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Introduction to Quantum Mechanics

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The quality of the equations is poor.
08 June 2020 (03:49) 
Mukhtar Musa Muhammad
Reading culture development
09 September 2020 (11:23) 
Awesome! This book is quite helpful for my study!
13 September 2020 (14:57) 
Damn griffiths is a gift from heavens xD A frikin good book and one of the best for beginners
10 November 2020 (18:44) 
Please give the download link
02 March 2021 (17:51) 
Griffiths often skips steps and does not bother to explain things that are “obvious”. For me this can sometimes be troublesome. Luckily you can find a full QM course (based on this book) on YouTube by Brant Carlson, who explains things with a little bit more details.
23 April 2021 (11:09) 
patel lav ishwarbhai
It's very useful app.?????
18 May 2021 (00:52) 
I am here to mark the beginning of journey.
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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.


Third edition
Reed College, Oregon


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DOI: 10.1017/9781316995433
Second edition © David Griffiths 2017
Third edition © Cambridge University Press 2018
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
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I Theory
1 The Wave Function
1.1 The Schrödinger Equation
1.2 The Statistical Interpretation
1.3 Probability
1.3.1 Discrete Variables
1.3.2 Continuous Variables
1.4 Normalization
1.5 Momentum
1.6 The Uncertainty Principle
Further Problems on Chapter 1
2 Time-Independent Schrödinger Equation
2.1 Stationary States
2.2 The Infinite Square Well
2.3 The Harmonic Oscillator
2.3.1 Algebraic Method
2.3.2 Analytic Method
2.4 The Free Particle
2.5 The Delta-Function Potential
2.5.1 Bound States and Scattering States
2.5.2 The Delta-Function Well
2.6 The Finite Square Well
Further Problems on Chapter 2
3 Formalism
3.1 Hilbert Space
3.2 Observables
3.2.1 Hermitian Operators
3.2.2 Determinate States
3.3 Eigenfunctions of a Hermitian Operator
3.3.1 Discrete Spectra
3.3.2 Continuous Spectra

3.4 Generalized Statistical Interpretation
3.5 The Uncertainty Principle
3.5.1 Proof of the Generalized Uncertainty Principle
3.5.2 The Minimum-Uncertainty Wave Packet
3.5.3 The Energy-Time Uncertainty Principle
3.6 Vectors and Operators
3.6.1 Bases in Hilbert Space
3.6.2 Dirac Notation
3.6.3 Changing Bases in Dirac Notation
Further Problems on Chapter 3
4 Quantum Mechanics in Three Dimensions
4.1 The Schröger Equation
4.1.1 Spherical Coordinates
4.1.2 The Angular Equation
4.1.3 The Radial Equation
4.2 The Hydrogen Atom
4.2.1 The Radial Wave Function
4.2.2 The Spectrum of Hydrogen
4.3 Angular Momentum
4.3.1 Eigenvalues
4.3.2 Eigenfunctions
4.4 Spin
4.4.1 Spin 1/2
4.4.2 Electron in a Magnetic Field
4.4.3 Addition of Angular Momenta
4.5 Electromagnetic Interactions
4.5.1 Minimal Coupling
4.5.2 The Aharonov–Bohm Effect
Further Problems on Chapter 4
5 Identical Particles
5.1 Two-Particle Systems
5.1.1 Bosons and Fermions
5.1.2 Exchange Forces
5.1.3 Spin
5.1.4 Generalized Symmetrization Principle
5.2 Atoms
5.2.1 Helium
5.2.2 The Periodic Table
5.3 Solids
5.3.1 The Free Electron Gas
5.3.2 Band Structure
Further Problems on Chapter 5


6 Symmetries & Conservation Laws
6.1 Introduction
6.1.1 Transformations in Space
6.2 The Translation Operator
6.2.1 How Operators Transform
6.2.2 Translational Symmetry
6.3 Conservation Laws
6.4 Parity
6.4.1 Parity in One Dimension
6.4.2 Parity in Three Dimensions
6.4.3 Parity Selection Rules
6.5 Rotational Symmetry
6.5.1 Rotations About the z Axis
6.5.2 Rotations in Three Dimensions
6.6 Degeneracy
6.7 Rotational Selection Rules
6.7.1 Selection Rules for Scalar Operators
6.7.2 Selection Rules for Vector Operators
6.8 Translations in Time
6.8.1 The Heisenberg Picture
6.8.2 Time-Translation Invariance
Further Problems on Chapter 6

II Applications
7 Time-Independent Perturbation Theory
7.1 Nondegenerate Perturbation Theory
7.1.1 General Formulation
7.1.2 First-Order Theory
7.1.3 Second-Order Energies
7.2 Degenerate Perturbation Theory
7.2.1 Two-Fold Degeneracy
7.2.2 “Good” States
7.2.3 Higher-Order Degeneracy
7.3 The Fine Structure of Hydrogen
7.3.1 The Relativistic Correction
7.3.2 Spin-Orbit Coupling
7.4 The Zeeman Effect
7.4.1 Weak-Field Zeeman Effect
7.4.2 Strong-Field Zeeman Effect
7.4.3 Intermediate-Field Zeeman Effect
7.5 Hyperfine Splitting in Hydrogen

