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Introduction to Quantum Mechanics
Third edition

Changes and additions to the new edition of this classic textbook include:
•
•
•
•
•
•

A new chapter on Symmetries and Conservation Laws
New problems and examples
Improved explanations
More numerical problems to be worked on a computer
New applications to solid state physics
Consolidated treatment of time-dependent potentials

David J. Griffiths received his BA (1964) and PhD (1970) from Harvard University. He taught
at Hampshire College, Mount Holyoke College, and Trinity College before joining the faculty
at Reed College in 1978. In 2001–2002 he was visiting Professor of Physics at the Five Colleges (UMass, Amherst, Mount Holyoke, Smith, and Hampshire), and in the spring of 2007 he
taught Electrodynamics at Stanford. Although his PhD was in elementary particle theory, most
of his research is in electrodynamics and quantum mechanics. He is the author of over fifty articles and four books: Introduction to Electrodynamics (4th edition, Cambridge University Press,
2013), Introduction to Elementary Particles (2nd edition, Wiley-VCH, 2008), Introduction to
Quantum Mechanics (2nd edition, Cambridge, 2005), and Revolutions in Twentieth-Century
Physics (Cambridge, 2013).
Darrell F. Schroeter is a condensed matter theorist. He received his BA (1995) from Reed
College and his PhD (2002) from Stanford University where he was a National Science Foundation Graduate Research Fellow. Before joining the Reed College faculty in 2007, Schroeter
taught at both Swarthmore College and Occidental College. His record of successful theoretical research with undergraduate students was recognized in 2011 when he was named as a
KITP-Anacapa scholar.

INTRODUCTION TO

Quantum Mechanics
Third edition
DAVID J. GRIFFITHS and DARRELL F. SCHROETER
Reed College, Oregon

i

Preface
Unlike Newton’s mechanics, or Maxwell’s electrodynamics, or Einstein’s relativity, quantum theory was not
created—or even definitively packaged—by one individual, and it retains to this day some of the scars of;  its
exhilarating but traumatic youth. There is no general consensus as to what its fundamental principles are, how
it should be taught, or what it really “means.” Every competent physicist can “do” quantum mechanics, but the
stories we tell ourselves about what we are doing are as various as the tales of Scheherazade, and almost as
implausible. Niels Bohr said, “If you are not confused by quantum physics then you haven’t really understood
it”; Richard Feynman remarked, “I think I can safely say that nobody understands quantum mechanics.”
The purpose of this book is to teach you how to do quantum mechanics. Apart from some essential
background in Chapter 1, the deeper quasi-philosophical questions are saved for the end. We do not believe
one can intelligently discuss what quantum mechanics means until one has a firm sense of what quantum
mechanics does. But if you absolutely cannot wait, by all means read the Afterword immediately after finishing
Chapter 1.

Not only is quantum theory conceptually rich, it is also technically difficult, and exact solutions to all but
the most artificial textbook examples are few and far between. It is therefore essential to develop special
techniques for attacking more realistic problems. Accordingly, this book is divided into two parts;1 Part I

covers the basic theory, and Part II assembles an arsenal of approximation schemes, with illustrative
applications. Although it is important to keep the two parts logically separate, it is not necessary to study the
material in the order presented here. Some instructors, for example, may wish to treat time-independent
perturbation theory right after Chapter 2.

This book is intended for a one-semester or one-year course at the junior or senior level. A one-semester
course will have to concentrate mainly on Part I; a full-year course should have room for supplementary
material beyond Part II. The reader must be familiar with the rudiments of linear algebra (as summarized in

the Appendix), complex numbers, and calculus up through partial derivatives; some acquaintance with Fourier
analysis and the Dirac delta function would help. Elementary classical mechanics is essential, of course, and a
little electrodynamics would be useful in places. As always, the more physics and math you know the easier it
will be, and the more you will get out of your study. But quantum mechanics is not something that flows
smoothly and naturally from earlier theories. On the contrary, it represents an abrupt and revolutionary
departure from classical ideas, calling forth a wholly new and radically counterintuitive way of thinking about
the world. That, indeed, is what makes it such a fascinating subject.
At first glance, this book may strike you as forbiddingly mathematical. We encounter Legendre,
Hermite, and Laguerre polynomials, spherical harmonics, Bessel, Neumann, and Hankel functions, Airy
functions, and even the Riemann zeta function—not to mention Fourier transforms, Hilbert spaces, hermitian
operators, and Clebsch–Gordan coefficients. Is all this baggage really necessary? Perhaps not, but physics is
like carpentry: Using the right tool makes the job easier, not more difficult, and teaching quantum mechanics
without the appropriate mathematical equipment is like having a tooth extracted with a pair of pliers—it’s
possible, but painful. (On the other hand, it can be tedious and diverting if the instructor feels obliged to give
elaborate lessons on the proper use of each tool. Our instinct is to hand the students shovels and tell them to

11

ii

start digging. They may develop blisters at first, but we still think this is the most efficient and exciting way to
learn.) At any rate, we can assure you that there is no deep mathematics in this book, and if you run into
something unfamiliar, and you don’t find our explanation adequate, by all means ask someone about it, or look
it up. There are many good books on mathematical methods—we particularly recommend Mary Boas,
Mathematical Methods in the Physical Sciences, 3rd edn, Wiley, New York (2006), or George Arfken and HansJurgen Weber, Mathematical Methods for Physicists, 7th edn, Academic Press, Orlando (2013). But whatever
you do, don’t let the mathematics—which, for us, is only a tool—obscure the physics.
Several readers have noted that there are fewer worked examples in this book than is customary, and that
some important material is relegated to the problems. This is no accident. We don’t believe you can learn
quantum mechanics without doing many exercises for yourself. Instructors should of course go over as many
problems in class as time allows, but students should be warned that this is not a subject about which anyone
has natural intuitions—you’re developing a whole new set of muscles here, and there is simply no substitute
for calisthenics. Mark Semon suggested that we offer a “Michelin Guide” to the problems, with varying
numbers of stars to indicate the level of difficulty and importance. This seemed like a good idea (though, like
the quality of a restaurant, the significance of a problem is partly a matter of taste); we have adopted the
following rating scheme:
an essential problem that every reader should study;

a somewhat more difficult or peripheral problem;
an unusually challenging problem, that may take over an hour.
(No stars at all means fast food: OK if you’re hungry, but not very nourishing.) Most of the one-star problems
appear at the end of the relevant section; most of the three-star problems are at the end of the chapter. If a
computer is required, we put a mouse in the margin. A solution manual is available (to instructors only) from
the publisher.
In preparing this third edition we have tried to retain as much as possible the spirit of the first and
second. Although there are now two authors, we still use the singular (“I”) in addressing the reader—it feels
more intimate, and after all only one of us can speak at a time (“we” in the text means you, the reader, and I,
the author, working together). Schroeter brings the fresh perspective of a solid state theorist, and he is largely
responsible for the new chapter on symmetries. We have added a number of problems, clarified many
explanations, and revised the Afterword. But we were determined not to allow the book to grow fat, and for
that reason we have eliminated the chapter on the adiabatic approximation (significant insights from that
chapter have been incorporated into Chapter 11), and removed material from Chapter 5 on statistical

mechanics (which properly belongs in a book on thermal physics). It goes without saying that instructors are
welcome to cover such other topics as they see fit, but we want the textbook itself to represent the essential
core of the subject.

12

iii

We have benefitted from the comments and advice of many colleagues, who read the original
manuscript, pointed out weaknesses (or errors) in the first two editions, suggested improvements in the
presentation, and supplied interesting problems. We especially thank P. K. Aravind (Worcester Polytech),
Greg Benesh (Baylor), James Bernhard (Puget Sound), Burt Brody (Bard), Ash Carter (Drew), Edward
Chang (Massachusetts), Peter Collings (Swarthmore), Richard Crandall (Reed), Jeff Dunham (Middlebury),
Greg Elliott (Puget Sound), John Essick (Reed), Gregg Franklin (Carnegie Mellon), Joel Franklin (Reed),
Henry Greenside (Duke), Paul Haines (Dartmouth), J. R. Huddle (Navy), Larry Hunter (Amherst), David
Kaplan (Washington), Don Koks (Adelaide), Peter Leung (Portland State), Tony Liss (Illinois), Jeffry
Mallow (Chicago Loyola), James McTavish (Liverpool), James Nearing (Miami), Dick Palas, Johnny Powell
(Reed), Krishna Rajagopal (MIT), Brian Raue (Florida International), Robert Reynolds (Reed), Keith Riles
(Michigan), Klaus Schmidt-Rohr (Brandeis), Kenny Scott (London), Dan Schroeder (Weber State), Mark
Semon (Bates), Herschel Snodgrass (Lewis and Clark), John Taylor (Colorado), Stavros Theodorakis
(Cyprus), A. S. Tremsin (Berkeley), Dan Velleman (Amherst), Nicholas Wheeler (Reed), Scott Willenbrock
(Illinois), William Wootters (Williams), and Jens Zorn (Michigan).
1

This structure was inspired by David Park’s classic text Introduction to the Quantum Theory, 3rd edn, McGraw-Hill, New York (1992).

13

University Printing House, Cambridge CB2 8BS, United Kingdom
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Cambridge University Press is part of the University of Cambridge.
It furthers the University’s mission by disseminating knowledge in the pursuit of
education, learning, and research at the highest international levels of excellence.
www.cambridge.org
Information on this title: www.cambridge.org/9781107189638
DOI: 10.1017/9781316995433
c David Griffiths 2017
Second edition 
c Cambridge University Press 2018
Third edition 
This publication is in copyright. Subject to statutory exception
and to the provisions of relevant collective licensing agreements,
no reproduction of any part may take place without the written
permission of Cambridge University Press.
This book was previously published by Pearson Education, Inc. 2004
Second edition reissued by Cambridge University Press 2017
Third edition 2018
Printed in the United Kingdom by TJ International Ltd. Padstow Cornwall, 2018
A catalogue record for this publication is available from the British Library.
Library of Congress Cataloging-in-Publication Data
Names: Griffiths, David J. | Schroeter, Darrell F.
Title: Introduction to quantum mechanics / David J. Griffiths (Reed College,
Oregon), Darrell F. Schroeter (Reed College, Oregon).
Description: Third edition. | blah : Cambridge University Press, 2018.
Identifiers: LCCN 2018009864 | ISBN 9781107189638
Subjects: LCSH: Quantum theory.
Classification: LCC QC174.12 .G75 2018 | DDC 530.12–dc23
LC record available at https://lccn.loc.gov/2018009864
ISBN 978-1-107-18963-8 Hardback
Additional resources for this publication at www.cambridge.org/IQM3ed
Cambridge University Press has no responsibility for the persistence or accuracy of
URLs for external or third-party internet websites referred to in this publication
and does not guarantee that any content on such websites is, or will remain,
accurate or appropriate.

