QM-Notes-SG

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Ten principles of Quantum Mechanics

Steven S. Gubser

A suggestion: Before reading the whole 46 pages, try reading just the bold-facedstatements of each principle at the beginning of each section.

Contents

1 Wave-particle duality 3

2 Einstein relation for the energy 7

3 De Broglie relation for the momentum 11

4 Heisenberg uncertainty relation 16

5 Probability and the wave-function 20

6 Fermions and bosons 24

7 Negative frequency and anti-matter 27

8 Spin 31

9 Entropy 34

10 Thermal occupation numbers 38

Preface

Your theory is crazy, but its not crazy enough to be true. Niels Bohr1

Quantum mechanics is unlike classical mechanics not only in substance but in style.

Whereas in classical mechanics, one is used to everything following more or less logically

from F = ma together with some knowledge of force laws, in quantum mechanics there are

a number of inter-related principles which do not seem to arise from a single underlying

1All quotations from Niels Bohr were copied verbatim from http://en.wikipedia.org/wiki/Niels Bohr.I did not check their correctness or authenticity.

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Figure 1: Left: A classical cat (actually drawn in a Baroque style). Right: A classical cat

at finite temperature. From http://www.stevenorton.com/shop/page1.html .

law. Whereas F = ma is a starting point with clear mathematical content, the principles

of quantum mechanics have a semi-quantitative, semi-empirical quality, and they do not

immediately seem to lend themselves to a unique mathematical implementation. And yet,

quantum mechanics has grown to be a precise mathematical framework which is highly

predictive and, as far as we can tell, correct. I aim to set forth in ten principles the main

ideas of quantum mechanics that inform that mathematical framework and which have some

relevance to the early history of the subject (early meaning up to about 1932, when thepositron was discovered).2 Two of these principles have to do more properly with quantum

statistical mechanics, where in addition to quantum uncertainty there is thermal randomness,

something like in figures 1 and 2.

I do not aim to formulate quantum mechanics precisely. The canonical presentation of

non-relativistic quantum mechanics captures all but principles 7, 9, and 10 pretty well. But

a full implementation of principles 7 and 8 requires relativistic quantum theory, which also

principle 5 is implemented in a more subtle fashion than non-relativistic quantum mechanics.

Principles 9 and 10 are the ones having to do with finite temperature, so they are alsooutside the purview of the simplest treatment of non-relativistic quantum mechanics. But

the fundamental physical constant describing quantum phenomena, Plancks constant, h =

6.626 1034 J s, was discovered in a finite temperature context. So, from a historicalperspective at least, quantum mechanics and quantum statistical mechanics are inseparable.

2If I went further forward in history, say up to 1980, other fundamental ideas would come in: secondquantization, gauge symmetry, and renormalization. The first two of these ideas are tucked into the currentdiscussion in small ways, mostly having to do with the description of photons.

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Figure 2: Left: A quantum cat (actually drawn in a Cubist style). Right: A quantum catat finite temperature. From http://www.stevenorton.com/shop/page1.html .

This adds to the difficulty of understanding either subject, because it seems necessary to

understand everything before you understand anything. Pity then the pioneers of quantum

mechanics, who not only had to start from the wrong end of the subject (blackbody radiation)

and work their way to the easy bits (Schrodingers equation), but also had to figure out along

the way which principles of classical mechanics should be retained and which abandoned.

1 Wave-particle duality

A triviality is a statement whose opposite is false. However, a great truth is a statement

whose opposite may well be another great truth. Niels Bohr

Particles like electrons and photons behave like waves in the sense of exhibitinginterference phenomena. But they behave like particles in the sense that they

are indivisible: for example, if an electron absorbs or emits a photon, it absorbs

or emits it entirely, and the change in energy of the electron equals the energy

of the photon.

A clean demonstration of the wave nature of light is the two-slit experiment, see figure 3.

Suppose a coherent green light source is incident on the two slits. For our purposes, coherent

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Figure 3: The setup of the double slit experiment. Upper left: Rays from the two slits traveldifferent distances to get to the screen. Lower left: A coherent source incident on the screenfrom the left produces interfering wave-fronts from the two slits. Upper right: A single slitproduces a broad peak as the central part of its diffraction pattern. Lower right: The two-slitinterference pattern.

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means that the light incident on the two slits is a plane wave such that a particular wave

crest (lets say crest A in the lower-left drawing in figure 3) enters through each slit at the

same moment. The resulting interference phenomenon observed on the screen has regularly

spaced maxima which are understood to be points of constructive interference between light

coming through slit 1 and light coming through slit 2. Between these minima one finds

minima corresponding to destructive interference. Constructive interference occurs when

the wave from slit 1 is in phase with the wave from slit 2. That will happen if the wave from

slit 2 has to travel the same distance as the wave from slit 1 in order to get to the screen,

or if it has to travel one wavelength further, or two wavelengths, or n times the wavelength

where n is an integer. Using a small angle approximation, which is justified in the limit

where is much greater than d and s, its straightforward to show that the angles at which

the maxima occur are given bysin n

d s

. (1)

The minima occur at angles such that sin (n + 1/2)/d because then the two waves are180 out of phase, so they destructively interfere.

So far, Ive summarized the double slit experiment at the level found in any introductory

physics text, for example chapter 22 of Knight. Now for the punchlines:

1. Electrons do the same thing. A double-slit apparatus capable of handling electrons was

not around in the early 20th century, but the Davisson-Germer experiment in 1927 isgenerally deemed an adequate stand-in. Electrons with momentum p 14 keV/cwere reflected off a nickel crystal, and the reflected wave was observed to exhibit an

interference pattern. The details are more complicated than figure 4 suggests partly

because nickel has a face-centered cubic crystal structure, not just a square lattice as

shown. (Face-centered cubic means a cubic lattice with an extra atom at the center

of the face of each side of each cube.) Simplifying a little, one type of scattering of

the electrons off the nickel crystal is like reflection of a wave of electrons from a line of

atoms with spacing a = 3.5 A. Then you expect maxima for angles such that

2a cos = n . (2)

The reason is that if the electron can hit the atom one below the surface, it travels

2a cos further than if it hit the atom right above it. The angle is arbitrary, but an

approximation is still being used: a.Maxima were observed corresponding to the result (2), where = h/p 0.14 A. The

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Figure 4: A cartoon of the Davisson-Germer experiment, where electrons are reflected offa nickel crystal. Top: The electrons reflect through an angle 2, with angle of incidenceequalling angle of reflection as measured from the normal to a crystal plane. Bottom: Elec-trons reflecting off atoms in different crystal planes travel different distances, so they caninterfere.

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Davisson-Germer result gives compelling support to de Broglies hypothesis, discussed

at greater length in section 3.

Maybe this means that we should understand electrons to be a wave as well, and give

up on all this matter-as-particles nonsense. But theres more...

2. Electrons in a double-slit experiment exhibit the standard interference pattern even

when they go through the apparatus one at a time. If you run the experiment for

a very short time, only one electron gets through, and you see only one dot on the

photographic plate behind the two slits. The top panel of figure 5 shows a situation

where only a few electrons have gotten through the apparatus. This seems like com-

pelling evidence that electrons are indeed particles. But if you wait a long time so

that many such dots accumulate, will form the usual interference pattern, as lowerpanels of figure 5 show. The explanation is that a single electron acts as a wave when

passing through the two slits, and this wave interferes with itself, but when the wave

is incident on the screen, the electron is forced to choose where to be, and it does so

probabilistically, with the probability proportional to the intensity of the wave. The

same thing happens with photons. If there were many photons (or electrons), then the

intensity of the wave can be understood as the energy delivered per unit area per unit

time. Loosely, this intensity is proportional to the brightness of the lines of constructive

interference in the original double-slit experiment.

2 Einstein relation for the energy

You know, what Mr. Einstein said is not so stupid.... Wolfgang Pauli3

A wave with frequency is composed of particles with energy , where = h/2.

