i

Invitation to
Quantum Mechanics

Daniel F. Styer
ii

Invitation to Quantum Mechanics

Daniel F. Styer
Schiffer Professor of Physics, Oberlin College

copyright c 9 February 2024 Daniel F. Styer

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and/or redistribute this work under the terms of the Creative Commons
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You may freely download this book in pdf format from

http://www.oberlin.edu/physics/dstyer/InvitationToQM.
It is formatted to print nicely on either A4 or U.S. Letter paper. You may
also purchase a printed and bound copy from World Scientific Publishing
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download/purchase: it is reward enough for him that you want to explore
quantum mechanics.
Love all God’s creation, the whole and every grain of sand in it.
Love the stars, the trees, the thunderstorms, the atoms.
The more you love, the more you will grow curious.
The more you grow curious, the more you will question.
The more you question, the more you will uncover.
The more you uncover, the more you will love.
And so at last you will come to love the entire universe with an
agile and resilient love founded upon facts and understanding.

— This improvisation by Dan Styer was inspired by

the first sentence, which appears in
Fyodor Dostoyevsky’s The Brothers Karamazov.

iii
iv

Dedicated to Linda Ong Styer, adventurer

 Contents

Synoptic Contents 1

Welcome 3

1. “Something Isn’t Quite Right” 9

1.1 Light in thermal equilibrium: Blackbody radiation . . . . 10

1.2 Photoelectric effect . . . . . . . . . . . . . . . . . . . . . . 23
1.3 Wave character of electrons . . . . . . . . . . . . . . . . . 30
1.4 How does an electron behave? . . . . . . . . . . . . . . . . 35
1.5 Quantization of atomic energies . . . . . . . . . . . . . . . 36
1.6 Quantization of magnetic moment . . . . . . . . . . . . . 42

2. What Is Quantum Mechanics About? 47

2.1 Quantization . . . . . . . . . . . . . . . . . . . . . . . . . 47
2.2 Interference . . . . . . . . . . . . . . . . . . . . . . . . . . 59
2.3 Aharonov-Bohm effect . . . . . . . . . . . . . . . . . . . . 67
2.4 Light on the atoms . . . . . . . . . . . . . . . . . . . . . . 70
2.5 Entanglement . . . . . . . . . . . . . . . . . . . . . . . . . 72
2.6 Quantum cryptography . . . . . . . . . . . . . . . . . . . 84
2.7 What is a qubit? . . . . . . . . . . . . . . . . . . . . . . . 88

v
vi Contents

3. Forging Mathematical Tools 89

3.1 What is a quantal state? . . . . . . . . . . . . . . . . . . . 89

3.2 Amplitude . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
3.3 Reversal-conjugation relation . . . . . . . . . . . . . . . . 99
3.4 Establishing a phase convention . . . . . . . . . . . . . . . 101
3.5 How can I specify a quantal state? . . . . . . . . . . . . . 103
3.6 States for entangled systems . . . . . . . . . . . . . . . . . 112
3.7 Are states “real”? . . . . . . . . . . . . . . . . . . . . . . . 116
3.8 What is a qubit? . . . . . . . . . . . . . . . . . . . . . . . 116

4. The Quantum Mechanics of Position 119

4.1 Probability and probability density: One particle in one

dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
4.2 Probability amplitude . . . . . . . . . . . . . . . . . . . . 124
4.3 How does wavefunction change with time? . . . . . . . . . 126
4.4 Wavefunction: Two particles . . . . . . . . . . . . . . . . . 127
4.5 Solving the Schrödinger time evolution equation for the
infinite square well . . . . . . . . . . . . . . . . . . . . . . 131
4.6 What did we learn by solving the Schrödinger time evolu-
tion equation for the infinite square well? . . . . . . . . . 139
4.7 Other potentials . . . . . . . . . . . . . . . . . . . . . . . 149
4.8 Energy loss . . . . . . . . . . . . . . . . . . . . . . . . . . 152
4.9 Mean values . . . . . . . . . . . . . . . . . . . . . . . . . . 153
4.10 The classical limit of quantum mechanics . . . . . . . . . 157
4.11 Transitions induced by light . . . . . . . . . . . . . . . . . 166
4.12 Position plus spin . . . . . . . . . . . . . . . . . . . . . . . 172
Contents vii

5. Solving the Energy Eigenproblem 177

5.1 Sketching energy eigenfunctions . . . . . . . . . . . . . . . 178

5.2 Scaled quantities . . . . . . . . . . . . . . . . . . . . . . . 198
5.3 Numerical solution of the energy eigenproblem . . . . . . 201

6. Identical Particles 205

6.1 Two or three identical particles . . . . . . . . . . . . . . . 205

6.2 Symmetrization and antisymmetrization . . . . . . . . . . 208
6.3 Consequences of the Pauli principle . . . . . . . . . . . . . 214
6.4 Consequences of the Pauli principle for product states . . 217
6.5 Energy states for two identical, noninteracting particles . 217
6.6 Spin plus space, two electrons . . . . . . . . . . . . . . . . 219
6.7 Spin plus space, three electrons, ground state . . . . . . . 225

7. Atoms 229

7.1 Central potentials in two dimensions . . . . . . . . . . . . 229

7.2 Central potentials in three dimensions . . . . . . . . . . . 236
7.3 The hydrogen atom . . . . . . . . . . . . . . . . . . . . . . 238
7.4 The helium atom . . . . . . . . . . . . . . . . . . . . . . . 247
7.5 The lithium atom . . . . . . . . . . . . . . . . . . . . . . . 252
7.6 All other atoms . . . . . . . . . . . . . . . . . . . . . . . . 253
7.7 The periodic table . . . . . . . . . . . . . . . . . . . . . . 256

8. The Vistas Open to Us 259

Appendix A Significant Figures 263

Appendix B Dimensions 269

viii Contents

Appendix C Complex Arithmetic 277

Appendix D Problem-Solving Tips and Techniques 279

Appendix E Catalog of Misconceptions 283

 Synoptic Contents

Welcome

What is quantum mechanics and why should I care about it?

1. “Something Isn’t Quite Right”

Historical experiments show that classical mechanics is flawed.

2. What Is Quantum Mechanics About?

If classical mechanics is wrong, then what is right? We explore, in the

context of modern experiments with qubits, the atomic phenomena that
quantum mechanics needs to explain.

3. Forging Mathematical Tools

We build a framework for the quantum mechanics of qubits, using a

mathematical tool called “amplitude”.

4. The Quantum Mechanics of Position

The framework, built to treat qubits, extends to treat continuum position

as well. Energy plays a central role here.

1
2 Synoptic Contents

5. Solving the Energy Eigenproblem

Since energy plays a central role, we devote a chapter to solving such
problems. We find that solving particular problems strengthens our
conceptual understanding, and that conceptual understanding strengthens
our skill in solving particular problems.

6. Identical Particles

This surprisingly subtle topic deserves a chapter of its own.

7. Atoms

We apply our new knowledge to physical (rather than model) systems.

8. The Vistas Open to Us

This book is an invitation. Where might you and quantum mechanics travel
together?
 Welcome

Why would anyone want to study quantum mechanics?

Starting in the year 1900, physicists exploring the newly discovered atom
found that the atomic world of electrons and protons is not just smaller than
our familiar world of trees, balls, and automobiles, it is also fundamentally
different in character. Objects in the atomic world obey different rules from
those obeyed by a tossed ball or an orbiting planet. These atomic rules are
so different from the familiar rules of everyday physics, so counterintuitive
and unexpected, that it took more than 25 years of intense research to
uncover them.
But it is really only since the year 1990 that physicists have come to
appreciate that the rules of the atomic world (now called “quantum mechan-
ics”) are not just different from the everyday rules (now called “classical
mechanics”). The atomic rules are also far richer. The atomic rules provide
for phenomena like particle interference and entanglement that are simply
absent from the everyday world. Every phenomenon of classical mechanics
is also present in quantum mechanics, but the quantum world provides for
many additional phenomena.
Here’s an analogy: Some films are in black-and-white and some are in
color. It does not malign any black-and-white film to say that a color film
has more possibilities, more richness. In fact, black-and-white films are
simply one category of color films, because black and white are both colors.
Anyone moving from the world of only black-and-white to the world of color
is opening up the door to a new world — a world ripe with new possibilities
and new expression — without closing the door to the old world.
This same flood of richness and freshness comes from entering the quan-
tum world. It is a difficult world to enter, because we humans have no expe-

3
4 Welcome

rience, no intuition, no expectations about this world. Even our language,

invented by people living in the everyday world, has no words for the new
quantal phenomena — just as a language among a race of the color-blind
would have no word for “red”.
Reading this book is not easy: it is like a color-blind student learning
about color from a color-blind teacher. The book is just one long argument,
building up the structure of a world that we can explore not through touch
or through sight or through scent, but only through logic. Those willing to
follow and to challenge the logic, to open their minds to a new world, will
find themselves richly rewarded.

The place of quantum mechanics in nature

Quantum mechanics is the framework for describing and analyzing small

things, like atoms and nuclei. Quantum mechanics also applies to big
things, like baseballs and galaxies, but when applied to big things, cer-
tain approximations become legitimate: taken together, these are called
the classical approximation to quantum mechanics, and the result is the
familiar classical mechanics.
Quantum mechanics is not only less familiar and less intuitive than
classical mechanics; it is also harder than classical mechanics. So whenever
the classical approximation is sufficiently accurate, we would be foolish not
to use it. This leads some to develop the misimpression that quantum
mechanics applies to small things, while classical mechanics applies to big
things. No. Quantum mechanics applies to all sizes, but classical mechanics
is a good approximation to quantum mechanics when it is applied to big
things.
For what size is the classical approximation good enough? That depends
on the accuracy desired. The higher the accuracy demanded, the more situ-
ations will require full quantal treatment rather than approximate classical
treatment. But as a rule of thumb, something as big as a DNA strand is
almost always treated classically, not quantum mechanically.
This situation is analogous to the relationship between relativistic me-
chanics and classical mechanics. Relativity applies always, but classical
mechanics is a good approximation to relativistic mechanics when applied
to slow things (that is, with speeds much less than light speed c). The speed
at which the classical approximation becomes legitimate depends upon the
Welcome 5

accuracy demanded, but as a rule of thumb particles moving less than a

quarter of light speed are treated classically.
The difference between the quantal case and the relativistic case is that
while relativistic mechanics is less familiar, less comforting, and less ex-
pected than classical mechanics, it is no more intricate than classical me-
chanics. Quantum mechanics, in contrast, is less familiar, less comforting,
less expected, and more intricate than classical mechanics. This intricacy
makes quantum mechanics harder than classical mechanics, yes, but also
richer, more textured, more nuanced. Whether to curse or celebrate this
intricacy is your choice.

speed

c -

fast relativistic relativistic

quantum mechanics
mechanics

quantum classical
slow mechanics mechanics
0 - size
small big

Finally, is there a framework that applies to situations that are both fast
and small? There is: it is called “relativistic quantum mechanics” and is
closely related to “quantum field theory”. Ordinary non-relativistic quan-
tum mechanics is a good approximation for relativistic quantum mechanics
when applied to slow things. Relativistic mechanics is a good approxima-
tion for relativistic quantum mechanics when applied to big things. And
classical mechanics is a good approximation for relativistic quantum me-
chanics when applied to big, slow things.
6 Welcome

What you can expect from this book

This book introduces quantum mechanics at the second-year American un-

dergraduate level. It assumes the reader knows about classical forces, po-
tential energy functions, and the simple harmonic oscillator. The reader
should know that wavelength is represented by λ, frequency by f , and that
for a wave moving at speed c, λf = c. S/he needs to know the meaning
and significance of “standard deviation”. Turning to mathematics, it as-
sumes the reader knows about complex numbers (see appendix C) and dot
products, knows the difference between an ordinary and a partial deriva-
tive, and can solve simple ordinary differential equations. It assumes that
the reader understands phrases like “orthonormal basis representation of a
position vector”.
This is a book about physics, not mathematics. The word “physics”
derives from the Greek word for “nature”, so the emphasis lies in nature,
not in the mathematics we use to describe nature. Thus the book starts
with experiments about nature, then builds mathematical machinery to
describe nature, and finally applies the machinery to atoms, where the
understanding of both nature and machinery is deepened.
The book never abandons its focus on nature. It provides a balanced,
interwoven treatment of concepts, techniques, and applications so that each
thread reinforces the other. There are many problems at many levels of
difficulty, but no problem is there just for “make-work”: each has a “moral
to the story”. Some problems are essential to the logical development of
the subject: these are labeled (unsurprisingly) “essential”. Other problems
promote learning far better than simple reading can: these are labeled
“recommended”. Sample problems build both mathematical technique and
physical insight.
The book does not merely convey correct ideas, it also refutes miscon-
ceptions. Just to get started, I list the most important and most pernicious
misconceptions about quantum mechanics: (a) An electron has a position
but you don’t know what it is. (b) The only states are energy states. (c) The
wavefunction ψ(~x, t) is “out there” in space and you could reach out and
touch it if only your fingers were sufficiently sensitive.
I do not provide summary lists of key ideas and difficult-to-remember
concepts and equations. That’s because an equation that I find easy to
remember might be hard for you to remember. I recommend instead that
Welcome 7

you write out for yourself, in your own words, a summary of the ideas and
equations that you consider most important and most difficult to remember.
The object of the biographical footnotes in this book is twofold: First, to
present the briefest of outlines of the subject’s historical development, lest
anyone get the misimpression that quantum mechanics arose fully formed,
like Aphrodite from sea foam. Second, to show that the founders of quan-
tum mechanics were not inaccessible giants, but people with foibles and
strengths, with interests both inside and outside of physics, just like you
and me.

Teaching tips

Most physics departments offer a second-year course titled Modern

Physics. The topics in this course vary widely from institution to institu-
tion: special relativity and elementary quantum mechanics are staples, but
the course might also cover classical waves, thermodynamics and elemen-
tary statistical mechanics, descriptive atomic, molecular, and solid state
physics. No textbook could cover all this variety, nor should any textbook
try: instead each institution should provide a mix of topics appropriate
for its own students. This book is devoted only to quantum mechanics at
the level of a Modern Physics course. You will want to add it to other
materials for the other topics in your own particular course.
In chapters 2 and 3, a surprising amount of student difficulty comes from
nothing more than getting straight which Stern-Gerlach analyzer is oriented
in which direction. I recommend that you mark up some cardboard boxes
to look like analyzers and analyzer loops, and use them as demonstrations
during your classes.
Chapter 5 presents two techniques for solving the energy eigenproblem:
one informal and one numerical. I discuss the first in class and assign
the second for reading, because the first benefits from a lot of blackboard
sketching, erasing, resketching, and gesturing. But your own priorities
might differ from mine, so you might take the opposite tack.
This text spends a lot of time on concepts before applying those concepts
to atoms. Atoms are mathematically intense, and it pays to get the concepts
straight first. If we jumped directly into atoms, that mathematical intensity
would completely obscure the conceptual issues. Some people like it that
way, because they don’t want to face the conceptual issues.
8 Welcome

Acknowledgments

I learned quantum mechanics from stellar teachers. My high school chem-

istry teacher Frank Dugan introduced me not only to quantum mechanics
but to the precept that science involves hard, fulfilling work in addition
to dreams and imagination. When I was an undergraduate, John Boccio
helped mold my understanding of quantum mechanics, and also molded the
shape of my life. In graduate school N. David Mermin, Vinay Ambegaokar,
Neil Ashcroft, and Michael Peskin pushed me without mercy but pushed
me in the direction of understanding and away from the mind-numbing at-
titude of “shut up and calculate”. My debt to my thesis adviser, Michael
Fisher, is incalculable. I’ve been inspired by research lectures from Tony
Leggett, Jürg Fröhlich, Jennifer and Lincoln Chayes, Shelly Goldstein, and
Chris Fuchs, among others.
I have taught quantum mechanics to thousands of students from the
general audience level through advanced undergraduates. Their questions,
confusions, triumphs, and despairs have infused my own understanding of
the discipline. I cannot name them all, but I would be remiss if I did not
thank my former students Paul Kimoto, Gail Welsh, K. Tabetha Hole, Gary
Felder, Sarah Clemmens, Dahyeon Lee, and Noah Morris.
This book has been in slow yet consistent development since 2010, and
many students have given me feedback over these years. In the fall 2020
and spring 2021 semesters I taught Modern Physics at Oberlin College
using a draft of this textbook. I received helpful corrections and suggestions
from several students, but especially from Ilana Meisler. Thank you.
My scientific prose style was developed by Michael Fisher and N. David
Mermin. In particular this book’s structure of “first lay out the phenomena
(chapters 1 and 2), then build mathematical tools to describe those phe-
nomena” echos the structure of Fisher’s 1964 essay “The Nature of Critical
Points”. I have also absorbed lessons in writing from John McPhee, Mau-
rice Forrester, and Terry Tempest Williams. My teaching style has been
influenced especially by Mark Heald, Tony French, Edwin Taylor, Arnold
Arons, and Robert H. Romer.
David Kaiser corrected some of my misconceptions concerning history.
Shelley Kronzek, my editor at World Scientific, kept faith in this project
through multiple delays. The book was skillfully copyedited by Matthew
Abbate (who was also my undergraduate roommate and who served as a
potential guardian to my children).
 Chapter 1

“Something Isn’t Quite Right”

We are used to things that vary continuously: An oven can take on any
temperature, a recipe might call for any quantity of flour, a child can grow to
a range of heights. If I told you that an oven might take on the temperature
of 172.1 ◦ C or 181.7 ◦ C, but that a temperature of 173.8 ◦ C was physically
impossible, you would laugh in my face.
So you can imagine the surprise of physicists on 14 December 1900,
when Max Planck announced that certain features of blackbody radiation
(that is, of light in thermal equilibrium) could be explained by assuming
that the energy of the light could not take on any value, but only certain
discrete values. Specifically, Planck found that light of frequency f could
take on only the energies of
E = nhf, where n = 0, 1, 2, 3, . . ., (1.1)
and where the constant h (now called the “Planck constant”) is
h = 6.626 070 15 × 10−34 J s. (1.2)
That is, light of frequency f can have an energy of 3.0 hf , and it can have
an energy of 4.0 hf , but it is physically impossible for this light to have an
energy of 3.8 hf . Any numerical quantity that can take on only discrete
values like this is called “quantized”. By contrast, a numerical quantity
that can take on any value is called “continuous”.
The photoelectric effect (section 1.2) supplies additional evidence that
the energy of light comes only in discrete values. And if the energy of
light comes in discrete values, then it’s a good guess that the energy of
an atom comes in discrete values too. This good guess was confirmed
through investigations of atomic spectra (where energy goes into or out of

9
10 Light in thermal equilibrium

an atom via absorption or emission of light) and through the Franck–Hertz

experiment (where energy goes into or out of an atom via collisions).
Furthermore, if the energy of an atom comes in discrete values, then
it’s a good guess that other properties of an atom — such as its magnetic
moment — also take on only discrete values. The story of this chapter is
that these good guesses have all proved to be correct.1

1.1 Light in thermal equilibrium: Blackbody radiation

You know that the logs of a campfire, or the coils of an electric stove, glow
orange. You might not know that objects at higher temperatures glow
white, although blacksmiths and glass blowers are quite familiar with this
fact and use it to judge the temperature of the metal or the molten glass
they work with. Objects at still higher temperatures, like the star Sirius,
glow blue. A nuclear bomb explosion glows with x-rays.
Going down the temperature scale, the tables, chairs, walls, trees, and
books around us glow with infrared radiation. (Many people are unaware
of this fact because our eyes can’t detect infrared light.) In fact, our own
bodies glow in the infrared — at a somewhat shorter wavelength than our
books, because our bodies are slightly warmer than our books. And the
bitter cold of outer space glows with the famous 3 K cosmic microwave
background radiation.
All these situations are examples of electromagnetic radiation — light
— in thermal equilibrium. What does that mean? The light streaming
from, say, a red neon tube is not in thermal equilibrium: for one thing, it
has only one color, for another all the light streams in the same direction.
Just as a stream of nitrogen molecules, each one with the same speed and
each one moving in the same direction, is not in thermal equilibrium, so
the red light, all the same wavelength and all moving in the same direction,
is not in thermal equilibrium. But after that light is absorbed by matter
at a given temperature, then re-emitted, then reabsorbed and then re-
emitted again, several times, that light relaxes into equilibrium at the same
1 This book is about physics, not the history of physics. In order to present the physical

ideas clearly they are sometimes developed ahistorically. For example in the next section
Planck’s 1900 radiation law is developed as a refinement of the 1905 Rayleigh–Jeans law.
In section 1.5.2 Bohr’s 1913 atomic theory is developed as a consequence of de Broglie’s
1924 concept of matter waves.
“Something Isn’t Quite Right” 11

temperature as the matter with which it’s been interacting. This light in
thermal equilibrium has a variety of wavelengths, and it moves with equal
probability in any direction. In exactly the same way, a stream of nitrogen
molecules will, after many collisions, have a variety of molecular speeds,
and the molecules move with equal probability in any direction.
If you want to do a high-accuracy experiment with light in thermal equi-
librium, you will want light that has been absorbed and re-emitted many
times. The worst possible object for putting light into thermal equilibrium
would be a mirror, which reflects rather than absorbs most incoming light.
Somewhat better would be matter painted white, which reflects much in-
coming light. Better still would be matter painted black. Best of all would
be a cavity surrounded by matter, like a cave, so that the light in the cavity
is absorbed by the walls and re-emitted many times. For these experimental
reasons, light in thermal equilibrium is often called “blackbody radiation”
or “cavity radiation”.
Qualitative arguments explain a number of familiar features of black-
body radiation. You know that the atoms in matter oscillate: as the tem-
perature increases, the oscillations become both farther and faster. The
“farther” oscillations suggest that high-temperature objects should glow
brighter; the “faster” oscillations suggest that they should glow with higher-
frequency (hence shorter-wavelength) light. Turning these qualitative ar-
guments into a quantitative prediction requires an understanding of elec-
trodynamics and of statistical mechanics beyond the needs of this book.
Here I summarize the reasoning involved. First, three principles from elec-
trodynamics:
(1) In all cases, the state of light within a cavity can be expressed
as a sum over the “normal modes” of light within that cavity. Normal
modes come about when one half-wavelength fits within a cubical cavity;
or two half-wavelengths; or three; or any integer. This principle states that
~ n (~x, t) denotes the electric field due to the light of the normal mode
if E
indexed n, then the electric field of an arbitrary state of light is
X
~ x, t) =
E(~ ~ n (~x, t)
an E (1.3)
n

where an sets the amplitude of that mode in the particular state. Each
mode is characterized by a particular wavelength.
12 Light in thermal equilibrium

energy (mJ)

6
1.5

1.0

T = 6000 K

0.5

T = 5000 K

T = 4000 K
0.0 -

0 500 1000 1500

ultraviolet visible infrared

wavelength λ (nm)

Energy within one cubic meter due to electromagnetic radiation in thermal

equilibrium with wavelength from λ to λ + (1 nm). At higher temperatures,
there is more energy, and more of that energy falls at shorter wavelengths.
“Something Isn’t Quite Right” 13

(2) The energy of any arbitrary state of light is just the sum of the
energies due to each normal mode.
(3) There are more normal modes at shorter wavelength. In May 1905,
Lord Rayleigh2 calculated that the number of modes with wavelength falling
within the window between λ and λ + dλ was
64 πV
dλ,
λ4
where V is the volume of the cavity (assuming a cavity large in the sense
that V  λ3 ). Two months later James Jeans3 pointed out that Rayleigh
had made a counting error: the correct result is
8πV
dλ. (1.4)
λ4
To these three electrodynamics principles, add one principle from clas-
sical statistical mechanics: In thermal equilibrium, the energy4 of each
mode is kB T , where T is the (absolute) temperature and kB is the Boltz-
mann constant. This principle is called “equipartition”: some modes have
short wavelengths and some have long, but the energy is equally partitioned
among the various different modes, independent of wavelength.
[[The Boltzmann5 constant kB comes up whenever drawing a connection
between energy and temperature. The U.S. National Institute of Standards
and Technology gives its value as
kB = 1.380 649 × 10−23 J/K, (1.5)
2 John William Strutt, the third Baron Rayleigh (1842–1919), of England, is usually
called just “Lord Rayleigh”. Although particularly interested in acoustics, he made con-
tributions throughout physics: he was the first to explain why the sky is blue and
how seabirds soar. The mode counting described here was published in Nature on
18 May 1905.
3 English physicist and astronomer (1877–1946). His 1930 popular book The Mysterious

Universe did much to introduce the new quantum mechanics to a wide audience. His
correct mode counting argument was published in Philosophical Magazine in July 1905.
Rayleigh acknowledged his error (“Mr. Jeans has just pointed out that I have introduced
a redundant factor 8. . . . I hasten to admit the justice of this correction.”) in Nature on
13 July 1905.
4 Technically the “average energy”, but in these situations the thermal fluctuations

about average energy are so small I’ll just call it the “energy”.
5 Ludwig Boltzmann (1844–1906) of Austria developed statistical mechanics, which ex-

plains the properties of matter in bulk (such as density, hardness, luster, viscosity, and
resistivity) in terms of the properties of atoms (such as mass, charge, and potential en-
ergy of interaction). His students include Paul Ehrenfest, who we will meet later, and
Lise Meitner, who discovered nuclear fission.
14 Light in thermal equilibrium

but this number is hard to remember. For one thing, the joule (J) is a huge
unit for measuring atomic energies — using it would be like measuring the
distance between screw threads in miles or kilometers. The energy unit
typically used in atomic discussions is instead the “electron volt”, where
1 eV = 1.60 × 10−19 J. (1.6)
Second, I like to remember kB through a product: at room temperature,
1
about 300 K, the value of kB T is close to 40 eV. (This knowledge has
rescued me several times during physics oral exams.) The famous ideal
gas constant R (as in pV = nRT ) is just kB times the Avogadro number,
6.02 × 1023 .]]
Putting these four principles together results in the “Rayleigh–Jeans
law”, which says that for light in thermal equilibrium at temperature T
within a volume V , the electromagnetic energy due to wavelengths from λ
to λ + dλ is
8πV
(kB T ) 4 dλ. (1.7)
λ
This formula has numerous admirable features: It is dimensionally consis-
tent, as required. Doubling the volume results in doubling the energy, as ex-
pected. Higher temperature results in higher energy, in agreement with our
previous expectation that “high-temperature objects should glow brighter”.
The very long wavelength modes are unimportant because there’s no signif-
icant amount of energy in them anyway, so we can disregard our previous
qualifier that the formula holds only when V  λ3 .
On the other hand, there is nothing in this formula supporting our
expectation and common experience that high-temperature objects should
glow with shorter-wavelength light. To the contrary, at any temperature
the light spectrum should have exactly the same 1/λ4 character! Even
worse: What is the total energy in blackbody radiation? It is
Z ∞    ∞
1 1 1
(kB T )8πV 4
dλ = (kB T )8πV − 3
= +∞. (1.8)
0 λ 3 λ 0
Infinite energy! In fact, infinite energy at any finite temperature! If this
were true, then every book and table and wall — not to mention every
person — would be as deadly as an exploding nuclear bomb. The infinite
energy arises from the short wavelengths of the spectrum, so this disastrous
feature of the Rayleigh–Jeans prediction is called the “ultraviolet catastro-
phe”.
“Something Isn’t Quite Right” 15

energy (µJ)

6
0.6
d
d
d
d
d
0.4 d
Rayleigh–Jeans law
d

d
d
d
0.2
d d
d
d d

0.0 -

0 2000 4000 6000 8000

wavelength λ (nm)

Energy within one cubic meter due to electromagnetic radiation at tempera-

ture 1259 K, with wavelength from λ to λ + (1 nm). Experiment6 compared
with the Rayleigh–Jeans law (1.7).
Experiment, and common experience, differ from the Rayleigh–Jeans
law. However the experimental results mimic the Rayleigh–Jeans prediction
at long wavelengths, so it’s not totally off base. What could be wrong
with the derivation? Perhaps the short-wavelength radiation is not really
at thermal equilibrium. Perhaps there’s some error in the mode counting
that is significant only at short wavelengths. Or perhaps there’s something
wrong with the equipartition result.

6 O. Lummer and E. Pringsheim, Verhandlungen der Deutschen Physikalischen

Gesellschaft 2 (1900) 163.

16 Light in thermal equilibrium

In the year 1900 Max Planck7 decided to pursue this third possibility.
He wondered what would happen if the energy of a normal mode with
frequency f couldn’t take on any possible energy, but only certain values
E = nhf, where n = 0, 1, 2, 3, . . . (1.9)
and where h is some constant to be determined by a fit to experiment.
(Except that, being a formal German professor, Planck didn’t say that he
“wondered”, he said that he “hypothesized”.) Because f λ = c for light,
Planck’s hypothesis is equivalent to
hc
E=n. (1.10)
λ
When Planck worked out the statistical mechanical consequences of his
hypothesis, he found that if it were correct then the energy of a mode
would not be the equipartition result kB T , but instead
hc/λ
. (1.11)
e(hc/λ)/(kB T )
−1
No longer would the energy be equally partitioned. . . instead it would de-
pend upon wavelength.
Before rushing into the laboratory to test Planck’s idea, let’s see if it
even makes sense. Planck’s hypothesis is that the energy is discrete, not
continuous, that it comes in packets of size hf . (Except that he didn’t call
them packets, he called them “quanta”, from the Latin word for “amount”.
The singular is “quantum”, the plural is “quanta”.) When the frequency is
small, that is, when the wavelength is long, these quanta are so small that
the continuous approximation ought to be excellent. What does Planck’s
formula say for for long wavelengths? If λ is large, then (hc/λ)/(kB T ) is
small. How does ex behave when x is small? Remember Taylor’s formula:
ex ≈ 1 + x when |x|  1.
Thus for long wavelengths
hc/λ
e(hc/λ)/(kB T ) ≈ 1 + when λ  hc .
kB T kB T
7 Max Karl Ernst Ludwig Planck (1858–1947) was a German theoretical physicist par-

ticularly interested in thermodynamics and radiation. Concerning his greatest discovery,

the introduction of quantization into physics, he wrote, “I can characterize the whole pro-
cedure as an act of desperation, since, by nature I am peaceable and opposed to doubtful
adventures.” [Letter from Planck to R.W. Wood, 7 October 1931, quoted in J. Mehra
and H. Rechenberg, The Historical Development of Quantum Theory (Springer–Verlag,
New York, 1982) volume 1, page 49.]
“Something Isn’t Quite Right” 17

It follows that, in this same long-wavelength regime,

hc/λ
e(hc/λ)/(kB T ) − 1 ≈
kB T
and
hc/λ hc/λ
≈ = kB T.
e(hc/λ)/(kB T ) −1 (hc/λ)/(kB T )
The λs have canceled and we have recovered the equipartition result!
This analysis tells us two things: First, we know that for long wave-
lengths the Planck result (1.11) is almost the same as the equipartition
result. Second, we know that the boundary between long and short wave-
lengths falls near the crossover wavelength
hc
λ× = . (1.12)
kB T
This doesn’t mean that Planck’s formula is right, but it’s not transparently
wrong.
Adding Planck’s result for energy to the same normal mode count re-
sult (1.4) that we used before results in the “Planck radiation law”, which
says that for light in thermal equilibrium at temperature T within a volume
V , the electromagnetic energy due to wavelengths from λ to λ + dλ is
 
hc/λ 8πV
dλ. (1.13)
e (hc/λ)/(kB T ) −1 λ4
This formula has the same admirable features possessed by the Rayleigh–
Jeans result and discussed immediately below equation (1.7). (Does the
electromagnetic energy increase with increasing temperature? See sample
problem 1.1.2 on page 20.)
We’re still not quite ready for the laboratory. Does Planck’s result suffer
from the same ultraviolet catastrophe that the Rayleigh–Jeans result did?
This question is investigated in sample problem 1.1.1 on page 18. The result
is: No, it doesn’t.
Now is a good time to go to the laboratory. The Planck radiation law fits
the data extraordinarily well, provided that one uses the value for h given in
equation (1.2). Planck’s hypothesis that an electromagnetic normal mode
can’t take on any energy, only certain values given by equation (1.9), seems
to be correct.
18 Light in thermal equilibrium

energy (µJ)

6
0.6
d
d
d
d
d
0.4 d
Rayleigh–Jeans law
d

d
d
d
0.2
d
Planck law d
d
d d

0.0 -

0 2000 4000 6000 8000

wavelength λ (nm)

Energy within one cubic meter due to electromagnetic radiation at tem-

perature 1259 K, with wavelength from λ to λ + (1 nm). Experiment and
Rayleigh–Jeans law compared with the Planck law (1.13).

1.1.1 Sample Problem: Stefan–Boltzmann law

We saw that the Rayleigh–Jeans formula (1.7) could not represent reality,
because it said that any body in thermal equilibrium would contain an
infinite amount of radiant energy. Does the Planck formula (1.13) suffer
from the same excruciating defect?
“Something Isn’t Quite Right” 19

Solution: If the Planck radiation law holds, then the energy in blackbody
radiation of all wavelengths is of course
Z ∞ 
hc/λ 8πV
dλ. (1.14)
0 e hc/(kB T λ) − 1 λ4
You will be tempted to jump immediately into evaluating this integral,
but I urge you to pause for a moment and find a good strategy before
executing it. Integration is a mathematical, not a physical, operation, so I
will first convert to a mathematical variable — that is, to a variable without
dimensions. A glance suggests that the proper dimensionless variable is
hc/λ
x= . (1.15)
kB T
(Comparison to the crossover wavelength λ× in definition (1.12) shows that
this variable is just the ratio x = λ× /λ.) Converting the integral to this
variable shows that the total energy is
(kB T )4 ∞ x3
Z
8πV dx. (1.16)
(hc)3 0 ex − 1
Even without evaluating the integral, this equation gives us a lot of in-
formation. The integral on the right is just a number (unless it diverges)
independent of T (or h, or c), so the total energy of light in thermal equi-
librium is proportional to T 4 . This fact, called the Stefan–Boltzmann law,
is the formal result corresponding to our common experience that objects
glow brighter at higher temperatures.
20 Light in thermal equilibrium

x3 /(ex − 1)
6
1.5

1.0

0.5

0.0 - x

0 2 4 6

Plot of the integrand x3 /(ex − 1). At small x (that is x  1) the integrand

behaves like x2 , at large x like x3 e−x .
Now that we’ve squeezed the physics out, it’s time to execute the math.
You can work out the integral yourself, or you can use a computer algebra
system, and in either case you’ll find that the integral evaluates to π 4 /15.
The total energy is therefore
8π 5 (kB T )4
V . (1.17)
15 (hc)3
which is safely finite.

1.1.2 Sample Problem: Characteristics of the Planck Ra-

diation Law

Your class has decided to write an “Underground Guide to Quantum Me-

chanics” for the benefit of next year’s students in this course. You have
volunteered to contribute about 500 words explicating the Planck radiation
law (1.13). What do you write?

Solution: Like most equations in physics, the Planck radiation law (1.13
in our textbook) is concise, but this conciseness hides a lot of information.
This essay unpacks some of this information to see what the equation is
trying to tell us about nature. I address two points:
“Something Isn’t Quite Right” 21

(1) Does the electromagnetic energy increase with increasing

temperature?
There are a lot of symbols in the equation, but the temperature depen-
dence comes in through only one term: The energy at temperature T is
proportional to
1
eT0 /T − 1
where I have defined the temperature-independent constant T0 = hc/λkB .
As T increases, T0 /T decreases, so eT0 /T decreases, so eT0 /T − 1 decreases,
so 1/(eT0 /T − 1) increases. Thus the electromagnetic energy increases with
increasing T , as expected.
[Note, however, that energy doesn’t increase linearly with temperature.
You might know that for a monatomic gas modeled as noninteracting clas-
sical point particles, the total energy for N particles in thermal equlibrium
at temperature T is 32 N kB T . From this, some people get the mistaken no-
tion that energy and temperature are always related linearly, or even that
this is the definition of temperature. Blackbody radiation provides a neat
counterexample.]
If you like calculus, you could find the same result by showing that the
slope of this curve is always positive. That slope is
 
d 1
dT eT0 /T − 1
1 d(eT0 /T )
=−
(eT0 /T − 1)2 dT
1 d(eT0 /T ) d(1/T )
=−
(eT0 /T − 1)2 d(1/T ) dT
 
1 T0 /T 1
= − T /T T0 e − 2
(e 0 − 1)2 T
eT0 /T T0
=
(eT0 /T − 1) T 2
2

which sure enough is always positive.

(2) What does the Planck law say about the energy in very
short wavelengths?
The graph on page 12 of our textbook shows very small energies at
small wavelengths. In fact all three energy curves hide behind the axis!
22 Light in thermal equilibrium

Having investigated the behavior of energy with temperature, what can we

say about the behavior of energy with wavelength?
Write the energy equation to emphasize the λ dependence using the con-
stant crossover wavelength λ× defined in equation (1.12) in the textbook.
The energy is proportional to
(λ× /λ)5
.
eλ× /λ − 1
As λ → 0 this approaches ∞/∞. You could resolve this indeterminate
form using l’Hôpital’s rule, but it’s more insightful to look at the rate of
approach to infinity. Define x = λ× /λ, and investigate the behavior as
x → ∞. Both the numerator x and the denominator ex − 1 ≈ ex approach
infinity, but ex goes to infinity faster than x, faster than x2 , faster than x3 ,
23
faster than x5 , even faster than x10 . As x → ∞, the quantity
x5
ex − 1
is not just small, it’s “exponentially small”, which explains why it hides
behind the axis.
Conclusions: An equation is not an inert blob of symbols awaiting
numbers to “plug in and chug through”. An equation is a troubadour
singing songs about nature. The songs are interesting only if you listen
for them. This essay has listened to two of the notes sung by the Planck
radiation law, equation (1.13).

Problems
1.1 How big is an atom?
How many times can a liter of water be cut in half until you’re left
with a single water molecule?
1.2 Units for atomic-sized energies
Physicists are fond of measuring typical atomic energies in
“eV/atom”, chemists are fond of measuring typical atomic energies
in “kJ/mole”. An energy of exactly 1 eV in an atom corresponds
to what energy, in kJ, in a mole of atoms? (In reporting your
numerical result, be sure to use significant figures and units: see
appendices A and B.)
1.2. Photoelectric effect 23

1.3 Behavior of integrand

Justify the claim, made in the caption of the graph on page 20,
that when x  1, x3 /(ex − 1) ≈ x2 . [Clue: ex = 1 + x + 12 x2 + · · · .]

1.4 Wien displacement law (recommended problem)

Use the Planck radiation law (1.13) to show that at any given
temperature, the wavelength holding maximum energy is λ̂ = b/T ,
where b is some constant that you don’t need to determine. This so-
called “Wien displacement law” is the formal result corresponding
to our common experience that objects glow with shorter wave-
lengths at higher temperatures. [Clue: You could take the deriva-
tive of equation (1.13) with respect to λ, set the result equal to zero,
and solve for λ̂. This would be grotesquely difficult. Instead, using
x as defined in equation (1.15), argue that the function x5 /(ex − 1)
has a maximum located at, say, x̂, and then derive an expression
for λ̂ in terms of x̂ and other quantities. You don’t need to find a
numerical value for x̂, just argue that it exists.]

1.2 Photoelectric effect

But does Planck’s hypothesis apply only to light in thermal equilibrium?

Does it apply only to normal modes? Here’s the relevant question: What if
we produce light of wavelength λ proceeding in a beam. This light is surely
not in thermal equilibrium, nor is it in a normal mode. Will the energy in
the beam still be restricted to the possible values
hc
E=n = nhf where n = 0, 1, 2, 3, . . .? (1.18)
λ
In 1905 Albert Einstein8 thought this question was worth pursuing. He
reasoned in four steps:
(1) The electrons in a metal can be used as light-energy detectors.
(2) The reason the electrons are inside the metal at all, instead of
roaming around free, is because it’s energetically favorable for an electron
to be within the metal. Some of the electrons will be bound tightly within
8 Although Albert Einstein (1879–1955) is most famous for his work on relativity, he

claimed that he had “thought a hundred times as much about the quantum problems as I
have about general relativity theory.” [Remark to Otto Stern, reported in Abraham Pais,
“Subtle Is the Lord. . . ”: The Science and the Life of Albert Einstein (Oxford University
Press, Oxford, UK, 1982) page 9.]
24 Photoelectric effect

the metal, some less so. Call the trapping energy of the least-well-bound
electron Ut .
(3) If an electron within the metal absorbs a certain amount of energy
from the light shining on it, then some of that energy will go into getting
the electron out of the metal. The rest will become kinetic energy of the
ejected electron. The electron ejected with maximum kinetic energy will
be the one with the minimum trapping energy, namely Ut . [The reasoning
up to this point has been purely classical.]
(4) Add to this the quantum hypothesis: the amount of light energy
available for absorption can’t take on any old value — it has to be hf .
Then the maximum kinetic energy of an ejected electron will depend upon
the frequency of the light shining on the metal:
KEmax = hf − Ut . (1.19)
If the light has low frequency, hf < Ut , then no electron will be ejected at
all.
This analysis is clearly oversimplified. A metal is a complex system
rather than a simple potential energy well, and it ignores the possibility
that the electron might absorb an energy of 2hf , or 3hf , or more, but it’s
worth an experimental test: Plot the measured maximum kinetic energy of
ejected electrons as a function of frequency f . Will the plot give a straight
line with slope h matching the slope determined through the completely
different blackbody radiation experiment?
The challenge was taken up by Robert A. Millikan.9 By 1916 he exper-
imentally verified Einstein’s prediction, writing that “Einstein’s photoelec-
tric equation has been subjected to very searching tests and it appears in
every case to predict exactly the observed results.” Nevertheless, he found
the quantum condition troubling: this same 1916 paper calls quantization
a “bold, not to say reckless, hypothesis” because it “flies in the face of the
thoroughly established facts of interference”, and in his 1923 Nobel Prize
acceptance speech he said that his confirmation of the Einstein equation
had come “contrary to my own expectation”.
9 American experimental physicist (1868–1953), famous also for measuring the charge

of the electron through his oil-drop experiment, and for his research into cosmic rays.
A graduate of Oberlin College and of Columbia University, he used his administrative
acumen to build the small vocational school called Throop College of Technology into
the research and teaching powerhouse known today as the California Institute of Tech-
nology. His photoelectric results were reported in “A direct photoelectric determination
of Planck’s ‘h’ ” Physical Review 7 (1 March 1916) 355–390.
“Something Isn’t Quite Right” 25

Despite Millikan’s reservations, the conclusion is clear: The energy of

electromagnetic waves of frequency f cannot take on just any old value; it
can take on only the values
nhf where n = 0, 1, 2, 3, . . ..
This result is counterintuitive and non-classical, but it is in accord with
blackbody and photoelectric experiments, and that’s what matters.
One way to picture this energy restriction is to imagine the light as
coming in noninteracting particles, each particle having energy hf . These
pictured particles are called “photons”. If no photons are present, the
electromagnetic energy is 0; if one photon is present the energy is hf , if
two photons are present the energy is 2hf , and so forth. Using this picture,
the blackbody and photoelectric experiments are said to demonstrate the
“particle nature of light”.
It is important to realize that this particle picture goes above and be-
yond what the experiments say. The experiments tell us that the energy
can take on only certain values; they say nothing about particles. Because
light, classically, consists of electric and magnetic fields, it is tempting to
picture a photon as a “ball of light”, a packet of classical electric and mag-
netic fields. We will see soon that the picture of a photon as a classical
particle with a precise position, a precise energy (hf ), and a precise speed
(c) is not tenable. We will also encounter energy restrictions that cannot
be interpreted through this picture at all: for example, the energy of a
hydrogen atom is restricted to the values
Ry
− where n = 1, 2, 3, . . .
n2
and where Ry is a constant. There is no way to picture this energy re-
striction through a collection of noninteracting hypothetical particles. The
picture of photons can be made quite precise and can be very valuable,
but only if you keep in mind that a photon does not behave exactly like a
familiar classical point particle.
Further evidence for the quantized character of electromagnetic energy
comes from the Compton effect (which involves the interaction of x-rays and
electrons; see problem 1.9 on page 28), from the discrete clicks produced
by a photomultiplier tube or any other highly sensitive detector of light
energy, and from photon anticoincidence experiments. I will not describe
these experiments in detail, but you should understand that the evidence for
26 Photoelectric effect

energy quantization in light is both wide and deep.10 It is worth your effort
to memorize that electromagnetic energy comes in lumps of magnitude
hc 1240 eV·nm
E= = . (1.20)
λ λ
[[This equation comes up frequently in the physics Graduate Record Exam
and in physics oral exams. I recommend that you remember the constant
hc in terms of the unit usually used for energy in atomic situations, namely
the electron volt, and the unit usually used for the wavelength of optical
light, namely the nanometer.]]

1.2.1 Sample Problem: Find the flaw

No one would write a computer program and call it finished without test-
ing and debugging their first attempt. Yet some approach physics problem
solving in exactly this way: they get to the equation that is “the solution”,
stop, and then head off to bed for some well-earned sleep without investi-
gating whether the solution makes sense. This is a loss, because the real
fun and interest in a problem comes not from our cleverness in finding “the
solution”, but from uncovering what that solution tells us about nature.
(Appendix D, “Problem-Solving Tips and Techniques”, calls this final step
“reflection”.) To give you experience in this reflection step, I’ve designed
“find the flaw” problems in which you don’t find the solution, you only test
it. Here’s an example.
This is a physics problem that you are not supposed to solve:

Blackbody radiation is largely infrared at room tem-

perature, largely red in a campfire, largely blue in the star
Sirius. What is the relationship between the wavelength
holding the peak energy, called λ̂, and the temperature?
10 A clear summary of the evidence that light energy is quantized, but that a photon is
not just like a small, hard version of a classical marble, is presented in section 2.1, “Do
photons exist?”, of George Greenstein and Arthur G. Zajonc, The Quantum Challenge
(Jones and Bartlett Publishers, Sudbury, Massachusetts, 2006). See also J.J. Thorn,
M.S. Neel, V.W. Donato, G.S. Bergreen, R.E. Davies, and M. Beck, “Observing the
quantum behavior of light in an undergraduate laboratory” American Journal of Physics
72 (September 2004) 1210–1219.
“Something Isn’t Quite Right” 27

(This relationship enables astronomers to find the temper-

atures of distant stars inaccessible to human-made ther-
mometers.)

Four friends work this problem independently. When they get together
afterwards to compare results, they find that they have produced four dif-
ferent answers! Their candidate answers are
hc2
(a) λ̂ = 0.201
kB T
hc
(b) λ̂ = 0.201
kB T
hc
(c) λ̂ = 0.201 × 10−3
kB T
kB T
(d) λ̂ = 0.201
hc
Provide simple reasons showing that three of these candidates must be
wrong.

Solution: Candidate (a) does not have the correct dimensions for wave-
length. There are no problems with candidate (b), which is in fact the
correct “Wien displacement law”. Candidate (c) claims that at room tem-
1
perature (kB T = 40 eV), the dominant wavelength would be λ̂ = 10 nm,
deep in the ultraviolet, whereas the problem statement tells you that it’s
actually in the infrared. (Recall that hc = 1240 eV·nm.) Candidate (d)
not only has incorrect dimensions, it also shows the dominant wavelength
increasing, not decreasing, with temperature.

Problems
1.5 Visible light photons (recommended problem)
The wavelength of visible light stretches from 700 nm (red) to
400 nm (violet). (Figures with one significant digit.) What is
the energy range of visible photons?
28 Photoelectric effect

1.6 Light bulb photons

Stand about 100 meters from a 60-watt light bulb and look at
the bulb. The pupil in your eye has a diameter of about 2 mm.
Estimate the number of photons entering one of your eyes each
second. You will need to make reasonable assumptions: be sure to
spell them out.
1.7 Rephrasing the Einstein relation (essential problem)
Using your knowledge of classical waves, rewrite the Einstein re-
lation for the energy of a photon, E = hc/λ, in terms of the an-
gular frequency ω = 2πf of light. Employ the shortcut notation
~ = h/2π and compare your result to
E = ~ω. (1.21)
1.8 Character of photons
Two classical particles, say two asteroids, interact with a gravi-
tational potential energy, and each particle can have any possible
non-negative kinetic energy. Write a paragraph or two contrasting
these characteristics of classical particles with the characteristics
of photons.
1.9 Compton scattering
(This problem requires background in relativity.)
When x-rays shine on a target, they are scattered in all directions.
Both the wave and photon pictures of light predict this scattering.
The wave picture, however, predicts that the scattered x-rays will
have the same wavelength as that of the incoming waves, whereas
the photon picture predicts that the wavelength of the scattered
x-rays will depend upon the direction of scattering. This problem
derives that dependence.
An x-ray photon of energy E0 strikes a stationary electron of mass
m. The photon scatters off with energy E at angle θ, the electron
recoils with momentum p at angle φ. Recall from your study of
relativity that, for both photon and electron, E 2 − (pc)2 = (mc2 )2 ,
but that the photon mass is zero.
“Something Isn’t Quite Right” 29

initial:

photon,
energy E0
c - s
stationary electron,
mass m

final:

3
c
 scattered photon,

θ energy E

@ φ
@s
@
scattered electron,@
@
momentum p @@R
@
@

a. Write down the expressions for energy conservation, for mo-

mentum conservation in the horizontal direction, and for mo-
mentum conservation in the vertical direction, using only the
variables in the figure above. Compare your result to
p
E0 + mc2 = E + (mc2 )2 + (pc)2
E0 /c = (E/c) cos θ + p cos φ
0 = (E/c) sin θ − p sin φ.
b. It is hard to detect electrons and (relatively) easy to detect
x-rays. Hence we set out to eliminate the quantities involving
the scattered electron, namely p and φ. There are three equa-
tions and we wish to eliminate two variables, so we expect to
end up with one equation. Begin by squaring and combining
the last two equations to eliminate φ, finding
E02 − 2E0 E cos θ + E 2 = (pc)2 .
30 Wave character of electrons

c. Meanwhile, show that the energy conservation equation is

equivalent to
E02 − 2E0 E + E 2 + 2(E0 − E)mc2 = (pc)2 .
Combine these two equations to find
(E0 − E)mc2 = E0 E(1 − cos θ).
d. So far, this has been a relativistic collision problem. It be-
comes a quantum mechanics problem when we note that, for
a photon, E = hc/λ. Show that the x-ray wavelength changes
by
h
λ − λ0 = (1 − cos θ).
mc
e. Let’s read meaning into this equation to make it something
more than a jumble of symbols. Does the photon picture
predict an increase or decrease in wavelength? For what angle
does the wave-picture prediction λ = λ0 hold true?
In 1923, Arthur Compton verified that scattered x-rays have ex-
actly this angle dependence. His experiment convinced many physi-
cists that the photon picture — strange though it may be — must
have some merit. Ever since then, a particle of mass m has been
said to have a “Compton wavelength” of h/mc.