Contents
Preface

I
1
1.1
1.2
1.3
1.3.1
1.3.2

page i

THEORY

2

The Wave Function

3

The Schrödinger Equation
The Statistical Interpretation
Probability

3
3
8

Discrete Variables

8

Continuous Variables

11

Normalization
Momentum
The Uncertainty Principle
Further Problems on Chapter 1

14
16
19
20

Time-Independent Schrödinger Equation

25

Stationary States
The Infinite Square Well
The Harmonic Oscillator

25
31
39

2.3.1

Algebraic Method

40

2.3.2

Analytic Method

48

The Free Particle
The Delta-Function Potential

55
61

2.5.1

Bound States and Scattering States

61

2.5.2

The Delta-Function Well

63

The Finite Square Well
Further Problems on Chapter 2

70
76

3

Formalism

91

3.1
3.2

Hilbert Space
Observables

91
94

3.2.1

Hermitian Operators

94

3.2.2

Determinate States

96

1.4
1.5
1.6

2
2.1
2.2
2.3

2.4
2.5

2.6

Contents

vi

3.3

Eigenfunctions of a Hermitian Operator

97

3.3.1

Discrete Spectra

98

3.3.2

Continuous Spectra

99

3.4
3.5

Generalized Statistical Interpretation
The Uncertainty Principle

102
105

3.5.1

Proof of the Generalized Uncertainty Principle

105

3.5.2

The Minimum-Uncertainty Wave Packet

108

3.5.3

The Energy-Time Uncertainty Principle

109

Vectors and Operators

113

3.6.1

Bases in Hilbert Space

113

3.6.2

Dirac Notation

117

3.6.3

Changing Bases in Dirac Notation

121

Further Problems on Chapter 3

124

Quantum Mechanics in Three Dimensions

131

3.6

4

The Schrödinger Equation

131

4.1.1

4.1

Spherical Coordinates

132

4.1.2

The Angular Equation

134

4.1.3

The Radial Equation

138

The Hydrogen Atom

143

4.2.1

The Radial Wave Function

144

4.2.2

The Spectrum of Hydrogen

155

Angular Momentum

157

4.2

4.3
4.3.1

Eigenvalues

157

4.3.2

Eigenfunctions

162

Spin

165

4.4
4.4.1

Spin 1/2

167

4.4.2

Electron in a Magnetic Field

172

4.4.3

Addition of Angular Momenta

176

Electromagnetic Interactions

181

Minimal Coupling

181

4.5
4.5.1
4.5.2

The Aharonov–Bohm Effect

182

Further Problems on Chapter 4

187

Identical Particles

198

Two-Particle Systems

198

5.1.1

Bosons and Fermions

201

5.1.2

Exchange Forces

203

5
5.1

Contents
5.1.3

Spin

206

5.1.4

Generalized Symmetrization Principle

207

5.2

Atoms

209

5.2.1

Helium

210

5.2.2

The Periodic Table

213

Solids

216

The Free Electron Gas

216

5.3
5.3.1
5.3.2

6
6.1
6.1.1

6.2
6.2.1
6.2.2

Band Structure

220

Further Problems on Chapter 5

225

Symmetries & Conservation Laws

232

Introduction

232

Transformations in Space

232

The Translation Operator

235

How Operators Transform

235

Translational Symmetry

238

Conservation Laws
Parity

242
243

6.4.1

Parity in One Dimension

243

6.4.2

Parity in Three Dimensions

244

6.4.3

Parity Selection Rules

246

Rotational Symmetry

248

Rotations About the z Axis

248

6.3
6.4

6.5
6.5.1
6.5.2

Rotations in Three Dimensions

249

Degeneracy
Rotational Selection Rules

252
255

6.7.1

Selection Rules for Scalar Operators

255

6.7.2

Selection Rules for Vector Operators

258

6.6
6.7

6.8

Translations in Time

262

6.8.1

The Heisenberg Picture

264

6.8.2

Time-Translation Invariance

266

Further Problems on Chapter 6

268

APPLICATIONS

277

Time-Independent Perturbation Theory

279

Nondegenerate Perturbation Theory

279

7.1.1

General Formulation

279

7.1.2

First-Order Theory

280

II
7
7.1

vii

Contents

viii

7.1.3

Second-Order Energies

284

Degenerate Perturbation Theory

286

7.2.1

Two-Fold Degeneracy

286

7.2.2

“Good” States

291

7.2.3

Higher-Order Degeneracy

294

The Fine Structure of Hydrogen

295

The Relativistic Correction

296

7.2

7.3
7.3.1
7.3.2

Spin-Orbit Coupling

299

7.4

The Zeeman Effect

304

7.4.1

Weak-Field Zeeman Effect

305

7.4.2

Strong-Field Zeeman Effect

307

7.4.3

Intermediate-Field Zeeman Effect

309

7.5

Hyperfine Splitting in Hydrogen
Further Problems on Chapter 7

311
313

8

The Variational Principle

327

8.1
8.2
8.3
8.4

Theory
The Ground State of Helium
The Hydrogen Molecule Ion
The Hydrogen Molecule
Further Problems on Chapter 8

327
332
337
341
346

9

The WKB Approximation

354

9.1
9.2
9.3

The “Classical” Region
Tunneling
The Connection Formulas
Further Problems on Chapter 9

354
358
362
371

10 Scattering
10.1

Introduction

376
376

10.1.1

Classical Scattering Theory

376

10.1.2

Quantum Scattering Theory

379

Partial Wave Analysis

380

10.2.1

Formalism

380

10.2.2

Strategy

383

Phase Shifts
The Born Approximation

385
388

10.2

10.3
10.4

Contents
10.4.1

Integral Form of the Schrödinger Equation

388

10.4.2

The First Born Approximation

391

10.4.3

The Born Series

395

Further Problems on Chapter 10

397

11 Quantum Dynamics
11.1

402

Two-Level Systems

403

11.1.1

The Perturbed System

403

11.1.2

Time-Dependent Perturbation Theory

405

11.1.3

Sinusoidal Perturbations

408

Emission and Absorption of Radiation

411

11.2.1

11.2

Electromagnetic Waves

411

11.2.2

Absorption, Stimulated Emission, and Spontaneous Emission

412

11.2.3

Incoherent Perturbations

413

11.3

Spontaneous Emission

416

11.3.1

Einstein’s A and B Coefficients

416

11.3.2

The Lifetime of an Excited State

418

11.3.3

Selection Rules

420

Fermi’s Golden Rule
The Adiabatic Approximation

422
426

11.5.1

Adiabatic Processes

426

11.5.2

The Adiabatic Theorem

428

Further Problems on Chapter 11

433

11.4
11.5

12 Afterword
12.1
12.2
12.3

446

The EPR Paradox
Bell’s Theorem
Mixed States and the Density Matrix

447
449
455

12.3.1

Pure States

455

12.3.2

Mixed States

456

12.3.3

Subsystems

458

The No-Clone Theorem
Schrödinger’s Cat

459
461

12.4
12.5

Appendix Linear Algebra
A.1
A.2

Vectors
Inner Products

464
464
466

ix

Contents

x

A.3
A.4
A.5
A.6

Matrices
Changing Bases
Eigenvectors and Eigenvalues
Hermitian Transformations

468
473
475
482

Index

486

Part I
Theory
◈

1 THE WAVE FUNCTION
1.1 THE SCHRÖDINGER EQUATION
Imagine a particle of mass m, constrained to move along the x axis, subject to some specified
force F(x, t) (Figure 1.1). The program of classical mechanics is to determine the position of
the particle at any given time: x(t). Once we know that,
 out the velocity (v =
 we can figure
d x/dt), the momentum ( p = mv), the kinetic energy T = (1/2)mv 2 , or any other dynamical
variable of interest. And how do we go about determining x(t)? We apply Newton’s second
law: F = ma. (For conservative systems—the only kind we shall consider, and, fortunately, the
only kind that occur at the microscopic level—the force can be expressed as the derivative of
a potential energy function,1 F = −∂ V /∂ x, and Newton’s law reads m d 2 x/dt 2 = −∂ V /∂ x.)
This, together with appropriate initial conditions (typically the position and velocity at t = 0),
determines x(t).
Quantum mechanics approaches this same problem quite differently. In this case what we’re
looking for is the particle’s wave function, (x, t), and we get it by solving the Schrödinger
equation:
i

2 ∂ 2 
∂
=−
+ V .
∂t
2m ∂ x 2

(1.1)

Here i is the square root of −1, and  is Planck’s constant—or rather, his original constant (h)
divided by 2π :
h
= 1.054573 × 10−34 J s.
=
(1.2)
2π
The Schrödinger equation plays a role logically analogous to Newton’s second law: Given
suitable initial conditions (typically, (x, 0)), the Schrödinger equation determines (x, t)
for all future time, just as, in classical mechanics, Newton’s law determines x(t) for all future
time.2

1.2 THE STATISTICAL INTERPRETATION
But what exactly is this “wave function,” and what does it do for you once you’ve got it? After
all, a particle, by its nature, is localized at a point, whereas the wave function (as its name
1 Magnetic forces are an exception, but let’s not worry about them just yet. By the way, we shall assume throughout

this book that the motion is nonrelativistic (v  c).
2 For a delightful first-hand account of the origins of the Schrödinger equation see the article by Felix Bloch in Physics

Today, December 1976.

4

CHAPTER 1

The Wave Function

m
F(x,t)

x

x(t)

Figure 1.1: A “particle” constrained to move in one dimension under the influence of a specified force.
⏐Ψ⏐2

a

b A

B

C

x

Figure 1.2: A typical wave function. The shaded area represents the probability of finding the particle
between a and b. The particle would be relatively likely to be found near A, and unlikely to be found
near B.

suggests) is spread out in space (it’s a function of x, for any given t). How can such an object
represent the state of a particle? The answer is provided by Born’s statistical interpretation,
which says that |(x, t)|2 gives the probability of finding the particle at point x, at time t—or,
more precisely,3


b


|(x, t)| d x =
2

a

probability of finding the particle
between a and b, at time t.


(1.3)

Probability is the area under the graph of ||2 . For the wave function in Figure 1.2, you would
be quite likely to find the particle in the vicinity of point A, where ||2 is large, and relatively
unlikely to find it near point B.
The statistical interpretation introduces a kind of indeterminacy into quantum mechanics,
for even if you know everything the theory has to tell you about the particle (to wit: its wave
function), still you cannot predict with certainty the outcome of a simple experiment to measure
its position—all quantum mechanics has to offer is statistical information about the possible results. This indeterminacy has been profoundly disturbing to physicists and philosophers
alike, and it is natural to wonder whether it is a fact of nature, or a defect in the theory.
Suppose I do measure the position of the particle, and I find it to be at point C.4 Question:
Where was the particle just before I made the measurement? There are three plausible answers
3 The wave function itself is complex, but ||2 =  ∗  (where  ∗ is the complex conjugate of ) is real and

non-negative—as a probability, of course, must be.
4 Of course, no measuring instrument is perfectly precise; what I mean is that the particle was found in the vicinity of

C, as defined by the precision of the equipment.

1.2 The Statistical Interpretation

to this question, and they serve to characterize the main schools of thought regarding quantum
indeterminacy:
1. The realist position: The particle was at C. This certainly seems
reasonable, and it is the response Einstein advocated. Note, however,
that if this is true then quantum mechanics is an incomplete theory, since the
particle really was at C, and yet quantum mechanics was unable to tell us so.
To the realist, indeterminacy is not a fact of nature, but a reflection of our
ignorance. As d’Espagnat put it, “the position of the particle was never
indeterminate, but was merely unknown to the experimenter.”5 Evidently
 is not the whole story—some additional information (known as a
hidden variable) is needed to provide a complete description of the
particle.
2. The orthodox position: The particle wasn’t really anywhere. It was the
act of measurement that forced it to “take a stand” (though how and why it
decided on the point C we dare not ask). Jordan said it most starkly:
“Observations not only disturb what is to be measured, they produce it . . . We
compel [the particle] to assume a definite position.”6 This view (the so-called
Copenhagen interpretation), is associated with Bohr and his followers.
Among physicists it has always been the most widely accepted position. Note,
however, that if it is correct there is something very peculiar about the act of
measurement—something that almost a century of debate has done precious
little to illuminate.
3. The agnostic position: Refuse to answer. This is not quite as silly
as it sounds—after all, what sense can there be in making assertions about the
status of a particle before a measurement, when the only way of knowing
whether you were right is precisely to make a measurement, in which case what
you get is no longer “before the measurement”? It is metaphysics (in the
pejorative sense of the word) to worry about something that cannot, by its
nature, be tested. Pauli said: “One should no more rack one’s brain about the
problem of whether something one cannot know anything about exists all the
same, than about the ancient question of how many angels are able to sit on the
point of a needle.”7 For decades this was the “fall-back” position of most
physicists: they’d try to sell you the orthodox answer, but if you were
persistent they’d retreat to the agnostic response, and terminate the
conversation.
Until fairly recently, all three positions (realist, orthodox, and agnostic) had their partisans. But in 1964 John Bell astonished the physics community by showing that it makes an
observable difference whether the particle had a precise (though unknown) position prior to
5 Bernard d’Espagnat, “The Quantum Theory and Reality” (Scientific American, November 1979, p. 165).
6 Quoted in a lovely article by N. David Mermin, “Is the moon there when nobody looks?” (Physics Today, April

1985, p. 38).
7 Ibid., p. 40.

