d

dtP (R) =

∫R

∂

∂t(Ψ(x, t)Ψ(x, t)) dxdydz =

∫R

(∂Ψ∗

∂tΨ+Ψ∗∂Ψ

∂t

)dxdydz.

(H.2)

The time-dependent Schrodinger equation gives

i~∂Ψ

∂t= − ~

2

2m∇2Ψ+ VΨ (H.3)

as well as its complex conjugate

−i~∂Ψ∗

∂t= − ~

2

2m∇2Ψ∗ + VΨ∗. (H.4)

303

304 APPENDIX H. PROBABILITY CURRENT

Therefore,

∂Ψ∗

∂tΨ =

1

−i~

(− ~

2

2m∇2Ψ∗ + VΨ∗

)Ψ =

~

i2m

(∇2Ψ∗

)Ψ− V

i~Ψ∗Ψ (H.5)

and

Ψ∗∂Ψ

∂t= Ψ∗ 1

i~

(− ~

2

2m∇2Ψ+ VΨ

)= − ~

i2mΨ∗∇2Ψ+

V

i~Ψ∗Ψ. (H.6)

So, the integrand of (H.2) is

∂Ψ∗

∂tΨ+Ψ∗∂Ψ

∂t=~

i2m

(Ψ∇2Ψ∗ −Ψ∗∇2Ψ

)= − i~

2m

(Ψ∇2Ψ∗ −Ψ∗∇2Ψ

).

(H.7)

Now we need the following identity.

Fact H.1 For differentiable functions f and g we have

∇ · (f∇g − g∇f) = f∇2g − g∇2f. (H.8)

ProofFor simplicity, we will introduce the following notation, which is used widelyin the physics literature.

fxi:=

∂f

∂xiand fxixj

:=∂2f

∂xj∂xi; where xi, xj = x, y, or z (H.9)

Then,

∇ · (f∇g − g∇f) = ∇ ·[f(gxı+ gy ȷ+ gzk

)− g

(fxı+ fy ȷ+ fzk

)]=

(ı∂

∂x+ ȷ

∂

∂y+ k

∂

∂z

)·[f(gxı+ gy ȷ+ gzk

)− g

(fxı+ fy ȷ+ fzk

)]=

∂

∂x(fgx − gfx) +

∂

∂y(fgy − gfy) +

∂

∂z(fgz − gfz)

= fxgx + fgxx − gxfx − gfxx + fygy + fgyy − gyfy − gfyy+ fzgz + fgzz − gzfz − gfzz

= fgxx − gfxx + fgyy − gfyy + fgzz − gfzz= f(gxx + gyy + gzz)− g(fxx + fyy + fzz)

305

= f∇2g − g∇2f.

Now, let f = Ψ and g = Ψ∗ in (H.8) to obtain

Ψ∇2Ψ∗ −Ψ∗∇2Ψ = ∇ · (Ψ∇Ψ∗ −Ψ∗∇Ψ) . (H.10)

If we define a vector j by

j :=i~

2m(Ψ∇Ψ∗ −Ψ∗∇Ψ) , (H.11)

then, from (H.10) we have

∇ · j =i~

2m

(Ψ∇2Ψ∗ −Ψ∗∇2Ψ

), (H.12)

and (H.2) becomes

d

dtP (R) =

∫R

(∂Ψ∗

∂tΨ+Ψ∗∂Ψ

∂t

)dxdydz

= −∫R

i~

2m

(Ψ∇2Ψ∗ −Ψ∗∇2Ψ

)dxdydz

= −∫R∇ · j dxdydz. (H.13)

Next, remember Stokes’ Theorem from multivariable calculus.

Theorem H.1 (Stokes’ Theorem: the Divergence Theorem)∫R∇ · j d3x =

∫∂R

j · n d2x; (H.14)

where n is the unit normal to the boundary or surface ∂R.

We finally have

d

dtP (R) = −

∫∂R

j · n d2x. (H.15)

What (H.15) is telling us is that the amount of j escaping through the surface∂R equals, in magnitude, the change in the amount of probability we havein the region R. This is why j is called the probability current.


Definition H.1 (Probability Current) Given a wavefuntion Ψ(x, t) wedefine the probability current j by

j :=i~

2m(Ψ∇Ψ∗ −Ψ∗∇Ψ) . (H.16)

In one dimension, this reduces to

j :=i~

2m

(Ψ∂

∂xΨ∗ −Ψ∗ ∂

∂xΨ

). (H.17)

In particular, when the potential, and hence the Hamiltonian, does not haveexplicit time dependence, j reduces to an expression in the spatial part of thewavefunction ψ(x) rather than the full wave function Ψ(x, t)1 such that

j =i~

2m

(ψ∂

∂xψ∗ − ψ∗ ∂

∂xψ

). (H.19)

Note that this is completely analogous in form and substance to the conti-nuity equation of fluid dynamics and electromagnetism.

dQ

dt= −

∫∂R

j · n dS (dS is the surface integral over ∂R, the boundary of R.)2

In fluid dynamics, Q is the amount of the fluid, and j is the flow of the fluid.1 The derivation is straightforward as below.

j :=i~

2m

(Ψ∂

∂xΨ∗ −Ψ∗ ∂

∂xΨ

)=

i~

2m

(ψ(x)e−iωt ∂

∂x

(ψ(x)e−iωt

)∗

−(ψ(x)e−iωt

)∗ ∂

∂x

(ψ(x)e−iωt

))=

i~

2m

(ψ(x)e−iωt ∂

∂xψ(x)∗eiωt − ψ(x)∗eiωt ∂

∂xψ(x)e−iωt

)=

i~

2m

(ψ∂

∂xψ∗ − ψ∗ ∂

∂xψ

)(H.18)

2This form is known as the integral form. Its differential counterpart, called the dif-ferential form quite appropriately, is given by

∂ρ(x, t)

∂t= −∇ · j(x, t) or ∂ρ(x, t)

∂t= − ∂

∂xj(x, t) in one dimension;

where ρ is the density.

307

In electromagnetism, Q is the amount of electric charge, and j is the electriccurrent. Finally, let us make a note of the fact that the equality does not holdif there is a “sink” or “source” in the region R, which is the difference betweenthe classical continuity equation and our quantum mechanical equation.

Appendix I

Chain Rules in PartialDifferentiation

I.1 One Independent VariableConsider a function f = f(x, y, z), where x = x(t), y = y(t), and z = z(t).Then, we have

df

dt=∂f

∂x

dx

dt+∂f

∂y

dy

dt+∂f

∂z

dz

dt. (I.1)

I.2 Three Independent VariablesConsider a function f = f(x, y, z), where x = x(r, θ, ϕ), y = y(r, θ, ϕ), andz = z(r, θ, ϕ). Then, we have

∂f

∂r=∂f

∂x

∂x

∂r+∂f

∂y

∂y

∂r+∂f

∂z

∂z

∂r, (I.2)

∂f

∂θ=∂f

∂x

∂x

∂θ+∂f

∂y

∂y

∂θ+∂f

∂z

∂z

∂θ, (I.3)

and∂f

∂ϕ=∂f

∂x

∂x

∂ϕ+∂f

∂y

∂y

∂ϕ+∂f

∂z

∂z

∂ϕ. (I.4)

Actually, the following operator forms may be more convenient in actual use.∂

∂r=∂x

∂r

∂

∂x+∂y

∂r

∂

∂y+∂z

∂r

∂

∂z(I.5)

309

310 APPENDIX I. CHAIN RULES IN PARTIAL DIFFERENTIATION

∂

∂θ=∂x

∂θ

∂

∂x+∂y

∂θ

∂

∂y+∂z

∂θ

∂

∂z(I.6)

∂

∂ϕ=∂x

∂ϕ

∂

∂x+∂y

∂ϕ

∂

∂y+∂z

∂ϕ

∂

∂z(I.7)

Of course, we also have

∂f

∂x=∂f

∂r

∂r

∂x+∂f

∂θ

∂θ

∂x+∂f

∂ϕ

∂ϕ

∂x, (I.8)

∂f

∂y=∂f

∂r

∂r

∂y+∂f

∂θ

∂θ

∂y+∂f

∂ϕ

∂ϕ

∂y, (I.9)

and∂f

∂z=∂f

∂r

∂r

∂z+∂f

∂θ

∂θ

∂z+∂f

∂ϕ

∂ϕ

∂z, (I.10)

or

∂

∂x=∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ, (I.11)

∂

∂y=∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂ϕ

∂y

∂

∂ϕ, (I.12)

and∂

∂z=∂r

∂z

∂

∂r+∂θ

∂z

∂

∂θ+∂ϕ

∂z

∂

∂ϕ. (I.13)

What we do not have is the following.

I.3 Reciprocal Relation is FalseUnlike the total derivative, we do not have

∂r

∂x=

(∂x

∂r

)−1

. FALSE (I.14)

Instead, we generally have

∂r

∂x,

(∂x

∂r

)−1

. TRUE (I.15)

I.4. INVERSE FUNCTION THEOREMS 311

This is easy to understand because ∂r∂x

means the derivative/change of r withrespect to x where y and z are held constant, while ∂x

∂ris the derivative/

change of x with respect to r where θ and ϕ are held fixed. An explicitcomputation follows.(∂r

∂x

)y,z: fixed

=

(∂√x2 + y2 + z2

∂x

)y,z: fixed

=1

2

2x√x2 + y2 + z2

=x

r= sin θ cosϕ

(I.16)

(∂x

∂r

)θ,ϕ: fixed

=

(∂r sin θ cosϕ

∂r

)θ,ϕ: fixed

= sin θ cosϕ (I.17)

Hence, the inequality (I.15) holds in this case. Needless to say, (I.15) is gen-erally true and is not limited to spherical and Cartesian coordinates.

Compare (I.15) with the following example where we have total deriva-tives.

y = x2 or x = ±√y (I.18)dy

dx= 2x and dx

dy=d± √ydy

= ±1

2

1√y= ±1

2

1

±x=

1

2x(I.19)

dy

dx=

(dx

dy

)−1

(I.20)

Note here that (I.20) really means

dy

dx

∣∣∣∣∣at x=x0

=

dxdy

∣∣∣∣∣at y=x20

−1

. (I.21)

See Theorem I.1 for a more general description.

I.4 Inverse Function TheoremsThe reciprocal relation (I.20) is nothing but a straightforward application ofthe Inverse Function Theorem.