1.3 Wave character of electrons

If light, a classical wave, has some sort of particle character, could it be

that an electron, a classical particle, has some sort of wave character?
The very idea seems absurd and meaningless: the word “particle” sug-
gests a point, the word “wavelength” requires a length extending beyond a
point. But in 1924 Louis de Broglie11 thought the question was worth pur-
suing. He thought about the relationship between energy and wavelength
for a photon:
hc
E= .
λ
11 Louis de Broglie (1892–1987) was born into the French nobility, and is sometimes

called “Prince de Broglie”, although I am told that he was actually a duke. He earned
an undergraduate degree in history, but then switched into physics and introduced the
concept of particle waves in his 1924 Ph.D. thesis.
“Something Isn’t Quite Right” 31

De Broglie realized that if he wanted to extrapolate from a photon, which

is intrinsically relativistic, to an electron, which might or might not be
relativistic, he’d have have to use relativistic, not classical, mechanics. I
hope you remember from your study of special relativity that for a photon
— or anything else moving at light speed — the energy and momentum
are related through E = pc. De Broglie stuck these two facts together and
hypothesized that any particle has an associated wavelength given through
h
p= . (1.22)
λ
This associated wavelength is today called the de Broglie wavelength.
Please realize that this is not a derivation; it’s more like a stab in the
dark. And it’s a difficult hypothesis to test, because particle wavelengths
are typically so short. Recall from your study of classical waves that we
test for wave character through interference experiments, and that such
experiments work best when the apparatus is about the size of a wavelength.
So what are the sizes of these de Broglie waves? In all cases λ = h/p, but
in the non-relativistic case,
√ a particle with kinetic energy K and mass m
has K = p2 /2m, so p = 2Km and thus the de Broglie wavelength is
h hc
λ= √ =√ , (1.23)
2Km 2Kmc2
where in the last step I have inserted a factor of c in the numerator and
the denominator to make the formula easier to remember.
[[“What?” you object, “How can those extra cs, which just cancel out in
the end, make anything easier?” First of all, I’ve already recommended on
page 26 that you not memorize the value of h in SI units; instead you should
remember that hc = 1240 eV·nm. Second, most people don’t remember the
mass of an electron me in kilograms; instead, they remember it through the
energy equivalent me c2 = 0.511 MeV. Similarly for all other elementary
particles: I would have to look up the mass of a proton in kilograms, but I
remember that mp c2 ≈ 1000 MeV right off the top of my head.]]
Let’s try this out in a calculation. Suppose a stationary electron is
accelerated through a potential difference of exactly 100 V, so it picks up
a kinetic energy of K = 100 eV. That electron will have a de Broglie
32 Wave character of electrons

wavelength of
hc
λ= √
2Kmc2
1240 eV·nm
= p
2 × (100 eV) × (511 000 eV)
1240 nm
= p
2 × (102 ) × (0.511 × 106 )
1240 nm
= √
1.022 × 108
= 0.123 nm.
This is a very short wavelength — if it were an electromagnetic wave, it
would be in the x-ray regime (10 nm to 0.01 nm). In 1924, it was impossible
to manufacture slits — or anything else — with a size near 0.123 nm.
(Higher-energy electrons, or protons of the same energy, would have even
shorter wavelengths, and their wave character would be even harder to
discern.)
This difficulty was overcome by using, not human-manufactured slits,
but the rows of atoms within a crystal of nickel (with an atomic spacing of
0.352 nm). In a series of experiments executed from 1923 to 1927, Clinton
Davisson and Lester Germer12 showed that electrons scattering from nickel
exhibit interference as predicted by de Broglie’s strange hypothesis.
Since 1927, the wave character of particles, as demonstrated through in-
terference experiments, has been tested time and again. One breakthrough
came in 1987, when Akira Tonomura13 and his colleagues at the Hitachi
Advanced Research Laboratory in Tokyo demonstrated interference in elec-
trons thrown one at a time through a classic two-slit apparatus. In 2013,
Markus Arndt and his colleagues at the University of Vienna,14 building on
12 Davisson (1881–1958) and Germer (1896–1971) were experimenting at Bell Telephone
Laboratories in Manhattan with the ultimate goal not of testing quantum mechanics,
but of building better telephone amplifiers. Earlier, Germer had served as a World War I
fighter pilot. Later, he would become an innovative rock climber.
13 A. Tonomura, J. Endo, T. Matsuda, T. Kawasaki, and H. Ezawa, “Demonstration

of single-electron buildup of an interference pattern” American Journal of Physics 57

(1989) 117–120. Tonomura (1942–2012) worked tirelessly to develop the full potential
of electron microscopy, and almost as a sideline used these developments to test the
fundamentals of quantum mechanics. If you search the Internet for “single electron two
slit interference Akira Tonomura”, you will probably find his stunning video showing
how an electron interference pattern builds up over time.
14 Sandra Eibenberger, Stefan Gerlich, Markus Arndt, Marcel Mayor, and Jens Tüxen,
“Something Isn’t Quite Right” 33

work by Anton Zeilinger, demonstrated quantal interference in a molecule

consisting of 810 atoms. Research in this direction continues.15 Exactly
what is meant by “wave character of a particle” needs to be elucidated, but
the effect is so well explored experimentally that it cannot be denied.

Problems
1.10 de Broglie wavelengths for various particles
We have found the de Broglie wavelength of an electron with energy
100 eV. What about for a neutron with that energy? An atom of
gold? (A neutron’s mass is 1849 times the mass of an electron,
and a gold nucleus consists of 197 protons and neutrons.) If these
de Broglie waves had instead been electromagnetic waves, would
such wavelengths be characterized as ultraviolet, x-rays, or gamma
rays?
1.11 de Broglie wavelength for a big molecule
The 2013 experiments by Markus Arndt, mentioned in the
text, used the molecule C284 H190 F320 N4 S12 with mass 10 118 amu
(where the “atomic mass unit” is very close to the mass of a hydro-
gen atom), and with a velocity of 85 m/s. What was its de Broglie
wavelength?
1.12 Rephrasing the de Broglie relation (essential problem)
Rewrite the de Broglie relation, p = h/λ, in terms of the wavenum-
ber k = 2π/λ. Employ the shortcut notation ~ = h/2π and com-
pare your result to
p = ~k. (1.24)

“Matter–wave interference of particles selected from a molecular library with masses

exceeding 10 000 amu” Physical Chemistry Chemical Physics 15 (2013) 14696–14700.
15 Jonas Wätzel, Andrew James Murray, and Jamal Berakdar, “Time-resolved buildup

of two-slit-type interference from a single atom” Physical Review A 100 (12 July 2019)
013407.
34 Wave character of electrons

1.13 de Broglie wavelength for relativistic particles

(This problem requires background in relativity.)
The kinetic energy of a relativistic particle with energy E and mass
m is defined as K = E − mc2 . Recall that E 2 − (pc)2 = (mc2 )2 .
a. Show that
(pc)2 = (E − mc2 )(E + mc2 ).
b. Combine the above result with the de Broglie relation λ = h/p
to show that the generalization of equation (1.23) applicable
to both relativistic and non-relativistic particles is
hc
λ= p .
K(K + 2mc2 )
c. Show that the above equation reduces to equation (1.23) in the
non-relativistic limit. [Clue: Given two positive numbers, s
and b, with s  b, then s+b ≈ b, but there’s no approximation
for sb.]
1.14 Hofstadter’s use of relativistic electrons
(This problem assumes you have completed the previous problem.)
The 1961 Nobel Prize in Physics was awarded to Robert Hofstadter
“for his pioneering studies of electron scattering in atomic nuclei
and for his thereby achieved discoveries concerning the structure
of the nucleons”. In one of these experiments he bombarded gold
nuclei with electrons of total relativistic energy 183 MeV.
a. What is the de Broglie wavelength of such an electron?
b. Compare that wavelength to 10−15 m, the typical size of an
atomic nucleus. (The experiment succeeded only because the
wavelength of the electron probe was smaller than or compa-
rable to the size of the nucleus. In exactly the same way, you
can obtain a good image of a person’s face using light — with
a wavelength near 600 nm — but not using FM radio waves
— with a wavelength near 3 m.)
c. Hofstadter reported his results in a paper16 that didn’t say
whether the electron energy “183 MeV” refers to the total
energy or just the kinetic energy. Is this a significant error on
Hofstadter’s part?
16 R. Hofstadter, B. Hahn, A.W. Knudsen, and J.A. McIntyre, “High-energy electron

scattering and nuclear structure determinations, II” Physical Review 95 (15 July 1954)
512–515.
1.4. How does an electron behave? 35

1.4 How does an electron behave?

We have seen that an electron has a particle-like character, yet somehow it

has a wave-like character as well. This seeming paradox invites the question:
“How does an electron behave: like a particle or like a wave?”
I approach an answer through an analogy17 drawn from another field
of physics: the theory of classical waves. When a classical wave (water
wave, sound wave, light wave, etc.) of wavelength λ passes through a slit
of width a, wave theory tells us how the wave behaves: If the slit is large
(a  λ; “geometrical optics limit”) then the wave acts almost like a ray,
which passes through the slit with no spreading. If the slit is small (a  λ;
“spherical wave limit”) then the wave acts like a Huygens wavelet, which
passes through the slit and then spreads throughout the half-circle on the far
side. For slits of intermediate size the wave acts in an intermediate manner.
The behavior of a classical wave is known exactly and can be calculated
with exquisite accuracy. Under some circumstances it behaves almost like
a ray, and in some circumstances it behaves almost like a Huygens wavelet,
although it takes on these behaviors exactly only in limiting cases.
As with classical waves, so with electrons. The theory of quantum
mechanics tells us with exquisite accuracy how an electron behaves in all
circumstances. Under some circumstances it behaves almost like a classical
particle. Under other circumstances it behaves almost like a classical wave.
The question “Does an electron behave like a classical particle or like a
classical wave?” is like the question “Does a classical wave behave like a
ray or like a Huygens wavelet?” It never behaves exactly like either. Instead,
it behaves in its own inimitable way, which you might call “typical quantum
mechanical behavior”.
It is the job of this book (and of the rest of your physics education) to
teach you “typical quantum mechanical behavior”. If you open your mind
to the idea that electrons behave in a manner unlike anything you have
previously encountered, then you can gain an appreciation of and build an
intuition for such “typical quantum mechanical behavior”. If you refuse to
admit this possibility, then you might be able to execute the problems, and
you might even get a good grade, but your mind will be forever closed to
one of the wonders of our universe.
17 Perhaps you will find that the parable of “the blind men and the elephant” makes a

more appealing analogy.

36 Quantization of atomic energies

1.5 Quantization of atomic energies

If the energy of light comes in quantized amounts, how about the energy
of an atom?

1.5.1 Experiments

It is not hard to change the energy of an atom: simply put a gas in a

high-voltage discharge tube to give the gas atoms a jolt. The atoms will
absorb energy from the discharge, and then will release light energy as they
fall back to their ground state. If the atoms can take on continuous energy
values, then they will emit light of continuous energy value and (through
equation 1.20) of continuous wavelength. But if the atoms can take on
only certain quantized energy values, then they will emit light of quantized
energy value and hence of only certain wavelengths. Perhaps you have
performed experiments sending the light from a discharge tube through
a diffraction grating and spreading it out by wavelength: if so, then you
know from your personal experience that the light comes out at only certain
wavelengths, not with continuous wavelengths. If energy is released from
an atom through light, it seems that the atom can take on only certain
quantized energy values.
But what if the energy goes into or out of an atom through some other
mechanism? In 1914 James Franck18 and Gustav Hertz19 figured out a way
to get energy into mercury atoms through collisions with electrons. There’s
no need to go through the details of the Franck–Hertz experiment, but the
conclusion is again that the atom cannot accept just any old amount of en-
ergy: it can only absorb energy in certain quantized amounts. Furthermore,
those amounts agreed with the amounts derived from spectral experiments.
18 German physicist (1882–1964) who left Germany in disgust after the Nazi Party came

to power. He went first to Denmark, then to the United States where he worked to build
the nuclear bomb. He authored a report recommending that U.S. nuclear bombs not be
used on Japanese cities without warning.
19 German physicist (1887–1975), nephew of Heinrich Hertz, for whom the unit of fre-

quency is named. Hertz was forced out of his career in Germany due to distant Jewish
ancestry. He went to the Soviet Union and there worked to build the Soviet nuclear
bomb.
“Something Isn’t Quite Right” 37

1.5.2 Bohr’s theory

Niels Bohr20 decided in 1913 not just to accept the quantization of atomic
energies as an experimental fact, but to find a theoretical underpinning.
He started with the simplest atom: hydrogen. Hydrogen consists of an
electron (mass me , charge −e) and a far more massive proton (charge +e).
Bohr made21 two assumptions: first that the electron orbits the proton
only in circular (never elliptical) orbits, second that the circumference of
the circular orbit holds an whole number of de Broglie wavelengths (one
or two or three or more but never 2.7). If you remember F = ma, and
the formula for centripetal acceleration (v 2 /r), and Coulomb’s law, you’ll
realize that the circular orbit assumption demands
e2 1 v2 p2
= m e = . (1.25)
4π0 r2 r me r
And if you remember the de Broglie formula λ = h/p you’ll realize that the
assumption of a whole number of wavelengths demands
2πr = nλ = nh/p where n = 1, 2, 3, . . .. (1.26)
Here are two different formulas connecting the two variables r and p, so we
can solve for these variables in terms of n, h, me , and e2 /4π0 . (Don’t get
distracted by the quantities λ and v . . . if we decide we want them later on
we can easily find them once we’ve solved for r and p.) Once our objectives
are clear it’s not hard to achieve them. Solve equation (1.26) for p,
h/2π
p=n , (1.27)
r
and then plug this into equation (1.25) giving
e2 /4π0 2 (h/2π)
2
= n .
r2 me r3
Solve this equation for r giving
(h/2π)2
r = n2 , (1.28)
me (e2 /4π0 )
20 Danish physicist (1885–1962), fond of revolutionary ideas. In 1924 and again in 1929

he suggested that the law of energy conservation be abandoned, but both suggestions
proved to be on the wrong track. Father of six children, all boys, one of whom won the
Nobel Prize in Physics and another of whom played in the 1948 Danish Olympic field
hockey team.
21 The treatment in this book captures the spirit but not all the nuance of Bohr’s argu-

ment.
38 Quantization of atomic energies

and then plug this back into (1.27) giving

1 me (e2 /4π0 )
p= . (1.29)
n h/2π
(Notice that the quantity e2 /4π0 makes a natural combination of quanti-
ties, so we keep it together as a packet and never rend it apart in the course
of our algebraic manipulations. Similarly for the quantity h/2π.)
If Bohr’s assumptions hold, then the orbital radius can’t be any old
value, it can only take on the values given in (1.28). And the momentum
can’t be any old value, it can only take on the values given in (1.29). This
is unexpected and curious and worthwhile, but in Bohr’s day, and still
today, our experimental apparatus is not so refined that it can determine
the radius or the momentum of a single electron orbiting a proton, so it’s
also sort of useless. We can find an experimentally accessible result by
calculating the energy
p2 e2 /4π0
E= − .
2me r
If Bohr’s assumptions hold, then the energy, too, can’t be any old value, it
can only take on the quantized values
1 me (e2 /4π0 )
E=− . (1.30)
n2 2(h/2π)2
And, as discussed in section 1.5.1, the energy values of an atom are exper-
imentally accessible.
Experiment shows this result for hydrogen to be correct. Yet the deriva-
tion clearly leaves much to be desired. The result (1.30) springs from two
equations: equation (1.25) assumes that the electron is a classical point
particle in a circular orbit; equation (1.26) assumes that the electron is a
de Broglie wave. Both equations are needed to produce the energy quanti-
zation result, yet the two assumptions cannot both be correct.

1.5.3 Visualizations

As with the photon, it is the job of the rest of this book to come up with
a description of an electron that is correct, but one thing is clear already:
the visualization of an electron as a classical point particle — a smaller
and harder version of a marble — cannot be correct. At this stage in your
quantal education I cannot yet give you a perfect picture of an electron, but
“Something Isn’t Quite Right” 39

you can see that the picture of an electron as a point particle with a position,
a speed, and an energy — the picture that appeared in, for example, the
seal of the U.S. Atomic Energy Commission — must be wrong.

An atom does not look like this.

An atom does not look like this, either.

The electron is not a small, hard marble with a position, a speed, and an
energy, and any intuition you hold in your mind based on that mistaken
visualization is likely to lead you astray.
40 Quantization of atomic energies

1.5.4 Multi-electron atoms

Bohr, of course, forged on to investigate atoms more complicated than

hydrogen. He found that he could indeed explain the spectra of more com-
plicated atoms — sometimes with high accuracy, sometimes only approx-
imately — but not by making the assumption about circular orbits that
worked so well for hydrogen. For example, here are the orbits required for
the eleven electrons in a sodium atom:

A sodium atom in the Bohr model.22

Atoms larger than sodium required even more elaborate schemes, and
each new atom required a new set of assumptions. Furthermore, even
for simple hydrogen, Bohr could never explain how the quantized ener-
gies changed when the atom was placed in a magnetic field (the “Zeeman23
effect”).
We must conclude that the Bohr model, despite its impressive prediction
concerning the energies of a hydrogen atom in the absence of magnetic field,
is wrong.

22 Reproduced from K.A. Kramers and Helge Holst, The Atom and the Bohr Theory of

Its Structure (Alfred A. Knopf, New York, 1923) rear endpaper.

23 Pieter Zeeman (1865–1943), Dutch physicist, discovered this effect in 1896, three years

after earning his Ph.D. The prestigious journal Nature had earlier described his obser-
vations of the meteoroid of 17 November 1882, made at age 17 years.
“Something Isn’t Quite Right” 41

Problem
1.15 Characteristic quantities (recommended problem)
[[It’s a good idea to develop a sense of typical, or “characteristic”,
sizes: if a problem in classical mechanics asks you to calculate the
mass of a squirrel, and you find 937 kg, then you know you’ve made
a mistake somewhere. In classical mechanics this sense of typical
quantities comes from everyday experience. In atomic physics this
sense has to be built.24 Although the Bohr model is not correct, it
does provide a reasonable picture of typical sizes for atomic quan-
tities, and this problem is your first step toward such a “tangible
picture”.]]
The “characteristic length” for atomic systems is the so-called
“Bohr radius”, the radius of the smallest allowed orbit, which is
(see equation 1.28)
(h/2π)2
a0 ≡ . (1.31)
me (e2 /4π0 )
a. Evaluate the Bohr radius numerically in nanometers. Com-
pare to a wavelength of blue light.
The “characteristic time” for atomic systems is conventionally de-
fined not as the period of the smallest allowed Bohr orbit, but as
this period divided by 2π. (Not the time for the electron to make
one orbit, but the time for it to sweep out an angle of one radian.)
b. Derive a formula for this time and evaluate it numerically in
femtoseconds. Compare to a period of blue light.
[[Characteristic quantities for atomic systems will be explored fur-
ther in problem 7.7, “Characteristic quantities for the Coulomb
problem”, on page 246. That problem shows why it makes sense
to divide the period by 2π.]]

24 When the mathematician Stanislaw Ulam became interested in nuclear physics he

“discovered that if one gets a feeling for no more than a dozen. . . nuclear constants, one
can imagine the subatomic world almost tangibly, and manipulate the picture dimen-
sionally and qualitatively, before calculating more precise relationships.” [Stanislaw M.
Ulam, Adventures of a Mathematician (Charles Scribner’s Sons, New York, 1976) pages
147–148. (From the chapter “Life among the Physicists: Los Alamos”.)]
42 Quantization of magnetic moment

1.6 Quantization of magnetic moment

An electric current flowing in a loop produces a magnetic moment, so it

makes sense that the electron orbiting (or whatever it does) an atomic
nucleus would produce a magnetic moment for that atom. And of course, it
also makes sense that physicists would be itching to measure that magnetic
moment.
It is not difficult to measure the magnetic moment of, say, a scout
compass. Place the magnetic compass needle in a known magnetic field
and measure the torque that acts to align the needle with the field. You
will need to measure an angle and you might need to look up a formula in
your magnetism textbook, but there is no fundamental difficulty.
Measuring the magnetic moment of an atom is a different matter. You
can’t even see an atom, so you can’t watch it twist in a magnetic field like a
compass needle. Furthermore, because the atom is very small, you expect
the associated magnetic moment to be very small, and hence very hard to
measure. The technical difficulties are immense.
These difficulties must have deterred but certainly did not stop Otto
Stern and Walter Gerlach.25 They realized that the twisting of a magnetic
moment in a uniform magnetic field could not be observed for atomic-sized
magnets, and also that the moment would experience zero net force. But
they also realized that a magnetic moment in a non-uniform magnetic field
would experience a net force, and that this force could be used to measure
the magnetic moment.

25 Otto Stern (1888–1969) was a Polish-German-Jewish physicist who made contributions

to both theory and experiment. He left Germany for the United States in 1933 upon
the Nazi ascension to power. Walter Gerlach (1889–1979) was a German experimental
physicist. During the Second World War he led the physics section of the Reich Research
Council and for a time directed the German effort to build a nuclear bomb.
“Something Isn’t Quite Right” 43

~
B

z
6 µ
~

A classical magnetic moment in a non-uniform magnetic field.

A classical magnetic moment µ ~ that

~ , situated in a magnetic field B
points in the z direction and increases in magnitude in the z direction, is
subject to a force
∂B
µz , (1.32)
∂z
where µz is the z-component of the magnetic moment or, in other words,
the projection of µ
~ on the z axis. (If this is not obvious to you, then work
problem 2.16, “Force on a classical magnetic moment”, on page 46.)
Stern and Gerlach used this fact to measure the z-component of the
magnetic moment of an atom. First, they heated silver in an electric “oven”.
The vaporized silver atoms emerged from a pinhole in one side of the oven,
and then passed through a non-uniform magnetic field. At the far side of
the field the atoms struck and stuck to a glass plate. The entire apparatus
had to be sealed within a good vacuum, so that collisions with nitrogen
molecules would not push the silver atoms around. The deflection of an
atom away from straight-line motion is proportional to the magnetic force,
and hence proportional to the projection µz . In this ingenious way, Stern
and Gerlach could measure the z-component of the magnetic moment of an
atom even though any single atom is invisible.
Before turning the page, pause and think about what results you would
expect from this experiment.
44 Quantization of magnetic moment

Here are the results that I expect: I expect that an atom which happens
to enter the field with magnetic moment pointing straight up (in the z
direction) will experience a large upward force. Hence it will move upward
and stick high up on the glass-plate detector. I expect that an atom which
happens to enter with magnetic moment pointing straight down (in the −z
direction) will experience a large downward force, and hence will stick far
down on the glass plate. I expect that an atom entering with magnetic
moment tilted upward, but not straight upward, will move upward but
not as far up as the straight-up atoms, and the mirror image for an atom
entering with magnetic moment tilted downward. I expect that an atom
entering with horizontal magnetic moment will experience a net force of
zero, so it will pass through the non-uniform field undeflected.
Furthermore, I expect that when a silver atom emerges from the oven
source, its magnetic moment will be oriented randomly — as likely to point
in one direction as in any other. There is only one way to point straight up,
so I expect that very few atoms will stick high on the glass plate. There are
many ways to point horizontally, so I expect many atoms to pass through
undeflected. There is only one way to point straight down, so I expect very
few atoms to stick far down on the glass plate.26
In summary, I expect that atoms would leave the magnetic field in any of
a range of deflections: a very few with large positive deflection, more with a
small positive deflection, a lot with no deflection, some with a small negative
deflection, and a very few with large negative deflection. This continuity of
deflections reflects a continuity of magnetic moment projections.

26 To be specific, this reasoning suggests that the number of atoms with moment tilted

at angle θ relative to the z direction is proportional to sin θ, where θ ranges from 0◦ to

180◦ . You might want to prove this to yourself, but we’ll never use this result so don’t
feel compelled.
“Something Isn’t Quite Right” 45

In fact, however, this is not what happens at all! The projection µz

does not take on a continuous range of values. Instead, it is quantized and
takes on only two values, one positive and one negative. Those two values
are called µz = ±µB where µB , the so-called “Bohr magneton”, has the
measured value of
µB = 9.274 010 078 × 10−24 J/T, (1.33)
with an uncertainty of 3 in the last decimal digit.

Distribution of µz
Expected: Actual:
µz µz

+µB

0 0

−µB

The Stern-Gerlach experiment was initially performed with silver atoms

but has been repeated with many other types of atoms. When nitrogen is
used, the projection µz takes on one of the four quantized values of +3µB ,
+µB , −µB , or −3µB . When sulfur is used, it takes on one of the five
quantized values of +4µB , +2µB , 0, −2µB , and −4µB . For no atom do the
values of µz take on the broad continuum of my classical expectation. For
all atoms, the projection µz is quantized.
 46 Quantization of magnetic moment

Problems
1.16 Force on a classical magnetic moment
The force on a classical magnetic moment is most easily calculated using
“magnetic charge fiction”: Consider the magnetic moment to consist of
two “magnetic charges” of magnitude +m and −m, separated by the
position vector d~ running from −m to +m. The magnetic moment is
then µ
~ = md.~
a. Use B+ for the magnitude of the magnetic field at +m, and B−
for the magnitude of the magnetic field at −m. Show that the net
force on the magnetic moment is in the z direction with magnitude
mB+ − mB− .
b. Use dz for the z-component of the vector d. ~ Show that to high
accuracy
∂B
B+ = B− + dz .
∂z
Surely, for distances of atomic scale, this accuracy is more than
adequate.
c. Derive expression (1.32) for the force on a magnetic moment.

1.17 Questions (recommended problem)

Answering questions is an important scientific skill and, like any skill,
it is sharpened through practice. This book gives you plenty of oppor-
tunities to develop that skill. Asking questions is another important
scientific skill.27 To hone that skill, write down a list of questions you
have about quantum mechanics at this point. Be brief and pointed:
you will not be graded for number or for verbosity. In future problems,
I will ask you to add to your list.
[[For example, one of my questions would be: “In ocean waves, the water
is doing the ‘waving’. In sound waves, it’s the air. In light waves, it’s
the abstract electromagnetic field. What is ‘waving’ in a de Broglie
wave?”]]

27 “The important thing is not to stop questioning,” said Einstein. “Never lose a holy

curiosity.” [Interview by William Miller, “Death of a Genius”, Life magazine, volume 38,
number 18 (2 May 1955) pages 61–64 on page 64.]
 Chapter 2

What Is Quantum Mechanics About?

The story of Planck’s discovery of the quantization of light energy is a

fascinating one, but it’s a difficult and elaborate story because it involves
not just quantization, but also thermal equilibrium and electromagnetic
radiation. The story of the discovery of atomic energy quantization is just
as fascinating, but again fraught with intricacies. In an effort to remove
the extraneous and dive deep to the heart of the matter, we focus on the
measurement of the magnetic moment of a silver atom. We will, to the
extent possible, do a quantum-mechanical treatment of an atom’s magnetic
moment while maintaining a classical treatment of all other aspects — such
as its energy and momentum and position. (In chapter 4, “The Quantum
Mechanics of Position”, we take up a quantum-mechanical treatment of
position and energy.)
The previous chapter attempted to apply classical pictures to atomic
entities — electrons pictured as small, hard marbles; magnetic moments
pictured as classical pointing arrows — and it found that the results were
untenable. So this chapter will not impose the classical pictures in our
minds onto nature. Instead we will perform experiments and let the atoms
themselves tell us how they behave.

2.1 Quantization

2.1.1 The conundrum of projections

The Stern-Gerlach result — that µz is quantized rather than continuous —

is counterintuitive and unexpected, but we can live with the counterintuitive
and unexpected. It happens all the time in politics.

47
48 Quantization

However, this fact of quantization appears to result in a logical con-

tradiction, because there are many possible axes upon which the magnetic
moment can be projected. The figures below make it clear that it is impos-
sible for any vector to have a projection of either ±µB on all axes!
Because if the projection of µ
~ on the z axis is +µB . . .

+µB
µ
~

. . . then the projection of µ

~ on this second axis must be more than +µB . . .

µ
~

. . . while the projection of µ

~ on this third axis must be less than +µB .

µ
~
What Is Quantum Mechanics About? 49

Whenever we measure the magnetic moment, projected onto any axis,

the result is either +µB or −µB . Yet is it impossible for the projection
of any classical arrow on all axes to be either +µB or −µB ! This seeming
contradiction is called “the conundrum of projections”. We can live with
the counterintuitive, the unexpected, the strange, but we cannot live with
a logical contradiction. How can we resolve it?
The resolution comes not from meditating on the question, but from
experimenting about it. Let us actually measure the projection on one
axis, and then on a second. To do this easily, we modify the Stern-Gerlach
apparatus and package it into a box called a “Stern-Gerlach analyzer”. This
box consists of a Stern-Gerlach apparatus followed by “pipes” that channel
the outgoing atoms into horizontal paths.1 This chapter treats only silver
atoms, so we use analyzers with two exit ports.

packaged into

An atom enters a vertical analyzer through the single hole on the left.
If it exits through the upper hole on the right (the “+ port”) then the
outgoing atom has µz = +µB . If it exits through the lower hole on the
right (the “− port”) then the outgoing atom has µz = −µB .
1 In general, the “pipes” will manipulate the atoms through electromagnetic fields, not

through touching. One way way to make such “pipes” is to insert a second Stern-Gerlach
apparatus, oriented upside-down relative to the first. The atoms with µz = +µB , which
had experienced an upward force in the first half, will experience an equal downward
force in the second half, and the net impulse delivered will be zero. But whatever their
manner of construction, the pipes must not change the magnetic moment of an atom
passing through them.
50 Quantization

µz = +µB

µz = −µB

2.1.2 Two vertical analyzers

In order to check the operation of our analyzers, we do preliminary exper-

iments. Atoms are fed into a vertical analyzer. Any atom exiting from the
+ port is then channeled into a second vertical analyzer. That atom exits
from the + port of the second analyzer. This makes sense: the atom had
µz = +µB when exiting the first analyzer, and the second analyzer confirms
that it has µz = +µB .

all
µz = +µB

none
µz = −µB
(ignore these)

Furthermore, if an atom exiting from the − port of the first analyzer

is channeled into a second vertical analyzer, then that atom exits from the
− port of the second analyzer.

2.1.3 One vertical and one upside-down analyzer

Atoms are fed into a vertical analyzer. Any atom exiting from the + port is
then channeled into a second analyzer, but this analyzer is oriented upside-
down. What happens? If the projection on an upward-pointing axis is +µB
(that is, µz = +µB ), then the projection on a downward-pointing axis is
−µB (we write this as µ(−z) = −µB ). So I expect that these atoms will
emerge from the − port of the second analyzer (which happens to be the
higher port). And this is exactly what happens.
What Is Quantum Mechanics About? 51

all
µz = +µB

none
µz = −µB
(ignore these)

Similarly, if an atom exiting from the − port of the first analyzer is

channeled into an upside-down analyzer, then that atom emerges from the
+ port of the second analyzer.

2.1.4 One vertical and one horizontal analyzer

Atoms are fed into a vertical analyzer. Any atom exiting from the + port is
then channeled into a second analyzer, but this analyzer is oriented horizon-
tally. The second analyzer doesn’t measure the projection µz , it measures
the projection µx . What happens in this case? Experiment shows that the
atoms emerge randomly: half from the + port, half from the − port.
z
y
x

half (µx = −µB )

µz = +µB
half (µx = +µB )

(ignore these)
µz = −µB

This makes some sort of sense. If a classical magnetic moment were

vertically oriented, it would have µx = 0, and such a classical moment
would go straight through a horizontal Stern-Gerlach analyzer. We’ve seen
that atomic magnetic moments never go straight through. If you “want” to
go straight but are forced to turn either left or right, the best you can do is
52 Quantization

turn left half the time and right half the time. (Don’t take this paragraph
literally. . . atoms have no personalities and they don’t “want” anything.
But it is a useful mnemonic.)

2.1.5 One vertical and one backwards horizontal analyzer

Perform the same experiment as above (section 2.1.4), except insert the
horizontal analyzer in the opposite sense, so that it measures the projection
on the negative x axis rather than the positive x axis. Again, half the atoms
emerge from the + port, and half emerge from the − port.
z
y
x

half (µ(−x) = +µB )

µz = +µB
half (µ(−x) = −µB )

(ignore these)
µz = −µB

2.1.6 One horizontal and one vertical analyzer

A +x analyzer followed by a +z analyzer is the same apparatus as above

(section 2.1.5), except that both analyzers are rotated as a unit by 90◦ about
the y axis. So of course it has the same result: half the atoms emerge from
the + port, and half emerge from the − port.
What Is Quantum Mechanics About? 53

z
y
x

µx = −µB µz = +µB

µx = +µB
µz = −µB

2.1.7 Three analyzers

Atoms are fed into a vertical analyzer. Any atom exiting from the + port
is then channeled into a horizontal analyzer. Half of these atoms exit from
the + port of the horizontal analyzer (see section 2.1.4), and these atoms
are channeled into a third analyzer, oriented vertically. What happens at
the third analyzer?
z

y
x

µx =−µB ?
µz =+µB
µx =+µB ?

µz =−µB

There are two ways to think of this: (I) When the atom emerged from
the + port of the first analyzer, it was determined to have µz = +µB .
When that same atom emerged from the + port of the second analyzer,
it was determined to have µx = +µB . Now we know two projections
of the magnetic moment. When it enters the third analyzer, it still has
µz = +µB , so it will emerge from the + port. (II) The last two analyzers
in this sequence are a horizontal analyzer followed by a vertical analyzer,
and from section 2.1.6 we know what happens in this case: a 50/50 split.
That will happen in this case, too.
54 Quantization

So, analysis (I) predicts that all the atoms entering the third analyzer
will exit through the + port and none through the − port. Analysis (II)
predicts that half the atoms will exit through the + port and half through
the − port.
Experiment shows that analysis (II) gives the correct result. But what
could possibly be wrong with analysis (I)? Let’s go through line by line:
“When the atom emerged from the + port of the first analyzer, it was
determined to have µz = +µB .” Nothing wrong here — this is what an
analyzer does. “When that same atom emerged from the + port of the
second analyzer, it was determined to have µx = +µB .” Ditto. “Now we
know two projections of the magnetic moment.” This has got to be the
problem. To underscore that problem, look at the figure below.

µ
~
+µB

x
+µB

If an atom did have both µz = +µB and µx = +µB , then the √ projection
◦
on an axis rotated 45 from the vertical would be µ45 = + 2 µB . But
◦

the Stern-Gerlach experiment assures us that√whenever µ45◦ is measured,

the result is either +µB or −µB , and never + 2 µB . In summary, it is not
possible for a moment to have a projection on both the z axis and on the
x axis. Passing to the fourth sentence of analysis (I) — “When the atom
enters the third analyzer, it still has µz = +µB , so it will emerge from the
+ port” — we immediately see the problem. The atom emerging from the
+ port of the second analyzer does not have µz = +µB — it doesn’t have
a projection on the z axis at all.
What Is Quantum Mechanics About? 55

Because it’s easy to fall into misconceptions, let me emphasize what I’m
saying and what I’m not saying:

I’m saying that if an atom has a value for µx , then it doesn’t have
a value for µz .
I’m not saying that the atom has a value for µz but no one knows
what it is.
I’m not saying that the atom has a value for µz but that value is
changing rapidly.
I’m not saying that the atom has a value for µz but that value is
changing unpredictably.
I’m not saying that a random half of such atoms have the value
µz = +µB and the other half have the value µz = −µB .
I’m not saying that the atom has a value for µz which will be
disturbed upon measurement.

The atom with a value for µx does not have a value for µz in the same way
that love does not have a color.
This is a new phenomenon, and it deserves a new name. That name
is “indeterminacy”. This is perhaps not the best name, because it might
suggest, incorrectly, that an atom with a value for µx has a value for µz and
we merely haven’t yet determined what that value is. The English language
was invented by people who didn’t understand quantum mechanics, so it is
not surprising that there are no perfectly appropriate names for quantum
mechanical phenomena. This is a defect in our language, not a defect in
quantum mechanics or in our understanding of quantum mechanics, and it
is certainly not a defect in nature.2
How can a vector have a projection on one axis but not on another? It
is the job of the rest of this book to answer that question, but one thing
is clear already: The visualization of an atomic magnetic moment as a
classical arrow must be wrong.
2 In exactly the same manner, the name “orange” applies to light within the wavelength

range 590–620 nm and the name“red” applies to light within the wavelength range 620–
740 nm, but the English language has no word to distinguish the wavelength range
1590–1620 nm from the wavelength range 1620–1740 nm. This is not because optical
light is “better” or “more deserving” than infrared light. It is due merely to the accident
that our eyes detect optical light but not infrared light.
56 Quantization

2.1.8 The upshot

We escape from the conundrum of projections through probability. If an

atom has µz = +µB , and if the projection on some other axis is measured,
then the result cannot be predicted with certainty: we instead give proba-
bilities for the various results. If the second analyzer is rotated by angle θ
relative to the vertical, the probability of emerging from the + port of the
second analyzer is called P+ (θ).
z

θ
µθ = +µB
µz = +µB

µθ = −µB
µz = −µB

We already know some special values: from section 2.1.2, P+ (0◦ ) = 1;

from section 2.1.4, P+ (90◦ ) = 12 ; from section 2.1.3, P+ (180◦ ) = 0; from
section 2.1.5, P+ (270◦ ) = 12 ; from section 2.1.2, P+ (360◦ ) = 1. It is not
hard to guess the curve that interpolates between these values:
P+ (θ) = cos2 (θ/2), (2.1)
and experiment confirms this guess.

P+ (θ)
1
2

0
0◦ 90◦ 180◦ 270◦ 360◦
θ
What Is Quantum Mechanics About? 57

Problems
2.1 Exit probabilities (essential problem)
a. An analyzer is tilted from the vertical by angle α. An atom leaving
its + port is channeled into a vertical analyzer. What is the proba-
bility that this atom emerges from the + port? The − port? (Clue:
Use the “rotate as a unit” concept introduced in section 2.1.6.)
z

b. An atom exiting the − port of a vertical analyzer behaves exactly

like one exiting the + port of an upside-down analyzer (see sec-
tion 2.1.3). Such an atom is channeled into an analyzer tilted from
the vertical by angle β. What is the probability that this atom
emerges from the + port? The − port?
z

(Problem continues on next page.)

58 Quantization

c. An analyzer is tilted from the vertical by angle γ. An atom leav-

ing its − port is channeled into a vertical analyzer. What is the
probability that this atom emerges from the + port? The − port?
z
γ

2.2 Multiple analyzers

An atom with µz = +µB is channeled through the following line of
three Stern-Gerlach analyzers.

β
α γ

- or -

A C

Find the probability that it emerges from (a) the − port of analyzer
A; (b) the + port of analyzer B; (c) the + port of analyzer C; (d) the
− port of analyzer C.
2.3 Properties of the P+ (θ) function
a. An atom exits the + port of a vertical analyzer; that is, it has
µz = +µB . Argue that the probability of this atom exiting from
the − port of a θ analyzer is the same as the probability of it
exiting from the + port of a (180◦ − θ) analyzer.
b. Conclude that the P+ (θ) function introduced in section 2.1.8 must
satisfy
P+ (θ) + P+ (180◦ − θ) = 1.
c. Does the experimental result (2.1) satisfy this condition?
2.2. Interference 59

2.2 Interference

There are more quantum mechanical phenomena to uncover. To support

our exploration, we build a new experimental device called the “analyzer
loop”.3 This is nothing but a Stern-Gerlach analyzer followed by “piping”
that channels the two exit paths together again.4

packaged into

The device must be constructed to high precision, so that there can be

no way to distinguish whether the atom passed through by way of the top
path or the bottom path. For example, the two paths must have the same
length: If the top path were longer, then an atom going through via the top
path would take more time, and hence there would be a way to tell which
way the atom passed through the analyzer loop.
In fact, the analyzer loop is constructed so precisely that it doesn’t
change the character of the atom passing through it. If the atom enters
3 We build it in our minds. The experiments described in this section have never been

performed exactly as described here, although researchers are getting close. [See Shi-
mon Machluf, Yonathan Japha, and Ron Folman, “Coherent Stern–Gerlach momentum
splitting on an atom chip” Nature Communications 4 (9 September 2013) 2424.] We
know the results that would come from these experiments because conceptually parallel
(but more complex!) experiments have been performed on photons, neutrons, atoms,
and molecules. (See page 33.)
4 If you followed the footnote on page 49, you will recall that these “pipes” manipulate

atoms through electromagnetic fields, not through touching. One way to make them
would be to insert two more Stern-Gerlach apparatuses, the first one upside-down and
the second one rightside-up relative to the initial apparatus. But whatever the manner of
their construction, the pipes must not change the magnetic moment of an atom passing
through them.
60 Interference

with µz = +µB , it exits with µz = +µB . If it enters with µx = −µB , it exits

with µx = −µB . If it enters with µ17◦ = −µB , it exits with µ17◦ = −µB .
It is hard to see why anyone would want to build such a device, because
they’re expensive (due to the precision demands), and they do absolutely
nothing!
Once you made one, however, you could convert it into something useful.
For example, you could insert a piece of metal blocking path a. In that case,
all the atoms exiting would have taken path b, so (if the analyzer loop were
oriented vertically) all would emerge with µz = −µB .
Using the analyzer loop, we set up the following apparatus: First, chan-
nel atoms with µz = +µB into a horizontal analyzer loop.5 Then, channel
the atoms emerging from that analyzer loop into a vertical analyzer. Ignore
atoms emerging from the + port of the vertical analyzer and look for atoms
emerging from the − port.

b
input ignore
µz = +µB output
a µz = −µB

We execute three experiments with this set-up: first we pass atoms

through when path a is blocked, then when path b is blocked, finally when
neither path is blocked.

2.2.1 Path a blocked

(1) Atoms enter the analyzer loop with µz = +µB .

(2) Half of them attempt path a, and end up impaled on the blockage.
(3) The other half take path b, and emerge from the analyzer loop with
µx = −µB .
(4) Those atoms then enter the vertical analyzer. Similar to the result
of section 2.1.6, half of these atoms emerge from the + port and are
ignored. Half of them emerge from the − port and are counted.
(5) The overall probability of passing through the set-up is 12 × 12 = 14 .

If you perform this experiment, you will find that this analysis is correct
and that these results are indeed obtained.
5 To make sure that all of these atoms have µ = +µ , they are harvested from the
z B
+ port of a vertical analyzer.
What Is Quantum Mechanics About? 61

2.2.2 Path b blocked

(1) Atoms enter the analyzer loop with µz = +µB .

(2) Half of them attempt path b, and end up impaled on the blockage.
(3) The other half take path a, and emerge from the analyzer loop with
µx = +µB .
(4) Those atoms then enter the vertical analyzer. Exactly as in sec-
tion 2.1.6, half of these atoms emerge from the + port and are ignored.
Half of them emerge from the − port and are counted.
(5) The overall probability of passing through the set-up is 12 × 12 = 14 .

Once again, experiment confirms these results.

2.2.3 Neither path blocked

Here, I have not just one, but two ways to analyze the experiment:
Analysis I:

(1) An atom passes through the set-up either via path b or via path a.
(2) From section 2.2.1, the probability of passing through via path b is 14 .
(3) From section 2.2.2, the probability of passing through via path a is 14 .
(4) Thus the probability of passing through the entire set-up is 14 + 14 = 21 .

Analysis II:

(1) Because “the analyzer loop is constructed so precisely that it doesn’t

change the character of the atom passing through it”, the atom emerges
from the analyzer loop with µz = +µB .
(2) When such atoms enter the vertical analyzer, all of them emerge
through the + port. (See section 2.1.2.)
(3) Thus the probability of passing through the entire set-up is zero.

These two analyses cannot both be correct. Experiment confirms the

result of analysis II, but what could possibly be wrong with analysis I?
Item (2) is already confirmed through the experiment of section 2.2.1,
item (3) is already confirmed through the experiment of section 2.2.2, and
don’t tell me that I made a mistake in the arithmetic of item (4). The only
thing left is item (1): “An atom passes through the set-up either via path b
62 Interference

or via path a.” This simple, appealing, common-sense statement must be

wrong !
Just a moment ago, the analyzer loop seemed like a waste of money and
skill. But in fact, a horizontal analyzer loop is an extremely clever way of
correlating the path through the analyzer loop with the value of µx : If the
atom has µx = +µB , then it takes path a. If the atom has µx = −µB , then
it takes path b. If the atom has µz = +µB , then it doesn’t have a value of
µx and hence it doesn’t take a path.
Notice again what I’m saying: I’m not saying the atom takes one path
or the other but we don’t know which. I’m not saying the atom breaks
into two pieces and each half traverses its own path. I’m saying the atom
doesn’t take a path. The µz = +µB atoms within the horizontal analyzer
loop do not have a position in the same sense that love does not have a
color. If you think of an atom as a smaller, harder version of a classical
marble, then you’re visualizing the atom incorrectly.
Once again, our experiments have uncovered a phenomenon that doesn’t
happen in daily life, so there is no word for it in conventional language.6
Sometimes people say that “the atom takes both paths”, but that phrase
does not really get to the heart of the new phenomenon. I have asked
students to invent a new word to represent this new phenomenon, and
my favorite of their many suggestions is “ambivate” — a combination of
ambulate and ambivalent — as in “an atom with µz = +µB ambivates
through both paths of a horizontal analyzer loop”. While this is a great
word, it hasn’t caught on. The conventional name for this phenomenon is
“quantal interference”.
The name “quantal interference” comes from a (far-fetched) analogy
with interference in wave optics. Recall that in the two-slit interference of
light, there are some observation points that have a light intensity if light
passes through slit a alone, and the same intensity if light passes through
slit b alone, but zero intensity if light passes through both slits. This is
called “destructive interference”. There are other observation points that
have a light intensity if the light passes through slit a alone, and the same
intensity if light passes through slit b alone, but four times that intensity if
6 In exactly the same way, there was no need for the word “latitude” or the word

“longitude” when it was thought that the Earth was flat. The discovery of the near-
spherical character of the Earth forced our forebears to invent new words to represent
these new concepts. Words do not determine reality; instead reality determines which
words are worth inventing.
What Is Quantum Mechanics About? 63

light passes through both slits. This is called “constructive interference”.

But in fact the word “interference” is a poor name for this phenomenon as
well. It’s adapted from a football term, and football players never (or at
least never intentionally) run “constructive interference”.
One last word about language: The device that I’ve called the “analyzer
loop” is more conventionally called an “interferometer”. I didn’t use that
name at first because that would have given away the ending.
Back on page 47 I said that, to avoid unnecessary distraction, this chap-
ter would “to the extent possible, do a quantum-mechanical treatment of
an atom’s magnetic moment while maintaining a classical treatment of all
other aspects — such as its energy and momentum and position”. You
can see now why I put in that qualifier “to the extent possible”: we have
found that within an interferometer, a quantum-mechanical treatment of
magnetic moment demands a quantum-mechanical treatment of position as
well.

2.2.4 Sample Problem: Constructive interference

Consider the same set-up as on page 60, but now ignore atoms leaving the
− port of the vertical analyzer and consider as output atoms leaving the
+ port. What is the probability of passing through the set-up when path
a is blocked? When path b is blocked? When neither path is blocked?

1 1 1
Solution: 4; 4; 1. Because 4 + 14 < 1, this is an example of constructive
interference.
 64 Interference

2.2.5 Sample Problem: Two analyzer loops

2a
1b
input output
µz = +µB
1a
2b

Atoms with µz = +µB are channeled through a horizontal analyzer loop

(number 1), then a vertical analyzer loop (number 2). If all paths are open,
100% of the incoming atoms exit from the output. What percentage of the
incoming atoms leave from the output if the following paths are blocked?

(a) 2a (d) 1b
(b) 2b (e) 1b and 2a
(c) 1a (f) 1a and 2b

Solution: Only two principles are needed to solve this problem: First,
an atom leaving an unblocked analyzer loop leaves in the same condition
it had when it entered. Second, an atom leaving an analyzer loop with
one path blocked leaves in the condition specified by the path that it took,
regardless of the condition it had when it entered. Use of these principles
gives the solution in the table on the next page. Notice that in changing
from situation (a) to situation (e), you add blockage, yet you increase the
output!
 paths input path taken intermediate path taken output probability of
blocked condition through # 1 condition through # 2 condition input → output
none µz = +µB “both” µz = +µB a µz = +µB 100%
2a µz = +µB “both” µz = +µB 100% blocked at a none 0%
What Is Quantum Mechanics About?

2b µz = +µB “both” µz = +µB a µz = +µB 100%

50% blocked at a
1a µz = +µB µx = −µB “both” µx = −µB 50%
50% pass through b
50% pass through a
1b µz = +µB µx = +µB “both” µx = +µB 50%
50% blocked at b
50% pass through a 25% blocked at a
1b and 2a µz = +µB µx = +µB µz = −µB 25%
50% blocked at b 25% pass through b
50% blocked at a 25% pass through a
1a and 2b µz = +µB µx = −µB µz = +µB 25%
50% pass through b 25% blocked at b
65
66 Interference

Problems
2.4 Tilted analyzer loop (recommended problem)

z
θ
a

input
µz =+µB output

An atom with µz = +µB enters the analyzer loop (interferometer)

shown above, tilted at angle θ to the vertical. The outgoing atom
enters a z-analyzer, and whatever comes out the − port is considered
output. What is the probability for passage from input to output when:
a. Paths a and b are both open?
b. Path b is blocked?
c. Path a is blocked?

2.5 Find the flaw: Tilted analyzer loop7

Five students — Aldo, Beth, Celine, Denzel, and Ellen — work the
above problem. All find the same answer for part (a), namely zero,
but for parts (b) and (c) they produce five different answers! Their
candidate answers are:
(b) (c)
Aldo cos4 (θ/2) sin4 (θ/2)
1 2 1 2
Beth 4 sin (θ) 4 sin (θ)
1 1
Celine 4 sin(θ) 4 sin(θ)
1
√ 1
√
Denzel 4 2 sin(θ/2) 4 2 sin(θ/2)
2
Ellen 1
2 sin (θ) 1
2 sin2 (θ)
Provide simple reasons showing that four of these candidates must be
wrong.

7 Background concerning “find the flaw” type problems is provided in sample prob-

lem 1.2.1 on page 26.

 2.3. Aharonov-Bohm effect 67

2.6 Three analyzer loops (recommended problem)

Atoms with µz = +µB are channeled into a horizontal analyzer loop,
followed by a vertical analyzer loop, followed by a horizontal analyzer
loop.

2a
1b 3b
µz =+µB output

1a 3a
2b

If all paths are open, 100% of the incoming atoms exit from the out-
put. What percent of the incoming atoms leave from the output if the
following paths are blocked?

(a) 3a (d) 2b (g) 1b and 3b

(b) 3b (e) 1b (h) 1b and 3a
(c) 2a (f) 2a and 3b (i) 1b and 3a and 2a
(Note that in going from situation (h) to situation (i) you get more
output from increased blockage.)

2.3 Aharonov-Bohm effect

We have seen how to sort atoms using a Stern-Gerlach analyzer, made

of a non-uniform magnetic field. But how do atoms behave in a uniform
magnetic field? In general, this is an elaborate question, and the answer
will depend on the initial condition of the atom’s magnetic moment, on the
magnitude of the field, and on the amount of time that the atom spends in
the field. But for one special case the answer, determined experimentally,
is easy. If an atom is exposed to uniform magnetic field B~ for exactly the
right amount of time [which turns out to be a time of h/(2µB B)], then the
atom emerges with exactly the same magnetic condition it had initially:
If it starts with µz = −µB , it ends with µz = −µB . If it starts with
µx = +µB , it ends with µx = +µB . If it starts with µ29◦ = +µB , it ends
with µ29◦ = +µB . Thus for atoms moving at a given speed, we can build a
box containing a uniform magnetic field with just the right length so that
any atom passing through it will spend just the right amount of time to
68 Aharonov-Bohm effect

emerge in the same condition it had when it entered. We call this box a
“replicator”.
If you play with one of these boxes you’ll find that you can build any
elaborate set-up of sources, detectors, blockages, and analyzers, and that
inserting a replicator into any path will not affect the outcome of any exper-
iment. But notice that this apparatus list does not include interferometers
(our “analyzer loops”)! Build the interference experiment of page 60. Do
not block either path. Instead, slip a replicator into one of the two paths a
or b — it doesn’t matter which.

b
µz =+µB ignore

output
a µz =−µB
replicator

Without the replicator no atom emerges at output. But experiment shows

that after inserting the replicator, all the atoms emerge at output.
How can this be? Didn’t we just say of a replicator that “any atom pass-
ing through it will. . . emerge in the same condition it had when it entered”?
Indeed we did, and indeed this is true. But an atom with µz = +µB doesn’t
pass through path a or path b — it ambivates through both paths.
If the atom did take one path or the other, then the replicator would
have no effect on the experimental results. The fact that it does have an
effect is proof that the atom doesn’t take one path or the other.
The fact8 that one can perform this remarkable experiment was pre-
dicted theoretically (in a different context) by Walter Franz. He announced
his result in Danzig (now Gdańsk, Poland) in May 1939, just months before
the Nazi invasion of Poland, and his prediction was largely forgotten in the
resulting chaos. The effect was rediscovered theoretically by Werner Ehren-
berg and Raymond Siday in 1949, but they published their result under the
opaque title of “The refractive index in electron optics and the principles of
dynamics” and their prediction was also largely forgotten. The effect was
rediscovered theoretically a third time by Yakir Aharonov and David Bohm
in 1959, and this time it sparked enormous interest, both experimental and
theoretical. The phenomenon is called today the “Aharonov-Bohm effect”.
8 See B.J. Hiley, “The early history of the Aharonov-Bohm effect” (17 April 2013)

https://arxiv.org/abs/1304.4736.
What Is Quantum Mechanics About? 69

Problem
2.7 Bomb-testing interferometer9 (recommended problem)
The Acme Bomb Company sells a bomb triggered by the presence of
silver, and claims that the trigger is so sensitive that the bomb explodes
when its trigger absorbs even a single silver atom. You have heard sim-
ilar extravagant claims from other manufacturers, so you’re suspicious.
You purchase a dozen bombs, then shoot individual silver atoms at
each in turn. The first bomb tested explodes! The trigger worked as
advertised, but now it’s useless because it’s blasted to pieces. The sec-
ond bomb tested doesn’t explode — the atom slips through a hole in
the trigger. This confirms your suspicion that not all the triggers are
as sensitive as claimed, so this bomb is useless to you as well. If you
continue testing in this fashion, at the end all your good bombs will be
blown up and you will be left with a stash of bad bombs.
So instead, you set up the test apparatus sketched here:
b
µz =+µB ?

a
?
bomb with trigger

An atom with µz = +µB enters the interferometer. If the bomb trigger

has a hole, then the atom ambivates through both paths, arrives at the
analyzer with µz = +µB , and exits the + port of the analyzer. But if
the bomb trigger is good, then either (a) the atom takes path a and
sets off the bomb, or else (b) the atom takes path b.
a. If the bomb trigger is good, what is the probability of option (a)?
Of option (b)?
b. If option (b) happens, what kind of atom arrives at the analyzer?
What is the probability of that atom exiting through the + port?
The − port?
Conclusion: If the atom exits through the − port, then the bomb is
good. If it exits through the + port then the bomb might be good or
bad and further testing is required. But you can determine that the
bomb trigger is good without blowing it up!
9 Avshalom C. Elitzur and Lev Vaidman, “Quantum mechanical interaction-free mea-

surements” Foundations of Physics 23 (July 1993) 987–997.

