Physics 451 - Statistical Mechanics II - Course Notespeople.ucalgary.ca/~dfeder/451/notes.pdfPHYS 451 - Statistical Mechanics II - Course Notes 5 This expression was rst derived by

Physics 451 - Statistical Mechanics II - Course Notes

David L. Feder

January 8, 2013

Contents

1 Boltzmann Statistics (aka The Canonical Ensemble) 31.1 Example: Harmonic Oscillator (1D) . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Example: Harmonic Oscillator (3D) . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Example: The rotor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 The Equipartition Theorem (reprise) . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4.1 Density of States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5 The Maxwell Speed Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.5.1 Interlude on Averages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5.2 Molecular Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Virial Theorem and the Grand Canonical Ensemble 142.1 Virial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.1.1 Example: ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.1.2 Example: Average temperature of the sun . . . . . . . . . . . . . . . . . . . . 15

2.2 Chemical Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2.1 Free energies revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2.2 Example: Pauli Paramagnet . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 Grand Partition Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.4 Grand Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Quantum Counting 223.1 Gibbs’ Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.2 Chemical Potential Again . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243.3 Arranging Indistinguishable Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.3.1 Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.3.2 Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.3.3 Anyons! . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.4 Emergence of Classical Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4 Quantum Statistics 324.1 Bose and Fermi Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.1.1 Fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334.1.2 Bosons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354.1.3 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.2 Quantum-Classical Transition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.3 Entropy and Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

1

PHYS 451 - Statistical Mechanics II - Course Notes 2

5 Fermions 435.1 3D Box at zero temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435.2 3D Box at low temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.3 3D isotropic harmonic trap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5.3.1 Density of States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465.3.2 Low Temperatures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475.3.3 Spatial Profile . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.4 A Few Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.4.1 Electrons in Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.4.2 Electrons in the Sun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.4.3 Ultracold Fermionic Atoms in a Harmonic Trap . . . . . . . . . . . . . . . . . 51

6 Bosons 526.1 Quantum Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526.2 Phonons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536.3 Blackbody Radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566.4 Bose-Einstein Condensation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

6.4.1 BEC in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 596.4.2 BEC in Lower Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606.4.3 BEC in Harmonic Traps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Chapter 1

Boltzmann Statistics (aka TheCanonical Ensemble)

This chapter covers the material in Ch. 6 of the PHYS 449 course notes that we didn’t get to lastterm. In particular, Sec. 1.1 here corresponds to Sec. 6.2.4 in the PHYS 449 course notes.

1.1 Example: Harmonic Oscillator (1D)

Before we can obtain the partition for the one-dimensional harmonic oscillator, we need to find thequantum energy levels. Because the system is known to exhibit periodic motion, we can again useBohr-Sommerfeld quantization and avoid having to solve Schrodinger’s equation. The total energyis

E =p2

2m+kx2

2=

p2

2m+mω2x2

2,

where ω =√k/m is the classical oscillation frequency. Inverting this gives p =

√2mE −m2ω2x2.

Insert this into the Bohr-Sommerfeld quantization condition:∮pdx =

∮ √2mE −m2ω2x2dx = nh,

where the integral is over one full period of oscillation. Let x =√

2E/mω2 sin(θ) so that m2ω2x2 =2mE sin2(θ). Then∮

pdx =

√2E

mω2

∫ 2π

0

√2mE cos2(θ)dθ =

2E

ω

1

22π =

2πE

ω= nh.

So, again making the switch E → εn, we obtain

εn = nhω

2π= nhω.

The full solution to Schrodinger’s equation (a lengthy process involving Hermite polynomials) givesεn = hω(n+ 1

2 ). Except for the constant factor, Bohr-Sommerfeld quantization has done a fine jobof determining the energy states of the harmonic oscillator.

3


Armed with the energy states, we can now obtain the partition function:

Z =∑n

exp(−εn/kBT ) =∑n

exp(−βεn) = 1 + exp(−βhω) + exp(−2βhω) + . . . .

But this is just a geometric series: if I make the substitution x ≡ exp(−βhω), then Z = 1 +x+x2 +x3 + . . .. But I also know that xZ = x+x2 +x3 + . . .. Since both Z and xZ have an infinite numberof terms, I can subtract them and all terms cancel except the first: Z − xZ = 1, which immediatelyyields Z = 1/(1− x), or

Z =1

1− exp(−βhω). (1.1)

Now I can calculate the mean energy:

U = NkBT2 ∂ ln(Z)

∂T=NkBT

2

Z

∂Z

∂T= NkBT

2 [1− exp(−βhω)]

[1− exp(−βhω)]2 (−1)

hω

kBT 2(−1) exp(−βhω)

= Nhωexp(−βhω)

1− exp(−βhω)=

Nhω

exp(βhω)− 1.

= Nhω〈n(T )〉, where 〈n(T )〉 ≡ 1

exp(hω/kBT )− 1is the occupation factor.

At very high temperatures T 1, exp(hω/kBT ) ≈ 1 + (hω/kBT ), so 〈n(T )〉 → kBT/hω and

U(T 0)→ NkBT and CV (T 0)→ NkB .

Notice that these high-temperature values are exactly twice those found for the one-dimensionalparticle in a box, even though the energy states themselves are completely different from each other.

1.2 Example: Harmonic Oscillator (3D)

By analogy to the three-dimensional box, the energy levels for the 3D harmonic oscillator are simply

εnx,ny,nz = hω(nx + ny + nz), nx, ny, nz = 0, 1, 2, . . . .

Again, because the energies for each dimension are simply additive, the 3D partition function canbe simply written as the product of three 1D partition functions, i.e. Z3D = (Z1D)

3. Because

almost all thermodynamic quantities are related to ln (Z3D) = ln (Z1D)3

= 3 ln (Z1D), almost allquantities will simply be mupltiplied by a factor of 3. For example, U3D = 3NkBT = 3U1D andCV (3D) = 3NkB = 3CV (1D).

One can think of atoms in a crystal as N point masses connected to each other with springs. Toa first approximation, we can think of the system as N harmonic oscillators in three dimensions.In fact, for most crystals, the specific heat is measured experimentally to be 2.76NkB at roomtemperature, accounting for 92% of this simple classical picture. It is interesting to consider theexpression for the specific heat at low temperatures. At low temperature, the mean energy goes toU → 3Nhω exp(−hω/kBT ), so that the specific heat approaches

CV → −3Nhω

kBT 2(−hω) exp

(− hω

kBT

)= 3NkB

(hω

kBT

)2

exp

(− hω

kBT

).


This expression was first derived by Einstein, and shows that the specific heat falls off exponentiallyat low temperature. It provided a tremendous boost to the field of statistical mechanics, becauseit was fully consistent with experimental observations of the day. Unfortunately, it turns out to bewrong: better experiments revealed that CV ∝ T 3 at low temperatures, not exponentially. This isbecause the atoms are not independent oscillators, but rather coupled oscillators, and the low-lyingexcitations are travelling lattice vibrations (now known at phonons). Actually, even CV ∝ T 3 iswrong at very low temperatures! The electrons that can travel around in crystals also contribute tothe specific heat, so in fact CV (T → 0) ∝ T .

1.3 Example: The rotor

Now let’s consider the energies associated with rotation. In classical mechanics, the rotational kineticenergy is

T =1

2~ω · I · ~ω,

where I is the moment of inertia tensor and ~ω is the angular velocity vector. In the inertial ellipsoid,this can be rewritten

T =L2x

2Ixx+

L2y

2Iyy+

L2z

2Izz,

where Lj is the angular momentum along direction and Ijj is the corresponding moment of inertia.Suppose that we have a spherical top, so that Ixx = Iyy = Izz = I:

T =1

2I

(L2x + L2

y + L2z

)=L2

2I.

In the quantum version, the kinetic energy is almost identical, except now the angular momentumis an operator, denoted by a little hat:

T =L2

2I.

The eigenvalues of this operator are `(` + 1)h2/2I, where ` = −L,−L + 1,−L + 2, . . . , L − 1, L sothat ` can take one of 2L+ 1 possible values.

For a linear molecule (linear top), the partition function for the rotor can then be written as

Z =

∞∑L=0

L∑`=−L

exp

(−`(`+ 1)h2

2IkBT

)≈∞∑L=0

(2L+ 1) exp

(−L(L+ 1)h2

2IkBT

),

where the second term assumes that the contributions from the different ` values are more or lessequal. This assumption should be pretty good at high temperatures where the argument of theexponential is small. In this case, there are simply 2L + 1 terms for each value of L. Again,because we are at high temperatures the discrete nature of the eigenvalues is not important, we canapproximate the sum by an integral:

Z ≈∫ ∞

0

(2L+ 1) exp

(−L(L+ 1)h2

2IkBT

)dL.


We can make the substitution x = L(L+ 1) so that dx = (2L+ 1)dL, which is just term already inthe integrand. So the partition function becomes

Z =

∫ ∞0

exp

(− h2x

2IkBT

)dx =

2IkBT

h2 .

Again, I can calculate the mean energy

U = NkBT2 ∂ ln(Z)

∂T=NkBT

2

Z

∂Z

∂T= NkBT

2 h2

2IkBT

2IkB

h2 = NkBT.

This is exactly the contribution that we expected from the equipartition theorem: there are twoways the linear top can rotate, so there should be two factors of (1/2)NkBT contributing to theenergy.

For a spherical top, each of the energy levels is (2L+ 1)-fold degenerate. The partition function forthe rotor can then be written as

Z =

∞∑L=0

L∑`=−L

(2L+ 1) exp

(−`(`+ 1)h2

2IkBT

)≈∞∑L=0

(2L+ 1)2

exp

(−L(L+ 1)h2

2IkBT

),

where the second term assumes that the contributions from the different ` values are more or lessequal. This assumption should be pretty good at high temperatures where the argument of theexponential is small. In this case, there are simply (2L + 1)2 terms for each value of L. Again,because we are at high temperatures so that the discrete nature of the eigenvalues is not important,we can approximate the sum by an integral:

Z ≈∫ ∞

0

(2L+ 1)2

exp

(−L(L+ 1)h2

2IkBT

)dL.

At high temperatures, one needs large values of L before the argument of the exponentials will besignificant, so it is reasonable to make the substitution L(L+ 1)→ L2 and (2L+ 1)2 → 4L2. Thisyields

Z =

∫ ∞0

4L2 exp

(− h2L2

2IkBT

)dL =

√π

(2IkBT

h2

)3/2

.

Again, I can calculate the mean energy

U = NkBT2 ∂ ln(Z)

∂T=NkBT

2

Z

∂Z

∂T= NkBT

2√π

(h2

2IkBT

)3/21√π

3

2

(2IkB

h2

)3/2

T 1/2 =3

2NkBT.

For the speherical top, there are now three contributions to the total energy, which accounts for theadditional factor of (1/2)NkBT over the linear top.

1.4 The Equipartition Theorem (reprise)

The examples presented in the previous section show that the high-temperature limits of the meanenergy for the particle-in-the-box and harmonic oscillator problems were very similar: the meanenergies were all some multiple of NkBT/2 and the specific heat some multiples of NkB/2. Notice


that for both the particle in the box and the harmonic oscillator, both quantities were three timeslarger going from 1D to 3D, i.e. where the number of degrees of freedom increased by a factor of 3.Perhaps U and CV at high temperatures provide some measure of the number of degrees of freedomof the particles in a given system.

To make further progress with this idea, we need to revisit the hokey derivation of the equation ofstate for an ideal gas presented in the PHYS 449 course notes. Recall that, in order to enumerateall the accessible states in a volume V , we subdivided the volume into little ‘volumelets’ of size∆V , each of which defined an accessible site that a particle can occupy. Then we stated that theentropy was given by S = kB ln(V/∆V ). But quantum mechanics tells us that we can’t know boththe exact position and momentum of a particle at the same time. The relationship between theuncertainty in the position ∆x and the momentum ∆p is quantified in the Heisenberg uncertaintyprinciple ∆x∆p ≥ h, where h again is Planck’s constant. For our purposes, this means that Planck’sconstant sets a fundamental limit on the size of the volumelets on can partition the system into. Ina sense, the Bohr-Sommerfeld quantization condition already stated this: the integral of momentumover space is minimally Planck’s constant.