A particle with energy can be described as a wave oscillating with frequency .

Frequency and wavelength for light are related by

= 2 = 2c . (3)

Visible light has a typical wavelength = 5500 A. (More precisely, this wavelength corre-

sponds to green light, which is in the middle of the visible spectrum.) Using (3) one finds

3All quotations from Pauli were copied verbatim from http://www.msu.edu/lewiska8/finalwebisp213h/pauli quotes.htm. I did not check their correctness or authenticity.

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Figure 5: Results from a double slit experiment using electrons. The number of electrons is:10 (a), 200 (b), 6000 (c), 40000 (d), 140000 (e). From A. Tanamura et al., Am. J. Phys. 57

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= 3.4 1015 Hz, or equivalent, = 5.5 1014 Hz. For some reason, frequencies of light areusually quoted in terms of rather than angular frequency (and to make matters worse,

people often use f for ordinary frequency rather than ), but is generally preferred in

quantum mechanics because there are fewer 2s. AM radio waves have typical frequency= 100 MHz, corresponding to = 3 m.

Using Einsteins relation,

E = , (4)

one finds E = 2.3 eV for a green photon, and E = 4.1 107 eV for an AM radio photon.Calculations like this are facilitated by remembering the value ofc in weird units:

c = 1973 eV A . (5)

Note that the photon energy has no clear relation to E = mc2: photons are massless and

travel at the speed of light, so = and m = 0 while E is finite.What prompted Einsteins study of (4) is the photo-electric effect. Lets start with a

simplified version of this experiment, where we have a source of photons whose frequency is

tunable and whose overall brightness is also tunable. We shine it on a metal, say gold, and

watch for electrons coming out. (That last bit is pretty sketchy: we have to assume that

weve learned how to observe single electrons and measure their energy.) The main features

of the photo-electric effect are:

1. The light has to have a certain minimum frequency 0 before any electrons are emitted.

The corresponding energy is W = 0, called the work function. Every metal has a

characteristic work function, and for typical metals W ranges from 4 to 5 eV. The work

function is lower for alkali metals like sodium, potassium, and cesium, on the order of

2.2eV,4 which means that visible light has a high enough frequency to produce photo-

electrons: see figure 6. Einsteins understanding of this minimum frequency is that the

electrons are bound inside the metal with binding energy W, and to kick an electron

loose, the photon has to deposit an energy > W into it.

2. When > 0, the electrons that are kicked loose (called photo-electrons) have a

4This makes sense because the outermost electron in their shell structure is more loosely held than inother metals. The same property makes alkali metals basic when dissolved in water (hence the name): theyare electron donors, like Lewis bases.

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Figure 6: The photo electric effect for potassium. Blue and green light producephoto-electrons, and red light doesnt. From http://hyperphysics.phy-astr.gsu.edu/hbase/mod1.html.

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maximum kinetic energy

K.E.max =1

2mv2max = W . (6)

(Its assumed here that the electrons are non-relativistic, otherwise we would have to

use K.E.max = ( 1)mc2.) Einsteins explanation of this is that all the photonsenergy goes into the electron that it hits: note that this is part of my earlier statement

of wave-particle duality in section 1. But it could happen that the electron scatters off

of other electrons in the metal before escaping, and if it does it can lose energy. Hence

its kinetic energy is often less than the bound (6).

3. When > 0, the number of photo-electrons emitted per unit time is proportional to

the intensity of the light source. Einsteins understanding of this is that intensity is

a measure of the number of photons that hit the metal per unit time. Each one has

some chance of producing a photo-electron, and there are enough electrons in the metal

and few enough photons hitting it per unit time that the individual photon-electron

collision events are independent.

Its worth noting that energy conservation is a cornerstone of Einsteins theory of the photo-

electric effect, which is what he won the Nobel Prize for. Excluding the rest energy of the

electron (E = mc2), its initial energy is

W, and after it absorbs the photon it is .

3 De Broglie relation for the momentum

A wave with wavelength is composed of particles with momentum p = k, where

k = 2/ is the wave-number. A particle with momentum p can be described as

a wave with wavelength = h/p.

The Davisson-Germer experiment was the experimental confirmation of de Broglies hy-

pothesis. Whats notable is that de Broglie came first, in 1925. He had two big hints. The

first was Einsteins relation (4) for the photo-electric effect (1905), which implies de Broglies

relation for photons:

p =E

c=

c= k =

h

for photons. (7)

In more detail: E = pc is a standard relation of special relativity (for photons), and = ck

is a standard relation for waves that have a k-independent phase velocity, d/dk = c, and

together they imply the equivalence of E = and p = k.

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The second big hint was Bohrs model of the hydrogen atom (1913) and the Bohr-

Sommerfeld quantization condition (c.a. 1915, reformulated by Einstein in 1917). Before

reviewing Bohrs model from de Broglies perspective, we need to remind ourselves about

the fundamental experimental observations that made Bohrs model compelling and yet con-

fusing. First, atoms have a very small nucleus (about (1.2fm)A1/3 in radius, where A is the

atomic number, compared to a typical size 1 A of the atom as a whole). This was discov-

ered by Rutherford in 1911 by bombarding gold with non-relativistic alpha particles (helium

atoms stripped of their electrons) and observing that they scatter off the nucleus in just the

way they should if the alpha particle and the nucleus are described as point particles with

a repulsive electrostatic interaction, V = 2Ze2/40r, where Z is the number of protons in

the nucleus and the 2 accounts for the two protons in an alpha particle.

Thinking that both protons and electrons are effectively point particles leads to the solarsystem model of the hydrogen atom: the electron orbits the proton in the way that the

earth orbits the sun. Even the force law is the same up to an overall factor as Newtonian

gravity:

V(r) = e2

40

1

rso F = e

2

40

r

r3, (8)

where the minus sign is because the electron has charge e while the proton has charge e. Butwhat about orbits where the electron falls radially inward toward the proton? Wouldnt there

be a collision of the two, and wouldnt the kinetic energy get arbitrarily large as V

?

No sign of such high-energy collisions is observed in a vessel of hydrogen. Worse yet, its

understood from the classical theory of radiation that an electron in a circular orbit radiates

energy. There is a formula for this, Larmors formula (1895) which has many quantitative

confirmations. Using it on the hydrogen atom should lead to the conclusion that the electron

spirals inward, radiating photons with a wide spectrum of energies. This should continue

without limit, and the total power going into radiated photons should increase with time,

without limit. This is very far from observations!5 A vessel of hydrogen at room temperature

scarcely radiates at all. Most of the atoms are in their ground state, meaning the state of

minimum possible energy.

Radiation from hydrogen atoms does occur when the hydrogen is raised to a high tem-

perature (without any oxygen present, please). The radiation has a number of characteristic

5It might seem odd that the same objection is not raised to the apparent stability of the solar system.Shouldnt the earth radiate gravitons as it goes around the sun and similarly spiral inward? In fact it does,but very gradually. Einsteins theory of General Relativity predicts precisely how fast this happens, andfor systems where a pulsar (a spinning neutron star) orbits around another compact massive object, thein-spiralling effect was observed by J. Taylor and collaborators, resulting in the 1993 Nobel Prize.

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wavelengths (all quoted in Anstroms):

mn m = 2 m = 3 m = 4 m = 5

Lyman series, n = 1 1216 1026 972 950Balmer series, n = 2 6565 4863 4342

Paschen series, n = 3 18760 12820

(9)

Remarkably, the corresponding frequencies, mn = 2c/mn, can be organized into a

single formula when numbered as indicated in (9):

mn =Em En

for m > n 1 , (10)

where

En = Rn2

and R = 13.6 eV . (11)

Already in (11) we see another justification for the idea that photons are emitted as particles,

all at once, with some definite energy. The energies En are interpreted as the possible energies

of the hydrogen atom, and transition between levels occurs via photon emission. See figure 7.