Further Problems on Chapter 7
8 The Varitional Principle
8.1 Theory
8.2 The Ground State of Helium
8.3 The Hydrogen Molecule Ion
8.4 The Hydrogen Molecule
Further Problems on Chapter 8
9 The WKB Approximation
9.1 The “Classical” Region
9.2 Tunneling
9.3 The Connection Formulas
Further Problems on Chapter 9
10 Scattering
10.1 Introduction
10.1.1 Classical Scattering Theory
10.1.2 Quantum Scattering Theory
10.2 Partial Wave Analysis
10.2.1 Formalism
10.2.2 Strategy
10.3 Phase Shifts
10.4 The Born Approximation
10.4.1 Integral Form of the Schrödinger Equation
10.4.2 The First Born Approximation
10.4.3 The Born Series
Further Problems on Chapter 10
11 Quantum Dynamics
11.1 Two-Level Systems
11.1.1 The Perturbed System
11.1.2 Time-Dependent Perturbation Theory
11.1.3 Sinusoidal Perturbations
11.2 Emission and Absorption of Radiation
11.2.1 Electromagnetic Waves
11.2.2 Absorption, Stimulated Emission, and Spontaneous Emission
11.2.3 Incoherent Perturbations
11.3 Spontaneous Emission
11.3.1 Einstein’s A and B Coefficients
11.3.2 The Lifetime of an Excited State
11.3.3 Selection Rules
11.4 Fermi’s Golden Rule
11.5 The Adiabatic Approximation
11.5.1 Adiabatic Processes


11.5.2 The Adiabatic Theorem
Further Problems on Chapter 11
12 Afterword
12.1 The EPR Paradox
12.2 Bell’s Theorem
12.3 Mixed States and the Density Matrix
12.3.1 Pure States
12.3.2 Mixed States
12.3.3 Subsystems
12.4 The No-Clone Theorem
12.5 Schrödinger’s Cat
Appendix Linear Algebra
A.1 Vectors
A.2 Inner Products
A.3 Matrices
A.4 Changing Bases
A.5 Eigenvectors and Eigenvalues
A.6 Hermitian Transformations



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

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.
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).

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


Part I


The Wave Function



The Schrödinger Equation

Imagine a particle of mass m, constrained to move along the x axis, subject to some specified force
(Figure 1.1). The program of classical mechanics is to determine the position of the particle at any given time:
. Once we know that, we can figure out the velocity
kinetic energy

, the momentum

, the

, or any other dynamical variable of interest. And how do we go about
? We apply Newton’s second law:

. (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












.) This, together with appropriate initial conditions (typically the position and
velocity at

), determines


Figure 1.1: A “particle” constrained to move in one dimension under the influence of a specified force.
Quantum mechanics approaches this same problem quite differently. In this case what we’re looking for
is the particle’s wave function,

, and we get it by solving the Schrödinger equation:

Here i is the square root of

, and

is Planck’s constant—or rather, his original constant (h) divided by



The Schrödinger equation plays a role logically analogous to Newton’s second law: Given suitable initial
conditions (typically,

), the Schrödinger equation determines

classical mechanics, Newton’s law determines

for all future



for all future time, just as, in


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 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
point x, at time t—or, more

gives the probability of finding the particle at


Probability is the area under the graph of

. For the wave function in Figure 1.2, you would be quite likely

to find the particle in the vicinity of point A, where

is large, and relatively unlikely to find it near point B.

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.
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 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


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 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

Figure 1.3: Collapse of the wave function: graph of

immediately after a measurement has found the

particle at point C.

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

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.
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







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


represent the number of people of age j, then

, for instance, is zero. The total number of people in the room is

(In the example, of course,

.) Figure 1.5 is a histogram of the data. The following are some questions

one might ask about this distribution.

Figure 1.5: Histogram showing the number of people,
Question 1

, with age j, for the example in Section 1.3.1.

If you selected one individual at random from this group, what is the probability that this

person’s age would be 15?
Answer One chance in 14, since there are 14 possible choices, all equally likely, of whom only one has that







, and so on. In general,






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:

Question 2 What is the most probable age?

25, obviously; five people share this age, whereas at most three have any other age. The most

probable j is the j for which

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?

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

) is

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?

You could get

, with probability 1/14, or

, with probability 1/14, or

, with probability 3/14, and so on. The average, then, is

In general, the average value of some function of j is given by

(Equations 1.6, 1.7, and 1.8 are, if you like, special cases of this formula.) Beware: The average of the squares,
, is not equal, in general, to the square of the average,
babies, aged 1 and 3, then

, but


. For instance, if the room contains just two

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 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,
and compute the average of

(Note that

. Trouble is, of course, that you get zero:

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

. But absolute values are nasty to work with; instead, we get around the sign problem by squaring

before averaging:
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


Figure 1.6: Two histograms with the same median, same average, and same most probable value, but different
standard deviations.
There is a useful little theorem on variances:

Taking the square root, the standard deviation itself can be written as

In practice, this is a much faster way to get σ than by direct application of Equation 1.11: simply calculate

, subtract, and take the square root. Incidentally, I warned you a moment ago that

general, equal to

. Since

is not, in

is plainly non-negative (from its definition 1.11), Equation 1.12 implies that

and the two are equal only when

, which is to say, for distributions with no spread at all (every member

having the same value).



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 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.)