Mathematically, this means that we can write the general partition function in one dimension as acontinuous function

Z1D →1

h

∫ ∞−∞

dx

∫ ∞−∞

dpx exp

(−ε(x, px)

kBT

),

where the accessible classical states ε are explicitly assumed to be functions of both position x andmomentum px. In three dimensions, it would be

Z3D →1

h3

∫d3r

∫d3p exp

(−ε(r,p)

kBT

).

The six dimensions (r,p) together constitute phase space, and the two three-dimensional integralsare denoted phase-space integrals.

To make these ideas more concrete, let’s again calculate the partition function for a particle in a 1Dbox of length L. The classical energy is entirely kinetic, so ε(p) = p2/2m, and

Zbox =1

h

∫ L/2

−L/2dx

∫ ∞−∞

dp exp

(− p2

2mkBT

)=

1

hL√π√

2mkBT = L

√mkBT

2πh2 =L

λD,

consistent with the result found last term. As found previously, U(T 0)→ NkBT/2 and CV (T 0)→ NkB/2. Let’s also consider the 1D harmonic oscillator, for which the total energy is ε(x, p) =p2/2m+mω2x2/2. Now the partition function is

Zh.o. =1

h

∫ ∞−∞

dx

∫ ∞−∞

dp exp

(−

p2

2m + mω2x2

2

kBT

)

=1

h

∫ ∞−∞

dx exp

(−mω

2x2

2kBT

)∫ ∞−∞

dp exp

(− p2

2mkBT

).

The second one of these we already did, giving us L/λD. The first one is equally easy to integrate,since it’s also a Gaussian integrand. So we obtain altogether

Zh.o. =

√π

2π

√2kBT

mω2

√π√

2mkBT =kBT

hω,


which is identical to the high-temperature limit of the 1D harmonic oscillator partition function(6.1):

limT0

1

1− exp(−βhω)=

1

1− (1− βhω)≈ kBT

hω.

The mean energy in this limit is

U(T 0) =NkBT

2

Z

∂Z

∂T≈ NkBT

2

kBT

hωkBhω

= NkBT.

But we already know that the second of the two integrals contributed NkBT/2 to the mean energyU at high temperature, because this was the result for the particle in the 1D box. This means thatthe first integral also contributed NkBT/2 to the mean energy. This is very much analogous to eachspatial integral contributing a factor of NkBT/2 to the mean energy going from 1D to 3D. If youthink of the kinetic and potential energy terms in the classical expression for the energy as eachstanding for one degree of freedom, then one can finally state the

D Equipartition Theorem: Each degree of freedom that contributes a term quadratic in positionor momentum to the classical single-particle energy contributes an average energy of kBT/2 perparticle.

Many examples showing how one can predict the value of U at high temperatures will be covered inclass.

1.4.1 Density of States

Recall that for a particle in a 3D box, the energy levels are given by

εn =h2π2

2mL2

(n2x + n2

y + n2z

).

We can think of the energy levels as giving us the coordinates of objects on the surface of a spherein Cartesian coordinates, except only in the first octant (because nx, ny, nz ≥ 0). So, instead, wecan think of the energies as continuous, ε = γn2, where γ = h2π2/2mL2 and n2 is the length of thevector in ‘energy space.’ So, n =

√ε/γ and therefore

dn =1

2√γ

dε√ε.

In spherical coordinates, we can write

dn = d3n = dnxdnydnz = n2 sin(θ)dndθdφ→(

1

8

)4πn2dn (integrating over angles)

=π

2

(ε

γ

)1

2√γ

dε√ε

=π√εdε

4γ3/2≡ g(ε)dε.

So, the density of states per unit energy g(ε) for a particle in a 3D box is given by

g(ε) ≡ dn

dε=

π

4γ3/2

√ε = V

m√

2m

2h3π2

√ε =

V (2m)3/2

4h3π2

√ε,


where in the last line I have put back in the explicit form for γ. The most important thing to noticeis that the density of states per unit energy for the 3D box goes like

√ε.

If you found this derivation confusing, here’s another one. One way to think about the energy levelsfor a particle in a box is that they are functions of k, which is proportional to the particle momentum(the Fourier transform of the coordinate):

εk =h2

2m

(πnL

)2

≡ h2k2

2m,

which is just the usual expression for the kinetic energy if you recognize that p ≡ hk. This is knownas the free-particle dispersion relation. Now, the energy sphere is in ‘k-space,’ or ‘Fourier space.’Then the density of states is the number of single-particle states in a volume element in k-space,times the density of points:

g(ε)dε = k2 sin(θ)dθdφdk

(L3

π3

).

Integrating over the angles gives

g(ε)dε =V

2π2k2dk ⇒ g3D(ε) =

V

2π2k2

(dk

dε

)for the 3D box. Now, ε = h2k2/2m so k =

√2mε/h. Putting it together we obtain

g(ε)dε =V

2π2k2dk =

V

2π2

2mε

h2

√2m

h2

1

2√εdε =

V

2π2

(2m)3/2√ε

2h3 dε,

as we obtained above.

It is straightforward to generalize the 3D box to the 2D and 1D cases. Going through the sameprocedure gives

g2D(ε) =Ak

2π

(dk

dε

); g1D(ε) =

L

π

(dk

dε

).

Explicitly using the free-particle dispersion relation as above shows that the density of states perunit volume is a constant (independent of energy) for 2D, and goes like 1/

√ε in 1D.

What is all this helping us to do? Let’s consider how many states are occupied at room temperature.The total number of states G(ε) is the integral of the density of states up to energy state ε:

G(ε) =

∫ ε

0

g(ε′)dε′ =V

2π2

(2m)3/2

2h3

√ε′dε′ =

V

6π2

(2m)3/2

h3 ε3/2 =π

6

(2mL2

h2π2ε

)3/2

.

At room temperature, ε = 32kBT where T = 300 K. So this means that the total number of states

accessible for nitrogen in a 1 m box at room temperatures is approximately G(ε) ≈ (π/6)(1020)3/2 ∼1030. But the density of air at sea level is about 1.25 kg/m3, or about 1025 molecules per m3. Soeven with these rough estimates it is clear that there are far more states accessible to the moleculesin the air at room temperature than there are particles. Alternatively, the probability that a givenenergy state has a particle in it is something like 10−5.


1.5 The Maxwell Speed Distribution

Let’s make the assumption that the energy levels are so closely spaced that we are left with amore-or-less continuous distribution of them, i.e.

N =∑i

ni ⇒∫dεn(ε) =

∫dεg(ε) exp(−βε)

=A(2m)3/2V

4h3π2

∫ ∞0

dε√ε exp(−βε), in three dimensions.

Because Z = N/A, this expression also quickly yields Z = V/λ3D in the usual way. Alternatively, we

can set A = N/Z and obtain

n(ε) = Ag(ε) exp(−βε) =N

V

(2πh2

mkBT

)3/2(2m)3/2V

4h3π2

√ε exp(−βε) =

2π√εN

(πkBT )3/2exp(−βε).

Now suppose that the energy levels were simply classical kinetic energy states: ε = mv2/2, so that√ε =

√m/2v and dε = mvdv. Then,

n(v)dv =

√2

πN

(m

kBT

)3/2

v2dv exp

(−βmv

2

2

).

This is the Maxwell-Boltzmann distribution of velocities. The important thing to notice is that forsmall velocities, the distribution increases quadratically, n(v)dv ∝ v2, while for large velocities itdecreases exponentially, n(v)dv ∝ exp(−βmv2/2). This means that the distribution is not even, i.e.is not symmetric around any given velocity. As shown below, this will have interesting consequencesfor the statistics.

Let’s obtain the mean velocity v first:

v =

∫vn(v)dv∫n(v)dv

=

∫∞0v3 exp

(−βmv

2

2

)dv∫∞

0v2 exp

(−βmv22

)dv,

where I’ve cancelled all the common constant terms. To turn these into integrals we can evaluate,let’s set x2 ≡ βmv2/2, so that v = x

√2kBT/m. Then

v =

√2kBT

m

∫∞0x3 exp

(−x2

)dx∫∞

0x2 exp (−x2) dx

.

Now it is very useful to know the following integrals:∫ ∞0

x2n exp(−a2x2)dx =(2n)!

√π

n!2(2a)2naand

∫ ∞0

x2n+1 exp(−a2x2)dx =n!

2a2n+2.

In the current case, we have a = 1 and

v =

√2kBT

m

1

2

8

2√π

=

√8kBT

πm≈ 1.596

√kBT

m.


Now let’s calculate the RMS velocity, ∆v ≡√v2:

v2 =

∫v2n(v)dv∫n(v)dv

=

∫∞0v4 exp

(−βmv

2

2

)dv∫∞

0v2 exp

(−βmv22

)dv

=2kBT

m

∫∞0x4 exp

(−x2

)dx∫∞

0x2 exp (−x2) dx

=2kBT

m

24√π

64

8

2√π

=3kBT

πm

⇒ ∆v =

√3kBT

πm≈ 1.73

√kBT

m.

The most probable speed v corresponds to the point at which the distribution is maximum:

∂

∂v

[√2

πN

(m

kBT

)3/2

v2dv exp

(−βmv

2

2

)]= 0

2v exp

(−βmv

2

2

)− v2

(m

2kBT

)2v exp

(−βmv

2

2

)= 0

⇒ v =

√2kBT

m≈ 1.414

√kBT

m.

The amazing thing about the Maxwell-Boltzmann distribution of velocities is that all the quantitiesv, ∆v, and v are different. This is in contrast to the Gaussian distribution seen early on in the term.The various values are in the ratio

∆v : v : v ≈ 1.224 : 1.128 : 1.

Why do you think that ∆v > v > v?

One factoid that you might find interesting is that the numbers you get for air are surprisingly large.Assuming that air is mostly nitrogen molecules, with m = 4.65×10−26 kg at a temperature of 273 K,one obtains v = 454 m/s, and ∆v = 493 m/s. But the speed of sound in air is 331 m/s at 273 K. Sothe molecules are moving considerably faster than the sound speed. Does this make sense?

There is an interesting application of this for the composition of the Earth’s atmosphere. First, let’sestimate the velocity a particle at the Earth’s surface would need to fully escape the gravitationalfield. The balance of kinetic and potential energies implies (1/2)mv2 = mgR, where R is the radiusof the Earth. (Actually I think that this implies that the particle is at the Earth’s center, where allof it’s mass would be concentrated?). In any case, we obtain vescape =

√2gR ≈ 11 000 m/s. The

mean velocity for hydrogen molecules (mass of 1.66×10−27 kg) from the formula v =√

8kBT/πm =1600 m/s. But the Maxwell-Boltzmann velocity distribution has a very long tail at high velocities,which means that there are approximately 2 × 109 hydrogen molecules that travel at more than 6times the average velocity. So there are lots of hydrogen molecules escaping forever all the time.Thankfully, our supply is continually replenished by protons bombarding us from the sun. Muchmore serious is helium, which is being lost irretrievably, with no replenishing. In fact, the U.S.Government has been stockpiling huge reserves of liquid helium for years in preparation of a world-wide shortage. But over the past few years the current administration has softened its policy on


helium conservation and these stockpiles are slowly dwindling. In any case, for oxygen and nitrogenmolecules, whose mean velocities are about a factor of 4 lower than that of hydrogen, very few ofthese can actually escape. Whew!

1.5.1 Interlude on Averages

When the various averages were calculated above, we made explicit use of the particular form of theenergy, ε = mv2/2. Of course, if the energy levels are different, or the dimension of the problem isnot 3D like it was above, then the way we take averages is going to be different. So how does onetake averages in general using the canonical ensemble?