The n = 1 energy level is thought to be the ground state, from which no further radiation is

possible.

Now we can start coming to the punchline. Consider circular orbits of the electron around

the proton, as shown in figure 7. Lets assume that the electron moves non-relativistically.

Then its velocity follows from using F = ma in the y direction of figure 9 and noting that

ay = v2/r:Fy = e

2

40

1

r2= may = mv

2

r= p

2

mr. (12)

We can solve for p:

p = mv =

me2

40r. (13)

De Broglie tells us that the electron is a wave with wavelength = h/p. This suggests thecriterion

2r = n , (14)

meaning that when the electron goes once around its orbit, it returns in phase with itself

because it traveled an integer number of wavelengths. A picture of what would happen if

this werent so is shown in figure 10.

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Figure 7: A cartoon of the Bohr model of the hydrogen atom, where a photon of def-inite energy is emitted during a transition from one energy level to a lower one. Fromhttp://encyclopedia.thefreedictionary.com/Bohr-Sommerfeld+quantization .

Using (13) and the de Broglie relation to find , we find that (14) becomes

2r = nh

40rme2

. (15)

This can be solved to find the radius:

rn = n2aB where aB =

240me2

0.529 A . (16)

Since (13) now tells us the momentum, its easy to evaluate the energy of the nth level:

En = K.E. + P.E. =

p2

2m + V(r) =

p2

2m e2

40r = e2

80rn , (17)

where in the last step we found (by calculation) that the kinetic energy of the circular orbit

is 1/2 times the potential energy, and we plugged in r = rn. Using (16) once more, onefinds

En = (e2/40)

2

2n2. (18)

Plugging in numbers, one recovers precisely (11). Bohr won the Nobel Prize for his theory

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Figure 8: The spectrum of hydrogen, showing Lyman, Paschen, and Balmer se-ries as well as energy levels in eV. From http://www.daviddarling.info/images/hydrogen spectrum.gif.

F

x

y

e

p

r

v

Figure 9: Free body diagram of an electron moving around a proton.

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Figure 10: An electron in a circular orbit represented as a wave. This is disallowed as aquantum orbit because the circumference isnt an integer multiple of the wavelength. So

the wave doesnt constructively interfere with itself after going once around the orbit. Notethat the notation for the radius of the orbit here is a instead of my preferred letter forradius, r. From http://www2.kutl.kyushu-u.ac.jp/seminar/MicroWorld2 E/2Part1 E/2P11 E/deBroglie wave E.htm.

of atomic spectra in 1922.

4 Heisenberg uncertainty relation

If quantum mechanics hasnt profoundly shocked you, you havent understood it yet.

Niels Bohr

xp /2, where x is the uncertainty in the position of a particle or systemof particles and p is the uncertainty in its momentum. Also, Et /2 whereE is the uncertainty in the energy of a particle or system of particles and t

is a finite time interval over which its properties are measured or altered.

The uncertainty principle is shocking because it says that the whole structure of Newto-

nian mechanics is based on a false premise. That premise is that a particles state at a given

time can be specified by giving its position and velocity. The discussion so far seems to rely

on this premise: especially the calculation (12) was squarely based on Newtonian mechanics,

although we quickly transitioned to a wave picture in (14). And yet we got a right answer

in (18).

According to the wave picture, the electron is spread out over its orbit, which is a standing

wave around the proton. Either it is in all places at once along this orbit, or it hasnt

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decided where to be along its orbit. In this sense (if it makes sense), we should regard the

uncertainty in its position to be comparable to the radius of its orbit: x r. Likewise,one cant be sure which way the momentum points, because it points in different directions

at different points along the orbit. But the magnitude of the momentum is fixed according

to (13). So p p, and the uncertainty relation tells us

pr >

2. (19)

We can check this against (14), which can be re-expressed as

pr = n , (20)

where n > 0 is an integer. So we see that the allowed quantum orbits according to the Bohr

model satisfy the Heisenberg inequality with about a factor of 2n to spare.6

We can arrive at a more precise manifestation of the uncertainty principle by considering

a wave-packet in a theory with a dispersion relation = (k). To keep things simple, lets

restrict to motion in one dimension:

(t, x) =

dk

2(k)eikxi(k)t . (21)

We saw wave-packets of this type when we considered a chain of coupled pendula. Moreprecisely, when we considered the continuum limit of such a system we found a dispersion

relation

(k) =

c2sk2 + 20 (22)

for some constants c2s and 0 expressible in terms of the properties of the pendula and their

coupling. And we remarked that if we set cs = c and 0 = mc2/, and use Einsteins relation

E = and de Broglies relation p = k, we recover one of the main relations of special

relativity:

E =

p2c2 + m2c4 =

c2k2 + m2c42

. (23)

6Actually, there are four subtleties that a fully quantum mechanical treatment would correct. First, wehave only estimated x and p in a reasonable way, and the inequality xp /2 is precise only whenwe give a precise definition ofx and p, as in the remainder of this section. Second, the real orbitals aresmeared out in radius, so r doesnt have a definite value. Third, the momentum is uncertain even in itsmagnitude, so p doesnt have a definite value. And fourth, there are multiple components of position andmomentum, and each separately satisfies the uncertainty principle: for example, xpx > /2. Even afterall the subtleties are properly accounted for, the uncertainty principle survives.

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Although excitations of coupled pendula have nothing obvious to do with electrons or pho-

tons, the proposal is to use wave-packets of precisely the form (21) to describe relativistic free

particles. For the photon we would set m = 0. For reasons to be explained in section 7, we

want to allow to be a complex variable when it describes an electron, but well take the real

part when it describes a photon. This simple construction is almost the whole story about

free particles in quantum mechanics: the only missing ingredients are spin and anti-matter.

It seems sensible to define the average position at time t as

x(t)

dx |(t, x)|2x

dx |(t, x)|2, (24)

and the average momentum to be

p k where k

dk2

|(k)|2k

dk2

|(k)|2. (25)

The average momentum doesnt change with time because (k) doesnt change. Physically,

this indicates that no forces act on the particle/wave: were describing free propagation. To

formulate precisely what we mean by x, recall the definition of the standard deviation of

a collection of numbers {xn}Nn=1:

x x2 x2 (26)where

x = 1N

Ni=1

xi =

Ni=1 xiNi=1 1

x2 = 1N

Ni=1

(xi)2 =

Ni=1(xi)

2Ni=1 1

. (27)

The identityN

i=1 1 = N is trivial, but it makes the expression for x look as similar aspossible to (24).7 Extending this analogy, we can define

x(t)2

dx |(t, x)|2x2

dx|(t, x)|2 k2

dk

2|(k)|2k2

dk2|(k)|2

. (28)

7If (26) seems unfamiliar, consider the deviation xi xi x of each point xi from the average, x.An equivalent definition to (26) is

x =

1N

Ni=1

(xi)2 .

x is sometimes called the root-mean-square deviation, or rms deviation, of the data from the average,because you first square the deviations, then average them, then take the square root.

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10 5 5 10k

0.2

0.4

0.6

0.8

1.0

10 5 5 10x

1.5

1.0

0.5

0.5

1.0

1.5

at t0

Figure 11: The wave-packet (31) with k = 4 and k = 1/2.

Then we can define

x(t)

x(t)2 x(t)2 p = k

k2 k2 . (29)

It must be admitted that we now have a lot of definitions. One good thing is that we

know exactly what we mean by average position and uncertainty for anywave-form. Another

good thing is that there is a deep theorem of Fourier analysis which says (for any t)

xk 12

, (30)

which evidently is the same as the uncertainty principle when we use p = k. The inequality

(30) is saturated for Gaussians, e.g.