5

6

CHAPTER 1

The Wave Function

⏐Ψ⏐2

C

x

Figure 1.3: Collapse of the wave function: graph of ||2 immediately after a measurement has found the
particle at point C.

the measurement, or not. Bell’s discovery effectively eliminated agnosticism as a viable option,
and made it an experimental question whether 1 or 2 is the correct choice. I’ll return to this
story at the end of the book, when you will be in a better position to appreciate Bell’s argument; for now, suffice it to say that the experiments have decisively confirmed the orthodox
interpretation:8 a particle simply does not have a precise position prior to measurement, any
more than the ripples on a pond do; it is the measurement process that insists on one particular number, and thereby in a sense creates the specific result, limited only by the statistical
weighting imposed by the wave function.
What if I made a second measurement, immediately after the first? Would I get C again,
or does the act of measurement cough up some completely new number each time? On
this question everyone is in agreement: A repeated measurement (on the same particle)
must return the same value. Indeed, it would be tough to prove that the particle was really
found at C in the first instance, if this could not be confirmed by immediate repetition of
the measurement. How does the orthodox interpretation account for the fact that the second measurement is bound to yield the value C? It must be that the first measurement
radically alters the wave function, so that it is now sharply peaked about C (Figure 1.3).
We say that the wave function collapses, upon measurement, to a spike at the point C (it
soon spreads out again, in accordance with the Schrödinger equation, so the second measurement must be made quickly). There are, then, two entirely distinct kinds of physical
processes: “ordinary” ones, in which the wave function evolves in a leisurely fashion under
the Schrödinger equation, and “measurements,” in which  suddenly and discontinuously
collapses.9

8 This statement is a little too strong: there exist viable nonlocal hidden variable theories (notably David Bohm’s), and

other formulations (such as the many worlds interpretation) that do not fit cleanly into any of my three categories.
But I think it is wise, at least from a pedagogical point of view, to adopt a clear and coherent platform at this stage,
and worry about the alternatives later.
9 The role of measurement in quantum mechanics is so critical and so bizarre that you may well be wondering what
precisely constitutes a measurement. I’ll return to this thorny issue in the Afterword; for the moment let’s take the
naive view: a measurement is the kind of thing that a scientist in a white coat does in the laboratory, with rulers,
stopwatches, Geiger counters, and so on.

1.2 The Statistical Interpretation

Example 1.1
Electron Interference. I have asserted that particles (electrons, for example) have a wave
nature, encoded in . How might we check this, in the laboratory?
The classic signature of a wave phenomenon is interference: two waves in phase
interfere constructively, and out of phase they interfere destructively. The wave nature
of light was confirmed in 1801 by Young’s famous double-slit experiment, showing
interference “fringes” on a distant screen when a monochromatic beam passes through
two slits. If essentially the same experiment is done with electrons, the same pattern
develops,10 confirming the wave nature of electrons.
Now suppose we decrease the intensity of the electron beam, until only one electron is
present in the apparatus at any particular time. According to the statistical interpretation
each electron will produce a spot on the screen. Quantum mechanics cannot predict the
precise location of that spot—all it can tell us is the probability of a given electron landing
at a particular place. But if we are patient, and wait for a hundred thousand electrons—one
at a time—to make the trip, the accumulating spots reveal the classic two-slit interference
pattern (Figure 1.4).11

a

b

c

d

Figure 1.4: Build-up of the electron interference pattern. (a) Eight electrons, (b) 270 electrons,
(c) 2000 electrons, (d) 160,000 electrons. Reprinted courtesy of the Central Research Laboratory,
Hitachi, Ltd., Japan.

10 Because the wavelength of electrons is typically very small, the slits have to be extremely close together. Histori-

cally, this was first achieved by Davisson and Germer, in 1925, using the atomic layers in a crystal as “slits.” For
an interesting account, see R. K. Gehrenbeck, Physics Today, January 1978, page 34.
11 See Tonomura et al., American Journal of Physics, Volume 57, Issue 2, pp. 117–120 (1989), and the amazing associated video at www.hitachi.com/rd/portal/highlight/quantum/doubleslit/. This experiment can now be done with
much more massive particles, including “Bucky-balls”; see M. Arndt, et al., Nature 40, 680 (1999). Incidentally,
the same thing can be done with light: turn the intensity so low that only one “photon” is present at a time and
you get an identical point-by-point assembly of the interference pattern. See R. S. Aspden, M. J. Padgett, and
G. C. Spalding, Am. J. Phys. 84, 671 (2016).

7

8

CHAPTER 1

The Wave Function

Of course, if you close off one slit, or somehow contrive to detect which slit each
electron passes through, the interference pattern disappears; the wave function of the
emerging particle is now entirely different (in the first case because the boundary conditions for the Schrödinger equation have been changed, and in the second because of
the collapse of the wave function upon measurement). But with both slits open, and
no interruption of the electron in flight, each electron interferes with itself; it didn’t
pass through one slit or the other, but through both at once, just as a water wave,
impinging on a jetty with two openings, interferes with itself. There is nothing mysterious about this, once you have accepted the notion that particles obey a wave equation.
The truly astonishing thing is the blip-by-blip assembly of the pattern. In any classical wave theory the pattern would develop smoothly and continuously, simply getting
more intense as time goes on. The quantum process is more like the pointillist painting
of Seurat: The picture emerges from the cumulative contributions of all the individual
dots.12

1.3 PROBABILITY
1.3.1

Discrete Variables
Because of the statistical interpretation, probability plays a central role in quantum mechanics, so I digress now for a brief discussion of probability theory. It is mainly a question
of introducing some notation and terminology, and I shall do it in the context of a simple
example.
Imagine a room containing fourteen people, whose ages are as follows:
one person aged 14,
one person aged 15,
three people aged 16,
two people aged 22,
two people aged 24,
five people aged 25.
If we let N ( j) represent the number of people of age j, then
N (14) = 1,
N (15) = 1,
N (16) = 3,
N (22) = 2,
N (24) = 2,
N (25) = 5,

12 I think it is important to distinguish things like interference and diffraction that would hold for any wave theory

from the uniquely quantum mechanical features of the measurement process, which derive from the statistical
interpretation.

1.3 Probability
N(j)

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26

j

Figure 1.5: Histogram showing the number of people, N ( j), with age j, for the example in Section 1.3.1.

while N (17), for instance, is zero. The total number of people in the room is
N=

∞


N ( j).

(1.4)

j=0

(In the example, of course, N = 14.) Figure 1.5 is a histogram of the data. The following are
some questions one might ask about this distribution.
Question 1 If you selected one individual at random from this group, what is the probability

that this person’s age would be 15?
One chance in 14, since there are 14 possible choices, all equally likely, of whom
only one has that particular age. If P( j) is the probability of getting age j, then P(14) =
1/14, P(15) = 1/14, P(16) = 3/14, and so on. In general,
Answer

P( j) =

N ( j)
.
N

(1.5)

Notice that the probability of getting either 14 or 15 is the sum of the individual probabilities
(in this case, 1/7). In particular, the sum of all the probabilities is 1—the person you select
must have some age:
∞


P( j) = 1.

(1.6)

j=0

Question 2 What is the most probable age?
Answer 25, obviously; five people share this age, whereas at most three have any other age.

The most probable j is the j for which P( j) is a maximum.
Question 3 What is the median age?
Answer 23, for 7 people are younger than 23, and 7 are older. (The median is that value of

j such that the probability of getting a larger result is the same as the probability of getting a
smaller result.)
Question 4 What is the average (or mean) age?
Answer

294
(14) + (15) + 3(16) + 2(22) + 2(24) + 5(25)
=
= 21.
14
14

9

10

CHAPTER 1

The Wave Function

In general, the average value of j (which we shall write thus:  j) is

∞
j N ( j) 
 j =
j P( j).
=
N

(1.7)

j=0

Notice that there need not be anyone with the average age or the median age—in this example
nobody happens to be 21 or 23. In quantum mechanics the average is usually the quantity of
interest; in that context it has come to be called the expectation value. It’s a misleading term,
since it suggests that this is the outcome you would be most likely to get if you made a single
measurement (that would be the most probable value, not the average value)—but I’m afraid
we’re stuck with it.
Question 5 What is the average of the squares of the ages?
Answer You could get 142 = 196, with probability 1/14, or 152 = 225, with probability 1/14,

or 162 = 256, with probability 3/14, and so on. The average, then, is
j2 =

∞


j 2 P( j).

(1.8)

j=0

In general, the average value of some function of j is given by
 f ( j) =

∞


f ( j)P( j).

(1.9)

j=0

(Equations 1.6, 1.7, and 1.8 are, if you like, special cases of this formula.) Beware: The average
of the squares, j 2 , is not equal, in general, to the square of the average,  j2 . For instance, if
the room contains just two babies, aged 1 and 3, then j 2 = 5, but  j2 = 4.
Now, there is a conspicuous difference between the two histograms in Figure 1.6, even
though they have the same median, the same average, the same most probable value, and the
same number of elements: The first is sharply peaked about the average value, whereas the
second is broad and flat. (The first might represent the age profile for students in a big-city
classroom, the second, perhaps, a rural one-room schoolhouse.) We need a numerical measure
N(j)

N(j)

1

2

3

4

5

6

7

8

9 10

j

1

2

3

4

5

6

7

8

9 10

j

Figure 1.6: Two histograms with the same median, same average, and same most probable value, but
different standard deviations.

1.3 Probability

of the amount of “spread” in a distribution, with respect to the average. The most obvious way
to do this would be to find out how far each individual is from the average,
j = j −  j ,

(1.10)

and compute the average of j. Trouble is, of course, that you get zero:



j =
( j −  j)P( j) =
j P( j) −  j
P( j)
=  j −  j = 0.
(Note that  j is constant—it does not change as you go from one member of the sample to
another—so it can be taken outside the summation.) To avoid this irritating problem you might
decide to average the absolute value of j. But absolute values are nasty to work with; instead,
we get around the sign problem by squaring before averaging:
σ 2 ≡ (j)2 .

(1.11)

This quantity is known as the variance of the distribution; σ itself (the square root of the average of the square of the deviation from the average—gulp!) is called the standard deviation.
The latter is the customary measure of the spread about  j.
There is a useful little theorem on variances:


(j)2 P( j) =
( j −  j)2 P( j)
σ 2 = (j)2 =

j 2 − 2 j  j +  j2 P( j)
=



=
j 2 P( j) − 2  j
P( j)
j P( j) +  j2
= j 2 − 2  j  j +  j2 = j 2 −  j2 .
Taking the square root, the standard deviation itself can be written as
σ =

j 2 −  j2 .

(1.12)

In practice, this is a much faster way to get σ than by direct application of Equation 1.11:
simply calculate j 2 and  j2 , subtract, and take the square root. Incidentally, I warned you a
moment ago that j 2 is not, in general, equal to  j2 . Since σ 2 is plainly non-negative (from
its definition 1.11), Equation 1.12 implies that
j 2 ≥  j2 ,

(1.13)

and the two are equal only when σ = 0, which is to say, for distributions with no spread at all
(every member having the same value).
1.3.2

Continuous Variables
So far, I have assumed that we are dealing with a discrete variable—that is, one that can take
on only certain isolated values (in the example, j had to be an integer, since I gave ages only
in years). But it is simple enough to generalize to continuous distributions. If I select a random
person off the street, the probability that her age is precisely 16 years, 4 hours, 27 minutes,
and 3.333. . . seconds is zero. The only sensible thing to speak about is the probability that
her age lies in some interval—say, between 16 and 17. If the interval is sufficiently short, this
probability is proportional to the length of the interval. For example, the chance that her age is
between 16 and 16 plus two days is presumably twice the probability that it is between 16 and

11

12

CHAPTER 1

The Wave Function

16 plus one day. (Unless, I suppose, there was some extraordinary baby boom 16 years ago,
on exactly that day—in which case we have simply chosen an interval too long for the rule to
apply. If the baby boom lasted six hours, we’ll take intervals of a second or less, to be on the
safe side. Technically, we’re talking about infinitesimal intervals.) Thus


probability that an individual (chosen
= ρ(x) d x.
(1.14)
at random) lies between x and (x + d x)
The proportionality factor, ρ(x), is often loosely called “the probability of getting x,” but this
is sloppy language; a better term is probability density. The probability that x lies between a
and b (a finite interval) is given by the integral of ρ(x):

Pab =

b

ρ(x) d x,

(1.15)

a

and the rules we deduced for discrete distributions translate in the obvious way:
 +∞
ρ(x) d x = 1,
−∞



x =

 f (x) =

(1.16)

+∞

−∞
+∞
−∞
2

xρ(x) d x,

(1.17)

f (x) ρ(x) d x,

(1.18)

σ 2 ≡ (x) = x 2 − x2 .

(1.19)

Example 1.2
Suppose someone drops a rock off a cliff of height h. As it falls, I snap a million photographs, at random intervals. On each picture I measure the distance the rock has fallen.
Question: What is the average of all these distances? That is to say, what is the time average
of the distance traveled?13
Solution: The rock starts out at rest, and picks up speed as it falls; it spends more time near
the top, so the average distance will surely be less than h/2. Ignoring air resistance, the
distance x at time t is
1
x(t) = gt 2 .
2
√
The velocity is d x/dt = gt, and the total flight time is T = 2h/g. The probability that
a particular photograph was taken between t and t + dt is dt/T , so the probability that it
shows a distance in the corresponding range x to x + d x is

dt
dx g
1
=
= √ d x.
T
gt 2h
2 hx
13 A statistician will complain that I am confusing the average of a finite sample (a million, in this case) with the

“true” average (over the whole continuum). This can be an awkward problem for the experimentalist, especially
when the sample size is small, but here I am only concerned with the true average, to which the sample average is
presumably a good approximation.