Recall that the general form for the average of something we can measure, call it B, is

B =∑i

piBi,

where the sum is over all the accessible states of the system, and pi are the probabilities of occupyingthose states. In the canonical ensemble, those probabilities are

pi =gi exp(−βεi)

Z,

where Z is the usual partition function, and I have explicitly inserted the degeneracy factor. So theaverage of B is defined as

B =

∑i giBi exp(−βεi)∑i gi exp(−βεi)

≈∫dk g(k)B(k) exp[−βε(k)]∫dk g(k) exp[−βε(k)]

=

∫∞0dεg(ε)B(ε) exp(−βε)∫∞0dεg(ε) exp(−βε)

.

Thus, the dimensionality, the dependence of the energy ε on the wavefector k or velocity v, and theinherent degeneracy of a given energy level are all buried in the density of states per unit energy,g(ε). In order to calculate any average, this must be done first.

For example, suppose that our fundamental excitations were ripples on the surface of water, whereε(k) = αk3/2. This is an effective 2D system, so we use the expression for the 2D density of states,

g(ε) =Ak

2π

dk

dε=Ak

2π

d

dε

( εα

)2/3

=A

2π

( εα

)2/3(

2

3α2/3ε1/3

)=

A

3πα4/3ε1/3.

So this is what we would use to evaluate averages. Of course, the constant terms would disappear,but the energy-dependence of the density of states would not. For example, the mean energy perparticle for this problem would be

U =

∫∞0dε ε4/3 exp(−βε)∫∞

0dε ε1/3 exp(−βε)

=(kBT )7/3Γ

(73

)(kBT )4/3Γ

(43

) =4kBT

3.

1.5.2 Molecular Beams

One of these is operational in Nasser Moazzen-Ahmadi’s lab, so it’s good that you’re learning aboutit! Suppose that we have an oven containing lots of hot molecules. There’s a small hole in oneend, out of which shoot the molecules. On the same table in front of the hole is a series of walls


with small holes lined up horizontally with the exit hole of the oven. The idea here is that mostof the molecules moving off the horizontal axis will hit the various walls, and only those moleculesmoving in a narrow cone around the horizontal axis will make it to the screen. The question is:what distribution of velocities do the molecules have that hit the screen?

Evidently, once the molecules leave the last pinhole, they spread out and form a cone whose base isarea A on the screen. The number of molecules in the cone with velocities between v and v + dv,and in angles between θ and θ + dθ and between φ and φ+ dφ is

number of particles = Avt cos(θ)n(v) dv dΩ = Avt cos(θ)n(v) dvdθ sin(θ)dφ

4π.

The flux of molecules f(v)dv is the number of molecules per unit area per unit time,

f(v)dv =vn(v)dv

4π

∫ π/2

0

dθ cos(θ) sin(θ)

∫ 2π

0

dφ =vn(v)dv

4.

For the first integral, set x = sin(θ) so that dx = cos(θ). The integral is therefore x2/2 =

sin2(θ)/2|π/20 = 1/2. And the second integral is 2π. Using the Maxwell-Boltzmann distributionof velocities, we obtain the flux density as

f(v) =Nπ

8

(2m

πkBT

)3/2

v3 exp

(− mv2

2kBT

)=Nλ3

D

8π2

(mh

)3

v3 exp

(− mv2

2kBT

).

What is the point? I’m not really sure, actually. Sometimes it’s good to know the distribution ofvelocities hitting a screen to interpret results of an experiment. One thing we can do right now is toderive the equation of state for an ideal gas using it! A crude way to do this is to assume that everytime the molecule strikes the surface with velocity v, and bounces off elastically, the screen picksup a momentum 2mv cos(θ), accounting for the angle off the horizontal axis. The mean pressure istherefore the integral over the pressure flux:

P =

∫dv

∫ π/2

0

dθ

∫ 2π

0

dφ2p cos(θ)v cos(θ)n(v)dvsin(θ)dθdφ

4πV

=1

3V

∫ ∞0

mvn(v)v dv =m

3V

∫ ∞0

v2n(v)dv =Nm

3Vv2 =

Nm

3V

3kBT

m=NkBT

V.

A slightly less hokey derivation (maybe) is to say that the pressure per unit volume is the meanforce along the horizontal per unit area:

P =F

A=

∫dvf(v)mv2

x =N

Vmv2

x.

But v2 = v2x + v2

y + v2z = 3v2

x. So

P =Nm

3Vv2 =

NkBT

V.

Chapter 2

Virial Theorem and the GrandCanonical Ensemble

2.1 Virial Theorem

Before launching into the theory of quantum gases, it’s useful to learn a powerful thing in statisticalmechanics, called the virial theorem of Clausius. Consider the quantity G ≡

∑i pi · ri, where the

sum is over the particles. The time-derivative of this quantity is

dG

dt=∑i

(pi · ri + ri · pi) =∑i

(pi · ri +mv2

i

)= 2K +

∑i

pi · ri = 2K +∑i

Fi · ri,

where I am using K to represent kinetic energy rather than T avoid confusion with temperature.Also, in the last equation I have used Newton’s equation F = p. Now, the time average of somefluctuating quantity (call it A(t)) is defined as 〈A(t)〉t:

〈A(t)〉t ≡1

τ

∫ τ

0

A(t)dt.

We can use this to define the following time average:⟨dG

dt

⟩t

=1

τ

∫ τ

0

dG

dtdt =

G(τ)−G(0)

τ.

Now here comes the crucial part. If G(t) is a periodic function, with period P , then clearly forτ = P the time average must be zero since G(τ) = G(0). Likewise, if G has finite values for alltimes, then as the elapsed time τ gets to be very large the average will also go to zero. This meansthat ⟨

dG

dt

⟩t

= 0

as long as G is finite. Finally we obtain the virial theorem

〈K〉t = −1

2

⟨∑i

Fi · ri

⟩i

. (2.1)

14


Apparently, the word ‘virial’ comes from the Latin word for ‘force’ though I remember from my highschool Latin that force was ‘fortis’ meaning strength. Oh well. You see anyhow where the forcecomes into the picture.

2.1.1 Example: ideal gas

As an example of the power of the virial theorem, let’s derive the equation of state for an idealgas (again! this probably won’t be the last time either). We’ve already seen from the equipartitiontheorem that for a particle in a 3D box, U = K = (3/2)NkBT . Neglecting particle interactions,the only forces at equilibrium are those of the particles on the box walls (and vice versa) when thecollide elastically, dFi = −PndA for all particles, where n is the normal to the wall surface (this isnothing but a restatement of P = dF/dV ). So∑

i

Fi · ri = −P∫n · rdA = −P

∫∇ · rdV = −3PV

where I used Stokes’ theorem in the last bit. Plugging into the virial theorem immediately gives(3/2)NkBT = (3/2)PV which is what we wanted.

2.1.2 Example: Average temperature of the sun

Now let’s use the virial theorem to estimate the temperature of the sun. The mass of the sun isM ≈ 3 · 1030 kg and its radius is R ≈ 3 · 107 m. Assuming that the sun is made up entirely ofhydrogen with mass 1.67 · 10−27 kg, then this gives N = 1.8 · 1057 atoms and therefore a densityρ = 1.6 × 1034 m−3. The force on a particle of mass m due to the gravity of the sun is F =−(GMm/R

2)r, where G ≈ 6.674×10−11 m3/kg/s2 is the gravitational constant, and M and R

are the mass and radius of the sun, respectively. The density of the sun (assuming that it’s constanteverywhere, which isn’t likely to be the case!) is ρ = M/V = M/(4πR

3/3). Now, we’d like to

sum up all the gravitional forces from the sun’s center to the surface. The radial-dependent mass istherefore

M(r) =M4πR3

3

4πr3

3=

(r

R

)3

M.

Also, for the test mass we have dm = ρdV = ρr2 sin(θ)dr dθ dφ. Putting it all together, we have

K = −1

2

∑i

Fi · ri =1

2

∫r2 sin(θ) dr dθ dφ

G

r2

r3

R3Mρr · r,

where K is the mean kinetic energy. But r · r = (r/r) · r = r, so

K =GMρ

2R3

4π

∫ R

0

r4dr =2πGMρR

5

5R3

=3

10

GM2

R,

where in the last part I used the expression for the density ρ. Because K = (3/2)NkBT by equipar-tition, we obtain T ≈ (1/5)(GM2

/NkBR) ≈ 5 × 106 K. In fact, except for the factor 5 I couldhave guessed this from dimensional analysis, because there is only one way to combine G, M, andR in a way that gives units of energy. In fact, the temperature of the sun varies from about 10million degrees at the center to about 6 000 degrees at the surface. So the virial theorem has donea pretty good job of estimating the average temperature.


2.2 Chemical Potential

Suppose that the energy of some small system (I’ll call it a subsystem) also depends on the (fluctu-ating) number of particles N in it. Then

dU = TdS − PdV +∂U

∂NdN ≡ TdS − PdV + µdN.

This can also be inverted to give

dS =dU + PdV − µdN

T.

So the chemical potential for the system is defined as

µ ≡(∂U

∂N

)S,V

= −T(∂S

∂N

)U,V

, N =∑

i∈subsystem

ni.

But remember to be careful when using the second definition of the chemical potential, becauseT = (∂U/∂S)V,N .

This reminds me about general forces: here S and T are conjugate variables linked through thetotal energy U . In a similar way, P and V are conjugate variables because P = − (∂U/∂V )S,N .Likewise in magnetic systems for the variables M and B. In a similar way, the first definition of thechemical potential above shows that µ and N are also conjugate variables: µ is the generalized forceassociated with the variable N , i.e. it is a ‘force’ that fixes the value of N for a given system.

What does the chemical potential mean? Consider the change in the entropy of the entire system(subsystem plus reservoir) with the number of particles in the subsystem Ns in terms of the changeof the entropies of the reservoir SR and subsystem Ss:

dSsystem =∂

∂Ns(SR + Ss) dNs =

∂SR∂Ns

dNs +∂Ss∂Ns

dNs = dNs

(∂SR∂NR

∂NR∂Ns

+∂Ss∂Ns

).

But NR = N −Ns so ∂NR/∂Ns = −1 giving

dSsystem = dNs

(∂Ss∂Ns

− ∂SR∂NR

)= −dNs

T(µs − µR) .

If you didn’t like this sloppy math, then how about this:

dSsystem =∂Ss∂Ns

dNs +∂SR∂NR

dNR.

Since dNR = −dNs this gives

dSsystem = dNs

(∂Ss∂Ns

− ∂SR∂NR

),

which is what we obtained above.

Equilibrium corresponds to maximizing entropy, or dS = 0. This means that the condition forequilibrium between the subsystem and the reservoir is that the chemical potentials for each should


be equal, µs = µR. But even more important, as equilibrium is being approached, the entropy ischanging with time like

dSsystem

dt= −dNs

dt

(µs − µR

T

)≥ 0

because the entropy must increase toward equilibrium (unless they are already at equilibrium). Ifinitially µR > µs, then clearly dNs/dt ≥ 0 to satisfy the above inequality. This means that in orderto reach equilibrium when the chemical potential for the reservoir is initially larger than that ofthe subsystem, particles must flow from the reservoir into the subsystem. So the chemical potentialprovides some measure of the number imbalance between two systems that are not at equilibrium.

2.2.1 Free energies revisited

If the number of particles in the system is not fixed, then we also need to include the chemicalpotential µ. Then the change in internal energy, and the changes in the enthalpy, Helmholtz, andGibbs free energies are given respectively by

dU = TdS − PdV + µdN ;

dH = TdS + V dP + µdN ;

dF = −SdT − PdV + µdN ;

dG = −SdT + V dP + µdN.

The first of these we saw back in Section 2.2. The rest of them are pretty trivially changed fromtheir cases seen above without allowing for changes to the total particle number! So we obtain thefollowing rules:

µ =

(∂U

∂N

)S,V

=

(∂H

∂N

)S,P

=

(∂F

∂N

)T,V

=

(∂G

∂N

)T,P

. (2.2)

If there is more than one kind of particle, then one would need to insert the additional conditions,i.e. the chemical potential for species 1 would be

µ1 =

(∂F

∂N1

)T,V,N2,N3,...