70 Light on the atoms

2.4 Light on the atoms

Our conclusion that, under some circumstances, the atom “does not have
a position” is so dramatically counterintuitive that you might — no, you
should — be tempted to test it experimentally. Set up the interference ex-
periment on page 60, but instead of simply allowing atoms to pass through
the interferometer, watch to see which path the atom takes through the
set-up. To watch them, you need light. So set up the apparatus with lamps
trained on the two paths a and b.
Send in one atom. There’s a flash of light at path a.
Another atom. Flash of light at b.
Another atom. Flash at b again.
Then a, then a, then b.
You get the drift. Always the light appears at one path or the other. (In
fact, the flashes come at random with probability 21 for a flash at a and 12
for a flash at b.) Never is there no flash. Never are there “two half flashes”.
The atom always has a position when passing through the interferometer.
“So much”, say the skeptics, “for this metaphysical nonsense about ‘the
atom takes both paths’.”
But wait. Go back and look at the output of the vertical analyzer.
When we ran the experiment with no light, the probability of coming out
the − port was 0. When we turn the lamps on, then the probability of
coming out the − port becomes 21 .
When the lamps are off, analysis II on page 61 is correct: the atoms
ambivate through both paths, and the probability of exiting from the − port
is 0. When the lamps are on and a flash is seen at path a, then the atom
does take path a, and now the analysis of section 2.2.2 on page 61 is correct:
the probability of exiting from the − port is 21 .
The process when the lamps are on is called “observation” or “measure-
ment”, and a lot of nonsense has come from the use of these two words.
The important thing is whether the light is present or absent. Whether
or not the flashes are “observed” by a person is irrelevant. To prove this
to yourself, you may, instead of observing the flashes in person, record the
flashes on video. If the lamps are on, the probability of exiting from the
− port is 12 . If the lamps are off, the probability of exiting from the − port
What Is Quantum Mechanics About? 71

is 0. Now, after the experiment is performed, you may either destroy the
video, or play it back to a human audience, or play it back to a feline au-
dience. Surely, by this point it is too late to change the results at the exit
port.
It’s not just light. Any method you can dream up for determining the
path taken will show that the atom takes just one path, but that method
will also change the output probability from 0 to 21 . No person needs to
actually read the results of this mechanism: as long as the mechanism is at
work, as long as it is in principle possible to determine which path is taken,
then one path is taken and no interference happens.
What happens if you train a lamp on path a but leave path b in the
dark? In this case a flash means the atom has taken path a. No flash means
the atom has taken path b. In both cases the probability of passage for the
atom is 12 .
How can the atom taking path b “know” that the lamp at path a is
turned on? The atom initially “sniffs out” both paths, like a fog creeping
down two passageways. The atom that eventually does take path b in
the dark started out attempting both paths, and that’s how it “knows”
the lamp at path a is on. This is called the “Renninger negative-result
experiment”.
It is not surprising that the presence or absence of light should affect an
atom’s motion: this happens even in classical mechanics. When an object
absorbs or reflects light, that object experiences a force, so its motion is
altered. For example, a baseball tossed upward in a gymnasium with the
overhead lamps off attains a slightly greater height that an identical baseball
experiencing an identical toss in the same gymnasium with the overhead
lamps on, because the downward-directed light beams push the baseball
downward. (This is the same “radiation pressure” that is responsible for
the tails of comets. And of course, the effect occurs whenever the lamps are
turned on: whether any person actually watches the illuminated baseball
is irrelevant.) This effect is negligible for typical human-scale baseballs
and tosses and lamps, but atoms are far smaller than baseballs and it is
reasonable that the light should alter the motion of an atom more than it
alters the motion of a baseball.
One last experiment: Look for the atoms with dim light. In this case,
some of the atoms will pass through with a flash. But — because of the
dimness — some atoms will pass through without any flash at all. For those
72 Entanglement

atoms passing through with a flash, the probability for exiting the − port
is 21 . For those atoms passing through without a flash, the probability of
exiting the − port is 0.

2.5 Entanglement

I have claimed that an atom with µz = +µB doesn’t have a value of µx ,

and that when such an atom passes through a horizontal interferometer, it
doesn’t have a position. You might say to yourself, “These claims are so
weird, so far from common sense, that I just can’t accept them. I believe
the atom does have a value of µx and does have a position, but something
else very complicated is going on to make the atom appear to lack a µx and
a position. I don’t know what that complicated thing is, but just because
I haven’t yet thought it up yet doesn’t mean that it doesn’t exist.”
If you think this, you’re in good company: Einstein thought it too. This
section introduces a new phenomenon of quantum mechanics, and shows
that no local deterministic mechanism, no matter how complex or how
fantastic, can give rise to all the results of quantum mechanics. Einstein
was wrong.

2.5.1 Flipping Stern-Gerlach analyzer

A new piece of apparatus helps us uncover this new phenomenon of nature.

Mount a Stern-Gerlach analyzer on a stand so that it can be oriented either
vertically (0◦ ), or tilted one-third of a circle clockwise (+120◦ ), or tilted
one-third of a circle counterclockwise (−120◦ ). Call these three orientations
V (for vertical), O (for out of the page), or I (for into the page). As an atom
approaches the analyzer, select one of these three orientations at random,
flip the analyzer to that orientation, and allow the atom to pass through as
usual. As a new atom approaches, again select an orientation at random,
flip the analyzer, and let the atom pass through. Repeat many times.
What Is Quantum Mechanics About? 73

V V

120◦

I O I O

Flipping Stern-Gerlach analyzer. The arrows V, O, and I, oriented 120◦

apart, all lie within the plane perpendicular to the atom’s approach path.
What happens if an atom with µz = +µB enters a flipping analyzer?
With probability 13 , the atom enters a vertical analyzer (orientation V), and
in that case it exits the + port with probability 1. With probability 31 , the
atom enters an out-of-the-page analyzer (orientation O), and in that case
(see equation 2.1) it exits the + port with probability
cos2 (120◦ /2) = 14 .
With probability 13 , the atom enters an into-the-page analyzer (orientation
I), and in that case it exits the + port with probability 14 . Thus the overall
probability of this atom exiting through the + port is
1 1 1 1 1
3 ×1+ 3 × 4 + 3 × 4 = 12 . (2.2)
A similar analysis shows that if an atom with µz = −µB enters the flipping
analyzer, it exits the + port with probability 12 .
You could repeat the analysis for an atom entering with µ(+120◦ ) = +µB ,
but you don’t need to. Because the three orientations are exactly one-third
of a circle apart, rotational symmetry demands that an atom entering with
µ(+120◦ ) = +µB behaves exactly as an atom entering with µz = +µB .
In conclusion, an atom entering in any of the six conditions µz = +µB ,
µz = −µB , µ(+120◦ ) = +µB , µ(+120◦ ) = −µB , µ(−120◦ ) = +µB , or
µ(−120◦ ) = −µB will exit through the + port with probability 12 .
74 Entanglement

2.5.2 EPR source of atom pairs

Up to now, our atoms have come from an oven. For the next experiments we
need a special source10 that expels two atoms at once, one moving to the left
and the other to the right. For the time being we call this an “EPR” source,
which produces an atomic pair in an “EPR” condition. The letters come
from the names of those who discovered this condition: Albert Einstein,
Boris Podolsky, and Nathan Rosen. After investigating this condition we
will develop a more descriptive name.
The following four experiments investigate the EPR condition:
(1) Each atom encounters a vertical Stern-Gerlach analyzer. The ex-
perimental result: the two atoms exit through opposite ports. To be precise:
with probability 21 , the left atom exits + and the right atom exits −, and
with probability 12 , the left atom exits − and the right atom exits +, but
it never happens that both atoms exit + or that both atoms exit −.

1
probability 2

1
probability 2

never

never

10 The question of how to build this special source need not concern us at the moment: it

is an experimental fact that such sources do exist. One way to make one would start with
a diatomic molecule with zero magnetic moment. Cause the molecule to disintegrate and
eject the two daughter atoms in opposite directions. Because the initial molecule had
zero magnetic moment, the pair of daughter atoms will have the properties of magnetic
moment described. In fact, it’s easier to build a source, not for a pair of atoms, but for
a pair of photons using a process called spontaneous parametric down-conversion.
What Is Quantum Mechanics About? 75

You might suppose that this is because for half the pairs, the left atom
is generated with µz = +µB while the right atom is generated with
µz = −µB , while for the other half of the pairs, the left atom is generated
with µz = −µB while the right atom is generated with µz = +µB . This
supposition seems suspicious, because it singles out the z axis as special,
but at this stage in our experimentation it’s possible.

(2) Repeat the above experiment with horizontal Stern-Gerlach analyz-

ers. The experimental result: Exactly the same as in experiment (1)! The
two atoms always exit through opposite ports.

Problem 2.9 on page 83 demonstrates that the results of this experiment

rule out the supposition presented at the bottom of experiment (1).

(3) Repeat the above experiment with the two Stern-Gerlach analyzers
oriented at +120◦ , or with both oriented at −120◦ , or with both oriented
at 57◦ , or for any other angle, as long as both have the same orientation.
The experimental result: Exactly the same for any orientation!
(4) In an attempt to trick the atoms, we set the analyzers to vertical,
then launch the pair of atoms, then (while the atoms are in flight) switch
both analyzers to, say, 42◦ , and have the atoms encounter these analyzers
both with switched orientation. The experimental result: Regardless of
what the orientation is, and regardless of when that orientation is set, the
two atoms always exit through opposite ports.
Here is one way to picture this situation: The pair of atoms has a total
magnetic moment of zero. But whenever the projection of a single atom
on any axis is measured, the result must be +µB or −µB , never zero.
The only way to insure that that total magnetic moment, projected on
any axis, sums to zero is the way described above. Do not put too much
weight on this picture: like the “wants to go straight” story of section 2.1.4
(page 51), this is a classical story that happens to give the correct result.
The definitive answer to any question is always experiment, not any picture
or story, however appealing it may be.
These four experiments show that it is impossible to describe the con-
dition of the atoms through anything like “the left atom has µz = +µB ,
the right atom has µz = −µB ”. How can we describe the condition of the
pair? This will require further experimentation. For now, we say it has an
EPR condition.
76 Entanglement

2.5.3 EPR atom pair encounters flipping Stern-Gerlach

analyzers

A pair of atoms leaves the EPR source, and each atom travels at the same
speed to vertical analyzers located 100 meters away. The left atom exits the
− port, the right atom exits the + port. When the pair is flying from source
to analyzer, it’s not correct to describe it as “the left atom has µz = −µB ,
the right atom has µz = +µB ”, but after the atoms leave their analyzers,
then this is a correct description.
Now shift the left analyzer one meter closer to the source. The left atom
encounters its analyzer before the right atom encounters its. Suppose the
left atom exits the − port, while the right atom is still in flight toward its
analyzer. We know that when the right atom eventually does encounter
its vertical analyzer, it will exit the + port. Thus it is correct to describe
the right atom as having “µz = +µB ”, even though that atom hasn’t yet
encountered its analyzer.
Replace the right vertical analyzer with a flipping Stern-Gerlach ana-
lyzer. (In the figure below, it is in orientation O, out of the page.) Suppose
the left atom encounters its vertical analyzer and exits the − port. Through
the reasoning of the previous paragraph, the right atom now has µz = +µB .
We know that when such an atom encounters a flipping Stern-Gerlach an-
alyzer, it exits the + port with probability 21 .

Similarly, if the left atom encounters its vertical analyzer and exits the
+ port, the right atom now has µz = −µB , and once it arrives at its flipping
analyzer, it will exit the − port with probability 21 . Summarizing these two
paragraphs: Regardless of which port the left atom exits, the right atom
will exit the opposite port with probability 12 .
Now suppose that the left analyzer were not vertical, but instead in
orientation I, tilted into the page by one-third of a circle. It’s easy to see
that, again, regardless of which port the left atom exits, the right atom will
exit the opposite port with probability 21 .
What Is Quantum Mechanics About? 77

Finally, suppose that the left analyzer is a flipping analyzer. Once again,
the two atoms will exit from opposite ports with probability 21 .
The above analysis supposed that the left analyzer was one meter closer
to the source than the right analyzer, but clearly it also works if the right
analyzer is one meter closer to the source than the left analyzer. Or one
centimeter. One suspects that the same result will hold even if the two
analyzers are exactly equidistant from the source, and experiment bears
out this suspicion.
In summary: Each atom from this EPR source enters a flipping Stern-
Gerlach analyzer.

(A) The atoms exit from opposite ports with probability 12 .

(B) If the two analyzers happen to have the same orientation, the atoms
exit from opposite ports.

This is the prediction of quantum mechanics, and experiment confirms this

prediction.

2.5.4 The prediction of local determinism

Suppose you didn’t know anything about quantum mechanics, and you
were told the result that “if the two analyzers have the same orientation,
the atoms exit from opposite ports.” Could you explain it?
I am sure you could. In fact, there are two possible explanations: First,
the communication explanation. The left atom enters its vertical analyzer,
and notices that it’s being pulled toward the + port. It calls up the right
atom with its walkie-talkie and says “If your analyzer has orientation I or O
then you might go either way, but if your analyzer has orientation V you’ve
got to go to the − port!” This is a possible explanation, but it’s not a local
explanation. The two analyzers might be 200 meters apart, or they might
be 200 light-years apart. In either case, the message would have to get from
the left analyzer to the right analyzer instantaneously. The walkie-talkies
would have to use not radio waves, which propagate at the speed of light,
but some sort of not-yet-discovered “insta-rays”. Physicists have always
been skeptical of non-local explanations, and since the advent of relativity
they have grown even more skeptical, so we set this explanation aside. Can
you find a local explanation?
78 Entanglement

Again, I am sure you can. Suppose that when the atoms are launched,
they have some sort of characteristic that specifies which exit port they will
take when they arrive at their analyzer. This very reasonable supposition,
called “determinism”, pervades all of classical mechanics. It is similar to
saying “If I stand atop a 131 meter cliff and toss a ball horizontally with
speed 23.3 m/s, I can predict the angle with which the ball strikes the
ground, even though that event will happen far away and long in the fu-
ture.” In the case of the ball, the resulting strike angle is encoded into the
initial position and velocity. In the case of the atoms, it’s not clear how the
exit port will be encoded: perhaps through the orientation of its magnetic
moment, perhaps in some other, more elaborate way. But the method of
encoding is irrelevant: if local determinism holds, then something within
the atom determines which exit port it will take when it reaches its ana-
lyzer.11 I’ll represent this “something” through a code like (+ + −). The
first symbol means that if the atom encounters an analyzer in orientation V,
it will exit through the + port. The second means that if it encounters an
analyzer in orientation O, it will exit through the + port. The third means
that if it encounters an analyzer in orientation I, it will exit through the
− port. The only way to ensure that “if the two analyzers have the same
orientation, the atoms exit from opposite ports” is to assume that when the
two atoms separate from each other within the source, they have opposite
codes. If the left atom has (+ − +), the right atom must have (− + −). If
the left atom has (− − −), the right atom must have (+ + +). This is the
local deterministic scheme for explaining fact (B) that “if the two analyzers
have the same orientation, the atoms exit from opposite ports”.
But can this scheme explain fact (A)? Let’s investigate. Consider first
the case mentioned above: the left atom has (+−+) and the right atom has
(− + −). These atoms will encounter analyzers set to any of 32 = 9 possible
pairs of orientations. We list them below, along with with exit ports taken
by the atoms. (For example, the third line of the table considers a left
analyzer in orientation V and a right analyzer in orientation I. The left
atom has code (+ − +), and the first entry in that code determines that
the left atom will exit from the V analyzer through the + port. The right
atom has code (− + −), and the third entry in that code determines that
the right atom will exit from the I analyzer through the − port.)
11 But remember that in quantum mechanics determinism does not hold. The infor-

mation can’t be encoded within the three projections of a classical magnetic moment
vector, because at any one instant, the quantum magnetic moment vector has only one
projection.
What Is Quantum Mechanics About? 79

left left right right opposite?

port analyzer analyzer port
+ V V − yes
+ V O + no
+ V I − yes
− O V − no
− O O + yes
− O I − no
+ I V − yes
+ I O + no
+ I I − yes

Each of the nine orientation pairs (VV, OI, etc.) are equally likely, five of
the orientation pairs result in atoms exiting from opposite ports, so when
atoms of this type emerge from the source, the probability of these atoms
exiting from opposite ports is 59 .
What about a pair of atoms generated with different codes? Suppose the
left atom has (− − +) so the right atom must have (+ + −). If you perform
the analysis again, you will find that the probability of atoms exiting from
opposite ports is once again 95 .
Suppose the left atom has (−−−), so the right atom must have (+++).
The probability of the atoms exiting from opposite ports is of course 1.
There are, in fact, just 23 = 8 possible codes:

code probability
for of exiting
left atom opposite
+++ 1
−++ 5/9
+−+ 5/9
++− 5/9
+−− 5/9
−+− 5/9
−−+ 5/9
−−− 1
80 Entanglement

If the source makes left atoms of only type (−−+), then the probability
of atoms exiting from opposite ports is 59 . If the source makes left atoms
of only type (+ + +), then the probability of atoms exiting from opposite
ports is 1. If the source makes left atoms of type (− − +) half the time,
and of type (+ + +) half the time, then the probability of atoms exiting
from opposite ports is halfway between 95 and 1, namely 79 . But no matter
how the source makes atoms, the probability of atoms exiting from opposite
ports must be somewhere between 59 and 1.
But experiment and quantum mechanics agree: That probability is ac-
tually 12 — and 12 is not between 95 and 1. No local deterministic scheme
— no matter how clever, or how elaborate, or how baroque — can give the
result 12 . There is no “something within the atom that determines which
exit port it will take when it reaches its analyzer”. If the magnetic moment
has a projection on axis V, then it doesn’t have a projection on axis O or
axis I.
There is a reason that Einstein, despite his many attempts, never pro-
duced a scheme that explained quantum mechanics in terms of some more
fundamental, local and deterministic mechanism. It is not that Einstein
wasn’t clever. It is that no such scheme exists.

2.5.5 The upshot

This is a new phenomenon — one totally absent from classical physics — so

it deserves a new name, something more descriptive than “EPR”. Einstein
called it “spooky action at a distance”.12 The phenomenon is spooky all
right, but this phrase misses the central point that the phenomenon involves
“correlations at a distance”, whereas the word “action” suggests “cause-
and-effect at a distance”. Schrödinger coined the term “entanglement” for
this phenomenon and said it was “not. . . one but rather the characteristic
trait of quantum mechanics, the one that enforces its entire departure from
classical lines of thought”.13 The world has followed Schrödinger and the
phenomenon is today called entanglement. We will later investigate en-
tanglement in more detail, but for now we will just call our EPR source a
12 Letter from Einstein to Max Born, 3 March 1947, The Born-Einstein Letters (Macmil-

lan, New York, 1971) translated by Irene Born.

13 Erwin Schrödinger, “Discussion of probability relations between separated systems”

Mathematical Proceedings of the Cambridge Philosophical Society 31 (October 1935)

555–563.
What Is Quantum Mechanics About? 81

“source of entangled atom pairs” and describe the condition of the atom
pair as “entangled”.
The failure of local determinism described above is a special case of
“Bell’s Theorem”, developed by John Bell14 in 1964. The theorem has
by now been tested experimentally numerous times in numerous contexts
(various different angles; various distances between the analyzers; various
sources of entangled pairs; various kinds of particles flying apart — gamma
rays, or optical photons, or ions). In every test, quantum mechanics has
been shown correct and local determinism wrong. What do we gain from
these results?
First, they show that nature does not obey local determinism. To our
minds, local determinism is common sense and any departure from it is
weird. Thus whatever theory of quantum mechanics we eventually develop
will be, to our eyes, weird. This will be a strength, not a defect, in the
theory. The weirdness lies in nature, not in the theory used to describe
nature.
Each of us feels a strong psychological tendency to reject the unfamil-
iar. In 1633, the Holy Office of the Inquisition found Galileo Galilei’s idea
that the Earth orbited the Sun so unfamiliar that they rejected it. The
inquisitors put Galileo on trial and forced him to abjure his position. From
the point of view of nature, the trial was irrelevant, Galileo’s abjuration
was irrelevant: the Earth orbits the Sun whether the Holy Office finds that
fact comforting or not. It is our job as scientists to change our minds to fit
nature; we do not change nature to fit our preconceptions. Don’t make the
inquisitors’ mistake.
Second, the Bell’s theorem result guides not just our calculations about
nature but also our visualizations of nature, and even the very idea of
what it means to “understand” nature. Lord Kelvin15 framed the situation
perfectly in his 1884 Baltimore lectures: “I never satisfy myself until I can
14 John Stewart Bell (1928–1990), a Northern Irish physicist, worked principally in accel-

erator design, and his investigation of the foundations of quantum mechanics was some-
thing of a hobby. Concerning tests of his theorem, he remarked that “The reasonable
thing just doesn’t work.” [Jeremy Bernstein, Quantum Profiles (Princeton University
Press, Princeton, NJ, 1991) page 84.]
15 William Thomson, the first Baron Kelvin (1824–1907), was an Irish mathematical

physicist and engineer who worked in Scotland. He is best known today for establishing
the thermodynamic temperature scale that bears his name, but he also made fundamen-
tal contributions to electromagnetism. He was knighted for his engineering work on the
first transatlantic telegraph cable.
82 Entanglement

make a mechanical model of a thing. If I can make a mechanical model

I can understand it. As long as I cannot make a mechanical model all
the way through I cannot understand, and this is why I cannot get the
electromagnetic theory.”16 If we take this as our meaning of “understand”,
then the experimental tests of Bell’s theorem assure us that we will never be
able to understand quantum mechanics.17 What is to be done about this?
There are only two choices. Either we can give up on understanding, or we
can develop a new and more appropriate meaning for “understanding”.
Max Born18 argued for the first choice: “The ultimate origin of the
difficulty lies in the fact (or philosophical principle) that we are compelled to
use the words of common language when we wish to describe a phenomenon,
not by logical or mathematical analysis, but by a picture appealing to the
imagination. Common language has grown by everyday experience and can
never surpass these limits.”19 Born felt that it was impossible to visualize
or “understand” quantum mechanics: all you could do was grind through
the “mathematical analysis”.
Humans are visual animals, however, and I have found that when we are
told not to visualize, we do so anyway. But we do so in an illicit and uncrit-
ical way. For example, many people visualize an atom passing through an
interferometer as a small, hard, marble, with a definite position, despite the
already-discovered fact that this visualization is untenable. Many people
visualize a photon as a “ball of light” despite the fact that a photon (as
conventionally defined) has a definite energy and hence can never have a
position.
It is possible to develop a visualization and understanding of quantum
mechanics. This can’t be done by building a “mechanical model all the
way through”. It must be done through both analogy and contrast: atoms
16 William Thomson, “Baltimore lectures on wave theory and molecular dynamics,” in

Robert Kargon and Peter Achinstein, editors, Kelvin’s Baltimore Lectures and Modern
Theoretical Physics (MIT Press, Cambridge, MA, 1987) page 206.
17 The first time I studied quantum mechanics seriously, I wrote in the margin of my

textbook “Good God they do it! But how?” I see now that I was looking for a mechanical
mechanism undergirding quantum mechanics. It doesn’t exist, but it’s very natural for
anyone to want it to exist.
18 Max Born (1882–1970) was a German-Jewish theoretical physicist with a particular in-

terest in optics. At the University of Göttingen in 1925 he directed Heisenberg’s research

which resulted in the first formulation of quantum mechanics. His granddaughter, the
British-born Australian actress and singer Olivia Newton-John, is famous for her 1981
hit song “Physical”.
19 Max Born, Atomic Physics, sixth edition (Hafner Press, New York, 1957) page 97.
 What Is Quantum Mechanics About? 83

behave in some ways like small hard marbles, in some ways like classical
waves, and in some ways like a cloud or fog of probability. Atoms don’t
behave exactly like any of these things, but if you keep in mind both the
analogy and its limitations, then you can develop a pretty good visualization
and understanding.
And that brings us back to the name “entanglement”. It’s an important
name for an important phenomenon, but it suggests that the two distant
atoms are connected mechanically, through strings. They aren’t. The two
atoms are correlated — if the left comes out +, the right comes out −, and
vice versa — but they aren’t correlated because of some signal sent back
and forth through either strings or walkie-talkies. Entanglement involves
correlation without causality.

Problems
2.8 An atom walks into an analyzer
Execute the “similar analysis” mentioned in the sentence below equa-
tion (2.2).
2.9 A supposition squashed (essential problem)
If atoms were generated according to the supposition presented below
experiment (1) on page 74, then would would happen when they en-
countered the two horizontal analyzers of experiment (2)?
2.10 A probability found through local determinism
Suppose that the codes postulated on page 78 did exist. Suppose also
that a given source produces the various possible codes with these prob-
abilities:
code probability
for of making
left atom such a pair
+++ 1/2
++− 1/4
+−− 1/8
−−+ 1/8
If this given source were used in the experiment of section 2.5.3 with
distant flipping Stern-Gerlach analyzers, what would be the probability
of the two atoms exiting from opposite ports?
 84 Quantum cryptography

2.11 A probability found through quantum mechanics

In the test of Bell’s inequality (the experiment of section 2.5.3), what
is the probability given by quantum mechanics that, if the orientation
settings are different, the two atoms exit from opposite ports?

2.6 Quantum cryptography

We’ve seen a lot of new phenomena, and the rest of this book is devoted
to filling out our understanding of these phenomena and applying that
understanding to various circumstances. But first, can we use them for
anything?
We can. The sending of coded messages used to be the province of
armies and spies and giant corporations, but today everyone does it. All
transactions through automatic teller machines are coded. All Internet
commerce is coded. This section describes a particular, highly reliable
encoding scheme and then shows how quantal entanglement may someday
be used to implement this scheme. (Quantum cryptography was used to
securely transmit voting ballots cast in the Geneva canton of Switzerland
during parliamentary elections held 21 October 2007. But it is not today
in regular use anywhere.)
In this section I use names conventional in the field of coded messages
(called cryptography). Alice and Bob wish to exchange private messages,
but they know that Eve is eavesdropping on their communication. How
can they encode their messages to maintain their privacy?

2.6.1 The Vernam cipher

The Vernam cipher or “one-time pad” technique is the only coding scheme
proven to be absolutely unbreakable (if used correctly). It does not rely on
the use of computers — it was invented by Gilbert Vernam in 1919 — but
today it is mostly implemented using computers, so I’ll describe it in that
context.
Data are stored on computer disks through a series of magnetic patches
on the disk that are magnetized either “up” or “down”. An “up” patch
is taken to represent 1, and a “down” patch is taken to represent 0. A
string of seven patches is used to represent a character. For example, by a
What Is Quantum Mechanics About? 85

convention called ASCII, the letter “a” is represented through the sequence
1100001 (or, in terms of magnetizations, up, up, down, down, down, down,
up). The letter “W” is represented through the sequence 1010111. Any
computer the world around will represent the message “What?” through
the sequence

1010111 1101000 1100001 1110100 0111111

This sequence is called the “plaintext”.

But Alice doesn’t want a message recognizable by any computer the
world around. She wants to send the message “What?” to Bob in such a
way that Eve will not be able to read the message, even though Eve has
eavesdropped on the message. Here is the scheme invented by Vernam:
Before sending her message, Alice generates a string of random 0s and 1s
just as long as the message she wants to send — in this case, 7 × 5 = 35
bits. She might do this by flipping 35 coins, or by flipping one coin 35
times. I’ve just done that, producing the random number

0100110 0110011 1010110 1001100 1011100

Then Alice gives Bob a copy of that random number – the “key”.
Instead of sending the plaintext, Alice modifies her plaintext into a
coded “ciphertext” using the key. She writes down her plaintext and writes
the key below it, then works through column by column. For each position,
if the key is 0 the plaintext is left unchanged; but if the key is 1 the plaintext
is reversed (from 0 to 1 or vice versa). For the first column, the key is 0, so
Alice doesn’t change the plaintext: the first character of ciphertext is the
same as the first character of plaintext. For the second column, the key is
1, so Alice does change the plaintext: the second character of ciphertext
is the reverse of the second character of plaintext. Alice goes through all
the columns, duplicating the plaintext where the key is 0 and reversing the
plaintext where the key is 1.

plaintext: 1010111 1101000 1100001 1110100 0111111

key: 0100110 0110011 1010110 1001100 1011100
ciphertext: 1110001 1011011 0110111 0111000 1100011

Then, Alice sends out her ciphertext over open communication lines.
86 Quantum cryptography

Now, the ciphertext that Bob (and Eve) receive translates to some mes-
sage through the ASCII convention – in fact, it translates to “q[78c” — but
because the key is random, the ciphertext is just as random. Bob deciphers
Alice’s message by carrying out the encoding process on the ciphertext,
namely, duplicating the ciphertext where the key is 0 and reversing the
ciphertext where the key is 1. The result is the plaintext. Eve does not
know the key, so she cannot produce the plaintext.
The whole scheme relies on the facts that the key is (1) random and
(2) unknown to Eve. The very name “one-time pad” underscores that a
key can only be used once and must then be discarded. If a single key is
used for two messages, then the second key is not “random” — it is instead
perfectly correlated with the first key. There are easy methods to break the
code when a key is reused.
Generating random numbers is not easy, and the Vernam cipher de-
mands keys as long as the messages transmitted. As recently as 1992,
high-quality computer random-number generators were classified by the
U.S. government as munitions, along with tanks and fighter planes, and
their export from the country was prohibited.
And of course Eve must not know the key. So there must be some way
for Alice to get the key to Bob securely. If they have some secure method
for transmitting keys, why don’t they just use that same secure method for
sending their messages?
In common parlance, the word “random” can mean “unimportant, not
worth considering” (as in “Joe made a random comment”). So it may
seem remarkable that a major problem for government, the military, and
commerce is the generation and distribution of randomness, but that is
indeed the case.

2.6.2 Quantum mechanics to the rescue

Since quantum mechanics involves randomness, it seems uniquely posi-

tioned to solve this problem. Here’s one scheme.
Alice and Bob set up a source of entangled atoms halfway between their
two homes. Both of them erect vertical Stern-Gerlach analyzers to detect
the atoms. If Alice’s atom comes out +, she will interpret it as a 1, if −,
a 0. Bob interprets his atoms in the opposite sense. Since the entangled
What Is Quantum Mechanics About? 87

atoms always exit from opposite ports, Alice and Bob end up with the
same random number, which they use as a key for their Vernam-cipher
communications over conventional telephone or computer lines.
This scheme will indeed produce and distribute copious, high-quality
random numbers. But Eve can get at those same numbers through the
following trick: She cuts open the atom pipe leading from the entangled
source to Alice’s home, and inserts a vertical interferometer.20 She watches
the atoms pass through her interferometer. If the atom takes path a, Eve
knows that when Alice receives that same atom, it will exit from Eve’s
+ port. If the atom takes path b, the opposite holds. Eve gets the key, Eve
breaks the code.
It’s worth looking at this eavesdropping in just a bit more detail. When
the two atoms depart from their source, they are entangled. It is not true
that, say, Alice’s atom has µz = +µB while Bob’s atom has µz = −µB
— the pair of atoms is in the condition we’ve called “entangled”, but the
individual atoms themselves are not in any condition. However, after Eve
sees the atom taking path a of her interferometer, then the two atoms are
no longer entangled — now it is true that Alice’s atom has the condition
µz = +µB while Bob’s atom has the condition µz = −µB . The key received
by Alice and Bob will be random whether or not Eve is listening in. To
test for evesdropping, Alice and Bob must examine it in some other way.
Replace Alice and Bob’s vertical analyzers with flipping Stern-Gerlach
analyzers. After Alice receives her random sequence of pluses and minuses,
encountering her random sequence of analyzer orientations, she sends both
these sequences to Bob over an open communication line. (Eve will in-
tercept this information but it won’t do her any good, because she won’t
know the corresponding information for Bob.) Bob now knows both the
results at his analyzer and the results at Alice’s analyzer, so he can test
to see whether the atom pairs were entangled. If he finds that they were,
then Eve is not listening in. If he finds that they were not entangled, then
he knows for certain that Eve is listening in, and they must not use their
compromised key.
Is there some other way for Eve to tap the line? No! If the atom pairs
pass the test for entanglement, then no one can know the values of their
20 Inspired by James Bond, I always picture Eve as exotic beauty in a little black dress

slinking to the back of an eastern European café to tap the diplomatic cable which
conveniently runs there. But in point of fact Eve would be a computer.
 88 What is a qubit?

µz projections because those projections don’t exist! We have guaranteed

that no one has intercepted the key by the interferometer method, or by
any other method whatsoever.
Once Alice has tested Bell’s theorem, she and Bob still have a lot of
work to do. For a key they must use only those random numbers produced
when their two analyzers happen to have the same orientations. There are
detailed protocols specifying how Alice and Bob must exchange information
about their analyzer orientations, in such a way that Eve can’t uncover
them. I won’t describe these protocols because while they tell you how
clever people are, they tell you nothing about how nature behaves. But
you should take away that entanglement is not merely a phenomenon of
nature: it is also a natural resource.

2.7 What is a qubit?

We’ve devoted an entire chapter to the magnetic moment of a silver atom.

Perhaps you find this inappropriate: do you really care so much about
silver atoms? Yes you do, because the phenomena and principles we’ve
established concerning the magnetic moment of a silver atom apply to a
host of other systems: the polarization of a light photon, the hybridization
of a benzene molecule, the position of the nitrogen atom within an ammonia
molecule, the neutral kaon, and more. Such systems are called “two-state
systems” or “spin- 12 systems” or “qubit systems”. The ideas we establish
concerning the magnetic moment of a silver atom apply equally well to all
these systems.
After developing these ideas in the next chapter, we will (in chapter 4,
“The Quantum Mechanics of Position”) generalize them to continuum sys-
tems like the position of an electron.

Problem
2.12 Questions (recommended problem)
Update your list of quantum mechanics questions that you started at
problem 1.17 on page 46. Write down new questions and, if you have un-
covered answers to any of your old questions, write them down briefly.
 Chapter 3

Forging Mathematical Tools

When you walked into your introductory classical mechanics course, you
were already familiar with the phenomena of introductory classical mechan-
ics: flying balls, spinning wheels, colliding billiard balls. Your introductory
mechanics textbook didn’t need to introduce these things to you, but in-
stead jumped right into describing these phenomena mathematically and
explaining them in terms of more general principles.
The last chapter of this book made you familiar with the phenomena
of quantum mechanics: quantization, interference, and entanglement —
at least, insofar as these phenomena are manifest in the behavior of the
magnetic moment of a silver atom. You are now, with respect to quan-
tum mechanics, at the same level that you were, with respect to classical
mechanics, when you walked into your introductory mechanics course. It
is now our job to describe these quantal phenomena mathematically, to
explain them in terms of more general principles, and (eventually) to inves-
tigate situations more complex than the magnetic moment of one or two
silver atoms.

3.1 What is a quantal state?

We’ve been talking about the state of the silver atom’s magnetic moment
by saying things like “the projection of the magnetic moment on the z axis
is µz = −µB ” or “µx = +µB ” or “µθ = −µB ”. This notation is clumsy.
First of all, it requires you to write down the same old µs time and time
again. Second, the most important thing is the axis (z or x or θ), and the
symbol for the axis is also the smallest and easiest to overlook.

89
90 What is a quantal state?

P.A.M. Dirac1 invented a notation that overcomes these faults. He

looked at descriptions like
µz = −µB or µx = +µB or µθ = −µB
and noted that the only difference from one expression to the other was
the axis subscript and the sign in front of µB . Since the only thing that
distinguishes one expression from another is (z, −), or (x, +), or (θ, −),
Dirac thought, these should be the only things we need to write down. He
denoted these three states as
|z−i or |x+i or |θ−i.
The placeholders | i are simply ornaments to remind us that we’re talking
about quantal states, just as the arrow atop ~r is simply an ornament to
remind us that we’re talking about a vector. States expressed using this
notation are sometimes called “kets”.
Simply establishing a notation doesn’t tell us much. Just as in classical
mechanics, we say we know a state when we know all the information needed
to describe the system now and to predict its future. In our universe the
classical time evolution law is
2
d ~r
F~ = m 2
dt
and so the state is specified by giving both a position ~r and a velocity ~v . If
nature had instead provided the time evolution law
3
d ~r
F~ = m 3
dt
then the state would have been specified by giving a position ~r, a velocity
~v , and an acceleration ~a. The specification of state is dictated by nature,
not by humanity, so we can’t know how to specify a state until we know the
laws of physics governing that state. Since we don’t yet know the laws of
quantal physics, we can’t yet know exactly how to specify a quantal state.
Classical intuition makes us suppose that, to specify the magnetic mo-
ment of a silver atom, we need to specify all three components µz , µx , and
µy . We have already seen that nature precludes such a specification: if the
magnetic moment has a value for µz , then it doesn’t have a value for µx ,
1 The Englishman Paul Adrien Maurice Dirac (1902–1984) in 1928 formulated a rela-
tivistically correct quantum mechanical equation that turns out to describe the electron.
In connection with this so-called Dirac equation, he predicted the existence of antimatter.
Dirac was painfully shy and notoriously cryptic.
3.2. Amplitude 91

and it’s absurd to demand a specification for something that doesn’t ex-
ist. As we learn more and more quantum physics, we will learn better and
better how to specify states. There will be surprises. But always keep in
mind that (just as in classical mechanics) it is experiment, not philosophy
or meditation, and certainly not common sense, that tells us how to specify
states.

3.2 Amplitude

b
input
|z+i output
a
|z−i

An atom in state |z+i ambivates through the apparatus above. We have

already seen that, when the atom ambivates in darkness,
probability to go from input to output 6=
probability to go from input to output via path a (3.1)
+ probability to go from input to output via path b.
On the other hand, it makes sense to associate some sort of “influence
to go from input to output via path a” with the path through a and a
corresponding “influence to go from input to output via path b” with the
path through b. This postulated influence is called “probability amplitude”
or just “amplitude”.2 Whatever amplitude is, its desired property is that
amplitude to go from input to output =
amplitude to go from input to output via path a (3.2)
+ amplitude to go from input to output via path b.
For the moment, the very existence of amplitude is nothing but a hopeful
surmise. Scientists cannot now and indeed never will be able to prove that
the concept of amplitude applies to all situations. That’s because new
situations are being investigated every day, and perhaps tomorrow a new
2 The name “amplitude” is a poor one, because it is also used for the maximum value of

a sinusoidal signal — in the function A sin(ωt), the symbol A represents the amplitude —
and this sinusoidal signal “amplitude” has nothing to do with the quantal “amplitude”.
One of my students correctly suggested that a better name for quantal amplitude would
be “proclivity”. But it’s too late now to change the word.
 92 Amplitude

situation will be discovered that cannot be described in terms of amplitudes.

But as of today, that hasn’t happened.
The role of amplitude, whatever it may prove to be, is to calculate
probabilities. We establish three desirable rules:

(1) From amplitude to probability. For every possible action there is an

associated amplitude, such that
probability for the action = |amplitude for the action|2 .
(2) Actions in series. If an action takes place through several successive
stages, the amplitude for that action is the product of the amplitudes
for each stage.
(3) Actions in parallel. If an action could take place in several possible
ways, the amplitude for that action is the sum of the amplitudes for
each possibility.

The first rule is a simple way to make sure that probabilities are al-
ways positive. The second rule is a natural generalization of the rule for
probabilities in series — that if an action happens through several stages,
the probability for the action as a whole is the product of the probabilities
for each stage. And the third rule simply restates the “desired property”
presented in equation (3.2).
We apply these rules to various situations that we’ve already encoun-
tered, beginning with the interference experiment sketched above. Recall
the probabilities already established (first column in table):

probability |amplitude| amplitude

go from input to output 0 0 0
1 1
go from input to output via path a 4 2 + 12
1 1
go from input to output via path b 4 2 − 12

If rule (1) is to hold, then the amplitude to go from input to output must
also be 0, while the amplitude to go via a path must have magnitude 12
(second column in table). According to rule (3), the two amplitudes to
go via a and via b must sum to zero, so they cannot both be represented
by positive numbers. Whatever mathematical entity is used to represent
amplitude, it must enable two such entities, each with non-zero magnitude,
to sum to zero. There are many such entities: real numbers, complex
Forging Mathematical Tools 93

numbers, hypercomplex numbers, and vectors in three dimensions are all

possibilities. For this particular interference experiment, it suffices to assign
real numbers to amplitudes: the amplitude to go via path a is + 12 , and the
amplitude to go via path b is − 12 . (Third column in table. The negative
sign could have been assigned to path a rather than to path b: this choice is
merely conventional.) For other interference experiments complex numbers
are required. It turns out that, for all situations yet encountered, one can
represent amplitude mathematically as a complex number. Once again,
this reflects the results of experiment, not of philosophy or meditation.
The second situation we’ll consider is a Stern-Gerlach analyzer.
z

θ
|θ+i
|z+i
|θ−i

The amplitude that an atom entering the θ-analyzer in state |z+i exits in
state |θ+i is called3 hθ+|z+i. That phrase is a real mouthful, so the symbol
hθ+|z+i is pronounced “the amplitude that |z+i is in |θ+i”, even though
this briefer pronunciation leaves out the important role of the analyzer.4
From rule (1), we know that
|hθ+|z+i|2 = cos2 (θ/2) (3.3)
2 2
|hθ−|z+i| = sin (θ/2). (3.4)
You can also use rule (1), in connection with the experiments described in
3 The states appear in the symbol in the opposite sequence from their appearance in

the description.
4 The ultimate source of such problems is that the English language was invented by

people who did not understand quantum mechanics, hence they never produced concise,
accurate phrases to describe quantal phenomena. In the same way, the ancient phrase
“search the four corners of the Earth” is still colorful and practical, and is used today
even by those who know that the Earth doesn’t have four corners.
94 Amplitude

problem 2.1, “Exit probabilities” (on page 57) to determine that

|hz+|θ+i|2 = cos2 (θ/2)
|hz−|θ+i|2 = sin2 (θ/2)
|hθ+|z−i|2 = sin2 (θ/2)
|hθ−|z−i|2 = cos2 (θ/2)
|hz+|θ−i|2 = sin2 (θ/2)
|hz−|θ−i|2 = cos2 (θ/2).

Clearly analyzer experiments like these find the magnitude of an am-

plitude. No analyzer experiment can find the phase of an amplitude.5 To
determine phases, we must perform interference experiments.
So the third situation is an interference experiment.

z
θ
a

input
|z+i output
|z−i
b

Rule (2), actions in series, tells us that the amplitude to go from |z+i to
|z−i via path a is the product of the amplitude to go from |z+i to |θ+i
times the amplitude to go from |θ+i to |z−i:
amplitude to go via path a = hz−|θ+ihθ+|z+i.
Similarly
amplitude to go via path b = hz−|θ−ihθ−|z+i.
And then rule (3), actions in parallel, tells us that the amplitude to go from
|z+i to |z−i is the sum of the amplitude to go via path a and the amplitude
to go via path b. In other words
hz−|z+i = hz−|θ+ihθ+|z+i + hz−|θ−ihθ−|z+i. (3.5)

5 The terms phase and magnitude are explained in appendix C, “Complex Arithmetic”.
Forging Mathematical Tools 95

We know the magnitude of each of these amplitudes from analyzer ex-

periments:

amplitude magnitude
hz−|z+i 0
hz−|θ+i | sin(θ/2)|
hθ+|z+i | cos(θ/2)|
hz−|θ−i | cos(θ/2)|
hθ−|z+i | sin(θ/2)|

The task now is to assign phases to these magnitudes in such a way that
equation (3.5) is satisfied. In doing so we are faced with an embarrassment
of riches: there are many consistent ways to make this assignment. Here
are two commonly used conventions:

amplitude convention I convention II

hz−|z+i 0 0
hz−|θ+i sin(θ/2) i sin(θ/2)
hθ+|z+i cos(θ/2) cos(θ/2)
hz−|θ−i cos(θ/2) cos(θ/2)
hθ−|z+i − sin(θ/2) −i sin(θ/2)

There are two things to notice about these amplitude assignments.

First, one normally assigns values to physical quantities by experiment, or
by calculation, but not “by convention”. Second, both of these conventions
show unexpected behaviors: Because the angle 0◦ is the same as the angle
360◦ , one would expect that h0◦ +|z+i would equal h360◦ +|z+i, whereas
in fact the first amplitude is +1 and the second is −1. Because the state
|180◦ −i (that is, |θ−i with θ = 180◦ ) is the same as the state |z+i, one
would expect that h180◦ −|z+i = 1, whereas in fact h180◦ −|z+i is either
−1 or −i, depending on convention. These two observations underscore
the fact that amplitude is a mathematical tool that enables us to calculate
physically observable quantities, like probabilities. It is not itself a physical
entity. No experiment measures amplitude. Amplitude is not “out there,
physically present in space” in the way that, say, a nitrogen molecule is.
A good analogy is that an amplitude convention is like a language. Any
language is a human convention: there is no intrinsic connection between a
physical horse and the English word “horse”, or the German word “pferd”,
96 Amplitude

or the Swahili word “farasi”. The fact that language is pure human con-
vention, and that there are multiple conventions for the name of a horse,
doesn’t mean that language is unimportant: on the contrary language is
an immensely powerful tool. And the fact that language is pure human
convention doesn’t mean that you can’t develop intuition about language:
on the contrary if you know the meaning of “arachnid” and the meaning
of “phobia”, then your intuition for English tells you that “arachnopho-
bia” means fear of spiders. Exactly the same is true for amplitude: it is a
powerful tool, and with practice you can develop intuition for it.
When I introduced the phenomenon of quantal interference on page 62,
I said that there was no word or phrase in the English language that ac-
curately represents what’s going on: It’s flat-out wrong to say “the atom
takes path a” and it’s flat-out wrong to say “the atom takes path b”. It
gives a wrong impression to say “the atom takes no path” or “the atom
takes both paths”. I introduced the phrase “the atom ambivates through
the two paths of the interferometer”. Now we have a technically correct
way of describing the phenomenon: “the atom has an amplitude to take
path a and an amplitude to take path b”.
Here’s another warning about language: If an atom in state |ψi enters
a vertical analyzer, the amplitude for it to exit from the + port is hz+|ψi.
(And of course the amplitude for it exit from the − port is hz−|ψi.) This is
often stated “If the atom is in state |ψi, the amplitude of it being in state
|z+i is hz+|ψi.” This is an acceptable shorthand for the full explanation,
which requires thinking about an analyzer experiment, even though the
shorthand never mentions the analyzer. But never say “If the atom is in
state |ψi, the probability of it being in state |z+i is |hz+|ψi|2 .” This gives
the distinct and incorrect impression that before entering the analyzer, the
atom was either in state |z+i or in state |z−i, and you just didn’t know
which it was. Instead, say “If an atom in state |ψi enters a vertical analyzer,
the probability of exiting from the + port in state |z+i is |hz+|ψi|2 .”
Forging Mathematical Tools 97

3.2.1 Sample Problem: Two paths

Find an equation similar to equation (3.5) representing the amplitude to

start in state |ψi at input, ambivate through a vertical interferometer, and
end in state |φi at output.

1a

|ψi |φi
input output

1b

Solution: Because of rule (2), actions in series, the amplitude for the
atom to take the top path is the product
hφ|z+ihz+|ψi.
Similarly the amplitude for it to take the bottom path is
hφ|z−ihz−|ψi.
Because of rule (3), actions in parallel, the amplitude for it to ambivate
through both paths is the sum of these two, and we conclude that
hφ|ψi = hφ|z+ihz+|ψi + hφ|z−ihz−|ψi. (3.6)

3.2.2 Sample Problem: Three paths

Stretch apart a vertical interferometer, so that the recombining rear end

is far from the splitting front end, and insert a θ interferometer into the
bottom path. Now there are three paths from input to output. Find an
equation similar to equation (3.5) representing the amplitude to start in
state |ψi at input and end in state |φi at output.
 98 Amplitude

1a
θ
2a
|ψi |φi
input output

1b

2b

Solution:
hφ|ψi = hφ|z+ihz+|ψi
+ hφ|z−ihz−|θ+ihθ+|z−ihz−|ψi (3.7)
+ hφ|z−ihz−|θ−ihθ−|z−ihz−|ψi

Problems
3.1 Talking about interference
An atom in state |ψi ambivates through a vertical analyzer. We say,
appropriately, that “the atom has an amplitude to take the top path
and an amplitude to take the bottom path”. For the benefit of students
in next year’s offering of this class (see page 20), find expressions for
those two amplitudes and describe, in ten sentences or fewer, why it is
not appropriate to say “the atom has probability |hz+|ψi|2 to take the
top path and probability |hz−|ψi|2 to take the bottom path”.
3.2 Other conventions
Two conventions for assigning amplitudes are given in the table on
page 95. Show that if hz−|θ+i and hz−|θ−i are multiplied by phase
factor eiα , and if hz+|θ+i and hz+|θ−i are multiplied by phase factor
eiβ (where α and β are both real), then the resulting amplitudes are
just as good as the original (for either convention I or convention II).
3.3. Reversal-conjugation relation 99

3.3 Peculiarities of amplitude

Page 95 pointed out some of the peculiarities of amplitude; this problem
points out another. Since the angle θ is the same as the angle 360◦ + θ,
one would expect that hθ+|z+i would equal h(360◦ + θ)+|z+i. Show,
using either of the conventions given in the table on page 95, that this
expectation is false. What is instead correct?