1.3 Probability

Thus the probability density (Equation 1.14) is
1
ρ(x) = √ ,
2 hx

(0 ≤ x ≤ h)

(outside this range, of course, the probability density is zero).
We can check this result, using Equation 1.16:
h
 h

1
1
1/2 
√ d x = √ 2x
 = 1.
2 h
0 2 hx
0
The average distance (Equation 1.17) is

h
 h
2 3/2 
1
h
1
x =
x √ dx = √
= ,
x

3
3
2 hx
2 h
0
0
which is somewhat less than h/2, as anticipated.
Figure 1.7 shows the graph of ρ(x). Notice that a probability density can be infinite,
though probability itself (the integral of ρ) must of course be finite (indeed, less than or
equal to 1).
ρ(x)

1
2h

h

Figure 1.7: The probability density in Example 1.2: ρ(x) = 1

∗

x



√
2 hx .

Problem 1.1 For the distribution of ages in the example in Section 1.3.1:
(a) Compute j 2 and  j2 .
(b) Determine j for each j, and use Equation 1.11 to compute the standard deviation.
(c) Use your results in (a) and (b) to check Equation 1.12.

Problem 1.2
(a) Find the standard deviation of the distribution in Example 1.2.
(b) What is the probability that a photograph, selected at random, would show a
distance x more than one standard deviation away from the average?

13

14

CHAPTER 1

∗

The Wave Function

Problem 1.3 Consider the gaussian distribution
ρ(x) = Ae−λ(x−a) ,
2

where A, a, and λ are positive real constants. (The necessary integrals are inside the
back cover.)
(a) Use Equation 1.16 to determine A.
(b) Find x, x 2 , and σ .
(c) Sketch the graph of ρ(x).

1.4 NORMALIZATION
We return now to the statistical interpretation of the wave function (Equation 1.3), which says
that |(x, t)|2 is the probability density for finding the particle at point x, at time t. It follows (Equation 1.16) that the integral of ||2 over all x must be 1 (the particle’s got to be
somewhere):


+∞

−∞

|(x, t)|2 d x = 1.

(1.20)

Without this, the statistical interpretation would be nonsense.
However, this requirement should disturb you: After all, the wave function is supposed to
be determined by the Schrödinger equation—we can’t go imposing an extraneous condition on
 without checking that the two are consistent. Well, a glance at Equation 1.1 reveals that if
(x, t) is a solution, so too is A(x, t), where A is any (complex) constant. What we must
do, then, is pick this undetermined multiplicative factor so as to ensure that Equation 1.20
is satisfied. This process is called normalizing the wave function. For some solutions to the
Schrödinger equation the integral is infinite; in that case no multiplicative factor is going to
make it 1. The same goes for the trivial solution  = 0. Such non-normalizable solutions
cannot represent particles, and must be rejected. Physically realizable states correspond to the
square-integrable solutions to Schrödinger’s equation.14
But wait a minute! Suppose I have normalized the wave function at time t = 0. How
do I know that it will stay normalized, as time goes on, and  evolves? (You can’t keep
renormalizing the wave function, for then A becomes a function of t, and you no longer
have a solution to the Schrödinger equation.) Fortunately, the Schrödinger equation has the
remarkable property that it automatically preserves the normalization of the wave function—
without this crucial feature the Schrödinger equation would be incompatible with the statistical
interpretation, and the whole theory would crumble.
This is important, so we’d better pause for a careful proof. To begin with,
 +∞
 +∞
∂
d
|(x, t)|2 d x.
|(x, t)|2 d x =
(1.21)
dt −∞
−∞ ∂t
14 Evidently (x, t) must go to zero faster than 1/√|x|, as |x| → ∞. Incidentally, normalization only fixes the

modulus of A; the phase remains undetermined. However, as we shall see, the latter carries no physical significance
anyway.

1.4 Normalization

(Note that the integral is a function only of t, so I use a total derivative (d/dt) on the left, but
the integrand is a function of x as well as t, so it’s a partial derivative (∂/∂t) on the right.) By
the product rule,
∂ ∗
∂
∂  ∗ 
∂
||2 =
  = ∗
+
.
(1.22)
∂t
∂t
∂t
∂t
Now the Schrödinger equation says that
∂
i ∂ 2 
i
=
− V ,
∂t
2m ∂ x 2

and hence also (taking the complex conjugate of Equation 1.23)
∂ ∗
i
i ∂ 2  ∗
+ V  ∗,
=−
∂t
2m ∂ x 2

so
∂
i
||2 =
∂t
2m






∂ ∗
∂ 2
∂ 2 ∗
∂ i
∗ ∂
∗ 2 −

−

.

=
∂ x 2m
∂x
∂x
∂x
∂x2

The integral in Equation 1.21 can now be evaluated explicitly:

+∞
 +∞

d
∂ ∗
i
2
∗ ∂
|(x, t)| d x =

−
 
.
dt −∞
2m
∂x
∂x
−∞

(1.23)

(1.24)

(1.25)

(1.26)

But (x, t) must go to zero as x goes to (±) infinity—otherwise the wave function would not
be normalizable.15 It follows that
 +∞
d
|(x, t)|2 d x = 0,
(1.27)
dt −∞
and hence that the integral is constant (independent of time); if  is normalized at t = 0, it
stays normalized for all future time. QED

Problem 1.4 At time t = 0 a particle is represented by the wave function
⎧
0 ≤ x ≤ a,
⎪
⎨ A(x/a),
(x, 0) = A(b − x)/(b − a), a ≤ x ≤ b,
⎪
⎩
0,
otherwise,
where A, a, and b are (positive) constants.
(a) Normalize  (that is, find A, in terms of a and b).
(b) Sketch (x, 0), as a function of x.
(c) Where is the particle most likely to be found, at t = 0?
(d) What is the probability of finding the particle to the left of a? Check your result in
the limiting cases b = a and b = 2a.
(e) What is the expectation value of x?

15 A competent mathematician can supply you with pathological counterexamples, but they do not arise in physics;

for us the wave function and all its derivatives go to zero at infinity.

15

16

CHAPTER 1

∗

The Wave Function

Problem 1.5 Consider the wave function
(x, t) = Ae−λ|x| e−iωt ,
where A, λ, and ω are positive real constants. (We’ll see in Chapter 2 for what potential
(V ) this wave function satisfies the Schrödinger equation.)
(a) Normalize .
(b) Determine the expectation values of x and x 2 .
(c) Find the standard deviation of x. Sketch the graph of ||2 , as a function of x, and
mark the points (x+σ ) and (x−σ ), to illustrate the sense in which σ represents
the “spread” in x. What is the probability that the particle would be found outside
this range?

1.5 MOMENTUM
For a particle in state , the expectation value of x is

x =

+∞

−∞

x |(x, t)|2 d x.

(1.28)

What exactly does this mean? It emphatically
does not mean that if you measure the position

of one particle over and over again, x ||2 d x is the average of the results you’ll get. On
the contrary: The first measurement (whose outcome is indeterminate) will collapse the wave
function to a spike at the value actually obtained, and the subsequent measurements (if they’re
performed quickly) will simply repeat that same result. Rather, x is the average of measurements performed on particles all in the state , which means that either you must find some
way of returning the particle to its original state after each measurement, or else you have to
prepare a whole ensemble of particles, each in the same state , and measure the positions
of all of them: x is the average of these results. I like to picture a row of bottles on a shelf,
each containing a particle in the state  (relative to the center of the bottle). A graduate student
with a ruler is assigned to each bottle, and at a signal they all measure the positions of their
respective particles. We then construct a histogram of the results, which should match ||2 ,
and compute the average, which should agree with x. (Of course, since we’re only using a
finite sample, we can’t expect perfect agreement, but the more bottles we use, the closer we
ought to come.) In short, the expectation value is the average of measurements on an ensemble of identically-prepared systems, not the average of repeated measurements on one and the
same system.
Now, as time goes on, x will change (because of the time dependence of ), and we might
be interested in knowing how fast it moves. Referring to Equations 1.25 and 1.28, we see that16




∂
∂ ∗
dx
∂
i
∂
= x ||2 d x =
∗
−
 d x.
(1.29)
x
dt
∂t
2m
∂x
∂x
∂x
16 To keep things from getting too cluttered, I’ll suppress the limits of integration (±∞).

1.5 Momentum

This expression can be simplified using integration-by-parts:17

 
i
∂ ∗
dx
∂
=−
∗
−
 d x.
dt
2m
∂x
∂x

(1.30)

(I used the fact that ∂ x/∂ x = 1, and threw away the boundary term, on the ground that 
goes to zero at (±) infinity.) Performing another integration-by-parts, on the second term, we
conclude:

i
∂
dx
=−
d x.
(1.31)
∗
dt
m
∂x
What are we to make of this result? Note that we’re talking about the “velocity” of the
expectation value of x, which is not the same thing as the velocity of the particle. Nothing
we have seen so far would enable us to calculate the velocity of a particle. It’s not even clear
what velocity means in quantum mechanics: If the particle doesn’t have a determinate position
(prior to measurement), neither does it have a well-defined velocity. All we could reasonably
ask for is the probability of getting a particular value. We’ll see in Chapter 3 how to construct
the probability density for velocity, given ; for the moment it will suffice to postulate that
the expectation value of the velocity is equal to the time derivative of the expectation value of
position:
dx
v =
.
(1.32)
dt
Equation 1.31 tells us, then, how to calculate v directly from .
Actually, it is customary to work with momentum ( p = mv), rather than velocity:
 p = m

dx
= −i
dt


 
∂
∗
d x.
∂x

Let me write the expressions for x and  p in a more suggestive way:

x =  ∗ [x]  d x,



 p =  ∗ −i (∂/∂ x)  d x.

(1.33)

(1.34)
(1.35)

We say that the operator18 x “represents” position, and the operator −i (∂/∂ x) “represents”
momentum; to calculate expectation values we “sandwich” the appropriate operator between
 ∗ and , and integrate.
17 The product rule says that

dg
df
d
( f g) = f
+
g,
dx
dx
dx
from which it follows that

b
 b

df
dg

dx = −
g d x + f g .
f

dx
a
a dx
a

 b

Under the integral sign, then, you can peel a derivative off one factor in a product, and slap it onto the other
one—it’ll cost you a minus sign, and you’ll pick up a boundary term.
18 An “operator” is an instruction to do something to the function that follows; it takes in one function, and spits
out some other function. The position operator tells you to multiply by x; the momentum operator tells you to
differentiate with respect to x (and multiply the result by −i).

17

18

CHAPTER 1

The Wave Function

That’s cute, but what about other quantities? The fact is, all classical dynamical variables
can be expressed in terms of position and momentum. Kinetic energy, for example, is
T =

p2
1 2
mv =
,
2
2m

and angular momentum is
L = r × mv = r × p
(the latter, of course, does not occur for motion in one dimension). To calculate the expectation
value of any such quantity, Q(x, p), we simply replace every p by −i (∂/∂ x), insert the
resulting operator between  ∗ and , and integrate:

Q(x, p) =



 ∗ Q(x, −i ∂/∂ x)  d x.

For example, the expectation value of the kinetic energy is

2
∂ 2
T  = −
 ∗ 2 d x.
2m
∂x

(1.36)

(1.37)

Equation 1.36 is a recipe for computing the expectation value of any dynamical quantity,
for a particle in state ; it subsumes Equations 1.34 and 1.35 as special cases. I have tried
to make Equation 1.36 seem plausible, given Born’s statistical interpretation, but in truth this
represents such a radically new way of doing business (as compared with classical mechanics)
that it’s a good idea to get some practice using it before we come back (in Chapter 3) and put
it on a firmer theoretical foundation. In the mean time, if you prefer to think of it as an axiom,
that’s fine with me.
Problem 1.6 Why can’t you do integration-by-parts directly on the middle expression in
Equation 1.29—pull the time derivative over onto x, note that ∂ x/∂t = 0, and conclude
that dx/dt = 0?
∗

Problem 1.7 Calculate d p/dt. Answer:



∂V
d p
= −
.
dt
∂x

(1.38)

This is an instance of Ehrenfest’s theorem, which asserts that expectation values obey the
classical laws.19

Problem 1.8 Suppose you add a constant V0 to the potential energy (by “constant” I mean
independent of x as well as t). In classical mechanics this doesn’t change anything, but
what about quantum mechanics? Show that the wave function picks up a time-dependent
phase factor: exp (−i V0 t/). What effect does this have on the expectation value of a
dynamical variable?