312 APPENDIX I. CHAIN RULES IN PARTIAL DIFFERENTIATION

Theorem I.1 (Inverse Function Theorem: Single Variable) If f is acontinuously differentiable function of x with nonzero derivative at the pointa, then, f is invertible in a neighborhood of a, the inverse f−1 is continuouslydifferentiable, and

(f−1

)′(b) =

1

f ′(a), (I.22)

where b = f(a).

Theorem I.2 (Inverse Function Theorem: Multiple Variables) If thetotal derivative of a continuously differentiable multivariable function F de-fined on an open subset O of Rn and into Rn is invertible at a point p, thatis, the Jacobian J of F at p is non-zero, then, F is an invertible functionwith the inverse F−1 near the point p. This means that the inverse F−1 isdefined in some neighborhood of F (p). Moreover, the inverse function F−1

is continuously differentiable. As for the Jacobian, we have

JF−1 (F (p)) = [JF (p)]−1 ; (I.23)

where JG(q) is the Jacobian of the function G at the point q, and [JF (p)]−1

is the matrix inverse.

Appendix J

Laplacian Operator inSpherical Coordinates

The Laplacian operator in Cartesian coordinates is

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2. (J.1)

We will use the following relations

r =√x2 + y2 + z2 x = r sin θ cosϕ

y = r sin θ sinϕ z = r cos θ

cos θ = z√x2 + y2 + z2

sin θ =√x2 + y2√

x2 + y2 + z2

cosϕ =x√

x2 + y2sinϕ =

y√x2 + y2

,

along with the chain rules below from p.309 of Appendix I.∂

∂x=∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ, (J.2)

∂

∂y=∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂ϕ

∂y

∂

∂ϕ, (J.3)

∂

∂z=∂r

∂z

∂

∂r+∂θ

∂z

∂

∂θ+∂ϕ

∂z

∂

∂ϕ. (J.4)

∂

∂r=∂x

∂r

∂

∂x+∂y

∂r

∂

∂y+∂z

∂r

∂

∂z(J.5)

313

314 APPENDIX J. LAPLACIAN OPERATOR IN SPHERICAL COORDINATES

∂

∂θ=∂x

∂θ

∂

∂x+∂y

∂θ

∂

∂y+∂z

∂θ

∂

∂z(J.6)

∂

∂ϕ=∂x

∂ϕ

∂

∂x+∂y

∂ϕ

∂

∂y+∂z

∂ϕ

∂

∂z(J.7)

The partial derivatives of r with respect to x, y, and z are as follows.

∂r

∂x=

1

2

2x√x2 + y2 + z2

=x

r= sin θ cosϕ

∂r

∂y=

1

2

2y√x2 + y2 + z2

=y

r= sin θ sinϕ

∂r

∂z=

1

2

2z√x2 + y2 + z2

=z

r= cos θ

Next, we will use z = r cos θ to compute ∂θ∂x

, ∂θ∂y

, and ∂θ∂z

.1

∂z

∂x=∂r

∂xcos θ + r

∂ cos θ∂x

= sin θ cosϕ cos θ + r

(∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ

)(cos θ)

= sin θ cosϕ cos θ + r∂θ

∂x(− sin θ) =⇒ 0 = sin θ cosϕ cos θ − r sin θ ∂θ

∂x

=⇒ r sin θ ∂θ∂x

= sin θ cosϕ cos θ

1There is a simpler, but possibly less comfortable way to compute ∂θ∂x , ∂θ

∂y , and ∂θ∂z using

the equalities

cos θ = z

rand sin θ =

√x2 + y2√

x2 + y2 + z2.

Namely,

cos θ = z

r=⇒ − sin θdθ = z(−x)

r3dx =⇒ ∂θ

∂x=

cos θ cosϕr

,

cos θ = z

r=⇒ − sin θdθ = z(−y)

r3dy =⇒ ∂θ

∂y=

cos θ sinϕr

,

and

sin θ =√x2 + y2√

x2 + y2 + z2=⇒ cos θdθ =

√x2 + y2(−z)

r3dz =⇒ ∂θ

∂z= − sin θ

r.

315

=⇒ ∂θ

∂x=

cos θ cosϕr

∂z

∂y=∂r

∂ycos θ + r

∂ cos θ∂y

= sin θ sinϕ cos θ + r

(∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂ϕ

∂y

∂

∂ϕ

)(cos θ)

= sin θ sinϕ cos θ + r∂θ

∂y(− sin θ) =⇒ 0 = sin θ sinϕ cos θ − r sin θ∂θ

∂y

=⇒ r sin θ∂θ∂y

= sin θ sinϕ cos θ

=⇒ ∂θ

∂y=

cos θ sinϕr

∂z

∂z=∂r

∂zcos θ + r

∂ cos θ∂z

= cos θ cos θ + r

(∂r

∂z

∂

∂r+∂θ

∂z

∂

∂θ+∂ϕ

∂z

∂

∂ϕ

)(cos θ)

= cos θ cos θ + r∂θ

∂z(− sin θ) =⇒ 1 = cos2 θ − r sin θ∂θ

∂z

=⇒ r sin θ∂θ∂z

= − sin2 θ

=⇒ ∂θ

∂z= −sin θ

r

Now, use x = r sin θ cosϕ to compute ∂ϕ∂x

, ∂ϕ∂y

, and ∂ϕ∂z

. 2

∂x

∂x=

(∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ

)(r sin θ cosϕ)

2There is an alternative way using the relations

cosϕ =x√

x2 + y2and sinϕ =

y√x2 + y2

.

Namely,

cosϕ =x√

x2 + y2=⇒ − sinϕdϕ =

(1√

x2 + y2+ x

(−12

)1

(x2 + y2)3/2(2x)

)dx

=

(1√

x2 + y2− x2

(x2 + y2)3/2

)dx =

x2 + y2 − x2

(x2 + y2)3/2dx =

y2

(x2 + y2)3/2dx


= sin θ cosϕ sin θ cosϕ+cos θ cosϕ

rr cos θ cosϕ− r sin θ sinϕ∂ϕ

∂x

= sin2 θ cos2 ϕ+ cos2 θ cos2 ϕ− r sin θ sinϕ∂ϕ∂x

= cos2 ϕ− r sin θ sinϕ∂ϕ∂x

=⇒1 = cos2 ϕ− r sin θ sinϕ∂ϕ∂x

=⇒ r sin θ sinϕ∂ϕ∂x

= − sin2 ϕ

=⇒∂ϕ

∂x= − sinϕ

r sin θ

∂x

∂y=

(∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂ϕ

∂y

∂

∂ϕ

)(r sin θ cosϕ)

= sin θ sinϕ sin θ cosϕ+cos θ sinϕ

rr cos θ cosϕ− r sin θ sinϕ∂ϕ

∂y

= sin2 θ sinϕ cosϕ+ cos2 θ sinϕ cosϕ− r sin θ sinϕ∂ϕ∂y

= sinϕ cosϕ− r sin θ sinϕ∂ϕ∂y

=⇒0 = sinϕ cosϕ− r sin θ sinϕ∂ϕ∂y

=⇒ r sin θ sinϕ∂ϕ∂y

= sinϕ cosϕ

=⇒∂ϕ

∂y=

cosϕr sin θ

∂x

∂z=

(∂r

∂z

∂

∂r+∂θ

∂z

∂

∂θ+∂ϕ

∂z

∂

∂ϕ

)(r sin θ cosϕ)

=(r sin θ sinϕ)2

(r sin θ)3 dx =sin2 ϕ

r sin θdx =⇒ ∂ϕ

∂x=

sin2 ϕ

r sin θ1

(− sinϕ) = − sinϕr sin θ ,

and

sinϕ =y√

x2 + y2=⇒ cosϕdϕ =

(1√

x2 + y2+ y

(−12

)1

(x2 + y2)3/2(2y)

)dy

=

(1√

x2 + y2− y2

(x2 + y2)3/2

)dy =

x2 + y2 − y2

(x2 + y2)3/2dy =

x2

(x2 + y2)3/2dy

=(r sin θ cosϕ)2

(r sin θ)3 dy =cos2 ϕr sin θ dy =⇒ ∂ϕ

∂y=

cos2 ϕr sin θ

1

cosϕ =cosϕr sin θ .

317

= cos θ sin θ cosϕ− sin θrr cos θ cosϕ− r sin θ sinϕ∂ϕ

∂z

= −r sin θ sinϕ∂ϕ∂z

=⇒∂ϕ

∂z= 0

Incidentally, we already knew that ∂ϕ∂z

= 0 from geometric considerations;that is, ϕ is determined completely by the x- and y-coordinates.

Let us take a deep breath at this point and summarize what we havefound so far.

∂r∂x

= sin θ cosϕ ∂r∂y

= sin θ sinϕ ∂r∂z

= cos θ∂θ∂x

= cos θ cosϕr

∂θ∂y

= cos θ sinϕr

∂θ∂z

= − sin θr

∂ϕ∂x

= − sinϕr sin θ

∂ϕ∂y

= cosϕr sin θ

∂ϕ∂z

= 0

Our next step is to compute ∂∂x

, ∂∂y

, and ∂∂z

.

∂

∂x=∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂ϕ

∂x

∂

∂ϕ= sin θ cosϕ ∂

∂r+

cos θ cosϕr

∂

∂θ− sinϕr sin θ

∂

∂ϕ(J.8)

∂

∂y=∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂ϕ

∂y

∂

∂ϕ= sin θ sinϕ ∂

∂r+

cos θ sinϕr

∂

∂θ+

cosϕr sin θ

∂

∂ϕ(J.9)

∂

∂z= cos θ ∂

∂r− sin θ

r

∂

∂θ+ 0 · ∂

∂ϕ= cos θ ∂

∂r− sin θ

r

∂

∂θ(J.10)

We are ready to compute ∇2 = ∂2

∂x2+ ∂2

∂y2+ ∂2

∂z2= ∂

∂x∂∂x

+ ∂∂y

∂∂y

+ ∂∂z

∂∂z

.