(k) = e(kk)2/42

k (0, x) =

xeikxx

2/42x , (31)

where x 1/2k. This wave-packet is precisely the one discussed starting on page VIII.13of the lecture notes; also see figure 11. It is readily demonstrated that this Gaussian has

k=

k

k=

k

x(0) = 0 x(0) = x .(32)

As we saw in our earlier wave-packet calculations, the width of the x-space Gaussian increases

both for t > 0 and t < 0, while of course (k) doesnt change (because no forces act on the

wave.)

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5 Probability and the wave-function

What we observe is not nature itself, but nature exposed to our method of questioning.

Werner Heisenberg

8

The probability of observing a particle at a given position is proportional to the

square of the amplitude of the corresponding wave at that position.

We already started using this in (24)-(32). Now lets formalize it a little with the notion

of a probability distribution function (PDF), which is simply a function (x) with two

properties:

(x) > 0

dx(x) = 1 . (33)

The probability to find a particle in a given region, say x (a, b), is

x (a, b) = ba

dx(x) (34)

The second condition in (33) says that the probability of finding the particle somewhere

anywhereis 1, which means its a sure thing. The average of any function f(x) is

f

dx(x)f(x) . (35)

Evidently, given any wave-function (x) which is square integrable,

(x) = |(x)|2

dx |(x)|2 (36)

is a PDF. We would define a different probability function (t, x) |(t, x)|2 at each timet. Time and space play a very different role in this discussion: its not that (t, x) is a

probability distribution function over t and x simultaneously; rather, for any fixed t, (t, x)

is a PDF over x.9

8All quotes from Werner Heisenberg were copied verbatim from http://www.brainyquote.com/quotes/authors/w/werner heisenberg.html. I did not check their correctness or authenticity.

9This difference between t and x appears to present some puzzles when we pass to relativistic quantumtheory: if we mix t and x with a Lorentz boost, then the PDF in the boosted frame seems to be makingstatements in the unboosted frame about the relative probability of a particle being here now versus therelater. Anti-matter also complicates the story. A full understanding requires a more powerful discussion usingHilbert spaces.

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Using an inverse Fourier transform

(k) =

dx(x)eikx (37)

to get (k) starting from (x), we can form a PDF in k-space:

(k) = |(k)|2

dk

2|(k)|2 . (38)

I generally insist on a factor 1/2 on the k-space integration measure, as you see in the

denominator of (38). This means that

dk2

(k) = 1 , (39)

expressing the fact that the probability that the particle has some momentumany momentum

is 1.

Two additional characteristic features of quantum mechanics are:

1. Quantum states are usually combined by superposing them. This means that if one

state has wave-function 1(t, x) and the other has wave-function 2(t, x), the super-

position has wave-function 1(t, x) + 2(t, x). This is very different from adding the

probability distribution functions, which are proportional to |1(t, x)|2 and |2(t, x)|

2.

(But see sections 9 and 10 for important exceptions to this rule!)

2. The overall phase of the wave-function doesnt enter into probabilities, but relative

phases between two superposed components do. For this reason, the overall phase is

said to be unobservable.

Heres a good example: lets consider the superposition

(t, x) = 1(t, x) + 2(t, x) where 1(t, x) = eikxit and 2(t, x) = eikxit ,

(40)

and =

c2k2 + m2c4/2. The wave-function (t, x) expresses the idea that an electron

either has momentum k or k. Its in a superposed state of the two possibilities. Recallingthat ei + ei = 2cos , we can rewrite (40) more simply as a single standing wave:

(t, x) = 2eit cos kx . (41)

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Now the PDF is

(t, x) |(t, x)|2 = 4cos2 kx . (42)

We notice right away that there are maxima (at x = n/k for integer n) and minima

(at x = (n + 1/2)/k). This is in contrast with a purely right-moving travelling wave,

(t, x) = eikxit, for which (t, x) |(t, x)|2 = 1, meaning that the particle is equallyprobable to be anywhere, with no maxima or minima.

Consider altering the wave-function (40) by multiplying by an overall phase:

(t, x) = ei1(t, x) + 2(t, x)

. (43)

Evidently, (t, x) is unchanged from (42) because |(t, x)|2 is unchanged. This is even true

if the phase depends on t and x.10 Now suppose we alter the wave-function by inserting arelative phase: to keep things simple, just consider

(t, x) = 1(t, x) 2(t, x) = 2eit sin kx . (44)

Now the PDF is different, being proportional to sin2 kx rather than cos2 kx. The maxima

and minima are in different places.

Theres a pathology in our discussion of traveling waves like 1(t, x) = eit+ikx, which

have |1(t, x)|2 = 1 everywhere. The trouble is, you cant have (x) = 1 over the whole

real line, because the integral of this function is infinite. You cant have (x) = for any

> 0, no matter how small, for the same reason. Nor can you have (x) = 4cos2 kx,

because this function also as an infinite integral. To solve this problem, you have to consider

wave-packets. Instead of (40), I should have superposed two wave-packets, like this:

(t, x) = 1(t, x) + 2(t, x) where

1(t, x) =

dk

2e(kk)

2/42keikxi(k)t

2(t, x) =

dk2

e(k+k)2/42keikxi(k)t .

(45)

1 is a right-moving wave-packet, and 2 is a left-moving wave-packet. The superposition of

10There is something odd about this last statement: it seems to say, for example, that eiteit+ikx isindistinguishable from eit+ikx, at least based on probabilities. But there is a difference in frequencybetween the two wave-functions, so there should be a difference E = in energy. A tentative conclusionfrom this is that the total energy is hard to define, and maybe this shouldnt bother us. But the same thinggoes through for momentum, and its strange not to know what p = 0 means. This problem has to do withhow you couple the electron to an electromagnetic field.

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5 5x

2

4

6

8

10

12

^2 at t5

5 5x

2

4

6

8

10

12

^2 at t2

5 5x

2

4

6

8

10

12

^2 at t0

5 5x

2

4

6

8

10

12

^2 at t5

Figure 12: The probability distribution function for colliding wave-packets with k = 4,k = 1/2, m = 4, and = c = 1. The smooth red lines in the plots at t = 2 and t = 0show the PDF one would get using (47).

the two does not describe two electrons; instead, it describes just one whose time-evolving

PDF is (t, x) |(t, x)|2. Either wave-packet by itself makes a smooth, node-free contri-bution to (t, x), but when we square the sum of the two, an inteference pattern emerges!

See figure 12. The interference pattern owes its existence to cross-terms in the expression

for the probability distribution function:

||2 = |1 + 2|2 = |1|2 + |2|2 + 12 + 12 (46)

If we were to omit these terms, we would obtain a different PDF:

wrong |1|2 + |2|2 1 + 2 . (47)

All expressions in (46) and (47) are functions of t and x. In figure 12 we show the difference

between (46) and (47).

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Interlude

The principles laid out so far provide most of the conceptual underpinning of introductory

non-relativistic quantum mechanics, i.e. PHY 208 and part of PHY 305. We could describethem as the five basic principles of quantum mechanics. They are not enough, however, to

understand the broad sweep of the subject. The remaining sections set forth three advanced

principles of quantum mechanics and two principles of quantum statistical mechanics. All

these notions were understood, in some form, by about 1932. Togetherin an appropriate

mathematical formulationthey underlie huge swathes of atomic physics, condensed matter

physics, high energy physics, and chemistry. We really need these additional principles: for

example, to understand the structure of the periodic table, the Pauli exclusion principle and

spin are indispensible.

At the same time, the five additional principles to be described below take us further and

further from classical intuitions, and they seem to me even more arbitrary and eclectic than

the first five. In the end, the whole of quantum mechanics, including spin, anti-matter, and

the relation to statistical mechanics, finds a beautiful unifying mathematical framework in

quantum field theory. Only gravity remains apart, requiring some new idea (string theory?)

to bring it fully within the purview of quantum phenomena.

6 Fermions and bosonsThis is the worst set of notes Ive ever seen! Wolfgang Pauli, in reference to notes

based on his lectures.