The proportionality factor,

, 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


and the rules we deduced for discrete distributions translate in the obvious way:




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

The velocity is

. Ignoring air resistance, the distance x at time t is

, and the total flight time is

. The probability that a

The velocity is

, and the total flight time is

particular photograph was taken between t and
distance in the corresponding range x to


. The probability that a
, so the probability that it shows a


Thus the probability density (Equation 1.14) is

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

The average distance (Equation 1.17) is

which is somewhat less than

, as anticipated.

Figure 1.7 shows the graph of

. 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).

Figure 1.7: The probability density in Example 1.2:



Problem 1.1 For the distribution of ages in the example in Section 1.3.1:
(a) Compute



(b) Determine

for each j, and use Equation 1.11 to compute the standard

(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.

What is the probability that a photograph, selected at random, would
show a distance x more than one standard deviation away from the


Problem 1.3 Consider the gaussian distribution

where A, a, and

are positive real constants. (The necessary integrals are

inside the back cover.)
(a) Use Equation 1.16 to determine A.
(b) Find


, and σ.

(c) Sketch the graph of





We return now to the statistical interpretation of the wave function (Equation 1.3), which says that
is the probability density for finding the particle at point x, at time t. It follows (Equation 1.16)
that the integral of

over all x must be 1 (the particle’s got to be somewhere):

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
two are consistent. Well, a glance at Equation 1.1 reveals that if

without checking that the

is a solution, so too is


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

. 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
will stay normalized, as time goes on, and

. How do I know that it

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,

(Note that the integral is a function only of t, so I use a total derivative
a function of x as well as t, so it’s a partial derivative

on the left, but the integrand is

on the right.) By the product rule,

Now the Schrödinger equation says that

and hence also (taking the complex conjugate of Equation 1.23)



The integral in Equation 1.21 can now be evaluated explicitly:


must go to zero as x goes to


infinity—otherwise the wave function would not be

It follows that

and hence that the integral is constant (independent of time); if

is normalized at

, it stays normalized

for all future time. QED

Problem 1.4 At time

a particle is represented by the wave function

where A, a, and b are (positive) constants.
(a) Normalize

(that is, find A, in terms of a and b).

(b) Sketch

, as a function of x.

(c) Where is the particle most likely to be found, at


(d) What is the probability of finding the particle to the left of a? Check your
result in the limiting cases



(e) What is the expectation value of x?


Problem 1.5 Consider the wave function

where A, , 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


(c) Find the standard deviation of x. Sketch the graph of
of x, and mark the points


, as a function

, 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?



For a particle in state


, the expectation value of x is

What exactly does this mean? It emphatically does not mean that if you measure the position of one particle
over and over again,

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.

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:

is the average of these results. I like to picture a row of bottles on a shelf, each containing a particle in the

(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

, and compute the average, which should agree with

. (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,

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

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

(I used the fact that

, 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:

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:

Equation 1.31 tells us, then, how to calculate

directly from

Actually, it is customary to work with momentum

, rather than velocity:

Let me write the expressions for


in a more suggestive way:

We say that the operator18 x “represents” position, and the operator
calculate expectation values we “sandwich” the appropriate operator between

“represents” momentum; to

, and integrate.

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

and angular momentum is

(the latter, of course, does not occur for motion in one dimension). To calculate the expectation value of any
such quantity,

, we simply replace every p by

, insert the resulting operator between

, and integrate:

For example, the expectation value of the kinetic energy is

Equation 1.36 is a recipe for computing the expectation value of any dynamical quantity, for a particle in

; 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
, and conclude that


Problem 1.7 Calculate


. Answer:

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

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:
the expectation value of a dynamical variable?


. What effect does this have on


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.

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

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


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,


is the standard deviation in x, and

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
just make


can be—

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

where A and a are positive real constants.
(a) Find A.

For what potential energy function,

, is this a solution to the

Schrödinger equation?
(c) Calculate the expectation values of



, and


. Is their product consistent with the uncertainty



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, … ).

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

, 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 dx (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 dx:


is the speed, and


This is perhaps the closest classical analog22 to
(a) Use conservation of energy to express

As an example, find
. Plot

in terms of E and


for the simple harmonic oscillator,
, and check that it is correctly normalized.

(c) For the classical harmonic oscillator in part (b), find



, and


Figure 1.10: Classical particle in a potential well.

Problem 1.12 What if we were interested in the distribution of momenta
, for the classical harmonic oscillator (Problem 1.11(b)).
(a) Find the classical probability distribution
(b) Calculate

(note that p ranges from


, and


What’s the classical uncertainty product,

, for this system? Notice

that this product can be as small as you like, classically, simply by sending
. But in quantum mechanics, as we shall see in Chapter 2, the
energy of a simple harmonic oscillator cannot be less than

, where

is the classical frequency. In that case what can you say about
the product


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
You might as well take

(that sets the scale for time) and


sets the scale for length). Make a plot of x at 10,000 random times, and
compare it with


Hint: In Mathematica, first define

then construct a table of positions:

and finally, make a histogram of the data:

Meanwhile, make a plot of the density function,

, and, using Show,

superimpose the two.
Problem 1.14 Let

be the probability of finding the particle in the range

, at time t.
(a) Show that


What are the units of

? Comment: J is called the probability

What are the units of

? Comment: J is called the probability

current, because it tells you the rate at which probability is “flowing” past
the point x. If

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

for any two (normalizable) solutions to the Schrödinger equation (with the




Problem 1.16 A particle is represented (at time

) by the wave function

(a) Determine the normalization constant A.
(b) What is the expectation value of x?