∂2

∂x2=

∂

∂x

∂

∂x=

(sin θ cosϕ ∂

∂r+

cos θ cosϕr

∂


∂

∂ϕ

)·(

sin θ cosϕ ∂∂r

+cos θ cosϕ

r

∂


∂

∂ϕ

)


=

(sin θ cosϕ ∂

∂r

)(sin θ cosϕ ∂

∂r

)+

(sin θ cosϕ ∂

∂r

)(cos θ cosϕ

r

∂

∂θ

)

−(

sin θ cosϕ ∂∂r

)(sinϕr sin θ

∂

∂ϕ

)

+

(cos θ cosϕ

r

∂

∂θ

)(sin θ cosϕ ∂

∂r

)+

(cos θ cosϕ

r

∂

∂θ

)(cos θ cosϕ

r

∂

∂θ

)

−(

cos θ cosϕr

∂

∂θ

)(sinϕr sin θ

∂

∂ϕ

)

−(

sinϕr sin θ

∂

∂ϕ

)(sin θ cosϕ ∂

∂r

)−(

sinϕr sin θ

∂

∂ϕ

)(cos θ cosϕ

r

∂

∂θ

)

+

(sinϕr sin θ

∂

∂ϕ

)(sinϕr sin θ

∂

∂ϕ

)

= sin2 θ cos2 ϕ ∂2

∂r2+ sin θ cos θ cos2 ϕ ∂

∂r

(1

r

∂

∂θ

)

− sinϕ cosϕ ∂∂r

(1

r

∂

∂ϕ

)

+cos θ cos2 ϕ

r

∂

∂θ

(sin θ ∂

∂r

)+

cos θ cos2 ϕr2

∂

∂θ

(cos θ ∂

∂θ

)

− cos θ sinϕ cosϕr2

∂

∂θ

(1

sin θ∂

∂ϕ

)

− sinϕr

∂

∂ϕ

(cosϕ ∂

∂r

)− cos θ sinϕ

r2 sin θ∂

∂ϕ

(cosϕ ∂

∂θ

)

+sinϕ

r2 sin2 θ

∂

∂ϕ

(sinϕ ∂

∂ϕ

)


∂r2+ sin θ cos θ cos2 ϕ

[− 1

r2∂

∂θ+

1

r

∂2

∂r∂θ

]

− sinϕ cosϕ[− 1

r2∂

∂ϕ+

1

r

∂2

∂r∂ϕ

]

+cos θ cos2 ϕ

r

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ

]

+cos θ cos2 ϕ

r2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2

]

319

− cos θ sinϕ cosϕr2 sin θ

[−cos θ

sin θ∂

∂ϕ+

∂2

∂θ∂ϕ

]

− sinϕr

[− sinϕ ∂

∂r+ cosϕ ∂2

∂r∂ϕ

]

− cos θ sinϕr2 sin θ

[− sinϕ ∂

∂θ+ cosϕ ∂2

∂θ∂ϕ

]

+sinϕ

r2 sin2 θ

[cosϕ ∂

∂ϕ+ sinϕ ∂2

∂ϕ2

]


∂r2+

cos2 θ cos2 ϕr2

∂2

∂θ2+

sin2 ϕ

r2 sin2 θ

∂2

∂ϕ2+

cos2 θ cos2 ϕ+ sin2 ϕ

r

∂

∂r

+ sin θ cos θ cos2 ϕ[− 1

r2∂

∂θ+

1

r

∂2

∂r∂θ

]− sinϕ cosϕ

[− 1

r2∂

∂ϕ+

1

r

∂2

∂r∂ϕ

]

+cos θ cos2 ϕ

r

[+ sin θ ∂2

∂r∂θ

]

+cos θ cos2 ϕ

r2

[− sin θ ∂

∂θ

]

− cos θ sinϕ cosϕr2 sin θ

[−cos θ

sin θ∂

∂ϕ+

∂2

∂θ∂ϕ

]

− sinϕr

[+ cosϕ ∂2

∂r∂ϕ

]


[− sinϕ ∂

∂θ+ cosϕ ∂2

∂θ∂ϕ

]

+sinϕ

r2 sin2 θ

[cosϕ ∂

∂ϕ

](J.11)

∂2

∂y2=

∂

∂y

∂

∂y=

(sin θ sinϕ ∂

∂r+

cos θ sinϕr

∂

∂θ+

cosϕr sin θ

∂

∂ϕ

)·(

sin θ sinϕ ∂∂r

+cos θ sinϕ

r

∂

∂θ+

cosϕr sin θ

∂

∂ϕ

)

=

(sin θ sinϕ ∂

∂r

)(sin θ sinϕ ∂

∂r

)+

(sin θ sinϕ ∂

∂r

)(cos θ sinϕ

r

∂

∂θ

)

+

(sin θ sinϕ ∂

∂r

)(cosϕr sin θ

∂

∂ϕ

)


+

(cos θ sinϕ

r

∂

∂θ

)(sin θ sinϕ ∂

∂r

)+

(cos θ sinϕ

r

∂

∂θ

)(cos θ sinϕ

r

∂

∂θ

)

+

(cos θ sinϕ

r

∂

∂θ

)(cosϕr sin θ

∂

∂ϕ

)

+

(cosϕr sin θ

∂

∂ϕ

)(sin θ sinϕ ∂

∂r

)+

(cosϕr sin θ

∂

∂ϕ

)(cos θ sinϕ

r

∂

∂θ

)

+

(cosϕr sin θ

∂

∂ϕ

)(cosϕr sin θ

∂

∂ϕ

)

= sin2 θ sin2 ϕ∂2

∂r2+ sin θ cos θ sin2 ϕ

∂

∂r

(1

r

∂

∂θ

)

+ sinϕ cosϕ ∂∂r

(1

r

∂

∂ϕ

)

+cos θ sin2 ϕ

r

∂

∂θ

(sin θ ∂

∂r

)+

cos θ sin2 ϕ

r2∂

∂θ

(cos θ ∂

∂θ

)

+cos θ sinϕ cosϕ

r2∂

∂θ

(1

sin θ∂

∂ϕ

)

+cosϕr

∂

∂ϕ

(sinϕ ∂

∂r

)+

cos θ cosϕr2 sin θ

∂

∂ϕ

(sinϕ ∂

∂θ

)

+cosϕr2 sin2 θ

∂

∂ϕ

(cosϕ ∂

∂ϕ

)

= sin2 θ sin2 ϕ∂2

∂r2+ sin θ cos θ sin2 ϕ

[− 1

r2∂

∂θ+

1

r

∂2

∂r∂θ

]

+ sinϕ cosϕ[− 1

r2∂

∂ϕ+

1

r

∂2

∂r∂ϕ

]

+cos θ sin2 ϕ

r

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ

]

+cos θ sin2 ϕ

r2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2

]

+cos θ sinϕ cosϕ

r2 sin θ

[−cos θ

sin θ∂

∂ϕ+

∂2

∂θ∂ϕ

]

+cosϕr

[cosϕ ∂

∂r+ sinϕ ∂2

∂r∂ϕ

]

321

+cos θ cosϕr2 sin θ

[cosϕ ∂

∂θ+ sinϕ ∂2

∂θ∂ϕ

]

+cosϕr2 sin2 θ

[− sinϕ ∂

∂ϕ+ cosϕ ∂2

∂ϕ2

](J.12)

∂2

∂z2=

∂

∂z

∂

∂z=

(cos θ ∂

∂r− sin θ

r

∂

∂θ

)·(

cos θ ∂∂r− sin θ

r

∂

∂θ

)

=

(cos θ ∂

∂r

)(cos θ ∂

∂r

)−(

cos θ ∂∂r

)(sin θr

∂

∂θ

)−(

sin θr

∂

∂θ

)(cos θ ∂

∂r

)

+

(sin θr

∂

∂θ

)(sin θr

∂

∂θ

)

= cos2 θ ∂2

∂r2− sin θ cos θ ∂

∂r

(1

r

∂

∂θ

)− sin θ

r

∂

∂θ

(cos θ ∂

∂r

)

+sin θr2

∂

∂θ

(sin θ ∂

∂θ

)

= cos2 θ ∂2

∂r2− sin θ cos θ

[− 1

r2∂

∂θ+

1

r

∂2

∂r∂θ

]− sin θ

r

[− sin θ ∂

∂r+ cos θ ∂2

∂r∂θ

]

+sin θr2

[cos θ ∂

∂θ+ sin θ ∂

2

∂θ2

](J.13)

Finally, we can compute ∇2, combining (J.11), (J.12), and (J.13). There aresome nice cancellations and simplifications. But, we have to be methodicaland systematic in deciding on our computational strategy, that is, whichterms to combine in what order.

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2

=[sin2 θ

(sin2 ϕ+ cos2 ϕ

)+ cos2 θ

] ∂2∂r2

+[sin θ cos θ

(sin2 ϕ+ cos2 ϕ

)− sin θ cos θ

] [− 1

r2∂

∂θ+

1

r

∂2

∂r∂θ

]

+ (sinϕ cosϕ− sinϕ cosϕ)[− 1

r2∂

∂ϕ+

1

r

∂2

∂r∂ϕ

]

+cos θ(sin2 ϕ+ cos2 ϕ)

r

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ

]


+cos θ(sin2 ϕ+ cos2 ϕ)

r2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2

]

+

(cos θ sinϕ cosϕ

r2 sin θ − cos θ sinϕ cosϕr2 sin θ

)[−cos θ

sin θ∂

∂ϕ+

∂2

∂θ∂ϕ

]The following three terms derive from ∂2

∂x2.

− sinϕr

[− sinϕ ∂

∂r+ cosϕ ∂2

∂r∂ϕ

]


[− sinϕ ∂

∂θ+ cosϕ ∂2

∂θ∂ϕ

]

+sinϕ

r2 sin2 θ

[cosϕ ∂

∂ϕ+ sinϕ ∂2

∂ϕ2

]The following three terms derive from ∂2

∂y2.

+cosϕr

[cosϕ ∂

∂r+ sinϕ ∂2

∂r∂ϕ

]

+cos θ cosϕr2 sin θ

[cosϕ ∂

∂θ+ sinϕ ∂2

∂θ∂ϕ

]

+cosϕr2 sin2 θ

[− sinϕ ∂

∂ϕ+ cosϕ ∂2

∂ϕ2

](J.14)

The following two terms derive from ∂2

∂z2.