19 Some authors limit the term to the pair of equations  p = m dx/dt and −∂ V /∂ x = d p/dt.

1.6 The Uncertainty Principle

10

20

30

40

50 x (feet)

Figure 1.8: A wave with a (fairly) well-defined wavelength, but an ill-defined position.

10

20

30

40

50 x (feet)

Figure 1.9: A wave with a (fairly) well-defined position, but an ill-defined wavelength.

1.6 THE UNCERTAINTY PRINCIPLE
Imagine that you’re holding one end of a very long rope, and you generate a wave by shaking it
up and down rhythmically (Figure 1.8). If someone asked you “Precisely where is that wave?”
you’d probably think he was a little bit nutty: The wave isn’t precisely anywhere—it’s spread
out over 50 feet or so. On the other hand, if he asked you what its wavelength is, you could
give him a reasonable answer: it looks like about 6 feet. By contrast, if you gave the rope a
sudden jerk (Figure 1.9), you’d get a relatively narrow bump traveling down the line. This time
the first question (Where precisely is the wave?) is a sensible one, and the second (What is its
wavelength?) seems nutty—it isn’t even vaguely periodic, so how can you assign a wavelength
to it? Of course, you can draw intermediate cases, in which the wave is fairly well localized
and the wavelength is fairly well defined, but there is an inescapable trade-off here: the more
precise a wave’s position is, the less precise is its wavelength, and vice versa.20 A theorem
in Fourier analysis makes all this rigorous, but for the moment I am only concerned with the
qualitative argument.
This applies, of course, to any wave phenomenon, and hence in particular to the quantum
mechanical wave function. But the wavelength of  is related to the momentum of the particle
by the de Broglie formula:21
p=

2π 
h
=
.
λ
λ

(1.39)

Thus a spread in wavelength corresponds to a spread in momentum, and our general observation
now says that the more precisely determined a particle’s position is, the less precisely is its
momentum. Quantitatively,
σx σ p ≥


,
2

(1.40)

20 That’s why a piccolo player must be right on pitch, whereas a double-bass player can afford to wear garden gloves.

For the piccolo, a sixty-fourth note contains many full cycles, and the frequency (we’re working in the time domain
now, instead of space) is well defined, whereas for the bass, at a much lower register, the sixty-fourth note contains
only a few cycles, and all you hear is a general sort of “oomph,” with no very clear pitch.
21 I’ll explain this in due course. Many authors take the de Broglie formula as an axiom, from which they then deduce
the association of momentum with the operator −i (∂/∂ x). Although this is a conceptually cleaner approach, it
involves diverting mathematical complications that I would rather save for later.

19

20

CHAPTER 1

The Wave Function

where σx is the standard deviation in x, and σ p is the standard deviation in p. This is Heisenberg’s famous uncertainty principle. (We’ll prove it in Chapter 3, but I wanted to mention it
right away, so you can test it out on the examples in Chapter 2.)
Please understand what the uncertainty principle means: Like position measurements,
momentum measurements yield precise answers—the “spread” here refers to the fact that measurements made on identically prepared systems do not yield identical results. You can, if you
want, construct a state such that position measurements will be very close together (by making
 a localized “spike”), but you will pay a price: Momentum measurements on this state will
be widely scattered. Or you can prepare a state with a definite momentum (by making  a
long sinusoidal wave), but in that case position measurements will be widely scattered. And,
of course, if you’re in a really bad mood you can create a state for which neither position nor
momentum is well defined: Equation 1.40 is an inequality, and there’s no limit on how big σx
and σ p can be—just make  some long wiggly line with lots of bumps and potholes and no
periodic structure.

∗

Problem 1.9 A particle of mass m has the wave function
(x, t) = Ae−a





mx 2 / +it

,

where A and a are positive real constants.
(a) Find A.
(b) For what potential energy function, V (x), is this a solution to the Schrödinger
equation?
(c) Calculate the expectation values of x, x 2 , p, and p 2 .
(d) Find σx and σ p . Is their product consistent with the uncertainty principle?

FURTHER PROBLEMS ON CHAPTER 1
Problem 1.10 Consider the first 25 digits in the decimal expansion of π (3, 1, 4, 1, 5, 9,
. . .).
(a) If you selected one number at random, from this set, what are the probabilities of
getting each of the 10 digits?
(b) What is the most probable digit? What is the median digit? What is the average
value?
(c) Find the standard deviation for this distribution.
Problem 1.11 [This problem generalizes Example 1.2.] Imagine a particle of mass m and
energy E in a potential well V (x), sliding frictionlessly back and forth between the
classical turning points (a and b in Figure 1.10). Classically, the probability of finding
the particle in the range d x (if, for example, you took a snapshot at a random time t)
is equal to the fraction of the time T it takes to get from a to b that it spends in the
interval d x:
(dt/d x) d x
1
dt
=
=
d x,
(1.41)
ρ(x) d x =
T
T
v(x) T

Further Problems on Chapter 1

V(x)

E
m

a

dx

x

b

Figure 1.10: Classical particle in a potential well.

where v(x) is the speed, and


T =

T


dt =

0

a

b

1
d x.
v(x)

(1.42)

Thus
ρ(x) =

1
.
v(x)T

(1.43)

This is perhaps the closest classical analog22 to ||2 .
(a) Use conservation of energy to express v(x) in terms of E and V (x).
(b) As an example, find ρ(x) for the simple harmonic oscillator, V (x) = kx 2 /2. Plot
ρ(x), and check that it is correctly normalized.
(c) For the classical harmonic oscillator in part (b), find x, x 2 , and σx .
∗∗

Problem 1.12 What if we were interested in the distribution of momenta ( p = mv), for
the classical harmonic oscillator (Problem 1.11(b)).
√
(a) Find√
the classical probability distribution ρ( p) (note that p ranges from − 2m E
to + 2m E).
(b) Calculate  p, p 2 , and σ p .
(c) What’s the classical uncertainty product, σx σ p , for this system? Notice that this
product can be as small as you like, classically, simply by sending E → 0. But in
quantum mechanics, as we shall see in Chapter 2, the energy of a simple harmonic
√
oscillator cannot be less than ω/2, where ω = k/m is the classical frequency.
In that case what can you say about the product σx σ p ?
Problem 1.13 Check your results in Problem 1.11(b) with the following “numerical
experiment.” The position of the oscillator at time t is
x(t) = A cos(ωt).

(1.44)

22 If you like, instead of photos of one system at random times, picture an ensemble of such systems, all with the same

energy but with random starting positions, and photograph them all at the same time. The analysis is identical, but
this interpretation is closer to the quantum notion of indeterminacy.

21

22

CHAPTER 1

The Wave Function

You might as well take ω = 1 (that sets the scale for time) and A = 1 (that sets the
scale for length). Make a plot of x at 10,000 random times, and compare it with ρ(x).
Hint: In Mathematica, first define
x[t_] := Cos[t]
then construct a table of positions:
snapshots = Table[x[π RandomReal[j]], {j, 10000}]
and finally, make a histogram of the data:
Histogram[snapshots, 100, PDF , PlotRange → {0,2}]
Meanwhile, make a plot of the density function, ρ(x), and, using Show, superimpose
the two.
Problem 1.14 Let Pab (t) be the probability of finding the particle in the range (a < x <
b), at time t.
(a) Show that
d Pab
= J (a, t) − J (b, t),
dt
where


∂ ∗
∂
i

− ∗
.
J (x, t) ≡
2m
∂x
∂x
What are the units of J (x, t)? Comment: J is called the probability current,
because it tells you the rate at which probability is “flowing” past the point x. If
Pab (t) is increasing, then more probability is flowing into the region at one end
than flows out at the other.
(b) Find the probability current for the wave function in Problem 1.9. (This is not
a very pithy example, I’m afraid; we’ll encounter more substantial ones in due
course.)
Problem 1.15 Show that
d
dt



∞
−∞

1∗ 2 d x = 0

for any two (normalizable) solutions to the Schrödinger equation (with the same
V (x)), 1 and 2 .
Problem 1.16 A particle is represented (at time t = 0) by the wave function
 

A a 2 − x 2 , −a ≤ x ≤ +a,
(x, 0) =
0,
otherwise.
(a) Determine the normalization constant A.
(b) What is the expectation value of x?
(c) What is the expectation value of p? (Note that you cannot get it from  p =
mdx/dt. Why not?)

Further Problems on Chapter 1

(d) Find the expectation value of x 2 .
(e) Find the expectation value of p 2 .
(f) Find the uncertainty in x (σ
 x ).
(g) Find the uncertainty in p σ p .
(h) Check that your results are consistent with the uncertainty principle.
∗∗

Problem 1.17 Suppose you wanted to describe an unstable particle, that spontaneously
disintegrates with a “lifetime” τ . In that case the total probability of finding the particle
somewhere should not be constant, but should decrease at (say) an exponential rate:
 +∞
|(x, t)|2 d x = e−t/τ .
P(t) ≡
−∞

A crude way of achieving this result is as follows. In Equation 1.24 we tacitly assumed
that V (the potential energy) is real. That is certainly reasonable, but it leads to the
“conservation of probability” enshrined in Equation 1.27. What if we assign to V an
imaginary part:
V = V0 − i ,
where V0 is the true potential energy and is a positive real constant?
(a) Show that (in place of Equation 1.27) we now get
2
dP
=−
P.
dt

(b) Solve for P(t), and find the lifetime of the particle in terms of .
Problem 1.18 Very roughly speaking, quantum mechanics is relevant when the de Broglie
wavelength of the particle in question (h/ p) is greater than the characteristic size of
the system (d). In thermal equilibrium at (Kelvin) temperature T , the average kinetic
energy of a particle is
3
p2
= kB T
2m
2
(where k B is Boltzmann’s constant), so the typical de Broglie wavelength is
λ= √

h
.
3mk B T

(1.45)

The purpose of this problem is to determine which systems will have to be treated
quantum mechanically, and which can safely be described classically.
(a) Solids. The lattice spacing in a typical solid is around d = 0.3 nm. Find the temperature below which the unbound23 electrons in a solid are quantum mechanical.
Below what temperature are the nuclei in a solid quantum mechanical? (Use silicon
as an example.)

23 In a solid the inner electrons are attached to a particular nucleus, and for them the relevant size would be the radius

of the atom. But the outer-most electrons are not attached, and for them the relevant distance is the lattice spacing.
This problem pertains to the outer electrons.

23

24

CHAPTER 1

The Wave Function

Moral: The free electrons in a solid are always quantum mechanical; the nuclei
are generally not quantum mechanical. The same goes for liquids (for which the
interatomic spacing is roughly the same), with the exception of helium below 4 K.
(b) Gases. For what temperatures are the atoms in an ideal gas at pressure P quantum
mechanical? Hint: Use the ideal gas law (P V = N k B T ) to deduce the interatomic
spacing.
3/5 2/5

P . Obviously (for the gas to show quantum
Answer: T < (1/k B ) h 2 /3m
behavior) we want m to be as small as possible, and P as large as possible. Put
in the numbers for helium at atmospheric pressure. Is hydrogen in outer space
(where the interatomic spacing is about 1 cm and the temperature is 3 K) quantum
mechanical? (Assume it’s monatomic hydrogen, not H2 .)

2

TIME-INDEPENDENT
SCHRÖDINGER
EQUATION

2.1 STATIONARY STATES
In Chapter 1 we talked a lot about the wave function, and how you use it to calculate various
quantities of interest. The time has come to stop procrastinating, and confront what is, logically,
the prior question: How do you get (x, t) in the first place? We need to solve the Schrödinger
equation,
i

2 ∂ 2 
∂
=−
+ V ,
∂t
2m ∂ x 2

(2.1)

for a specified potential1 V (x, t). In this chapter (and most of this book) I shall assume that
V is independent of t. In that case the Schrödinger equation can be solved by the method of
separation of variables (the physicist’s first line of attack on any partial differential equation):
We look for solutions that are products,
(x, t) = ψ(x) ϕ(t),

(2.2)

where ψ (lower-case) is a function of x alone, and ϕ is a function of t alone. On its face, this is
an absurd restriction, and we cannot hope to obtain more than a tiny subset of all solutions in
this way. But hang on, because the solutions we do get turn out to be of great interest. Moreover
(as is typically the case with separation of variables) we will be able at the end to patch together
the separable solutions in such a way as to construct the most general solution.
For separable solutions we have
dϕ
∂
=ψ
,
∂t
dt

∂ 2
d 2ψ
=
ϕ
∂x2
dx2

(ordinary derivatives, now), and the Schrödinger equation reads
iψ

2 d 2 ψ
dϕ
=−
ϕ + V ψϕ.
dt
2m d x 2

1 It is tiresome to keep saying “potential energy function,” so most people just call V the “potential,” even though this

invites occasional confusion with electric potential, which is actually potential energy per unit charge.