What is the expectation value of p? (Note that you cannot get it from
. Why not?)

(d) Find the expectation value of


(e) Find the expectation value of


(f) Find the uncertainty in


(g) Find the uncertainty in


(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:

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:


is the true potential energy and Γ is a positive real constant?

(a) Show that (in place of Equation 1.27) we now get

(b) Solve for

, and find the lifetime of the particle in terms of Γ.

Problem 1.18 Very roughly speaking, quantum mechanics is relevant when the de
Broglie wavelength of the particle in question
characteristic size of the system

is greater than the

. In thermal equilibrium at (Kelvin)

temperature T, the average kinetic energy of a particle is


is Boltzmann’s constant), so the typical de Broglie wavelength is

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
the temperature below which the unbound23

nm. Find

electrons in a solid are

quantum mechanical. Below what temperature are the nuclei in a solid
quantum mechanical? (Use silicon as an example.)
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


deduce the interatomic spacing.

. Obviously (for the gas to show

quantum 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 H .)


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



For a delightful first-hand account of the origins of the Schrödinger equation see the article by Felix Bloch in Physics Today, December


The wave function itself is complex, but


is the complex conjugate of

) is real and non-negative—as a

probability, of course, must be.

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.


Bernard d’Espagnat, “The Quantum Theory and Reality” (Scientific American, November 1979, p. 165).


Quoted in a lovely article by N. David Mermin, “Is the moon there when nobody looks?” (Physics Today, April 1985, p. 38).



Ibid., p. 40.
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.


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.


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


See Tonomura et al., American Journal of Physics, Volume 57, Issue 2, pp. 117–120 (1989), and the amazing associated video at 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).


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.


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.



must go to zero faster than

, as

. Incidentally, normalization only fixes the modulus of A; the phase

remains undetermined. However, as we shall see, the latter carries no physical significance anyway.

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.


To keep things from getting too cluttered, I’ll suppress the limits of integration


The product rule says that


from which it follows that

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.

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


Some authors limit the term to the pair of equations


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.

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

. Although this is a conceptually cleaner approach, it involves diverting mathematical

complications that I would rather save for later.

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.


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.


Time-Independent Schrödinger Equation



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

in the first place? We need to solve the Schrödinger equation,

for a specified potential1

. 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,

(lower-case) is a function of x alone, and

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

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

Or, dividing through by


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





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

, but we might as well absorb the constant C into

interest is the product


(since the quantity of


The second (Equation 2.5) is called the time-independent Schrödinger equation; we can go no further with it
until the potential

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
. I offer three answers—two of them physical, and one mathematical:
1. They are stationary states. Although the wave function itself,
does(obviously) depend on t, the probability density,
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

Every expectation value is constant in time; we might as well drop the factor

in place of

. (Indeed, it is common to refer to

altogether, and simply

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,

is constant, and hence (Equation 1.33)

. 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:

The corresponding Hamiltonian operator, obtained by the canonical substitution

, is

The corresponding Hamiltonian operator, obtained by the canonical substitution

, is


Thus the time-independent Schrödinger equation (Equation 2.5) can be written
and the expectation value of the total energy is

(Notice that the normalization of

entails the normalization of

.) Moreover,

and hence

So the variance of H is
But remember, if

, 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.)

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

, which we write as

, each with its associated separation constant

; thus there is a different wave function for each allowed energy:

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

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

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 time-independent Schrödinger equation, you’re essentially done; getting from there

to the general solution of the time-dependent Schrödinger equation is, in principle, simple and
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
; your job is to find the wave function,

, and the starting wave function

, 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 timeindependent Schrödinger equation (Equation 2.5); this yields, in general, an infinite set of solutions,
each with its own associated energy,

. To fit


you write down the general linear combination of

these solutions:

the miracle is that you can always match the specified initial state7 by appropriate choice of the constants
To construct


you simply tack onto each term its characteristic time dependence (its “wiggle factor”),

The separable solutions themselves,
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


Example 2.1
Suppose a particle starts out in a linear combination of just two stationary states:

(To keep things simple I’ll assume that the constants
wave function

and the states

are real.) What is the

at subsequent times? Find the probability density, and describe its motion.

Solution: The first part is easy:



are the energies associated with


. It follows that

The probability density oscillates sinusoidally, at an angular frequency

; 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

You may be wondering what the coefficients

represent physically. I’ll tell you the answer, though the

explanation will have to await Chapter 3:

A competent measurement will always yield one of the “allowed” values (hence the name), and
probability of getting the particular value


is the

Of course, the sum of these probabilities should be 1:

and the expectation value of the energy must be

We’ll soon see how this works out in some concrete examples. Notice, finally, that becausethe constants
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


and Γ real), and show

that if Equation 1.20 is to hold for all t, Γ must be zero.
(b) The time-independent wave function

can always be taken to be real

, 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

s that are real. Hint: If

satisfies Equation

2.5, for a given E, so too does its complex conjugate, and hence also the
real linear combinations



is an even function (that is,


always be taken to be either even or odd. Hint: If
2.5, for a given E, so too does
linear combinations


satisfies Equation

, and hence also the even and odd




Problem 2.2 Show that E must exceed the minimum value of

, 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


, then

and its second derivative always have the same sign—argue

that such a function cannot be normalized.