− sin θr

[− sin θ ∂

∂r+ cos θ ∂2

∂r∂θ

]

+sin θr2

[cos θ ∂

∂θ+ sin θ ∂

2

∂θ2

]

=∂2

∂r2+

cos θr

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ

]+

cos θr2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2

]

+(sin2 ϕ+ cos2 ϕ+ sin2 θ

) 1r

∂

∂r+

(cos θr2 sin θ

)(sin2 ϕ+ cos2 ϕ

) ∂

∂θ+

sin θ cos θr2

∂

∂θ

+1

r2 sin2 θ(sinϕ cosϕ− sinϕ cosϕ) ∂

∂ϕ+

sin2 θ

r2∂2

∂θ2− sin θ cos θ

r

∂2

∂r∂θ

+sinϕ cosϕ− sinϕ cosϕ

r

∂2

∂r∂ϕ+

cos θr2 sin θ (sinϕ cosϕ− sinϕ cosϕ) ∂2

∂θ∂ϕ

323

+sin2 ϕ+ cos2 ϕ

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

cos θr

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ

]+

cos θr2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2

]

+(1 + sin2 θ

) 1r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

sin θ cos θr2

∂

∂θ

+sin2 θ

r2∂2

∂θ2− sin θ cos θ

r

∂2

∂r∂θ

+1

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

cos θr

[cos θ ∂

∂r+ sin θ ∂2

∂r∂θ− sin θ ∂2

∂r∂θ

]

+cos θr2

[− sin θ ∂

∂θ+ cos θ ∂

2

∂θ2+ sin θ ∂

∂θ

]

+(1 + sin2 θ

) 1r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ

+sin2 θ

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

cos θr

[cos θ ∂

∂r

]+

cos θr2

[cos θ ∂

2

∂θ2

]

+(1 + sin2 θ

) 1r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

sin2 θ

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

cos θr2

[cos θ ∂

2

∂θ2

]+(1 + sin2 θ + cos2 θ

) 1r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ

+sin2 θ

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

2

r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

sin2 θ + cos2 θr2

∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2

=∂2

∂r2+

2

r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

1

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2

Therefore, we have shown

∇2 =∂2

∂r2+

2

r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

1

r2∂2

∂θ2+

1

r2 sin2 θ

∂2

∂ϕ2. (J.15)


Alternatively, we can also write

∇2 =1

r2∂

∂rr2∂

∂r+

1

r2 sin θ∂

∂θsin θ ∂

∂θ+

1

r2 sin2 θ

∂2

∂ϕ2. (J.16)

Verification is straightforward as below.

1

r2∂

∂rr2∂

∂r+

1

r2 sin θ∂

∂θsin θ ∂

∂θ

=1

r2

(2r

∂

∂r+ r2

∂2

∂r2

)+

1

r2 sin θ

(cos θ ∂

∂θ+ sin θ ∂

2

∂θ2

)

=∂2

∂r2+

2

r

∂

∂r+

(cos θr2 sin θ

)∂

∂θ+

1

r2∂2

∂θ2(J.17)

Appendix K

Legendre and AssociatedLegendre Polynomials

K.1 Legendre PolynomialsThe n-th degree Legendre polynomial Pn(x) is a solution to the followingLegendre’s differential equation, where n = 0, 1, 2, . . ..

d

dx

[(1− x2) d

dxPn(x)

]+ n(n+ 1)Pn(x) = 0 (K.1)

Each Pn(x) can be obtained using Rodrigues’ formula

Pn(x) =1

2nn!

dn

dxn

[(x2 − 1)n

], (K.2)

which can be verified by differentiating both sides of the identity

(x2 − 1)d

dx(x2 − 1)n = 2nx(x2 − 1)n (K.3)

(n+ 1) times. When the condition Pn(1) = 1 is imposed, we have

P0(x) = 1

P1(x) = x

P2(x) =1

2(3x2 − 1)

P3(x) =1

2(5x3 − 3x)

325

326 APPENDIX K. LEGENDRE AND ASSOCIATED LEGENDRE POLYNOMIALS

P4(x) =1

8(35x4 − 30x2 + 3)

P5(x) =1

8(63x5 − 70x3 + 15x)

P6(x) =1

16(231x6 − 315x4 + 105x2 − 5)

K.2 Associated Legendre PolynomialsThe associated Legendre polynomial of degree l and order m, denoted byPml (x), satisfies the following general Legendre equation.

(1− x2) d2

dx2Pml (x)− 2x

d

dxPml (x) +

[l(l + 1)− m2

1− x2

]Pml (x) = 0 (K.4)

Pml (x) can be expressed in the following Rodrigues type form.

Pml (x) =

(−1)l

2ll!(1− x2)m/2 d

l+m

dxl+m(1− x2)l (K.5)

In particular, for m > 0, we have

Pml (x) = (1− x2)m/2 d

m

dxmPl(x). (K.6)

Fact K.1 We have the following identity.

P−ml (x) = (−1)m (l −m)!

(l +m)!Pml (x) (K.7)

Proof (taken from a lecture note prepared by Professor Hitoshi Murayamaof UC Berkeley)We will first compute how

dl+m

dxl+m(1− x2)l is related to dl−m

dxl−m(1− x2)l.

We need two lemmas for this proof.

K.2. ASSOCIATED LEGENDRE POLYNOMIALS 327

Lemma K.1 (Combinatorial Identity) We have the following identity.(k + 1r

)=

(k

r − 1

)+

(kr

)(K.8)

Proof of Lemma K.1It is nothing but a straightforward computation.(

kr − 1

)+

(kr

)=

k!

(k − r + 1)!(r − 1)!+

k!

(k − r)!r!

=k!r

(k + 1− r)!r(r − 1)!+

k!(k − r + 1)

(k − r + 1)(k − r)!r!=k!(r + k − r + 1)

(k + 1− r)!r!

=k!(k + 1)

(k + 1− r)!r!=

(k + 1)!

(k + 1− r)!r!=

(k + 1r

)(K.9)

Lemma K.2 Let F (x) and G(x) be two functions which are at least n-timesdifferentiable. Then,

dn

dxnF (x)G(x) =

n∑r=0

(nr

)[dr

dxrF (x)

] [dn−r

dxn−rG(x)

]. (K.10)

Proof of Lemma K.2Let us introduce a short-hand notation

drF :=dr

dxrF (x); (K.11)

where d0

dx0F (x) = F (x) by definition. Our proof is by mathematical induc-

tion. First, when n = 1, we have

d(FG) = (dF )G+ F (dG) =

(10

)(dF )G+

(11

)F (dG)

=1∑r=0

(1r

)[dr

dxrF (x)

] [d1−r

dx1−rG(x)

], (K.12)

and (K.10) holds for n = 1. Now, suppose (K.10) holds for n = k ≥ 1, sothatdk

dxkF (x)G(x) =

k∑r=0

(kr

)[dr

dxrF (x)

] [dk−r

dxk−rG(x)

]=

k∑r=0

(kr

)drFdk−rG.

(K.13)


Then, for n = k + 1, we get

dk+1

dxk+1F (x)G(x) =

d

dx

[dk

dxkF (x)G(x)

]

=d

dx

k∑r=0

(kr

)[dr

dxrF (x)

] [dk−r

dxk−rG(x)

]

=d

dx

[k∑r=0

(kr

)drFdk−rG

]=

k∑r=0

(kr

)d(drFdk−rG

)

=k∑r=0

(kr

)(dr+1Fdk−rG+ drFdk−r+1G

)

=k∑r=0

(kr

)dr+1Fdk−rG+

k∑r=0

(kr

)drFdk−r+1G. (K.14)

Now, let u = r + 1. Then,

k∑r=0

(kr

)dr+1Fdk−rG =

k+1∑u=1

(k

u− 1

)duFdk+1−uG

=k+1∑r=1

(k

r − 1

)drFdk+1−rG. (K.15)

Plug this into (K.14) and use Lemma K.1.

dk+1

dxk+1F (x)G(x) =

k∑r=0

(kr

)dr+1Fdk−rG+

k∑r=0

(kr

)drFdk−r+1G

=k+1∑r=1

(k

r − 1

)drFdk+1−rG+

k∑r=0

(kr

)drFdk−r+1G

=

(kk

)dk+1Fd0G+

k∑r=1

(k

r − 1

)drFdk+1−rG+

(k0

)d0Fdk+1G

+k∑r=1

(kr

)drFdk+1−rG

=

(k + 10

)d0Fdk+1G+

k∑r=1

[(k

r − 1

)+

(kr

)]drFdk+1−rG

+

(k + 1k + 1

)dk+1Fd0G


=

(k + 10

)d0Fdk+1G+

k∑r=1

(k + 1r

)drFdk+1−rG+

(k + 1k + 1

)dk+1Fd0G

=k+1∑r=0

(k + 1r

)drFdk+1−rG (K.16)

So, K.10 holds for n = k+1, and we have proved the lemma by mathematicalinduction.

Let us now resume the proof of Fact K.1. We have

dl+m

dxl+m(1− x2)l = dl+m

dxl+m(1− x)l(1 + x)l

=l+m∑r=0

(l +mr

)(dr

dxr(1− x)l

)(dl+m−r

dxl+m−r (1 + x)l). (K.17)

However, because (1− x)l and (1 + x)l are both l-th degree polynomials,

dr

dxr(1− x)l = 0 if r > l

anddl+m−r

dxl+m−r (1 + x)l = 0 if r < m.

Therefore, the only nonzero terms are for m ≤ r ≤ l, and

dl+m

dxl+m(1− x2)l =

l∑r=m

(l +mr

)(dr

dxr(1− x)l

)(dl+m−r

dxl+m−r (1 + x)l).

(K.18)

Now, observe the following.

d

dx(1− x)l = d(1− x)l

d(1− x)d(1− x)dx

= l(1− x)l−1(−1),

d2

dx2(1− x)l = d

dx

[d

dx(1− x)l

]= (−1)l d

dx(1− x)l−1

= (−1)l d(1− x)l−1

d(1− x)d(1− x)dx

= (−1)l(l − 1)(1− x)l−2(−1)

= (−1)2l(l − 1)(1− x)l−2,


and more generallydk

dxk(1− x)l = (−1)kl(l − 1) . . . (l − k + 1)(1− x)l−k = (−1)k l!

(l − k)!(1− x)l−k

(K.19)

Similarly, we also have

dk

dxk(1 + x)l =

l!

(l − k)!(1 + x)l−k. (K.20)

Therefore, (K.18) can be expressed as follows.

dl+m

dxl+m(1− x2)l =

l∑r=m

(l +mr

)(−1)rl!(l − r)!

(1− x)l−r l!

(r −m)!(1 + x)r−m

(K.21)

At this point, let s = r −m to obtain

dl+m

dxl+m(1− x2)l =

l−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!(1− x)l−m−s l!

s!(1 + x)s.