3.3 Reversal-conjugation relation

The “complex conjugate” of any complex number is the same number but
with every “i” changed to “−i”: if x and y are real numbers, then
z = x + iy has complex conjugate z ∗ = x − iy. (3.8)
A useful theorem says that the amplitude to go from state |ψi to state
|φi and the amplitude to go in the opposite direction are related through
complex conjugation:
∗
hφ|ψi = hψ|φi . (3.9)
The proof below works for states of the magnetic moment of a silver atom
— the kind of states we’ve worked with so far — but in fact the result holds
for any quantal system.
The proof relies on three facts: First, the probability for one state to
be analyzed into another depends only on the magnitude of the angle be-
tween the incoming magnetic moment and the analyzer, and not on the
sense of that angle. (An atom in state |z+i has the same probability of
leaving the + port of an analyzer whether it is rotated 17◦ clockwise or 17◦
counterclockwise.) Thus
|hφ|ψi|2 = |hψ|φi|2 . (3.10)
Second, an atom exits an interferometer in the same state in which it en-
tered, so
hφ|ψi = hφ|θ+ihθ+|ψi + hφ|θ−ihθ−|ψi. (3.11)
Third, an atom entering an analyzer comes out somewhere, so
1 = |hθ+|ψi|2 + |hθ−|ψi|2 . (3.12)
100 Reversal-conjugation relation

The proof also relies on a mathematical result called “the triangle in-
equality for complex numbers”: If a and b are real numbers with a + b = 1,
and in addition eiα a + eiβ b = 1, with α and β real, then α = β = 0. You
can find very general, very abstract, proofs of the triangle inequality, but
the complex plane sketch below encapsulates the idea:

imaginary

eiα a eiβ b

real
a b 1

From the first fact (3.10), the two complex numbers hφ|ψi and hψ|φi
have the same magnitude, so they differ only in phase. Write this statement
as
∗
hφ|ψi = eiδ hψ|φi (3.13)
where the phase δ is a real number that might depend on the states |φi and
|ψi. Apply this general result first to the particular state |φi = |θ+i:
∗
hθ+|ψi = eiδ+ hψ|θ+i , (3.14)
and then to the particular state |φi = |θ−i:
∗
hθ−|ψi = eiδ− hψ|θ−i , (3.15)
where the two real numbers δ+ and δ− might be different. Our objective is
to prove that δ+ = δ− = 0.
Apply the second fact (3.11) with |φi = |ψi, giving
1 = hψ|θ+ihθ+|ψi + hψ|θ−ihθ−|ψi
∗ ∗
= eiδ+ hψ|θ+ihψ|θ+i + eiδ− hψ|θ−ihψ|θ−i
= eiδ+ |hψ|θ+i|2 + eiδ− |hψ|θ−i|2
= eiδ+ |hθ+|ψi|2 + eiδ− |hθ−|ψi|2 . (3.16)
Compare this result to the third fact (3.12)
1 = |hθ+|ψi|2 + |hθ−|ψi|2 (3.17)
and use the triangle inequality with a = |hθ+|ψi|2 and b = |hθ−|ψi|2 . The
two phases δ+ and δ− must vanish, so the “reversal-conjugation relation”
is proven.
3.4. Establishing a phase convention 101

3.4 Establishing a phase convention

Although there are multiple alternative phase conventions for amplitudes

(see problem 3.2 on page 98), we will from now on use only phase conven-
tion I from page 95:
hz+|θ+i = cos(θ/2)
hz−|θ+i = sin(θ/2)
(3.18)
hz+|θ−i = − sin(θ/2)
hz−|θ−i = cos(θ/2)
In particular, for θ = 90◦ we have
√
hz+|x+i = 1/√2
hz−|x+i = 1/√2
(3.19)
hz+|x−i = −1/√2
hz−|x−i = 1/ 2
This convention has a desirable special case for θ = 0◦ , namely
hz+|θ+i = 1
hz−|θ+i = 0
(3.20)
hz+|θ−i = 0
hz−|θ−i = 1
but an unexpected special case for θ = 360◦ , namely
hz+|θ+i = −1
hz−|θ+i = 0
(3.21)
hz+|θ−i = 0
hz−|θ−i = −1
This is perplexing, given that the angle θ = 0◦ is the same as the angle θ =
360◦ ! Any convention will have similar perplexing cases. Such perplexities
underscore the fact that amplitudes are important mathematical tools used
to calculate probabilities, but are not “physically real”.
Given these amplitudes, we can use the interference result (3.6) to cal-
culate any amplitude of interest:
hφ|ψi = hφ|z+ihz+|ψi + hφ|z−ihz−|ψi
∗ ∗ (3.22)
= hz+|φi hz+|ψi + hz−|φi hz−|ψi
where in the last line we have used the reversal-conjugation relation (3.9).
102 Establishing a phase convention

Problems
3.4 Other conventions, other peculiarities
Write what this section would have been had we adopted convention II
rather than convention I from page 95. In addition, evaluate the four
amplitudes of equation (3.18) for θ = +180◦ and θ = −180◦ .
3.5 Finding amplitudes (recommended problem)
Using the interference idea embodied in equation (3.22), calculate the
amplitudes hθ+|54◦ +i and hθ−|54◦ +i as a function of θ. Do these
amplitudes have the values you expect for θ = 54◦ ? For θ = 234◦ ?
Plot hθ+|54◦ +i for θ from 0◦ to 360◦ . Compare the result for θ = 0◦
and θ = 360◦ .
3.6 Rotations
Use the interference idea embodied in equation (3.22) to show that
hx+|θ+i = √12 [cos(θ/2) + sin(θ/2)]
hx−|θ+i = − √12 [cos(θ/2) − sin(θ/2)]
(3.23)
hx+|θ−i = √12 [cos(θ/2) − sin(θ/2)]
hx−|θ−i = √12 [cos(θ/2) + sin(θ/2)]
If and only if you enjoy trigonometric identities, you should then show
that these results can be written equivalently as
hx+|θ+i = cos((θ − 90◦ )/2)
hx−|θ+i = sin((θ − 90◦ )/2)
(3.24)
hx+|θ−i = − sin((θ − 90◦ )/2)
hx−|θ−i = cos((θ − 90◦ )/2)
This makes perfect geometric sense, as the angle relative to the x axis
is 90◦ less than the angle relative to the z axis:

x
3.5. How can I specify a quantal state? 103

3.5 How can I specify a quantal state?

We introduced the Dirac notation for quantal states on page 90, but haven’t
yet fleshed out that notation by specifying a state mathematically. Start
with an analogy:

3.5.1 How can I specify a position vector?

We are so used to writing down the position vector ~r that we rarely stop
to ask ourselves what it means. But the plain fact is that whenever we
measure a length (say, with a meter stick) we find not a vector, but a single
number! Experiments measure never the vector ~r but always a scalar —
the dot product between ~r and some other vector, call it ~s for “some other”.
If we know the dot product between ~r and every vector ~s, then we know
everything there is to know about ~r. Does this mean that to specify ~r, we
must keep a list of all possible dot products ~s · ~r ? Of course not. . . such a
list would be infinitely long!
You know that if you write ~r in terms of an orthonormal basis {î, ĵ, k̂},
namely
~r = rx î + ry ĵ + rz k̂ (3.25)
where rx = î · ~r, ry = ĵ · ~r, and rz = k̂ · ~r, then you’ve specified the vector.
Why? Because if you know the triplet (rx , ry , rz ) and the triplet (sx , sy , sz ),
then you can easily find the desired dot product
 

rx
~s · ~r = sx sy sz  ry  = sx rx + sy ry + sz rz . (3.26)
rz
It’s a lot more compact to specify the vector through three dot products
— namely î · ~r, ĵ · ~r, and k̂ · ~r — from which you can readily calculate an
infinite number of desired dot products, than it is to list all infinity dot
products themselves!

3.5.2 How can I specify a quantal state?

Like the position vector ~r, the quantal state |ψi cannot by itself be mea-
sured. But if we determine (through some combination of analyzer exper-
iments, interference experiments, and convention) the amplitude hσ|ψi for
104 How can I specify a quantal state?

every possible state |σi, then we know everything there is to know about
|ψi. Is there some compact way of specifying the state, or do we have to
keep an infinitely long list of all these amplitudes?
This nut is cracked through the interference experiment result
hσ|ψi = hσ|θ+ihθ+|ψi + hσ|θ−ihθ−|ψi, (3.27)
which simply says, in symbols, that the atom exits an interferometer in the
same state in which it entered (see equation 3.11). It gets hard to keep
track of all these symbols, so I’ll introduce the names
hθ+|ψi = ψ+
hθ−|ψi = ψ−
and
hθ+|σi = σ+
hθ−|σi = σ− .
From the reversal-conjugation relation, this means
∗
hσ|θ+i = σ+
∗
hσ|θ−i = σ− .
In terms of these symbols, the interference result (3.27) is
 
∗ ∗ ∗ ∗

ψ+
hσ|ψi = σ+ ψ+ + σ− ψ− = σ+ σ− . (3.28)
ψ−
And this is our shortcut! By keeping track of only two amplitudes, ψ+ and
ψ− , for each state, we can readily calculate any amplitude desired. We
don’t have to keep an infinitely long list of amplitudes.
This dot product result for computing amplitude is so useful and so
convenient that sometimes people say the amplitude is a dot product. No.
The amplitude reflects analyzer experiments, plus interference experiments,
plus convention. The dot product is a powerful mathematical tool for com-
puting amplitudes. (A parallel situation: There are many ways to find the
latitude and longitude coordinates for a point on the Earth’s surface, but
the easiest is to use a GPS device. Some people are so enamored of this
ease that they call the latitude and longitude the “GPS coordinates”. But
in fact the coordinates were established long before the Global Positioning
System was built.)
Forging Mathematical Tools 105

3.5.3 What is a basis?

For vectors in three-dimensional space, an orthonormal basis such as

{î, ĵ, k̂} is a set of three vectors of unit magnitude perpendicular to each
other. As we’ve seen, the importance of a basis is that every vector ~r can
be represented as a sum over these basis vectors,
~r = rx î + ry ĵ + rz k̂,
and hence any vector ~r can be conveniently represented through the triplet
î · ~r
   
rx
 ry  =  ĵ · ~r  .
rz k̂ · ~r

For quantal states, we’ve seen that a set of two states such as
{|θ+i, |θ−i} plays a similar role, so it too is called a basis. For the magnetic
moment of a silver atom, two states |ai and |bi constitute a basis when-
ever ha|bi = 0, and the analyzer experiment of section 2.1.3 shows that
the states |θ+i and |θ−i certainly satisfy this requirement. In the basis
{|ai, |bi} an arbitrary state |ψi can be conveniently represented through
the pair of amplitudes
 
ha|ψi
.
hb|ψi

3.5.4 Hilbert space

We have learned to express a physical state as a mathematical entity —

namely, using the {|ai, |bi} basis, the state |ψi is represented as a column
matrix of amplitudes
 
ha|ψi
.
hb|ψi
This mathematical entity is called a “state vector in Hilbert6 space”.
For example, in the basis {|z+i, |z−i} the state |θ+i is represented by
   
hz+|θ+i cos(θ/2)
= . (3.29)
hz−|θ+i sin(θ/2)
6 The German mathematician David Hilbert (1862–1943) made contributions to func-

tional analysis, geometry, mathematical physics, and other areas. He formalized and
extended the concept of a vector space. Hilbert and Albert Einstein raced to uncover
the field equations of general relativity, but Einstein beat Hilbert by a matter of weeks.
106 How can I specify a quantal state?

Whereas (in light of equation 3.23) in the basis {|x+i, |x−i} that same
state |θ+i is represented by the different column matrix
!
√1 [cos(θ/2) + sin(θ/2)]
 
hx+|θ+i 2
= . (3.30)
hx−|θ+i − √12 [cos(θ/2) − sin(θ/2)]

Write down the interference experiment result twice

ha|ψi = ha|z+ihz+|ψi + ha|z−ihz−|ψi
hb|ψi = hb|z+ihz+|ψi + hb|z−ihz−|ψi
and then write these two equations as one using column matrix notation
     
ha|ψi ha|z+i ha|z−i
= hz+|ψi + hz−|ψi.
hb|ψi hb|z+i hb|z−i
Notice the column matrix representations of states |ψi, |z+i, and |z−i, and
write this equation as
|ψi = |z+ihz+|ψi + |z−ihz−|ψi. (3.31)
And now we have a new thing under the sun. We never talk about adding
together two classical states, nor multiplying them by numbers, but this
equation gives us the meaning of such state addition in quantum mechan-
ics. This is a new mathematical tool, it deserves a new name, and that
name is “superposition”. Superposition is the mathematical reflection of
the physical phenomenon of interference, as in the sentence: “When an
atom ambivates through an interferometer, its state is a superposition of
the state of an atom taking path a and the state of an atom taking path b.”
Superposition is not familiar from daily life or from classical mechanics,
but there is a story7 that increases understanding: “A medieval European
traveler returns home from a journey to India, and describes a rhinoceros
as a sort of cross between a dragon and a unicorn.” In this story the
rhinoceros, an animal that is not familiar but that does exist, is described
as intermediate (a “sort of cross”) between two fantasy animals (the dragon
and the unicorn) that are familiar (to the medieval European) but that do
not exist.
Similarly, an atom in state |z+i ambivates through both paths of a
horizontal interferometer. This action is not familiar but does happen, and
it is characterized as a superposition (a “sort of cross”) between two actions
7 Invented by John D. Roberts, but first published in Robert T. Morrison and Robert

N. Boyd, Organic Chemistry, second edition (Allyn & Bacon, Boston, 1966) page 318.
Forging Mathematical Tools 107

(“taking path a” and “taking path b”) that are familiar (to all of us steeped
in the classical approximation) but that do not happen.
In principle, any calculation performed using the Hilbert space rep-
resentation of states could be performed by considering suitable, cleverly
designed analyzer and interference experiments. But it’s a lot easier to use
the abstract Hilbert space machinery. (Similarly, any result in electrostatics
could be found using Coulomb’s Law, but it’s a lot easier to use the ab-
stract electric field and electric potential. Any calculation involving vectors
could be performed graphically, but it’s a lot easier to use abstract compo-
nents. Any addition or subtraction of whole numbers could be performed
by counting out marbles, but it’s a lot easier to use abstract mathematical
tools like carrying and borrowing.)

3.5.5 Peculiarities of state vectors

Because state vectors are built from amplitudes, and amplitudes have pe-
culiarities (see pages 95 and 101), it is natural that state vectors have
similar peculiarities. For example, since the angle θ is the same as the an-
gle θ + 360◦ , I would expect that the state vector |θ+i would be the same
as the state vector |(θ + 360◦ )+i.
But in fact, in the {|z+i, |z−i} basis, the state |θ+i is represented by
   
hz+|θ+i cos(θ/2)
= , (3.32)
hz−|θ+i sin(θ/2)
so the state |(θ + 360◦ )+i is represented by
hz+|(θ + 360◦ )+i cos((θ + 360◦ )/2)
   
= (3.33)
hz−|(θ + 360◦ )+i sin((θ + 360◦ )/2)
cos(θ/2 + 180◦ )
   
− cos(θ/2)
= = .
sin(θ/2 + 180◦ ) − sin(θ/2)
So in fact |θ+i = −|(θ + 360◦ )+i. Bizarre!
This bizarreness is one facet of a general rule: If you multiply any state
vector by a complex number with magnitude unity — a number such as
−1, or i, or √12 (−1 + i), or e2.7i — a so-called “complex unit” or “phase
factor” — then you get a different state vector that represents the same
state. This fact is called “global phase freedom” — you are free to set the
overall phase of your state vector for your own convenience. This general
108 How can I specify a quantal state?

rule applies only for multiplying both elements of the state vector by the
same complex unit: if you multiply the two elements with different complex
units, you will obtain a vector representing a different state (see problem 3.8
on page 110).

3.5.6 Names for position vectors

The vector ~r is specified in the basis {î, ĵ, k̂} by the three components
î · ~r
   
rx
 ry  =  ĵ · ~r  .
rz k̂ · ~r
Because this component specification is so convenient, it is sometimes said
that the vector ~r is not just specified, but is equal to this triplet of numbers.
That’s false.
Think of the vector ~r = 5î + 5ĵ. It is represented in the basis {î, ĵ, k̂} by
the triplet (5,√5, 0). But this is√not the only basis that exists. In the basis
{î0 = (î+√
ĵ)/ 2, ĵ 0 = (−î+ ĵ)/ 2, k̂}, that same vector is represented
√ by the
triplet (5 2, 0, 0). If we had said that ~r = (5, 5, 0)√ and that ~
r = (5 2, 0, 0),
then we would be forced to conclude that 5 = 5 2 and that 5 = 0!

ĵ
6
0
ĵ î0
@
I 
@
@
@
@
@ - î

To specify a position vector ~r, we use the components of ~r in a particular

basis, usually denoted (rx , ry , rz ). We often write “~r = (rx , ry , rz )” but in
fact that’s not exactly correct. The vector ~r represents a position — it is
independent of basis. The row matrix (rx , ry , rz ) represents the components
of that position vector in a particular basis — it is the “name” of the
position in a particular basis. Instead of using an equals sign = we use
. .
the symbol = to mean “represented by in a particular basis”, as in “~r =
Forging Mathematical Tools 109

(5, 5, 0)” meaning “the vector ~r = 5î + 5ĵ is represented by the triplet
(5, 5, 0) in the basis {î, ĵ, k̂}”.
Vectors are physical things: a caveman throwing a spear at a mam-
moth was performing addition of position vectors, even though the caveman
didn’t understand basis vectors or Cartesian coordinates. The concept of
“position” was known to cavemen who did not have any concept of “basis”.

3.5.7 Names for quantal states

We’ve been specifying a state like |ψi = |17◦ +i by stating the axis upon
which the projection of µ~ is definite and equal to +µB — in this case, the
axis tilted 17◦ from the vertical.
Another way to specify a state |ψi would be to give the amplitude
that |ψi is in any possible state: that is, to list hθ+|ψi and hθ−|ψi for
all values of θ: 0◦ ≤ θ < 360◦ . One of those amplitudes (in this case
h17◦ +|ψi) will have value 1, and finding this one amplitude would give
us back the information in the specification |17◦ +i. In some ways this is a
more convenient specification because we don’t have to look up amplitudes:
they’re right there in the list. On the other hand it is an awful lot of
information to have to carry around.
The Hilbert space approach is a third way to specify a state that com-
bines the brevity of the first way with the convenience of the second way.
Instead of listing the amplitude hσ|ψi for every state |σi we list only the
two amplitudes ha|ψi and hb|φi for the elements {|ai, |bi} of a basis. We’ve
already seen (equation 3.28) how quantal interference then allows us to
readily calculate any amplitude.
Just as we said “the position vector ~r is represented in the basis {î, ĵ, k̂}
as (1, 1, 0)” or
.
~r = (1, 1, 0),
so we say “the quantal state |ψi is represented in the basis {|z+i, |z−i} as
 
. hz+|ψi
|ψi = .”
hz−|ψi
110 How can I specify a quantal state?

Problems
3.7 Superposition and interference (recommended problem)
On page 106 I wrote that “When an atom ambivates through an in-
terferometer, its state is a superposition of the state of an atom taking
path a and the state of an atom taking path b.”
a. Write down a superposition equation reflecting this sentence for
the interference experiment sketched on page 91.
b. Do the same for the interference experiment sketched on page 94.
3.8 Representations (recommended problem)
In the {|z+i, |z−i} basis the state |ψi is represented by
 
ψ+
.
ψ−
(In other words, ψ+ = hz+|ψi and ψ− = hz−|ψi.)
a. If ψ+ and ψ− are both real, show that there is one and only one
axis upon which the projection of µ
~ has a definite, positive value,
and find the angle between that axis and the z axis in terms of
ψ+ and ψ− .
b. What would change if you multiplied both ψ+ and ψ− by the same
phase factor (complex unit)?
c. What would change if you multiplied ψ+ and ψ− by different phase
factors?
This problem invites the question “What if the ratio of ψ+ /ψ− is not
pure real?” When you study more quantum mechanics, you will find
that in this case the axis upon which the projection of µ
~ has a definite,
positive value is not in the x-z plane, but instead has a component in
the y direction as well.
3.9 Addition of states
Some students in your class wonder “What does it mean to ‘add two
quantal states’ ? You never add two classical states.” For the Under-
ground Guide to Quantum Mechanics (see page 20) you decide to write
four sentences interpreting the equation
|ψi = a|z+i + b|z−i (3.34)
describing why you can add quantal states but can’t add classical states.
Your four sentences should include a formula for the amplitude a in
terms of the states |ψi and |z+i.
 Forging Mathematical Tools 111

3.10 Names of six states, in two bases

Write down the representations (the “names”) of the states |z+i, |z−i,
|x+i, |x−i, |θ+i, and |θ−i in (a) the basis {|z+i, |z−i} and in (b) the
basis {|x+i, |x−i}.
3.11 More peculiarities of states
Because a vector pointing down at angle θ is the same as a vector point-
ing up at angle θ − 180◦ , I would expect that |θ−i = |(θ − 180◦ )+i.
Show that this expectation is false by uncovering the true relation be-
tween these two state vectors.
3.12 Translation matrix
(This problem requires background knowledge in the mathematics of
matrix multiplication.)
Suppose that the representation of |ψi in the basis {|z+i, |z−i} is
   
ψ+ hz+|ψi
= .
ψ− hz−|ψi
The representation of |ψi in the basis {|θ+i, |θ−i} is just as good, and
we call it
 0   
ψ+ hθ+|ψi
0 = .
ψ− hθ−|ψi
Show that you can “translate” between these two representations using
the matrix multiplication
 0    
ψ+ cos(θ/2) sin(θ/2) ψ+
0 = .
ψ− − sin(θ/2) cos(θ/2) ψ−
112 States for entangled systems

3.6 States for entangled systems

In the Einstein-Podolsky-Rosen experiment (1) on page 74, with two ver-

tical analyzers, the initial state is represented by |ψi, and various possible
final states are represented by | ↑↓ i and so forth, as shown below. (In this
section all analyzers will be vertical, so we adopt the oft-used convention
that writes |z+i as | ↑ i and |z−i as | ↓ i.)

|ψi

| ↑↓ i

| ↓↑ i

| ↑↑ i

| ↓↓ i

The experimental results tell us that

|h ↑↓ |ψi|2 = 1
2
|h ↓↑ |ψi|2 = 1
2 (3.35)
2
|h ↑↑ |ψi| = 0
|h ↓↓ |ψi|2 = 0.
Additional analysis (sketched in problem 6.8, “Normalization of singlet spin
state”) is needed to assign phases to these amplitudes. The results are
h ↑↓ |ψi = + √12
h ↓↑ |ψi = − √12 (3.36)
h ↑↑ |ψi = 0
h ↓↓ |ψi = 0.
Forging Mathematical Tools 113

Using the generalization of equation (3.31) for a four-state basis, these

results tell us that
|ψi = | ↑↓ ih↑↓ |ψi + | ↓↑ ih↓↑ |ψi + | ↑↑ ih↑↑ |ψi + | ↓↓ ih↓↓ |ψi
= √1 (| ↑↓ i − | ↓↑ i). (3.37)
2

A simple derivation, with profound implications.

3.6.1 State pertains to system, not to atom

In this entangled situation there is no such thing as an “amplitude for the

right atom to exit from the + port,” because the probability for the right
atom to exit from the + port depends on whether the left atom exits the
+ or the − port. The pair of atoms has a state, but the right atom by
itself doesn’t have a state, in the same way that an atom passing through
an interferometer doesn’t have a position and that love doesn’t have a color.
Leonard Susskind8 puts it this way: If entangled states existed in auto
mechanics as well as quantum mechanics, then an auto mechanic might tell
you “I know everything about your car but . . . I can’t tell you anything
about any of its parts.”

3.6.2 “Collapse of the state vector”

Set up this EPR experiment with the left analyzer 100 kilometers from the
source, and the right analyzer 101 kilometers from the source. As soon as
the left atom comes out of its − port, then it is known that the right atom
will come out if its + port. The system is no longer in the entangled state
√1 (| ↑↓ i − | ↓↑ i); instead the left atom is in state | ↓ i and the right atom
2
is in state | ↑ i. The state of the right atom has changed (some say it has
“collapsed”) despite the fact that it is 200 kilometers from the left analyzer
that did the state changing!
This fact disturbs those who hold the misconception that states are
physical things located out in space like nitrogen molecules, because it
seems that information about state has made an instantaneous jump across
200 kilometers. In fact no information has been transferred from left to
right: true, Alice at the left interferometer knows that the right atom will
8 LeonardSusskind and Art Friedman, Quantum Mechanics: The Theoretical Minimum
(Basic Books, New York, 2014) page xii.
114 States for entangled systems

exit the + port 201 kilometers away, but Bob at the right interferome-
ter doesn’t have this information and won’t unless she tells him in some
conventional, light-speed-or-slower fashion.9
If Alice could in some magical way manipulate her atom to ensure that
it would exit the − port, then she could send a message instantaneously.
But Alice does not possess magic, so she cannot manipulate the left-bound
atom in this way. Neither Alice, nor Bob, nor even the left-bound atom
itself knows from which port it will exit. Neither Alice, nor Bob, nor even
the left-bound atom itself can influence from which port it will exit.

3.6.3 Measurement and entanglement

Back in section 2.4, “Light on the atoms” (page 70), we discussed the
character of “observation” or “measurment” in quantum mechanics. Let’s
bring our new machinery concerning quantal states to bear on this situation.
The figure on the next page shows, in the top panel, a potential mea-
surement about to happen. An atom (represented by a black dot) in state
|z+i approaches a horizontal interferometer at the same time that a photon
(represented by a white dot) approaches path a of that interferometer.
We employ a simplified model in which the photon either misses the
atom, in which case it continues undeflected upward, or else the photon
interacts with the atom, in which case it is deflected outward from the
page. In this model there are four possible outcomes, shown in the bottom
four panels of the figure.
After this potential measurement, the system of photon plus atom is
in an entangled state: the states shown on the right must list both the
condition of the photon (“up” or “out”) and the condition of the atom (+
or −).
If the photon misses the atom, then the atom must emerge from the +
port of the analyzer: there is zero probability that the system has final state
|up; −i. But if the photon interacts with the atom, then the atom might
emerge from either port: there is non-zero probability that the system has
9 If you are familiar with gauges in electrodynamics, you will find quantal state similar

to the Coulomb gauge. In the Coulomb gauge, the electric potential at a point in
space changes the instant that any charged particle moves, regardless of how far away
that charged particle is. This does not imply that information moves instantly, because
electric potential by itself is not measurable. The same applies for quantal state.
Forging Mathematical Tools 115

final state |out; −i. These two states are exactly the same as far as the
atom is concerned; they differ only in the position of the photon.
If we focus only on the atom, we would say that something strange has
happened (a “measurement” at path a) that enabled the atom to emerge
from the − port which (in the absence of “measurement”) that atom would
never do. But if we focus on the entire system of photon plus atom, then
it is an issue of entanglement, not of measurement.

b
|z+i
|ψi
a

|up; +i
a

|up; −i
a

|out; +i
a

|out; −i
a
 116 What is a qubit?

Problem
3.13 Amplitudes for “Measurement and entanglement”
Suppose that, in the “simplified model” for measurement and entan-
glement, the probability for photon deflection is 15 . Find the four prob-
abilities |hup; +|ψi|2 , |hup; −|ψi|2 , |hout; +|ψi|2 , and |hout; −|ψi|2 .

3.7 Are states “real”?

This is a philosophical question for which there is no specific meaning and

hence no specific answer. But in my opinion, states are mathematical tools
that enable us to efficiently and accurately calculate the probabilities that
can be found through repeated analyzer experiments, interference experi-
ments, and indeed all experiments. They are not physically “real”.
Indeed, it is possible to formulate quantum mechanics in such a way that
probabilities and amplitudes are found without using the mathematical tool
of “state” at all. Richard Feynman and Albert Hibbs do just this in their
1965 book Quantum Mechanics and Path Integrals. States do not make an
appearance until deep into their book, and even when they do appear they
are not essential. The Feynman “sum over histories” formulation described
in that book is, for me, the most intuitively appealing approach to quantum
mechanics. There is, however, a price to be paid for this appeal: it’s very
difficult to work problems in the Feynman formulation.

3.8 What is a qubit?

At the end of the last chapter (on page 88) we listed several so-called “two-
state systems” or “spin- 12 systems” or “qubit systems”. You might have
found these terms strange: There are an infinite number of states for the
magnetic moment of a silver atom: |z+i, |1◦ +i, |2◦ +i, and so forth. Where
does the name “two-state system” come from? You now see the answer:
it’s short for “two-basis-state system”.
The term “spin” originated in the 1920s when it was thought that an
electron was a classical charged rigid sphere that created a magnetic mo-
ment through spinning about an axis. A residual of that history is that
Forging Mathematical Tools 117

people still call10 the state |z+i by the name “spin up” and by the symbol
| ↑ i, and the state |z−i by “spin down” and | ↓ i. (Sometimes the associa-
tion is made in the opposite way.) Meanwhile the state |x+i is given the
name “spin sideways” and the symbol | → i.
Today, two-basis-state systems are more often called “qubit” systems
from the term used in quantum information processing. In a classical com-
puter, like the ones we use today, a bit of information can be represented
physically by a patch of magnetic material on a disk: the patch magnetized
“up” is interpreted as a 1, the patch magnetized “down” is interpreted as
a 0. Those are the only two possibilities. In a quantum computer, a qubit
of information can be represented physically by the magnetic moment of a
silver atom: the atom in state |z+i is interpreted as |1i, the atom in state
|z−i is interpreted as |0i. But the atom might be in any (normalized) su-
perposition a|1i + b|0i, so rather than two possibilities there are an infinite
number.
Furthermore, qubits can interfere with and become entangled with other
qubits, options that are simply unavailable to classical bits. With more
states, and more ways to interact, quantum computers can only be faster
than classical computers, and even as I write these possibilities are being
explored.
In today’s state of technology, quantum computers are hard to build,
and they may never live up to their promise. But maybe they will.
Chapters 2 and 3 have focused on two-basis-state systems, but of course
nature provides other systems as well. For example, the magnetic moment
of a nitrogen atom (mentioned on page 45) is a “four-basis-state” system,
where one basis is
|z; +2i, |z; +1i, |z; −1i, |z; −2i. (3.38)
In fact, the next chapter shifts our focus to a system with an infinite number
of basis states.

10 The very most precise and pedantic people restrict the term “spin” to elementary

particles, such as electrons and neutrinos. For composite systems like the silver atom
they speak instead of “the total angular momentum J~ of the silver atom in its ground
state, projected on a given axis, and divided by ~.” For me, the payoff in precision is
not worth the penalty in polysyllables.
 118 What is a qubit?

Problem
3.14 Questions (recommended problem)
Update your list of quantum mechanics questions that you started at
problem 1.17 on page 46. Write down new questions and, if you have un-
covered answers to any of your old questions, write them down briefly.

[[For example, one of my questions would be: “I’d like to see a proof
that the global phase freedom mentioned on page 107, which obviously
changes the amplitudes computed, does not change any experimentally
accessible result.”]]
 Chapter 4

The Quantum Mechanics of Position

In the last two chapters we’ve studied the quantum mechanics of a silver
atom’s magnetic moment, and we got a lot out of it: we uncovered the
phenomena of quantization and interference and entanglement; we found
how to use amplitude as a mathematical tool to predict probabilities; we
learned about quantum mechanical states. If all of this makes you feel weak
and dizzy, that’s a good thing: Niels Bohr pointed out that “those who are
not shocked when they first come across quantum theory cannot possibly
have understood it.”1 As profitable as this has been, we knew from the start
(page 47) that eventually we would need to treat the quantum mechanics
of position. Now is the time.
Chapters 2 and 3 treated the atom’s magnetic moment but (to the extent
possible) ignored the atom’s position. This chapter starts off with the
opposite approach: it treats only position and ignores magnetic moment.
Section 4.12, “Position plus spin”, at the end of this chapter welds the two
aspects together.

4.1 Probability and probability density:

One particle in one dimension

A single particle ambivates in one dimension. You know the story of quan-
tum mechanics: The particle doesn’t have a position. Yet if we measure
the position (say, by shining a lamp), then we will find that it has a sin-
gle position. However, because the particle started out without a position,
1 Recalled by Werner Heisenberg in Physics and Beyond (Harper and Row, New York,

1971) page 206.

119
 120 Probability and probability density: One particle in one dimension

there is no way to predict the position found beforehand: instead, quantum

mechanics predicts probabilities.
But what exactly does this mean? There are an infinite number of points
along any line, no matter how short. If there were a finite probability at
each of these points, the total probability would be infinity. But the total
probability must be one! To resolve this essential technical issue, we look
at a different situation involving probability along a line.
You are studying the behavior of ants in an ant farm. (Ant farm: two
panes of glass close together, with sand and ants and ant food between the
panes.) The ant farm is 100.0 cm long. You paint one ant red, and 9741
times you look at the ant farm and measure (to the nearest millimeter) the
distance of the red ant from the left edge of the farm.
You are left with 9741 raw numbers, and a conundrum: how should you
present these numbers to help draw conclusions about ant behavior?
The best way is to conceptually divide the ant farm into bins. Start
with five equal bins: locations from 0.0 cm to 20.0 cm are in the first bin,
from 20.0 cm to 40.0 cm in the second, from 40.0 cm to 60.0 cm in the third,
from 60.0 cm to 80.0 cm in the fourth, and from 80.0 cm to 100.0 cm in
the fifth. Find the number of times the ant was in the first bin, and divide
by the total number of observations (9741) to find the probability that the
ant was in the first bin. Similarly for the other bins. You will produce a
graph like this:

probability of finding ant in bin

0.3

0.2

0.1

0.0
00 20 40 60 80 100
position (cm)

The five probabilities sum to 1, as they must.

 The Quantum Mechanics of Position 121

Now you want more detail about the ant’s location. Instead of dividing
the ant farm into five conceptual bins each of width 20.0 cm, divide it into
ten bins each of width 10.0 cm. The probabilities now look like:

probability of finding ant in bin

0.3

0.2

0.1

0.0
00 10 20 30 40 50 60 70 80 90 100
position (cm)

There are now ten probabilities, yet they still sum to 1, so the probabilities
are each smaller. (For example, the first graph shows a probability of 0.28
for the ant appearing between 0.0 cm and 20.0 cm. The second graph
shows probability 0.18 for the ant appearing between 0.0 cm and 10.0 cm
and probability 0.10 for the ant appearing between 10.0 cm and 20.0 cm.
Sure enough 0.28 = 0.18 + 0.10.)
 122 Probability and probability density: One particle in one dimension

If you want still more detail, you can divide the ant farm into fifty bins,
each of width 2.0 cm, as in:

probability of finding ant in bin

0.3

0.2

0.1

0.0
00 10 20 30 40 50 60 70 80 90 100
position (cm)

These fifty probabilities must still sum to 1, so the individual probabilities

are smaller still.
You could continue this process, making the bins smaller and smaller,
and every bin probability would approach zero. In symbols, if the proba-
bility for appearing in the bin surrounding point x0 is called P0 , and the
width of each bin is called ∆x, we have that
lim P0 = 0
∆x→0
for all points x0 . This is true but provides no information whatsoever about
ant behavior!
To get information, focus not on the bin probability but on the so-called
probability density, defined as
P0
lim ≡ ρ(x0 ).
∆x→0 ∆x
In terms of probability density, we say that “the probability of finding
the ant in a small window of width w centered on x0 is approximately
ρ(x0 )w, and this approximation grows better and better as the window
grows narrower and narrower.” And that “the probability of finding the
ant between xA and xB is
Z xB
ρ(x) dx.”
xA
The Quantum Mechanics of Position 123

The fact that the bin probabilities, summed over all bins, is unity, or in
symbols
X
Pi = 1,
i

becomes, in the limit ∆x → 0,

X X Z 100.0 cm
Pi ≈ ρ(xi )∆x → ρ(x) dx = 1.
i i 0.0 cm

This property of probability densities is called “normalization”.

Problem
4.1 Mean and standard deviation for an ant (essential problem)
The mean2 ant position hxi is given by summing all the position mea-
surements and dividing that sum by the number of measurements (in
this case 9741). Using Pi for the probability of the ant appearing in
bin i and xi for the position of the center of bin i, argue that this mean
position is given approximately by
X
xi P i
bin i

and that the approximation grows better and better as the bins grow
narrower and narrower. The formula becomes exact when ∆x → 0, so
show that the mean value is given by
Z 100.0 cm
hxi = xρ(x) dx. (4.1)
0.0 cm
In the same way, argue that the standard deviation of ant position is
s
Z 100.0 cm
(x − hxi)2 ρ(x) dx. (4.2)
0.0 cm

2 The “mean value” is also called the “average value” and sometimes the “expectation

value”. The latter name is particularly poor. If you toss a die, the mean value of the
number facing up is 3.5. Yet no one expects to toss a die and find the number 3.5 facing
up!
124 Probability amplitude

4.2 Probability amplitude

The probability considerations for one ant walking in one dimension are
directly analogous to the probability considerations for one quantal parti-
cle ambivating in one dimension. A graph with five bins like the one on
page 120 approximates the quantal particle as a five-state system. A graph
with ten bins like the one on page 121 approximates the quantal particle
as a ten-state system. A graph with fifty bins like the one on page 122
approximates the quantal particle as a fifty-state system. You know the
drill of quantum mechanics: in all these cases the bin probability P0 will
be related to some sort of bin amplitude ψ0 through P0 = |ψ0 |2 . How does
bin amplitude behave as ∆x → 0? Because
P0 |ψ0 |2 ψ
= → ρ(x0 ), √ 0 → ψ(x0 ),
we will have
∆x ∆x ∆x
where, for any point x0 , the probability density is
ρ(x0 ) = |ψ(x0 )|2 . (4.3)

What would be a good name for this function ψ(x)? I like the name
“amplitude density”. It’s not really a density: a density
p would have di-
mensions 1/[length], whereas ψ(x) has dimensions 1/ [length]. But it’s
closer to a density than it is to anything else. Unfortunately, someone else
(namely Schrödinger3 ) got to name it before I came up with this sensible
name, and that name has stuck. It’s called “wavefunction”.
The normalization condition for wavefunction is
Z +∞
|ψ(x)|2 dx = 1. (4.4)
−∞

You should check for yourself that this equation is dimensionally consistent.
The global phase freedom described for qubit systems on page 107 ap-
plies for wavefunctions as well: If you multiply any wavefunction by a com-
plex number with magnitude unity — called a “complex unit” or “phase
3 Erwin Schrödinger (1887–1961) had interests in physics, biology, philosophy, and East-

ern religion. Born in Vienna, he held physics faculty positions in Germany, Poland, and
Switzerland. In 1926 he developed the concept of wavefunction and discovered the quan-
tum mechanical time evolution equation (4.12) that now bears his name. This led, in
1927, to a prestigious appointment in Berlin. In 1933, disgusted with the Nazi regime, he
left Berlin for Oxford, England. He held several positions in various cities before ending
up in Dublin. There, in 1944, he wrote a book titled What is Life? which stimulated
interest in what had previously been a backwater of science: biochemistry.
The Quantum Mechanics of Position 125

factor” — then you get a different wavefunction that represents the same
state. You must multiply by a number with magnitude unity, not a function
with magnitude unity: the wavefunctions ψ(x) and eiδ ψ(x) represent the
same state, but the wavefunction eiδ(x) ψ(x) represents a different state.
Keep in mind that to specify a quantal state we must know amplitudes
(wavefunction) rather than merely probabilities (probability density). Just
as in classical mechanics (see page 90) we say that we know a state when we
know all the information needed to describe the system now and to predict
its future. The probability density ρ(x) alone tells you a lot about the
state right now, but cannot predict how the state will change in the future.
Knowing probability density alone in quantum mechanics is like knowing
the position alone in classical mechanics: The probability density gives a
lot of information about now, but not enough information to predict the
future.

Problems
4.2 Mean and standard deviation for a quantal particle
(essential problem)
For any function f (x), define the mean value
Z +∞
hf (x)i = f (x)|ψ(x)|2 dx. (4.5)
−∞
Show that the mean position is hxi and that the standard deviation of
position ∆x is given through
2
(∆x)2 = hx2 i − hxi . (4.6)

4.3 What we say makes no sense. What do we mean?

It sounds strange to say “The particle with wavefunction ψ(x) doesn’t
have a position and its mean
Z position is
+∞
x|ψ(x)|2 dx.”
−∞
Write two or three sentences unpacking what this sentence really means.

4.4 Bump wavefunction

The parabolic bump wavefunction
 is defined as
 0 x<0
ψ(x) = −ax(x − L) 0 ≤ x ≤ L .

0 L<x
Find the mean position and standard deviation of position.
126 How does wavefunction change with time?

4.3 How does wavefunction change with time?

I’m going to throw down three equations. First, the classical formula for
energy,
p2
E= + V, (4.7)
2m
where V is the potential energy. Second, the Einstein and de Broglie rela-
tions for energy and momentum (1.21) and (1.24)
E = ~ω and p = ~k. (4.8)
Third, the particular wavefunction
ψ(x, t) = Aei(kx−ωt) . (4.9)

Plugging equations (4.8) mindlessly into equation (4.7) we obtain

~2 k 2
~ω = +V
2m
and multiplying both sides by ψ(x, t) gives
~2 k 2
~ωψ(x, t) = ψ(x, t) + V ψ(x, t). (4.10)
2m
Meanwhile, the particular wavefunction (4.9) satisfies
1 ∂ψ 1 ∂ψ 1 ∂2ψ ∂2ψ
ωψ(x, t) = ; kψ(x, t) = ; k 2 ψ(x, t) = = − .
−i ∂t i ∂x i2 ∂x2 ∂x2
Plugging these into equation (4.10) gives
~2
   2 
1 ∂ψ ∂ ψ
~ = − 2 + V ψ(x, t), (4.11)
−i ∂t 2m ∂x
which rearranges to
~2 ∂ 2 ψ(x, t)
 
∂ψ(x, t) i
=− − + V (x)ψ(x, t) . (4.12)
∂t ~ 2m ∂x2

I concede from the very start that this is a stupid argument, and that if
you had proposed it to me I would have gone ballistic. First, equation (4.7)
is a classical fact plopped mindlessly into a quantal argument. Second,
the Einstein relation (4.8) applies to photons, not to massive particles.
Third, there are many possible wavefunctions other than equation (4.9).
The unjustified change of potential energy value V in equation (4.11) to
potential energy function V (x) in equation (4.12) merely adds insult to
4.4. Wavefunction: Two particles 127

injury. The only good thing I can say about this equation is that it’s
dimensionally consistent.
Oh, and one more thing. The equation is correct. Despite its dubious
provenance, experimental tests have demonstrated to everyone’s satisfac-
tion that wavefunction really does evolve in time this way. (I must qualify:
wavefunction evolves this way in a wide range of situations: non-relativistic,
no magnetic field, no friction or any other non-conservative force, and where
the particle’s magnetic moment is unimportant.)
This equation for time evolution in quantal systems plays the same cen-
tral role in quantum mechanics that F~ = m~a does in classical mechanics.
And just as F~ = m~a cannot be derived, only motivated and then tested
experimentally, so this time-evolution result cannot be derived. The mo-
tivation is lame, but the experimental tests are impressive and cannot be
ignored.
This time evolution equation has a name: it is “the Schrödinger equa-
tion”.

4.4 Wavefunction: Two particles in one or three dimensions

We will soon work on solving the Schrödinger equation for one particle
in one dimension, but first we ask how to describe two particles in one
dimension.
Two particles, say an electron and a neutron, ambivate in one dimension.
As before, we start with a grid of bins in one-dimensional space:

- ∆x 
 - x
i j

We ask for the probability that the electron will be found in bin i and
the neutron will be found in bin j, and call the result Pi,j . Although
our situation is one-dimensional, this question generates a two-dimensional
array of probabilities.
128 Wavefunction: Two particles

bin of neutron
6

Pi,j
j

 - bin of electron
i

?
To produce a probability density, we must divide the bin probability Pi,j
by (∆x)2 , and then take the limit as ∆x → 0, resulting in
Pi,j
→ ρ(xe , xn ).
(∆x)2
So the probability of finding an electron within a narrow window of width w
centered on xe = 5 and finding the neutron within a narrow window of width
u centered on xn = 9 is approximately ρ(5, 9)wu, and this approximation
grows better and better as the two windows grow narrower and narrower.
2
p Pi,j = |ψi,j | . To turn a bin amplitude
The bin amplitude is ψi,j with
2
into a wavefunction, divide by (∆x) = ∆x and take the limit
ψi,j
lim = ψ(xe , xn ). (4.13)
∆x→0 ∆x
This wavefunction has dimensions 1/[length].
The generalization to more particles and higher dimensionality is
straightforward. For a single electron in three-dimensional space, the wave-
function ψ(~x) has dimensions 1/[length]3/2 . For an electron and a neu-
tron in three-dimensional space, the wavefunction ψ(~xe , ~xn ) has dimensions
1/[length]3 . Note carefully: For a two-particle system, the state is speci-
fied by one function ψ(~xe , ~xn ) of six variables. It is not specified by two
functions of three variables, with ψe (~x) giving the state of the electron and
The Quantum Mechanics of Position 129

ψn (~x) giving the state of the neutron. There are four consequences of this
simple yet profound observation.
First, the wavefunction (like amplitude in general) is a mathematical
tool for calculating the results of experiments; it is not physically “real”.
I have mentioned this before, but it particularly stands out here. Even
for a system as simple as two particles, the wavefunction does not exist in
ordinary three-dimensional space, but in the six-dimensional “configuration
space”, as it is called. I don’t care how clever or talented an experimentalist
you are: you cannot insert an instrument into six-dimensional space in order
to measure wavefunction.
Second, wavefunction is associated with a system, not with a particle. If
you’re interested in a single electron and you say “the wavefunction of the
electron”, then you’re technically incorrect — you should say “the wave-
function of the system consisting of a single electron” — but no one will go
ballistic and say that you are in thrall of a deep misconception. However,
if you’re interested in a pair of particles (an electron and a neutron, for
instance) and you say “the wavefunction of the electron”, then someone
(namely me) will go ballistic because you are in thrall of a deep misconcep-
tion.
Third, it might happen that the wavefunction factorizes:
ψ(~xe , ~xn ) = ψe (~xe )ψn (~xn ).
In this case the electron has state ψe (~xe ) and the neutron has state ψn (~xn ).
Such a peculiar case is called “non-entangled”. But in all other cases the
state is called “entangled” and the individual particles making up the sys-
tem do not have states. The system has a state, namely ψ(~xe , ~xn ), but
there is no state for the electron and no state for the neutron, in exactly
the same sense that there is no position for a silver atom ambivating through
an interferometer.
Fourth, quantum mechanics is intricate. To understand this point, con-
trast the description needed in classical versus quantum mechanics.
How does one describe the state of a single classical particle moving
in one dimension? It requires two numbers: a position and a velocity.
Two particles moving in one dimension require merely that we specify the
state of each particle: four numbers. Similarly specifying the state of three
particles require six numbers and N particles require 2N numbers. Exactly
the same specification counts hold if the particle moves relativistically.
130 Wavefunction: Two particles

How, in contrast, does one describe the state of a single quantal par-
ticle ambivating in one dimension? Here an issue arises at the very start,
because the specification is given through a complex-valued wavefunction
ψ(x). Technically the specification requires an infinite number of numbers!
Let’s approximate the wavefunction through its value on a grid of, say, 100
points. This suggests that a specification requires 200 real numbers, a com-
plex number at each grid point, but global phase freedom means that we
can always set one of those numbers to zero through an overall phase factor,
and one number is not independent through the normalization requirement.
The specification actually requires 198 independent real numbers.
How does one describe the state of two quantal particles ambivating
in one dimension? Now the wavefunction is a function of two variables,
ψ(xe , xn ). The wavefunction of the system is a function of two-dimensional
configuration space, so an approximation of the accuracy established previ-
ously requires a 100×100 grid of points. Each grid point carries one complex
number, and again overall phase and normalization reduce the number of
real numbers required by two. For two particles the specification requires
2 × (100)2 − 2 = 19 998 independent real numbers.
Similarly, specifying the state of N quantal particles moving in one
dimension requires a wavefunction in N -dimensional configuration space
which (for a grid of the accuracy we’ve been using) is specified through
2 × (100)N − 2 independent real numbers.
The specification of a quantal state not only requires more real numbers
than the specification of the corresponding classical state, but that number
increases exponentially rather than linearly with particle number N .
The fact that a quantal state holds more information than a classical
state is the fundamental reason that a quantal computer can be (in prin-
ciple) faster than a classical computer, and the basis for much of quantum
information theory.
Relativity is different from classical physics, but no more complicated.
Quantum mechanics, in contrast, is both different from and richer than
classical physics. You may refer to this richness using terms like “splendor”,
or “abounding”, or “intricate”, or “ripe with possibilities”. Or you may
refer to it using terms like “complicated”, or “messy”, or “full of details
likely to trip the innocent”. It’s your choice how to react to this richness,
but you can’t deny it.
4.5. Solving the Schrödinger equation for the I.S.W. 131

4.5 Solving the Schrödinger time evolution equation

for the infinite square well

Setup. A single particle is restricted to one dimension. In classical me-

chanics, the state of the particle is given through position and velocity:
that is, we want to know the two functions of time
x(t); v(t).
These functions stem from the solution to the ordinary differential equation
P~
(ODE) F = m~a, or, in this case,
d2 x(t) 1
2
= F (x(t)) (4.14)
dt m
subject to the given initial conditions
x(0) = x0 ; v(0) = v0 .

In quantum mechanics, the state of the particle is given through the

wavefunction: that is, we want to know the two-variable function
ψ(x, t).
This is the solution of the Schrödinger partial differential equation (PDE)
~2 ∂ 2 ψ(x, t)
 
∂ψ(x, t) i
=− − + V (x)ψ(x, t) , (4.15)
∂t ~ 2m ∂x2
subject to the given initial condition
ψ(x, 0) = ψ0 (x).

[[The classical time evolution equation (4.14) is second order in time,

so there are two initial conditions: initial position and initial velocity. The
quantal time evolution equation (4.15) is first order in time, so there is only
one initial condition: initial wavefunction.]]
Infinite square well. Since this is our first quantal time evolution
problem, let’s start out cautiously by choosing the easiest potential energy
function: the so-called infinite square well4 or “particle in a box”:

 ∞ for x ≤ 0
V (x) = 0 for 0 < x < L
∞ for L ≤ x


4 Any potential energy function with a minimum is called a “well”.

132 Solving the Schrödinger equation for the I.S.W.

This is an approximate potential energy function for an electron added to

a hydrocarbon chain molecule (a “conjugated polymer”), or for an atom
trapped in a capped carbon nanotube.
The infinite square well is like the “perfectly rigid cylinder that rolls
without slipping” in classical mechanics. It does not exactly exist in reality:
any cylinder will be dented or cracked if hit hard enough. But it is a good
model for some real situations. And it’s certainly better to work with this
model than it is to shrug your shoulders and say “I have no idea.”

V (x)

ψ(x)
x
0 L

The infinite square well potential energy function V (x) in olive green, and
a possible wavefunction ψ(x) in red.
It is reasonable (although not rigorously proven) that for the infinite
square well

0 for x ≤ 0
ψ(x, t) = something for 0 < x < L
for L ≤ x

0
and we adopt these conditions.
Strategy. The PDE is linear, so if we find some special solutions
f1 (x, t), f2 (x, t), f3 (x, t), . . . , then we can generate many more solutions
through
X
Dn fn (x, t),
n
where D1 , D2 , D3 , . . . represent constants. Because any Dn can be any
possible complex number, this is a big set of solutions; indeed it might be
a big enough set to be the most general solution. Once we have the most
general solution, we will need to find the values of Dn that correspond to
the particular initial condition ψ(x, 0).
The Quantum Mechanics of Position 133

Casting about for special solutions: separation of variables. So,

how do we find even one solution of the PDE? Let’s try a solution f (x, t)
that is a product of a function X(x) of position alone and a function T (t)
of time alone, that is, try a solution of the form
f (x, t) = X(x)T (t).
Plugging this guess into the PDE, we find
~2 d2 X(x)
 
dT (t) i
X(x) =− − T (t) + V (x)X(x)T (t) ,
dt ~ 2m dx2
where the partial derivatives have become ordinary derivatives because they
now act upon functions of a single variable. Divide both sides by X(x)T (t)
to find
~2 1 d2 X(x)
 
1 dT (t) i
=− − + V (x) .
T (t) dt ~ 2m X(x) dx2
In this equation, there is a function of time alone on the left-hand side, and
a function of position alone on the right-hand side. But time and position
are independent variables. It seems as if the left-hand side will vary with
time, even while the position is held constant so the right-hand side stays
constant! Similarly the other way around. There is only one way a function
of t alone can always be equal to a function of x alone, and that’s if both
sides are equal to the same constant.
We don’t yet know what that constant is, or how many such constants
there might be. To allow for the possibility that there might be many such
constants, we call the constant value of the quantity in square brackets by
the name En . (This name suggests, correctly, that this constant must have
the dimensions of energy.) We conclude that
1 dT (t) i
= − En (4.16)
T (t) dt ~
~2 1 d2 X(x)
− + V (x) = En .
2m X(x) dx2
We started with one partial differential equation in two variables but (for
solutions of the form f (x, t) = X(x)T (t)) ended with two ordinary differen-
tial equations. And we know a lot about how to solve ordinary differential
equations! This technique for finding special solutions of the PDE is called
“separation of variables”.
Solving the first ODE. When faced with solving two equations, I
always solve the easy one first. That way, if the result is zero, I won’t have
to bother solving the second equation.
134 Solving the Schrödinger equation for the I.S.W.