26

CHAPTER 2

Time-Independent Schrödinger Equation

Or, dividing through by ψϕ:
i

1 dϕ
2 1 d 2 ψ
+ V.
=−
ϕ dt
2m ψ d x 2

(2.3)

Now, the left side is a function of t alone, and the right side is a function of x alone.2 The
only way this can possibly be true is if both sides are in fact constant—otherwise, by varying
t, I could change the left side without touching the right side, and the two would no longer be
equal. (That’s a subtle but crucial argument, so if it’s new to you, be sure to pause and think
it through.) For reasons that will appear in a moment, we shall call the separation constant E.
Then
1 dϕ
= E,
i
ϕ dt
or
iE
dϕ
= − ϕ,
dt


(2.4)

and
−

2 1 d 2 ψ
+ V = E,
2m ψ d x 2

or
−

2 d 2 ψ
+ V ψ = Eψ.
2m d x 2

(2.5)

Separation of variables has turned a partial differential equation into two ordinary differential equations (Equations 2.4 and 2.5). The first of these is easy to solve (just multiply through
by dt and integrate); the general solution is C exp (−i Et/), but we might as well absorb the
constant C into ψ (since the quantity of interest is the product ψϕ). Then3
ϕ(t) = e−i Et/ .

(2.6)

The second (Equation 2.5) is called the time-independent Schrödinger equation; we can go
no further with it until the potential V (x) is specified.
The rest of this chapter will be devoted to solving the time-independent Schrödinger equation, for a variety of simple potentials. But before I get to that you have every right to ask:
What’s so great about separable solutions? After all, most solutions to the (time dependent)
Schrödinger equation do not take the form ψ(x) ϕ(t). I offer three answers—two of them
physical, and one mathematical:
2 Note that this would not be true if V were a function of t as well as x.
3 Using Euler’s formula,

eiθ = cos θ + i sin θ,
you could equivalently write
ϕ(t) = cos(Et/) + i sin(Et/);
the real and imaginary parts oscillate sinusoidally. Mike Casper (of Carleton College) dubbed ϕ the “wiggle
factor”—it’s the characteristic time dependence in quantum mechanics.

2.1 Stationary States

1. They are stationary states. Although the wave function itself,
(x, t) = ψ(x)e−i Et/ ,

(2.7)

does (obviously) depend on t, the probability density,
|(x, t)|2 =  ∗  = ψ ∗ e+i Et/ ψe−i Et/ = |ψ(x)|2 ,

(2.8)

does not—the time-dependence cancels out.4 The same thing happens in calculating the
expectation value of any dynamical variable; Equation 1.36 reduces to
 


d
Q(x, p) = ψ ∗ Q x, −i
ψ d x.
(2.9)
dx
Every expectation value is constant in time; we might as well drop the factor ϕ(t) altogether, and simply use ψ in place of . (Indeed, it is common to refer to ψ as “the
wave function,” but this is sloppy language that can be dangerous, and it is important to
remember that the true wave function always carries that time-dependent wiggle factor.)
In particular, x is constant, and hence (Equation 1.33)  p = 0. Nothing ever happens
in a stationary state.
2. They are states of definite total energy. In classical mechanics, the total energy (kinetic
plus potential) is called the Hamiltonian:
p2
+ V (x).
(2.10)
2m
The corresponding Hamiltonian operator, obtained by the canonical substitution p →
−i(∂/∂ x), is therefore5
H (x, p) =

Ĥ = −

2 ∂ 2
+ V (x).
2m ∂ x 2

(2.11)

Thus the time-independent Schrödinger equation (Equation 2.5) can be written
Ĥ ψ = Eψ,

(2.12)

and the expectation value of the total energy is



∗
2
H  = ψ Ĥ ψ d x = E |ψ| d x = E ||2 d x = E.

(2.13)

(Notice that the normalization of  entails the normalization of ψ.) Moreover,
Ĥ 2 ψ = Ĥ Ĥ ψ = Ĥ Eψ = E Ĥ ψ = E 2 ψ,
and hence


H2 =

ψ ∗ Ĥ 2 ψ d x = E 2


|ψ|2 d x = E 2 .

So the variance of H is
σ H2 = H 2 − H 2 = E 2 − E 2 = 0.

(2.14)

4 For normalizable solutions, E must be real (see Problem 2.1(a)).
5 Whenever confusion might arise, I’ll put a “hat” (ˆ) on the operator, to distinguish it from the dynamical variable it

represents.

27

28

CHAPTER 2

Time-Independent Schrödinger Equation

But remember, if σ = 0, then every member of the sample must share the same value
(the distribution has zero spread). Conclusion: A separable solution has the property
that every measurement of the total energy is certain to return the value E. (That’s why
I chose that letter for the separation constant.)
3. The general solution is a linear combination of separable solutions. As we’re about
to discover, the time-independent Schrödinger equation (Equation 2.5) yields an infinite collection of solutions (ψ1 (x), ψ2 (x), ψ3 (x), . . . , which we write as {ψn (x)}),
each with its associated separation constant (E 1 , E 2 , E 3 , . . . = {E n }); thus there is a
different wave function for each allowed energy:
1 (x, t) = ψ1 (x)e−i E 1 t/ ,

2 (x, t) = ψ2 (x)e−i E 2 t/ , . . . .

Now (as you can easily check for yourself) the (time-dependent) Schrödinger equation
(Equation 2.1) has the property that any linear combination6 of solutions is itself a solution. Once we have found the separable solutions, then, we can immediately construct
a much more general solution, of the form
(x, t) =

∞


cn ψn (x)e−i E n t/ .

(2.15)

n=1

It so happens that every solution to the (time-dependent) Schrödinger equation can be
written in this form—it is simply a matter of finding the right constants (c1 , c2 , . . .)
so as to fit the initial conditions for the problem at hand. You’ll see in the following sections how all this works out in practice, and in Chapter 3 we’ll put it into
more elegant language, but the main point is this: Once you’ve solved the timeindependent Schrödinger equation, you’re essentially done; getting from there to the
general solution of the time-dependent Schrödinger equation is, in principle, simple and
straightforward.
A lot has happened in the past four pages, so let me recapitulate, from a somewhat different perspective. Here’s the generic problem: You’re given a (time-independent) potential
V (x), and the starting wave function (x, 0); your job is to find the wave function, (x, t),
for any subsequent time t. To do this you must solve the (time-dependent) Schrödinger equation (Equation 2.1). The strategy is first to solve the time-independent Schrödinger equation
(Equation 2.5); this yields, in general, an infinite set of solutions, {ψn (x)}, each with its own
associated energy, {E n }. To fit (x, 0) you write down the general linear combination of these
solutions:
(x, 0) =

∞


cn ψn (x);

n=1

6 A linear combination of the functions f (z), f (z), . . . is an expression of the form
1
2

f (z) = c1 f 1 (z) + c2 f 2 (z) + · · · ,
where c1 , c2 , . . . are (possibly complex) constants.

(2.16)

2.1 Stationary States

the miracle is that you can always match the specified initial state7 by appropriate choice of
the constants {cn }. To construct (x, t) you simply tack onto each term its characteristic time
dependence (its “wiggle factor”), exp (−i E n t/):8

(x, t) =

∞


cn ψn (x)e−i E n t/ =

n=1

∞


cn n (x, t).

(2.17)

n=1

The separable solutions themselves,
n (x, t) = ψn (x)e−i E n t/ ,

(2.18)

are stationary states, in the sense that all probabilities and expectation values are independent
of time, but this property is emphatically not shared by the general solution (Equation 2.17):
the energies are different, for different stationary states, and the exponentials do not cancel,
when you construct ||2 .
Example 2.1
Suppose a particle starts out in a linear combination of just two stationary states:
(x, 0) = c1 ψ1 (x) + c2 ψ2 (x).
(To keep things simple I’ll assume that the constants cn and the states ψn(x) are real.) What
is the wave function (x, t) at subsequent times? Find the probability density, and describe
its motion.
Solution: The first part is easy:
(x, t) = c1 ψ1 (x)e−i E 1 t/ + c2 ψ2 (x) e−i E 2 t/ ,
where E 1 and E 2 are the energies associated with ψ1 and ψ2 . It follows that
c1 ψ1 e−i E 1 t/ + c2 ψ2 e−i E 2 t/


= c12 ψ12 + c22 ψ22 + 2c1 c2 ψ1 ψ2 cos (E 2 − E 1 ) t/ .

|(x, t)|2 = c1 ψ1 ei E 1 t/ + c2 ψ2 ei E 2 t/

The probability density oscillates sinusoidally, at an angular frequency ω = (E 2 − E 1 ) /;
this is certainly not a stationary state. But notice that it took a linear combination of
stationary states (with different energies) to produce motion.9

7 In principle, any normalized function (x, 0) is fair game—it need not even be continuous. How you might actually

get a particle into that state is a different question, and one (curiously) we seldom have occasion to ask.
8 If this is your first encounter with the method of separation of variables, you may be disappointed that the solution

takes the form of an infinite series. Occasionally it is possible to sum the series, or to solve the time-dependent
Schrödinger equation without recourse to separation of variables—see, for instance, Problems 2.49, 2.50, and 2.51.
But such cases are extremely rare.
9 This is nicely illustrated in an applet by Paul Falstad, at www.falstad.com/qm1d/.

29

30

CHAPTER 2

Time-Independent Schrödinger Equation

You may be wondering what the coefficients {cn } represent physically. I’ll tell you the
answer, though the explanation will have to await Chapter 3:
|cn |2

is the probability that a measurement of the
energy would return the value E n .

(2.19)

A competent measurement will always yield one of the “allowed” values (hence the name),
and |cn |2 is the probability of getting the particular value E n .10 Of course, the sum of these
probabilities should be 1:
∞


|cn |2 = 1,

(2.20)

n=1

and the expectation value of the energy must be

H  =

∞


|cn |2 E n .

(2.21)

n=1

We’ll soon see how this works out in some concrete examples. Notice, finally, that because the
constants {cn } are independent of time, so too is the probability of getting a particular energy,
and, a fortiori, the expectation value of H . These are manifestations of energy conservation
in quantum mechanics.
∗

Problem 2.1 Prove the following three theorems:
(a) For normalizable solutions, the separation constant E must be real. Hint: Write E
(in Equation 2.7) as E 0 + i (with E 0 and real), and show that if Equation 1.20
is to hold for all t, must be zero.
(b) The time-independent wave function ψ(x) can always be taken to be real (unlike
(x, t), which is necessarily complex). This doesn’t mean that every solution to
the time-independent Schrödinger equation is real; what it says is that if you’ve got
one that is not, it can always be expressed as a linear combination of solutions (with
the same energy) that are. So you might as well stick to ψs that are real. Hint: If
ψ(x) satisfies Equation 2.5, for a given E, so too does its complex conjugate, and
hence also the real linear combinations (ψ + ψ ∗ ) and i (ψ − ψ ∗ ).
(c) If V (x) is an even function (that is, V (−x) = V (x)) then ψ(x) can always be taken
to be either even or odd. Hint: If ψ(x) satisfies Equation 2.5, for a given E, so too
does ψ(−x), and hence also the even and odd linear combinations ψ(x) ± ψ(−x).

10 Some people will tell you that |c |2 is “the probability that the particle is in the nth stationary state,” but this is bad
n

language: the particle is in the state , not n , and anyhow, in the laboratory you don’t “find the particle to be in
a particular state,” you measure some observable, and what you get is a number, not a wave function.

2.2 The Infinite Square Well
V(x)

a

x

Figure 2.1: The infinite square well potential (Equation 2.22).

∗

Problem 2.2 Show that E must exceed the minimum value of V (x), for every normalizable
solution to the time-independent Schrödinger equation. What is the classical analog to this
statement? Hint: Rewrite Equation 2.5 in the form
2m
d 2ψ
= 2 [V (x) − E] ψ;
2
dx

if E < Vmin , then ψ and its second derivative always have the same sign—argue that such
a function cannot be normalized.