Particles are either fermions or bosons. Identical fermions cannot occupy the

same quantum state (Pauli exclusion). Identical bosons can occupy the same

state. If two identical particles are in two different quantum states, there is no

way of telling which one is in which state.

Consider energy eigenstates, like the levels of the hydrogen atom. Although hydrogen has

only one electron, helium has two, and if we neglect the interations between the two, each can

occupy energy levels which are similar in structure to hydrogens. The exclusion principle

says that the two electrons cant occupy the same state. But there are two subtleties:

1. We forgot about spin! An electron can be spin up or spin down in any given energy

eigenstate. So two electrons (but no more) can occupy a given energy eigenstate, and

when they do, one has to be spin up and the other spin down.

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2. There can be several states with the same energy: these are called degenerate en-

ergy levels. Electron orbitals happen to have some interesting degeneracies. Solving

Schrodingers equation shows that the n = 1 energy level of Bohrs model is unique

(apart from the spin degeneracy mentioned above), but the n = 2 level is four-fold

degenerate (again without accounting for spin degeneracy), and the n = 3 level is

nine-fold degenerate, and so on.

The first two periods of the periodic table can now be explained. Each of Bohrs energy

levels is referred to as a shell, because a heuristic picture is that each successive level is a

little further out from the nucleus. Electrons fill the lower shells before starting to fill the

upper shells. An atom with its outermost shell filled is a noble gas: for instance, neon has 10

electrons, of which two occupy the n = 1 shell and 8 = 4

2 occupy the n = 2 shell. Adding

one more electron (and correspondingly increasing the charge of the nucleus to maintain

overall charge neutrality), one gets sodium, where in addition to filling the n = 1 and n = 2

levels, theres one electron in the n = 3 level. It seems intuitive that this additional electron

is easier to scoop out of sodium than any of neons electrons are, because E3 in Bohrs model

is less negative than E2. And indeed, the chemistry of sodium is all about losing that last,

lonely electron.

Working out the number of electrons that give filled shells correctly predicts the atomic

number of the first two noble gases: see figure 13.

Z = 2 helium

Z = 2 (1 + 4) = 10 neonZ = 2 (1 + 4 + 9) = 28 WRONG! Argon has 18

(48)

We should ask what went wrong with with argon. The answer is that when you include the

interactions among electrons (which in practice can only be done approximately), the energy

eigenstates turn out not to be as degenerate as in the Bohr model. And it turns out that

energy eigenstates with less angular momentum have lower energy, to an extent that ten of

the level 3 states with high angular momentum are more energetic than two of the level 4

states with no angular momentum (to be more precise: 3d states are more energetic than

4s).11

11Zero angular momentum may seem odd from the perspective of the Bohr model: the only way to achievethis is to have the electron moving radially, which seems to mean its going to hit the nucleus. In a fullycorrect quantum treatment, the uncertainty principle spreads out the electron enough so that it (mostly)avoids the nucleus. In particular, ground state of hydrogen (1s) has no angular momentum (ignoring spin),

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Figure 13: Periodic table of the elements. From http://www.bpc.edu/mathscience/chemistry/history of the periodic table.html.

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C A

21

C

BA

21

B

Figure 14: Two ways of distributing three identical bosons among two quantum states.Because the occupation numbers of each state are the same, there is no distinction betweenthe two ways of distributing the bosons among them.

In contrast to fermions, two identical bosons are allowed to be in the same quantum state.

The notion of identical particles is very precise in quantum mechanics. Ignoring interactions

between particles, what it means is that to describe a state of several identical particles,

you shouldnt inquire which quantum state each one is in; rather you should enumerate all

the states and ask how many particles are in each state. The difference is perhaps best

illustrated with an example of three identical bosons, lets call them A, B, and C, with two

quantum states available to each, call them states 1 and 2. For example, these states could

be orbitals of the bosons in some central potential similar to the hydrogen atom. Then the

punchline is that theres no difference between the state where A and B are in state 1 while

C is in state 2, and the state where A and C are in state 1 while B is in state 2: see figure 14.

In fact, it was misleading to give separate designations to the bosons in the first place. The

only meaningful statement is that state 1 is occupied by two bosons, and state 2 is occupiedby one boson. If you had N bosons to distribute among the two states, all could be in state

1, or N 1, or whatever. If instead we considered fermions, the allowed occupation numbersof each quantum state are 0 and 1 (ignoring spin degeneracies).

7 Negative frequency and anti-matter

Particles with charge (like electrons) have complex wave functions, and so there

is a difference between time dependence eit and ei

t. When > 0, the former

describes an electron of energy > 0. The latter solution necessitates the

existence of anti-electrons, or positrons. Particles without charge (like photons)

have real wave functions, and there is no distinction between a photon and its

anti-particle: that is, there are only photons, and an anti-photon is a photon.

whereas in the Bohr model one incorrectly predicts that it does have angular momentum.

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Figure 15: Cartoon depiction of the Dirac sea. Two electrons occupy each negative energylevel because one is spin up and one is spin down. From http://www.phys.ualberta.ca/gingrich/phys512/latex2html/node63.html .

At a classical level, we might inquire whether the standard relation

E2 = p2c2 + m2c4 (49)

of special relativity (often called the mass-shell relation) could be solved to give not only

E =

p2c2 + m2c4, but also E =

p2c2 + m2c4. At first sight, the second solution seems

unphysical, because how can a particle have negative energy? Dirac suggested that such

solutions imply the existence of positrons. More precisely, he suggested that all solutions

eit+ikx to the Klein-Gordon equation with 2 = c2k2 + m2c4/2 are allowed, but when

describing electrons, the solutions with < 0 correspond to states that are already occupied,

in the sense that there is an electron already present in this negative-energy state in the

state that we experience as the vacuum. This hypothesis is described as the Dirac sea. The

exclusion principle forbids any additional electrons from being in negative energy states.

That is why the electrons of our experience (as contrasted with the ones in the sea) have

positive energy. But we could also make a positive-energy state by removing one of the

negative-energy electrons from the sea. If the electron we remove has energy E < 0 and

momentum p, then the state we end up with has energy E > 0 and momentum p,relative to the original vacuum state where all negative energy states are occupied.

The notion of the Dirac sea seems fanciful, and its not clear to me that its required in

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the mathematical implementation. But positrons are not fanciful! They were discovered in

1932, a year after Dirac predicted they should exist.

People objected to the Dirac sea on grounds that it implies that there is an infinite

negative charge density in the vacuum. Later it was realized that the Fermi sea of electrons

that exists in metals and semi-conductors provides a concrete realization of the Dirac sea,

except that instead of populating all energy levels with E < 0, the Fermi sea has electrons

in all states with 0 < E < Ef.12 The Fermi energy Ef is by definition the energy of the

highest state thats occupied. The electrons of the Fermi sea are real objects, understood

approximately as the electrons donated from the outermost shell of each atom in the metal

or semi-conductor. If you add one more electron to the system with E > Ef, its analogous

to creating an electron with energy E Ef > 0 above the Dirac sea. If you take awayone electron from the system, it has to have energy E < Ef (because only such electronsexist in the ground state of the system before you disturb it), and its analogous to creating

a positron with energy Ef E > 0 in Diracs theory. These two types of excitations arecommonly referred to as electrons and holes, or as particles and holes.

Simplifying a little, if we start with pure silicon and dope it by replacing a few silicon

atoms with phosphorus, which has one additional electron, then the additional electrons

act as particle excitations on top of the Fermi sea. But if instead of phosphorus we used

aluminum, which has one fewer electrons than silicon, then the absences of a few electrons

act as hole excitations of the Fermi sea.13

Using phosphorus gives an n-type semi-conductor,so named because the extra electrons are negatively charged, and using aluminum gives a

p-type semi-conductor. The holes in a p-type semi-conductor carry electrical current, so they

are quite tangible even though their existence is more properly an absence. If you like, holes

are only a matter of careful bookkeeping for the electrons in the Fermi sea.