The Infinite Square Well


(Figure 2.1). A particle in this potential is completely free, except at the two ends


, 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.)

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

(the probability of finding the particle there is zero). Inside the well, where

, the time-independent Schrödinger equation (Equation 2.5) reads


(By writing it in this way, I have tacitly assumed that

; we know from Problem 2.2 that


work.) Equation 2.24 is the classical simple harmonic oscillator equation; the general solution is
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

? Ordinarily, both



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
Continuity of

, in Section 2.5; for now I hope you will trust me.)

requires that


so as to join onto the solution outside the well. What does this tell us about A and B? Well,


, and hence


, so either


(in which case we’re left with the trivial—non-normalizable—

, or else

, which means that


is no good (again, that would imply


, and the negative solutions give nothing new,

and we can absorb the minus sign into A. So the distinct solutions are
Curiously, the boundary condition at

does not determine the constant A, but rather the constant

k, and hence the possible values of E:

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


This only determines the magnitude of A, but it is simplest to pick the positive real root:


phase of A carries no physical significance anyway). Inside the well, then, the solutions are

As promised, the time-independent Schrödinger equation has delivered an infinite set of solutions (one
for each positive integer

. The first few of these are plotted in Figure 2.2. They look just like the standing

waves on a string of length a;

, which carries the lowest energy, is called the ground state, the others, whose

energies increase in proportion to

, are called excited states. As a collection, the functions

have some

interesting and important properties:
1. They are alternately even and odd, with respect to the center of the well:
even, and so

is even,

is odd,



2. As you go up in energy, each successive state has one more node (zero-crossing):
points don’t count),

has one,

has none (the end

has two, and so on.

3. They are mutually orthogonal, in the sense that15


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

Note that this argument does not work if

. (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:


(the so-called Kronecker delta) is defined by

We say that the

s are orthonormal.

They are complete, in the sense that any other function,

, can be expressed as a linear

combination of them:

I’m not about to prove the completeness of the functions

, but if you’ve studied

advanced calculus you will recognize that Equation 2.35 is nothing but the Fourier series for
the fact that “any” function can be expanded in this way is sometimes called Dirichlet’s
The coefficients

can be evaluated—for a given

beautifully exploits the orthonormality of

, and


—by a method I call Fourier’s trick, which

: Multiply both sides of Equation 2.35 by

, and


(Notice how the Kronecker delta kills every term in the sum except the one for which

.) Thus the

(Notice how the Kronecker delta kills every term in the sum except the one for which
nth coefficient in the expansion of

.) Thus the


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

I claimed (Equation 2.17) that the most general solution to the (time-dependent) Schrödinger equation is a
linear combination of stationary states:

(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,

The completeness of the

by appropriate choice of the coefficients


s (confirmed in this case by Dirichlet’s theorem) guarantees that I can always

in this way, and their orthonormality licenses the use of Fourier’s trick to determine the

actual coefficients:

That does it: Given the initial wave function,

, we first compute the expansion coefficients

using Equation 2.40, and then plug these into Equation 2.39 to obtain


. 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

s and the equation for the allowed energies.

Example 2.2
A particle in the infinite square well has the initial wave function

for some constant A (see Figure 2.3). Outside the well, of course,


. Find


Figure 2.3: The starting wave function in Example 2.2.
Solution: First we need to determine A, by normalizing



The nth coefficient is (Equation 2.40)

Thus (Equation 2.39):

Example 2.3

Check that Equation 2.20 is satisfied, for the wave function in Example 2.2. If you measured the

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
Solution: The starting wave function (Figure 2.3) closely resembles the ground state
This suggests that



(Figure 2.2).

and in fact

The rest of the coefficients make up the difference:20

The most likely outcome of an energy measurement is

—more than 99.8% of all

measurements will yield this value. The expectation value of the energy (Equation 2.21) is

As one would expect, it is very close to

(5 in place of

—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


s are independent of time, so I’m going to do the proof for

bothers you, you can easily generalize the argument to arbitrary

(Again, the Kronecker delta picks out the term

; if this


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


Problem 2.3 Show that there is no acceptable solution to the (time-independent)
Schrödinger equation for the infinite square well with


. (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

, and

, 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?


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:

(a) Normalize

. (That is, find A. This is very easy, if you exploit the

orthonormality of


. Recall that, having normalized



you can rest assured that it stays normalized—if you doubt this, check it
explicitly after doing part (b).)
(b) Find


. Express the latter as a sinusoidal function

of time, as in Example 2.1. To simplify the result, let



. 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
(d) Compute

, go directly to jail.)

. (As Peter Lorre would say, “Do it ze kveek vay, Johnny!”)

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



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

where ϕ is some constant. Find


in Problem 2.5:

, and

, and compare your

where ϕ is some constant. Find

, and

, and compare your

results with what you got before. Study the special cases



(For a graphical exploration of this problem see the applet in footnote 9 of this


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

(a) Sketch

, and determine the constant A.