.

Let’s explore this last expression a bit more:

µ =

(∂F

∂N

)T,V

= −kBT(∂N ln(Z)

∂N

)T,V

= −kBT ln(Z)−NkBT(∂ ln(Z)

∂N

)T,V

.

Because Z = N/A, this last relation can be solved directly: ∂ ln(N/A)/∂N = ∂ ln(N)/∂N = 1/N sothat µ = −kBT [ln(Z) + 1]. At high temperatures, the arguments of the exponentials in Z will allbe small, which implies that Z > 0. This means that the chemical potential for classical particles isalways negative. At low temperatures it might approach zero or get positive. This possibility willbe explored a bit later in the term when we discuss quantum statistics.


2.2.2 Example: Pauli Paramagnet

To clarify the various meanings of the chemical potential, let’s return first to the Pauli paramagnet.Recall that in the microcanonical ensemble, the entropy for the spin- 1

2 case was

S = −NkB[(n1

N

)ln(n1

N

)+(

1− n1

N

)ln(

1− n1

N

)].

Assuming that we have some subsystem with number N in contact with a reservoir at temperatureT , one might naıvely assume that the chemical potential is defined as

µ = −T ∂S∂N

= −kBT ln

(N

N − n1

)= kBT ln

(1− n1

N

)= kBT ln

(n2

N

)after a bit of algebra. But something is fishy about this result. In the limit of low temperature,either n2 → 0 or n2 → N , depending on whether n2 represents the number of spin-up or spin-downatoms. But we don’t know from the entropy expression above which spin direction corresponds tothe lower energy. This means that the answer should have been symmetric in n1 and n2, which itis definitely not. In fact, there is a fundamental error. As N varies, so also do the values of n1 andn2. In other words, n1 = n1(N) and n2 = n2(N). When taking this derivative we have to rememberthat the items kept constant are the volume (in this case the external magnetic field) and the totalenergy U .

So to calculate µ properly we need to convert n1 and n2 into explicit functions of U and N . Recallythat n1 + n2 = N and −n1ε + n2ε = U . The second of these gives n2 − n1 = U/ε and combiningwith the first gives 2n2 = N + U/eε or n2/N = (1 + U/Nε)/2. We immediately then obtainn1/N = (1 + U/Nε)/2. The expression for the entropy is then:

S = −NkB

1

2

(1− U

Nε

)ln

[1

2

(1− U

Nε

)]+

1

2

(1 +

U

Nε

)ln

[1

2

(1 +

U

Nε

)].

Now we can calculate the chemical potential using the formula above. After some straightforwardalgebra one obtains

µ = kBT ln(n1n2

N2

).

This result looks better, because it is symmetric under the exchange of labels n1 ↔ n2. At hightemperatures n1 = n2 = N/2, and the chemical potential becomes large and negative, just as inthe example for the 3D box. At low temperatures, n1 → 0 or n2 → 0, which makes the ln divergelogarithmically, but it’s o.k. because the temperature is going to zero linearly (i.e. fundamentallyfaster than a log). So µ→ 0 as T → 0.

What does µ = 0 mean? Suppose that the total number of particles is not zero. Then a zerochemical potential means that the change in the number of particles in the reservoir is not relatedto the change in the number of particles in the subsystem; alternatively, the entropy is invariantunder changes in the number of particles. This implies that a zero chemical potential meansthat the system doesn’t conserve the number of particles. For the Pauli paramagnet, I cankeep increasing the number of atoms with spin down, and as long as I don’t create a single spin up,then the system’s entropy doesn’t change: it remains exactly zero.


2.3 Grand Partition Function

Recall that in the derivation of the Boltzmann distribution in the canonical ensemble, we maximizedthe entropy (or the number of microstates) subject to the two constraints N =

∑i ni and U =∑

i niεi. So we had

d ln(Ω) + α

(dN −

∑i

dni

)+ β

(dU −

∑i

εidni

)= 0,

where α and β were the Lagrange multipliers fixing the two constraints (alternatively, they areunknown multiplicative constants in front of terms that are zero). So we immediately have thefollowing relations:

α =∂ ln(Ω)

∂N; β =

∂ ln(Ω)

∂U.

But above we have ∂S/∂N = −µ/T = kB(∂ ln(Ω)/∂N) because S = kB ln(Ω). So we immediatelyobtain α = −µ/kBT = −βµ. So the chemical potential is in fact the Lagrange multiplier that fixesthe number of particles in the total system. Putting these together, we have

µ = −kBT∂ ln(Ω)

∂N;

∂ ln(Ω)

∂U=

1

kBT.

⇒ ln(Ω) = βU − βµN + const.

⇒ Ω = A exp [β (U − µN)] ,

where A = exp(const.).

So far, this enumeration has been for the reservoir, which contains the fixed temperature. To findΩ for the subsystem, we note that UR = U − Es and NR = N −Ns. So

ΩR = A exp [β (UR − µNR)] = A exp [β (U − Es)− βµ (N −Ns)]= A exp [β (U − µN)] exp [−β (Es − µNs)] = ΩsystemΩsubsystem.

Finally we obtainΩs = A exp [−β (Es − µNs)] .

The physical system is actually comprised of a very large number of these subsystems, all of whichhave the same temperature and chemical potential at equilibrium, but all of which have differentenergies Ei and number of particles Ni. The total number of ways of distributing the particlesΩ is therefore the sum of all of the subsystems’ contributions, Ω =

∑i Ωs. So the probability of

occupying a given subsystem is the fraction of the distribution of a given subsystem over all of them,

pi =ΩiΩ

=exp [−β (Ei − µNi)]∑i exp [−β (Ei − µNi)]

≡ exp [−β (Ei − µNi)]Ξ

,

where Ξ is the grand partition function,

Ξ ≡∑i

exp [−β (Ei − µNi)] .


Here’s an alternate derivation if you didn’t like that one. The ratio of probabilities for two macrostatesis the same as the ratio of their number of microstates,

P (s2)

P (s1)=

ΩR(s2)

ΩR(s1)=eSR(s2)/kB

eSR(s1)/kB= exp

1

kB[SR(s2)− SR(s1)]

.

We know that

dSR =1

T(dUR + PdVR − µdNR) = − 1

T(dUs + PdVs − µdNs)

= − 1

TE(s2)− E(s1)− µ [N(s2)−N(s1)] .

So we obtain

P (s2)

P (s1)=

exp− 1kBT

[Es(s2)− µNs(s2)]

exp− 1kBT

[Es(s1)− µNs(s1)] .

The probability of occupying any state s is then

P (s) =exp

− 1kBT

[Es(s)− µNs(s)]

Ξ,

where Ξ is given above.

2.3.1 Examples

It is important to note that the way one obtains the grand partition function is different from theway it was done in the canonical ensemble. Suppose that the subsystem contains three accessibleenergy levels 0, ε, and 2ε. Now suppose that there is only one atom in the total system. How manyways can I distribute atoms in my energy states? I might have no atoms in any energy state (Es = 0,Ns = 0). I might have one atom in state 0 (Es = 0, Ns = 1). I might have one atom in state ε(Es = ε, Ns = 1), or I might have one atom in state 2ε (Es = 2ε, Ns = 1). So my grand partitionfunction is:

Ξ = exp[−β(0− µ0)] + exp[−β(0− µ1)] + exp[−β(ε− µ1)] + exp[−β(2ε− µ1)]

= 1 + exp(βµ) [1 + exp(−βε) + exp(−2βε)] .

Suppose instead that there are an unknown number of particles in the total system, but that eachof these three energy levels in my subsystem can only accommodate up to two particles. Then thegrand partition function is

Ξ = 1 + exp(βµ) [1 + exp(−βε) + exp[−2βε)]

+ exp(2βµ) [1 + exp(−βε) + 2 exp(−2βε) + exp(−3βε) + exp(−4βε)] ,

where in the second line I have recognized that with two particles we can have Es = 0 (both in state0), Es = ε (one in state 0, the other in state ε), Es = 2ε (either both in state ε, or one in state 0while the other is in state 2ε, thus the factor of two out front of this term), Es = 3ε (one in state ε,the other in state 2ε), and Es = 4ε (both in state 2ε).


Now suppose that the number of particles in our system is totally unknown, and also there is norestriction on the number of particles that can exist in a particular energy level. Then the grandpartition function is

Ξ = 1 + exp(βµ) [1 + exp(−βε) + exp[−2βε)] + exp(2βµ) [1 + exp(−βε) + exp[−2βε)]2

= 1 + eβµZ + e2βµZ2 + . . .

=

∞∑n=0

(eβµZ

)n=

1

1− eβµZ,

where I have used Euler’s solution in the last line. In other words, the grand canonical ensemble forthe fully unrestricted case is simply a linear combination of the canonical partition functions for noparticles, one particle, two particles, etc., suitably weighted by their fugacities. In fact, this classicalgrand partition function leads to slightly paradoxical results, which will be discussed at length nextterm. Anyhow, that’s how one constructs the grand partition function in practice!

2.4 Grand Potential

As we did for the canonical ensemble, we can obtain thermodynamic quantities using the grandpartition function instead of the regular partition function. The entropy is defined as

S = −kB∑i

pi ln(pi) = −kB∑i

pi [−β(Ei − µNi)− ln(Ξ)] =U

T− µN

T+ kB ln(Ξ),

where U =∑i piEi and N =

∑i piNi are the mean energy and mean particle number for the

subsystem, respectively. This expression can be inverted to yield

U − µN − TS = −kBT ln(Ξ) ≡ ΦG,

where ΦG is the grand potential. Sometimes this is also written as ΩG in statistical mechanicsbooks. Recall that in the canonical ensemble we had U − TS = F = −kBT ln(ZN ), where F is theHelmholtz free energy. So the grand potential is simply related to the free energy by ΦG = F −µN .Anyhow, the following thermodynamic relations immediately follow:

S = −∂ΦG∂T

; N = −∂ΦG∂µ

= kBT∂ ln(Ξ)

∂µ.

Two other relations that follow in analogy with the results for the canonical ensemble are:

P = −∂ΦG∂V

; U = −∂ ln(Ξ)

∂β= kBT

2 ∂ ln(Ξ)

∂T.

Chapter 3

Quantum Counting

3.1 Gibbs’ Paradox

It turns out that much of what we have done so far is fundamentally wrong. One of the first people torealize this was Gibbs (of Free Energy fame!), so it is called Gibbs’ Paradox. Basically, he showedthat there was a problem with the definition of entropy that we have been using so far. The only wayto resolve the paradox is using quantum mechanics, which we’ll see later in this chapter. Of course,we have been using quantum mechanics already in order to define the accessible energy levels thatparticles can occupy. But so far, we haven’t been concerned with the particles themselves. In thissection we’ll see the paradox, and over the next sections we’ll resolve it using quantum mechanics.

Recall that the entropy in the canonical ensemble is defined as

S = NkB ln(Z) +U

T= NkB ln(Z) +NkBT

∂ ln(Z)

∂T.

As we saw last term, the partition function for a monatomic ideal gas in a 3D box is Z = V/λ3D,

where V is the volume and λD =√

2πh2/mkBT is the de Broglie wavelength. Plugging this in, we

obtain

ln(Z) = ln(V )− 3

2ln

(2πh2

mkB

)+

3

2ln(T ).

Evidently, ∂ ln(Z)/∂T = (1/Z)∂Z/∂T = 3/2T so

S = NkB

[ln(V )− 3

2ln

(2πh2

mkB

)+

3

2ln(T ) +

3

2

]≡ NkB

[ln(V ) +

3

2ln(T ) + σ

],

where

σ =3

2

[1− ln

(2πh2

mkB

)]is some constant we don’t really care about. Now to the paradox.