So first we try to solve

1 dT (t) i
= − En
T (t) dt ~
dT (t) i
= − En dt
T (t) ~
Z Z
dT i
= − En dt
T ~
i
ln T = − En (t + constant )
~
Tn (t) = T0 e−(i/~)En t
Well, that went well. I don’t know about you, but it was easier than I
expected. In the last step, I changed the name of T (t) to Tn (t) to reflect
the fact that we get different solutions for different values of En .
Solving the second ODE. We move on to
~2 1 d2 X(x)
− + V (x) = En .
2m X(x) dx2
Remembering the form of the infinite square well potential, and the bound-
ary conditions ψ(0, t) = 0 plus ψ(L, t) = 0, the problem to solve is
~2 d2 X(x)
− = En X(x) with X(0) = 0; X(L) = 0. (4.17)
2m dx2
Perhaps you regard this sort of ordinary differential equation as unfair.
After all, you don’t yet know the permissible values of En . I’m not just
asking you to solve an ODE with given coefficients, I’m asking you find find
out what the coefficients are! Fair or not, we plunge ahead.
You are used to solving differential equations of this form. If I wrote
d2 f (t)
M = −kf (t),
dt2
you’d respond: “Of course, this is the ODE for a classical mass on a spring!
The solution is
p
f (t) = C cos(ωt) + D sin(ωt) where ω = k/M .”
Well, then, the solution for X(x) has to be
p
Xn (x) = Cn cos(ωx) + Dn sin(ωx) where ω = 2mEn /~2 ,
where again I have taken to calling X(x) by the name Xn (x) to reflect the
fact there there are different solutions for different values of En . Writing
this out neatly,
p p
Xn (x) = Cn cos(( 2mEn /~)x) + Dn sin(( 2mEn /~)x). (4.18)
The Quantum Mechanics of Position 135

When you solved the classical problem of a mass on a spring, you had to
supplement the ODE solution with the initial values f (0) = x0 , f 0 (0) = v0 ,
to find the constants C and D. This is called an “initial value problem”. For
the problem of a particle in a box, we don’t have an initial value problem;
instead we are given Xn (0) = 0 and Xn (L) = 0, which is called a “boundary
value problem”.
Plugging x = 0 into equation (4.18) will be easier than plugging in
x = L, so I’ll do that first. The result gives
Xn (0) = Cn cos(0) + Dn sin(0) = Cn ,
so the boundary value Xn (0) = 0 means that Cn = 0 — for all values of n!
Thus
p
Xn (x) = Dn sin(( 2mEn /~)x). (4.19)

Now plug x = L into equation (4.19), giving

p
Xn (L) = Dn sin(( 2mEn /~)L),
so the boundary value Xn (L) = 0 means that
√
2mEn
L = nπ where n = 0, ±1, ±2, ±3, . . .
~
and it follows that
Xn (x) = Dn sin((nπ/L)x).
If you think about it for a minute, you’ll realize that n = 0 gives rise to
X0 (x) = 0. True, this is a solution to the differential equation, but it’s not
an interesting one. Similarly, the solution for n = −3 is just the negative
of the solution for n = +3, so we get the same effect by changing the sign
of D3 . We don’t have to worry about negative or zero values for n.
In short, the solutions for the boundary value problem are
Xn (x) = Dn sin(nπx/L) where n = 1, 2, 3, . . .
and with
π 2 ~2
En = n2 .
2mL2
We have accomplished the “unfair”: we have not only solved the differential
equation, we have also determined the permissible values of En .
136 Solving the Schrödinger equation for the I.S.W.

Pulling things together. We now know that a solution to the

Schrödinger time evolution equation for the infinite square well of width
L is
∞
X
ψ(x, t) = Dn e−(i/~)En t sin(nπx/L),
n=1
where
π 2 ~2
En = n2 .
2mL2
This is a lot of solutions — there are an infinite number of adjustable
parameters Dn , after all! — but the question is whether it is the most
general solution. In fact it is the most general solution, although that’s
not obvious. The branch of mathematics devoted to such questions, called
Sturm-Liouville5 theory, is both powerful and beautiful, but this is not the
place to explore it.
Fitting the initial conditions. Remember that our problem is not
simply to solve a PDE, it is to find how a given initial wavefunction
ψ(x, 0) = ψ0 (x)
changes with time. To do this, we fit our solution to the given initial
conditions.
To carry out this fitting, we must find Dn such that
X∞
ψ(x, 0) = Dn sin(nπx/L) = ψ0 (x).
n=1
The problem seems hopeless at first glance, because there are an infinite
number of unknowns Dn , yet only one equation! But there’s a valuable
trick, worth remembering, that renders it straightforward.
The trick relies on the fact that, for n, m integers,
Z L 
L/2 for n = m
sin(nπx/L) sin(mπx/L) dx = . (4.20)
0 0 for n 6= m
You can work this integral out for yourself, using either
sin A sin B = 21 [cos(A − B) − cos(A + B)]
5 Charles-François Sturm (1803–1855), French mathematician, also helped make the first

experimental determination of the speed of sound in water. Joseph Liouville (1809–1882),

another French mathematician, made contributions in complex analysis, number theory,
differential geometry, and classical mechanics. He was also a public servant elected to
the French Constituent Assembly of 1848, which established the Second Republic.
The Quantum Mechanics of Position 137

or else
e+iθ − e−iθ
sin θ = ,
2i
whichever you like better. (Or you can look at problem 4.5, “Informal inte-
gration”, on page 138, for an informal but easily remembered treatment.)
To employ this fact, start with
∞
X
Dn sin(nπx/L) = ψ0 (x),
n=1

multiply both sides by sin(mπx/L), and integrate from 0 to L:

X∞ Z L Z L
Dn sin(nπx/L) sin(mπx/L) dx = ψ0 (x) sin(mπx/L) dx.
n=1 0 0

This looks even worse, until you realize that all but one of the terms on the
left vanish! Once you do make that realization, you find
Z L
Dm (L/2) = ψ0 (x) sin(mπx/L) dx
0
and you have a formula for Dm .
Pulling all things together. For a particle of mass m ambivating
in an infinite square well of width L, how does the quantal wave function
change (“evolve”) with time? If the initial wavefunction is ψ0 (x), then the
wavefunction at time t is
X∞
ψ(x, t) = Dn e−(i/~)En t sin(nπx/L), (4.21)
n=1

where
π 2 ~2
En = n2 (4.22)
2mL2
and
Z L
2
Dn = ψ0 (x) sin(nπx/L) dx. (4.23)
L 0
138 Solving the Schrödinger equation for the I.S.W.

Problem
4.5 Informal integration (recommended problem)
The integral (4.20) undergirds the Fourier sine series technique, and
it’s useful to remember. Here’s how I do it. If n 6= m the integrand is
sometimes positive, sometimes negative over its range from 0 to L, so
it’s plausible that the two signs cancel out and result in a zero integral.
If n = m the integrand is always positive, so it must not be zero. But
what is it?
a. Below is a graph of sin2 (3πx/L).

sin2 (3πx/L)
6
1

0 - x

0 L

What is the area within the dashed box? Does it look like the area
above the curve within the box is the same as the area below the
curve within the box? What can you conclude about the value of
the integral
Z L
sin2 (3πx/L) dx ?
0
b. Make the above argument rigorous using the relationship
sin2 θ + cos2 θ = 1.
Your result should be that the integral
Z L
sin2 (nπx/L) dx
0
has the same value whenever n is a non-zero integer.
4.6. What did we learn? 139

4.6 What did we learn by solving

the Schrödinger time evolution equation
for the infinite square well?

In one sense, we learned that the time evolution of a particle of mass m in

an infinite square well of width L, with initial wavefunction ψ0 (x), is given
by equation (4.21). But we should delve more deeply than simply saying
“There’s the answer, now let’s go to sleep.”
Quantal revivals. Does this jumble of symbols tell us anything about
nature? Does it have any peculiar properties? Here’s one. Suppose there
were a time Trev such that
e−(i/~)En Trev = 1 for n = 1, 2, 3, . . .. (4.24)
What would the wavefunction ψ(x, t) look like at time t = Trev ? It would
be exactly equal to the initial wavefunction ψ0 (x)! If there is such a time,
it’s called the “revival time”.
But it’s not clear that such a revival time exists. After all, equa-
tion (4.24) lists an infinite number of conditions to be satisfied for revival
to occur. Let’s investigate. Because e−i 2π integer = 1 for any integer, the
revival conditions (4.24) are equivalent to
(1/~)En Trev = 2π(an integer) for n = 1, 2, 3, . . ..
Combined with the separation constant values (4.22), these conditions are
π~
n2 Trev = (an integer) for n = 1, 2, 3, . . ..
4mL2
And, looked at this way, it’s clear that yes, there is a time Trev that satisfies
this infinite number of conditions. The smallest such time is
4mL2
Trev = . (4.25)
π~
Cute and unexpected! This behavior is packed into equations (4.21) and
(4.22), but no one would have uncovered this from a glance.
140 What did we learn?

Moving across a node. Think about the wavefunction

D sin(3πx/L).
This wavefunction and corresponding probability density are graphed below
the infinite square well potential energy function.

V (x)

η(x)
x
0 L

|η(x)|2

This particular wavefunction has two interior zeros, also called nodes. A
common question is “There is zero probability of finding the particle at
the node, so how can it move from one side of the node to the other?”
People who ask this question suffer from the misconception that the particle
is an infinitely small, infinitely hard version of a classical marble, which
hence has a definite position. They think that the definite position of
this infinitely small marble is changing rapidly, or changing erratically, or
changing unpredictably, or changing subject to the slings and arrows of
outrageous fortune. In truth, the quantal particle in this state doesn’t have
a definite position: it doesn’t have a position at all! The quantal particle
in the state above doesn’t, can’t, change its position from one side of the
node to the other, because the particle doesn’t have a position.
Investigating the solution technique. But I want to do more than
investigate the properties of the solution, I want to investigate the charac-
teristics of the solution technique. In his book Mathematics in Action,
O. Graham Sutton writes that “A technique succeeds in mathematical
The Quantum Mechanics of Position 141

physics, not by a clever trick, or a happy accident, but because it expresses

some aspect of a physical truth.” What aspect of physical truth is exposed
through the techniques we developed to solve this time evolution problem?
First, let’s review the problem we solved, then the techniques we used.
The problem was solving the partial differential equation
~2 ∂ 2 ψ(x, t)
 
∂ψ(x, t) i
=− − + V (x)ψ(x, t) ,
∂t ~ 2m ∂x2
subject to the initial condition
ψ(x, 0) = ψ0 (x).
The three techniques used were:

(1) Finding many particular solutions of the PDE that happen to factorize:
f (x, t) = X(x)T (t) (“separation of variables”).
(2) Summing all of these particular solutions to find a more general (and,
P
as it turns out, the most general) PDE solution: Dn Xn (x)Tn (t)
(“superposition”).
(3) Finding the coefficients Dn that match up to initial value ψ0 (x)
(“Fourier6 sine series”).

Fourier sine series. Let’s look at the last step first. The technique of
Fourier sine series is generally powerful. Any function f (x) with f (0) = 0
and f (L) = 0 can be expressed as
∞
2 L
X Z
f (x) = fn sin(nπx/L) where fn = f (x) sin(nπx/L) dx.
n=1
L 0
This seems paradoxical: complete information about the function is ob-
tained through knowing f (x) at every real number 0 ≤ x ≤ L. Alterna-
tively, complete information about the function is obtained through know-
ing the coefficients fn for every positive integer n. But there are more real
numbers between 0 and L than there are positive integers! I have no res-
olution for this paradox — I’ll just remark that in knowing the function
through its Fourier coefficients fn , it seems that we’re getting something
for nothing.
6 Joseph Fourier (1768–1830) was French, so his name is pronounced “Four - e - a” with

a silent “r”. He arrived at the series which today bears his name through studies of heat
flow. He was the first to propose the phenomenon that we today call “the greenhouse
effect”. (So much for the climate-change denialist claim that the greenhouse effect is a
modern day liberal/Chinese hoax.)
142 What did we learn?

Well, there are lots of times when we want to get something for nothing!
Fourier sine series are useful in data compression. For example, suppose
you want to record a sound that starts with silence at time 0, proceeds
through several notes, then ends with silence at time L. You could do this
by keeping track of the air pressure f (t) at every instant from 0 to L, or
you could do it by keeping track of the corresponding Fourier coefficients
fn . In either case an infinite amount of data are required, so some will have
to be thrown out to let it fit within a finite computer. It is more efficient to
store this information in the form of fn than in the form of f (t): for a given
amount of storage space, the fn provide a more accurate reproduction of
the sound than the f (t). There are many schemes for the details of exactly
when the Fourier series should be truncated: one such scheme is called
“MP3”.
Or, for pictures rather than sounds: A black-and-white photograph is a
two-dimensional intensity function f (x, y). You could store the image on a
computer by breaking space (x, y) into a grid (“pixels”) and storing a value
for the intensity at each grid point (the so-called bitmap or BMP format)
or you could store the information through Fourier coefficients fn,m (the
so-called JPEG format). For a given level of image quality, the JPEG file
is considerably smaller than the BMP file.
Stationary states. Okay, this is fun and profitable, but it tells us
about how clever humans are; it doesn’t tell us anything about nature. I’m
going to probe in another direction: We see that, as far as time evolution is
concerned, functions like sin(nπx/L) play a special role. What if the initial
wavefunction ψ0 (x) happens to have this form? We investigate n = 3.
Once you see how things work in this case, you can readily generalize to
any positive integer n.
So the initial wavefunction is
ψ0 (x) = A sin(3πx/L).
We need the constant A so that the initial wavefunction will (1) have di-
mensions and (2) be normalized. For all wavefunctions, the probability of
being somewhere is 1, that is
Z +∞
|ψ(x)|2 dx = 1.
−∞

This requirement is called “normalization”. Applying the general normal-

ization requirement to this initial wavefunction for our particle in a box
The Quantum Mechanics of Position 143

results in
Z L
A2 sin2 (3πx/L) dx = 1,
0
whence (remembering the sine-squared integral 4.20)
p
A2 (L/2) = 1 so A = 2/L.
Notice that then r
2
ψ0 (x) = sin(3πx/L)
L
has the proper dimensions.
Well, for this initial wavefunction, what are the values of
2 L
Z
Dn = ψ0 (x) sin(nπx/L) dx ?
L 0
They are
r Z
2 2 L
Dn = sin(3πx/L) sin(nπx/L) dx
L L 0
r 
2 2 L/2 for n = 3
= ×
L L 0 for n 6= 3
r 
2 1 for n = 3
= × ,
L 0 for n 6= 3
so r
2 −(i/~)E3 t
ψ(x, t) = e sin(3πx/L). (4.26)
L
That’s it! For this particular initial wavefunction, the system remains al-
ways in that same wavefunction, except multiplied by an time-dependent
phase factor of e−(i/~)E3 t . This uniform phase factor has no effect whatso-
ever on the probability density! Such states are called “stationary states”.
Generic states. Contrast the time evolution of stationary states with
the time evolution of generic states. For example, suppose the initial wave-
function were r r
4 2 3 2
ψ0 (x) = sin(3πx/L) + sin(7πx/L).
5 L 5 L
How does this state change with time? You should check two things: First,
the wavefunction ψ0 (x) given here is normalized. Second, it evolves in time
to r r
4 2 −(i/~)E3 t 3 2 −(i/~)E7 t
ψ(x, t) = e sin(3πx/L) + e sin(7πx/L).
5 L 5 L
(4.27)
144 What did we learn?

Although it takes a little effort to see exactly how the probability density
changes with time, it’s clear from a glance that it does change with time.
This is not a stationary state.
Let’s go back to the Einstein relation
E = ~ω.
Neither Einstein nor de Broglie was ever clear about what it was that was
“oscillating” with frequency ω, but now we have a better idea. In stationary
state (4.26), the amplitude at every point oscillates with frequency E3 /~.
Using the Einstein relation, we say this state has energy E3 .
In contrast, the amplitude in generic state (4.27) has no single oscilla-
tion: there’s a combination of frequency E3 /~ and frequency E7 /~. This
state doesn’t have an energy, in the same way that a silver atom with
µx = +µB doesn’t have a value of µz , in the same way that an atom in
state |z+i passing through a horizontal interferometer doesn’t have a posi-
tion, in the same way that love doesn’t have a color. Instead, this state has
amplitude 54 to have energy E3 and amplitude 35 to have energy E7 .
We have uncovered the “aspect of physical truth” expressed by the
separation constant En .
Energy eigenstates. How did the remarkable stationary states come
about? Remember how they arose mathematically: we looked for solutions
to
~2 d2 Xn (x)
− + V (x)Xn (x) = En Xn (x),
2m dx2
and the solutions we found (for the infinite square well) were those functions
Xn (x) = sin(nπx/L)
that we later used as building blocks to build up any wavefunction. These
now seem important enough that they warrant their own name. Because
each is associated with a particularly energy En we call them “energy
states”. Because wavefunctions are usually represented by Greek letters
we give them the name ηn (x) where the Greek letter η (eta) suggests “en-
ergy” through alliteration. We write
~2 d2 ηn (x)
− + V (x)ηn (x) = En ηn (x), (4.28)
2m dx2
and recognize this as one of those “unfair” problems where you must find not
only the ODE solution ηn (x), but you must also find the value of En . Such
The Quantum Mechanics of Position 145

problems are given a half-German, half-English name: eigenproblems. We

say that the “energy eigenfunction” ηn (x) represents a “stationary state”,
or an “energy state”, or an “energy eigenstate”, and that En is an “energy
eigenvalue”. (The German word eigen derives from the same root as the
English word “own”, as in “my own state”. It means “characteristic of” or
“peculiar to” or “belonging to”. The eigenstate η3 (x) is the state “belonging
to” energy E3 .)
The normalized energy eigenstate η3 (x) is
r
2
η3 (x) = sin(3πx/L).
L
We saw at equation (4.26) that this energy eigenstate evolves in time to
ψ(x, t) = e−(i/~)E3 t η3 (x). (4.29)
This state “belongs to” the energy E3 . In contrast, the state
ψ(x, t) = 45 e−(i/~)E3 t η3 (x) + 35 e−(i/~)E7 t η7 (x) (4.30)
does not “belong to” any particular energy, because it involves both E3 and
E7 . Instead, this state has amplitude 45 to have energy E3 and amplitude
3
5 to have energy E7 . We say that this state is a “superposition” of the
energy states η3 (x) and η7 (x).
A particle trapped in a one-dimensional infinite square well cannot have
any old energy: the only energies possible are the energy eigenvalues E1 ,
E2 , E3 , . . . given in equation (4.22).
From the very first page of the very first chapter of this book we have
been talking about quantization. But always before it has been a supple-
ment added to the theory to make the results come out right: Max Planck
added energy quantization to the otherwise-classical theory of blackbody
radiation (equation 1.9); Niels Bohr supplemented the theory of a point-like
classical electron orbiting a point-like classical proton with the requirement
that the circular orbit contain a quantized (integer) number of de Broglie
wavelengths (equation 1.26). Here, for the first time, quantization comes
out of the theory, rather than being shoehorned into the beginning of the
theory. Here, for the first time, the theory predicts that the only possible
energies are those listed in equation (4.22). We have reached a milestone
in our development of quantum mechanics.
146 What did we learn?

Because the only possible energies are the energy eigenvalues E1 , E2 , E3 ,

. . ., some people get the misimpression that the only possible states are the
energy eigenstates η1 (x), η2 (x), η3 (x), . . .. That’s false. The state (4.30),
for example, is a superposition of two energy states with different energies.
Analogy. A silver atom in magnetic moment state |z+i enters a vertical
interferometer. It passes through the upper path. While traversing the
interferometer, this atom has a position.
A different silver atom in magnetic moment state |x−i enters that same
vertical interferometer. It ambivates through both paths. In more detail
(see equation 3.19), it has amplitude hz+|x−i = − √12 to take the upper
path and amplitude hz−|x−i = √12 to take the lower path, but it doesn’t
take a path. While traversing the interferometer, this atom has no position
in the same way that love has no color.
A particle trapped in an infinite square well has state η6 (x). This par-
ticle has energy E6 .
A different particle trapped in that same infinite square well has state
√1 η3 (x) − √1 η4 (x).
2 2

This particle does not have an energy. In more detail, it has amplitude √12
to have energy E3 and amplitude − √12 to have energy E4 , but it doesn’t
have an energy in the same way that love doesn’t have a color.
Summary. Our journey into quantum mechanics started with the ex-
perimental fact of quantized energies of blackbody radiation or of an atom.
This inspired a search for quantized values of µz , which in turn prompted
discovery of the new phenomena of interference and entanglement. In-
terference experiments suggested the mathematical tool of amplitude, and
generalizing amplitude from magnetic moment to position prompted the
mathematical tool of wavefunction. We asked the obvious question of how
wavefunction changed with time, and answering that question brought us
back to energy quantization with deeper insight. As T.S. Eliot wrote,

We shall not cease from exploration

And the end of all our exploring
Will be to arrive where we started
And know the place for the first time.
The Quantum Mechanics of Position 147

Problems
4.6 Revival
In an infinite square well, any wavefunction comes back to itself after
the revival time given in equation (4.25) has passed. What happens
after one-half of this time has passed?
4.7 Normalized for all time
Show that the wavefunction (4.27) is normalized for all values of the
time t.
4.8 Zero-point energy
The lowest possible energy for a classical infinite square well is zero.
The lowest possible energy for a quantal infinite square well is E1 as
given in equation (4.22). This difference is called the “zero-point en-
ergy” or the “vacuum energy”.
Your grandparents are intelligent and thoughtful but have little back-
ground in science. They hold a bank trust fund for your eventual
benefit. They have learned that on 27 May 2008, U.S. Patent 7379286
for “Quantum Vacuum Energy Extraction” was issued to the Jovion
Corporation and they know that, if zero-point energy could be har-
nessed, it would produce enormous societal and financial gains. Your
grandparents are thinking of withdrawing the trust fund money from
the bank and investing it in the Jovion Corporation, but they want
your advice before making the investment. What do you tell them?
(Remember their intelligence: they want not just advice but concise,
cogent reasoning behind that advice.)
4.9 Fourier sine series for tent (recommended problem)
Suppose the initial wavefunction is a pure real tent:


 0 x<0
Ax 0 ≤ x ≤ L/2

ψ0 (x) = .

 A(L − x) L/2 ≤ x ≤ L

0 L<x
a. Sketch this initial wavefunction.
b. Show that, to insure normalization, we must use
r
2 3
A= .
L L
c. Verify that ψ0 (x) has the proper dimensions.
 148 What did we learn?

d. The Fourier expansion coefficients are

2 L
Z
Dn = ψ0 (x) sin(nπx/L) dx
L 0
"Z #
L/2 Z L
2A
= x sin(nπx/L) dx + (L − x) sin(nπx/L) dx .
L 0 L/2

Before rushing in to evaluate this integral, pause to think! With-

out evaluating the integrals, show that when n is even the second
integral within square brackets is the negative of the first integral,
whereas when n is odd these two integrals are equal. Consequently,
for this particular ψ0 (x), when n is even Dn = 0, while when n is
odd
4 L/2
Z
Dn = ψ0 (x) sin(nπx/L) dx.
L 0
e. Now go ahead and evaluate Dn for n odd. (I used the substitution
θ = nπx/L, but there are other ways to do it.)
f. Write out the Fourier sine series representation for ψ0 (x). Check
that it has the proper dimensions, that it satisfies ψ0 (0) = ψ0 (L) =
0, and that it satisfies ψ0 (L/2) = AL/2. For this last check, use
the result
1 1 1 π2
1+ 2
+ 2 + 2 + ··· = .
3 5 7 8

4.10 Find the flaw: Fourier sine series for ramp7

After working the above problem four students — Aldo, Beth, Celine,
and Denzel — decide to find the Fourier sine series representation of
the ramp wavefunction

0 x<0
ψ0 (x) = Ax 0 ≤ x < L .
L≤x

0
They split up to work independently, and when they get together af-
terwards they find that they have produced four different answers!
7 Background concerning “find the flaw” type problems is provided in sample prob-

lem 1.2.1 on page 26.

4.7. Other potentials 149

r ∞
2 3 X (−1)n
Aldo: − sin(nπx/L)
π L3 n=1 n
r ∞
2 3 X (−1)n
Beth: − cos(nπx/L)
π L n=1 n
r
2 3 X (−1)n
Celine: − sin(nπx/L)
π L n=1,3,5,··· n
r ∞
2 3 X (−1)n
Denzel: − sin(nπx/L)
π L n=1 n

Provide simple reasons showing that three of these candidates must be

wrong.

4.7 Other potentials

Is quantization peculiar to the infinite square well? No. At this stage in

your mathematical education you don’t have the tools to prove it, but in
fact the infinite square well is entirely generic.
For any one-dimensional potential energy function that has V (x) → ∞
when x → ±∞, there are energy eigenstates ηn (x) with discrete energy
eigenvalues En such that
~2 d2 ηn (x)
− + V (x)ηn (x) = En ηn (x), n = 1, 2, 3, . . . (4.31)
2m dx2
and with the property that
Z +∞ 
∗ 1 for n = m
ηm (x)ηn (x) dx = . (4.32)
−∞ 0 for n 6= m
A side note on vocabulary is that this property is called “orthonormal-
ity”. A side note on notation is that the symbol on the right-hand side of
equation (4.32) defines the “Kronecker8 delta”:

1 for n = m
δn,m ≡ . (4.33)
0 for n 6= m
8 Leopold Kronecker (1823–1891), German mathematician. After earning his Ph.D. he

spent a decade managing a farm, which made him financially comfortable enough that
he could pursue mathematics research for the rest of his life as a private scholar without
university position.
150 Other potentials

Any wavefunction can be expressed as a “superposition” or “linear com-

bination” of energy states:
∞
X
ψ0 (x) = cn ηn (x) (4.34)
n=1

where
Z +∞
cn = ηn∗ (x)ψ0 (x) dx. (4.35)
−∞

We say that the energy states “span the set of wavefunctions” or that
they constitute a “basis”, but don’t let these fancy terms bamboozle you:
they just mean that starting from the energy states, you can use linear
combinations to build up any wavefunction. The basis states in quantum
mechanics play the same role as building blocks in a child’s construction
set. Just as a child can build castles, roadways, or trees — anything she
wants — out of building blocks, so you can build any wavefunction you
want out of energy states.
Superposition is the mathematical reflection of the physical phenomenon
of interference. An example of quantal interference is “the atom passing
through an interferometer doesn’t take either path; instead it has amplitude
ca to take path a and amplitude cb to take path b”. An example of super-
position is “the particle with wavefunction ψ0 (x) given in equation (4.34)
doesn’t have an energy; instead it has amplitude cn to have energy En ,
with n = 1, 2, 3, . . .”. It requires no great sophistication to see that these
are parallel statements.
In the infinite square well, each energy eigenstate has a different en-
ergy. This is not always true: it might happen that two energy eigenvalues,
perhaps E8 and E9 , are equal. In this situation any linear combination
α η8 (x) + β η9 (x) (4.36)
is again an energy eigenstate with energy E8 = E9 . (Although, to insure
normalization, we must select |α|2 +|β|2 = 1.) The two eigenvalues are then
said to be “degenerate”. I don’t know how such a disparaging term9 came
to be attached to such a charming result, but it has been. If two eigenstates
9 According to George F. Simmons, “This terminology follows a time-honored tradition

in mathematics, according to which situations that elude simple analysis are dismissed
by such pejorative terms as ‘improper’, ‘inadmissible’, ‘degenerate’, ‘irregular’, and so
on.” [Differential Equations with Applications and Historical Notes, third edition (CRC
Press, Boca Raton, Florida, 2017) page 220.]
 The Quantum Mechanics of Position 151

have the same energy, that energy is called “two-fold degenerate”. If three
have the same energy, it is “three-fold degenerate”. And so forth.
Finally, solving the energy eigenproblem opens the door to solving the
time evolution problem, because the wavefunction ψ0 (x) evolves in time to
∞
X
ψ(x, t) = cn e−(i/~)En t ηn (x). (4.37)
n=1

Because the energy eigenproblem (4.31) tells you the resulting quantized
energy values, and because energy quantization is one of the easiest aspects
of quantum mechanics to access experimentally, some people develop the
mistaken impression that finding energy values is all there is to quantum
mechanics. No. It’s actually just like a classical mechanics problem: “You
stand atop a cliff 97 meters tall and hurl a ball horizontally at 12 m/s.
(a) How far from the base of the cliff does it land? (b) At what speed does
it strike the ground?” Using energy techniques alone you can answer ques-
tion (b). But to answer both questions you need to solve the time evolution
problem. The same holds in quantum mechanics. Some questions can be
answered knowing only the energy eigenvalues, but to answer any question
you must solve the time evolution problem. Often the easiest way to do
this is by first solving the energy eigenproblem (finding both eigenvalues
and eigenfunctions) and then employing the time evolution equation (4.37).
The energy eigenvalues are important — no doubt about that — but they’re
not the full story.

Problem
4.11 Basis with degeneracy (essential problem) n o
You know that if the two orthonormal vectors î, ĵ constitute a basis
for position vectors in two dimensions, then rotating the pair by angle
θ produces two different orthonormal vectors
î0 = cos θ î + sin θ ĵ
0
ĵ = − sin θ î + cos θ ĵ (4.38)
that constitute a basis just as good as the original basis.
Show that if
ηn (x) n = 1, 2, 3, . . .
152 Energy loss

constitutes an orthonormal basis of energy eigenstates (an “energy

eigenbasis”), with a degeneracy so that E8 = E9 , then if η8 (x) and
η9 (x) are replaced by
η80 (x) = cos θ η8 (x) + sin θ η9 (x)
η90 (x) = − sin θ η8 (x) + cos θ η9 (x) (4.39)
this new basis is just as good an orthonormal energy eigenbasis as the
original.

4.8 Energy loss

I said earlier [at equations (4.26) and (4.29)] that any energy eigenstate
η6 (x) is a “stationary state”: that if the system started off in state η6 (x),
it would remain in that state forever (with a time-dependent phase factor
in front). This seems to contradict the experimental fact that most of the
atoms we find lying about are in their ground states.10 Why don’t they
just stay in state η6 (x) for ever and ever?
Furthermore: If the system starts off in the state c3 η3 (x)+c7 η7 (x), then
for all time the probability of measuring energy E3 is |c3 |2 , the probability
of measuring energy E7 is |c7 |2 , and the probability of measuring the ground
state energy is zero. Again, how can this conclusion be consistent with the
experimental observation that most atoms are in the ground state?
The answer is that if time evolution were given exactly by equa-
tion (4.12),
~2 ∂ 2 ψ(x, t)
 
∂ψ(x, t) i
=− − + V (x)ψ(x, t) , (4.40)
∂t ~ 2m ∂x2
then the atom would stay in that stationary state forever. But real atoms
are subject to collisions and radiation meaning that the time-evolution
equation above is not exactly correct. These phenomena, unaccounted for
in the equation above, cause the atom to fall into its ground state.
Because collisions and radiation are small effects, an atom starting off
in state η6 (x) stays in that stationary state for a “long” time — but that
means long relative to typical atomic times, such as the characteristic time
10−17 seconds generated at problem 1.15 on page 41. If you study more
10 The energy eigenstate with lowest energy eigenvalue has a special name: the ground

state.
4.9. Mean values 153

quantum mechanics,11 you will find that a typical atomic excited state
lifetime is 10−9 seconds. So the excited state lifetime is very short by
human standards, but very long by atomic standards. (To say “very long”
is an understatement: it is 100 million times longer; by contrast the Earth
has completed only 66 million orbits since the demise of the dinosaurs.)

4.9 Mean values

Recall the interference experiment. A single atom ambivates through

the two paths of an interferometer.

b xb = −w/2

a xa = +w/2

Say the interferometer has width w, so that path a has position xa = +w/2
while path b has position xb = −w/2.
You know the drill: the atom has amplitude ca of taking path a, am-
plitude cb of taking path b. If a lamp is turned on while an interference
experiment is proceeding, the probability of the atom appearing in path a
is |ca |2 , the probability of the atom appearing in path b is |cb |2 . In other
words, if the atom’s position is measured while the interference experiment
is proceeding, the result would be +w/2 with probability |ca |2 , and it would
be −w/2 with probability |cb |2 . Hence the mean position measured would
be
+ (w/2)|ca |2 − (w/2)|cb |2 . (4.41)

It seems weird to say “The atom doesn’t have a position but its mean
position is given by equation (4.41)” — sort of like saying “Unicorns don’t
exist but their mean height is 1.3 meters.” Indeed, it would be more accu-
rate to say “The atom doesn’t have a position but if a light were turned on
11 See for example David J. Griffiths and Darrell F. Schroeter, Introduction to Quan-

tum Mechanics, third edition (Cambridge University Press, Cambridge, UK, 2018) sec-
tion 11.3.2, “The Lifetime of an Excited State”.
154 Mean values

— or if position could be determined in some other way — then it would

have a position, and the mean position found in that way would be given
by equation (4.41).” This more accurate sentence is such a mouthful that
it’s rarely said: people say the first, inaccurate sentence as shorthand for
the second, correct sentence.
A particle in a potential. A particle is in state
∞
X
ψ(x, t) = cn e−(i/~)En t ηn (x). (4.42)
n=1
Here cn is the amplitude that the particle has energy En , so |cn |2 is the
probability that, if the energy were measured, the result En would be found.
The mean energy is thus clearly
∞
X
hEi = |cn |2 En . (4.43)
n=1

Once again, it seems weird to have a formula for the mean energy of
a particle that doesn’t have an energy. The meaning is that if the energy
were measured, hEi is the mean of the energy that would be found.
New expression for mean energy. The above expression for mean
energy is correct but difficult to use. Suppose a particle with wavefunction
ψ(x) is subject to a potential energy function V (x). To find the mean
energy you must: (1) Solve the energy eigenproblem to find the energy
eigenvalues En and eigenfunctions ηn (x). (2) Write the wavefunction ψ(x)
P
in the form ψ(x) = n cn ηn (x). (3) Now that you know the energies En and
the amplitudes (expansion coefficients) cn , execute the sum n |cn |2 En .
P

Whew! Isn’t there an easier way?

There is. The wavefunction (4.42) has time derivative
∞  
∂ψ(x, t) X i
= cn − En e−(i/~)En t ηn (x) (4.44)
∂t n=1
~
and complex conjugate
∞
X
ψ ∗ (x, t) = c∗m e+(i/~)Em t ηm
∗
(x). (4.45)
m=1
Thus
∂ψ(x, t)
ψ ∗ (x, t) (4.46)
∂t
∞ X ∞  
X i
= c∗m cn − En e−(i/~)(En −Em )t ηm
∗
(x)ηn (x)
m=1 n=1
~
The Quantum Mechanics of Position 155

and
Z +∞
∂ψ(x, t)
ψ ∗ (x, t) dx (4.47)
−∞ ∂t
∞ X ∞   Z +∞
X
∗ i −(i/~)(En −Em )t ∗
= cm cn − En e ηm (x)ηn (x) dx.
m=1 n=1
~ −∞

But (see equation 4.32) the integral on the right is zero unless m = n, in
which case it is 1. Thus
Z +∞ ∞  
∗ ∂ψ(x, t) X
∗ i i
ψ (x, t) dx = cn cn − En = − hEi (4.48)
−∞ ∂t n=1
~ ~
or
Z +∞  
i ∂ψ(x, t)
− hEi = ψ ∗ (x, t) dx. (4.49)
~ −∞ ∂t

Now, according to the Schrödinger equation

~2 ∂ 2 ψ(x, t)
 
∂ψ(x, t) i
=− − + V (x)ψ(x, t) , (4.50)
∂t ~ 2m ∂x2
so
+∞
~2 ∂ 2 ψ(x, t)
Z  
hEi = ψ ∗ (x, t) − + V (x)ψ(x, t) dx. (4.51)
−∞ 2m ∂x2
This expression for mean energy does not require solving the energy eigen-
problem or expanding ψ(x) in energy eigenstates.
This expression parallels the expressions already determined in prob-
lem 4.2 on page 125: For example the mean position is
Z +∞
hxi = ψ ∗ (x, t) [x] ψ(x, t) dx, (4.52)
−∞

the mean of position squared is

Z +∞
ψ ∗ (x, t) x2 ψ(x, t) dx,
2
 
hx i = (4.53)
−∞

and indeed for any function of position f (x), the mean is

Z +∞
hf (x)i = ψ ∗ (x, t) [f (x)] ψ(x, t) dx. (4.54)
−∞
 156 Mean values

Problems
4.12 Mean position vs. “expected position”
For the infinite square well energy eigenstate η2 (x), what is the mean
position? What is the probability density at that point? Is this mean
position really the “expected position”?
4.13 Wavefunction vs. probability density (recommended problem)
The wavefunction ψ(x) is not directly measurable, but can be inferred
(up to an overall phase) through a number of position and interference
experiments. The probability density |ψ(x)|2 is measurable through
a number of position experiments alone. These facts lead some to
the misconception that the probability density tells the “whole story”
of a quantal state. This problem demonstrates the falsehood of that
misconception by presenting a series of wavefunctions, all with the same
probability density, but each with a different mean energy. (And hence
each with different behavior in the future.) The so-called Gaussian
wavefunctions are
2
/2σ 2 ikx
ψ(x) = Ae−x e ,
where A is a normalization constant.
a. (Mathematical preliminary.) Use integration by parts to show that
Z +∞ Z +∞
2 2
e−t dt = 2 t2 e−t dt,
−∞ −∞
where t is any dimensionless variable.
b. When the particle is free, V (x) = 0, find the mean energy. (In this
case the mean kinetic energy, since there is no potential energy.) If
you use the above result, you will not need to evaluate any integral
nor find the normalization constant.

4.14 Mean of function vs. function of mean (recommended problem)

Show that hf (x)i might not equal f (hxi) by using the function f (x) =
x2 and the so-called Gaussian wavefunction
2
/2σ 2 ikx
ψ(x) = Ae−x e .
a. What is the normalization constant A? Does your answer have
the proper dimensions?
b. What is the mean position hxi for this wavefunction?
c. What is the mean of the function hf (x)i?
d. What is the function of the mean f (hxi)?
4.10. The classical limit of quantum mechanics 157

4.10 The classical limit of quantum mechanics

I told you way back on page 4 that when quantum mechanics is applied
to big things, it gives the results of classical mechanics. It’s hard to see
how my claim could possibly be correct: the whole structure of quantum
mechanics differs so dramatically from the structure of classical mechanics
— the character of a “state”, the focus on potential energy function rather
than on force, the emphasis on energy eigenproblems instead of initial value
problems, the fact that the quantal time evolution equation involves a first
derivative with respect to time while the classical time evolution equation
involves a second derivative with respect to time.

4.10.1 How does mean position change with time?

This nut is cracked by focusing, not on the full quantal state ψ(x, t), but
on the mean position
Z +∞
hxi = ψ ∗ (x, t)xψ(x, t) dx, (4.55)
−∞
How does this mean position change with time?
The answer depends on the classical force function F (x) — i.e., the
classical force that would be exerted on a classical particle if it were at
position x. (I’m not saying that the particle is at x, I’m not even saying
that the particle has a position; I’m saying that’s what the force would be
if the particle were classical and at position x.)
The answer is that
d2 hxi
hF (x)i = m , (4.56)
dt2
a formula that certainly plucks our classical heartstrings! This result is
called Ehrenfest’s theorem.12 We will prove this theorem later (in sec-
tion 4.10.4 on page 162), but first discuss its significance.
Although the theorem is true in all cases, it is most useful when the
spread in position ∆x is in some sense small, so the wavefunction is rel-
ativity compact. Such wavefunctions are called “wavepackets”. In this
12 Paul Ehrenfest (1880–1933) contributed to relativity theory, quantum mechanics, and

statistical mechanics, often by pointing out concrete difficulties in these theories. As a

result, several telling arguments have names like “Ehrenfest’s paradox” or “Ehrenfest’s
urn” or “the Ehrenfest dog-flea model”.
158 The classical limit of quantum mechanics

situation we might hope for a useful approximation — the classical limit

— by ignoring the quantal indeterminacy of position and focusing solely on
mean position.
If the force function F (x) varies slowly on the scale of ∆x, then our
hopes are confirmed: the spread in position is small, the spread in force is
small, and to a good approximation the mean force hF (x)i is equal to the
force at the mean position F (hxi).

hF (x)i
F (x)

F (hxi)

|ψ(x)|2
x
hxi

∆x

But if the force function varies rapidly on the scale of ∆x, then our
hopes are dashed: the spread in position is small, but the spread in force
is not, and the classical approximation is not appropriate.

F (x)
hF (x)i
F (hxi)

|ψ(x)|2
x
hxi

∆x
The Quantum Mechanics of Position 159

To head off a misconception, I emphasize that Ehrenfest’s theorem is

not that
d2 hxi
F (hxi) = m 2 .
dt
If this were true, then the mean position of a quantal particle would in
all cases move exactly as a classical particle does. But (see problem 4.14,
“Mean of function vs. function of mean”, on page 156) it’s not true.

4.10.2 Is the classical approximation good enough?

If the quantal position indeterminacy ∆x is small compared to the exper-

imental uncertainty of your position-locating experimental apparatus, for
the entire duration of your experiment, then the classical approximation is
usually appropriate. So the central question is: How big is the quantal ∆x
in my situation? This will of course vary from case to case and from time to
time within a given case. But there’s an important theorem that connects
the indeterminacy of position ∆x with the indeterminacy of momentum
∆p: in all situations
∆x∆p ≥ 21 ~. (4.57)
This theorem is called the Heisenberg indeterminacy principle. Because
this book has focused on position and not momentum, we cannot prove the
theorem at this time: you’ll have to read a more advanced book. But you
should know about the result for two reasons: First, because it’s important
for determining whether the classical limit is appropriate in a given case.
Second, because it was important in the historical development of quantum
mechanics.
Quantum mechanics has a long and intricate (and continuing!) history,
but one of the keystone events occurred in the spring of 1925. Werner
Heisenberg,13 a freshly minted Ph.D., had obtained a position as assistant
13 German theoretical physicist (1901–1976) who nearly failed his Ph.D. oral exam due

to his fumbling in experimental physics. He went on to discover quantum mechanics as

we know it today. Although attacked by Nazis as a “white Jew”, he became a princi-
pal scientist in the German nuclear program during World War II, where he focused on
building nuclear reactors rather than nuclear bombs. After the war he worked to re-
build German science, and to extend quantum theory into relativistic and field theoretic
domains. He enjoyed hiking, particularly in the Bavarian Alps, and playing the piano.
After a three-month whirlwind romance, Heisenberg married Elisabeth Schumacher, sis-
ter of the Small Is Beautiful economist E.F. Schumacher, and they went on to parent
seven children.
160 The classical limit of quantum mechanics

to Max Born at the University of Göttingen. There he realized that the

key to formulating quantum mechanics was to develop a theory that fit the
experiments described in chapter 1, and that also had the correct classical
limit. He was searching for such a theory when he came down with a bad
case of allergies to spring pollen from the “mass of blooming shrubs, rose
gardens and flower beds”14 of Göttingen. He decided to travel to Helgoland,
a rocky island and fishing center in the North Sea, far from pollen sources,
arriving there by ferry on 8 June 1925.
Once his health returned, Heisenberg reproduced his earlier work, clean-
ing up the mathematics and simplifying the formulation. He worried that
the mathematical scheme he invented might prove to be inconsistent, and
in particular that it might violate the principle of energy conservation. In
Heisenberg’s own words:15

One evening I reached the point where I was ready to determine

the individual terms in the energy table, or, as we put it today, in
the energy matrix, by what would now be considered an extremely
clumsy series of calculations. When the first terms seemed to ac-
cord with the energy principle, I became rather excited, and I began
to make countless arithmetical errors. As a result, it was almost
three o’clock in the morning before the final result of my compu-
tations lay before me. The energy principle had held for all the
terms, and I could no longer doubt the mathematical consistency
and coherence of the kind of quantum mechanics to which my cal-
culations pointed. At first, I was deeply alarmed. I had the feeling
that, through the surface of atomic phenomena, I was looking at a
strangely beautiful interior, and felt almost giddy at the thought
that I now had to probe this wealth of mathematical structures na-
ture had so generously spread out before me. I was far too excited
to sleep, and so, as a new day dawned, I made for the southern tip
of the island, where I had been longing to climb a rock jutting out
into the sea. I now did so without too much trouble, and waited
for the sun to rise.

Because the correct classical limit was essential in producing this theory,
it was easy to fall into the misconception that an electron really did behave
classically, with a single position, but that this single position is disturbed
14 Werner Heisenberg, Physics and Beyond (Harper and Row, New York, 1971) page 37.
15 Physics and Beyond, page 61.
The Quantum Mechanics of Position 161

by the measuring apparatus used to determine position. Indeed, Heisenberg

wrote as much:16

observation of the position will alter the momentum by an unknown

and undeterminable amount.

But Neils Bohr repeatedly objected to this “disturbance” interpretation.

For example, at a 1938 conference in Warsaw,17 he

warned specifically against phrases, often found in the physical lit-

erature, such as “disturbing of phenomena by observation.”

Today, interference and entanglement experiments make clear that Bohr

was right and that “measurement disturbs the system” is not a tenable
position.18 In an interferometer, there is no local way that a photon at
path a can physically disturb an atom taking path b. For an entangled pair
of atoms, there is no local way that an analyzer measuring the magnetic
moment of the left atom can physically disturb the right atom. It is no
defect in our measuring apparatus that it cannot determine what does not
exist.
And this brings us to one last terminology note. What we have called
the “Heisenberg indeterminacy principle” is called by some the “Heisenberg
uncertainty principle”.19 The second name is less accurate because it gives
the mistaken impression that an electron really does have a position and
we are just uncertain as to what that position is. It also gives the mistaken
impression that an electron really does have a momentum and we are just
uncertain as to what that momentum is.
16 Werner Heisenberg, The Physical Principles of the Quantum Theory, translated by
Carl Eckart and F.C. Hoyt (University of Chicago Press, Chicago, 1930) page 20.
17 Niels Bohr, “Discussion with Einstein on epistemological problems in atomic physics,”

in Albert Einstein, Philosopher–Scientist, edited by Paul A. Schilpp (Library of Living

Philosophers, Evanston, Illinois, 1949) page 237.
18 To be completely precise, “measurement disturbs the system locally” is not a tenable

position. The “de Broglie–Bohm pilot wave” formulation of quantum mechanics can be
interpreted as saying that “measurement disturbs the system”, but the measurement at
one point in space is felt instantly at points arbitrarily far away. When this formulation is
applied to a two-particle system, a “pilot wave” situated in six-dimensional configuration
space somehow physically guides the two particles situated in ordinary three-dimensional
space.
19 Heisenberg himself, writing in German, called it the “Genauigkeit Beziehung” — ac-

curacy relationship. See “Über den anschaulichen Inhalt der quantentheoretischen Kine-
matik und Mechanik” Zeitschrift für Physik 43 (March 1927) 172–198.
162 The classical limit of quantum mechanics

4.10.3 Sample Problem

For the “Underground Guide to Quantum Mechanics” (described on

page 20), you decide to write a passionate persuasive paragraph or two con-
cerning the misconception that “measurement disturbs the system”. What
do you write?

Solution: For those of us who know and love classical mechanics, there’s
a band-aid, the idea that “measurement disturbs the system”. This idea is
that fundamentally classical mechanics actually holds, but that quantum
mechanics is a mask layered over top of, and obscuring the view of, the
classical mechanics because our measuring devices disturb the underlying
classical system. That’s not possible. It is no defect of our measuring
instruments that they cannot determine what does not exist, just as it is
no defect of a colorimeter that it cannot determine the color of love.
This idea that “measurement disturbs the system” is a psychological
trick to comfort us, and at the same time to keep us from exploring, fully
and openly, the strange world of quantum mechanics. I urge you, I implore
you, to discard this security blanket, to go forth and discover the new world
as it really is rather than cling to the familiar classical world. Like Miranda
in Shakespeare’s Tempest, take delight in this “brave new world, that has
such people in’t”.
Unlike most band-aids, this band-aid does not protect or cover up. In-
stead it exposes a lack of imagination.

4.10.4 Proof of Ehrenfest’s theorem

I’m not going to kid you: this derivation is long, difficult, and, frankly,
unenlightening. It is necessary to show the coherence of the entire quantal
scheme we’ve been building, and if you follow it critically you will learn
some tricks of the trade, but if you decide to skip this section I won’t be
offended.
The Quantum Mechanics of Position 163

Because quantum mechanics emphasizes potential energy V (x), and

classical mechanics emphasizes force F (x), let’s recall how they’re related.
The definition of potential energy (in one
Z dimension) is
x
V (x) − V (x0 ) = − F (x0 ) dx0 , (4.58)
x0
where F (x) is the classical force function. Taking the derivative of both
sides with respect to x (and using the fundamental theorem of calculus on
the right-hand side) gives
∂V (x)
= −F (x). (4.59)
∂x
Remember that the classical time evolution equation is
d2 x(t)
F (x(t)) = m
dt2
which is second-order with respect to time and which, of course, contains
no reference to ~.
We ask how the mean position moves in quantum mechanics:
Z +∞
hxi = xψ ∗ (x, t)ψ(x, t) dx, (4.60)
−∞
so Z +∞ Z +∞
dhxi ∂ψ ∗ (x, t) ∂ψ(x, t)
= x ψ(x, t) dx + xψ ∗ (x, t) dx. (4.61)
dt −∞ ∂t −∞ ∂t
But the Schrödinger time evolution equation tells us how wavefunction
ψ(x, t) changes with time:
~2 ∂ 2 ψ(x, t)
 
∂ψ(x, t) i
=− − + V (x)ψ(x, t) (4.62)
∂t ~ 2m ∂x2
and
∂ψ ∗ (x, t) ~2 ∂ 2 ψ ∗ (x, t)
 
i ∗
=+ − + V (x)ψ (x, t) . (4.63)
∂t ~ 2m ∂x2
(From here on I’m going to write ψ(x, t) as ψ and V (x) as V .) Thus
Z +∞ 
~2 ∂ 2 ψ ∗

dhxi i ∗
= x − + V ψ ψ dx
dt ~ −∞ 2m ∂x2
Z +∞
~2 ∂ 2 ψ
  
− xψ ∗ − + V ψ dx
−∞ 2m ∂x2
Z +∞ 2 ∗ Z +∞
~2 2
 
i ∂ ψ ∗∂ ψ
= − x ψ dx − xψ dx
~ 2m −∞ ∂x2 −∞ ∂x2
Z +∞ Z +∞ 
+ xV ψ ∗ ψ dx − xψ ∗ V ψ dx
−∞ −∞
+∞ 2 ∗ Z +∞
∂2ψ
Z 
~ ∂ ψ
= −i x ψ dx − xψ ∗ dx . (4.64)
2m −∞ ∂x2 −∞ ∂x2
164 The classical limit of quantum mechanics

It seems odd that the two terms involving potential energy cancel, so no
explicit dependence on V (x) appears in this result, but we’ll just push on.
Can we say anything about integrals such as the second integral in
square brackets above? Surprisingly, the answer is yes. If we define
∂ψ
f (x) = xψ ∗ and g(x) = (4.65)
∂x
then
Z +∞ Z +∞
∂2ψ
xψ ∗ 2 dx = f (x)g 0 (x) dx (4.66)
−∞ ∂x −∞
which suggests integration by parts:
Z +∞  +∞ Z +∞
0
f (x)g (x) dx = f (x)g(x) − f 0 (x)g(x) dx. (4.67)
−∞ −∞ −∞
Now remember that the wavefunction is normalized, so it has to fall to
zero at both infinity and negative infinity. Typically the slope ∂ψ/∂x also
falls to zero at both infinity and negative infinity, and does so very rapidly
— much more rapidly than linearly. (There are exceptions to these typical
behaviors, such as scattering wavefunctions, and in these atypical cases this
argument has to be rethought.) The upshot is that in typical situations
 +∞
f (x)g(x) =0 (4.68)
−∞
so
+∞ Z +∞
∂2ψ ∂(xψ ∗ ) ∂ψ
Z
xψ ∗2
dx = − dx. (4.69)
−∞ ∂x −∞ ∂x ∂x
We’ll use this trick several times. . . I’ll just call it the “integration-by-parts
trick”.
Applying this trick to both integrals of equation (4.64) gives
 Z +∞ Z +∞
∂(xψ) ∂ψ ∗ ∂(xψ ∗ ) ∂ψ

dhxi ~
= −i − dx + dx
dt 2m −∞ ∂x ∂x −∞ ∂x ∂x
 Z +∞ ∗ Z +∞
~ ∂ψ ∂ψ ∂ψ ∗
= −i − x dx − ψ dx
2m −∞ ∂x ∂x −∞ ∂x
Z +∞ Z +∞
∂ψ ∗ ∂ψ

∗ ∂ψ
+ x dx + ψ dx
−∞ ∂x ∂x −∞ ∂x
 Z +∞ Z +∞
∂ψ ∗

~ ∂ψ
= −i − ψ dx + ψ∗ dx
2m −∞ ∂x −∞ ∂x
Z +∞ 
~ ∂ψ
= =m ψ∗ dx . (4.70)
m −∞ ∂x
The Quantum Mechanics of Position 165

Notice that dhxi/dt is pure real, as it must be. And notice that the di-
mensions are the same on both sides. (This isn’t proof that we’ve made no
algebra errors, but if our expression for dhxi/dt had been complex, or if it
had been dimensionally incorrect, then that would have been proof that we
had made algebra errors.)
All this is fine and good, but it takes us only part way to our goal.
This is clearly not a classical equation. . . it contains ~ right there! Since the
classical F = ma involves the second derivative of position with respect to
time, we take one more time derivative of hxi, finding
Z +∞ Z +∞
d2 hxi ∂ψ ∗ ∂ψ

~ ∗ ∂ ∂ψ
= =m dx + ψ dx . (4.71)
dt2 m −∞ ∂t ∂x −∞ ∂x ∂t
The second-order derivative on the right looks particularly grotesque, so we
use the integration-by-parts trick to get rid of it:
Z +∞ Z +∞
d2 hxi ∂ψ ∗ ∂ψ ∂ψ ∗ ∂ψ

~
= =m dx − dx
dt2 m −∞ ∂t ∂x −∞ ∂x ∂t
Z +∞ ∗

2~ ∂ψ ∂ψ
= − =m dx . (4.72)
m −∞ ∂x ∂t
Now use the Schrödinger equation:
Z +∞
d2 hxi ∂ψ ∗ ~2 ∂ 2 ψ
   
2~ i
= − =m − − +Vψ dx
dt2 m −∞ ∂x ~ 2m ∂x2
Z +∞
∂ψ ∗ ~2 ∂ 2 ψ
  
2
= <e − + V ψ dx . (4.73)
m −∞ ∂x 2m ∂x2
Look at that. . . two of the ~s have canceled out! We’re not home yet because
there’s still an ~ within the square brackets, but we’re certainly making
progress. We have that
Z +∞ Z +∞
d2 hxi ~2 ∂ψ ∗ ∂ 2 ψ ∂ψ ∗
 
2
= <e − dx + V ψ dx , (4.74)
dt2 m 2m −∞ ∂x ∂x2 −∞ ∂x
but let’s apply the integration-by-parts trick to the first integral:
Z +∞ Z +∞ 2 ∗
∂ψ ∗ ∂ 2 ψ ∂ ψ ∂ψ
2
dx = − dx. (4.75)
−∞ ∂x ∂x −∞ ∂x2 ∂x
Think about this for a minute: if the integral on the left is z, this equation
says that z = −z ∗ , whence z is pure imaginary or <e{z} = 0. Thus
Z +∞
d2 hxi ∂ψ ∗

2
= <e V ψ dx , (4.76)
dt2 m −∞ ∂x
166 Transitions induced by light

an expression devoid of ~s! Apply the integration-by-parts trick to this

integral:
Z +∞ Z +∞
∂ψ ∗ ∂(V ψ)
V ψ dx = − ψ∗ dx
−∞ ∂x −∞ ∂x
Z +∞ Z +∞ Z +∞
∂ψ ∗ ∂ψ ∂V
V ψ dx = − ψ∗ V dx − ψ∗ ψ dx
−∞ ∂x −∞ ∂x −∞ ∂x
Z +∞ Z +∞ Z +∞
∂ψ ∗ ∂ψ ∂V
V ψ dx + ψ∗ V dx = − ψ∗ ψ dx
−∞ ∂x −∞ ∂x −∞ ∂x
Z +∞ Z +∞
∂ψ ∗

∂V
2<e V ψ dx = − ψ∗ ψ dx. (4.77)
−∞ ∂x −∞ ∂x
Plugging this result back into equation (4.76) gives
d2 hxi 1 +∞ ∗ ∂V
Z
= − ψ ψ dx. (4.78)
dt2 m −∞ ∂x
But the force function is F (x) = −∂V /∂x, so
d2 hxi 1 +∞ ∗
Z
1
= ψ (x, t)F (x)ψ(x, t) dx = hF (x)i. (4.79)
dt2 m −∞ m
There it is —
d2 hxi
hF (x)i = m . (4.80)
dt2
We have proven Ehrenfest’s theorem.