(b) Find


(c) What is the probability that a measurement of the energy would yield the


(d) Find the expectation value of the energy, using Equation 2.21.21

Problem 2.8 A particle of mass m in the infinite square well (of width

starts out

in the state

for some constant A, so it is (at

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

would yield the value


Problem 2.9 For the wave function in Example 2.2, find the expectation value of
H, at time

, the “old fashioned” way:

Compare the result we got in Example 2.3. Note: Because
time, there is no loss of generality in using



is independent of


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,

(ignoring friction), and the solution is


is the (angular) frequency of oscillation. The potential energy is

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
in a Taylor series about the minimum:


(you can add a constant to

recognize that
long as


with impunity, since that doesn’t change the force),

is a minimum), and drop the higher-order terms (which are negligible as

stays small), we get

which describes simple harmonic oscillation (about the point

, with an effective spring constant

. 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


Figure 2.4: Parabolic approximation (dashed curve) to an arbitrary potential, in the neighborhood of a local
The quantum problem is to solve the Schrödinger equation for the potential

(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:

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.



Algebraic Method

To begin with, let’s rewrite Equation 2.45 in a more suggestive form:
is the momentum operator.24 The basic idea is to factor the Hamiltonian,



If these were numbers, it would be easy:

Here, however, it’s not quite so simple, because

is not the same as

and x are operators, and operators do not, in general,

, 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

(the factor in front is just there to make the final results look nicer).
Well, what is the product


As anticipated, there’s an extra term, involving

. We call this the commutator of x and ; it is a

measure of how badly they fail to commute. In general, the commutator of operators


(written with

square brackets) is
In this notation,

We need to figure out the commutator of x and . 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,”

, 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:


Dropping the test function, which has served its purpose,

This lovely and ubiquitous formula is known as the canonical commutation relation.25
With this, Equation 2.50 becomes


Evidently the Hamiltonian does not factor perfectly—there’s that extra
ordering of


is important here; the same argument, with

on the right. Notice that the

on the left, yields

In particular,
Meanwhile, the Hamiltonian can equally well be written

In terms of

, then, the Schrödinger equation26 for the harmonic oscillator takes the form

(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:


satisfies the Schrödinger equation with energy E (that is:

Schrödinger equation with energy


, then

satisfies the



(I used Equation 2.56 to replace



in the second line. Notice that whereas the

(I used Equation 2.56 to replace
ordering of



in the second line. Notice that whereas the

does matter, the ordering of

and any constants—such as

, and E—does not;

an operator commutes with any constant.)
By the same token,

is a solution with energy


Here, then, is a wonderful machine for generating new solutions, with higher and lower energies—if we
could just find one solution, to get started! We call
down in energy;

is the raising operator, and

ladder operators, because they allow us to climb up and
the lowering operator. The “ladder” of states is illustrated

in Figure 2.5.

Figure 2.5: The “ladder” of states for the harmonic oscillator.
But wait! What if I apply the lowering operator repeatedly? Eventually I’m going to reach a state with
energy less than zero, which (according to the general theorem in Problem 2.3) does not exist! At some point
the machine must fail. How can that happen? We know that


is a new solution to the Schrödinger

equation, but there is no guarantee that it will be normalizable—it might be zero, or its square-integral might be
infinite. In practice it is the former: There occurs a “lowest rung” (call it

such that

We can use this to determine



This differential equation is easy to solve:


We might as well normalize it right away:


, and hence

To determine the energy of this state we plug it into the Schrödinger equation (in the form of Equation 2.58),
, and exploit the fact that


With our foot now securely planted on the bottom rung (the ground state of the quantum oscillator), we
simply apply the raising operator (repeatedly) to generate the excited states,27 increasing the energy by
with each step:


is the normalization constant. By applying the raising operator (repeatedly) to

principle) construct


, then, we can (in

the stationary states of the harmonic oscillator. Meanwhile, without ever doing that

explicitly, we have determined the allowed energies!


Example 2.4
Find the first excited state of the harmonic oscillator.
Solution: Using Equation 2.62,

We can normalize it “by hand”:

so, as it happens,


I wouldn’t want to calculate

this way (applying the raising operator fifty times!), but never

mind: In principle Equation 2.62 does the job—except for the normalization.

You can even get the normalization algebraically, but it takes some fancy footwork, so watch closely. We
know that

is proportional to


but what are the proportionality factors,


? First note that for “any”29 functions



In the language of linear algebra,

is the hermitian conjugate (or adjoint) of



and integration by parts takes


the reason indicated in footnote 29), so

In particular,

But (invoking Equations 2.58 and 2.62)


(the boundary terms vanish, for


But since


are normalized, it follows that

, and hence30




and so on. Clearly

which is to say that the normalization factor in Equation 2.62 is

(in particular,


confirming our result in Example 2.4).
As in the case of the infinite square well, the stationary states of the harmonic oscillator are orthogonal:

This can be proved using Equation 2.66, and Equation 2.65 twice—first moving


, then,



must be zero. Orthonormality means that we can again use Fourier’s trick

(Equation 2.37) to evaluate the coefficients
states (Equation 2.16). As always,

and then moving

, when we expand

as a linear combination of stationary

is the probability that a measurement of the energy would yield the



Example 2.5
Find the expectation value of the potential energy in the nth stationary state of the harmonic

There’s a beautiful device for evaluating integrals of this kind (involving powers of x or
definition (Equation 2.48) to express x and

: Use the

in terms of the raising and lowering operators:

In this example we are interested in




is (apart from normalization)
, which is proportional to

, which is orthogonal to

, and the same goes for

. So those terms drop out, and we can use Equation 2.66 to

evaluate the remaining two:

As it happens, the expectation value of the potential energy is exactly half the total (the other half, of
course, is kinetic). This is a peculiarity of the harmonic oscillator, as we’ll see later on (Problem 3.37).