Consider a box of volume V containing N atoms. Now, suppose that a barrier is inserted thatdivides the box into two regions of equal volume V/2, each containing N/2 atoms. Now the total

22


entropy is the sum of the entropies for each region,

S =N

2kB

[ln

(V

2

)+

3

2ln(T ) + σ

]+N

2kB

[ln

(V

2

)+

3

2ln(T ) + σ

]= NkB

[ln

(V

2

)+

3

2ln(T ) + σ

]6= NkB

[ln(V ) +

3

2ln(T ) + σ

].

In other words, simply putting in a barrier seems to have reduced the entropy of the particles inthe box by a factor of NkB ln(2). Recall that this is the same as the high-temperature entropy forthe two-level Pauli paramagnet (a.k.a. the coin). So it seems to suggest that the two sides of thepartition once the barrier is up take the place of ‘heads’ or ‘tails’ in that there are two kinds of statesatoms can occupy: the left or right partitions. But here’s the paradox: putting in the barrier hasn’tdone any work on the system, or added any heat, so the entropy should be invariant! And simplyremoving the barrier brings the entropy back where it was before. Simply reducing the capabilityof the particles in the left partition from accessing locations in the right partition (and vice versa)shouldn’t change the entropy, because in reality you can’t tell the difference between atomson the left or on the right.

Clearly, we are treating particles on the left and right as distinguishable objects, which has given riseto this paradox. How can we fix things? The cheap fix is to realize that because all of the particles arefundamentally indistinguishable, the N ! permutations of the particles among themselves shouldn’tlead to physically distinct realizations of the system. So, our calculation of Z must be too large bya factor of N !:

ZN ≡ ZNcorrect =ZNold

N !,

where I explicitly make use of the fact that the N -particle partition function is N -fold product ofthe single-particle partition function Zold. With this ‘ad hoc’ solution,

ln(ZN ) = N ln(Zold)− ln(N !) = N ln(Zold)−N ln(N) +N.

Now the entropy of the total box is

S = NkB

[ln(V ) +

3

2ln(T ) + σ − ln(N) + 1

]= NkB

[ln

(V

N

)+

3

2ln(T ) + σ + 1

].

Now, if we partition the box into two regions of volume V/2, each withN/2 particles, then ln(V/N)→ln(V/N) and the combined entropy of the two regions is identical to the original entropy. So Gibbs’paradox is resolved.

The equations for the mean energy, entropy, and free energy in terms of the proper partition functionZN are now

U = kBT2 ∂ ln(ZN )

∂T; S + kB ln(ZN ) + kBT

∂ ln(ZN )

∂T; F = −kBT ln(ZN ).

That is, the expression are identical to the ones we’ve seen before, except the N factors in fronthave disappeared, and the Z is replaced by a ZN .


3.2 Chemical Potential Again

Now let’s return to the particle in the 3D box. Including the Gibbs’ correction, the entropy is

S = NkB

[ln

(V

N

)+

3

2ln

(mkBT

2πh2

)+

5

2

].

Now, U = 32NkBT , so kBT = 2U/3N . Inserting this into the above expression for the entropy gives

S = NkB

[ln

(V

N

)+

3

2ln

(mU

3πh2N

)+

5

2

].

Now we can take the derivative with respect to N to obtain the chemical potential:

µ = −kBT[ln

(V

N

)+

3

2ln

(mU

3πh2N

)+

5

2

]+ kBT

(1 +

3

2

).

µ = −kBT[ln

(V

N

)+

3

2ln

(mkBT

2πh2

)]= kBT ln

(nλ3

D

),

where n = N/V is the density. Note that the subsystem stops conserving the number of particleswhen nλ3

D = 1, or when the mean interparticle separation ` becomes comparable to the de Brogliewavelength, ` ≡ n−1/3 ∼ λD. But we already know that something quantum happens on the lengthscale of the de Broglie wavelength. We’ll come back to this a bit later. For very classical systems,` λD which means that the chemical potential is usually large and negative for the particle in the3D box.

How can you calculate the chemical potential from the free energy? I’m sure that you were burningto find this out. So let’s do it. The free energy is F = U − TS so dF = dU − TdS − SdT =TdS − PdV + µdN − TdS − SdT , so

µ =∂F

∂N= −kBT

∂ ln (ZN )

∂N.

Let’s check that this gives the right answer for the particle in the 3D box. In that case, ZN =(V/λ3

D)N/N ! so ln(ZN ) = N ln(V/λ3D)−N ln(N)+N . So ∂ ln(ZN )/∂N = ln(V/λ3

D)− ln(N)−1+1,or µ = kBT ln(nλ3

D). Suppose that the energies of the particles were larger by a constant factor of∆, so that εk = (h2k2/2m) + ∆. Then

Z =∑k

exp

[−β(h2k2

2m+ ∆

)]= exp(−β∆)

∑k

exp

(−βh

2k2

2m

)=

V

λ3D

exp(−β∆).

So, the chemical potential is nowµ = kBT ln(nλ3

D) + ∆.

Clearly, the shift in the energies has led to exactly the same shift in the chemical potential. Thisadditional piece can be thought of as an ‘external chemical potential’ that increases the total value.If the shift had depended on position or momentum, though, then we would have needed to use theequipartition theorem to evaluate the contribution of the additional piece to the chemical potential.Also, if we had included vibration and rotation for molecules, say, then the chemical potential wouldhave increased as well. These would be ‘internal chemical potential’ contributions. Would thesetend to increase or decrease the chemical potential?


3.3 Arranging Indistinguishable Particles

The way we have derived statistical mechanics so far was to count the ways of arranging variousoutcomes into bins, and call the most-occupied bin the equilibrium state of the system. In themicrocanonical ensemble, these outcomes tended to be things like ‘headness’ or ‘tailness’, and thebins were the number of heads and tails after N trials. In the canonical ensemble, the outcomesare usually designated as the accessible (quantum) energy levels of a given system, and the bins arethe various ways of N particles occupying these levels. But both of these approaches assumed thatwe could tell the difference between various outcomes (call them sides of a coin, or the energy level,etc.). With completely indistinguishable particles, things get a bit more problematic.

As an example, consider a three-sided coin. Or even better a three-level system like a spin-1 particlethat can be in the states ↑, , and ↓, with energies 1, 0, and −1, respectively. Suppose we havethree of these particles. Then we obtain the following table, where the ‘ways’ column denotes thenumber of ways we could have obtained the number of ↑, , and ↓ in the state (i.e. the number ofmicrostates in the macrostate).

state energy ways| ↑↑↑〉 3 1| ↑↑ 〉 2 3| ↑↑↓〉 1 3| ↑ 〉 1 3| ↑ ↓〉 0 6| 〉 0 1| ↑↓↓〉 −1 3| ↓〉 −1 3| ↓↓〉 −2 3| ↓↓↓〉 −3 1

We started with 10 distinguishable macrostates. But if the macrostates are the various values ofthe total energy of the macrostates, then we actually have only 7 macrostates, labeled by energies3, 2, 1, 0,−1,−2,−3, with 1, 3, 6, 7, 6, 3, 1 microstates in each, respectively. In fact, we’ve brieflyrun into this problem of degeneracy before. Now, the state with zero energy is seven-fold degenerate(g = 7) when there are three particles. How do we count states properly taking the degeneracy ginto account?

3.3.1 Bosons

Suppose that we want to enumerate the number of ways to arrange two particles that you can’tdistinguish into four degenerate levels. How many ways are there to do this? Here’s another table,where each column represents one of the four degenerate levels, and the number tells you how manyparticles are in that level:In this example, there are 10 combinations. If you remember your binomial coefficients, this number

corresponds to

(52

). Why would this be? Let’s do another example to find some kind of trend.

How about three particles also with g = 4:


0 0 0 20 0 1 10 1 0 11 0 0 10 0 2 00 1 1 01 0 1 00 2 0 01 1 0 02 0 0 0

This gives me 20 combinations, or

(63

). O.k., this is enough for a trend. The lower number of the

‘choose’ bracket is definitely turning out to be the number of particles. What is the upper number?Well, the number N + g − 1 seems to work! So it looks like the number of ways of arranging niparticles into gi degenerate levels of energy state εi is

Ω(b)i =

(ni + gi − 1

ni

)=

(ni + gi − 1)!

ni!(gi − 1)!.

I haven’t imposed any kind of restrictions about how many particles can exist in one of the degeneratelevels. These kinds of particles are called ‘bosons,’ because they obey ‘Bose-Einstein statistics,’ afterthe two guys who figured it out; thus the (b) in the superscript of Ωi above.

If there are M energy levels in total, each with their own degeneracy factors gi, then the totalnumber of ways of distributing all the bosons is

Ω(b) =∏i

Ω(b)i =

∏i

(ni + gi − 1)!

ni!(gi − 1)!. (3.1)

Note that if gi = 1 for a given level εi, so that there is only one state per energy level, then Ω(b)i = 1,

i.e. that there is only one way to arrange the ni particles. How do we reconcile this with the countingmethod we used before? If all we get is one microstate in every macrostate, how do we maximizethe entropy and obtain our equilibrium state? Let’s postpone this quandary for a moment, and firstintroduce fermions.

3.3.2 Fermions

Suppose that there is another kind of particle that refuses to share its quantum state with anotherparticle; that is, each state within a degenerate energy level can only hold one particle. Particleswith this restrictions are called ‘fermions,’ because the satisfy ‘Fermi-Dirac statistics.’ How manyways of arranging the particles are there now? Let’s consider the same examples as was done abovefor bosons. First, N = 2, and g = 4:

This gives Ω = 6 =

(42

). And the case with N = 3 and g = 2 is:


0 0 0 30 0 1 20 1 0 21 0 0 20 0 2 10 1 1 11 0 1 10 2 0 11 1 0 12 0 0 10 0 3 00 1 2 01 0 2 00 2 1 01 1 1 02 0 1 00 3 0 01 2 0 02 1 0 03 0 0 0

0 0 1 10 1 0 11 0 0 10 1 1 01 0 1 01 1 0 0

Now we only have four combinations, or

(43

). Again, this is enough to see the trend. The top

number is clearly the degeneracy, because it hasn’t changed. The bottom number looks like thenumber of particles. So for fermions we obtain

Ω(f) =∏i

Ω(f)i =

∏i

gi!

ni!(gi − ni)!. (3.2)

For fermions, Ω(f)i = 1 when gi = ni, i.e. when there is exactly one particle in each of the degenerate

states within the energy level εi.


0 1 1 11 0 1 11 1 0 11 1 1 0

3.3.3 Anyons!

As discussed in the previous two subsections, the number of ways of arranging ni bosons and fermionsinto an energy level with degeneracy gi are respectively given by

Ω(b)i =

(ni + gi − 1)!

ni!(gi − 1)!; Ω

(f)i =

gi!

ni!(gi − ni)!.

These two forms for Ωi look fairly similar to one another. In fact, we can write only one expressionfor both of them, that depends only on a parameter I’ll call α:

Ω(a)i =

[gi + (ni − 1)(1− α)]!

ni![gi − 1− α(ni − 1)]!,

where clearly for α = 0 we recover the expression for bosons, and for α = 1 we obtain that forfermions. But what about values of α in between? As discussed above, an unlimited number ofbosons can occupy any state, independently of whether that state is already occupied by anotherparticle or not; whereas only a single fermion can occupy a given state. So, intuitively, there mightexist particles where only a finite number of them could occupy a given state, i.e. 1 < nmax

i < ∞.In fact, we’ve already discussed these: particles that have repulsive interactions will probably ‘push’other particles away from the state they already occupy.

In the late 1980’s, Duncan Haldane came up with the idea of ‘fractional exclusion statistics’ (FES),where the occupation of a given state depends on the number of particles already in that state. Intwo dimensions, these particles correspond to ‘anyons,’ because they obey ‘any’ statistics between(and also including) bosonic and fermionic statistics. These exotic particles are known to be theones responsible for the fractional quantum Hall effect, for example. The ‘anyon’ is the reason for

the ‘a’ superscript in the Ω(a)i expression above.