(K.22)

Next, we will multiply the right-hand side by (1−x2)m

(1−x2)m .

dl+m

dxl+m(1− x2)l =

l−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!(1− x)l−m−s l!

s!(1 + x)s

=(1− x2)m

(1− x2)ml−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!(1− x)l−m−s l!

s!(1 + x)s

=(1− x)m(1 + x)m

(1− x2)ml−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!

(1− x)l−m−s l!

s!(1 + x)s

=1

(1− x2)ml−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!

(1− x)l−m−s(1− x)m l!s!(1 + x)s(1 + x)m


=1

(1− x2)ml−m∑s=0

(l +mm+ s

)(−1)m+sl!

(l −m− s)!

(1− x)l−s l!s!(1 + x)m+s

=1

(1− x2)ml−m∑s=0

(l +m)!

(l − s)!(m+ s)!

(−1)m+sl!

(l −m− s)!

(1− x)l−s l!s!(1 + x)m+s

=1

(1− x2)ml−m∑s=0

(l +m)!

s!(l −m− s)!(−1)m+sl!

(l − s)!

(1− x)l−s l!

(m+ s)!(1 + x)m+s

=(l +m)!

(l −m)!

1

(1− x2)ml−m∑s=0

(l −m)!

s!(l −m− s)!(−1)m+s

[(−1)s(−1)s l!

(l − s)!(1− x)l−s

][

l!

(l − (l −m− s))!(1 + x)l−(l−m−s)

]

=(l +m)!

(l −m)!

1

(1− x2)ml−m∑s=0

(l −m)!

s!(l −m− s)!(−1)m+s

[(−1)s(−1)s l!

(l − s)!(1− x)l−s

] [l!

(m+ s)!(1 + x)m+s

]

=(l +m)!

(l −m)!

(−1)m

(1− x2)ml−m∑s=0

(l −ms

)[(−1)s l!

(l − s)!(1− x)l−s

] [l!

(m+ s)!(1 + x)m+s

]

=(l +m)!

(l −m)!

(−1)m

(1− x2)ml−m∑s=0

(l −ms

)[ds

dxs(1− x)l

] [dl−m−s

dxl−m−s (1 + x)l]

=(l +m)!

(l −m)!

(−1)m

(1− x2)mdl−m

dxl−m(1− x2)l; (K.23)


where the last equality follows from (K.17) with l +m replaced by l −m.Then, from (K.5) and (K.23), we get

Pml (x) =

(−1)l

2ll!(1− x2)m/2 d

l+m

dxl+m(1− x2)l

=(−1)l

2ll!(1− x2)m/2 (l +m)!

(l −m)!

(−1)m

(1− x2)mdl−m

dxl−m(1− x2)l

=(l +m)!

(l −m)!(−1)m (−1)l

2ll!(1− x2)−m/2 d

l−m

dxl−m(1− x2)l

=(l +m)!

(l −m)!(−1)mP−m

l (x)

=⇒ P−ml (x) = (−1)m (l −m)!

(l +m)!Pml (x) (K.24)

Appendix L

Laguerre and AssociatedLaguerre Polynomials

For a nonnegative integer m, i.e. m = 0, 1, 2, 3, . . ., Laguerre’s equation isthe linear second order ordinary differential equation

xy′′(x) + (1− x)y′(x) +my(x) = 0, (L.1)

whose polynomial solutions, denoted by Lm(x) are known as Laguerre poly-nomials. The general term is given by the Rodrigues formula

Ln(x) =ex

n!

dn

dxn

(e−xxn

), (L.2)

or the closed form1

Lk(x) =k∑i=0

(k

i

)(−1)i

i!xi or

k∑i=0

kCi(−1)i

i!xi. (L.3)

The first few polynomials are given below.

L0(x) = 1

L1(x) = −x+ 1

L2(x) =1

2(x2 − 4x+ 2)

1Note that(

ki

)and kCi mean the same thing. Depending on where you went to school,

you may be familiar with either.

333

334 APPENDIX L. LAGUERRE AND ASSOCIATED LAGUERRE POLYNOMIALS

L3(x) =1

6(−x3 + 9x2 − 18x+ 6)

L4(x) =1

24(x4 − 16x3 + 72x2 − 96x+ 24)

L5(x) =1

120(−x5 + 25x4 − 200x3 + 600x2 − 600x+ 120)

L6(x) =1

720(x6 − 36x5 + 450x4 − 2400x3 + 5400x2 − 4320x+ 720)

The Laguerre polynomials satisfy a three-term recurrence relation

(k + 1)Lk+1(x) = (2k + 1− x)Lk(x)− kLk−1(x). (L.4)

According to Favard’s theorem, the Laguerre polynomials form a family oforthogonal polynomials.

Theorem L.1 (Favard’s Theorem) If a system of polynomials Pn(x) ofdegree n = 0, 1, 2, . . . satisfies

xPn(x) = anPn+1(x) + bnPn(x) + cnPn−1(x) n ≥ 1, (L.5)xP0(x) = a0P1(x) + b0P0(x) (L.6)

where an, bn (n ≥ 0), and cn (n ≥ 1) are real constants, and ancn+1 > 0,then, there exists a positive measure µ such that the polynomials Pn(x) areorthogonal with respect to µ.2

Indeed, Lk’s are orthogonal with respect to an inner product defined by

⟨f |g⟩ =∫ ∞

0f(x)g(x)e−x dx. (L.7)

2The following holds for µ.

1. The measure µ may not be unique even if a scaling factor is ignored.

2. If µ is unique, the polynomials are dense in L2 with respect to µ.

3. If there is a µ with bounded support, then, µ is unique.

Bibliography

[Anton, 2010] Anton, H. (2010). Elementary Linear Algerbra. Wiley, Hobo-ken, NJ, 10th edition.

[Bocher, 1901] Bocher, M. (1901). Certain cases in which the vanishing ofthe wronskian is a sufficient condition for linear dependence. Transactionsof the American Mathematical Society, 2(2):139–149.

[Bostan and Dumas, 2010] Bostan, A. and Dumas, P. (2010). Wronskiansand linear independence. American Mathematical Monthly, 117(8):722–727.

[Carlip, 2013] Carlip, S. (2013). Probability current.

[Gerlach and Stern, 1922] Gerlach, V. W. and Stern, O. (1922). Der experi-mentelle nachweis der richtungsquantelung im magnetfeld. Zeitschrift furPhysik, 9:349–352.

[Gordy and Cook, 1984] Gordy, W. and Cook, R. L. (1984). Microwavemolecular spectra. Wiley, New York, 3rd edition.

[Kwa, nd] Kwa, m. (n.d.). Linear independence and the wronskian.

[Wikipedia contributors, nd] Wikipedia contributors (n.d.). Degenerate en-ergy levels.

[Reed and Simon, 1980] Reed, M. and Simon, B. (1980). Functional analysis:Revised and enlarged edition. Methods of modern mathematical physics.Acaemic Press, New York.

[Shankar, 1980] Shankar, R. (1980). Principles of quantum mechanics.Plenum, New York.

335

336 BIBLIOGRAPHY

[Zettili, 2009] Zettili, N. (2009). Quantum mechanics: Concepts and appli-cations. John Wiley & Sons, West Sussex, UK, second edition.

Answers to Exercises

This section is a work in progresswith some dummy place-holdersat this point.

Chapter 21. (a) Ω is Hermitian because (Ωt)

∗= Ω as shown below.

[ 0 11 0

]t∗

=

([0 11 0

])∗

=

[0 11 0

]

(b)[0 11 0

] [ab

]= 1 ·

[ab

]=⇒

[ba

]=

[ab

]=⇒ a = b[

0 11 0

] [ab

]= −1 ·

[ab

]=⇒

[ba

]=

[−a−b

]=⇒ a = −b

Hence, |1⟩ = 1√2

[11

], and |−1⟩ = 1√

2

[1−1

].

(c)

⟨1| − 1⟩ = 1√2

[1 1

] 1√2

[1−1

]= 0

337

338 ANSWERS TO EXERCISES

(d) Noting that

|1⟩+ |−1⟩√2

=

[10

]and

|1⟩ − |−1⟩√2

=

[01

],

we get

v =

[v1v2

]= v1

[10

]+ v2

[01

]= v1

(|1⟩+ |−1⟩√

2

)+ v2

(|1⟩ − |−1⟩√

2

)

=v1 + v2√

2|1⟩+ v1 − v2√

2|−1⟩

(e)

⟨1|Ω|1⟩ = 1√2

[1 1

] [ 0 11 0

]1√2

[11

]=

1

2

[1 1

] [ 11

]= 1

⟨1|Ω| − 1⟩ = 1√2

[1 1

] [ 0 11 0

]1√2

[1−1

]=

1

2

[1 1

] [ −11

]= 0

⟨−1|Ω|1⟩ = 1√2

[1 1

] [ 0 11 0

]1√2

[1−1

]=

1

2

[1 1

] [ −11

]= 0

⟨−1|Ω| − 1⟩ = 1√2

[1 −1

] [ 0 11 0

]1√2

[1−1

]=

1

2

[1 −1

] [ −11

]= −1

Hence,

[Ωij] =

[1 00 −1

].

2. (a) As M is Hermitian, we have to have M t∗ =M . So,[ 6 4

α 0

]t∗

=

[6 α4 0

]∗

=

[6 α∗

4 0

]=

[6 4α 0

]=⇒ α = 4.

CHAPTER 2 339

(b) [6 44 0

] [ab

]=

[6a+ 4b4a

]= 8

[ab

]=⇒ a = 2b =⇒ |8⟩ = 1√

5

[21

][6 44 0

] [ab

]=

[6a+ 4b4a

]= −2

[ab

]=⇒ b = −2a =⇒ |−2⟩ = 1√

5

[1−2

]

(c)

M11 =1√5

[2 1

] [ 6 44 0

]1√5

[21

]=

1

5

[2 1

] [ 168

]=

1

5· 40 = 8

M12 =1√5

[2 1

] [ 6 44 0

]1√5

[1−2

]=

1

5

[2 1

] [ −24

]= 0

M21 =1√5

[1 −2

] [ 6 44 0

]1√5

[21

]=

1

5

[1 −2

] [ −24

]= 0

M22 =1√5

[1 −2

] [ 6 44 0

]1√5

[1−2

]=

1

5

[1 −2

] [ −24

]= −2

Therefore,

M = [Mij] =

[8 00 −2

].

3. (a)

[A,B] = AB −BA =

[0 21 1

] [−1 21 0

]−[−1 21 0

] [0 21 1

]

=

[2 00 2

]−[2 00 2

]= 0

(b) [0 21 1

] [ab

]= 2

[ab

]=⇒

[2ba+ b

]=

[2a2b

]=⇒ a = b

So,

|2⟩A =1√2

[11

].