Problem 2.10
(a) Construct
(b) Sketch

, and


(c) Check the orthogonality of

, and

, by explicit integration. Hint:

If you exploit the even-ness and odd-ness of the functions, there is really
only one integral left to do.


Problem 2.11
(a) Compute

, and

, for the states

(Equation 2.60) and

(Equation 2.63), by explicit integration. Comment: In this and other
problems involving the harmonic oscillator it simplifies matters if you







(b) Check the uncertainty principle for these states.
(c) Compute


for these states. (No new integration allowed!) Is

their sum what you would expect?


Problem 2.12 Find

, and

, for the nth stationary state of

the harmonic oscillator, using the method of Example 2.5. Check that the
uncertainty principle is satisfied.

Problem 2.13 A particle in the harmonic oscillator potential starts out in the state

(a) Find A.



. Don’t get too excited if

oscillates at exactly the classical frequency; what would it have been had I
(c) Find

, instead of


. Check that Ehrenfest’s theorem (Equation 1.38) holds,

for this wave function.

If you measured the energy of this particle, what values might you get,
and with what probabilities?



Analytic Method

We return now to the Schrödinger equation for the harmonic oscillator,

and solve it directly, by the power series method. Things look a little cleaner if we introduce the dimensionless

in terms of ξ the Schrödinger equation reads

where K is the energy, in units of


Our problem is to solve Equation 2.73, and in the process obtain the “allowed” values of K (and hence of
To begin with, note that at very large ξ (which is to say, at very large


completely dominates over

the constant K, so in this regime

which has the approximate solution (check it!)
The B term is clearly not normalizable (it blows up as

; the physically acceptable solutions, then,

have the asymptotic form
This suggests that we “peel off” the exponential part,
in hopes that what remains,

, has a simpler functional form than




itself.32 Differentiating Equation

so the Schrödinger equation (Equation 2.73) becomes

I propose to look for solutions to Equation 2.79 in the form of power series in ξ:33

Differentiating the series term by term,


Putting these into Equation 2.80, we find

It follows (from the uniqueness of power series expansions34 ) that the coefficient of each power of ξ must

and hence that

This recursion formula is entirely equivalent to the Schrödinger equation. Starting with

, it generates

all the even-numbered coefficients:

and starting with

, it generates the odd coefficients:

We write the complete solution as


is an even function of ξ, built on

is an odd function, built on

, and

. Thus Equation 2.82 determines

in terms of two arbitrary constants

—which is just what we would expect, for a second-order differential equation.
However, not all the solutions so obtained are normalizable. For at very large j, the recursion formula

becomes (approximately)

with the (approximate) solution

for some constant C, and this yields (at large ξ, where the higher powers dominate)

Now, if h goes like

, then


?—that’s what we’re trying to calculate) goes like

(Equation 2.78), which is precisely the asymptotic behavior we didn’t want.35 There is only one
way to wiggle out of this: For normalizable solutions the power series must terminate. There must occur some
“highest” j (call it
or the series

, such that the recursion formula spits out
; the other one must be zero from the start:

(this will truncate either the series
if n is even, and

if n is odd).

For physically acceptable solutions, then, Equation 2.82 requires that

for some positive integer n, which is to say (referring to Equation 2.74) that the energy must be

Thus we recover, by a completely different method, the fundamental quantization condition we found
algebraically in Equation 2.62.
It seems at first rather surprising that the quantization of energy should emerge from a technical detail in
the power series solution to the Schrödinger equation, but let’s look at it from a different perspective.
Equation 2.71 has solutions, of course, for any value of E (in fact, it has two linearly independent solutions for

. But almost all of these solutions blow up exponentially at large x, and hence are not normalizable.

Imagine, for example, using an E that is slightly less than one of the allowed values (say,
plotting the solution: Figure 2.6(a). Now try an E slightly larger (say,

, and

; the “tail” now blows up in the

other direction (Figure 2.6(b)). As you tweak the parameter in tiny increments from 0.49 to 0.51, the graph


“flips over” at precisely the value 0.5—only here does the solution escape the exponential asymptotic growth
that renders it physically unacceptable.36

Figure 2.6: Solutions to the Schrödinger equation for (a)

, and (b)


For the allowed values of K, the recursion formula reads


, there is only one term in the series (we must pick


to kill

, and

in Equation 2.85

we take

,37 and Equation


and hence

(which, apart from the normalization, reproduces Equation 2.60). For
2.85 with


, so

and hence


(confirming Equation 2.63). For


, and


, so


and so on. (Compare Problem 2.10, where this last result was obtained by algebraic means).
In general,

will be a polynomial of degree n in ξ, involving even powers only, if n is an even

integer, and odd powers only, if n is an odd integer. Apart from the overall factor
called Hermite polynomials,


they are the so-

The first few of them are listed in Table 2.1. By tradition, the arbitrary

multiplicative factor is chosen so that the coefficient of the highest power of ξ is


. With this convention, the

stationary states for the harmonic oscillator are

They are identical (of course) to the ones we obtained algebraically in Equation 2.68.
Table 2.1: The first few Hermite polynomials,

In Figure 2.7(a) I have plotted


for the first few ns. The quantum oscillator is strikingly different

from its classical counterpart—not only are the energies quantized, but the position distributions have some
bizarre features. For instance, the probability of finding the particle outside the classically allowed range (that
is, with x greater than the classical amplitude for the energy in question) is not zero (see Problem 2.14), and in
all odd states the probability of finding the particle at the center is zero. Only at large n do we begin to see
some resemblance to the classical case. In Figure 2.7(b) I have superimposed the classical position distribution
(Problem 1.11) on the quantum one (for

; if you smoothed out the bumps, the two would fit pretty



Figure 2.7: (a) The first four stationary states of the harmonic oscillator. (b) Graph of

, with the

classical distribution (dashed curve) superimposed.