Recall that for bosons, Ω(b)i = 1 only if gi = 1, and for fermions, Ω

(f)i = 1 only if gi = ni; in other

words, the distribution saturates for these values of the degeneracy gi. For what values of gi do

particles obeying F.E.S. saturate? For Ω(a)i = 1, this means that we require gi+(n1−1)(1−α) = ni,

and gi− 1−α(ni− 1) = 0 because 0! = 1 (notice that both of these equations state the same thing).So we obtain the critical value of α for saturation:

αc =gi − 1

ni − 1,

which means that α must be a rational fraction. Here’s a table of results for F.E.S. particles:In other words, as the denominator of the fraction α increases, more and more particles can fit intothe state with degeneracy gi, so that anyons with F.E.S. approach Fermi and Bose statistics in thetwo limits.


α gi ni gi ni gi ni gi ni1 1 1 2 2 3 3 4 412 1 0 2 3 3 5 4 713 1 0 2 4 3 7 4 1014 1 0 2 5 3 9 4 13

0 1 * 2 * 3 * 4 *

3.4 Emergence of Classical Statistics

Another way to think about the much larger number of energies than particles at everyday temper-atures is this. Suppose you were to define an accessible energy state as an energy region that spanssomething like 105 quantum energy states. Then there would be on average 1 atoms per region, butthere would be a huge degeneracy for this region, because the atom could be sitting in any of the105 nearby energy levels. (I’m assuming that they are so closely-spaced as to be degenerate for ourpurposes). Then, we’re in the situation where gi 1. Now let’s go back to our expressions for thenumber of ways of distributing ni particles in gi states:

Ω(a)i =

[gi + (ni − 1)(1− α)]!

ni![gi − 1− α(ni − 1)]!,

If gi 1, then

Ω(a)i ≈ [gi + ni(1− α)]!

ni!(gi − αni)!≈ [gi + ni − niα)]!

ni!(gi − αni)!

=(gi + ni − αni)(gi + ni − αni − 1) · · · (gi − αni + 1)(gi − αni)(gi − αni − 1) · · · (1)

ni!(gi − αni)(gi − αni − 1) · · · (1)

≈ gnii

ni!,

because not only is gi 1 but also gi ni. Notice that this final result doesn’t depend on alphaand therefore on the kind of statistics. So the quantum way of counting at high temperatures reducesto the simple universal result

Ω =∏i

gnii

ni!=

∏i gnii∏

i ni!.

In fact, this isn’t very different from the microcanonical result for a situation where there are variousoutcomes of N trials, Ωmicro = N !/

∏i ni! except that the factor of N ! is replaced by gni

i . But as wewill see now, this new factor resolves Gibbs’ paradox:

ln(Ω) = ln

(∏i

gnii

)− ln

(∏i

ni!

)=∑i

[ni ln(gi)− ni ln(ni) + ni] =∑i

[ni ln

(gini

)+ ni

].

This means that

d ln(Ω) = 0 =∑i

∂

∂ni

[ni ln

(gini

)]dni =

∑i

[ln

(gini

)− nini

)

]dni.


⇒∑i

ln

(gini

)dni = 0.

This equation isn’t much different from the maximization of entropy condition used previously toderive the Boltzmann distribution! Again, we have the supplementary conditions N =

∑i ni or∑

i dni = 0 and that of constant mean energy dU =∑i εidni = 0. Again, I use the method of

Lagrange multipliers to obtain the single equation∑i

[ln

(gini

)− α− βεi

]dni = 0

from which I getgini

= exp [α+ βεi]

orni = gi exp[−βεi − α] = giA exp(−βεi).

Now the degeneracy factor has appeared more naturally than the ad hoc addition we did last time.In the continuous notation of the previous subsection, the distribution is

n(ε) = Ag(ε) exp(−βε),

where g(ε) is just the density of states per unit energy, rather than the degeneracy. The total particlenumber for a 3D box is then

N =A(2m)3/2V

4h3π2

∫ √ε exp(−βε)dε

which determines the unknown constant A in terms of N and the box volume V .

Now the entropy is written

S = kB ln(Ω) = kB

[∑i

ni ln

(gini

)+N

]= kB

∑i

ni [ln(gi)− ln(A)− ln(gi) + βεi] + kBN

= kB [βU −N ln(A) +N ] .

But N = AZ which means that ln(A) = ln(N)− ln(Z) and the entropy becomes

S = kB [βU −N ln(N) +N ln(Z) +N ] = kB

[βNkBT

2 ∂

∂Tln(Z) +N ln(Z)− ln(N !)

]= NkB ln(Z) +

U

T− kB ln(N !) = kB ln

(ZN

N !

)+U

T,

which is different from the expression assuming distinguishable particles by a factor of ln(N !). Thisis exactly the term that we needed to resolve Gibbs’ paradox! Note that the expression for U is thesame:

U =∑i

giAεi exp(−βεi) = N

∑i giεi exp(−βεi)∑i gi exp(−βεi)

= NkBT2 ∂

∂T

ln

[∑i

gi exp(−βεi)

]

= NkBT2 ∂

∂Tln(Z) = kBT

2 ∂

∂Tln

(ZN

N !

)


where I got away with adding the N ! term in the log because the derivative with respect to T wouldmake this vanish! And of course I also have

F = −kBT ln

(ZN

N !

).

Chapter 4

Quantum Statistics

Recall that the way one counts the number of ways quantum particles can be distributed in accessiblequantum states is very different from the way classical particles are distributed in their quantumstates. But we more or less dropped that story for a while in order to resolve Gibbs’ paradox. Atlarge temperatures, when the number of states that are occupied is much larger than the number ofparticles, we recover the usual Boltmann distribution. But you might not remember a subtle point.We assumed that the effective degeneracy of a single level is large: we grouped many very closelyspaced singly-degenerate quantum states into another state, which we called the highly degeneratequantum state. But we also found that at high temperatures, the mean number of levels occupiedis also much larger than the number of particles in the system. In this chapter, we’ll relax thesecond assumption about high temperatures, so that the number of particles can begin to approachthe degeneracy of one of these compound levels. In this case, we shouldn’t recover the classicalBoltzmann distribution, but rather the quantum distributions for bosons and fermions.

4.1 Bose and Fermi Distributions

In Section 5.1.3 we saw that the number of ways of arranging ni quantum particles in a single energylevel εi with degeracy gi is given by

Ω(a)i =

[gi + (ni − 1)(1− α)]!

ni![gi − 1− α(ni − 1)]!,

where α = 0 for bosons and α = 1 for fermions, and any other α corresponds to particles withfractional exclusion statistics. The total number of ways of arranging the particles is then

Ω =∏i

[gi + (ni − 1)(1− α)]!

ni![gi − 1− α(ni − 1)]!.

Let’s follow the same procedure as we did previously in deriving the Boltzmann distribution, bymaximing the entropy (or ln(Ω)) subject to the constraints N =

∑i ni and U =

∑i εini. Now,

in the grand canonical ensemble we know that the Lagrange multiplier fixing the first condition isnothing but the chemical potential µ. Putting this together we have∑i

(1−α) ln[gi+(ni−1)(1−α)]+(1−α)− ln(ni)−1+α ln[gi−1−α(ni−1)]+α+βµ−βεidni = 0

32


⇒ (1− α) ln[gi + (ni − 1)(1− α)]− ln(ni) + α ln[gi − 1− α(ni − 1)] + βµ− βεi = 0

because the variations of ni are all independent and arbitrary.

4.1.1 Fermions

For fermions (α = 1) we have

− ln(ni) + ln[gi − ni] + βµ− βεi = 0

⇒ ln

(gini− 1

)= β(εi − µ).

This immediately gives the Fermi-Dirac (FD) distribution function

nFDi =

giexp [β(εi − µ)] + 1

, (4.1)

otherwise known as the Fermi-Dirac occupation factor.

A simpler way to derive the FD distribution is to use the grand canonical partition function. Becauseeach state within a degenerate level can hold either no particles (with energy zero), or one particle(with energy Ei = εi and particle number Ni = 1), the FD partition function is

Ξi = 1 + exp[−β(εi − µ)]gi , Ξ =∏i

1 + exp[−β(εi − µ)]gi .

The single-particle partition function is taken to the power of gi, because each of the gi states isassumed to be independent. The grand potential is therefore

ΦG = −kBT ln(Ξ) = −kBT∑i

gi ln 1 + exp[−β(εi − µ)] . (4.2)

Now, the mean number of particles in the system is

N = −∂ΦG∂µ

= kBT∂ ln(Ξ)

∂µ= kBT

∑i

giβ exp[−β(εi − µ)]

1 + exp[−β(εi − µ)]=∑i

giexp[β(εi − µ)] + 1

.

Because N =∑i ni, we immediately obtain

nFDi =

giexp [β(εi − µ)] + 1

in agreement with the expression above.

The Fermi-Dirac distribution function for various inverse temperatures β is shown in Fig. 4.1. Astemperature increases (lower β), the distribution becomes flatter and shows a longer exponentialtail, which begins to look like the Boltzmann distribution. At lower temperature (larger β), thedistribution approaches a step-like function, signifying full occupation for levels ε < µ and zerooccupation for levels larger than the chemical potential.


0 5 10 15 20ε

0

0.2

0.4

0.6

0.8

1

n

FD

(ε)

Figure 4.1: The fermion distribution function for various temperatures. The solid, dashed, anddot-dashed lines correspond to β = 10, 1, and 0.5, respectively. The chemical potential is set at 10and g = 1.

In order to get a better feeling for the Fermi distribution (4.1), let’s modify the way it looks a little.Consider ni − gi

2 :

ni −gi2

= gi1− 1

2 exp [β(εi − µ)]− 12

exp [β(εi − µ)] + 1=(gi

2

) 1− exp [β(εi − µ)]

exp [β(εi − µ)] + 1.

Pulling a factor of exp [β(εi − µ)/2] out of both the top and the bottom gives

ni −gi2

=(gi

2

) exp [−β(εi − µ)/2]− exp [β(εi − µ)/2]

exp [β(εi − µ)/2] + exp [−β(εi − µ)/2]

= −(gi

2

)tanh

[β

2(εi − µ)

].

So we are finally left with a slightly more intuitive expression (I hope!) for the Fermi distributionfunction that now involves a tanh:

ni =gi2

1− tanh

[β

2(εi − µ)

]. (4.3)

Recall that the tanh function has the following property:

tanh(ax) =

−1 x 0+1 x 00 x = 0

Also, tanh(ax) ∼ ax when x ∼ 0. So the value of a determines how quickly the function changesfrom its value of −1 to +1 as x passes through zero. In our case, x = εi − µ and a = β. This meansthat the Fermi distribution is equal to gi for εi µ and is exactly zero when εi µ. Also, as the


temperature gets lower (β increases), the slope through εi = µ gets larger and larger, approachinga vertical line when T = 0. So the Fermi distribution at zero temperature looks just like a step(Heaviside-theta) function:

ni(T = 0) = giΘ(εi − µ) ≡ gi εi < µ

0 εi > µ.

This means that exactly gi particles occupy each energy level εi up to the Fermi energy EF = µ,and no energy levels are occupied above this.

4.1.2 Bosons

For bosons (α = 0) we have

ln(gi − 1 + ni)− ln(ni) + βµ− βεi = 0.

Now, the degeneracy of a given level is still assumed to be large, so that gi 1. In this case weobtain

ln

(gini

+ 1

)= β(εi − µ).

This can be inverted to givegini

= exp [β(εi − µ)]− 1,

or finally the Bose-Einstein (BE) distribution:

nBEi =

giexp [β(εi − µ)]− 1

, (4.4)

otherwise known as the Bose-Einstein occupation factor. Note that the BE and FD distributionsdiffer only in the sign of the ‘1’ in the denominator! As we’ll see later, this small difference leads toprofoundly different kinds of behaviour. But one thing is immediately clear: the chemical potentialfor bosons cannot ever exceed the value of the lowest energy level ε0. If it did, the value of thedistribution function for this level n0 would be a negative number, which is unphysical. So forbosons we must have µ < 0.