[0 21 1

] [cd

]= −1

[cd

]=⇒

[2dc+ d

]=

[−c−d

]=⇒ c = −2d

So,

|−1⟩A =1√5

[2−1

]. (12.8)

(c)

B |2⟩A =

[−1 21 0

]1√2

[11

]= 1 · 1√

2

[11

]

B |−1⟩A =

[−1 21 0

]1√5

[2−1

]=

1√5

[−42

]= −2 · 1√

5

[2−1

]

Therefore, |2⟩A is an eigenvector of B with the associated eigen-value 1, and |−1⟩ is an eigenvector of B with the associated eigen-value −2.

(d)√5 |−1⟩A +

√2 |2⟩A

3=

1

3

([2−1

]+

[11

])=

1

3

[30

]=

[10

]√5 |−1⟩A − 2

√2 |2⟩A

−3=

1

−3

([2−1

]−[22

])=

1

−3

[0−3

]=

[01

]

Therefore, |2⟩A , |−1⟩A indeed forms a basis for C2.(e) Remembering that Ωij = ⟨i|Ω|j⟩, we can compute the entries as

follows.

A11 =A ⟨2|A|2⟩A =[

1√2

1√2

] [ 0 21 1

] [ 1√2

1√2

]

=[

1√2

1√2

] [ √2√2

]= 2; where A ⟨2| is the adjoint of |2⟩A

A12 =[

1√2

1√2

] [ 0 21 1

] [ 2√5

−1√5

]=[

1√2

1√2

] [ −2√5

1√5

]=−1√10

A21 =[

2√5

−1√5

] [ 0 21 1

] [ 1√2

1√2

]=[

2√5

−1√5

] [ √2√2

]=

√2

5

CHAPTER 2 341

A22 =[

2√5

−1√5

] [ 0 21 1

] [ 2√5

−1√5

]=[

2√5

−1√5

] [ −2√5

1√5

]= −1

B11 =[

1√2

1√2

] [ −1 21 0

] [ 1√2

1√2

]=[

1√2

1√2

] [ 1√2

1√2

]= 1

B12 =[

1√2

1√2

] [ −1 21 0

] [ 2√5

−1√5

]=[

1√2

1√2

] [ −4√5

2√5

]=−2√10

B21 =[

2√5

−1√5

] [ −1 21 0

] [ 1√2

1√2

]=[

2√5

−1√5

] [ 1√2

1√2

]=

1√10

B22 =[

2√5

−1√5

] [ −1 21 0

] [ 2√5

−1√5

]=[

2√5

−1√5

] [ −4√5

2√5

]= −2

4. (a) As M = I −N , [M,N ] = [I −N,N ] = 0 is obvious.(b)

det

(λI −

[0 11 0

])= det

[λ −1−1 λ

]= λ2 − 1 = (λ+ 1)(λ− 1) = 0

=⇒ λ = ±1

So, the eigenvalues of N are ±1.[0 11 0

] [ab

]=

[ba

]= ±1 ·

[ab

]=⇒ a = ±b

=⇒ |1⟩N =1√2

[11

]and |−1⟩N =

1√2

[1−1

]As we already know M and N commute, they share the sameeigenvectors. [

1 −1−1 1

] [11

]=

[00

]= 0 ·

[11

][

1 −1−1 1

] [1−1

]=

[2−2

]= 2 ·

[1−1

]So, the eigenvalues of M are 0 and 2, and corresponding normal-ized eigenvectors are

|0⟩M =1√2

[11

]and |2⟩M =

1√2

[1−1

].


(c) ( 1√2

[1 1−1 1

])t∗ [1 −1−1 1

](1√2

[1 1−1 1

])

=1

2

[1 −11 1

] [1 −1−1 1

] [1 1−1 1

]

=1

2

[1 −11 1

] [2 0−2 0

]=

1

2

[4 00 0

]=

[2 00 0

]and( 1√

2

[1 1−1 1

])t∗ [0 11 0

](1√2

[1 1−1 1

])

=1

2

[1 −11 1

] [0 11 0

] [1 1−1 1

]

=1

2

[1 −11 1

] [−1 11 1

]=

1

2

[−2 00 2

]=

[−1 00 1

]

Chapter 31. [3 pts]

(a) 0.4772 [1 pt](b) 0.5− 0.4495 = 0.0505 [1 pt](c) P (|z| > 1.96) = 2× (0.5− 0.4750) = 0.05 [1 pt]

Chapter 51.

Ψ(x+ vt0, t+ t0) = A sin[2π

λ(x+ vt0 − v(t+ t0))

]= A sin

[2π

λ(x+ vt0 − vt− vt0)

]= A sin

[2π

λ(x− vt)

]= Ψ(x, t)

Ψ(x, t) is the displacement at location x at time t. As the speed is v,this moves to the position “x+ vt0” time “t0” later or at time “t+ t0”.

CHAPTER 5 343

2.− ~

2

2m

∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂t

3. We have

− ~2

2m

∂2Ψ1(x, t)

∂x2+ V (x, t)Ψ1(x, t) = i~

∂Ψ1(x, t)

∂tand

− ~2

2m

∂2Ψ2(x, t)

∂x2+ V (x, t)Ψ2(x, t) = i~

∂Ψ2(x, t)

∂t.

Noting that ∂2

∂x2, multiplication by V (x, t), and ∂

∂tare all linear such

that

∂2

∂x2[αΨ1(x, t) + βΨ2(x, t)] = α

∂2

∂x2Ψ1(x, t) + β

∂2

∂x2Ψ2(x, t)

V (x, t) [αΨ1(x, t) + βΨ2(x, t)] = αV (x, t)Ψ1(x, t) + βV (x, t)Ψ2(x, t)

∂

∂t[αΨ1(x, t) + βΨ2(x, t)] = α

∂

∂tΨ1(x, t) + β

∂

∂tΨ2(x, t),

we get

− ~2

2m

∂2 [αΨ1(x, t) + βΨ2(x, t)]

∂x2+ V (x, t) [αΨ1(x, t) + βΨ2(x, t)]

= − ~2

2m

[α∂2

∂x2Ψ1(x, t) + β

∂2

∂x2Ψ2(x, t)

]+ αV (x, t)Ψ1(x, t) + βV (x, t)Ψ2(x, t)

= α

[− ~

2

2m

∂2

∂x2Ψ1(x, t) + V (x, t)Ψ1(x, t)

]

+ β

[− ~

2

2m

∂2

∂x2Ψ2(x, t) + V (x, t)Ψ2(x, t)

]

= α

[i~∂Ψ1(x, t)

∂t

]+ β

[i~∂Ψ2(x, t)

∂t

]

= i~∂ [αΨ1(x, t) + βΨ2(x, t)]

∂t

=⇒ − ~2

2m

∂2 [αΨ1(x, t) + βΨ2(x, t)]

∂x2+ V (x, t) [αΨ1(x, t) + βΨ2(x, t)]


= i~∂ [αΨ1(x, t) + βΨ2(x, t)]

∂t.

Therefore, αΨ1(x, t) + βΨ2(x, t) is also a solution.

4. (a) eiθeiϕ = (cos θ+i sin θ)(cosϕ+i sinϕ) = (cos θ cosϕ−sin θ sinϕ)+i(sin θ cosϕ+ cos θ sinϕ) = cos(θ + ϕ) + i sin(θ + ϕ) = ei(θ+ϕ)

(b) Our proof is by mathematical induction.Consider (cos θ + i sin θ)k. For k = 1, we have (cos θ + i sin θ)1 =ei·1·θ = eiθ. Hence, (cos θ + i sin θ)k = eikθ holds fro k = 1.Now suppose (cos θ+ i sin θ)k = eikθ = cos(kθ)+ i sin(kθ) for somek. Then, (cos θ + i sin θ)k+1 = (cos θ + i sin θ)k(cos θ + i sin θ) =[cos(kθ)+ i sin(kθ)](cos θ+ i sin θ) = cos(kθ) cos θ− sin(kθ) sin θ+i[sin(kθ) cos θ + sin θ cos(kθ)] = cos[(k + 1)θ] + i sin[(k + 1)θ] =ei(k+1)θ.Therefore,

(eiθ)n

= einθ by induction.

5. (a) ∫ +a2

− a2

Ψ∗Ψ dx =∫ +a

2

− a2

A∗ cos(πx

a

)e+iEt/~A cos

(πx

a

)e−iEt/~ dx

=∫ +a

2

− a2

A∗A cos2(πx

a

)dx = |A|2

∫ +a2

− a2

1 + cos(2πxa

)2

dx

= |A|2[x

2+

a

2 · 2πsin

(2πx

a

)]+a2

− a2

= |A|2[a

4+

a

4πsinπ − −a

4− a

4πsin(−π)

]= |A|2a

2= 1

=⇒ A = eiθ√2

a

(b)

− ~2

2m

∂2

∂x2

[A cos

(πx

a

)e−iEt/~

]= i~

∂[A cos

(πxa

)e−iEt/~

]∂t

=⇒ − ~2

2mA

[−(π

a

)2]

cos(πx

a

)e−iEt/~

= i~A cos(πx

a

)(−iE~

)e−iEt/~

CHAPTER 5 345

=⇒ − ~2

2m

[−(π

a

)2]= i~

(−iE~

)= E =⇒ E =

π2~2

2ma2(12.9)

(c)

⟨P ⟩ =∫ +a

2

− a2

A∗ cos(πx

a

)e+iEt/~

(−i~ ∂

∂x

)A cos

(πx

a

)e−iEt/~ dx

=2

a(−i~)

∫ +a2

− a2

cos(πx

a

) −πa

sin(πx

a

)dx

=iπ~

a2

∫ +a2

− a2

2 sin(πx

a

)cos

(πx

a

)dx

=iπ~

a2

∫ +a2

− a2

sin(2πx

a

)dx =

iπ~

a2

[−a2π

cos(2πx

a

)]+a2

− a2

=iπ~

a2−a2π

[cos π − cos (−π)] = iπ~

a2−a2π

[−1− (−1)] = 0

(d)

⟨x2⟩ =∫ +a

2

− a2

A∗ cos(πx

a

)e+iEt/~x2A cos

(πx

a

)e−iEt/~ dx

=2

a

∫ +a2

− a2

x2 cos2(πx

a

)dx

Let y =π

ax, so that dx =

a

πdy.