4.11 Transitions induced by light

This section is more intricate than other sections of this book, and it takes
many steps to reach its conclusion. Furthermore, it is not needed as back-
ground for any following section, so you might want to skip over it. But the
steps are valuable and the conclusion itself is one of the most fascinating
and useful relations in all of physics.
The problem. An electron in the ground state of a symmetric potential
well is exposed to light. What is the probability that it transitions through
light absorption to a particular excited state?
The Quantum Mechanics of Position 167

Expectation. I expect that the light would induce transitions from

the ground state to the excited state. If a collection of atoms is exposed to
light for two seconds, there will be twice as many transitions as there were
when those atoms were exposed to light for one second.

transition probability
6




 

 

  - t
0
0

Setup. Call the ground state wavefunction ηg (x) with energy Eg and
the excited state wavefunction ηe (x) with energy Ee . In light of the Einstein
relation (1.21) we define the frequency characteristic of this transition
ω0 = (Ee − Eg )/~. (4.81)
For most symmetric potential wells the mean position for any energy eigen-
state vanishes,
hxig = 0 and hxie = 0, (4.82)
and we will assume this. (If this assumption is wrong, the derivation needs
to be rethought.)
When no light shines, the Schrödinger time evolution equation is
~2 ∂ 2 ψ(x, t)
 
∂ψ(x, t) i
=− − + Vwell (x)ψ(x, t)
∂t ~ 2m ∂x2
i
≡ − [Hwell ψ(x, t)] . (4.83)
~
In this section we abbreviate the term in square brackets as “Hwell ψ(x, t)”.
For example, when ψ(x, t) = ηg (x) we have
Hwell ηg (x) = Eg ηg (x), (4.84)
168 Transitions induced by light

and when ψ(x, t) = ηe (x) we have

Hwell ηe (x) = Ee ηe (x). (4.85)
You know how this time evolution behaves: the initial wavefunction
cg ηg (x) + ce ηe (x) (4.86)
evolves in time to
cg e−(i/~)Eg t ηg (x) + ce e−(i/~)Ee t ηe (x). (4.87)

When light does shine, the electron is subject not only to the well’s
potential energy function, but also to the potential energy function due to
the light. If the electric field at the center of the well is E0 cos(ωt), then
that additional potential energy function is
eE0 cos(ωt) x, (4.88)
where the charge on an electron is −e. In this circumstance the wavefunc-
tion no longer evolves like (4.87), but instead like
ψ(t) = cg (t)e−(i/~)Eg t ηg (x) + ce (t)e−(i/~)Ee t ηe (x). (4.89)

Our job is to find the probability of starting in the ground state and
ending in the excited state. That is, assuming cg (0) = 1 and ce (0) = 0, we
need to find ce (t). The transition probability is then |ce (t)|2 .
Time evolution when the light shines. When light shines, the
potential energy function changes from that of the well alone, Vwell (x),
to Vwell (x) + eE0 cos(ωt) x. Thus the Schrödinger time evolution equation
changes from equation (4.83) to
d i
ψ(t) = − [Hwell ψ(t) + eE0 cos(ωt) xψ(t)] . (4.90)
dt ~
First, look at the left-hand side:
d d h i
ψ(t) = cg (t)e−(i/~)Eg t ηg (x) + ce (t)e−(i/~)Ee t ηe (x) (4.91)
dt hdt i
= ċg (t)e−(i/~)Eg t ηg (x) + ċe (t)e−(i/~)Ee t ηe (x)
ih i
− Eg cg (t)e−(i/~)Eg t ηg (x) + Ee ce (t)e−(i/~)Ee t ηe (x) .
~
Meanwhile, on the right-hand side
Hwell ψ(t) = cg (t)e−(i/~)Eg t Hwell ηg (x) + ce (t)e−(i/~)Ee t Hwell ηe (x)
= cg (t)e−(i/~)Eg t Eg ηg (x) + ce (t)e−(i/~)Ee t Ee ηe (x). (4.92)
The Quantum Mechanics of Position 169

Putting these three equations together gives

ċg (t)e−(i/~)Eg t ηg (x) + ċe (t)e−(i/~)Ee t ηe (x) (4.93)
i h i
= − eE0 cos(ωt) x cg (t)e−(i/~)Eg t ηg (x) + ce (t)e−(i/~)Ee t ηe (x) .
~

Multiply the above equation by ηe∗ (x) and integrate over all values of x.
Because of orthonormality
Z +∞ Z +∞
ηe∗ (x)ηg (x) dx = 0 and ηe∗ (x)ηe (x) dx = 1, (4.94)
−∞ −∞

while because hxie = 0

Z +∞
ηe∗ (x) x ηe (x) dx = 0. (4.95)
−∞

Hence we find
Z +∞
i
ċe (t)e−(i/~)Ee t = − eE0 cos(ωt)cg (t)e−(i/~)Eg t ηe∗ (x) x ηg (x) dx.
~ −∞
(4.96)
The integral on the right is just a (complex) number — not a function
of x, not a function of t — and we’ll call that number he|x|gi. Recalling
definition (4.81), we write
i
ċe (t) = − eE0 he|x|gi cos(ωt)e+iω0 t cg (t). (4.97)
~

If we had multiplied (4.93) instead by ηg∗ (x) and integrated we would

have found
i
ċg (t) = − eE0 hg|x|ei cos(ωt)e−iω0 t ce (t). (4.98)
~
So far, we have made assumptions but no approximations.
Approximate solution of the time evolution equations. Our
job is to find the transition probability |ce (t)|2 with the initial conditions
cg (0) = 1 and ce (0) = 0. This coupled problem is difficult. However, we
can make progress under conditions where cg (t) and ce (t) change slowly. In
this case replace cg (t) on the right-hand side of (4.97) with its initial value,
namely 1. (This approximation is called “first order perturbation theory
for the time evolution problem”.) This gives
i
ċe (t) = − eE0 he|x|gieiω0 t cos(ωt) (4.99)
~
170 Transitions induced by light

which integrates with respect to time to

Z t
i 0
ce (t) = − eE0 he|x|gi eiω0 t cos(ωt0 ) dt0 . (4.100)
~ 0
Performing the integral gives
Z t Z t
0 0 0 0
eiω0 t cos(ωt0 ) dt0 = 1
2 eiω0 t (e+iωt + e−iωt ) dt0
0 0
Z th
1 0 0
i
= ei(ω0 +ω)t + ei(ω0 −ω)t dt0
2 0
" 0 0
#t
1 ei(ω0 +ω)t ei(ω0 −ω)t
= +
2 i(ω0 + ω) i(ω0 − ω)
0
1 ei(ω0 +ω)t − 1 ei(ω0 −ω)t − 1
 
= + .
2i ω0 + ω ω0 − ω

This full expression is formidable, to put it mildly. But Enrico Fermi20

noticed something about the magnitudes concerned. Look at
ei(ω0 +ω)t − 1
.
ω0 + ω
The numerator is a complex number with magnitude between 0 and 2. The
denominator is a real number involving ω which, for light, is about 1015 s−1 .
So this fraction will be numerically tiny. Similarly for the piece
ei(ω0 −ω)t − 1
ω0 − ω
except that when ω ≈ ω0 , the denominator is near zero so the fraction can
be large indeed. In this way Fermi realized that the transition probability
is almost always tiny. It is more than tiny only when ω ≈ ω0 , and in that
regime the approximation
Z t
0 1 ei(ω0 −ω)t − 1
eiω0 t cos(ωt0 ) dt0 ≈ (4.101)
0 2i ω0 − ω
20 Enrico Fermi (1901–1954) of Italy and the United States excelled in both experimental

and theoretical physics. He directed the building of the first nuclear reactor and produced
the first theory of the weak nuclear interaction. The Fermi surface in the physics of metals
was named in his honor. He elucidated the statistics of what are now called fermions in
1926. He produced so many thoughtful conceptual and estimation problems that such
problems are today called “Fermi problems”. I never met him (he died before I was
born) but I have met several of his students, and each of them speaks of him in that
rare tone reserved for someone who is not just a great scientist and a great teacher and
a great leader, but also a great human being.
 The Quantum Mechanics of Position 171

is highly accurate.
Furthermore,
1 ei(ω0 −ω)t − 1 1 [ei(ω0 −ω)t/2 − e−i(ω0 −ω)t/2 ]ei(ω0 −ω)t/2
= (4.102)
2i ω0 − ω 2i ω0 − ω
which seems like a step backwards, until you remember that eiθ − e−iθ =
2i sin θ, so
1 ei(ω0 −ω)t − 1 sin[(ω0 − ω)t/2] i(ω0 −ω)t/2
= e . (4.103)
2i ω0 − ω ω0 − ω
So, at this excellent level of approximation,
i sin[(ω0 − ω)t/2] i(ω0 −ω)t/2
ce (t) = − eE0 he|x|gi e (4.104)
~ ω0 − ω
and the transition probability is
e2 E02 |he|x|gi|2 sin2 [(ω0 − ω)t/2]
|ce (t)|2 = . (4.105)
~2 (ω0 − ω)2
Given all the assumptions and approximations we introduced to derive this
result, you might think it’s an obscure equation of limited applicability.
You’d be wrong. It is used so often that Fermi called it the “golden rule”.
Reflection. The transition probability result, graphed below as a func-
tion of time, shows oscillatory behavior called “Rabi21 flopping”. This is
the beat at the heart of an atomic clock.

transition probability
6
e2 E02 |he|x|gi|2
~2 (ω0 − ω)2

0 - t

0 2π 4π 6π
|ω0 − ω| |ω0 − ω| |ω0 − ω|
21 Isidor Isaac Rabi (1898–1988), Polish-Jewish-American physicist. His fascinating life

cannot be summarized in a few sentences: I recommend John Rigden’s biography Rabi:

Scientist and Citizen (Basic Books, New York, 1987).
 172 Position plus spin

I have made bad guesses in my life, but none worse than the difference
between my expectation graphed on page 167 and the real behavior graphed
above. It’s as if, while hammering a nail into a board, the first few strikes
drive the nail deeper and deeper into the board, but additional strikes make
the nail come out of the board. And one strike (at time 2π/|ω0 − ω|) makes
the nail pop out of the board altogether! Is there any way to account for
this bizarre result other than shrugging that “It comes out of the math”?
There is. This is a form of interference22 where the particle moves not
from point to point through two possible slits, but from energy state to
energy state with two possible intermediate states. The initial state is the
ground state and the final state is the ground state. The two possible inter-
mediates are the excited state and the ground state. There is an amplitude
to go from ground state to ground state via the excited state, and an am-
plitude to go from ground state to ground state via the ground state. At
time π/|ω0 − ω| those two amplitudes interfere destructively so there is a
small probability of ending up in the ground state and hence a large prob-
ability of ending up in the excited state. At time 2π/|ω0 − ω| those two
amplitudes interfere constructively so there is a large probability of ending
up in the ground state and hence a small probability of ending up in the
excited state.

Problem
4.15 Explore some more
There’s a lot more to say to flesh out the story told by equation (4.105),
but I’ll restrict myself to one question: The denominator vanishes when
ω = ω0 , so you might think that the probability goes to infinity there.
Bad idea. Show that the probability is instead
e2 E02 |he|x|gi|2 t2
.
~2 4

4.12 Position plus spin

In chapters 2 and 3 we investigated particles with magnetic moment, like

the silver atom, doing our best to treat the quantum mechanics of mag-
22 This point of view is expounded by R.P. Feynman and A.R. Hibbs in section 6-5 of

Quantum Mechanics and Path Integrals, emended edition (Dover Publications, Mineola,
NY, 2010).
The Quantum Mechanics of Position 173

netic moment while ignoring the quantum mechanics of position. Then in

chapter 4 we investigated the quantum mechanics of position, ignoring the
quantum mechanics of magnetic moment. For historical reasons, the mag-
netic moment is said to reflect the intrinsic “spin” of a particle. It’s time
to weld the two pieces of spin and position together.
This is achieved in a straightforward way. Think back to our discussion
of bin amplitudes for a single particle in one dimension. We asked the
question “What is the amplitude for the particle to be found in bin i?” But
if the particle has a two-basis-state spin, like a silver atom, we have to ask
the question “What is the amplitude for the particle to be found in bin i
with spin up?” Or “What is the amplitude for the particle to be found in
bin i with spin down?” (Alternatively we might ask for x-spin positive or
x-spin negative, or 17◦ -spin positive or negative, or for the projection on
any other axis, but it’s conventional to focus on the vertical axis.) We’ll
call the answer to the first question ψi,+ , the answer to the second question
ψi,− , and we’ll write the two answers together as
ψi (+ 21 ) = ψi,+ and ψi (− 21 ) = ψi,−
so that the answer to both questions at once is ψi (ms ), where ms = ± 12 .
(The choice ms = ± 12 instead of ms = ±1 or even ms = ±3π is again for
historical reasons.)
Now do the standard thing: divide by the square root of bin size and take
the limit as bin size shrinks to zero. This quantity becomes an amplitude
density (wavefunction)
ψ(x, ms ) where −∞ < x < +∞ and ms = ± 21 .
Sometimes people write this as two separate wavefunctions:
ψ+ (x) = ψ(x, + 21 ) and ψ− (x) = ψ(x, − 21 ).
And sometimes they write it as a “spatial part” times a “spin part”:
ψ+ (x) = φ(x)χ+ and ψ− (x) = ξ(x)χ− .
But don’t let the notation fool you: all of these expressions represent the
same thing.
It might happen that the two spatial parts are equal, φ(x) = ξ(x), in
which case we can say
ψ(x, ms ) = φ(x)χ(ms ) where χ(+ 21 ) = χ+ and χ(− 21 ) = χ− .
174 Position plus spin

But not all wavefunctions factorize in this way.

If the atom were nitrogen, with four possible spin projections (see
page 45), then we would have to ask “What is the amplitude for the ni-
trogen atom to be found in bin i with vertical spin projection + 32 ?” or + 12
or − 12 or − 32 . (Alternatively we might ask “What is the amplitude for the
nitrogen atom to be found in bin i with a projection on the 67◦ axis of
− 23 ?”) After asking these questions and taking the appropriate limits, the
relevant wavefunction will be
ψ(x, ms ) where −∞ < x < +∞ and ms = + 23 , + 12 , − 12 , − 23 .
In the same way for an atom of sulfur, with five possible spin projections,
the relevant wavefunction will be
ψ(x, ms ) where −∞ < x < +∞ and ms = +2, +1, 0, −1, −2.
If the atom moves in three dimensions, the wavefunction will take the
form
ψ(x, y, z, ms ) ≡ ψ(x), (4.106)
where the single sans serif symbol x stands for the four variables x, y, z, ms .
[Because the variables x, y, and z are continuous, while the variable ms is
discrete, one sometimes sees the dependence on ms written as a subscript
rather than as an argument: ψms (x, y, z). This is a bad habit: ms is a
variable not a label, and it should not be notated as a second-class variable
just because it’s discrete.]
The electron is a spin- 21 particle. So are the positron and the muon,
and all the quarks. Protons and neutrons are composite particles, made up
of quarks, but the quarks combine in such a way that the proton and the
neutron are spin- 21 particles. The He3 atom is also a composite particle,
made up of protons, neutrons, and electrons, but in its ground state it has
spin 21 . The same applies, as we’ve seen extensively in chapters 3 and 4,
for the silver atom.
The Higgs boson is a spin-0 particle. So is the composite He4 atom (in
its ground state). The photon is a spin-1 particle, but it does not behave
like a typical massive spin-1 particle because it is intrinsically relativistic.

Epilogue

The quantum mechanics of position is very strange, yes.

And it’s very difficult, yes.
But it’s also very wonderful.
 The Quantum Mechanics of Position 175

Problems
4.16 Spin- 12 electron in a potential well, I (essential problem)
All electrons are spin- 21 particles. Using the χ+ , χ− notation, write
down the wavefunction for an electron ambivating in a potential well
with energy eigenfunctions ηn (x):
a. with amplitude 45 of being in the spatial ground state (n = 1) with
spin up and amplitude 35 of being in the spatial n = 3 state with
spin down;
b. with amplitude 54 of being in the spatial ground state (n = 1) with
spin up and amplitude 35 of being in the spatial n = 3 state with
negative spin projection on the x axis (see equation 3.19).
4.17 Spin- 21 electron in a potential well, II
An electron ambivates in a potential well with with energy eigenvalues
En and energy eigenfunctions ηn (x). The electron’s wavefunction is
3 χ
5 η3 (x) + + 45 η4 (x)χ− .
a. What is the mean energy?
b. If the vertical spin projection is measured, what is the probability
of finding +? (That is, of finding ms = + 21 .)
c. The vertical spin projection is measured and found to be +. Now
what is the mean energy?
4.18 Questions (recommended problem)
Update your list of quantum mechanics questions that you started at
problem 1.17 on page 46. Write down new questions and, if you have un-
covered answers to any of your old questions, write them down briefly.

[[For example, one of my questions would be: “What are the detailed
mechanisms for the energy loss outlined in section 4.8?”]]
 Chapter 5

Solving the Energy Eigenproblem

Energy eigenproblems are important: they determine the “allowed” energy

eigenvalues, and, as chapter 1 made clear, such energy quantization is the
most experimentally accessible facet of quantum mechanics. Also, the eas-
iest way to solve the time evolution problem is to first solve the energy
eigenproblem. This chapter focuses only on the spatial part of the wave-
function and ignores any spin part. For particles with spin, the two parts
can be welded together using the techniques of section 4.12 on page 172.
In fact, Erwin Schrödinger discovered the energy eigenproblem first (in
December 1925) and five months later discovered the time evolution equa-
tion, which he called “the true wave equation”. Today, both equations
carry the name “Schrödinger equation”, which can result in confusion.
There are large numbers of analytic and numerical techniques for solving
eigenproblems. Most of these are effective but merely technical: they find
the answer, but don’t give any insight into the character of the resulting
energy eigenfunctions. For example, if you study more quantum mechanics
you will find that for the simple harmonic oscillator, V (x) = 21 kx2 , the
energy eigenfunctions are
√ !1/4
km/~ √ 2 √
ηn (x) = 2n 2
e−( km/2~)x Hn (( km/~)1/2 x) (5.1)
2 (n!) π
for n = 0, 1, 2, 3, . . .
where the Hermite polynomials are defined through
2
dn e−z
2
Hn (z) = (−1)n ez .
dz n
Yikes! This is true, but provides little insight.

177
178 Sketching energy eigenfunctions

This chapter presents two of the many solution techniques available.

First we investigate an informal, rough-and-ready technique for sketching
energy eigenfunctions that doesn’t give rigorous solutions, but that does
provide a lot of insight. Second comes a numerical technique of wide appli-
cability.
Put both of these techniques into your problem-solving toolkit. You’ll
find them valuable not only in quantum mechanics, but whenever you need
to solve a second-order ordinary differential equation.

5.1 Sketching energy eigenfunctions

Since this chapter is more mathematical than physical in character, I start

off by writing the energy eigenequation (4.31) in the mathematically sug-
gestive form
d2 η(x) 2m 2m
= − 2 [E − V (x)]η(x) = − 2 Kc (x)η(x) (5.2)
dx2 ~ ~
which defines the “classical kinetic energy function” Kc (x). This parallels
the potential energy function: V (x) is the potential energy that the classical
system would have if the particle were located at x. I’m not saying that
the particle is classical nor that it does have a location; indeed a quantal
particle might not have a location. But V (x) is the potential energy that the
system would have if it were classical with the particle located at point x.
In the same way Kc (x) is the kinetic energy that a classical particle would
have if the particle were located at x and total energy were E. Whereas
no classical particle can ever have a negative kinetic energy, it is perfectly
permissible for the classical kinetic energy function to be negative: in the
graph that follows, Kc (x) is negative on the left, positive in the center, and
strongly negative on the right.
Solving the Energy Eigenproblem 179

V (x)

Kc (x)
E

classically classically classically

prohibited allowed prohibited
region: region: region:
Kc negative Kc positive Kc negative

A region were Kc (x) is positive or zero is called a “classically allowed re-

gion”; otherwise it is a “classically prohibited region”.
Remember that
dη d2 η
represents slope; represents curvature.
dx dx2
When curvature is positive, the slope increases as x increases (e.g. from neg-
ative to positive, or from positive small to positive large). When curvature
is negative, the slope decreases as x increases.
Start off by thinking of a classically allowed region where Kc (x) is
constant and positive. Equation (5.2) says that if η(x) is positive, then
the curvature is negative, whereas if η(x) is negative, then the curvature
is positive. Furthermore, the size of the curvature depends on the size of
η(x):

when η(x) is. . . curvature is. . .

strongly positive strongly negative
weakly positive weakly negative
zero zero
weakly negative weakly positive
strongly negative strongly positive

These observations allow us to find the character of η(x) without finding a

formal solution. If at one point η(x) is positive with positive slope, then
moving to the right η(x) will grow because of the positive slope, but that
growth rate will decline because of the negative curvature. Eventually
180 Sketching energy eigenfunctions

the slope becomes zero and then negative, but the curvature continues
negative. Because of the negative slope, η(x) eventually plunges through
η(x) = 0 (where its curvature is zero) and into regions where η(x) is negative
and hence the curvature is positive. The process repeats to produce the
following graph:

strong
negative
curvature

η(x) etc.

weak
negative zero
curvature curvature x

weak
positive
curvature

strong
positive
curvature

[[You can solve differential equation (5.2) formally to obtain

p
η(x) = A sin(( 2mKc /~)x + φ) (5.3)
where A and φ are adjusted to fit the initial or boundary conditions. In
fact, this is exactly the equation that we already solved at (4.17). The
formal approach has the advantage of finding an exact expression for the
wavelength. The informal approach has the advantage of building your
intuition.]]
Solving the Energy Eigenproblem 181

The direct way of keeping track of curvature in this classically allowed

region is
negative curvature when η(x) is positive;
positive curvature when η(x) is negative.
But this is sort of clunky: to keep track of curvature, you have to keep
track of height. A compact way of keeping track of the signs is that
in a classically allowed region, (5.4)
the eigenfunction curves toward the axis.
It doesn’t slope toward the axis, as you can see from the graph, it curves
toward the axis. Draw a tangent to the energy eigenfunction: in a classically
allowed region, the eigenfunction will fall between that tangent line and the
axis.
In fact, the informal approach uncovers more than just the oscillatory
character of η(x). Equation (5.2) shows that when Kc is large and positive,
the “curving toward” impetus is strong; when Kc is small and positive,
that impetus is weak. Thus when Kc is large, the wavefunction takes tight
turns and snaps back toward the axis; when Kc is small, it lethargically
bends back toward the axis. And sure enough the formal approach at
equation (5.3) shows that the wavelength λ depends on Kc through
2π~
λ= √ , (5.5)
2mKc
so a large Kc results in a short wavelength — a “tight turn” toward the
axis.
Now turn your attention to a classically prohibited region where
Kc (x) is constant and negative. Equation (5.2) says that if η(x) is positive,
then the curvature is positive. Once again we can uncover the character
of η(x) without finding a formal solution. If at one point η(x) is positive
with positive slope, then moving to the right η(x) will grow because of
the positive slope, and that growth rate increases because of the positive
curvature. The slope becomes larger and larger and η(x) rockets to infinity.
Or, if η(x) starts out negative with negative slope, then it rockets down to
negative infinity. Or, if η(x) starts out positive with negative slope, it
might cross the axis before rocketing down to negative infinity, or it might
dip down toward the axis without crossing it, before rocketing up to positive
infinity.
182 Sketching energy eigenfunctions

strong
positive
curvature

η(x)

weak
positive
zero
curvature curvature x

[[You can solve differential equation (5.2) formally to obtain

√ √
η(x) = Ae+( 2m|Kc |/~)x + Be−( 2m|Kc |/~)x
where A and B are adjusted to fit the initial or boundary conditions.]]
The direct way of keeping track of curvature in this classically prohibited
region is
positive curvature when η(x) is positive;
negative curvature when η(x) is negative.
But a compact way is remembering that
in a classically prohibited region, (5.6)
the eigenfunction curves away from the axis.
Draw a tangent to the energy eigenfunction: in a classically prohibited
region, that tangent line will fall between the eigenfunction and the axis.
Let’s apply all these ideas to finding the character of energy eigenfunc-
tions in a finite square well. Solve differential equation (5.2) for an energy
E just above the bottom of the well. (I will draw the potential energy func-
tion in olive green, the energy E in blue, and the solution η(x) in red.)
Solving the Energy Eigenproblem 183

Suppose the wavefunction starts out on the left small and just above the
axis. The region is strongly prohibited, that is Kc (x) is strongly negative,
so η(x) curves strongly away from the axis. Then (at the dashed vertical
line) the solution moves into a classically allowed region. But Kc (x) is only
weakly positive, so η(x) curves only weakly toward the axis. By the time
the solution gets to the right-hand classically prohibited region at the next
dashed vertical line, η(x) has only a weakly negative slope. In the prohib-
ited region the slope increases as η(x) curves strongly away from the axis
and rockets off to infinity.

curve strongly
away from
axis

curve weakly
toward
curve strongly axis
away from
axis x

You should check that the curvatures and tangents of this energy eigen-
function strictly obey the rules set down at (5.4) and (5.6). What happens
when η(x) crosses a dashed vertical line, the boundary between a classically
prohibited and a classically allowed region?
If you have studied differential equations you know that for any value
of E, equation (5.2) has two linearly independent solutions. We’ve just
sketched one of them. The other is the mirror image of it: small to the
right and rocketing to infinity toward the left. Because of the “rocketing off
to infinity” neither solution is normalizable. So these two solutions don’t
correspond to any physical energy eigenstate. To find such a solution we
have to try a different energy.
184 Sketching energy eigenfunctions

So we try an energy slightly higher. Now the region on the left is not so
strongly prohibited as it was before, so η(x) curves away from the axis less
dramatically. Then when it reaches the classically allowed region it curves
more sharply toward the axis, so that it’s strongly sloping downward when
it reaches the right-hand prohibited region. But not strongly enough: it
curves away from the axis and again rockets off to infinity — although this
time not so dramatically.

Once again we find a solution (and its mirror image is also a solution), but
it’s a non-physical, unnormalizable solution.
As we try energies higher and higher, the “rocketing to infinity” happens
further and further to the right, until at one special energy it doesn’t happen
at all. Now the wavefunction is normalizable, and now we have found an
energy eigenfunction.
Solving the Energy Eigenproblem 185

What happens when we try an energy slightly higher still? At the

right-hand side the wavefunction now rockets off to negative infinity! With
increased energies, the wavefunction rockets down to negative infinity with
increased drama. But then at some point, the drama decreases: as the
energy rises the wavefunction continues to go to negative infinity, but it does
so more and more slowly. Finally at one special energy the wavefunction
settles down exactly to zero as x → ∞, and we’ve found a second energy
eigenfunction.
186 Sketching energy eigenfunctions

(The misconception concerning “pointlike particles moving across a node”,

discussed on page 140, applies to this state as well.)
The process continues: with still higher values of E, the wavefunction
η(x) diverges to positive infinity as x → ∞ until we reach a third special
energy eigenvalue, then to negative infinity until we reach a fourth. Higher
and higher energies result in higher and higher values of Kc and hence
stronger and stronger snaps back toward the axis. The first (lowest) eigen-
function has no nodes, the second has one node, the third will have two
nodes, and in general the nth energy eigenfunction will have n − 1 nodes.
(See also the discussion on page 190.)
Notice that for a potential energy function symmetric about a point, the
energy eigenfunction is either symmetric or antisymmetric about that point.
The energy eigenfunction does not need to possess the same symmetry as
the potential energy function. (See also problem 5.7, “Parity”.)
Solving the Energy Eigenproblem 187

What about a “lopsided” square well that lacks symmetry? In the

case sketched below the energy is strongly prohibited to the left, weakly
prohibited to the right. Hence the wavefunction curves away sharply to the
left, mildly to the right. The consequence is that the tail is short on the
left, long on the right.

weakly prohibited
curve weakly away from axis
strongly prohibited
curve strongly away from axis x

In some way it makes sense that the wavefunction tail should be longer
where the classical prohibition is milder.
188 Sketching energy eigenfunctions

Now try a square well with two different floor levels:

Within the deep left side of the well, Kc is relatively high, so the tendency
for η to curve toward the axis is strong; within the shallow right side Kc is
relatively low, so the tendency to curve toward the axis is weak. Thus within
the deep side of the well, η(x) snaps back toward the axis, taking the curves
like an expertly driven sports car; within the shallow side η(x) leisurely
curves back toward the axis, curving like a student driver in a station
wagon. Within the deep side, wavelength will be short and amplitude will
be small; within the shallow side, wavelength will be longer and amplitude
will be large (or at least the same size). One finds smaller amplitude at
the deeper side of the well, and hence, all other things being equal, smaller
probability for the particle to be in the deep side of the well.
Solving the Energy Eigenproblem 189

This might seem counterintuitive: Shouldn’t it be more probable for the

particle to be in the deep side? After all, if you throw a classical marble
into a bowl it comes to rest at the deepest point and spends most of its
time there. The problem with this analogy is that it compares a classical
marble rolling with friction to a quantal situation without friction. Imagine
a classical marble rolling instead in a frictionless bowl: it never does come to
rest at the deepest point of the bowl. In fact, at the deepest point it moves
fastest: the marble spends little time at the deepest point and a lot of time
near the edges, where it moves slowly. The classical and quantal pictures
don’t correspond exactly (there’s no such thing as an energy eigenstate
in classical mechanics, the classical marble always has a position, and its
description never has a node), but the two pictures agree that the particle
has high probability of appearing where the potential energy function is
shallow, not deep.
190 Sketching energy eigenfunctions

Similar results hold for three-level square wells, for four-level square
wells, and so forth. And because any potential energy function can be
approximated by a series of steps, similar results hold for any potential
energy function.

Number of nodes. For the infinite square well, the energy eigen-
function ηn (x) has n − 1 interior nodes. The following argument1 shows
that same holds for any one-dimensional potential energy function V (x).
Imagine a modified potential

∞ x ≤ −a
Va (x) = V (x) −a < x < +a .
∞ +a ≤ x


When a is very small this is virtually an infinite square well, whose en-
ergy eigenfunctions we know. As a grows larger and larger, this potential
becomes more and more like the potential of interest V (x). During this
expansion, can an extra node pop into an energy eigenfunction? If it does,
then at the point xp where it pops in the wavefunction vanishes, η(xp ) = 0,
and its slope vanishes, η 0 (xp ) = 0. But the energy eigenproblem is a second-
order ordinary differential equation: the only solution with η(xp ) = 0 and
η 0 (xp ) = 0 is η(x) = 0 everywhere. This is not an eigenfunction. This can
never happen.

1 M. Moriconi, “Nodes of wavefunctions” American Journal of Physics 75 (March 2007)

284–285.
Solving the Energy Eigenproblem 191

Summary

In classically prohibited regions, the eigenfunction magnitude de-

clines while stepping away from the well: the stronger the pro-
hibition, the more rapid the decline.
In classically allowed regions, the eigenfunction oscillates: in re-
gions that are classically fast, the oscillation has small ampli-
tude and short wavelength; in regions that are classically slow,
the oscillation has large amplitude and long wavelength.
If the potential energy function is symmetric under reflection about
a point, the eigenfunction will be either symmetric or antisym-
metric under the same reflection.
The nth energy eigenfunction has n − 1 nodes.

Quantum mechanics involves situations (very small) and phenomena

(interference, entanglement) remote from daily experience. And the energy
eigenproblem, so central to quantum mechanics, does not arise in classical
mechanics at all. Some people conclude from these facts that one cannot
develop intuition about quantum mechanics, but that is false: the tech-
niques of this section do allow you to develop a feel for the character of
energy eigenstates. Just as chess playing or figure skating must be stud-
ied and practiced to develop proficiency, so quantum mechanics must be
studied and practiced to develop intuition. If people don’t develop intu-
ition regarding quantum mechanics, it’s not because quantum mechanics is
intrinsically fantastic; it’s because these people never try.
192 Sketching energy eigenfunctions

Problems
5.1 Would you buy a used eigenfunction from this man?
(recommended problem)
The four drawings below and on the next pages show four one-
dimensional potential energy functions V (x) (in olive green) along with
candidate energy eigenfunctions η(x) (in red) that purport to associate
with those potential energy functions. There is something wrong with
every candidate. Using the letter codes below, identify all eigenfunc-
tion errors, and sketch a qualitatively correct eigenfunction for each
potential.
The energy eigenfunction is drawn incorrectly because:
A. Wrong curvature. (It curves toward the axis in a classi-
cally prohibited region or away from the axis in a classi-
cally allowed region.)
B. Its wavy part has the wrong number of nodes.
C. The amplitude of the wavy part varies incorrectly.
D. The wavelength of the wavy part varies incorrectly.
E. One or more of the declining tails has the wrong length.

a.

E3

η3 (x)
x
Solving the Energy Eigenproblem 193

b.

E4

η4 (x)
x

c.

E5

η5 (x)
x
194 Sketching energy eigenfunctions

d.

E6

η6 (x)
x

5.2 Simple harmonic oscillator energy eigenfunctions

Here are sketches of the three lowest-energy eigenfunctions for the po-
tential energy function V (x) = 12 kx2 (called the “simple harmonic os-
cillator”). In eight sentences or fewer, describe how these energy eigen-
functions do (or don’t!) display the characteristics discussed in the
summary on page 191.
Solving the Energy Eigenproblem 195

5.3 de Broglie wavelength

Compare equation (5.5) to the formula for de Broglie wavelength. Does
this shed any light on the question (see page 46) of “what is waving”
in a de Broglie wave?
5.4 Wavelength as a function of Kc
Before equation (5.5) we provided an informal argument that the wave-
length λ would decrease with increasingpKc . This argument didn’t say
whether λ would vary as 1/Kc , or as 1/ Kc , or even as e−Kc /(constant) .
Produce a dimensional argument showing
p that if λ depends only on ~,
m, and Kc , then it must vary as ~/ mKc .
5.5 “At least the same size amplitude”
Page 188 claims that in the two-level square well, the amplitude of η(x)
on the right would be larger “or at least the same size” as the amplitude
on the left. Under what conditions will the amplitude be the same size?
196 Sketching energy eigenfunctions

5.6 Placement of nodes

Let ηn (x) and ηm (x) be solutions to
~2 00
− η (x) + V (x)ηm (x) = Em ηm (x) (5.7)
2M m
~2 00
− η (x) + V (x)ηn (x) = En ηn (x) (5.8)
2M n
with Em > En . The Sturm comparison theorem states that between
any two nodes of ηn (x) there exists at least one node of ηm (x). Prove
the theorem through contradiction by following these steps:
a. Multiply (5.7) by ηn , multiply (5.8) by ηm , and subtract to show
that
~2 0
− [η (x)ηn (x)−ηm (x)ηn0 (x)]0 = (Em −En )ηm (x)ηn (x). (5.9)
2M m
b. Call two adjacent nodes of ηn (x) by the names x1 and x2 . Argue
that we can select ηn (x) to be always positive for x1 < x < x2 ,
and show that with this selection ηn0 (x1 ) > 0 while ηn0 (x2 ) < 0.
c. Integrate equation (5.9) from x1 to x2 , producing
~2
− [−ηm (x2 )ηn0 (x2 ) + ηm (x1 )ηn0 (x1 )]
2M Z x2
= (Em − En ) ηm (x)ηn (x) dx. (5.10)
x1

d. If ηm (x) does not have a zero within x1 < x < x2 , then argue that
we can select ηm (x) always positive on the same interval, including
the endpoints.
The assumption that “ηm (x) does not have a zero” hence implies that
the left-hand side of (5.10) is strictly negative, while the right-hand
side is strictly positive. This assumption, therefore, must be false.
5.7 Parity
a. Think of an arbitrary potential energy function V (x). Now think
of its mirror image potential energy function U (x) = V (−x) Show
that if η(x) is an eigenfunction of V (x) with energy E, then σ(x) =
η(−x) is an eigenfunction of U (x) with the same energy.
b. If V (x) is symmetric under reflection about the origin, that is
U (x) = V (x), you might think that σ(x) = η(x). But no! This
identification ignores global phase freedom (pages 107 and 124).
Show that in fact σ(x) = rη(x) where the “overall phase factor”
r is a complex number with magnitude 1.
Solving the Energy Eigenproblem 197

c. The overall phase factor r is a number, not a function of x: the

same phase factor r applies at x = 2 (η(−2) = rη(2)), at x = 7
(η(−7) = rη(7)), and at x = −2 (η(2) = rη(−2)). Conclude that
r can’t be any old complex number with magnitude 1, it must be
either +1 or −1.
Energy eigenfunctions symmetric under reflection, η(x) = η(−x), are
said to have “even parity” while those antisymmetric under reflection,
η(x) = −η(−x), are said to have “odd parity”.
5.8 Scaling
Think of an arbitrary potential energy function V (x), for example per-
haps the one sketched on the left below. Now think of another po-
tential energy function U (y) that is half the width and four times the
depth/height of V (x), namely U (y) = 4V (x) where y = x/2. Without
solving the energy eigenproblem for either V (x) or U (y), I want to find
how the energy eigenvalues of U (y) relate to those of V (x).

V (x) U (y)
x y

Show that if η(x) is an eigenfunction of V (x) with energy E, then

σ(y) = η(x) is an eigenfunction of U (y). What is the corresponding
energy? After working this problem for the scale factor 2, repeat for a
general scale factor s so that U (y) = s2 V (x) where y = x/s.
[[This problem has a different cast from most: instead of giving you a
problem and asking you to solve it, I’m asking you to find the relation-
ship between the solutions of two different problems, neither of which
you’ve solved. My thesis adviser, Michael Fisher, called this “Juicing
an orange without breaking its peel.”]]
198 Scaled quantities

5.2 Scaled quantities

Look again at the quantal energy eigenproblem (5.2)

d2 η(x) 2m
= − 2 [E − V (x)]η(x). (5.11)
dx2 ~
Suppose you want to write a computer program to solve this problem for
the lopsided square well with potential energy function

 V1 x < 0
V (x) = 0 0<x<L . (5.12)

V2 L < x
The program would have to take as input the particle mass m, the energy
E, the potential well length L, and the potential energy values V1 and
V2 . Five parameters! Once the program is written, you’d have to spend a
lot of time typing in these parameters and exploring the five-dimensional
parameter space to find interesting values. Furthermore, these parameters
have inconvenient magnitudes like the electron’s mass 9.11 × 10−31 kg or
the length of a typical carbon nanotube 1.41 × 10−10 m. Isn’t there an
easier way to set up this problem?
There is. The characteristic length for this problem is L. If you try to
combine the parameters L, m, and ~ to form a quantity with the dimensions
of energy (see sample problem 5.2.1 on page 200) you will find that there
is only one way: this problem’s characteristic energy is Ec = ~2 /mL2 .
Define the dimensionless length variable x̃ = x/L, the dimensionless energy
parameter Ẽ = E/Ec , and the dimensionless potential energy function
Ṽ (x̃) = V (x̃L)/Ec = V (x)/Ec .
In terms of these new so-called “scaled quantities” the quantal energy
eigenproblem is
d2 η(x̃) 1 2m ~2
 
=− 2 [Ẽ − Ṽ (x̃)]η(x̃)
dx̃2 L2 ~ mL2
or
d2 η(x̃)
= −2[Ẽ − Ṽ (x̃)]η(x̃) (5.13)
dx̃2
where

 Ṽ1 x̃ < 0
Ṽ (x̃) = 0 0 < x̃ < 1 . (5.14)

Ṽ2 1 < x̃
Solving the Energy Eigenproblem 199

The scaled problem has many advantages. Instead of five there are only
three parameters: Ẽ, V˜1 , and V˜2 . And those parameters have nicely sized
values like 1 or 0.5 or 6. But it has the disadvantage that you have to write
down all those tildes. Because no one likes to write down tildes, we just
drop them, writing the problem as
d2 η(x)
= −2[E − V (x)]η(x) (5.15)
dx2
where

 V1 x<0
V (x) = 0 0<x<1 (5.16)

V2 1<x
and saying that these equations are written down “using scaled quantities”.
When you compare these equations with equations (5.11) and (5.12),
you see that we would get the same result if we had simply said “let ~ =
m = L = 1”. This phrase as stated is of course absurd: ~ is not equal to
1; ~, m, and L do have dimensions. But some people don’t like to explain
what they’re doing so they do say this as shorthand. Whenever you hear
this phrase, remember that it covers up a more elaborate — and more
interesting — truth.
200 Scaled quantities

5.2.1 Sample Problem: Characteristic energy

Show that there is only one way to combine the quantities L, m, and ~ to
form a quantity with the dimensions of energy, and find an expression for
this so-called characteristic energy Ec .

Solution:

quantity dimensions
L [length]
m [mass]
2
~ [mass] × [length] /[time]
2 2
Ec [mass] × [length] /[time]

If we are to build Ec out of L, m, and ~, we must start with ~, because

that’s the only source of the dimension [time]. And in fact we must start
2
with ~2 , because that’s the only way to make a [time] .

quantity dimensions
L [length]
m [mass]
2 4 2
~2 [mass] × [length] /[time]
2 2
Ec [mass] × [length] /[time]

But ~2 has too many factors of [mass] and [length] to make an energy.
There is only one way to get rid of them: to divide by m once and by L
twice.

quantity dimensions
2 2
~2 /mL2 [mass] × [length] /[time]
2 2
Ec [mass] × [length] /[time]

There is only one possible characteristic energy, and it is Ec = ~2 /mL2 .

 5.3. Numerical solution of the energy eigenproblem 201

Problems
5.9 Characteristic time
Find the characteristic time for the square well problem by combining
the parameters L, m, and ~ to form a quantity with the dimensions
of time. Compare this characteristic time to the infinite square well
revival time found at equation (4.25).
5.10 Scaling for the simple harmonic oscillator
(recommended problem)
Execute the scaling strategy for the simple harmonic oscillator poten-
tial energy function V (x) = 12 kx2 . What is the characteristic length in
terms of k, ~, and m? What is the resulting scaled energy eigenprob-
lem? If you didn’t like to explain what you were doing, how would you
use shorthand to describe the result of this scaling strategy?

5.3 Numerical solution of the energy eigenproblem

Now that the quantities are scaled, we return to our task of writing a
computer program to solve, numerically, the energy eigenproblem. In order
to fit the potential energy function V (x) and the energy eigenfunction η(x)
into a finite computer, we must of course approximate those continuum
functions through their values on a finite grid. The grid points are separated
by a small quantity ∆. It is straightforward to replace the function V (x)
with grid values Vi and the function η(x) with grid values ηi . But what
should we do with the second derivative d2 η/dx2 ?
202 Numerical solution of the energy eigenproblem

Start with a representation of the grid function ηi :

ηi
6
ηi+1
6
ηi−1
6

- x
i−1 i i+1

The slope at a point halfway between points i − 1 and i (represented by the

left dot in the figure below) is approximately
ηi − ηi−1
,
∆
while the slope half way between the points i and i + 1 (represented by the
right dot) is approximately
ηi+1 − ηi
.
∆

ηi+1 − ηi
ηi − ηi−1 ""a
6aa ∆
∆ "" aa
a
" 6
"
6

- x
u u
i−1 i i+1

An approximation for the second derivative at point i is the change in slope

divided by the change in distance
ηi+1 − ηi ηi − ηi−1
−
∆ ∆
∆
 Solving the Energy Eigenproblem 203

so at point i we approximate
d2 η ηi+1 − 2ηi + ηi−1
2
≈ . (5.17)
dx ∆2

The discretized version of the energy eigenproblem (5.15) is thus

ηi+1 − 2ηi + ηi−1
= −2[E − Vi ]ηi (5.18)
∆2
which rearranges to
ηi+1 = 2[1 + ∆2 (Vi − E)]ηi − ηi−1 . (5.19)
The algorithm then proceeds from left to right. Start in a classically pro-
hibited region and select η1 = 0, η2 = 0.001. Then find
η3 = 2[1 + ∆2 (V2 − E)]η2 − η1 .
Now that you know η3 , find
η4 = 2[1 + ∆2 (V3 − E)]η3 − η2 .
Continue until you know ηi at every grid point.
For most values of E, this algorithm will result in a solution that rockets
to ±∞ at the far right. When you pick a value of E where the solution
approaches zero at the far right, you’ve found an energy eigenvalue. The
algorithm is called “shooting”, because it resembles shooting an arrow at a
fixed target: your first shot might be too high, your second too low, so you
try something between until you home in on your target.

Problems
5.11 Program
a. Implement the shooting algorithm using a computer spreadsheet,
your favorite programming language, or in any other way. You
will have to select reasonable values for ∆ and η2 .
b. Check your implementation by solving the energy eigenproblem
for a free particle and for an infinite square well.
c. Find the three lowest-energy eigenvalues for a square well with
V1 = V2 = 30. Do the corresponding eigenfunctions have the
qualitative character you expect?
d. Repeat for a square well with V1 = 50 and V2 = 30.
 204 Numerical solution of the energy eigenproblem

5.12 Algorithm parameter

Below equation (5.19) I suggested that you start the stepping algorithm
with η1 = 0, η2 = 0.001. What would have happened had you selected
η1 = 0, η2 = 0.003 instead?
5.13 Simple harmonic oscillator
(Work problem 5.10 on page 201 before working this one.)
Implement the algorithm for a simple harmonic oscillator using scaled
quantities. Find the five lowest-energy eigenvalues, and compare them
to the analytic results 0.5, 1.5, 2.5, 3.5, and 4.5.
5.14 Questions (recommended problem)
Update your list of quantum mechanics questions that you started at
problem 1.17 on page 46. Write down new questions and, if you have un-
covered answers to any of your old questions, write them down briefly.