Again, we can derive the BE distribution within the grand canonical ensemble instead. There is norestriction to how many bosons can occupy a single state within a degenerate level. So the grandpartition function for a given level is now a sum over contributions with no particles, one particle(with energy Ei = εi and number Ni = 1), two particles (with energy Ei = 2εi and Ni = 2), etc.:

Ξi = 1 + exp[−β(εi − µ)] + exp[−2β(εi − µ)] + exp[−3β(εi − µ)] + . . .gi .

Obviously, this is an infinite geometric series, 1 + x+ x2 + . . . = 1/(1− x):

Ξi =

1

1− exp[−β(εi − µ)]

gi.

The grand potential is

ΦG = −kBT ln(Ξ) = kBT∑i

gi ln 1− exp[−β(εi − µ)] . (4.5)


10 12 14 16 18 20ε

0

2

4

6

8

10

n

FD

(ε)

Figure 4.2: The bosons distribution function for various temperatures. The solid and dashed linescorrespond to β = 0.1 and 1, respectively. The chemical potential is set at 10 and g = 1.

The mean number of particles is

N = −∂ΦG∂µ

= kBT∑i

giβ exp[−β(εi − µ)]

1− exp[−β(εi − µ)]=∑i

giexp[β(εi − µ)]− 1

.

Again, we know that N =∑i ni so

nBEi =

giexp[β(εi − µ)]− 1

,

also in agreement with the expression above.

The Bose-Einstein distribution function for various inverse temperatures β is shown in Fig. 4.2. Astemperature increases (lower β), the distribution shows a longer exponential tail, which again beginsto look like the Boltzmann distribution. At lower temperature (larger β), the distribution tightensup, with more probability near the chemical potential.

Incidentally, for particles with fractional exclusion statistics, we can write the distribution as

nFESi =

giw exp[β(εi − µ)]+ α

,

where w(x) is a function of x = exp[β(εi − µ)] satisfying the following equation:

w(x)α[1 + w(x)]1−α = x.

If α = 1 (fermions), then w = x as expected. For α = 0 (bosons), we obtain 1 + w = x, againreproducing the expected result. For rational fractions α = p/q with p and q mutually prime, this


0 5 10 15 20ε

0

0.5

1

1.5

2

n

(Bo

ltz,F

D,B

E) (ε

)

Figure 4.3: The Boltzmann, Fermi-Dirac, and Bose-Einstein distribution functions, shown as solid,dashed, and dot-dashed lines, respectively, are plotted at the same temperature (β = 1) and chemicalpotential (µ = 10). The degeneracy factor is g = 1.

implies a polynomial of order q. The distribution for semions, which satisfy α = 1/2, is

nsemioni =

gi√exp[2β(εi − µ)] + 1

4

.

The Boltzmann, Fermi-Dirac, and Bose-Einstein distributions are plotted together in Fig. 4.3. Noticethat they all coincide for large energies levels ε, and also at high temperature.

4.1.3 Entropy

As discussed at the end of last term and in the review chapter above, the entropy can be obtainedin terms of the grand potential using

S = −∂ΦG∂T

.

The grand potential for bosons and fermions, derived above, are

ΦG = ∓kBT∑i

gi ln 1± exp[−β(εi − µ)] ,

where the upper sign corresponds to fermions, and the lower sign to bosons. The derivative is carriedout more easily in terms of β, so we’ll define the entropy instead as

S = kBβ2 ∂ΦG∂β

= ∓kBβ2 ∂

∂β

1

β

∑i

gi ln 1± exp[−β(εi − µ)]

=∑i

gi

±kB ln 1± exp[−β(εi − µ)] ∓ kBβ

∓(εi − µ) exp[−β(εi − µ)]

1± exp[−β(εi − µ)]


= kB∑i

gi

± ln 1± exp[−β(εi − µ)]+

β(εi − µ)

exp[β(εi − µ)]± 1

.

This is a pretty ugly expression, but it can be made to look nicer using the expressions for the FDand BE distributions

ni =gi

exp[β(εi − µ)]± 1.

This can be inverted to give

gi ∓ ni =gi exp[β(εi − µ)]± gi ∓ gi

exp[β(εi − µ)]± 1=

gi1± exp[−β(εi − µ)]

which yields

ln 1± exp[−β(εi − µ)] = − ln

(gi ∓ nigi

).

We’re getting there! We also know that

exp[β(εi − µ)] =gini∓ 1 =

gi ∓ nini

⇒ β(εi − µ)] = ln(gi ∓ ni)− ln(ni).

Putting all of these into the expression for the entropy gives

S = kB∑i

gi

∓ ln(gi ∓ ni)± ln(gi) +

nigi

ln(gi ∓ ni)−nigi

ln(ni)

= kB

∑i

(ni ∓ gi) ln(gi ∓ ni)− ni ln(ni)± gi ln(gi) .

For fermions, the resulting entropy takes the form:

SFD = −kB∑i

ni ln(ni) + (gi − ni) ln(gi − ni) + gi ln(gi) , (4.6)

which reduces so something intuitive when gi = 1:

SFD(gi = 1) = −kB∑i

ni ln(ni) + (1− ni) ln(1− ni) .

In other words, it is the sum of two contributions, corresponding to each level either empty, orcontaining one particle. This is very reminiscent of the entropy for a two-state system (coins orspins)! Note that the entropy is zero if each state is occupied by one particle (ni = 1). This is theground state of the system at zero temperature, and it reflects the Pauli exclusion principle. Forbosons, the form of the entropy is less intuitive looking, but still nice:

SBE = kB∑i

(gi + ni) ln(gi + ni)− ni ln(ni)− gi ln(gi) . (4.7)

Note that for bosons, the entropy is zero only if all states are unoccupied, i.e. ni = 0 ∀i. But thismust be nonsense, because we have a finite number of particles in our system! Where have all theparticles gone? This is the origin of Bose-Einstein condensation, which we’ll get to below. Noticealso that there is a complete symmetry in the Bose-Einstein case between ni and gi.


4.2 Quantum-Classical Transition

Last term we found that the chemical potential at high temperatures is large and negative. Thismeans that one might expect exp[β(εi−µ)] 1 at high temperatures, but only if |µ| kBT at hightemperatures as well. Supposing that the magnitude of the chemical potential gets large faster thankBT gets large (checked below), then both the Bose and Fermi distributions become the Boltzmanndistribution:

nBE,FDi (T 0) ≈ gi exp(βµ) exp(−βεi),

where A = exp(βµ) is called the fugacity. But remember that we defined A = N/Z in the canonicalensemble. So this immediately gives

µ = kBT ln

(N

Z

),

where Z is understood as the high-temperature limit of the canonical partition function. This isconsistent with the results of the examples (Pauli paramagnet and particles in a 3D box) discussedlast term. If µ/kBT −1, then this implies that N Z or Z N . For a 3D box, this means

V

Nλ3D

1 or n 1

λ3D

at high temperatures. But the mean distance between particles d = (1/n)1/3 so this criterion isequivalent to saying that the classical limit is equivalent to stating that the mean distance betweenparticles must be much larger than the de Broglie wavelength, d λD.

What does this criterion for the world to be classical come from? Suppose we use the equipartitiontheorem, which states that the mean kinetic energy of each particle is U/N = p2/2m = (3/2)kBT .Then p =

√3mkBT h/d, where h/d is a measure of the largest quantum wavevector in the system.

The condition is equivalent to

d

√h2

3mkBT=

λD√6π

or d λD.

So, the transition from the classical to the quantum regimes occurs when the particles get sufficientlyclose together to begin to detect each others’ wave-like natures.

Now let’s tie up that important loose end, and check if |µ| kBT at high temperature. Considerthe mean particle (fermion or boson) density for a 3D box:

n =N

V=

1

2π2

∫ ∞0

k2dk

exp[β(ε− µ)]± 1,

where εk = h2k2/2m. Making the change of variables k2 = (2mkBT/h2)x2, we obtain

n =1

2π2

(2mkBT

h2

)3/2 ∫ ∞0

x2dx

exp(x2 − η)± 1, (4.8)

where η ≡ µ/kBT . We assume that the mean density in the box does not change as a functionof temperature. So let’s suppose that we start increasing the temperature. Clearly, the integral


has to decrease like 1/T if n is conserved; at very large temperatures, the integral must approachzero. But the only thing in the integral that knows about the temperature is η = µ/kBT . For thisto be true, we clearly require η → −∞ at high temperatures, or |µ|/kBT 1. This justifies theconnection between the grand canonical function for indistinguishable quantum particles and thecanonical distribution function for classical particles discussed above. At large (but not infinite)temperature, the integral becomes

n ≈ 1

2π2

(2mkBT

h2

)3/2 ∫ ∞0

dxx2 exp[−(x2 − η)] =1

2π2

(2mkBT

h2

)3/2

exp

(µ

kBT

) √π

4.

Inverting this gives

µ

kBT= ln(n) +

3

2ln

(2πh2

mkBT

)⇒ µ = kBT ln

(nλ3

D

)as we found last term.

4.3 Entropy and Equations of State

To obtain the equations of state for quantum particles, let’s calculate the pressure. Obviously, for a3D box at high temperature we’re expecting the result to be P = nkBT , where n is the density. Whatabout at lower temperatures? How do you think the quantum statistics will affect the pressure?

The pressure is defined in terms of the grand potential as

P = −∂ΦG∂V

,

and the grand potential for fermions and bosons is

ΦG = ∓kBT∑i

gi ln 1± exp[−β(εi − µ)] ≈ ∓kBTV2π2

∫ ∞0

k2dk ln 1± exp[−β(εk − µ)] .

In the second line I have explicitly assumed a 3D box, and that the energy levels εk = h2k2/2mare sufficiently closely spaced relative to the temperature that the energy levels are more or lesscontinuously distributed. The pressure is therefore

P = ±kBT2π2

∫ ∞0

k2dk ln 1± exp[−β(εk − µ)] .

To make progress, let’s integrate this expression by parts once. Recall to integrate by parts we usethe rule ∫ b

a

udv = uv∣∣ba−∫ b

a

vdu.

The first term on the right hand side is called the surface contribution. In the pressure expression,we’ll let u = ln 1± exp[−β(εk − µ)] and dv = k2dk. Now, v = k3/3 which is zero for k = 0, andu = 0 when k =∞ because the argument of the ln becomes one. So the surface term vanishes. Thepressure expression becomes

P = ∓kBT2π2

∫ ∞0

h2k4

3m(∓β)

exp[−β(εk − µ)]

1± exp[−β(ε− µ)]dk =

1

2π2

∫ ∞0

h2k4

3mnkdk.


Now you see why integrating by parts makes such an improvement! Notice that this expression iscorrect for either bosons or fermions; you only have to use the appropriate expression for nk.

We can divide the factor of exp[−β(εk − µ)] in the numerator into the denominator, which makesthe integrand explicitly depend on the distribution functions nk. Making the same substitutionx2 = (h2/2mkBT )k2 as was done in the evaluation of the mean particle density, and simplifying abit, one obtains

P =kBT

3π2

(2mkBT

h2

)3/2 ∫ ∞0

x4

exp(x2 − η)± 1dx,

where as before η = µ/kBT . Note that the prefactor is almost exactly the same as the right handside of the expression for the particle density, Eq. (4.8). So finally we have

P =2kBTn

3

∫∞0

x4

exp(x2−η)±1dx∫∞0

x2

exp(x2−η)±1dx. (4.9)

Let’s check first if this reduces to the known equation of state at high temperatures, when η → −∞:

P (T 1) ≈ 2nkBT

3

∫∞0x4e−x

2

dx∫∞0x2e−x2dx

=2nkBT

3

3√π

8

(√π

4

)−1

= nkBT,

as expected.

For arbitrary values of η, the pressure cannot be evaluated analytically, but it is relatively straight-forward to obtain the first-order correction from the ideal gas expression. Because η −1, one canwrite

1

exp(x2 − η)± 1=

(1

exp(x2 − η)

)1

1± exp(−x2 + η)≈ exp(−x2 + η)

[1∓ exp(−x2 + η)

].