=2

a

∫ +π2

− π2

(a

π

)2

y2 cos2 y aπdy =

2

a

(a

π

)3 ∫ +π2

− π2

y2 cos2 y dy

=2

a

(a

π

)3[y3

6+

(y2

4− 1

8

)sin 2y +

y cos 2y4

]+π2

− π2

=2

a

(a

π

)3[y3

6+y cos 2y

4

]+π2

− π2

=2

a

(a

π

)3

2(π2

)36

+ 2π2

cosπ4

=2a2

π3

[π3

24− π

4

]

=a2

12

(1− 6

π2

);


where we used the formula∫y2 cos2 y dy =

y3

6+

(y2

4− 1

8

)sin 2y +

y cos 2y4

+ C.

(e)

⟨P 2⟩ =∫ +a

2

− a2

A∗ cos(πx

a

)e+iEt/~

(−i~ ∂

∂x

)2

A cos(πx

a

)e−iEt/~ dx

= −~2[−(π

a

)2] ∫ +a

2

− a2

A∗ cos(πx

a

)e+iEt/~A cos

(πx

a

)e−iEt/~ dx.

But, the normalization condition gives∫ +a2

− a2

A∗ cos(πx

a

)e+iEt/~A cos

(πx

a

)e−iEt/~ dx = 1.

Therefore, we get

⟨P 2⟩ =(π~

a

)2

.

6. (a) As eiθA for any real number θ can replace A, the wavefunction isnot unique in that sense.

(b) The normalization constant eiθA enters into the computations ofΨ∗(x, t)Ψ(x, t), ⟨P ⟩, ⟨P 2⟩, and ⟨x2⟩ as

(eiθA

)∗ (eiθA

). However,(

eiθA)∗ (

eiθA)= e−iθA∗eiθA = A∗A, and the value of θ does not

affect these physical observables.

Chapter 61.

− ~2

2m

d2

dx2ψ(x) + V (x)ψ(x) = Eψ(x)

2. Plugging V (x, t) = V (x) and Ψ(x, t) = ψ(x)e−iEt/~ into the time-dependent Schrodinger Equation, we get

− ~2

2m

∂2Ψ(x, t)

∂x2+V (x, t)Ψ(x, t) = i~

∂Ψ(x, t)

∂t

CHAPTER 7 347

=⇒ − ~2

2m

∂2ψ(x)e−iEt/~

∂x2+V (x)ψ(x)e−iEt/~ = i~

∂ψ(x)e−iEt/~

∂t

=⇒ − ~2

2me−iEt/~∂

2ψ(x)

∂x2+V (x)ψ(x)e−iEt/~ = i~(−iE/~)e−iEt/~ψ(x)

=⇒ − ~2

2m

∂2ψ(x)

∂x2+V (x)ψ(x) = Eψ(x).

3. The first derivative of sinϕ is cosϕ. Both sinϕ and cosϕ are continuous,and sin2 ϕ is obviuosly integrable over the interval [0, 2π]. Furthermore,sin(ϕ+2mπ) = sinϕ and cos(ϕ+2mπ) = cosϕ for all integersm. Hence,the condition of single-valuedness is also satisfied.On the other hand, if ψ(ϕ) = ϕ, we have ψ(0) = 0 and ψ(2π) = 2π ,0 = ψ(0) despite the fact that ϕ = 0 and ϕ = 2π signify the sameposition in space. Hence, this function is not single-valued and cannotbe a wavefunction as such.

Chapter 71. Same as Chapter 5 Problem 5. So, see the answers to Chapter

5 Problem 5.

2. Same as Chapter 5 Problem 6. So, see the answers to Chapter5 Problem 6.

3.

− ~2

2m

d2

dx2ψ(x) = Eψ(x) =⇒ d2

dx2ψ(x) = −2mE

~2ψ(x) =⇒ ψ(x) = Ae±i

√2mE~

x

4. Let k =√2mE~

. Then, the general solution is

ψ(x) = C sin kx+D cos kx.

Taking the boundary conditions into account,

ψ(±a2

)= 0 =⇒

C sin ka2+D cos ka

2= 0

−C sin ka2+D cos ka

2= 0

=⇒D cos ka

2= 0

C sin ka2= 0

.


Because ka2> 0, we have

D cos ka2

= 0⇐⇒ D = 0 or ka

2=(n+

1

2

)π for n = 0, 1, 2, . . .

and

C sin ka2

= 0⇐⇒ C = 0 or ka

2= nπ for n = 1, 2, 3, . . . .

Note that C = D = 0 leads to a zero wavefunction, and ka2=(n+ 1

2

)π

and ka2= nπ cannot hold simultanepously

Hence, we have either

ka

2=(n+

1

2

)π (n = 0, 1, 2, . . .) and C = 0

orka

2= nπ (n = 1, 2, 3, . . .) and D = 0.

When, C = n = 0, k = πa

and

ψ(x) = D cos πxa.

When D = 0 and n = 1, k = 2πa

and

ψ(x) = C sin 2πx

a.

Let us now calculate the real and positive normalization constants.For D,∫ +a

2

− a2

ψ∗(x)ψ(x) dx =∫ +a

2

− a2

|D|2 cos2 πxadx = |D|2

∫ +a2

− a2

1 + cos 2πxa

2dx

=|D|2

2

[x+

a

2πsin 2πx

a

]+a2

− a2

=|D|2

2· a = 1 =⇒ D =

√2

a.

For C,∫ +a

2

− a2

ψ∗(x)ψ(x) dx =∫ +a

2

− a2

|C|2 sin2 2πx

adx = |C|2

∫ +a2

− a2

1− cos 4πxa

2dx

CHAPTER 7 349

=|C|2

2

[x− a

4πsin 4πx

a

]+a2

− a2

=|C|2

2· a = 1 =⇒ C =

√2

a.

So, the functions we have found are

ψ1(x) =

√2

acos πx

aand ψ2(x) =

√2

asin 2πx

a.

5. Because H(x) = K(t) for all (x, t) ∈ R2 := R × R, H(0) = K(t) andH(x) = K(0) for all (x, t) ∈ R2. Combining this with H(x) = K(t), weget H(0) = K(0), which in turn implies H(x) = K(t) = H(0)(= K(0))for all (x, t) ∈ R2.

6. Here is a summary of the full solution given in Section 7.2. TheSchrodinger equations are

−~2

2m

d2ψ(x)

dx2= Eψ(x) x ≤ 0

−~2

2m

d2ψ(x)

dx2+ V0ψ(x) = Eψ(x) x > 0

with the corresponding general solutions

ψ−(x) = Aeik1x +Be−ik1x where k1 =√2mE

~and x < 0

and

ψ+(x) = Cek2x +De−k2x where k2 =

√2m(V0 − E)~

and x > 0.

As ψ+ has to be normalizable/integrable in the region x > 0, we setC = 0 to obtain

ψ+(x) = De−k2x where k2 =

√2m(V0 − E)~

and x > 0.

Now use two continuity conditions

ψ+(0) = ψ−(0) and ψ′+(0) = ψ′

−(0)


to obtain ik2k1D = A−BD = A+B

with the solution

A = 12

(1 + ik2

k1

)D

B = 12

(1− ik2

k1

)D.

This gives

ψ(x) =

D2(1 + ik2/k1)e

ik1x + D2(1− ik2/k1)e−ik1x x ≤ 0

De−k2x x > 0.

7.

8.

9.

10. (a)

− ~2

2m

d2ψ(x)

dx2= Eψ(x) x < 0

− ~2

2m

d2ψ(x)

dx2+ V0ψ(x) = Eψ(x) x > 0

(b) Continuity of the wavefunction ψ(x) at x = 0 gives

Aeik1x +Be−ik1x = Ceik2x (at x = 0) or A+B = C.

Continuity of the first derivative dψ(x)dx

gives

ik1Aeik1x−ik1Be−ik1x = ik2Ce

ik2x (at x = 0) or k1(A−B) = k2C.

The rest is Arithmetic, and we get

C =2k1

k1 + k2A.

(c)

R =B∗B

A∗A=

(k1 − k2k1 + k2

)2

CHAPTER 7 351

11.

12. (a) −~22m

d2ψ(x)dx2

= Eψ(x) x < 0−~22m

d2ψ(x)dx2

+ V0ψ(x) = Eψ(x) x > 0

(b) Aeik1x +Be−ik1x x < 0

Ceik2x +De−ik2x x > 0

(c) A = 0

(d) ψ(x)|0− = ψ(x)|0+ψ′(x)|0− = ψ′(x)|0+

=⇒

B = C +D−ik1B = ik2(C −D)

=⇒C +D = BC −D = −k1

k2B

=⇒C = k2−k1

2k2B

D = k2+k12k2

B.

(e)

R =C∗C

D∗D=

(k2−k12k2

)2B∗B(

k2+k12k2

)2B∗B

=

(k1 − k2k1 + k2

)2

13. (a)

− ~2

2m

d2ψ(x)

dx2= Eψ(x) x < −a and x > 0

− ~2

2m

d2ψ(x)

dx2+ V0ψ(x) = Eψ(x) − a < x < 0

(b) i. Since there is nothing that reflects the wave in the regionx > 0, there should not be a wave traveling to the left inthis region. This means H = 0 as e−ik1xe−iωt = e−i(kx+ωt)

represents a wave traveling toward the left.ii. Continuity of the wavefunction ψ(x) at x = 0 gives

Cek2x +De−k2x = Geik1x (at x = 0) =⇒ C +D = G.

Continuity of the first derivative dψ(x)dx

at x = 0 gives

k2Cek2x−k2De−k2x = ik1Ge

ik1x (at x = 0) =⇒ k2(C−D) = ik1G.


iii. In the region x > 0, ψ∗(x)ψ(x) = G∗e−ik1xGeik1x = |G|2, anddψ∗(x)ψ(x)

dx

∣∣∣0+

= 0. Hence, we also need dψ∗(x)ψ(x)dx

∣∣∣0−

= 0.Suppose C = 0. Then, we have ψ(x) = De−k2x in the region−a < x < 0. This gives dψ∗(x)ψ(x)

dx

∣∣∣0−

= dD∗De−2k2x

dx

∣∣∣0−

=

(−2k2)|D|2e−2k2x∣∣∣0−

= −2k2|D|2 = 0 =⇒ D = 0, and both C

and D are zero. Similarly, D = 0 forces C to be zero.If ψ(x) is zero inside the barrier, the continuity conditions onψ(x) and dψ(x)

dxat x = −a, 0 implies ψ(x) = 0 everywhere,

which is a physically meaningless solution. Therefore, neitherC nor D can be zero.