2.2 THE INFINITE SQUARE WELL
Suppose


V (x) =

0,
∞,

0 ≤ x ≤ a,
otherwise

(2.22)

(Figure 2.1). A particle in this potential is completely free, except at the two ends (x = 0 and
x = a), where an infinite force prevents it from escaping. A classical model would be a cart
on a frictionless horizontal air track, with perfectly elastic bumpers—it just keeps bouncing
back and forth forever. (This potential is artificial, of course, but I urge you to treat it with
respect. Despite its simplicity—or rather, precisely because of its simplicity—it serves as a
wonderfully accessible test case for all the fancy machinery that comes later. We’ll refer back
to it frequently.)
Outside the well, ψ(x) = 0 (the probability of finding the particle there is zero). Inside the
well, where V = 0, the time-independent Schrödinger equation (Equation 2.5) reads
−
or

2 d 2 ψ
= Eψ,
2m d x 2

(2.23)

√
d 2ψ
2m E
2
.
(2.24)
= −k ψ, where k ≡
2

dx
(By writing it in this way, I have tacitly assumed that E ≥ 0; we know from Problem 2.2 that
E < 0 won’t work.) Equation 2.24 is the classical simple harmonic oscillator equation; the
general solution is

31

32

CHAPTER 2

Time-Independent Schrödinger Equation

ψ(x) = A sin kx + B cos kx,

(2.25)

where A and B are arbitrary constants. Typically, these constants are fixed by the boundary
conditions of the problem. What are the appropriate boundary conditions for ψ(x)? Ordinarily,
both ψ and dψ/d x are continuous,11 but where the potential goes to infinity only the first
of these applies. (I’ll justify these boundary conditions, and account for the exception when
V = ∞, in Section 2.5; for now I hope you will trust me.)
Continuity of ψ(x) requires that
ψ(0) = ψ(a) = 0,

(2.26)

so as to join onto the solution outside the well. What does this tell us about A and B? Well,
ψ(0) = A sin 0 + B cos 0 = B,
so B = 0, and hence
ψ(x) = A sin kx.

(2.27)

Then ψ(a) = A sin ka, so either A = 0 (in which case we’re left with the trivial—nonnormalizable—solution ψ(x) = 0), or else sin ka = 0, which means that
ka = 0, ±π, ±2π, ±3π, . . . .

(2.28)

But k = 0 is no good (again, that would imply ψ(x) = 0), and the negative solutions give
nothing new, since sin(−θ ) = − sin(θ ) and we can absorb the minus sign into A. So the
distinct solutions are
nπ
, with n = 1, 2, 3, . . . .
(2.29)
kn =
a
Curiously, the boundary condition at x = a does not determine the constant A, but rather
the constant k, and hence the possible values of E:
En =

n 2 π 2 2
2 kn2
=
.
2m
2ma 2

(2.30)

In radical contrast to the classical case, a quantum particle in the infinite square well cannot
have just any old energy—it has to be one of these special (“allowed”) values.12 To find A, we
normalize ψ:13
 a
a
2
|A|2 sin2 (kx) d x = |A|2 = 1, so |A|2 = .
2
a
0
This only determines the magnitude of A, but it is simplest to pick the positive real root:
√
A = 2/a (the phase of A carries no physical significance anyway). Inside the well, then, the
solutions are

nπ
2
sin
x .
ψn (x) =
(2.31)
a
a
11 That’s right: ψ(x) is a continuous function of x, even though (x, t) need not be.
12 Notice that the quantization of energy emerges as a rather technical consequence of the boundary conditions on

solutions to the time-independent Schrödinger equation.
13 Actually, it’s (x, t) that must be normalized, but in view of Equation 2.7 this entails the normalization of ψ(x).

2.2 The Infinite Square Well
Ψ1(x)

Ψ2(x)

a

Ψ3(x)

x

a

x

a

x

Figure 2.2: The first three stationary states of the infinite square well (Equation 2.31).

As promised, the time-independent Schrödinger equation has delivered an infinite set of
solutions (one for each positive integer n). The first few of these are plotted in Figure 2.2.
They look just like the standing waves on a string of length a; ψ1 , which carries the lowest
energy, is called the ground state, the others, whose energies increase in proportion to n 2 , are
called excited states. As a collection, the functions ψn (x) have some interesting and important
properties:
1. They are alternately even and odd, with respect to the center of the well: ψ1 is even, ψ2
is odd, ψ3 is even, and so on.14
2. As you go up in energy, each successive state has one more node (zero-crossing): ψ1
has none (the end points don’t count), ψ2 has one, ψ3 has two, and so on.
3. They are mutually orthogonal, in the sense that15


ψm (x)∗ ψn (x) d x = 0,

(m = n) .

(2.32)

Proof:


nπ
mπ
2 a
x sin
x dx
sin
ψm (x)∗ ψn (x) d x =
a 0
a
a



  
1 a
m−n
m+n
=
cos
π x − cos
πx
dx
a 0
a
a

a




1
m−n
m+n
1
πx −
π x 
sin
sin
=
a
a
(m − n) π
(m + n) π
0


1 sin [(m − n) π ] sin [(m + n) π ]
−
= 0.
=
π
(m − n)
(m + n)
Note that this argument does not work if m = n. (Can you spot the point at which it
fails?) In that case normalization tells us that the integral is 1. In fact, we can combine
orthogonality and normalization into a single statement:


ψm (x)∗ ψn (x) d x = δmn ,

(2.33)

14 To make this symmetry more apparent, some authors center the well at the origin (running it now from −a to +a).

The even functions are then cosines, and the odd ones are sines. See Problem 2.36.
15 In this case the ψs are real, so the complex conjugation (*) of ψ is unnecessary, but for future purposes it’s a
m

good idea to get in the habit of putting it there.

33

34

CHAPTER 2

Time-Independent Schrödinger Equation

where δmn (the so-called Kronecker delta) is defined by

0, m = n,
δmn =
1, m = n.

(2.34)

We say that the ψs are orthonormal.
4. They are complete, in the sense that any other function, f (x), can be expressed as a
linear combination of them:
 ∞
∞

2
nπ
x .
(2.35)
cn ψn (x) =
cn sin
f (x) =
a
a
n=1

n=1

√
I’m not about to prove the completeness of the functions 2/a sin(nπ x/a), but if
you’ve studied advanced calculus you will recognize that Equation 2.35 is nothing but
the Fourier series for f (x), and the fact that “any” function can be expanded in this
way is sometimes called Dirichlet’s theorem.16
The coefficients cn can be evaluated—for a given f (x)—by a method I call Fourier’s
trick, which beautifully exploits the orthonormality of {ψn }: Multiply both sides of
Equation 2.35 by ψm (x)∗ , and integrate.


ψm (x)∗ f (x) d x =

∞



cn

ψm (x)∗ ψn (x) d x =

n=1

∞


cn δmn = cm .

(2.36)

n=1

(Notice how the Kronecker delta kills every term in the sum except the one for which
n = m.) Thus the nth coefficient in the expansion of f (x) is17

cn =

ψn (x)∗ f (x) d x.

(2.37)

These four properties are extremely powerful, and they are not peculiar to the infinite square
well. The first is true whenever the potential itself is a symmetric function; the second is
universal, regardless of the shape of the potential.18 Orthogonality is also quite general—I’ll
show you the proof in Chapter 3. Completeness holds for all the potentials you are likely to
encounter, but the proofs tend to be nasty and laborious; I’m afraid most physicists simply
assume completeness, and hope for the best.
The stationary states (Equation 2.18) of the infinite square well are

nπ
2
2 2
2
sin
x e−i(n π /2ma )t .
(2.38)
n (x, t) =
a
a
I claimed (Equation 2.17) that the most general solution to the (time-dependent) Schrödinger
equation is a linear combination of stationary states:
16 See, for example, Mary Boas, Mathematical Methods in the Physical Sciences, 3rd edn (New York: John Wiley,

2006), p. 356; f (x) can even have a finite number of finite discontinuities.
17 It doesn’t matter whether you use m or n as the “dummy index” here (as long as you are consistent on the two sides

of the equation, of course); whatever letter you use, it just stands for “any positive integer.”
18 Problem 2.45 explores this property. For further discussion, see John L. Powell and Bernd Crasemann, Quantum

Mechanics (Addison-Wesley, Reading, MA, 1961), Section 5-7.

2.2 The Infinite Square Well

(x, t) =

∞



cn

n=1

2
nπ
2 2
2
sin
x e−i(n π /2ma )t .
a
a

(2.39)

(If you doubt that this is a solution, by all means check it!) It remains only for me to demonstrate that I can fit any prescribed initial wave function, (x, 0) by appropriate choice of the
coefficients cn :
(x, 0) =

∞


cn ψn (x) .

n=1

The completeness of the ψs (confirmed in this case by Dirichlet’s theorem) guarantees that I
can always express (x, 0) in this way, and their orthonormality licenses the use of Fourier’s
trick to determine the actual coefficients:
 
2 a
nπ
x (x, 0) d x.
(2.40)
cn =
sin
a 0
a
That does it: Given the initial wave function, (x, 0), we first compute the expansion coefficients cn , using Equation 2.40, and then plug these into Equation 2.39 to obtain (x, t).
Armed with the wave function, we are in a position to compute any dynamical quantities of
interest, using the procedures in Chapter 1. And this same ritual applies to any potential—the
only things that change are the functional form of the ψs and the equation for the allowed
energies.
Example 2.2
A particle in the infinite square well has the initial wave function
(x, 0) = Ax (a − x) ,

(0 ≤ x ≤ a) ,

for some constant A (see Figure 2.3). Outside the well, of course,  = 0. Find (x, t).
Ψ(x, 0)
Aa2
4

a

x

Figure 2.3: The starting wave function in Example 2.2.

Solution: First we need to determine A, by normalizing (x, 0):
 a
 a
a5
2
2
|(x, 0)| d x = |A|
1=
x 2 (a − x)2 d x = |A|2 ,
30
0
0
so

30
A=
.
a5
The nth coefficient is (Equation 2.40)

35

36

CHAPTER 2

Time-Independent Schrödinger Equation

 

2 a
30
nπ
cn =
sin
x (a − x) d x
x
a 0
a
a5
√   a

 a
2 15
nπ
nπ
2
a
x dx −
x dx
=
x sin
x sin
a
a
a3
0
0
√  
 a

a 2
ax
nπ
2 15
nπ
a
x −
cos
x 
=
sin
3
nπ
a
nπ
a
a
0

a 
2

a 2
nπ
nπ
(nπ x/a) − 2
x −
x 
− 2
x sin
cos
3
nπ
a
a
(nπ/a)
0
√ 
2
2 15
a3
(nπ )2 − 2
cos (nπ ) + a 3
=
cos (nπ ) + a 3
cos (0)
−
3
nπ
a
(nπ )3
(nπ )3
√
4 15
[cos (0) − cos (nπ )]
=
(nπ )3

0,
n even,
= √
8 15/ (nπ )3 , n odd.
Thus (Equation 2.39):
(x, t) =



30
a

 3
2
π


n=1,3,5...

1
nπ
2 2
2
x e−in π t/2ma .
sin
a
n3

Example 2.3
Check that Equation 2.20 is satisfied, for the wave function in Example 2.2. If you measured the energy of a particle in this state, what is the most probable result? What is the
expectation value of the energy?
Solution: The starting wave function (Figure 2.3) closely resembles the ground state ψ1
(Figure 2.2). This suggests that |c1 |2 should dominate,19 and in fact
! √ "2
8 15
|c1 |2 =
= 0.998555 . . . .
π3
The rest of the coefficients make up the difference:20

19 Loosely speaking, c tells you the “amount of ψ that is contained in .”
n
n
20 You can look up the series

1
1
1
π6
+ 6 + 6 + ··· =
6
960
1
3
5
and
1
1
π4
1
+
+
+
·
·
·
=
96
14
34
54
in math tables, under “Sums of Reciprocal Powers” or “Riemann Zeta Function.”

2.2 The Infinite Square Well
∞

n=1

! √ "2
8 15
|cn |2 =
π3

∞

n=1,3,5,...

1
= 1.
n6

The most likely outcome of an energy measurement is E 1 = π 2 2 /2ma 2 —more than
99.8% of all measurements will yield this value. The expectation value of the energy
(Equation 2.21) is
! √ "2
∞
∞


8 15
n 2 π 2 2
1
4802
52
H  =
=
=
.
n3π 3
2ma 2
π 4 ma 2
n4
ma 2
n=1,3,5,...

n=1,3,5,...

As one would expect, it is very close to E 1 (5 in place of π 2 /2 ≈ 4.935)—slightly larger,
because of the admixture of excited states.
Of course, it’s no accident that Equation 2.20 came out right in Example 2.3. Indeed, this
follows from the normalization of  (the cn s are independent of time, so I’m going to do the
proof for t = 0; if this bothers you, you can easily generalize the argument to arbitrary t).
"∗ ! ∞
"
 !