Problem 2.14 In the ground state of the harmonic oscillator, what is the
probability (correct to three significant digits) of finding the particle outside the
classically allowed region? Hint: Classically, the energy of an oscillator is
, where a is the amplitude. So the “classically
allowed region” for an oscillator of energy E extends from


. Look in a math table under “Normal Distribution” or “Error
Function” for the numerical value of the integral, or evaluate it by computer.

Problem 2.15 Use the recursion formula (Equation 2.85) to work out


. Invoke the convention that the coefficient of the highest power of ξ is
to fix the overall constant.


Problem 2.16 In this problem we explore some of the more useful theorems
(stated without proof) involving Hermite polynomials.
(a) The Rodrigues formula says that

Use it to derive



The following recursion relation gives you

in terms of the two

preceding Hermite polynomials:
Use it, together with your answer in (a), to obtain



If you differentiate an nth-order polynomial, you get a polynomial of

. For the Hermite polynomials, in fact,

Check this, by differentiating


is the nth z-derivative, at

, of the generating function

; or, to put it another way, it is the coefficient of
in the Taylor series expansion for this function:

Use this to obtain

, and




The Free Particle

We turn next to what should have been the simplest case of all: the free particle


Classically this would just be motion at constant velocity, but in quantum mechanics the problem is
surprisingly subtle. The time-independent Schrödinger equation reads


So far, it’s the same as inside the infinite square well (Equation 2.24), where the potential is also zero; this
time, however, I prefer to write the general solution in exponential form (instead of sines and cosines), for
reasons that will appear in due course:
Unlike the infinite square well, there are no boundary conditions to restrict the possible values of k (and hence

; the free particle can carry any (positive) energy. Tacking on the standard time dependence,
Now, any function of x and t that depends on these variables in the special combination

some constant

represents a wave of unchanging shape, traveling in the


-direction at speed v: A fixed

point on the waveform (for example, a maximum or a minimum) corresponds to a fixed value of the argument,
and hence to x and t such that

Since every point on the waveform moves with the same velocity, its shape doesn’t change as it propagates.
Thus the first term in Equation 2.94 represents a wave traveling to the right, and the second represents a wave
(of the same energy) going to the left. By the way, since they only differ by the sign in front of k, we might as
well write
and let k run negative to cover the case of waves traveling to the left:

Evidently the “stationary states” of the free particle are propagating waves; their wavelength is
and, according to the de Broglie formula (Equation 1.39), they carry momentum



The speed of these waves (the coefficient of t over the coefficient of


On the other hand, the classical speed of a free particle with energy E is given by
kinetic, since


, so

Apparently the quantum mechanical wave function travels at half the speed of the particle it is supposed to
represent! We’ll return to this paradox in a moment—there is an even more serious problem we need to
confront first: This wave function is not normalizable:

In the case of the free particle, then, the separable solutions do not represent physically realizable states. A free
particle cannot exist in a stationary state; or, to put it another way, there is no such thing as a free particle with a
definite energy.
But that doesn’t mean the separable solutions are of no use to us. For they play a mathematical role that is
entirely independent of their physical interpretation: The general solution to the time-dependent Schrödinger
equation is still a linear combination of separable solutions (only this time it’s an integral over the continuous
variable k, instead of a sum over the discrete index


(The quantity

is factored out for convenience; what plays the role of the coefficient

2.17 is the combination

in Equation

.) Now this wave function can be normalized (for appropriate

. But it necessarily carries a range of ks, and hence a range of energies and speeds. We call it a wave
In the generic quantum problem, we are given

, and we are asked to find

particle the solution takes the form of Equation 2.101; the only question is how to determine

. For a free
so as to

match the initial wave function:

This is a classic problem in Fourier analysis; the answer is provided by Plancherel’s theorem (see Problem


is called the Fourier transform of
difference is the sign in the
integrals have to




is the inverse Fourier transform of

(the only

There is, of course, some restriction on the allowable functions: The

For our purposes this is guaranteed by the physical requirement that

itself be

normalized. So the solution to the generic quantum problem, for the free particle, is Equation 2.101, with

Example 2.6
A free particle, which is initially localized in the range

where A and a are positive real constants. Find
Solution: First we need to normalize

Next we calculate

, is released at time




, using Equation 2.104:

Finally, we plug this back into Equation 2.101:

Unfortunately, this integral cannot be solved in terms of elementary functions, though it can of course
be evaluated numerically (Figure 2.8). (There are, in fact, precious few cases in which the integral for
(Equation 2.101) can be carried out explicitly; see Problem 2.21 for a particularly beautiful


Figure 2.8: Graph of