14.

15. (a)∫ +a/2

−a/2ψ∗ψdx = 2|A|2

∫ a/2

0sin2 kx dx = 1 =⇒ A = ±

√2

ai

=⇒ ψ2(x) = ±√2

ai sin 2πx

a(one of these)

(b) ∫ +a/2

−a/2Ψ∗xΨ dx =

∫ +a/2

−a/2A∗

2 sin (k2x)eiE2t/~·x·A2 sin (k2x)e

−iE2t/~ dx

= |A2|2∫ +a/2

−a/2x sin2 (k2x) dx = 0 (We have an odd function.)

(c)

−~2

2m

d2

dx2A2 sin (k2x) =

2π2~2

ma2A2 sin (k2x) =⇒ E2 =

2π2~2

ma2

16. (a) − ~22m

d2ψ(x)dx2

= Eψ(x)

(b) i. d2ψ(x)dx2

= ddx

(AB cos(Bx)) = −AB2 sin(Bx)=⇒ − ~2

2m· (−AB2) sin(Bx) = AE sin(Bx)

=⇒ B2 = 2mE~2

=⇒ B =√2mE~

CHAPTER 7 353

ii. sin(±a

2B)= 0 =⇒ a

2B = π =⇒ B = 2π

a

iii. B2 = 4π2

a2= 2mE

~2=⇒ E = 2π2~2

ma2

iv.∫ a/2

−a/2A sin(Bx) · A sin(Bx)dx = 1 =⇒ A =√2/a

17. (a) − ~22m

d2ψ(x)dx2

= Eψ(x)

(b) i. ψ(0) = A sin 0 +B cos 0 = B = 0

ii. ψ(L) = A sin kL = 0 =⇒ kL = nπ =⇒ k = nπL

iii.∫ L0 ψ

∗ψdx =∫ L0 |A|2 sin2 knxdx =

[x2− sin(2knx)

4kn

]L0= |A|2L

2=

1 =⇒ An =√

2L

(c) kn = nπL

=√2mEn

~=⇒ En =

(~nπL

)212m

= n2h2

8mL2

18.df(x)

dx=

∞∑j=1

jajxj−1 =

∞∑j=0

(j + 1)aj+1xj

=⇒ df(x)

dx+ f(x) =

∞∑j=0

[aj + (j + 1)aj+1]xj = 3x2 + 8x+ 3

=⇒ aj + (j + 1)aj+1 = 0 for j ≥ 3.

19.

20. The full wave function is ψ0e−iE0t/~.

⟨x⟩ =∫ +∞

−∞(ψ0e

−iE0t/~)∗x(ψ0e−iE0t/~)dx

=∫ +∞

−∞ψ∗0xψ0dx

=∫ +∞

−∞xA∗

0e−u2/2A0e

−u2/2dx

= |A0|2∫ +∞

−∞xe−[(Cm)1/2/~]x2︸︷︷︸

an odd functiondx

= 0

21. (a) The full ground state wavefuntion Ψ0 is given by Ψ0 = ψ0e−iEt/~.

Therefore,

< x >=∫ +∞

−∞Ψ∗

0xΨ0 dx =∫ +∞

−∞ψ∗0eiEt/~xψ0e

−iEt/~ dx =∫ +∞

−∞ψ∗0xψ0 dx


∫ +∞

−∞A∗

0e−u2/2xA0e

−u2/2 dx = A∗0A0

∫ +∞

−∞xe−u2 dx

= A∗0A0

∫ +∞

−∞xe−[(Cm)1/2~−1]x2 dx = 0

as the integrand is an odd function.(b)

∫ +∞

−∞ψ∗0ψ1 dx =

∫ +∞

−∞A∗

0e−u2/2A1ue

−u2/2 dx = A∗0A1

∫ +∞

−∞ue−u2 dx

= A∗0A1

1√α

∫ +∞

−∞ue−u2 du = 0

because ue−u2 is an odd function.(c) Because the Hamiltonian is a Hermitian operator, its eigenvectors/

functions/kets associated with different eigenvalues are orthogonalto each other.

(d) Vibration of diatomic molecules can be approximated by the sim-ple harmonic oscillator. As the lowest allowed energy is strictlypositive, there is always some residual motion. This nonzero ki-netic energy prevents the system from reaching zero degrees kelvin.

22.

23. Some examples are as follows.

• Energy levels are continuous in classical mechanics, while they arediscrete in quantum mechanics.• There is barrier penetration only in quantum mechanics.• The position of the particle is probabilistically determined in quan-

tum mechanics. But, it takes one unique value in classical mechan-ics.

Chapter 81.

CHAPTER 9 355

Chapter 91. Proceeding as in (9.12),

[Lz, Lx] = [xpy − ypx, ypz − zpy]= [xpy, ypz]− [xpy, zpy]− [ypx, ypz] + [ypx, zpy]

= x[py, y]pz + y[x, pz]py + yx[py, pz] + [x, y]pypz

− (x[py, z]py + z[x, py]py + zx[py, py] + [x, z]pypy)

− (y[px, y]pz + y[y, pz]px + yy[px, pz] + [y, y]pxpz)

+ y[px, z]py + z[y, py]px + zy[px, py] + [y, z]pxpy

= x(−i~I)pz + y · 0 · py + yx · 0 + 0 · pypz− (x · 0 · py + z · 0 · py + zx · 0 + 0 · pypy)− (y · 0 · pz + y · 0 · px + yy · 0 + 0 · pxpz)+ y · 0 · py + z(i~I)px + zy · 0 + 0 · pxpy

= i~(zpx − xpz) = i~Ly.

In spherical coordinates as in (9.11),

[Lz, Lx] = LzLx − LxLz= −i~∂ϕi~ (sinϕ∂θ + cot θ cosϕ∂ϕ)− i~ (sinϕ∂θ + cot θ cosϕ∂ϕ) (−i~)∂ϕ

= −(i~)2(cosϕ∂θ + sinϕ∂ϕ∂θ − cot θ sinϕ∂ϕ + cot θ cosϕ∂2ϕ

)+ (i~)2

(sinϕ∂θ∂ϕ + cot θ cosϕ∂2ϕ

)= i~ [i~ (− cosϕ∂θ + cot θ sinϕ∂ϕ)]

= i~

[i~

(− cosϕ ∂


∂ϕ

)]= i~Ly.

Chapter 101.

2.

3.


4. (a)l = 0, 1, 2

(b)ml = 0 for l = 0,

ml = −1, 0, 1 for l = 1,

ml = −2,−1, 0, 1, 2 for l = 2

(c) 9 dgenerate states

5.

E3−E1 = −µe4

(4πε0)22~232−(− µe4

(4πε0)22~212

)=

8µe4

(4πε0)22~29=

µe4

36(πε0)2~2

6. (a) eik(2πm) = eik(2πn) for all (m,n). Consider (m,n) = (1, 0). Then,we have

|eik2π| = |ei(a+bi)2π| = |eia2π|e−b2π = 1 =⇒ b = 0.

Now consider m = n+ 1.

eik2π = 1 =⇒ k = 0,±1,±2, · · · (necessary condition).

But, this is also sufficient.(b) For each n, l = 0, 1, 2, · · · , n− 1, and ml = −l,−l + 1, · · · , l − 1, l︸︷︷︸

2l+1

n−1∑l=0

(2l + 1) = 2n−1∑l=0

l +n−1∑l=0

1 = 2 · n(n− 1)

2+ n = n2 Q.E.D.

7. (a) i. l = 3

ii. ∥L∥ =√3(3 + 1)~ = 2

√3~

iii. 2l + 1 = 2 · 3 + 1 = 7

iv. 3~

CHAPTER 10 357

(b) i.

P (0 ≤ r < a0) =

a0∫0

π∫0

2π∫0

(1√π

(1

a0

)3/2

e−r/a0

)∗

(1√π

(1

a0

)3/2

e−r/a0

)r2dr sin θdθdϕ

=1

π

1

a30

a0∫0

π∫0

2π∫0

e−2r/a0r2dr sin θdθdϕ =1

π

1

a30(4π)

a0∫0

r2e(−2/a0)r dr

=4

a30

e−2r/a0

[r2

−2/a0− 2r

(−2/a0)2+

2

(−2/a0)3

]a00

=4

a30

e−2

[−1

2a30 −

1

2a30 −

1

4a30

]−[−1

4a30

]= 4e−2

[−2− 2− 1

4

]+ 1 = 1− 5e−2

ii.

< r >=

∞∫0

π∫0

2π∫0

(1√π

(1

a0

)3/2

e−r/a0

)∗

r

(1√π

(1

a0

)3/2

e−r/a0

)r2dr sin θdθdϕ

=1

πa30

∞∫0

π∫0

2π∫0

e−2r/a0r3dr sin θdθdϕ =1

πa30(4π)

∞∫0

r3e(−2/a0)r dr

=4

a30

e−2r/a0

[r3

−2/a0− 3r2

(−2/a0)2+

6r

(−2/a0)3− 6

(−2/a0)4

]∞

0

=4

a30· 1 · 6a

40

4 · 4=

3

2a0

(c)

EZ,n = − Z2µe4

(4πε0)22~2n2= −

(Z2µe4

32π2ε20~2

)1

n2=⇒ En,3 = −

(9µe4

32π2ε20~2

)1

n2

Hence,

E3,3 − E1,3 = −(

9µe4

32π2ε20~2

)(1

9− 1

)= −

(9µe4

32π2ε20~2

)(−8

9

)=

µe4

4π2ε20~2


Chapter 111. (a) The probability of the first device getting an up-spin is

(1√2

)2=

0.5. After the measurement, the state is

|↑⟩ = 1√2|↑⟩y +

1√2|↓⟩y .

Hence, the probability that the second device measures a down-spin in the y-direction is also

(1√2

)2= 0.5. Therefore, the answer

is 0.5× 0.5 = 0.25.(b) The final spin state is |↓⟩y due to the fundamental postulates of

quantum mechanics.

Chapter 121.

Subject Index

This is a work in progress. A very slow progress it is.

EErwin Schrodinger . . . . . . . . . . . . 99

NNewtonian Mechanics . . . . . . . . . 11

Ttraveling wave . . . . . . . . . . . . . . . . 99

359

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