Now lets pass to a wave equation description and ask how things simplify when we

consider matter only, and not anti-matter. Recall that in our treatment of wave-phenomena

in section VIII of the lecture notes, we recovered the mass-shell relation E2 = p2c2 + m2c4

from a wave-perspective by finding the normal modes of the Klein-Gordon equation:

2

t2+

2

x2 m

2c4

2

= 0 . (50)

12Often the allowed electron energies have a band structure, meaning, for example, that states withE1,i < E < E1,f are allowed, and then states with E2,i < E < E2,f are allowed, but states in between thesetwo bands are disallowed by the underlying crystal structure.

13For reasons I dont understand, boron seems to be a more common dopant than aluminum. It has thesame effect because it has the same number of valence electrons, namely three, as opposed to four in siliconand five in phosphorus.

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In writing = eit+ikx, it was a choice to make positive, and then we got (22) and

then (using Einstein and de Broglies relations) (23). There are lots of situations where

we can be fairly sure were dealing with electrons, not positrons: in particular, in non-

relativistic quantum mechanics, if you start with electrons, youre not going to accidentally

make positrons, because the kinetic energies are much less than the rest mass of an electron-

positron pair: K.E. 2mc2 is essentially the same statement as v c.We would like to inquire whether there is a more explicit way of stipulating in the

differential equation that we want only the positive frequency solutions. To see that there

is, consider the analogies

E i t

p

k

i

x.

(51)

The first step in each line of (51) is just Einstein or de Broglie. The second is based on

noting that

i

teit+ikx = eit+ikx and i

xeit+ikx = k eit+ikx . (52)

These equations remind me of eigenvalue equations, Mv = v, where M is, for example,

i /t, v is eit+ikx, and = = E. Briefly, energy is an eigenvalue of the differential

operator i /t, and momentum is an eigenvalue of

i /x. Lets be bold and consider

the identifications

E = i

tp = i

x. (53)

If we use these identifications in the mass-shell relation (49), we recover almost the Klein-

Gordon equation:

2 2

t2+ 2c2

2

x2 m2c4 = 0 . (54)

This last equation doesnt quite mean anything, because it is just a differential operator that

isnt acting on a function. But if we formally multiply both sides on the right by (t,x), we

get the Klein-Gordon equation.

What wed like to do, then, is to translate

E =

p2c2 + m2c4 (55)

into a differential equation. But it doesnt work, at least not straightforwardly: you wind up

with 2/x2 under a square root, and thats hard to make any mathematical sense of even

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when you do allow yourself to multiply in some wave function. (Actually, the Dirac equation

is a subtle way of taking this square root using matrices, but it turns out that it doesnt help

you escape the E < 0 solutionshence Diracs prediction of the positron.) What we can do

is expand the square root, making the non-relativistic assumption p mc:

E = mc2

1 +p2

m2c2 mc2 + p

2

2m. (56)

Now lets use the identifications (53) to get

i

t=

2

2m

2

x2+ mc2 . (57)

As before, this equation isnt quite sensible yet, but if we continue at a formal level by

multiplying from the right by the wave function, now denoted to accord with Schrodingers

preferred notation, we get

i

t(t, x) =

2

2m

2

x2+ mc2

(t, x) . (58)

This is almost Schrodingers equation. Schrodinger replaced mc2 by the potential energy

V(x), which is allowed to depend non-trivially on xso we are no longer describing free

particles! Although this may not be how Schrodinger came to his equation,14 it seems

reasonable to think of it as a slight modification of the non-relativistic limit of the Klein-

Gordon equation. Admittedly, we are not used to incorporating mc2 into the potential energy

V(r) of an electron; but including it or excluding just corresponds to including or excluding an

overall factor eimc2t/ in the wave-function. This has no effect on the probability distribution

functions: it is an example of an unobservable phase (see footnote 10).

8 Spin

Most particles have intrinsic angular momentum, the spin. Fermions, such as

electrons, have half integer spin, meaning that the spin along a specified axis

can only be times a half-integer, like /2 (which are the only values allowed

for an electron). Bosons, such as photons, have integer spin, meaning times

14I am told that he had previously made an intensive study of the Fokker-Planck equation, which isapproximately Schrodingers equation without the i on the left-hand side and has to do with diffusion. Soit seems at least plausible that he came at the formulation of his equation from this completely differentperspective

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an integer. Photons have an additional special property: their spin is always

aligned or anti-aligned with their momentum, and its magnitude is always .

Lets start with an apparent digression, namely a calculation of the angular momentumof the circular orbits in the Bohr model. Assume that the orbit is counter-clockwise in the

x-y plane. Then the only non-zero component of L = r p is

Lz = pr =h

r = n . (59)

In the second equality we used de Broglies relation, p = h/. In the third equality we used

the criterion (14) that the orbital should correspond to a standing wave with an integer

number n wavelengths around the circumference. Its worth noting that we didnt need to

assume anything about the potential V(r) except that it was a central force potential, so

that angular momentum is conserved. Thus it should be a fairly general conclusion that Lz,

or any other component of L, has to be some integer multiple of.

As claimed above, it turns out that electrons by themselves have spin 1/2, meaning that

any component of their intrinsic angular momentum, say Sz, takes values /2. I dont

know of any semi-classical calculation that leads to this conclusion. You would think there

might be: an electron could be visualized as uniform ball of negative charge that literally

spins around some axis. This does not seem to work. Instead, the mathematics underlying

spin 1/2 is the group representation theory of rotations in three dimensionsa somewhat

abstract algebraic construction! But there is a reasonably concrete demonstration that a

given component Sz has two possible values. It is the Stern-Gerlach experiment, pictured in

figure 16. Electrons are shot through a magnetic field and then observed on a screen. 15 The

magnetic field couples to the spin like this:

H =p2

2m+

e

mS B(x,y,z) =

p2

2m+

e

mSzB(x,y,z) , (60)

where in the last expression I assumedB = Bz. The electron is like a little bar magnetic

whose south-to-north axis points opposite its spin. The factor of e/m in (60) is called the

gyromagnetic ratio. Understanding why this particular factor must appearas opposed to

e/2m or some other multipleis subtle, but it can be understood given a sufficiently precise

understanding of relativistic quantum mechanics.16

15Actually, Stern and Gerlach used silver, not free electrons. This is easier, and the physics is nearly thesame because all but one of the electrons in silver are paired into zero-angular-momentum states.

16Actually, there are quantum field theoretic corrections to the gyromagnetic ratio which have been cal-

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N x

xS = h/2

z

y

xe

S

e

S = h/2z

zS = h/2

S =

h/2

S

N

Figure 16: The Stern-Gerlach experiment. Electrons are passed through a magnetic field andthen observed when they hit a screen. Only two trajectories are observed, corresponding toSz = /2. If the magnets are rotated 90

around the y axis, and nothing else is changed,then it is still true that only two trajectories are observed, but now they correspond toSx = /2.

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We can understand classical trajectories of electrons in the Stern-Gerlach apparatus by

using one of Hamiltons equations:

py = H

y = e

2mSzB

y . (61)

Evidently, the last expression is the force on the electron due to a gradient in the magnetic

field. This force can be either up or down according to whether Sz = /2 or Sz = /2.The two dots observed in the experiment support the existence of only two possible choices

for Sz, but unless we believe in the factor of e/2m and have a very precise understanding of

our apparatus, its hard to claim that Stern-Gerlach determines the particular values Sz is

allowed to take.

There is something odd about our description of spin so far: weve claimed that any

component of the spin, like Sz, can take only two values. Naively, that seems to mean that

S can take one of eight values, namely

2

1

1

1

2

1

1

1

2

1

11

2

111

. (62)

But this is wrong! There are only two spin states for the electron, not eight. One basis

for these states is the state | with Sz = /2 and the state | with Sz = /2. Thenotation |state is an abstract notation to indicate a quantum state with some property.Representation theory leads to funny relations like this:

|Sx=/2 = | + | . (63)

Translating (63) into words: spin in the positive x direction is a specific superposition of spin

in the positive and negative z directions. Other superpositions of | and | correspond tospin in other directions: for example, |

+ i |

points in the positive y direction.