∞

2
cm ψm (x)
cn ψn (x) d x
1 = |(x, 0)| d x =
=
=

∞ 
∞

m=1 n=1
∞ 
∞


∗
cm
cn



m=1

ψm (x)∗ ψn (x) d x

∗
cm
cn δmn =

n=1 m=1

n=1

∞


|cn |2 .

n=1

(Again, the Kronecker delta picks out the term m = n in the summation over m.) Similarly, the expectation value of the energy (Equation 2.21) can be checked explicitly: The
time-independent Schrödinger equation (Equation 2.12) says
Ĥ ψn = E n ψn ,
so



(2.41)





∗
 Ĥ  d x =
cm ψm Ĥ
cn ψn d x



∗
|cn |2 E n .
=
cm
cn E n ψm∗ ψn d x =

H  =

∗

Problem 2.3 Show that there is no acceptable solution to the (time-independent)
Schrödinger equation for the infinite square well with E = 0 or E < 0. (This is a special
case of the general theorem in Problem 2.2, but this time do it by explicitly solving the
Schrödinger equation, and showing that you cannot satisfy the boundary conditions.)

∗

Problem 2.4 Calculate x, x 2 ,  p, p 2 , σx , and σ p , for the nth stationary state of the
infinite square well. Check that the uncertainty principle is satisfied. Which state comes
closest to the uncertainty limit?

37

38

CHAPTER 2

∗

Time-Independent Schrödinger Equation

Problem 2.5 A particle in the infinite square well has as its initial wave function an even
mixture of the first two stationary states:
(x, 0) = A [ψ1 (x) + ψ2 (x)] .
(a) Normalize (x, 0). (That is, find A. This is very easy, if you exploit the orthonormality of ψ1 and ψ2 . Recall that, having normalized  at t = 0, you can rest
assured that it stays normalized—if you doubt this, check it explicitly after doing
part (b).)
(b) Find (x, t) and |(x, t)|2 . Express the latter as a sinusoidal function of time, as
in Example 2.1. To simplify the result, let ω ≡ π 2 /2ma 2 .
(c) Compute x. Notice that it oscillates in time. What is the angular frequency of the
oscillation? What is the amplitude of the oscillation? (If your amplitude is greater
than a/2, go directly to jail.)
(d) Compute  p. (As Peter Lorre would say, “Do it ze kveek vay, Johnny!”)
(e) If you measured the energy of this particle, what values might you get, and what is
the probability of getting each of them? Find the expectation value of H . How does
it compare with E 1 and E 2 ?

Problem 2.6 Although the overall phase constant of the wave function is of no physical
significance (it cancels out whenever you calculate a measurable quantity), the relative
phase of the coefficients in Equation 2.17 does matter. For example, suppose we change
the relative phase of ψ1 and ψ2 in Problem 2.5:
#
$
(x, 0) = A ψ1 (x) + eiφ ψ2 (x) ,
where φ is some constant. Find (x, t), |(x, t)|2 , and x, and compare your results
with what you got before. Study the special cases φ = π/2 and φ = π . (For a graphical
exploration of this problem see the applet in footnote 9 of this chapter.)

∗

Problem 2.7 A particle in the infinite square well has the initial wave function

Ax,
0 ≤ x ≤ a/2,
(x, 0) =
A (a − x) , a/2 ≤ x ≤ a.
(a)
(b)
(c)
(d)

Sketch (x, 0), and determine the constant A.
Find (x, t).
What is the probability that a measurement of the energy would yield the value E 1 ?
Find the expectation value of the energy, using Equation 2.21.21

21 Remember, there is no restriction in principle on the shape of the starting wave function, as long as it is normal-

izable.
In particular, (x, 0) need not have a continuous derivative. However, if you try to calculate H  using

(x, 0)∗ Ĥ (x, 0) d x in such a case, you may encounter technical difficulties, because the second derivative of
(x, 0) is ill defined. It works in Problem 2.9 because the discontinuities occur at the end points, where the wave
function is zero anyway. In Problem 2.39 you’ll see how to manage cases like Problem 2.7.

2.3 The Harmonic Oscillator

Problem 2.8 A particle of mass m in the infinite square well (of width a) starts out in the
state

A, 0 ≤ x ≤ a/2,
(x, 0) =
0, a/2 ≤ x ≤ a,
for some constant A, so it is (at t = 0) equally likely to be found at any point in the left
half of the well. What is the probability that a measurement of the energy (at some later
time t) would yield the value π 2 2 /2ma 2 ?

Problem 2.9 For the wave function in Example 2.2, find the expectation value of H , at time
t = 0, the “old fashioned” way:

H  = (x, 0)∗ Ĥ (x, 0) d x.
Compare the result we got in Example 2.3. Note: Because H  is independent of time,
there is no loss of generality in using t = 0.

2.3 THE HARMONIC OSCILLATOR
The paradigm for a classical harmonic oscillator is a mass m attached to a spring of force
constant k. The motion is governed by Hooke’s law,
F = −kx = m

d2x
dt 2

(ignoring friction), and the solution is
x(t) = A sin(ωt) + B cos(ωt) ,
where



k
m
is the (angular) frequency of oscillation. The potential energy is
ω≡

V (x) =

1 2
kx ;
2

(2.42)

(2.43)

its graph is a parabola.
Of course, there’s no such thing as a perfect harmonic oscillator—if you stretch it too far the
spring is going to break, and typically Hooke’s law fails long before that point is reached. But
practically any potential is approximately parabolic, in the neighborhood of a local minimum
(Figure 2.4). Formally, if we expand V (x) in a Taylor series about the minimum:
1
V (x) = V (x0 ) + V (x0 ) (x − x0 ) + V (x0 ) (x − x0 )2 + · · · ,
2
subtract V (x0 ) (you can add a constant to V (x) with impunity, since that doesn’t change the
force), recognize that V (x0 ) = 0 (since x0 is a minimum), and drop the higher-order terms
(which are negligible as long as (x − x0 ) stays small), we get

39

40

CHAPTER 2

Time-Independent Schrödinger Equation
V(x)

x0

x

Figure 2.4: Parabolic approximation (dashed curve) to an arbitrary potential, in the neighborhood of a
local minimum.

1
V (x0 ) (x − x0 )2 ,
2
which describes simple harmonic oscillation (about the point x0 ), with an effective spring
constant k = V (x0 ). That’s why the simple harmonic oscillator is so important: Virtually any
oscillatory motion is approximately simple harmonic, as long as the amplitude is small.22
The quantum problem is to solve the Schrödinger equation for the potential
V (x) ≈

1
mω2 x 2
(2.44)
2
(it is customary to eliminate the spring constant in favor of the classical frequency, using
Equation 2.42). As we have seen, it suffices to solve the time-independent Schrödinger
equation:
V (x) =

−

1
2 d 2 ψ
+ mω2 x 2 ψ = Eψ.
2
2m d x
2

(2.45)

In the literature you will find two entirely different approaches to this problem. The first is
a straightforward “brute force” solution to the differential equation, using the power series
method; it has the virtue that the same strategy can be applied to many other potentials (in
fact, we’ll use it in Chapter 4 to treat the hydrogen atom). The second is a diabolically clever
algebraic technique, using so-called ladder operators. I’ll show you the algebraic method
first, because it is quicker and simpler (and a lot more fun);23 if you want to skip the power
series method for now, that’s fine, but you should certainly plan to study it at some stage.
2.3.1

Algebraic Method
To begin with, let’s rewrite Equation 2.45 in a more suggestive form:
$
1 # 2
p̂ + (mωx)2 ψ = Eψ,
2m

(2.46)

22 Note that V (x ) ≥ 0, since by assumption x is a minimum. Only in the rare case V (x ) = 0 is the oscillation
0
0
0

not even approximately simple harmonic.
23 We’ll encounter some of the same strategies in the theory of angular momentum (Chapter 4), and the technique

generalizes to a broad class of potentials in supersymmetric quantum mechanics (Problem 3.47; see also Richard
W. Robinett, Quantum Mechanics (Oxford University Press, New York, 1997), Section 14.4).

2.3 The Harmonic Oscillator

where p̂ ≡ −i d/d x is the momentum operator.24 The basic idea is to factor the Hamiltonian,
$
1 # 2
p̂ + (mωx)2 .
(2.47)
Ĥ =
2m
If these were numbers, it would be easy:
u 2 + v 2 = (iu + v) (−iu + v) .
Here, however, it’s not quite so simple, because p̂ and x are operators, and operators do not,
in general, commute (x p̂ is not the same as p̂x, as we’ll see in a moment—though you might
want to stop right now and think it through for yourself). Still, this does motivate us to examine
the quantities


1
∓i p̂ + mωx
(2.48)
â± ≡ √
2mω
(the factor in front is just there to make the final results look nicer).
Well, what is the product â− â+ ?
1
2mω
1
=
2mω

â− â+ =




i p̂ + mωx −i p̂ + mωx
#

$
p̂ 2 + (mωx)2 − imω x p̂ − p̂x .



As anticipated, there’s an extra term, involving x p̂ − p̂x . We call this the commutator of
x and p̂; it is a measure of how badly they fail to commute. In general, the commutator of
operators  and B̂ (written with square brackets) is
#
$
Â, B̂ ≡ Â B̂ − B̂ Â.
(2.49)
In this notation,
â− â+ =

$

i 
1 # 2
p̂ + (mωx)2 −
x, p̂ .
2mω
2

(2.50)

We need to figure out the commutator of x and p̂. Warning: Operators are notoriously
slippery to work with in the abstract, and you are bound to make mistakes unless you give
them a “test function,” f (x), to act on. At the end you can throw away the test function, and
you’ll be left with an equation involving the operators alone. In the present case we have:






df
df
d
d
−x
− f
x, p̂ f (x) = x (−i)
( f ) − (−i)
(x f ) = −i x
dx
dx
dx
dx
= i f (x) .

(2.51)

Dropping the test function, which has served its purpose,



x, p̂ = i.

(2.52)

This lovely and ubiquitous formula is known as the canonical commutation relation.25
24 Put a hat on x, too, if you like, but since x̂ = x we usually leave it off.
25 In a deep sense all of the mysteries of quantum mechanics can be traced to the fact that position and momentum

do not commute. Indeed, some authors take the canonical commutation relation as an axiom of the theory, and use
it to derive p̂ = −i d/d x.

41

42

CHAPTER 2

Time-Independent Schrödinger Equation

With this, Equation 2.50 becomes
â− â+ =
or

1
1
Ĥ + ,
ω
2



1
Ĥ = ω â− â+ −
.
2

(2.53)

(2.54)

Evidently the Hamiltonian does not factor perfectly—there’s that extra −1/2 on the right.
Notice that the ordering of â+ and â− is important here; the same argument, with â+ on the
left, yields
1
1
(2.55)
â+ â− =
Ĥ − .
ω
2
In particular,


â− , â+ = 1.
(2.56)
Meanwhile, the Hamiltonian can equally well be written


1
.
Ĥ = ω â+ â− +
2

(2.57)

In terms of â± , then, the Schrödinger equation26 for the harmonic oscillator takes the form


1
ψ = Eψ
(2.58)
ω â± â∓ ±
2
(in equations like this you read the upper signs all the way across, or else the lower signs).
Now, here comes the crucial step: I claim that:
If ψ satisfies the Schrödinger equation with energy E (that is: Ĥ ψ =
equation with energy (E +
Eψ), then

 â+ψ satisfies the Schrödinger
ω): Ĥ â+ ψ = (E + ω) â+ ψ .
Proof:








1 
1
â+ ψ = ω â+ â− â+ + â+ ψ
Ĥ â+ ψ = ω â+ â− +
2 
2

 
 
1
1
= ωâ+ â− â+ +
ψ = â+ ω â+ â− + 1 +
ψ
2
2


= â+ Ĥ + ω ψ = â+ (E + ω)ψ = (E + ω) â+ ψ . QED


(I used Equation 2.56 to replace â− â+ by â+ â− + 1 in the second line. Notice that
whereas the ordering of â+ and â− does matter, the ordering of â± and any constants—
such as , ω, and E—does not; an operator commutes with any constant.)
By the same token, â− ψ is a solution with energy (E − ω):







1 
1
Ĥ â− ψ = ω â− â+ −
â− ψ = ωâ− â+ â− −
ψ
2
2
 
 
1
= â− ω â− â+ − 1 −
ψ = â− Ĥ − ω ψ = â− (E − ω)ψ
2


= (E − ω) â− ψ .

26 I’m getting tired of writing “time-independent Schrödinger equation,” so when it’s clear from the context which

one I mean, I’ll just call it the “Schrödinger equation.”

2.3 The Harmonic Oscillator
E

E+3h ω

â +3ψ

E+2h ω

â +2ψ
â +ψ

E+ h ω

ψ

E

E− h ω
E−2h ω

â+

â_ψ
â _2ψ

â_
E0

ψ0

Figure 2.5: Th