[[For example, one of my questions would be: “For any value of E —

energy eigenvalue or no — equation (5.2) has two linearly independent
solutions. We saw on page 183 that often the two linearly independent
solutions are mirror images, one rocketing off to infinity as x → +∞
and the other rocketing off to infinity as x → −∞. But what about the
energy eigenfunctions, which go to zero as x → ±∞? What does the
other linearly independent solution look like then?”]]
 Chapter 6

Identical Particles

6.1 Two or three identical particles

Please review section 4.4, “Wavefunction: Two particles in one or three

dimensions”, on page 127. In that section we talked about two different
particles, say an electron and a neutron. We set up a grid, discussed bin
amplitudes ψi,j , and talked about the limit as the width of each bin shrank
to zero.
There is a parallel development for two identical particles, but with one
twist. Here is the situation when one particle is found in bin 5, the other
in bin 8:

x
5 8

And here is the situation when one particle is found in bin 8, the other in
bin 5:

x
5 8

No difference, of course. . . that’s the meaning of “identical”. And of course

this holds not only for bins 5 and 8, but for any pair of bins i and j, even if
i = j. (If the two particles don’t interact, it is perfectly plausible for both
of them to occupy the same bin at the same time.)
What does this mean for the state of a system with two identical par-
ticles? Suppose that, by hook or by crook, we come up with a set of bin

205
206 Two or three identical particles

amplitudes ψi,j that describes the state of the system. Then the set of
amplitudes φi,j = ψj,i describes that state just as well as the original set
ψi,j . Does this mean that φi,j = ψi,j ? Not at all. Remember global phase
freedom (pages 107 and 124): If every bin amplitude is multiplied by the
same “overall phase factor” — a complex number with magnitude unity
— then the resulting set of amplitudes describes the state just as well as
the original set did. Calling that overall phase factor s, we conclude that
φi,j = sψi,j .
But, because φi,j = ψj,i , the original set of amplitudes must satisfy
ψj,i = sψi,j . The variable name s comes from “swap”: when we swap
subscripts, we introduce a factor of s. The quantity s is a number. . . not
a function of i or j. For example, the same value of s must work for
ψ8,5 = sψ5,8 , for ψ7,3 = sψ3,7 , for ψ5,8 = sψ8,5 , . . . . Wait. What was that
last one? Put together the first and last examples:
ψ8,5 = sψ5,8 = s(sψ8,5 ) = s2 ψ8,5 .
Clearly, s2 = 1, so s can’t be any old complex number with magnitude
unity: it can be only s = +1 or s = −1.
Execute the now-familiar program of turning bin amplitudes into am-
plitude density, that is wavefunction, to find that
ψ(xA , xB ) = +ψ(xB , xA ) or ψ(xA , xB ) = −ψ(xB , xA ). (6.1)
The first kind of wavefunction is called “symmetric under coordinate swap-
ping”, the second is called “antisymmetric under coordinate swapping”.
This requirement for symmetry or antisymmetry under coordinate swap-
ping is called the Pauli1 principle.
It might distress you to see variables like xA : doesn’t xA mean the
position of particle “A” while xB means the position of particle “B”? So
doesn’t this terminology label the particles as “A” and “B”, which would
1 Wolfgang Pauli (1900–1958), Vienna-born Swiss physicist, was one of the founders of

quantum mechanics. In 1924 he proposed the “exclusion principle”, ancestor of today’s

symmetry/antisymmetry requirement; in 1926 he produced the first solution for the
energy eigenproblem for atomic hydrogen; in 1930 he proposed the existence of the
neutrino, a prediction confirmed experimentally in 1956; in 1934 he and “Viki” Weisskopf
discovered how to make sense of relativistic quantum mechanics by realizing that the
solutions to relativistic quantal equations do not give an amplitude for a single particle to
have a position (technically, a wavefunction), but rather an amplitude for an additional
particle to be created at a position or for an existing particle to be annihilated at a
position (technically, a creation or annihilation operator). He originated the insult,
applied to ideas that cannot be tested, that they are “not even wrong”.
Identical Particles 207

violate our initial requirement that the particles be identical? The answer
is that this terminology does not label one particle “A” and the other
particle “B”. Instead, it labels one point “A” and the other point “B”.
Look back to the figures on page 205: the numbers 5 and 8 label bins,
not particles, so when these bins shrink to zero the variables xA and xB
apply to points, not particles. That’s why I like to call these wavefunctions
“(anti)symmetric under swap of coordinates”. But you’ll hear people using
terms like “(anti)symmetric under particle swapping” or “. . . under particle
interchange” or “. . . under particle exchange”.
What if the two particles are in three-dimensional space, and what if
they have spin? In that case, the swap applies to all the coordinates: using
the sans serif notation of equation (4.106),
ψ(xA , xB ) = +ψ(xB , xA ) or ψ(xA , xB ) = −ψ(xB , xA ). (6.2)

From our argument so far, two identical electrons might be in a sym-

metric or an antisymmetric state. Similarly for two identical neutrons,
two identical silver atoms, etc. But it’s an empirical fact that the swap
symmetry depends only on the kind of particle involved: Two electrons
are always antisymmetric under swapping. Two 4 He atoms (both in their
ground state) are always symmetric. Particles that are always symmet-
ric under swapping are called bosons;2 those always antisymmetric under
swapping are called fermions.3

2 Named for Satyendra Bose (1894–1974) of India. He contributed to fields ranging

from chemistry to school administration, but his signal contribution was elucidating the
statistics of photons. Remarkably, he made this discovery in 1922, three years before
Schrödinger developed the concept of wavefunction.
3 Named for Enrico Fermi. See footnote on page 170.
208 Symmetrization and antisymmetrization

What about three particles? The wavefunction of three identical bosons

must be completely symmetric, that is, symmetric under swaps of any co-
ordinate pair:
+ψ(xA , xB , xC )
= +ψ(xA , xC , xB )
= +ψ(xC , xA , xB )
= +ψ(xC , xB , xA )
= +ψ(xB , xC , xA )
= +ψ(xB , xA , xC ). (6.3)
(These 6 = 3! permutations are listed in the sequence called “plain changes”
or “the Johnson-Trotter sequence”. This sequence has the admirable prop-
erty that each permutation differs from its predecessor by a single swap of
adjacent letters.4 ) Whereas the wavefunction of three identical fermions
must be completely antisymmetric, that is, antisymmetric under swaps of
any coordinate pair:
+ψ(xA , xB , xC )
= −ψ(xA , xC , xB )
= +ψ(xC , xA , xB )
= −ψ(xC , xB , xA )
= +ψ(xB , xC , xA )
= −ψ(xB , xA , xC ). (6.4)

As you would expect, Pauli’s requirement of complete symmetry or

antisymmetry under swaps of any coordinate pair holds for any number of
identical particles.

6.2 Symmetrization and antisymmetrization

Given the importance of wavefunctions symmetric or antisymmetric un-

der coordinate swaps, it makes sense to investigate the mathematics of
such “permutation symmetry”. This section treats systems of two or three
4 Donald Knuth, The Art of Computer Programming, volume 4A, “Combinatorial Al-

gorithms, Part 1” (Addison-Wesley, Boston, 1997) section 7.2.1.2, “Generating all per-
mutations”.
Identical Particles 209

particles; the generalization to systems of four or more particles is straight-

forward.
Start with any two-variable garden-variety function f (xA , xB ), not nec-
essarily symmetric or antisymmetric. Can that function be used as a “seed”
to build a symmetric or antisymmetric function? It can. The function
s(xA , xB ) = f (xA , xB ) + f (xB , xA ) (6.5)
is symmetric under swapping while the function
a(xA , xB ) = f (xA , xB ) − f (xB , xA ) (6.6)
is antisymmetric. If you don’t believe me, try it out:
s(5, 2) = f (5, 2) + f (2, 5)
s(2, 5) = f (2, 5) + f (5, 2)
so clearly s(5, 2) = s(2, 5). Meanwhile
a(5, 2) = f (5, 2) − f (2, 5)
a(2, 5) = f (2, 5) − f (5, 2)
so just as clearly a(5, 2) = −a(2, 5).
Can this be generalized to three variables? Start with a three-variable
garden-variety function f (xA , xB , xC ). The function
s(xA , xB , xC ) = f (xA , xB , xC )
+f (xA , xC , xB )
+f (xC , xA , xB )
+f (xC , xB , xA )
+f (xB , xC , xA )
+f (xB , xA , xC ) (6.7)
is completely symmetric while the function
a(xA , xB , xC ) = f (xA , xB , xC )
−f (xA , xC , xB )
+f (xC , xA , xB )
−f (xC , xB , xA )
+f (xB , xC , xA )
−f (xB , xA , xC ) (6.8)
210 Symmetrization and antisymmetrization

is completely antisymmetric. Once again, if you don’t believe me I invite

you to try it out with xA = 5, xB = 2, and xC = 7.
This trick is often used when the seed function is a product,
f (xA , xB , xC ) = f1 (xA )f2 (xB )f3 (xC ), (6.9)
in which case you may think of the symmetrization/antisymmetrization
machinery as being the sum over all permutations of the coordinates xA ,
xB , and xC , as above, or as the sum over all permutations of the functions
f1 (x), f2 (x), and f3 (x): the function
s(xA , xB , xC ) = f1 (xA )f2 (xB )f3 (xC )
+f1 (xA )f3 (xB )f2 (xC )
+f3 (xA )f1 (xB )f2 (xC )
+f3 (xA )f2 (xB )f1 (xC )
+f2 (xA )f3 (xB )f1 (xC )
+f2 (xA )f1 (xB )f3 (xC ) (6.10)
is completely symmetric while the function
a(xA , xB , xC ) = f1 (xA )f2 (xB )f3 (xC )
−f1 (xA )f3 (xB )f2 (xC )
+f3 (xA )f1 (xB )f2 (xC )
−f3 (xA )f2 (xB )f1 (xC )
+f2 (xA )f3 (xB )f1 (xC )
−f2 (xA )f1 (xB )f3 (xC ) (6.11)
is completely antisymmetric. Some people write this last expression as the
determinant of a matrix
f1 (xA ) f2 (xA ) f3 (xA )
a(xA , xB , xC ) = f1 (xB ) f2 (xB ) f3 (xB ) , (6.12)
f1 (xC ) f2 (xC ) f3 (xC )
and call it the “Slater5 determinant”. I personally think this terminol-
ogy confuses the issue (the expression works only if the seed function is a
product of one-variable functions, it suppresses the delightful and useful
“plain changes” sequence of permutations, plus I never liked determinants6
to begin with), but it’s widely used.
5 John C. Slater (1900–1976), American theoretical physicist who made major contribu-

tions to our understanding of atoms, molecules, and solids. Also important as a teacher,
textbook author, and administrator.
6 I am not alone. See Sheldon Axler, “Down with determinants!” American Mathemat-

ical Monthly 102 (February 1995) 139–154.

Identical Particles 211

6.2.1 Sample Problem: Try it out

Find the symmetric and antisymmetric functions generated by the seed

functions
f (xA , xB ) = xA x2B g(xA , xB , xC ) = xA x2C + 2xB xC .
What happens if the resulting functions are multiplied by 5, or by −1, or
by i?

Solution: For the two-variable function f (xA , xB ),

s(xA , xB ) = f (xA , xB ) + f (xB , xA ) = xA x2B + xB x2A = xA x2B + x2A xB ,
a(xA , xB ) = f (xA , xB ) − f (xB , xA ) = xA x2B − xB x2A = xA x2B − x2A xB .
And sure enough, just to try some particular cases, s(5, 2) = s(2, 5) = 70,
a(5, 2) = −30, a(2, 5) = +30, s(3, 3) = 54, a(3, 3) = 0. If you multiplied
s(xA , xB ) by 5, or by −1, or by i, or by any number, you would still have a
function symmetric under coordinate swaps. Similarly for a(xA , xB ). Note
particularly the multiplication by −1: in definition (6.6) we could have
swapped the + and − signs.
For the three-variable function g(xA , xB , xC ), equation (6.7) becomes
s(xA , xB , xC ) = g(xA , xB , xC )
+g(xA , xC , xB )
+g(xC , xA , xB )
+g(xC , xB , xA )
+g(xB , xC , xA )
+g(xB , xA , xC )
= xA x2C + 2xB xC
+xA x2B + 2xC xB
+xC x2B + 2xA xB
+xC x2A + 2xB xA
+xB x2A + 2xC xA
+xB x2C + 2xA xC
= x2A (xB + xC ) + x2B (xA + xC ) + x2C (xA + xB )
+ 4xA xB + 4xA xC + 4xB xC ,
212 Symmetrization and antisymmetrization

while equation (6.8) becomes

a(xA , xB , xC ) = g(xA , xB , xC )
−g(xA , xC , xB )
+g(xC , xA , xB )
−g(xC , xB , xA )
+g(xB , xC , xA )
−g(xB , xA , xC )
= xA x2C + 2xB xC
−xA x2B − 2xC xB
+xC x2B + 2xA xB
−xC x2A − 2xB xA
+xB x2A + 2xC xA
−xB x2C − 2xA xC
= x2A (xB − xC ) + x2B (−xA + xC ) + x2C (xA − xB ).
Trying out some particular cases: s(1, 2, 3) = s(2, 1, 3) = s(2, 3, 1) = 92,
a(1, 2, 3) = −a(2, 1, 3) = a(2, 3, 1) = −2, s(1, 1, 3) = s(1, 3, 1) = 54,
a(1, 1, 3) = a(1, 3, 1) = 0. As in the two-variable case, if we multiply
either of these functions by a constant the symmetry or antisymmetry will
be unaffected. In the definition (6.8) for the antisymmetrization process we
could have swapped the + and − signs.
Identical Particles 213

Problems
6.1 Product functions
Can a product function F (xA )G(xB ) ever be symmetric under swap-
ping? Antisymmetric?
6.2 Special case
Implement the symmetrization and antisymmetrization proce-
dures (6.7) and (6.8) for the garden-variety function f (xA , xB , xC ) =
xA x2B x3C . Evaluate the resulting functions s(xA , xB , xC ) and
a(xA , xB , xC ) first at xA = 1, xB = 2, and xC = 3, then at xA = 3,
xB = 2, and xC = 1. [[Results: s(1, 2, 3) = s(3, 2, 1) = 288,
a(1, 2, 3) = 12, a(3, 2, 1) = −12.]]
6.3 Antisymmetrizing the symmetric
a. There is one function that is both completely symmetric and com-
pletely antisymmetric. What is it?
b. Suppose the seed function is symmetric under a swap of the first
two coordinates
f (xA , xB , xC ) = f (xB , xA , xC )
and the antisymmetrization process (6.8) is executed. What is the
result?
c. Repeat part (b) for a seed function symmetric under a swap of the
last two coordinates.
d. Repeat part (b) for a seed function symmetric under a swap of the
first and third coordinates.
e. Suppose the seed function is a product as in equation (6.9), and
two of the functions happen to be equal. What is the result of the
antisymmetrization process?
214 Consequences of the Pauli principle

6.3 Consequences of the Pauli principle

Does the requirement of symmetry or antisymmetry under coordinate swap-

ping have any consequences? Here’s an immediate one for fermions: Take
both xA = X and xB = X. Now when these coordinates are swapped, you
get back to where you started:
ψ(X, X) = −ψ(X, X) so ψ(X, X) = 0. (6.13)
Thus, the probability density for two identical fermions to have all the same
coordinates is zero.
And here’s a consequence for both bosons and fermions. Think about
space only, no spin. The (unnormalized) seed function
2
+(xB +0.3σ)2 ]/2σ 2
f (xA , xB ) = e−[(xA −0.5σ)
has a maximum when xA = 0.5σ and when xB = −0.3σ. This shows up
as one hump in the two-variable plots below (drawn taking σ = 1), which
show the normalized probability density proportional to |f (xA , xB )|2 .
xB

xB

xA
xA

Depending on your background and preferences, you might find it easier

to read either the surface plot on the left or the contour plot on the right:
both depict the same two-variable function. (And both were drawn using
Paul Seeburger’s applet CalcPlot3D.)
Identical Particles 215

But what of the symmetric and antisymmetric combinations generated

from this seed? Here are surface plots of the normalized probability densi-
ties associated with the symmetric (left) and antisymmetric (right) combi-
nations:

xB xB

xA xA

And here are the corresponding contour plots:

xB xB

xA xA

The seed function has no special properties on the xA = xB diagonal axis.

But, as required by equation (6.13), the antisymmetric combination van-
ishes there. And the symmetric combination is high there!
216 Consequences of the Pauli principle

The “vanishing on diagonal requirement” and this particular example

are but two facets of the more general rule of thumb that:

In a symmetric spatial wavefunction, the particles tend to huddle

together.
In an antisymmetric spatial wavefunction, the particles tend to
spread apart.

This rule is not a theorem and you can find counterexamples,7 but such
exceptions are rare.
In everyday experience, when two people tend to huddle together or
spread apart, it’s for emotional reasons. In everyday experience, when
two particles tend to huddle together or spread apart, it’s because they’re
attracted to or repelled from each other through a force. This quantal
case is vastly different. The huddling or spreading is of course not caused
by emotions and it’s also not caused by a force — it occurs for identical
particles even when they don’t interact. The cause is instead the symme-
try/antisymmetry requirement: not a force like a hammer blow, but a piece
of mathematics!
Therefore it’s difficult to come up with terms for the behavior of identical
particles that don’t suggest either emotions or forces ascribed to particles:
congregate, avoid; gregarious, loner; attract, repel; flock, scatter. “Huddle
together” and “spread apart” are the best terms I’ve been able to devise,
but you might be able to find better ones.

Problem
6.4 Symmetric and antisymmetric combinations
Two identical particles ambivate in a one-dimensional infinite square
well. Take as a seed function the product of energy eigenstates
η2 (xA )η3 (xB ). Use your favorite graphics package to plot the proba-
bility densities associated with the symmetric and antisymmetric com-
binations generated from this seed. Does the “huddle together/spread
apart” rule hold?
7 See D.F. Styer, “On the separation of identical particles in quantum mechanics” Eu-

ropean Journal of Physics 41 (14 October 2020) 065402.

6.4. Consequences of the Pauli principle for product states 217

6.4 Consequences of the Pauli principle for product states

A commonly encountered special case comes when the many-particle seed

function is a product of one-particle functions — we glanced at this special
case in equation (6.9). What happens if two of these one-particle functions
are the same? Nothing special happens for the symmetrization case. But
the answer for antisymmetrization is cute. It pops out of equation (6.11):
If f1 (x) = f2 (x), then the last line cancels the first line, the second cancels
the fifth, and the fourth cancels the third. The antisymmetric combination
vanishes everywhere!
Unlike the “huddle together/spread apart” rule of thumb, this result is
a theorem: the antisymmetric combination vanishes if any two of the one-
particle functions are the same. It is a partner to the xA = xB theorem of
equation (6.13): just as the two particles can’t have the same coordinates,
so their wavefunction can’t be built from the same one-particle functions.

6.5 Energy states for two identical, noninteracting particles

A single particle ambivates subject to some potential energy function.

There are M energy eigenstates (where usually M = ∞)
η1 (x), η2 (x), η3 (x), . . . , ηM (x). (6.14)

Now two non-identical particles ambivate subject to the same po-

tential energy. They have the same mass, and do not interact with each
other. You can see what the energy eigenstates are: state η3 (xA )η8 (xB ), for
example, has energy E3 + E8 . There’s necessarily a degeneracy, as defined
on page 150, because the different state η8 (xA )η3 (xB ) has the same energy.
The basis of energy eigenstates has M 2 elements, and they are normalized.
Any state can be represented as a linear combination of these elements. I
could go on, but the picture is clear: the fact that there are two particles
rather than one is unimportant; this basis of energy eigenstates has all the
properties you expect of an energy eigenbasis.
This basis consists of product states, but of course that’s just a coinci-
dence. You could replace the two basis states
η3 (xA )η8 (xB ) and η8 (xA )η3 (xB )
with, for example [using equation (4.39) with cos θ = 54 ],
+ 45 η3 (xA )η8 (xB ) + 35 η8 (xA )η3 (xB ) and − 53 η3 (xA )η8 (xB ) + 45 η8 (xA )η3 (xB ).
218 Energy states for two identical, noninteracting particles

Now two identical, noninteracting particles ambivate subject to

the same potential energy. In this case we don’t want a basis from which
we can build any wavefunction: we want a basis from which we can build
any symmetric wavefunction, or a basis from which we can build any anti-
symmetric wavefunction.
The basis for antisymmetric wavefunctions has elements like
√1 [−η3 (xA )η8 (xB ) + η8 (xA )η3 (xB )] . (6.15)
2

The basis for symmetric wavefunctions has elements like

√1 [+η3 (xA )η8 (xB ) + η8 (xA )η3 (xB )] (6.16)
2

plus elements like

η8 (xA )η8 (xB ). (6.17)
You should convince yourself that there are 21 M (M − 1) elements in the an-
tisymmetric basis and 21 M (M + 1) elements in the symmetric basis. Notice
that non-product8 states are a strict necessity in these bases.
The basis for antisymmetric wavefunctions united with the basis for
symmetric wavefunctions produces a basis for any wavefunction. This is
a peculiarity of the two-particle case, and reflects the fact that any two-
variable function is the sum of a completely symmetric function and a com-
pletely antisymmetric function. It is not true that any three-variable func-
tion is the sum of a completely symmetric and a completely antisymmetric
function. For three noninteracting particles, the general basis has M 3 ele-
ments, the basis for antisymmetric wavefunctions has 16 M (M − 1)(M − 2)
elements, and the basis for symmetric wavefunctions has 16 M (M +1)(M +2)
elements. If you enjoy mathematical puzzles, you will enjoy proving these
statements for yourself. But we won’t need them for this book.
Before proceeding, I introduce some terminology. The phrase “one-
particle states multiplied together then permuted through the symmetriza-
tion/antisymmetrization machinery of equations (6.10) and (6.11) to build
a many-particle state” is a real mouthful. A “many-particle state” like
(6.15) or (6.16) or (6.17) is called just a “state”, while a building block
8 Because the states (6.15 and (6.16) are clearly not in the form of a product

ψA (xA )ψB (xB ), the definition of entangled state on page 129 suggests that these states
should be called entangled. However the correct definition of entanglement for identical
particles remains unsettled, so I use the term “non-product state” instead.
6.6. Spin plus space, two electrons 219

one-particle state like η3 (x) or η8 (x) is called a “level”.9 This terminology

relieves us of the need say “one-particle” or “many-particle” or “antisym-
metrization machinery” over and over again.
The basis for bosons is bigger than the basis for fermions because you
can combine levels 1 and 7 to build either a boson state or a fermion state,
but you can combine levels 7 and 7 to build a boson state but not a fermion
state. As detailed in the previous section, “Consequences of the Pauli
principle for product states”, the levels combined to make a fermion state
must all be different.

Problem
6.5 Building three-particle basis states
Suppose you had three particles and three “building block” levels (say
the orthonormal levels η1 (x), η3 (x), and η7 (x)). Construct normalized
three-particle basis states for the case of
a. three non-identical particles
b. three identical bosons
c. three identical fermions
How many states are there in each basis? Repeat for three particles
with four one-particle levels, but in this case simply count and don’t
write down all the three-particle states.

6.6 Spin plus space, two electrons

Electrons are spin-half fermions. Two of them ambivate subject to the

same potential. Energy doesn’t depend on spin. Pretend the two electrons
don’t interact. (Perhaps a better name for this section would be “Spin
plus space, two noninteracting spin- 12 fermions”, but yikes, how long do
you want this section’s title to be? Should I add “non-relativistic” and
“ignoring collisions” and “ignoring radiation”?)
The spatial energy levels for one electron are ηn (~x) for n = 1, 2, . . . , M/2.
Thus the full (spin plus space) energy levels for one electron are the M levels
9 Some people, particularly chemists referring to atomic systems, use the term “orbital”

rather than “level”. This term unfortunately suggests a circular Bohr orbit. An electron
with an energy does not execute a circular Bohr orbit at constant speed. Instead it
ambivates without position or velocity.
220 Spin plus space, two electrons

ηn (~x)χ+ and ηn (~x)χ− . Now the question: What are the energy eigenstates
for the two noninteracting electrons?
Well, what two-particle states can we build from the one-particle spatial
levels with, say, n = 1 and n = 3? (Once you see how to do it for n = 1 and
n = 3, you can readily generalize to any two values of n.) These correspond
to four levels:
η1 (~x)χ+ , (6.18)
η1 (~x)χ− , (6.19)
η3 (~x)χ+ , (6.20)
η3 (~x)χ− . (6.21)
What states mixing n = 1 with n = 3 can be built from these four levels?
The antisymmetric combination of (6.18) with itself vanishes. The an-
tisymmetric combination of (6.18) with (6.19) is a combination of n = 1
with n = 1, not of n = 1 with n = 3. The (unnormalzed) antisymmetric
combination of (6.18) with (6.20) is
η1 (~xA )χ+ (A)η3 (~xB )χ+ (B) − η3 (~xA )χ+ (A)η1 (~xB )χ+ (B). (6.22)
The antisymmetric combination of (6.18) with (6.21) is
η1 (~xA )χ+ (A)η3 (~xB )χ− (B) − η3 (~xA )χ− (A)η1 (~xB )χ+ (B). (6.23)
The antisymmetric combination of (6.19) with (6.20) is
η1 (~xA )χ− (A)η3 (~xB )χ+ (B) − η3 (~xA )χ+ (A)η1 (~xB )χ− (B). (6.24)
The antisymmetric combination of (6.19) with (6.21) is
η1 (~xA )χ− (A)η3 (~xB )χ− (B) − η3 (~xA )χ− (A)η1 (~xB )χ− (B). (6.25)
Finally, the antisymmetric combination of (6.20) with (6.21) is a combina-
tion of n = 3 with n = 3, not of n = 1 with n = 3.
All four of these states are energy eigenstates with energy E1 + E3 .
State (6.22) factorizes into a convenient space-times-spin form:
η1 (~xA )χ+ (A)η3 (~xB )χ+ (B) − η3 (~xA )χ+ (A)η1 (~xB )χ+ (B)
 
= η1 (~xA )η3 (~xB ) − η3 (~xA )η1 (~xB ) χ+ (A)χ+ (B). (6.26)

The space part of the wavefunction is antisymmetric under coordinate swap.

The spin part is symmetric. Thus the total wavefunction is antisymmetric.
Identical Particles 221

Before proceeding I confess that I’m sick and tired of writing all these
ηs and χs and As and Bs that convey no information. I always write the
η in front of the χ. I always write the As in front of the Bs. You’ll never
confuse an η with a χ, because the ηs are labeled 1, 3 while the χs are
labeled +, −. Dirac introduced a notation (see page 90) that takes all this
for granted, so that neither you nor I have to write the same thing out over
and over again. This notation usually replaces + with ↑ and − with ↓ (see
page 112). In this notation, equation (6.26) is written
 
|1 ↑, 3 ↑i − |3 ↑, 1 ↑i = |1, 3i − |3, 1i | ↑↑ i. (6.27)

In this new notation the states (6.22) through (6.25) are written
 
|1, 3i − |3, 1i | ↑↑ i (6.28)

|1 ↑, 3 ↓i − |3 ↓, 1 ↑i (6.29)
|1 ↓, 3 ↑i − |3 ↑, 1 ↓i (6.30)
 
|1, 3i − |3, 1i | ↓↓ i. (6.31)

Well, this is cute. Two of the four states have this convenient space-times-
spin form. . . and furthermore these two have the same spatial wavefunction!
Two other states, however, don’t have this convenient form.
One thing to do about this is nothing. There’s no requirement that
states have a space-times-spin form. But in this two-electron case there’s a
slick trick that enables us to put the states into space-times-spin form.
Because all four states (6.28) through (6.31) have the same energy,
namely E1 + E3 , I can make linear combinations of the states to form other
equally good energy states. Can I make a combination of states (6.29) and
(6.30) that does factorize into space times spin? Nothing ventured, nothing
gained. Let’s try it:
   
α |1 ↑, 3 ↓i − |3 ↓, 1 ↑i + β |1 ↓, 3 ↑i − |3 ↑, 1 ↓i
   
= |1, 3i α| ↑↓ i + β| ↓↑ i − |3, 1i α| ↓↑ i + β| ↑↓ i .

This will factorize only if the left term in square brackets is proportional
to the right term in square brackets:
   
α| ↑↓ i + β| ↓↑ i = c β| ↑↓ i + α| ↓↑ i ,
222 Spin plus space, two electrons

that is only if
α = cβ and β = cα.
Combining these two equations results in c = ±1. If c = +1 then the
combination results in the state
   
|1, 3i − |3, 1i α | ↑↓ i + | ↓↑ i , (6.32)

whereas when c = −1 the result is

   
|1, 3i + |3, 1i α | ↑↓ i − | ↓↑ i . (6.33)

Putting all this together and, for the sake of good form, insuring normal-
ized states, we find that the two-electron energy states in equations (6.28)
through (6.31) can be recast as
 
√1 (|1, 3i − |3, 1i) | ↑↑ i (6.34)
2
  
√1 (|1, 3i − |3, 1i) √1 (| ↑↓ i + | ↓↑ i) (6.35)
2 2
 
√1 (|1, 3i − |3, 1i) | ↓↓ i (6.36)
2
  
√1 (|1, 3i + |3, 1i) √1 (| ↑↓ i − | ↓↑ i) . (6.37)
2 2

The first three of these states have spatial wavefunctions antisymmetric

under coordinate swaps and spin wavefunctions symmetric under coordinate
swaps — these are called “ortho states” or “a triplet”. The last one has a
symmetric spatial wavefunction and an antisymmetric spin wavefunction —
these are called “para states” or “a singlet”. Our discussion in section 6.3,
“Consequences of the Pauli principle”, demonstrates that in ortho states,
the two electrons tend to spread apart in space; in para states, they tend
to huddle together.
I write out the singlet spin state
√1 [| ↑↓ i − | ↓↑ i] (6.38)
2
using the verbose terminology
√1 [χ+ (A)χ− (B) − χ− (A)χ+ (B)] (6.39)
2
to make it absolutely clear that coordinate A is associated with both spin +
and spin −, as is coordinate B. It is impossible to say that “one electron
has spin up and the other has spin down”.
Identical Particles 223

This abstract machinery might seem purely formal, but in fact it has
tangible experimental consequences. In the sample problem below, the
machinery suggests that the ground state of the hydrogen atom is two-fold
degenerate, while the ground state of the helium atom is non-degenerate.
And this prediction is borne out by experiment!

6.6.1 Sample Problem:

Ground state degeneracy for one and two electrons

A certain potential energy function has two spatial energy eigenstates:

η1 (~x) with energy E1 and η2 (~x) with a higher energy E2 . These energies
are independent of spin.

a. A single electron (spin- 12 ) ambivates in this potential. Write out the

four energy eigenstates and the energy eigenvalue associated with each.
What is the ground state degeneracy?
b. Two non-interacting electrons ambivate in this same potential. Write
out the six energy eigenstates and the energy eigenvalue associated with
each. What is the ground state degeneracy?

Solution: (a) For the single electron:

energy eigenstate energy eigenvalue

η1 (~x)χ+ E1
η1 (~x)χ− E1
η2 (~x)χ+ E2
η2 (~x)χ− E2

The first two states listed are both ground states, so the ground state is
two-fold degenerate.
(b) For the two electrons, we build states from levels just as we did
in this section. The first line below is the antisymmetrized combination of
η1 (~x)χ+ with η1 (~x)χ− . This state has energy 2E1 . The next four lines are
built up exactly as equations (6.34) through (6.37) were. Each of these four
224 Spin plus space, two electrons

states has energy E1 +E2 . The last line is the antisymmetrized combination
of η2 (~x)χ+ with η2 (~x)χ− . This state has energy 2E2 .
η1 (~xA )η1 (~xB ) √1 [χ+ (A)χ− (B) − χ− (A)χ+ (B)]
2
√1 [η1 (~xA )η2 (~xB ) − η2 (~xA )η1 (~xB )] [χ+ (A)χ+ (B)]
2
√1 [η1 (~xA )η2 (~xB ) − η2 (~xA )η1 (~xB )] √12 [χ+ (A)χ− (B) + χ− (A)χ+ (B)]
2
√1 [η1 (~xA )η2 (~xB ) − η2 (~xA )η1 (~xB )] [χ− (A)χ− (B)]
2
√1 [η1 (~xA )η2 (~xB ) + η2 (~xA )η1 (~xB )] √12 [χ+ (A)χ− (B) − χ− (A)χ+ (B)]
2
η2 (~xA )η2 (~xB ) √12 [χ+ (A)χ− (B) − χ− (A)χ+ (B)] .
The ground state of the two-electron system is the first state listed: it is
non-degenerate.

Problems
6.6 Combining a spatial one-particle level with itself
What two-particle states can we build from the one-particle spatial level
with n = 3? How many of the resulting states are ortho, how many
para?
6.7 Change of basis through abstract rotation
Show that, in retrospect, the process of building states (6.35) and (6.37)
from states (6.29) and (6.30) is nothing but a “45◦ rotation” in the style
of equation (4.39).
6.8 Normalization of singlet spin state
Justify the normalization constant √12 that enters in moving from equa-
tion (6.33) to equation (6.37). Compare this singlet spin state to the
entangled state (3.37). (Indeed, one way to produce an entangled pair
of electrons is to start in a singlet state and then draw the two electrons
apart.)
6.9 Ortho and para accounting
Show that in our case with M/2 spatial energy levels, the two-electron
energy basis has 21 M (M − 1) elements, of which
3
2 (M/2)[(M/2) − 1] are ortho
(antisymmetric in space and symmetric in spin) and
1
2 (M/2)[(M/2) + 1] are para
(symmetric in space and antisymmetric in spin).
 6.7. Spin plus space, three electrons, ground state 225

6.10 Intersystem crossing

A one-electron system has a ground level ηg (~x) and an excited level
ηe (~x), for a total of four basis levels:
ηg (~x)χ+ , ηg (~x)χ− , ηe (~x)χ+ , ηe (~x)χ− .
A basis for two-electron states is then the six states:
ηg (~xA )ηg (~xB ) √12 [χ+ (A)χ− (B) − χ− (A)χ+ (B)]
√1 [ηg (~xA )ηe (~xB ) − ηe (~xA )ηg (~xB )] [χ+ (A)χ+ (B)]
2
√1 [ηg (~xA )ηe (~xB ) − ηe (~xA )ηg (~xB )] √12 [χ+ (A)χ− (B) + χ− (A)χ+ (B)]
2
√1 [ηg (~xA )ηe (~xB ) − ηe (~xA )ηg (~xB )] [χ− (A)χ− (B)]
2
√1 [ηg (~xA )ηe (~xB ) + ηe (~xA )ηg (~xB )] √12 [χ+ (A)χ− (B) − χ− (A)χ+ (B)]
2
ηe (~xA )ηe (~xB ) √12 [χ+ (A)χ− (B) − χ− (A)χ+ (B)] .
A transition from the second state listed above to the first is called an
“intersystem crossing”. One sometimes reads, in association with the
diagram below, that in an intersystem crossing “the spin of the excited
electron is reversed”. In five paragraphs or fewer, explain why this
phrase is inaccurate, perhaps even grotesque, and suggest a replace-
ment.

6.7 Spin plus space, three electrons, ground state

Three electrons are in the situation described in the first paragraph of sec-
tion 6.6 (energy independent of spin, electrons don’t interact). The full
listing of energy eigenstates has been done, but it’s an accounting night-
mare, so I ask a simpler question: What is the ground state?
Call the one-particle spatial energy levels η1 (~x), η2 (~x), η3 (~x), . . . . The
ground state will be the antisymmetrized combination of the three levels
η1 (~xA )χ+ (A) η1 (~xB )χ− (B) η2 (~xC )χ+ (C)
226 Spin plus space, three electrons, ground state

or the antisymmetrized combination of the three levels

η1 (~xA )χ+ (A) η1 (~xB )χ− (B) η2 (~xC )χ− (C).
The two states so generated are degenerate:10 both have energy 2E1 + E2 .
Write out the first state in detail. It is
√1 [ η1 (~xA )χ+ (A) η1 (~xB )χ− (B) η2 (~xC )χ+ (C)
6
−η1 (~xA )χ+ (A) η2 (~xB )χ+ (B) η1 (~xC )χ− (C)
+η2 (~xA )χ+ (A) η1 (~xB )χ+ (B) η1 (~xC )χ− (C)
−η2 (~xA )χ+ (A) η1 (~xB )χ− (B) η1 (~xC )χ+ (C)
+η1 (~xA )χ− (A) η2 (~xB )χ+ (B) η1 (~xC )χ+ (C)
−η1 (~xA )χ− (A) η1 (~xB )χ+ (B) η2 (~xC )χ+ (C) ]. (6.40)
This morass is another good argument for the abbreviated Dirac notation
introduced on page 221. I’m not concerned with normalization for the
moment, so I’ll write this first state as
|1 ↑, 1 ↓, 2 ↑i
−|1 ↑, 2 ↑, 1 ↓i
+|2 ↑, 1 ↑, 1 ↓i
−|2 ↑, 1 ↓, 1 ↑i
+|1 ↓, 2 ↑, 1 ↑i
−|1 ↓, 1 ↑, 2 ↑i (6.41)
and the second one (with 2 ↓ replacing 2 ↑) as
|1 ↑, 1 ↓, 2 ↓i
−|1 ↑, 2 ↓, 1 ↓i
+|2 ↓, 1 ↑, 1 ↓i
−|2 ↓, 1 ↓, 1 ↑i
+|1 ↓, 2 ↓, 1 ↑i
−|1 ↓, 1 ↑, 2 ↓i. (6.42)

Both of these states are antisymmetric, but neither factorizes into a

neat “space part times spin part”. If, following the approach used with two
electrons, you attempt to find a linear combination of these two that does
so factorize, you will fail: see problem 6.11. The ground state wavefunction
cannot be made to factor into a space part times a spin part.
10 See the definition on page 150 and problem 4.11 on page 151.
 Identical Particles 227

Problems
6.11 A doomed attempt (essential problem)
Any linear combination of state (6.41) with state (6.42) has the form
 
|1, 1, 2i α| ↑↓↑i + β| ↑↓↓i
 
−|1, 2, 1i α| ↑↑↓i + β| ↑↓↓i
 
+|2, 1, 1i α| ↑↑↓i + β| ↓↑↓i
 
−|2, 1, 1i α| ↑↓↑i + β| ↓↓↑i
 
+|1, 2, 1i α| ↓↑↑i + β| ↓↓↑i
 
−|1, 1, 2i α| ↓↑↑i + β| ↓↑↓i . (6.43)

Show that this form can never be factorized into a space part times a
spin part.
6.12 Questions (recommended problem)
Update your list of quantum mechanics questions that you started at
problem 1.17 on page 46. Write down new questions and, if you have un-
covered answers to any of your old questions, write them down briefly.
 Chapter 7

Atoms

During the months following these discussions [in the autumn of 1926] an
intensive study of all questions concerning the interpretation of quantum
theory in Copenhagen finally led to a complete and, as many physicists
believe, satisfactory clarification of the situation. But it was not a so-
lution which one could easily accept. I remember discussions with Bohr
which went through many hours till very late at night and ended almost
in despair; and when at the end of the discussion I went alone for a
walk in the neighboring park I repeated to myself again and again the
question: Can nature possibly be as absurd as it seemed to us in these
atomic experiments?
— Werner Heisenberg, Physics and Philosophy
(Harper, New York, 1958) page 42

All this is fine and good — lovely, in fact. But we have to apply quan-
tum mechanics to experimentally accessible systems, and while things like
carbon nanotubes exist, the most readily accessible systems are atoms.

7.1 Central potentials in two dimensions

Before jumping directly to three-dimensional atoms, we test out the math-

ematics in two dimensions.
In one dimension, the energy eigenproblem is
~2 d2 η(x)
− + V (x)η(x) = Eη(x). (7.1)
2M dx2
The generalization to two dimensions is straightforward:
~2 ∂ 2 η(x, y) ∂ 2 η(x, y)
 
− + + V (x, y)η(x, y) = Eη(x, y). (7.2)
2M ∂x2 ∂y 2

229
230 Central potentials in two dimensions

The part in square brackets is called “the Laplacian of η(x, y)” and is
represented by the symbol “∇2 ” as follows
 2
∂ f (x, y) ∂ 2 f (x, y)

+ ≡ ∇2 f (x, y). (7.3)
∂x2 ∂y 2
Thus the “mathematical form” of the energy eigenproblem is
2M
∇2 η(~r) + [E − V (~r)]η(~r) = 0. (7.4)
~2

Suppose V (x, y) is a “central potential” — that is, a function of distance

from the origin r only. Then it makes sense to use polar coordinates r and
θ rather than Cartesian coordinates x and y. What is the expression for the
Laplacian in polar coordinates? This can be uncovered through the chain
rule, and it’s pretty hard to do. Fortunately, you can look up the answer:
1 ∂ 2 f (r, θ)
   
2 1 ∂ ∂f (r, θ)
∇ f (~r) = r + 2 . (7.5)
r ∂r ∂r r ∂θ2
Thus, the partial differential equation to be solved is
1 ∂ 2 η(r, θ)
   
1 ∂ ∂η(r, θ) 2M
r + 2 2
+ 2 [E − V (r)]η(r, θ) = 0 (7.6)
r ∂r ∂r r ∂θ ~
or
∂ 2 η(r, θ)
 
∂ ∂η(r, θ) 2M
+r r + 2 r2 [E − V (r)]η(r, θ) = 0. (7.7)
∂θ2 ∂r ∂r ~

Use the “separation of variables” strategy introduced on page 133 : look

for solutions of the product form
η(r, θ) = R(r)Θ(θ), (7.8)
and hope against hope that all the solutions (or at least some of them) will
be of this form. Plugging this product form into the PDE gives
   
d dR(r) 2M
R(r)Θ00 (θ) + Θ(θ) r r + 2 r2 [E − V (r)]R(r) = 0
dr dr ~
00
   
Θ (θ) r d dR(r) 2M
+ r + 2 r2 [E − V (r)] = 0. (7.9)
Θ(θ) R(r) dr dr ~
Through the usual separation-of-variables argument, we recognize that if a
function of θ alone plus a function of r alone sum to zero, where θ and r are
independent variables, then both functions must be equal to a constant:
Θ00 (θ)
 
r d dR(r) 2M
r + 2 r2 [E − V (r)] = − = const. (7.10)
R(r) dr dr ~ Θ(θ)
Atoms 231

First, look at the angular part:

Θ00 (θ) = −const Θ(θ). (7.11)
This is the differential equation for a mass on a spring! We’ve already
examined it at equations (4.17) and (5.2). The two linearly independent
solutions are
√ √
Θ(θ) = sin( const θ) or Θ(θ) = cos( const θ). (7.12)
Now, the boundary condition for this ODE is just that the function must
come back to itself if θ increases by 2π:
Θ(θ) = Θ(2π + θ). (7.13)
√
If you think about this for a minute, you’ll see that this means const must
be an integer. You’ll also see that negative integers don’t give us anything
new, so we’ll take
√
const = ` where ` = 0, 1, 2, . . . . (7.14)

In summary, the solution to the angular problem is

`=0 `=1 `=2 `=3 ···

Θ(θ) 1 sin θ or cos θ sin 2θ or cos 2θ sin 3θ or cos 3θ ···

Now examine the radial part of the problem:

 
r d dR(r) 2M
r + 2 r2 [E − V (r)] = const = `2 (7.15)
R(r) dr dr ~
or, after some manipulation,
~2 `2
   
1 d dR(r) 2M
r + 2 E − V (r) − R(r) = 0. (7.16)
r dr dr ~ 2M r2
Compare this differential equation with another one-variable differential
equation, namely the one for the energy eigenproblem in one dimension:
d2 η(x) 2M
+ 2 [E − V (x)] η(x) = 0. (7.17)
dx2 ~
The parts to the right are rather similar, but the parts to the left — the
derivatives — are rather different. In addition, the one-dimensional energy
eigenfunction satisfies the normalization
Z ∞
|η(x)|2 dx = 1, (7.18)
−∞
232 Central potentials in two dimensions

whereas the two-dimensional energy eigenfunction satisfies the normaliza-

tion
Z
|η(x, y)|2 dx dy = 1
Z ∞ Z 2π
dr r dθ |R(r) sin(`θ)|2 = 1
0 0
Z ∞
π dr r|R(r)|2 = 1. (7.19)
0

This suggests that the true analog of the one-dimensional η(x) is not
R(r), but rather
√
u(r) = rR(r). (7.20)
Furthermore,
√
   
1 d dR(r) 1 1 u(r)
00
if u(r) = rR(r), then r =√u (r) + .
r dr dr r 4 r2
(7.21)
Using this change of function, the radial equation (7.16) becomes
d2 u(r) 1 u(r) 2M ~2 `2
 
+ + E − V (r) − u(r) = 0,
dr2 4 r2 ~2 2M r2
d2 u(r) 2M ~2 (`2 − 41 ) 1
 
+ 2 E − V (r) − u(r) = 0. (7.22)
dr2 ~ 2M r2

In this form, the radial equation is exactly like a one-dimensional energy

eigenproblem, except that where the one-dimensional problem has the func-
tion V (x), the radial problem has the function V (r) + ~2 (`2 − 14 )/(2M r2 ).
These two functions play parallel mathematical roles in the two problems.
To emphasize these similar roles, we define an “effective potential energy
function” for the radial problem, namely
~2 (`2 − 41 ) 1
Veff (r) = V (r) + . (7.23)
2M r2
Don’t read too much into the term “effective potential energy”. No actual
potential energy function will depend upon ~, still less upon the separation
constant `! I’m not saying that Veff (r) is a potential energy function, merely
that it plays the mathematical role of one in solving this one-dimensional
eigenproblem.
Now that the radial equation (7.22) is in exact correspondence with
the one-dimensional equation (7.17), we can solve this eigenproblem using
Atoms 233

either of the techniques described in chapter 5, “Solving the Energy Eigen-

problem”. (Or any other technique that works for the one-dimensional
problem.) The resulting eigenfunctions and eigenvalues will, of course, de-
pend upon the value of the separation constant `, because the effective
potential depends upon `. And as always, for each ` there will be many
eigenfunctions and eigenvalues, which we will label by index n = 1, 2, 3, . . .
calling them un,` (r) with eigenvalue En,` .
So we see how to find an infinite number of solutions to the partial
differential eigenproblem (7.7). The question is, did we get all of them?
The answer is in fact “yes,” although that’s not at all obvious. If you
want to learn more, you will need to read up on PDEs and Sturm-Liouville
theory!
Summary: To solve the two-dimensional energy eigenproblem for a
radially symmetric potential energy function V (r), namely
~2 2
∇ η(~r) + V (r)η(~r) = Eη(~r),
− (7.24)
2M
first solve the one-dimensional radial energy eigenproblem
~2 d2 u(r) ~2 (`2 − 14 ) 1
 
− + V (r) + u(r) = Eu(r) (7.25)
2M dr2 2M r2
for ` = 0, 1, 2, . . .. For a given `, call the resulting energy eigenfunctions and
eigenvalues un,` (r) and En,` for n = 1, 2, 3, . . .. Then the two-dimensional
solutions are
un,0 (r)
for ` = 0: η(r, θ) = √ with energy En,0 (7.26)
r
and
un,` (r)
η(r, θ) = √ sin(`θ)
r
for ` = 1, 2, 3, . . .: and with energy En,` .
un,` (r)
η(r, θ) = √ cos(`θ)
r
(7.27)
Remark 1: For ` 6= 0, there are two different eigenfunctions attached
to the same eigenvalue, a situation called degeneracy.1 Degeneracy is not
merely an abstraction concocted by air-head theorists. It can be uncovered
experimentally through the intensity — although not the wavelength — of
1 See the definition on page 150 and problem 4.11 on page 151.
234 Central potentials in two dimensions

spectral lines, through statistical mechanical effects, and through Zeeman

splitting when the atom is placed in a magnetic field.
Remark 2: The energy eigenvalues En,` come about from solving the
one-variable energy eigenproblem with effective potential
~2 (`2 − 41 ) 1
Veff (r) = V (r) + .
2M r2
Now, it’s clear from inspection that for any value of r, Veff (r) increases with
increasing `. It’s reasonable then that the energy eigenvalues also increase
with increasing `: that the fifth eigenvalue, for example, will always satisfy
E5,0 < E5,1 < E5,2 and so forth. This guess is in fact correct, and it can
be proven mathematically, but it’s so reasonable that I won’t interrupt this
story to prove it.
Remark 3: The conventional choice of zero level for a potential energy
function is to set V (r) = 0 as r → ∞. Hence all of the bound-state energy
eigenvalues are expected to be negative.
 Atoms 235

In summary, the energy eigenvalues for some generic two-dimensional

radially symmetric potential will look sort of like this (showing only the
four lowest energy eigenvalues for each value of `):

`=0 `=1 `=2 `=3

degen = 1 degen = 2 degen = 2 degen = 2

energy eigenvalue

Problem
7.1 Normalization condition
What is the normalization condition for un,` (r)? Be sure to distinguish
the cases ` = 0 and ` 6= 0.
236 Central potentials in three dimensions

7.2 Central potentials in three dimensions

The method used for central potentials in two dimensions works in three
dimensions as well. The details are (as expected) messier: you have to
use three spherical coordinates (r, θ, φ) rather than two polar coordinates
(r, θ), so you have to use separation of variables with a product of three one-
variable functions rather than a product of two one-variable functions. Thus
there are two separation constants rather than one. Instead of presenting
these messy details, I’ll just quote the result:
To solve the three-dimensional energy eigenproblem for a spherically
symmetric potential energy function V (r), namely
~2 2
− ∇ η(~r) + V (r)η(~r) = Eη(~r), (7.28)
2M
first solve the one-dimensional radial energy eigenproblem
~2 d2 u(r) ~2 `(` + 1) 1
 
− + V (r) + u(r) = Eu(r) (7.29)
2M dr2 2M r2
for ` = 0, 1, 2, . . .. For a given `, call the resulting energy eigenfunctions and
eigenvalues un,` (r) and En,` for n = 1, 2, 3, . . .. Then the three-dimensional
solutions are
un,` (r) m
ηn,`,m (r, θ, φ) = Y` (θ, φ) with energy En,` , (7.30)
r
where the “spherical harmonics” Y`m (θ, φ) are particular special functions
of the angular variables that you could look up if you needed to. The integer
separation constant m takes on the 2` + 1 values
−`, −` + 1, . . . , 0, . . . , ` − 1, `.
Notice that the 2` + 1 different solutions for a given n and `, but with
different m, are degenerate.
In addition, there’s a strange terminology that you need to know. You’d
think that the states with ` = 0 would be called “` = 0 states”, but in fact
they’re called “s states”. You’d think that the states with ` = 1 would be
called “` = 1 states”, but in fact they’re called “p states”. States with ` = 2
are called “d states” and states with ` = 3 are called “f states”. (I am told
that these names come from a now-obsolete system for categorizing atomic
spectral lines as “sharp”, “principal”, “diffuse”, and “fundamental”. States
with ` ≥ 4 are not frequently encountered, but they are called g, h, i, k, l,
m, . . . states. For some reason j is omitted. “Sober physicists don’t find
giraffes hiding in kitchens.”)
 Atoms 237

In summary, the energy eigenvalues for some generic three-dimensional

radially symmetric potential will look sort of like this:

` = 0 (s) ` = 1 (p) ` = 2 (d) ` = 3 (f)

m=0 m = −1, 0, +1 m = −2 . . . + 2 m = −3 . . . + 3
degen = 1 degen = 3 degen = 5 degen = 7

energy eigenvalue

This graph shows only the four lowest energy eigenvalues for each value of `.
A single horizontal line in the “` = 0 (s)” column represents a single energy
eigenfunction, whereas a single horizontal line in the “` = 2 (d)” column
represents five linearly independent energy eigenfunctions, each with the
same energy (“degenerate states”).
238 The hydrogen atom

Problem
7.2 Dimensions of η(~r) and of u(r)
In equation (7.27) for the two-dimensional central potential problem,
what are the dimensions of η(~r) and of u(r)? In equation (7.30) for the
three-dimensional central potential problem, what are the dimensions
of η(~r) and of u(r)? [This result helps motivate the definitions u(r) =
√
rR(r) in two dimensions and u(r) = rR(r) in three dimensions.]

7.3 The hydrogen atom

7.3.1 The model

An electron (of mass M ) and a proton interact through the classical elec-
trostatic potential energy function — called the “Coulomb potential” —
1 e2
V (r) = − , (7.31)
4π0 r
so you might think that the energy eigenproblem for the hydrogen atom is
~2 2 1 e2
− ∇ η(~r) − η(~r) = Eη(~r). (7.32)
2M 4π0 r
That’s not exactly correct. This eigenproblem treats the proton as station-
ary while the electron does all the moving: in fact, although the proton is
almost 2000 times more massive than the electron, it’s not infinitely massive
and it does do some moving. This eigenproblem assumes the proton is a
point particle: in fact, although the nucleus is small compared to an atom,
it does have some size. This eigenproblem is non-relativistic and it treats
the electromagnetic field as purely classical: both false. This eigenproblem
ignores the electron’s spin. All of these are good approximations, but this
is a model for a hydrogen atom, not the exact thing.2
But let’s work with the approximation we have, rather than holding out
for an exact solution of an exact eigenproblem that will never come.3 What
happens if we solve the three-dimensional central potential problem with
the model potential energy function (7.31)? We don’t yet have the mathe-
matical tools to actually perform this solution, but we are in a position to
appreciate the character of the solution.
2 The corrections to the energy eigenvalues produced by equation (7.32) due to these

effects are called “fine structure” and “hyperfine structure”.

3 Everyone knows that weather prediction is inexact. But you’d still rather know a

prediction that a hurricane has an 80% chance of arriving at about 7:00 pm than be
totally clueless about a hurricane bearing down on your home.
 Atoms 239

7.3.2 The energy eigenvalues

First of all, because the Coulomb potential is a particular kind of central

potential, it will have all the properties listed in the last section for three-
dimensional central potentials: Each energy eigenstate will be characterized
by an ` and an m, where ` = 0, 1, 2, . . . and where m = −`, . . . , +`. The
energy eigenvalues will be independent of m, resulting in degeneracy. And
for a given n, the energy eigenvalue will increase with increasing `.
The energy eigenvalues for the Coulomb potential turn out to be:

` = 0 (s) ` = 1 (p) ` = 2 (d) ` = 3 (f)

m=0 m = −1, 0, +1 m = −2 . . . + 2 m = −3 . . . + 3
degen = 1 degen = 3 degen = 5 degen = 7

n=4
n=3

n=2

n=1

energy eigenvalue
240 The hydrogen atom

What a surprise! The energy eigenvalues for ` = 1 are pushed up so much

that they exactly line up with all but the lowest energy eigenvalues for
` = 0. The energy eigenvalues for ` = 2 are pushed up so much that they
exactly line up with all but the lowest energy eigenvalues for ` = 1. And
so forth. This surprising line-up is called “accidental degeneracy”.
Normally eigenfunctions are labeled by n = 1, 2, 3, . . .. But this sur-
prising line-up of energies suggests a different notation for the Coulomb
problem. For “` = 0 (s)” the eigenfunctions are labeled as usual by
n = 1, 2, 3, 4, . . .. But for “` = 1 (p)” the eigenfunctions are labeled by
n = 2, 3, 4, . . .. For “` = 2 (d)” they are labeled by n = 3, 4, . . .. And so
forth. With this labeling scheme the energy eigenvalue turn out to be given
by
 2 2
M e 1
En = − 2 . (7.33)
2~ 4π0 n2
The coefficient in this equation is called the “Rydberg4 energy”, and the
equation is usually written
Ry
En = − , where Ry = 13.6 eV. (7.34)
n2
(I recommend that you memorize this energy 13.6 eV, the ionization energy
for hydrogen, which sets the scale for typical energies in atomic physics.
Much to my embarrassment, I forgot it during my graduate qualifying
oral exam. Problem 7.4, “Dimensional analysis for energy eigenvalues”,
on page 245 presents a way to help understand and remember this result.)