9 Entropy

The entropy of a system is S = kB log W, where W is the number of quantum

states accessible to it, or consistent with its macroscopic properties, assuming

that all these states are equally probable.

culated and experimentally verified to about 12 decimal places.

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0.2 0.4 0.6 0.8 1.0p

0.35

0.30

0.25

0.20

0.15

0.10

0.05

p log p

Figure 17: The function log as a function of probability .

state accessible to it, namely the one we know its in. So indeed S = kB log W = 0 because

W = 1. If we know that it must be in one of two states but are completely unsure which

one, then W = 2 and S = kB log 2. But what about a case where the system is 95% likely to

be in state 1 and 5% likely to be in state 2? This goes beyond the definition S = kB log W,

because to apply this definition we stipulated that all the accessible states should be equally

probable. It turns out one can argue that in a system which has probability i to be in any

of several states i, the entropy is

S = kBi

i logi . (67)

You can easily check that if there are W states, and 1 = 2 = = W = 1/W, then (67)

reproduces S = kB log W. If1 = 1 and all other i vanish, then S = 0 because the function

log vanishes as 0 and as 1: see figure 17.Lets try to apply the thinking above to a spin-1/2 particle. If we know nothing about

the system, then S = kB log 2, because the system could be either in the state | withSz = /2 or in the state |

with Sz =

/2. But waitit could also be in the superposed

state |, which I claimed was the state where Sx = /2. The most general superposed stateof the spin-1/2 particle is

| = Z1| + Z2| (68)

for arbitrary complex numbers Z1 and Z2.18 This seems to show that there are infinitely

18Actually, only one of Z1 and Z2 should really be regarded as independent, because the overall normal-ization of a wave-function is not important. Alternatively, one often says that the wave-function has to haveunit norm, and its overall phase is not important.

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many states. Should we say S = for a single spin? The standard answer is No: insteadwe should count a new state as distinct only if it cannot be represented as a superposition

of states counted previously. So W = 2 for a spin 1/2 particle, and for more complicated

systems, W is the dimension of the vector space of all possible wave-functions for the system.

This could still be infinite: it is, for example, for the hydrogen atom, where there are an

unlimited number of energy eigenstates. But if we stipulate that the hydrogen atom has

energy E < 1 eV, then only finitely many of the energy eigenstates are accessible to it, soW becomes finite.

The relationship between quantum mechanics and entropy actually is quite twisty, as

exemplified by the entropy of a pair of electrons, of which one is spin up and the other is

spin down, like the electrons in the ground state of a helium atom. This is a unique state, so

S = 0. But what if we considered the electrons, call them A and B, separately? Theres nodifference between them, so each one has to be equally likely to be in a spin up state or a spin

down state. So we should say WA = 2 and WB = 2, right? But together, the electrons are in

a unique state, so WAB = 1 = WAWB. That violates our basic premise, (64). Maybe its notso bad because the electrons interact with one another. What seems really uncomfortable is

that we could in principle prepare two spins in an up-down state and then, without disturbing

their spins, separate them a long way apart so that they cant interact. Then (keeping track

only of spin) should we still say WA = WB = 2 and WAB = 1? Experimentally, if you make

such states over and over, and then measure only spin A, you will find that its up in halfthe cases and down in the other half. So A =

B = 1/2. If you measure both spins, you

always find one up and one down, only its impossible to predict which one will be up and

which down. Maybe then we should say WAB = 2 for the whole system, since there are

only two possible outcomes when you measure both spins? But this runs afoul of our notions

of identical particles, where we said that the only thing there is to know is how many spins

are up and how many down. From the identical particle perspective the spin state of the

electrons is unique: one up, one down.19 This is the Einstein-Padolsky-Rosen paradox.20

The consensus view of practitioners of quantum mechanics and quantum statistical me-chanics is that indeed WA = WB = 2 while WAB = 1. The spins arent really independent

19In fact, it is misleading even to given the electrons different labels when they are identical. But we couldimagine replaying the discussion where one electron is replaced by a proton, which also has spin 1 /2. Thestory about spin states still goes through with the proton as particle A and the electron as particle B.

20More precisely, EPR were bothered by the claim that if you measure spin A and find spin up, thenno matter how shortly afterward you measure spin B, you must find it to be spin down. How does spin Bknow the outcome of the measurement of spin A? No signal, not even a light wave, could tell spin B howthe spin A experiment came out if we measured B sufficiently shortly afterward. So quantum measurementseems acausal in some sense.

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because we stipulated from the start that one is spin up and one is spin down. They are

said to be correlated, or entangled. Independent systems, to which (64) applies, are by

definition systems where no correlations existthat is, systems whose states can be chosen

without reference to one another or to the totality of A B.

10 Thermal occupation numbers

If a system has quantum states with definite energies En, then raising it to finite

temperature T corresponds to a probability proportional to eEn/kBT that the

system will be in the quantum state with energy En, where kB = 1.381 1023 J/Kis Boltzmanns constant.

There are two thorny issues here. First, the probability that were talking about here

is different from the one that comes from probability distribution functions associated with

superpositions of different states. And second, we seem to have pulled the functional form

eEn/kBT out of a hat.

Considering a spin-1/2 particle in a constant magnetic field helps clarify both issues.

The Hamiltonian is the same as in (60), but now we additionally assume that B is constant.

Also we imagine somehow trapping the particle spatially and keeping it always in the

same trapped state: only its spin is allowed to fluctuate. So the only important part of the

Hamiltonian, for the purposes of the example I want to describe, is

H =eB

m(Sz + /2) . (69)

Really I should have written H = eBm Sz, but adding the constant eB/2m corresponds to

an unobservable overall phase in the evolution of the wave-function. The energy eigenstates

are the state with Sz = /2 (spin up, |) and the state with Sz = /2 (spin down, |).The corresponding energies are

E = E0 = 0

E = E1 =eB

m.

(70)

So spin down is the ground state, and our adjustment of the energy was contrived so that

its energy E0 vanishes. Intuitively, it seems reasonable that at zero temperature, the system

stays always in its ground state, and at non-zero temperature, the spin can fluctuate in

some sense. If the temperature is low, then usually the spin is down, but sometimes it is up.

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If the temperature is sufficiently high, then the spin scarcely cares about the magnetic field,

so it will be up as often as down.

Here is how we would naturally attempt to implement such states with a superposition:

| = | + eE1/kBT| . (71)

This seems like a good idea because when we form the corresponding probability distribution

function by squaring the coefficients, we get

=1

Z =

eE1/kBT

Z, (72)

where Z = 1 + eE1/kBT is a normalization factor chosen so that + = 1. By design,

s eEs/kBT where Es is the energy for spin s. Were pleased to observe that as T 0, 1 while 0, so S 0; and as T , and both approach 1/2, soS kB log 2, which is the largest entropy that a single spin can have.

The problem is that if we take T big, then according to (71),

| | + | , (73)

which corresponds to a spin that points in the +x direction. Thats not what we wanted!

The problem is that when you form a superposition, you know too much about the statefor it to truly describe thermal fluctuations.21 In (73), for example, in the limit T , weknow that the spin is in the Sx = /2 state, which is unique, so S = 0. We followed the

rules laid out in section 5 in forming the probability distribution function , but the trouble

is that the superposition (71) contains extra information that we do not want, namely a

definite direction for the spin for any choice of T. So we have to give something up, and

the thing we give up is the superposition. Thermal randomness is different from quantum

randomness: it has to do only with probabilities, not superpositions.