4 Johannes Rydberg (1854–1919), Swedish spectroscopist, discovered a closely related

formula empirically in 1888. Do not confuse the Rydberg energy Ry = 13.6 eV with the
Rydberg constant R∞ = 1.097 × 107 m−1 .
Atoms 241

7.3.3 The energy eigenfunctions

It’s a triumph to know the energy eigenvalues, but we should know also
something about the energy eigenfunctions, which are labeled ηn,`,m (~r). A
terminology note is that an energy eigenfunction with n = 3, ` = 2, and
any value of m — that is η3,2,m (x) — is called a “3d state”.
To gain this knowledge we need to first understand the effective potential
energy function falling within square brackets in equation (7.29):
1 e2 ~2 `(` + 1) 1
Veff (r) = − + . (7.35)
4π0 r 2M r2
This function is sketched schematically on the next page. For large values
of r, to the right in the sketch, 1/r is bigger than 1/r2 , so Veff (r) is almost
the same as the 1/r Coulomb potential energy alone. For small values of
r, to the left in the sketch, 1/r is smaller than 1/r2 , so Veff (r) is almost
the same as the 1/r2 part alone. For intermediate values of r, the function
Veff (r) has to swing between these two limits, as sketched.
The result of this swinging will of course depend upon the value of `,
and the results for four values of ` are sketched schematically on page 243.
242 The hydrogen atom

Veff (r)

~2 `(` + 1) 1
+
2M r2

1 e2
−
4π0 r
Atoms 243

Veff (r)

`=3

`=2

`=1
r

`=0
244 The hydrogen atom

This graph suggests that for a given value of n, the states with larger
` will have larger mean values for r, the distance from the proton to the
electron.

7.3.4 Transitions

If the energy eigenequation (7.32) for the hydrogen atom were exactly cor-
rect, then a hydrogen atom starting in the excited energy state η3,2,−1 (~r)
would remain in that state forever. Furthermore, a hydrogen atom start-
ing in a linear combination with probability 0.6 of being in energy state
η3,2,−1 (~r) and probability 0.4 of being in energy state η2,1,0 (~r) would main-
tain those probabilities forever.
But the energy eigenequation (7.32) is not exactly correct. It ignores
collisions, it ignores external electromagnetic field (e.g., incident light), and
it ignores coupling to the electromagnetic field (e.g., radiated light). These
effects mean that the state η3,2,−1 (~r) is a stationary state of the model
eigenproblem, but it is not a stationary state of the exact eigenproblem. In
other words, these effects result in transitions between stationary states of
the model eigenproblem.
To understand these transitions you need to understand the transition-
causing effects, and at this point in your education you’re not ready to
do that. But I’ll tell you one thing right now: a transition involving a
single photon (either absorbing or emitting a single photon) will result in
a transition with ∆` = ±1. So, for example, a hydrogen atom in a 2p state
(that is, one with n = 2, ` = 1, and any legal value of m) could transition
to the 1s ground state by emitting a single photon. A hydrogen atom in a
2s state (that is, one with n = 2, ` = 0, and m = 0) cannot transition to the
ground state by emitting a single photon. (It could do so by emitting two
photons, or through a collision.) An atom in the 1s ground state, exposed
to a light source with photons of energy 43 Ry, can be excited to 2p state
by absorbing a single photon, but it cannot be excited to the 2s state by
absorbing a single photon.
I regard this fact (which, by the way, holds not only for the hydrogen
atom but for any central potential) as a picky detail appropriate for an
advanced course, but the people who write the Graduate Record Exam in
physics seem to think it’s important so you should probably remember it.
(Or at least review this page the evening before you take the physics GRE.)
Atoms 245

Problems
7.3 Counting hydrogen states
Show that the degeneracy for states characterized by n is n2 .
7.4 Dimensional analysis for energy eigenvalues
The eigenproblem (7.32) contains only two parameters:
~2 e2
and .
M 4π0
Use dimensional analysis to show that these two parameters can come
together to form an energy in only one way. [[I remember the Rydberg
energy as
1 (e2 /4π0 )2
Ry = (7.36)
2 ~2 /M
using this dimensional analysis trick.]]
7.5 Which states are distant, which are close? (essential problem)
Argue, on the basis of the graph on page 242, that for a given value of
`, states with larger n will have larger mean values for r.
7.6 Energy eigenvalues for the He+ ion (essential problem)
A helium atom with one electron stripped away is called a He+ ion.
This situation is simply one electron ambivating in the potential energy
established by a highly-massive nucleus: it is just like the hydrogen
atom, except that the nuclear change is +2e rather than +e. At the
level of approximation used in equation (7.32), the energy eigenproblem
for the He+ ion is
~2 2 1 2e2
− ∇ η(~r) − η(~r) = Eη(~r). (7.37)
2M 4π0 r
Show that (at this level of approximation) the energy eigenvalues for
the He+ ion are
Ry
En = −4 2 . (7.38)
n
The situation of a single electron ambivating in the potential energy
established by a highly-massive carbon nucleus of charge +6e is called
the C5+ ion. Show that (at this level of approximation) the energy
eigenvalues are
Ry
En = −36 2 . (7.39)
n
246 The hydrogen atom

7.7 Characteristic quantities for the Coulomb problem

The time evolution Schrödinger equation for the Coulomb problem is
~2 2 e2 1
 
∂ψ(~r, t) i
=− − ∇ ψ(~r, t) − ψ(~r, t) . (7.40)
∂t ~ 2M 4π0 r
There are only three parameters in this equation: ~, M , and e2 /4π0 .
Using the techniques of sample problem 5.2.1 on page 200, find the
characteristic time and length for the Coulomb problem. Define the
scaled quantities
t ~r
t̃ = and ~r̃ = ,
characteristic time characteristic length
and write the time evolution equation (7.40) in terms of these variables.
If you didn’t like to explain what you were doing (or if you wanted to
sound cryptic to impress the uninitiated) how would you use shorthand
to describe the result of this scaling strategy?
7.8 Hybridization
For some chemical applications, it is useful to define the four “sp3
hybrid states”
1
φ1 (~r) = 2 [η2,0,0 (~r) + η2,1,+1 (~r) + η2,1,0 (~r) + η2,1,−1 (~r)]
1
φ2 (~r) = 2 [η2,0,0 (~r) + η2,1,+1 (~r) − η2,1,0 (~r) − η2,1,−1 (~r)]
1
φ3 (~r) = 2 [η2,0,0 (~r) − η2,1,+1 (~r) − η2,1,0 (~r) + η2,1,−1 (~r)]
1
φ4 (~r) = 2 [η2,0,0 (~r) − η2,1,+1 (~r) + η2,1,0 (~r) − η2,1,−1 (~r)] .
a. Which, if any, of these are energy eigenstates? What is the energy
eigenvalue associated with each such eigenstate?
b. The eigenstates ηn,`,m (~r) are “orthonormal” in the sense that the
integral over all space satisfies
1 if n0 = n, `0 = `, and m0 = m
Z 
ηn∗ 0 ,`0 ,m0 (~r)ηn,`,m (~r) d3 r = .
0 otherwise
(7.41)
Evaluate the sixteen integrals
Z
φ∗i (~r)φj (~r) d3 r. (7.42)

Clue: This is not a difficult problem. If you’re working hard, then

you’re working too hard.
7.4. The helium atom 247

7.4 The helium atom

Here’s the energy eigenproblem for the helium atom, at the same level of
approximation as the eigenproblem (7.32) for the hydrogen problem:
~2 2 ~2 2
− ∇A η(~rA , ~rB ) − ∇ η(~rA , ~rB )
2M 2M B
1 2e2 1 2e2 1 e2
− η(~rA , ~rB ) − η(~rA , ~rB ) + η(~rA , ~rB )
4π0 rA 4π0 rB 4π0 |~rA − ~rB |
= Eη(~rA , ~rB ). (7.43)
We have no chance whatsoever of solving this “two electron plus one nu-
cleus” problem exactly. Even the classical problem of three particles inter-
acting through 1/r potentials, first posed by Isaac Newton in 1687, has not
yet been solved exactly. (And probably never will be, because the result-
ing behavior is known to be chaotic.) Since classical mechanics is a subset
of quantum mechanics, an exact solution to this helium problem would
contain within it an exact solution to the unsolved classical “three-body
problem”.
Does this mean we should give up? Not at all. We should instead look
for approximate solutions that are not exact, but highly accurate for the
bound-state regime of interest.
Our approach will involve solving the one-electron problem for a differ-
ent potential, and then using those one-electron levels as building blocks
for the two-electron problem through the antisymmetrization machinery
of equation (6.11). The strategy may seem crude, but in practice it can
produce highly accurate results.
Instead of focusing on two electrons, interacting with the nucleus and
with each other, focus on one electron interacting with the nucleus and with
the average of the other electron. I don’t yet know exactly how the “other”
electron is averaged, but I assume it spreads out in a spherically symmetric
cloud-like fashion.
Finding the potential. Remember, from your electrostatics course,
the shell theorem for spherically symmetric charge distributions: When the
electron under focus is close to the nucleus, it feels only the electric field
due to the nucleus, so the potential energy is
1 2e2
for small r, V (r) ≈ − . (7.44)
4π0 r
248 The helium atom

Whereas when the electron under focus is far from the nucleus, it feels the
electric field due to the nucleus, plus the electric field due to the cloud
collapsed into the nucleus, so the potential energy is
1 e2
for large r, V (r) ≈ − . (7.45)
4π0 r
The potential energy felt by the electron under focus will interpolate be-
tween these two limits, something like the solid line graphed below.

V (r)

1 e2
−
4π0 r 1 2e2
−
4π0 r

This phenomenon is called “shielding”. The shielded potential interpo-

lates between the known limits of small r and large r behavior. The exact
character of that interpolation is unclear: if you were doing high-accuracy
calculations, you would need to find it.5 For our purposes it will be enough
just to know the two limits.
5 Using a technique called the Hartree-Fock approximation.
Atoms 249

This approach is called a “mean field approximation” and is said to

“ignore correlations” between the two electrons. This approach is not exact
and cannot be made exact, but it enables progress to be made.
Finding the one-electron eigenvalues. What will the one-electron
energy eigenvalues be for a shielded potential energy? If the potential
energy were
1 e2
V (r) = − , (7.46)
4π0 r
then the system would be a hydrogen atom (H atom) and the energy eigen-
values would be
Ry
En = − 2 . (7.47)
n
If the potential energy were
1 2e2
V (r) = − , (7.48)
4π0 r
then the system would be a positively charged helium ion (He+ ion; a
helium atom with one electron stripped away); equation (7.38) shows that
the energy eigenvalues would be
Ry
En = −4 . (7.49)
n2
But in fact, the potential energy interpolates between these two forms,
so the energy eigenvalues interpolate between these two possibilities. Let’s
examine this interpolation, first for the s levels:
 250 The helium atom

energy eigenvalues for s levels

He+ ion shielded H atom

n=4
n=3

n=2

n=1

energy eigenvalue

To the left, under the heading “He+ ion”, are the s state eigenvalues (7.49),
which are four times deeper than those under the heading “H atom”, the
s state eigenvalues (7.47). I’ve drawn a vertical line midway between them
and dashed lines connecting the the two sets of eigenvalues. If the eigenval-
ues for the shielded potential were exactly halfway between the eigenvalues
for the two limits, then they would fall where the dashed lines cross the ver-
tical line. But they don’t fall exactly there. For states that are mostly near
the nucleus, the energies are closer to (7.49). For states that are mostly far
from the nucleus, the energies are closer to (7.47). We have already seen
(problem 7.5 on page 245) that, for a given `, the eigenfunctions with larger
 Atoms 251

n are farther from the nucleus. Chemists like to say that the eigenfunctions
that are mostly close to the nucleus — those with smaller n — have more
“penetration”.
This process can be repeated for p states, d states, and f states. Because,
for a given n, the eigenfunction with larger ` is farther from the nucleus
(less “penetration”, see page 244), the eigenfunction with larger ` will have
higher energy. Thus a shielded potential energy function will give rise to a
set of energy eigenvalues like this:

` = 0 (s) ` = 1 (p) ` = 2 (d) ` = 3 (f)

m=0 m = −1, 0, +1 m = −2 . . . + 2 m = −3 . . . + 3
degen = 1 degen = 3 degen = 5 degen = 7

n=4
n=3

n=2

n=1

energy eigenvalue
252 The lithium atom

Building two-electron states from one-electron levels. Now that

we have one-electron eigenfunctions (the “levels”), we can combine them
through the antisymmetrization machinery to produce two-electron eigen-
functions (the “states”). This process was described in section 6.6 on
page 219.
Misconception. Perhaps you learned in high-school chemistry that
the ground state of the helium atom has “one electron in the ‘1s, spin
up’ level, and one electron in the ‘1s, spin down’ level”. That’s just plain
wrong — if it were right, then you’d be able to distinguish between the
two electrons (the one with spin up, the one with spin down) and then the
two electrons wouldn’t be identical. What’s correct is that the individual
electrons don’t have states: instead the pair of electrons is in one state,
namely the antisymmetric non-product state
η1,0,0 (~xA )η1,0,0 (~xB ) √1 [χ+ (A)χ− (B) − χ− (A)χ+ (B)].
2
(Compare problem 6.10, “Intersystem crossing”, on page 225.)

Problem
7.9 The hydrogen molecule ion
If a hydrogen molecule H2 is stripped of one electron, the result is
the “hydrogen molecule ion” consisting of two nuclei and one electron.
Show that if we had solved the helium problem exactly we would have
also solved the hydrogen molecule ion problem. (But we have not solved
the problem exactly: instead we found approximations appropriate for
the case of atomic helium. A completely different set of approximations
are appropriate for the hydrogen molecule ion.)

7.5 The lithium atom

This situation (within the shielded potential approximation) was discussed

in section 6.7 on page 225. In summary:
The ground state of hydrogen is two-fold degenerate. The ground state
of helium is non-degenerate. The ground state of lithium is (within the
shielded potential approximation) two-fold degenerate.
The ground states of hydrogen and helium can (within the shielded
potential approximation) be written as a spatial part times a spin part.
The ground state of lithium cannot be so written.
7.6. All other atoms 253

7.6 All other atoms

Atoms larger than lithium are difficult. Specialists have examined them in
exquisite detail, but in this book we’re not going to try to find the energy
spectrum, we’re not going to try to find the ground state degeneracy, we’re
not even going to try to write down a ground state. Instead, we’re only
going to list the one-electron levels that are thrown together through the
antisymmetrization machinery (6.11) to make the many-electron ground
state.
Try this buliding-up machinery for carbon, with a nucleus and six elec-
trons.6 In the figure on page 250, pertaining to helium, the left-hand ener-
gies are four times deeper than the right-hand energies. If I were to draw a
parallel figure for carbon (see equation 7.39), the left-hand energies would
be 36 times deeper than the right-hand energies! The net result is that,
while the figure on page 251 shows a modest increase in energy En,` for
a given n as you move right to higher values of `, for carbon the energy
increase will be dramatic: something like the figure on the next page. For
atoms bigger than carbon, the increase will be still more dramatic.
6 Carbon is a seven-body problem, not a three-body problem like helium, so of course

it is correspondingly less tractable.

 254 All other atoms

` = 0 (s) ` = 1 (p) ` = 2 (d) ` = 3 (f)

m=0 m = −1, 0, +1 m = −2 . . . + 2 m = −3 . . . + 3
degen = 1 degen = 3 degen = 5 degen = 7

6d 5f
7s 6p 5d 4f
6s
5p 4d
5s
4p
4s 3d

3p
3s

2p
2s

1s

energy eigenvalue

Schematic energy levels for a generic atom, about the size of carbon.

I underscore once more that these levels come from an approximation

that ignores relativity, electron spin, nuclear size, nuclear motion, and the
quantal character of the electromagnetic field. Most damning of all, it
replaces electron-electron repulsion with a shielded potential. Since this
shielded potential is spherically symmetric, the language of ` and m and
2p levels and so forth can be used. But this is only an approximation.
Atoms 255

With this understanding, we ask which one-electron levels will go into

the antisymmetrization machinery to make the carbon six-electron ground
state. The ground state will be constructed from the six lowest possible
energy levels. There will be a “1s, spin up” level, plus a “1s, spin down”
level, plus a “2s, spin up” level, plus a “2s, spin down” level. In addition
there will be two 2p levels, but it’s not clear which ones they will be: Will
they be the 2p level with m = 0 and spin up, plus the 2p level with m = +1
and spin up? Or will they be the 2p level with m = 0 and spin up, plus
the 2p level with m = 0 and spin down? Or will they be some other
superposition?
At the level of approximation used here, all such combinations have
exactly the same energy: they are degenerate. If you study more atomic
physics you’ll learn Hund’s7 rules for figuring out how the degeneracy is
broken at a more accurate level of approximation. But for the purposes of
this book, it’s only necessary to list the ns and the `s of the one-electron lev-
els that go into making the six-electron ground state. This list is called the
“electronic configuration” and it’s represented through a special notation:
the configuration of carbon is written 1s2 2s2 2p2 .
For still larger atoms, the shielding effect is more dramatic and the
energy levels shift still further, but still, usually, the one-electron levels
fall within the energy sequence shown on page 254. This is the so-called
“Madelung8 sequence” of level energies:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p
< 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d . . .

It would be a miracle indeed if the theoretical calculations for all the

different atoms resulted in exactly the same qualitative sequence of energy
levels. And it would be more miraculous still if the approximations used
were close enough to reality that the prediction of the approximation was
always accurate. And indeed neither of these miracles occur.9 For ex-
ample in chromium, atomic number 24, the configuration predicted by the
7 FriedrichHund (1896–1997), German physicist who applied quantum mechanics to
atoms and molecules, and who discovered quantum tunneling.
8 Erwin Madelung (1881–1972), German physicist with interests in crystal structure,

atomic physics, and quantum mechanics. He produced a set of equations equivalent to

the Schrödinger equation but which emphasized the flow of probability density rather
than of amplitude density.
9 See W.H. Eugen Schwarz and Ronald L. Rich, “Theoretical basis and correct ex-

planation of the periodic system: Review and update” Journal of Chemical Education
87 (April 2020) 435–443; Gregory Anderson, Ravi Gomatam, and Laxmidhar Behera,
 256 The periodic table

Madelung sequence is 1s2 2s2 2p6 3s2 3p6 4s2 3d4 whereas experiment shows the
actual configuration ends instead with 4s1 3d5 .

Problem
7.10 Ground state degeneracy
At the level of approximation of the diagram on page 254, find the
degeneracy of the ground state of boron, of carbon, of nitrogen, of
oxygen, of fluorine, and of neon.

7.7 The periodic table

While the Madelung sequence is not perfect (few things are), it makes sense
to see what it has to say (“better to light a single candle, no matter how
faint, than to curse the darkness”).
Compare the levels that go into building up carbon (atomic number
6) with those that go into building up silicon (atomic number 14). For
carbon they are 1s2 2s2 2p2 ; for silicon they are 1s2 2s2 2p6 3s2 3p2 . Note the
similarities of those last, highest energy levels: carbon ends with 2s2 2p2 ,
silicon ends with 3s2 3p2 . It’s possible that for carbon the nucleus (charge
+6) and the 1s and 2s electron levels (charge −4) act together to form an
atom core of net charge +2. Meanwhile it’s just as possible that for silicon
the nucleus (charge +14) and the 1s, 2s, 2p, and 3s electron levels (charge
−12) similarly act together to form an atom core again with net charge +2.
If this possibility is correct, then you would expect carbon and silicon to
have similar chemical behavior. Sure enough each of them bonds with four
other atoms: methane (CH4 ) is chemically analogous to silane (SiH4 ).
Through parallel reasoning you would expect neon, 1s2 2s2 2p6 , to behave
similarly to argon, 1s2 2s2 2p6 3s2 3p6 . Again this expectation holds: both
neon and argon are noble gases that react reluctantly with other atoms.
Chemists have an ingenious system for showing these chemical similarities
through a graphic called “the periodic table”, shown on the next page.

“Contradictions in the quantum mechanical explanation of the periodic table” Journal

of Physics: Conference Series 490 (2014) 012197.
 Atoms 257

1
H He
1 2

2
Li Be B C N O F Ne
3 4 5 6 7 8 9 10

3
Na Mg Al Si P S Cl Ar
11 12 13 14 15 16 17 18

4
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

5
Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe
37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

6
Cs Ba ∗ Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn
55 56 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

7
Fr Ra ∗ Lr Rf Db Sg Bh Hs Mt Ds Rg Cn
87 88 ∗ 103 104 105 106 107 108 109 110 111 112

∗ La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb
57 58 59 60 61 62 63 64 65 66 67 68 69 70

∗ Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No
∗ 89 90 91 92 93 94 95 96 97 98 99 100 101 102

Let’s carry out this building-up scheme systematically. Hydrogen and

helium we have already discussed. Lithium adds one 2s level to the mix,10
and beryllium one more. The next six elements each add one 2p level,
ending at neon. Now pass on to row 3 of the periodic table. Sodium (atomic
number 11, symbol Na from the Latin “natrium”) has one 3s level just as
lithium has one 2s level. Sure enough lithium and sodium are both highly
reactive, chemically, and they react in similar ways. (They are called “alkali
metals”.) As we add electron levels, the march right on row 3 parallels the
march right on row 2, ending in the noble gas argon, atomic number 18.
Row 4 starts out just like rows 2 and 3: potassium (atomic number 19,
symbol K from the Latin “kalium”) is chemically similar to lithium and
sodium; calcium is chemically similar to beryllium and magnesium. But
scandium, atomic number 21, is not at all like boron, atomic number 5.
That’s because the highest energy level in boron is a 2p level. The highest
10 To write this out in detail, the three-electron wavefunction for the lithium ground state

comes from feeding a 1s spin up level, plus a 1s spin down level, plus now a 2s level, into
the “multiply and antisymmetrize” machinery at equation (6.11) in order to generate
a three-electron state. This is a real mouthful, so we simply write “add one 2s level”.
In some books you will see this written as “add one 2s electron”, but that language
reinforces the misconception that the electron in the 2s level can be distinguished from
the other two electrons, in which case the three electrons would not be identical. See
the warnings on page 252 and in problem 6.10, “Intersystem crossing”, on page 225.
 258 The periodic table

energy level in scandium is not a p level at all, it’s a 3d level. There are
ten such levels, accounting for the ten elements from scandium through
zinc. The element beyond zinc is gallium, which sure enough shares a lot of
properties with its vertical neighbor aluminum. Row 4 continues by adding
4p levels until ending with the noble gas krypton.
Row 5 is very much like row 4.
Row 6 adds a new twist after barium, atomic number 56. The extra
level in lanthanum, atomic number 57, is a 4f level. The fourteen 4f levels
account for the fourteen elements from lanthanum through ytterbium. (The
page in this book is not wide enough to hold row 6 of the periodic table, so
I shoehorn in these fourteen elements using an asterisk.) Then the ten 5d
levels account for the ten elements from lutetium through mercury (symbol
Hg from the Greek “hydrargyrum”, meaning “liquid silver”). Finally the
six 6p levels account for the six elements from thallium through the noble
gas radon.
Row 7 is very much like row 6. Many of the elements in row 7 have
short-lived nuclei, but that’s a different story.
Any chemist will tell you, correctly, that this lightning tour of the pe-
riodic table leaves out a lot of fascinating detail. But its very briefness
means that you have not been distracted by detail and have kept sight of
the central fact that the entire structure of the periodic table — and hence
all of chemistry — follows from the Pauli requirement for antisymmetry
under fermion coordinate swaps.

Problem
7.11 Questions (recommended problem)
Update your list of quantum mechanics questions that you started at
problem 1.17 on page 46. Write down new questions and, if you have un-
covered answers to any of your old questions, write them down briefly.

[[For example, one of my questions would be: “The text claims

on page 256 that the experimentally determined configuration for
chromium, element 24, is 1s2 2s2 2p6 3s2 3p6 4s1 3d5 . How can experiment
determine such a thing?”]]
 Chapter 8

The Vistas Open to Us

I reckon I got to light out for the territory ahead. . .

— Mark Twain (last sentence of Huckleberry Finn)

This is the last chapter of the book, but this book itself is an invita-
tion only, so it is not the last chapter of quantum mechanics. There are
many fascinating topics that this book hasn’t even touched on. Quantum
mechanics will — if you allow it — surprise and delight and mystify you
for the rest of your life.

This book devotes two chapters to qubits, also called spin- 12 systems.
Plenty remains to investigate: “which path” interference experiments,
delayed-choice interference experiments, many different entanglement situ-
ations. For example, we developed entanglement through a situation where
the quantal probability was 21 while the local deterministic probability was
5
9 or more (page 80). Different, to be sure, but not dramatically differ-
ent. In the Greenberger–Horne–Zeilinger entanglement situation the quan-
tal probability is 1 and the local deterministic probability is 0. You can’t
find probabilities more different than that! If you find these situations as
fascinating as I do, then I recommend George Greenstein and Arthur G.
Zajonc, The Quantum Challenge: Modern Research on the Foundations of
Quantum Mechanics.
For many decades, research into qubits yielded insight and understand-
ing, but no practical applications. All that changed with the advent of
quantum computing. This is a rapidly changing field, but the essay “Quan-
tum Entanglement: A Modern Perspective” by Barbara M. Terhal, Michael
M. Wolf, and Andrew C. Doherty (Physics Today, April 2003) contains core

259
260 The territory ahead

insights that will outlive any transient. From the abstract: “It’s not your
grandfather’s quantum mechanics. Today, researchers treat entanglement
as a physical resource: Quantum information can now be measured, mixed,
distilled, concentrated, and diluted.”
Because quantum mechanics is both intricate and unfamiliar, a
formidable yet beautiful mathematical formalism has developed around
it: position wavefunctions, momentum wavefunctions, Fourier transforms,
operators, Wigner functions. These are powerful precision tools, so mag-
nificent that some confuse the tools with nature itself. Every quantum
mechanics textbook develops this formalism to a greater or lesser extent,
but I also recommend the cute book by Leonard Susskind and Art Fried-
man, Quantum Mechanics: The Theoretical Minimum.
Here are two formal problems to whet your appetite: Back on page 140
I quoted O. Graham Sutton that “A technique succeeds in mathematical
physics, not by a clever trick, or a happy accident, but because it expresses
some aspect of a physical truth.” Thus inspired, we asked about the mean-
ing of the separation constant (4.16), and that inquiry led us to the whole
structure of stationary states and the energy eigenproblem. (And to a poem
by T.S. Eliot, page 146.) But when we faced a similar separation constant
at equation (7.10) we were too busy to follow up and ask what it was telling
us. If you study more quantum mechanics, you will learn that this sepa-
ration constant is related to angular momentum, that angular momentum
is related to rotations, and that the conservation of angular momentum is
related to rotational symmetry!
The second problem involves the simple harmonic oscillator, that is,
the potential energy function V (x) = 12 kx2 . As with any one-dimensional
potential well there are energy eigenstates. If any one of these states is
shifted by a distance, it is of course no longer an energy eigenstate, so it
does change with time. The remarkable thing is how it changes with time:
the probability density does not spread, nor compact, nor change shape.
Instead it rigidly slides back and forth with the same period that a classical
particle would have in that same potential well.1 When I first did the math
to show that this is so, I was so astounded that I wrote a computer program
to check out the math. This remarkable fact is true, and I have the feeling
1 M.E. Marhic, “Oscillating Hermite-Gaussian wave functions of the harmonic oscilla-

tor” Lettere al Nuovo Cimento 22 (1978) 376–378, and C.C. Yan, “Soliton like solutions
of the Schrödinger equation for simple harmonic oscillator” American Journal of Physics
62 (1994) 147–151.
The Vistas Open to Us 261

that it “expresses some aspect of a physical truth”, but I have no idea of

what that physical truth might be.
We have applied quantum mechanics to cryptography, to model systems,
and to atoms. Applications continue to molecules and to solids, to nuclei
and to elementary particles, to superfluids, superconductors, and lasers,
to liquid crystals, polymers, and membranes; the list is endless. Indeed,
sunlight itself is generated through a quantal tunneling process! White
dwarf stars work because of quantum mechanics, so do transistors and
light-emitting diodes. In 1995 a new state of matter, the Bose-Einstein
condensate, came into existence in a laboratory in Boulder, Colorado. In
2003 an even more delicate state, the fermionic condensate, was produced,
again in Boulder. Both of these states of matter exist because of the Pauli
principle, applied over and over again to millions of atoms.
Way back on page 5 we mentioned the need for a relativistic quantum
mechanics and its associate, quantum field theory. The big surprise is that
these theories don’t just treat particles moving from place to place. They
predict that particles can be created and destroyed, and sure enough that
happens in nature under appropriate conditions.
There’s plenty more to investigate: quantal chaos and the classical
limit of quantum mechanics, friction and the transition to ground state,
applications to astrophysics and cosmology and elementary particles.
But I want to close with one important yet rarely mentioned item:
it’s valuable to develop your intuition concerning quantum mechanics.
We saw on page 39 that two common visualizations are flawed. Then on
page 82 we found that no picture drawn with classical ink could successfully
capture all aspects of quantum mechanics. How, then, can one develop a
visualization or intuition for quantum mechanics? This is a lifelong journey
which you have already begun. A good next step is to read the slim but
profound book by Richard Feynman titled QED: The Strange Theory of
Light and Matter.
The story of quantum mechanics began with the glowing logs of a camp-
fire. It continued through atomic spectra; quantization, interference, and
entanglement; Fourier sine series and partial differential equations. The
story is not finished, and I invite you to add to it yourself.
262 The territory ahead

Problem
8.1 Questions (recommended problem)
This is the end of the book, not the end of quantum mechanics. Write
down any questions you have concerning quantum mechanics. Perhaps
you will answer some of these through future study. Others might
suggest future research directions for you.
 Appendix A

Significant Figures

The impossibility of certainty

What is the pattern of thought most characteristic of science? Is it know-

ing that “net force causes acceleration rather than velocity”? Is it knowing
that“momentum is conserved in the absence of external forces”? Is it know-
ing that “quantum mechanics has a classical limit”? No, it is none of these
three facts — important though they are. The pattern of thought most
characteristic of science is knowing that “every measurement is imperfect
and thus every observed number comes with an uncertainty”.
As with any facet of science, the proper approach to uncertainty is not
“plug into a formula for error propagation” but instead “think about the
issues involved”. For example, in the course of building a tree house I
measured a plank with a meter stick and found it to be 187.6 cm long.
A more accurate measurement would of course provide a more accurate
length: perhaps 187.64722031 cm. I don’t know the plank’s exact length, I
only know an approximate value. In a math class, 187.6 cm means the same
as 187.60000000 cm. But in a physics class, 187.6 cm means the same as
187.6??????? cm, where the question marks represent not zeros, but digits
that you don’t know. The digits that you do know are called “significant
digits” or “significant figures”.
This has an important philosophical consequence: Because all scientific
conclusions are based on measurements, and all measurements contain some
uncertainty, no scientific conclusion can be absolutely certain. All science is
tentative; all those worshiping at the altar of science are fooling themselves.
This book is more concerned with the day-to-day practicalities of uncer-
tainty than with the grand philosophical consequences. How do we express

263
264 Significant Figures

our lack of certainty? How do we work with (add, subtract, multiply, take
logarithms of, and so forth) quantities that aren’t certain?

Expressing uncertainty

The convention for expressing uncertain quantities is simple: any digit writ-
ten down is a significant digit. A plank measured to the nearest millimeter
has a length expressed as, say, 103.7 cm or 91.5 cm or 135.0 cm. Note
particularly the trailing zero in 135.0 cm: this final digit is significant.
The quantity “135.0 cm” is different from “135 cm”. The former means
“135.0?? cm”, the latter means “135.??? cm”. In the former, the digit in
the tenths place is 0, while in the latter, the digit in the tenths place is
unknown.
This convention gives rise to a problem for representing large numbers.
Suppose the distance between two stakes is 45.6 meters. What is this
distance expressed in centimeters? The answer 4560 cm is unsatisfactory,
because the trailing zero is not significant and so, according to our rule,
should not be written down. This quandary is resolved using exponential
notation: 45.6 meters is the same as 4.56 × 103 cm. (This is, unfortunately,
one of the world’s most widely violated rules.)

Working with uncertainty

Addition and subtraction. I measured a plank with a meter stick and

found it to be 187.6 cm long. Then I measured a dowel with a micrometer
and found it to be 2.3405 cm in diameter. If I place the dowel next to the
plank, how long is the dowel plus plank assembly? You might be tempted
to say

187.6
+ 2.3405
--------
189.9405

But no! This is treating the unknown digits in 187.6 cm as if they were
zeros, when in fact they’re question marks. The proper way to perform the
 265

sum is

187.6?????
+ 2.3405??
----------
189.9?????

So the correct answer, with only significant figures written down, is

189.9 cm.
Multiplication and division. The same question mark technique
works for multiplication and division, too. For example, a board measuring
124.3 cm by 5.2 cm has an area given through

1243?
x 52?
--------
?????
2486?
6215?
--------
64????

Adjusting the decimal point gives an answer of 6.4 × 102 cm2

Although the question mark technique works, it’s very tedious. (It’s
even more tedious for division.) Fortunately, the following rule of thumb
works as well as the question mark technique and is a lot easier to apply:

When multiplying or dividing two numbers, round the answer

to the number of significant digits in the least certain of the two
numbers.

For example, when multiplying a number with four significant digits by a

number with two significant digits, the result should be rounded to two
significant digits (as in the example above).
266 Significant Figures

Rounding up vs. down. Mathematical operations result in an infinite

number of digits, and they should be rounded to the appropriate number
of significant digits. For example in the mathematical quotient
749.1
= 4.928289 . . .
152
the result “4.928” should be rounded up to the physical result 4.93 with
three significant digits. In contrast for the mathematical quotient
742.2
= 4.882894 . . .
152
the result “4.882” should be rounded down to the physical result 4.88. But
what about
731.9
= 4.815131 . . . ?
152
When the leading non-significant digit is five, should it be rounded down
to 4.81 or up to 4.82? It’s permissible to round either way.
I was told in high school that when the leading non-significant digit
was five, you could keep that digit, so that the result of the division above
would be 4.815. I always loved it when the result came out this way: I was
getting four significant digits for the price of three! You should not do this,
because that fourth digit is in fact unknown. You’re not getting four for
the price of three, you’re substituting fiction at the expense of fact.
Evaluating functions. How many significant figures does sin(87.2◦ )
contain? We know that the real angle is somewhere between 87.200 . . .◦
and 87.300 . . .◦ , so the real sine is somewhere between
sin(87.200 . . .◦ ) = 0.9988061 . . .
sin(87.300 . . .◦ ) = 0.9988898 . . .
The usual rule is to make the last significant digit in the result to be the
first one from the left that changes when you repeat the calculation. In this
case the first digit that changed was the “0” that changed to an “8”, so we
round the result to four significant figures, namely
sin(87.2◦ ) = 0.9988.
 267

Numbers that are certain

Any measured number is uncertain, but counted and defined quantities can
be certain. If there are seven people in a room, there are 7.0000000 . . .
people. There are never 7.00395 people in a room. And the inch is defined
to be exactly 2.5400000 . . . centimeters — there’s no uncertainty in this
conversion factor, either.

Conclusions

For most problems in this book the answer is an equation or a graph or a

paragraph. But for a problem whose answer is a number, you must present
that number with appropriate use of significant figures.
 Appendix B

Dimensions

What does “dimensions” mean?

Suppose a table is two-hundred-thirty-six centimeters long or, in symbols,

`T = 236 cm,
where `T represents the length of the table. This means that the table is
236 times as long as the length of the standard centimeter:
`T = 236 cm means `T = 236×(the length of the standard centimeter).
In other words, the symbol “cm” used in the equations above represents
“the length of the standard centimeter”.
The symbol “`T ” stands for “236 cm”. That is, it stands for the number
“236” times the length of the standard centimeter, or in other words, for
the number “236” times the unit “cm”. If I wrote “`T = 236” instead of
“`T = 236 cm”, I’d be dead wrong. . . just as wrong as if the solution to
an algebra problem were “y = 236 x” and I wrote “y = 236”, or if the
solution to an arithmetic problem were “236 × 7” and I wrote “236”. In all
three cases, my answer would be wrong because it omitted a factor. (These
are, respectively, the factor of “the length of the standard centimeter”, the
factor of x, and the factor of 7.) The length of the table is not 236 —
rather, the ratio of the length of the table to the length of the standard
centimeter is 236.
Ignoring the units of a measurement results in practical as well as
conceptual error. On 23 September 1999 the “Mars Climate Orbiter”
spaceprobe crashed into the surface of Mars, dashing the hopes and dreams
of dozens of scientists and resulting in the waste of $125 million. This

269
270 Dimensions

spacecraft had survived perfectly the long and perilous trip from Earth to
Mars. How could it have failed so spectacularly in the final phase of its
journey? The manufacturer, Lockheed Martin Corporation, had told the
the spacecraft controllers, at NASA’s Jet Propulsion Laboratory, the thrust
that the probe’s rockets could produce. But the Lockheed engineers gave
the thruster performance data in pounds (the English unit of force), and
they didn’t specify which units they used. The NASA controllers assumed
that the data were in newtons (the SI unit of force).

New York Times, 1 October 1999, page 1, an embarrassing place to have

your blunders published.
 271

Modern information technology actually encourages mistakes like this.

When you use a calculator, a spreadsheet, or a computer program, you
enter pure numbers like “1.79”, rather than quantities like “1.79 cm”. So
it’s especially important to be on your guard and document your units when
using computers. Keep the units in your mind, even if you can’t keep them
in your calculator!
A nitpicky distinction is that the length of the table has the units of
“centimeters” and the dimensions of “length”. If the length of the table
were measured in centimeters or meters or even in cubits it would still have
the dimensions of length. But in everyday language the terms “units” and
“dimensions” are often used interchangeably.

Unit conversions

It is standard usage to refer to the length of the standard centimeter by

the symbol “cm”. But in this appendix I’ll also refer to the length of the
standard centimeter by the symbol `cm . Similarly I’ll call the length of the
standard meter either “m” or `m .
You know that if a table is 236 centimeters long it is also 2.36 meters
long:
`T = 236 cm = 236 `cm = 2.36 m = 2.36 `m .
How can this be? It’s certainly not true that 236 = 2.36! It’s true instead
that 236 cm = 2.36 m because the length of a meterstick is 100 times the
length of a standard centimeter:
`m = 100 `cm .
This tells us that
2.36 m = 2.36 `m = 2.36 × (100 `cm ) = 236 `cm = 236 cm,
or, in the opposite direction,
 
`m
236 cm = 236 `cm = 236 `cm (1) = 236 `cm
100 `cm
 
`m
= 236 `\cm = 2.36 `m = 2.36 m.
100 `\cm

In short, the symbol “cm” can be manipulated exactly like the symbol
“`cm ”, because that’s exactly what it means!
272 Dimensions

Incompatible dimensions

If I walk for 4.00 m, and then for 59 cm, how far did I go? The answer is
459 cm or 4.59 m, but not 4.00 + 59 = 63 of anything!
If I walk for 4.00 m, and then pause for 42 seconds, how far did I go?
Certainly not 4.00 m + 42 s. The number 46 has no significance in this
problem. For example, it can’t be converted into minutes.
In general, you can’t add quantities with different units.
This rule can be quite helpful. Suppose you’re working a problem that
involves a speed v and a time t, and you’re asked to find a distance d.
Someone approaches you and whispers: “Here’s a hint: use the equa-
tion d = vt + 21 vt2 .” You know that this hint is wrong: The quantity
vt has the dimensions of [length], but the quantity 12 vt2 has the dimen-
sions of [length]×[time]. You can’t add a quantity with the dimensions
of [length]×[time] to a quantity with the dimensions of [length], any more
than you could add 42 seconds to 4.00 meters.

A famous use of dimensional analysis

Dimensional analysis sometimes makes it possible to uncover a lot about

complicated situations if you can only ferret out the essential features of
the situation. For example the fluid flow expert G.I. Taylor was able to
deduce the explosive yield of the first nuclear bomb from a sequence of
photographs of the expanding fireball published in Life magazine.2

2 Details presented in Michael B.A. Deakin, “G.I. Taylor and the Trinity test” In-

ternational Journal of Mathematical Education in Science and Technology 42 (2011)

1069–1079.
 273

Taylor realized that this sequence of photos showed a shock wave ex-
panding into an undisturbed medium, and he knew from his previous expe-
rience that the radius of the fireball r could depend upon only three things:
the density of undisturbed air ρ, the energy released through the explosion
E, and the time since the explosion t.

quantity dimensions
3
ρ [mass]/[length]
2 2
E [mass] × [length] /[time]
t [time]
r [length]

Taylor knew that to build r out of ρ, E, and t, he had to cancel out the
[mass] that appears in ρ and E but that cannot enter into r. Thus he had
to build r out of E/ρ and t.

quantity dimensions
5 2
E/ρ [length] /[time]
t [time]
r [length]

2
Now Taylor had to cancel the [time] from the denominator of E/ρ using
the variable t:

quantity dimensions
2 5
Et /ρ [length]
r [length]

Taylor concluded that

p
5
r(t) = C Et2 /ρ
where C is some dimensionless constant like π or 7 but not a number with
dimensions like 9.8 m/s2 .
Sure enough, a plot of r as a function of t2/5 , with data taken from the
magazine photo sequence, yielded a straight line. The energy released by
274 Dimensions

the explosion was

1 r5
E= ρ .
C 5 t2
Taylor had experimental data suggesting that 1/C 5 = 1.033, and from this
he was able to find the energy output of the nuclear bomb explosion at a
time when this precise number was a closely guarded secret.

Conclusions

For most problems in this book the answer is an equation or a graph or a

paragraph. But for a problem whose answer is a number, you must present
that number with units attached.

Problems

B.1 A new law of nature?

It has been proposed that the speed of sound vs and the speed of light c
√
are related through vs = 12 3 c. Check the accuracy of this formula using
speeds expressed in meters/second, then recheck its accuracy using speeds
expressed in kilometers/second. (According to the book U.S. Standard
Atmosphere, 1976, the standard speed of sound at sea level is 340.29 m/s.)
Is this proposal a new and surprising law of nature, or merely a coincidence?
Explain.
B.2 Sound speed
The speed of sound vs in air, in a given room, could reasonably depend on
three things: the air density ρ, the air pressure p, and the room volume V .
In other words
vs = Cρx py V z
where C is some dimensionless number. Find the exponents x, y, and z.
 275

B.3 Physics in film

Alfred Hitchcock’s 1935 film The 39 Steps is one of the great spy thrillers
of all time. In the film, several men and women crisscross England and
Scotland in pursuit of an important but unspecified secret document. Only
in the final minute of the film does the audience find that the the document
contains the specifications for a completely silent aircraft engine, and that
these specifications hinge upon “the secret formula
 γ
1
r−
r
where r represents the ratio of compression and γ the axis of the fluid line
of the cylinder.”
Show that this formula is not worth the pursuit of a cadre of spies, and
in fact is entirely without meaning.
Clue: You know that 53 means “5 times itself three times” or 53 =
5 × 5 × 5. What does 5(3 feet) mean? How is it related to 5(1 yard) ? [[Note
for non-Americans: The “foot” and the “yard” are archaic units of length
used in the United States of America. A foot is about 30 cm long, and a
yard is defined as exactly three feet.]]
 Appendix C

Complex Arithmetic

√
The familiar numbers like 17, 35 , 2, −π and so forth are called “real
numbers”. The square of any real number is non-negative. But we can
imagine a new category of numbers that have negative squares. We first
imagine the number i, with i2 = −1. Then we can imagine the number 3
times i, with (3i)2 = −9. These are called “imaginary numbers”.
The names “real” and “imaginary” are unfortunate. Numbers are use-
ful abstractions that exist in our minds: you’ve seen two hands, and two
fingers, and two apples; you’ve seen the Arabic numeral “2” and the Roman
numeral “II”, which are made of ink; you’ve seen the English word “two”,
the German word “zwei”, and the Somali word “laba”, again made of ink;
but you’ve never seen the number 2, which is made of pure thought. In
the usual sense of the words “real” and “imaginary”, no number is real; all
numbers are imaginary.
The sum of a real number and an imaginary number is called a “complex
number” (another unfortunate name). Just as the real number x can be
profitably visualized as a point on the one-dimensional real line, so the
complex number z = x+iy can be profitably visualized as the point (x, y) on
the two-dimensional “complex plane”. This book assumes some background
in complex arithmetic. If your knowledge is rusty, these problem should
grease your mental gears.

Problems

C.1 Complex sum and product

Find the sum (2 +3i) + (−3 + 5i), the product (5+ 7i)(2−i), and the square
(3 + i)2 .

277
278 Complex Arithmetic

C.2 Cartesian and polar forms of a complex number

The “Cartesian form” of a complex number z is x + iy, where x and y are
real. (The quantity x is called the “real part of z”: x = <e{z}, while the
quantity y is called the “imaginary part of z”: y = =m{z}.) The “polar
form” of a complex number is reiθ , where r and θ are real and r is non-
negative. (The quantity r is called the “magnitude”, while the quantity θ
is called the “phase”.) Using the Euler relation eiθ = cos θ + i sin θ, show
that these forms are related through
p
r = x2 + y 2 and tan θ = y/x.

C.3 Express in polar form

√ √
Express in polar form: 2 + 12 i and −1 + 3 i.
C.4 Multiplication√of complex√numbers
Find the product (2 + 12 i)(−1 + 3 i) using both Cartesian and polar
forms.
C.5 Polar form of i and 1
1
Show that i = ei( 2 π+2πn) , where n is any integer (positive, zero, or neg-
ative). Similarly, find an infinite number of polar representations of the
number 1.
C.6 Complex conjugate
If z = x + iy, where x and y are real, then the “complex conjugate” of z
is defined as z ∗ = x − iy. Show that r2 = zz ∗ . (The magnitude r is also
called |z|, so this result is often written |z|2 = zz ∗ .)
 Appendix D

Problem-Solving Tips and Techniques

A physicist can wax eloquent about concepts like interference and entangle-
ment, but can also use those concepts to solve problems about the behavior
of nature and the results of experiments. This appendix gives general ad-
vice on problem solving, then lists the problem-solving tools introduced and
elaborated upon in this book.
You have heard that “practice makes perfect”, but in fact practice makes
permanent. If you practice slouchy posture, sloppy reasoning, or inefficient
problem-solving technique, these bad habits will become second nature to
you. For proof of this, just consider the career of [[insert here the name of
your least favorite public figure, current or historical, foreign or domestic]].
So I urge you to start now with straight posture, dexterous reasoning, and
facile problem-solving technique, lest you end up like [[insert same name
here]].

An approach to problem solving

Suppose you want to travel from San Diego to Boston. You start by deciding
whether to fly, take a bus, or drive a car. If you decide to fly, you then
make subsidiary decisions like choice of airline. If you decide to drive, you
make different subsidiary decisions: Should you first change your car’s oil?
Should you take a side trip to the Grand Canyon? Or to visit your friends
in Boulder, Colorado? What you don’t do is just step out of your front door
and walk northeast: you make a plan before taking that very first step.
And just as your journey begins before you take your first step, so it
extends after you take your last step. Any journey, properly considered,
includes reflection upon that journey. This might be merely technical (“I’ll

279
280 Problem-Solving Tips and Techniques

never travel on that bus line again.”) or it might open a door to future
travel (“The Grand Canyon was so spectacular! Next time I’ll hike from
the rim down to the Colorado River.”) or it might be deeper still (“My
friends in Boulder seem so happy together. I need to rethink my plan of
remaining single all my life.”).
As with travel, so with physics problem solving. The first step is to un-
derstand the problem. What is given? What is asked for? For problems
in classical mechanics and electromagnetism, one tool for understanding
the problem is to sketch the situation. In relativity, if often helps to
make two sketches: one for each reference frame. Some quantum mechan-
ics problems are so abstract that a sketch doesn’t help you understand the
problem, but often doodling plays that same role.
The next step is to select a strategy — a key idea to employ — before
rushing in to make detailed derivations. Is this a time evolution problem?
An energy eigenproblem? An interference or entanglement problem?
Once you pass on to implementing your strategy, keep your goal in
mind to avoid deriving endless numbers of equations that are true but that
don’t help you reach your goal.
Finally, once you’re reached that goal, reflect on your final result.
What is nature trying to tell you through this problem? Is the result in
accord with your expectations? A good example of this stage of problem
solving appears on page 171. Instead of reaching equation (4.105), saying
“That’s the end”, and heading to bed for some well-earned sleep, we spent
two paragraphs on the consequences of that equation, found them remark-
able and unexpected, and used them to illuminate the role of interference
in quantum mechanics. I.I. Rabi reflected on the consequences even more
deeply, and used those reflections to invent the atomic clock. Another ex-
ample concerns the Planck radiation law (1.13): pages 16–23, plus sample
problem 1.2.1, are ten pages of reflection upon that single brief equation.
As with travel, such reflection might be merely technical (“Why did I
have to work out that integral in detail? I should have seen from symmetry
that the result would be zero.”) or it might be deeper (“When Styer said
an atom might not have a position, I thought he was spouting bullshit. But
after working problem 2.7, ‘Bomb-testing interferometer’, I realize that I
have to rethink my ideas about how atoms behave.”).
Problem-Solving Tips and Techniques 281

List of problem-solving tools

average of sine squared function, 138
avoid finding the normalization constant, 156
check for divergence, 17
check limiting cases, 16–17
dimensional analysis, 195, 245
easy part first, 133
everyone makes errors, 13, 160, 240
Fourier sine series, 136–137, 141–142, 147–149
list known quantities and desired quantities, 37
look for physically significant combinations of quantities, don’t rend
them apart, 38
ODE, informal solution of, 178–197
scaled quantities, 198–201, 246
scaling, 197
test and reflect on your solution, 14, 16–17, 20–22, 26–27, 30, 66,
139–146, 148–149, 171–172, 279–280
 Appendix E

Catalog of Misconceptions

Effective teaching does not merely instruct on what is correct — it also

guards against beliefs that are not correct. There are a number of preva-
lent misconceptions concerning quantum mechanics. This catalog presents
misconceptions mentioned in this book, together with the page number
where that misconception is pointed out and corrected.

a “wheels and gears” mechanism undergirds quantum mechanics,

82, 159–162
all states are energy states, 6, 146
amplitude is physically “real”, 6, 95–96, 101, 113–114, 129
atom behaves like a miniature solar system, 38–39
“collapse of the quantal state” involves (or permits) instantaneous
communication, 113–114
electron is a small, hard marble, 38–39, 62
energy and temperature are always related linearly, 21
energy eigenfunction has the same symmetry as the potential en-
ergy function, 186
energy eigenvalues alone solve the quantal problem in full, 151
generic quantal state time-evolves into an energy eigenstate, 143,
152
identical particles attract/repel through a force, 216
identical particles in ground state of helium, 252
identical particles reside in different levels, 222, 225, 252, 257
identical particles, label particles vs. coordinates, 206–207
identical particles, three particles, 218

283
284 Catalog of Misconceptions

indeterminate quantity exists but changes rapidly, 55, 140

indeterminate quantity exists but changes unpredictably, 55, 140
indeterminate quantity exists but is disturbed upon measurement,
55, 159–162
indeterminate quantity exists but knowledge is lacking, 6, 55, 80,
96, 161–162
indeterminate quantity exists in random shares, 55
magnetic moment behaves like a classical arrow, 55
mean position moves classically, 159
particle has no probability of being in classically prohibited region,
187
particle is likely to be where potential energy is low, 189
photon as ball of light, 25, 26, 28, 82
photon is a small, hard marble, 25, 26, 28, 82
pointlike particles shimmy across nodes, 140, 186
probability density is all that matters, 125, 156
quantum mechanics applies only to small things, 4
quantum mechanics is just classical mechanics supplemented with
a veneer of uncertainty, 159–162
state of a two-particle system, 129
state of system given through states of each constituent, 113, 129–
130
transition to ground state, 152, 244
wavefunction associated not with system but with particle, 129
wavefunction exists in position space, 129
wavefunction is dimensionless, 124, 143
wavefunction must factorize into space × spin, 173–174, 226