I still havent explained why I chose the function eEn/kBT

for the probabilities. To justifythis, imagine that we have a large number N of spins, each in the same B field, and coupled

very weakly to each other so that they can exchange energy but dont appreciably affect

21It is this inherent incompleteness of knowledge about a thermal state that I tried to depict on the righthand sides of figures 1 and 2. But whereas lack of knowledge of a classical system is continuous in character,as exemplified by a thermal uncertainty in the position or the momentum, lack of knowledge of a quantumsystem has itself a discrete character, because it (usually) amounts to assigning probabilities to a discretenumber of possible quantum states. Thats what I was thinking of when I made the quantum cat pixelatedinstead of blurred.

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each others energy levels. Thermal equilibrium can be reached by dumping a certain total

amount Etot of energy into the system and then waiting for it to equilibrate among the spins.

We assume that all ways of sharing this energy among the spins are equally probable. How

many are there? Well, Etot clearly has to be some multiple of E1, say Etot = ME1 for some

integer M, and 0 M N because the smallest possible amount of energy is 0 (all spinsdown) and the largest is NE1 (all spins up). For some intermediate M, the number of ways

of distributing the energy is

W =

N

M

=

N!

M!(N M)! NNeN

MMeM(N M)NM = eNlogNMlogM(NM) log(NM) ,

(74)

where in the second equality we used Stirlings approximation, N!

NN/eN.22 Im cheating

a little by ignoring the possibility of superposed states: each spin is either up or down in this

discussion. Spins of this type are referred to as Ising spins, and they are not fully quantum

mechanical because superpositions are disallowed. The full quantum treatment is a little

trickier, but the final result is essentially the same.

Based on (74) we can calculate the entropy of the N spins at energy Etot = ME1:

Stot = kB log W kB

Nlog N Mlog M (N M)log(N M)

. (75)

From this we extract the average entropy S per spin,

S

kB Stot

NkB= M

Nlog

M

N

1 MN

log

1 M

N

= EE1

logE

E1

1 EE1

log

1 E

E1

(76)

in terms of the average energy per spin,

E

Etot

N

=M

N

E1 . (77)

Recall now the first law of thermodynamics:

dE = T dS . (78)

22A more accurate formula is N!

2N NNeN

1 + 112N

+ O(N2)

, but for present purposes therather crude estimate I quoted in the main text is sufficient.

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(Usually you say dE = Q + W with Q = T dS and W = P dV, but in this spin system,there is no notion of volume or pressure, so T dS is all there can be on the right-hand side

of (78).) We can manipulate (78) into a form that allows us to compute the temperature:

1

T=

dS

dE=

kBE1

log

E1E

1

. (79)

We can now solve for the energy:

E =E1e

E1/kBT

1 + eE1/kBT. (80)

On the other hand, we can calculate the expectation value for the energy from the probability

distribution function (72) derived by assuming s

eEs/kBT:

E = E + E = E1 eE1/kBT

Z= E1

eE1/kBT

1 + eE1/kBT. (81)

Comparing (80) with (81), we see that they match! Furthermore, there is only way to

simultaneously have + = 1 and make the average E match with E as computed in(80). So we conclude that s eEs/kBT is the only possible choice.

That was quite a derivation, and things only get harder when there are more than two

states. So let me just assume that n

eEn/kBT is true in general and pass to an example

of considerable historical as well as modern interest: blackbody radiation. Planck knew that

an evacuated cavity heated to a very high temperature glows with a light whose intensity

depends on frequency and temperature in a very specific way (see figure 18):

I(, T) =3

43c21

e/kBT 1 . (82)

It was well understood that (82) is equivalent to an energy density

u(, T) =4

c I(, T) . (83)

Its difficult for us to understand all the prefactors just now, but the dependence 1/(e/kBT1) is something we can work out, as follows. Photons with frequency have energy . They

are bosons, so any number of them can be in a given state, and that number is the only

information we can have about photons of a particular frequency. That means that theres

one state, call it |0, with no photons, another, call it |1, with one photon, yet another, call

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Figure 18: The spectrum of blackbody radiation quantified in terms of intensity per wave-length. From http://en.wikipedia.org/wiki/Plancks law of black body radiation.

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it |2, with two photons, and so on. Photons do not interact with one another, so the energyof the state |n must be

En = n . (84)

Now we construct a thermal probability distribution function based on these energy eigen-

states:23

0 =1

Z1 =

e/kBT

Z2 =

e2/kBT

Z. . . n =

en/kBT

Z. . . , (85)

where the normalization factor that ensures

n=0 n = 1 is

Z = 1 + e/kBT + e2/kBT + e3/kBT + . . . + en/kBT + . . .

=n=0

en/kBT = 11 en/kBT ,

(86)

where in the last step we noticed that the infinite sum is a geometric series. Just as in the

calculation (81) for a spin system, we can now compute the average energy E contributedby photons in definite state of frequency :

E = E00 + E11 + E22 + . . . = 1Z

n=0

nen/kBT . (87)

That last series may seem hard to sum, because its not geometric. Its a little easier to look

at if we define x = /kBT: then

E = Z

n=0

nenx and Z =n=0

enx =1

1 ex (88)

The trick is to notice that

dZ

dx=

n=0 nenx =

ex

(1 ex)2. (89)

23The same cheat is in place here that I mentioned earlier: I am assigning probabilities to states with adefinite number of photons without telling you why the system should choose to be in one of these statesinstead of a superposition of states with different particle numbers. In fact, superpositions are allowed, buta fully correct treatment in quantum statistical mechanics will reproduce the same result for u(, T) as thecurrent one.

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Using this, we arrive finally at

E = ex

1

ex

= 1

ex

1

= 1

e/kBT

1

. (90)

This still differs from the energy density (83) by a factor 2/2c3. This factor has to do

with the number of different states of frequency : the wave-number k must have |k| = /c

because the photon is massless, but k can point in different directions, and different directions

correspond to different states for a photon. Suffice it to say here that a sufficiently careful

treatment of photon modes, together with the ideas summarized in (90), allow one to derive

precisely the formulas (82)-(83) which fit the data.

Three final notes:

1. Planck somehow figured all this out without previous knowledge of Einsteins relation

E = , or of photons, or of bosons, or of identical particles, or of quantized energy

levels. He did know statistical mechanics quite well, and he invented quantized energy

levels with spacing as well as establishing as a fundamental constant of Nature.

Must have been a clever guy! He won the Nobel Prize in 1918 for his theory of

blackbody radiation.

2. The term blackbody is appropriate because the radiation spectrum (82), shown in

figure 18, is much less featured than the spectrum from, say, the hydrogen atom,

where instead of a single hump there are many intense lines. The latter spectrum can

reasonably be described as colorful, because each line has a specific color. Perhaps

a more conventional justification for the term blackbody is that the more perfectly

a surface absorbs light, the better it is at emitting it at sufficiently high temperatures,

and the more closely its emitted light follows the blackbody form. Perfectly absorbing

light means something is black, hence the term.

3. The cosmic microwave background radiation conforms spectacularly well to a black-

body curve with T 2.73 K. See figure 19. This is understood as arising from atransition when the universe was about 400, 000 years old, in which photons decoupled

from electrons. This happens at quite a high temperature (about 3000 K), and the

2.73 K observed today results from the cooling of the photons due to the expansion of

the universe over the intervening 14 billion years or so. The surviving leaders of the

COBE team, John Mather and George Smoot, won the Nobel Prize in 2006.

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Figure 19: Results from the COBE satellite, showing a perfect fit to the blackbody formula.From http://en.wikipedia.org/wiki/COBE .

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Quantum mechanics is certainly imposing. But an inner voice tells me that it is not yet the

real thing. The theory says a lot, but does not really bring us any closer to the secret of the

old one. I, at any rate, am convinced that He is not playing at dice. Albert Einstein

It is very difficult to make an accurate prediction, especially about the future. Niels

Bohr

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