Inserting this into the expression for the pressure gives

P ≈ 2kBTn

3

∫∞0x4(e−x

2 ∓ eηe−2x2)dx∫∞

0x2(e−x2 ∓ eηe−2x2

)dx

=2kBTn

3

3√π

8 ∓ eη 3√π

32√

2√π

4 ∓ eη√π

8√

2

≈ nkBT1∓ eη 1

4√

2

1∓ eη 12√

2

.

The results below Eq. (4.8) suggest that we simply make the identification eη = eβµ = nλ3D. The

pressure then becomes

P ≈ nkBT(

1± nλ3D

4√

2

).

The two terms in this expression are called the first and second virial coefficients, respectively.The first virial coefficient simply reflects the ideal gas law, as we found above anyhow. The secondis a density-dependent correction that arises strictly from the particle statistics. The contributionto the pressure is positive for fermions, but negative for bosons. For fermions, the pressure increasesas the temperature is lowered (i.e. nλ3

D increases) because the Pauli exclusion principle effectivelypushes the particle apart: the Pauli principle acts like a repulsive two-body interaction. In contrast,the second virial coefficient for bosons is negative, indicating that they prefer to stay closer togetherat lower temperature; this presages the idea of Bose-Einstein condensation, discussed in detail below.


In general, the deviation of the ideal gas law for non-interacting particles is one important signaturefor the onset of quantum degeneracy.

In exactly two dimensions, the second virial coefficient for particles with arbitrary exclusions statis-tics can be calculated analytically, and the resulting equation of state is exactly

P = nkBT

[1 +

2α− 1

4nλ2

D

].

For fermions (α = 1), the second virial coefficient is clearly positive, while for bosons (α = 0), it isnegative. Interestingly, it is exactly zero for semions (α = 1

2 ), which are halfway between the twolimits.

The high-temperature correction to the total energy can be calculated in a similar fashion. We have

U = kBT2 ∂ ln(Ξ)

∂T= kBT

2 ∂

∂T

(− ΦGkBT

)= ΦG − T

∂ΦG∂T

.

We could go through the same procedure as above, integrating by parts, etc. But we can save lotsof time by observing that U =

∑i niεi, so that

U =V

2π2

∫ ∞0

h2k4

2mnkdk =

3

2PV

when comparing to the expression for the pressure above. So one obtains

U ≈ 3

2NkBT

(1± 1

25/2nλ3

D

).

Chapter 5

Fermions

5.1 3D Box at zero temperature

For a three-dimensional box, the highest energy level is EF = h2k2F /2m, where kF is the Fermi

wavevector, or the radius of the Fermi sphere in momentum space (recally p = hk). What is theFermi energy at zero temperature in this case? We know that the mean number of particles istherefore

N =

iF∑i=0

ni =gV

2π2

∫ kF

0

k2dk,

where I’ve kept a factor of g out front in case each energy level has any intrinsic degeneracy overand above the usual density of states. For example, if I assume that each fermion has spin- 1

2 , thenI could place exactly two particles (g = 2) in each energy level (one with spin up, one with spindown) and still satisfy the Pauli exclusion principle. Integrating, and making use of the fact thatEF = h2k2

F /2m, I obtain

N =gV

6π2

(2mEF

h2

)3/2

, (5.1)

which can be inverted to give

EF =h2

2m

(6π2n

g

)2/3

. (5.2)

Here, n = N/V is the mean density. This expression immediately gives the value for the Fermiwavector at zero temperature:

kF =

(6π2n

g

)1/3

.

Let’s also calculate the mean energy for the fermions at zero temperature:

U =

iF∑i=0

giεini =gV

2π2

∫ kF

0

k2

(h2k2

2m

)dk =

gV h2k5F

10π22m=

6

10

(gV k3

F

6π2

)(h2k2

F

2m

)=

3

5NEF . (5.3)

The pressure for a Fermi gas is meanwhile

P =g

2π2

∫ ∞0

h2k4

3mnkdk =

gh2

6mπ2

∫ kF

0

k4dk =

(gV k3

F

6π2

)(h2k2

F

2m

)2

5V=

2nEF5

.

43


But wait a second! These fermions are non-interacting, i.e. are ideal gases. So why is there anypressure at zero temperature? It is called Fermi pressure, and arises solely from the Pauli exclusionprinciple: you can only squeeze some number of fermions into a given volume before you will getsome resistance!

5.2 3D Box at low temperature

At low but finite temperature, we need to evaluate expressions involving the full Fermi occupationfactors, i.e.

X =∑i

Xini =

∫d3k g(k)n(k)X(k) =

∫ ∞−∞

g(ε)n(ε)X(ε)dε =

∫ ∞−∞

n(ε)Y (ε)dε, (5.4)

where Y (ε) ≡ g(ε)X(ε). Notice that I have explicitly inserted the density of states into the integralexpressions, but that the inherent degeneracy of states g (for example originating in the spin degreeof freedom) is already included in the ni. Here it is also assumed that Y (ε) vanishes when ε→ −∞and is at most a power of ε for ε→∞. If we define K(ε) in terms of Y (ε) as

K(ε) ≡∫ ε

−∞Y (ε′)dε′,

so that Y (ε) = dK(ε)/dε, then one can integrate Eq. (5.4) by parts to obtain∫ ∞−∞

n(ε)Y (ε)dε =

∫ ∞−∞

n(ε)dK(ε) =

∫ ∞−∞

K(ε)

(−∂n∂ε

)dε.

The surface term is zero because K(ε = −∞) = 0 and n(ε = ∞) is exponentially suppressedcompared with K(ε =∞). Now, n(ε) at low temperatures is constant for most ε, and varies rapidlyonly near ε = µ. So it is reasonable to expand K(ε) in a Taylor series around ε = µ, so that we have∫ ∞

−∞n(ε)Y (ε)dε =

∫ ∞−∞

[K(µ) +

∞∑n=1

(ε− µ)n

n!

∂nK(ε)

∂εn

∣∣∣∣ε=µ

](−∂n∂ε

)dε.

The leading term is simply gK(µ), and only even terms will appear at higher-order because ∂n/∂εis an even function of ε− µ. So we have∫ ∞

−∞n(ε)Y (ε)dε = g

∫ µ

−∞Y (ε′)dε′ +

∞∑n=1

∫ ∞−∞

(ε− µ)2n

(2n)!

(−∂n∂ε

)dε∂2n−1Y (ε)

∂ε2n−1

∣∣∣∣ε=µ

.

Finally, making the substitution x ≡ (ε− µ)/kBT , one obtains∫ ∞−∞

n(ε)Y (ε)dε = g

∫ µ

−∞Y (ε)dε+ g

∞∑n=1

an(kBT )2n ∂2n−1

∂ε2n−1Y (ε)

∣∣∣∣ε=µ

,

where

an ≡∫ ∞−∞

x2n

(2n)!

(− d

dx

1

ex + 1

)dx =

(2− 1

22(n−1)

)ζ(2n),

with the Riemann zeta functions defined as

ζ(n) ≡ 1 +1

2n+

1

3n+

1

4n+ . . . . (5.5)


In practice we don’t care about anything except the first correction term, n = 1, for which a1 =ζ(2) = π2/6. So ∫ ∞

−∞n(ε)Y (ε)dε ≈ g

∫ µ

−∞Y (ε)dε+ g

π2

6(kBT )2Y ′(µ). (5.6)

That was a lot of work to obtain a very simple correction!

We are finally able to calculate the finite-temperature expressions for the mean number, the meanenergy, and the pressure. Using the 3D density of states seen last term, the mean number is

N =

∫ ∞0

g(ε)n(ε)dε =gV

2π2

(2m)3/2

2h3

∫ ∞0

ε1/2n(ε)dε ≈ gV

2π2

(2m)3/2

2h3

[∫ µ

0

ε1/2dε+π2

6(kBT )2 1

2µ−1/2

]=

gV

6π2

(2m)3/2

h3 µ3/2

[1 +

π2

8

(kBT

µ

)2]. (5.7)

This means that as we raise the temperature from zero at constant chemical potential, the meanparticle number will increase. Alternatively, one can assume that the mean number of particles Nremains constant as temperature is increased from zero, so that the Fermi energy EF (5.2) remainswell-defined at all temperatures, then the chemical potential must also vary:

N =gV

6π2

(2mEF

h2

)3/2

=gV

6π2

(2m)3/2

h3 µ3/2

[1 +

π2

8

(kBT

µ

)2].

⇒ µ ≈ EF

[1− π2

12

(kBT

EF

)2]. (5.8)

That is, at fixed Fermi energy, the chemical potential decreases from EF at zero temperature. Thisis consistent with the picture of the chemical potential we found last term, where the chemicalpotential becomes increasingly negative as the temperature is increased. But it is important topoint out that in the fermion case, the chemical potential isn’t zero at zero temperature; rather itis the highest-occupied energy level.

Next let’s calculate the mean energy:

U =

∫ ∞0

g(ε)εn(ε)dε =gV

2π2

(2m)3/2

2h3

∫ ∞0

ε3/2n(ε)dε ≈ gV

2π2

(2m)3/2

2h3

[∫ µ

0

ε3/2dε+π2

6(kBT )2 3

2

√µ

]=

gV

10π2

(2m)3/2

2h3 µ5/2

[1 +

5π2

8

(kBT

µ

)2]. (5.9)

While correct, this expression isn’t so nice because it is difficult to compare it to the zero-temperatureresult (5.3). Substituting into Eq. (5.7) gives

U ≈ 3

5Nµ

1 + 5π2

8

(kBTµ

)2

1 + π2

8

(kBTµ

)2 ≈3

5Nµ

[1 +

π2

2

(kBT

µ

)2].

Finally, inserting the expression (5.8) gives

U ≈ 3

5NEF

[1− π2

12

(kBT

EF

)2][

1 +π2

2

(kBT

µ

)2]

=3

5NEF

[1 +

5π2

12

(kBT

EF

)2].


This means that the specific heat at low temperatures is proportional to the temperature, CV =∂U/∂T ∼ T . This is in contrast to the classical (Boltzmann) gas, where the specific heat at lowtemperature was zero. It’s also the reason for the linear-temperature specific heat observed inclean crystals at very low temperatures, because electrons are fermions. Again, using the relationP = 2U/3V gives the correction to the pressure:

P ≈ 2nEF5

[1 +

5π2

12

(kBT

EF

)2]. (5.10)

The finite-temperature correction term is exactly the same as for the mean energy above. This isnothing but the second virial coefficient, which is clearly positive for all finite temperatures. Recallthat at very high temperatures, the pressure expression is instead

P (T →∞) = nkBT

[1 +

1

25/2nλ3

D

].

5.3 3D isotropic harmonic trap

5.3.1 Density of States

Almost all of the spin-polarized fermionic atoms that have been cooled to ultralow temperatureshave been trapped by magnetic fields or focused laser beams. The confining potentials are generally3D harmonic traps. So let’s consider this case in more detail. You might be interested to notethat Fermi’s original paper on fermionic particles considered this case, not the 3D box case above.As we saw previously, ignoring the zero-point energy in each dimenion the eigenvalues (accessibleenergy states) are given by ε(nx, ny, nz) = nxhωx+nyhωy +nzhωz. In order to evaluate the variousintegrals, we first need to obtain the density of states per unit energy. A rough way to do this is tosimply set ki = ni, so that ε2 = k2

x(hωx)2 +k2y(hωy)2 +k2

z(hωz)2 ≡ k2(hω)2, where ω = (ωxωyωz)

1/3

is the mean frequency, and dki/εi = 1/hω. Because ki = ni now rather than ki = πni/L, the 3Ddensity of states is given by

g(ε) =k2

2

dk

dε=

ε2

2(hω)3. (5.11)

Another (more rigorous) way to obtain the density of states is to ask how many states G(ε) areenclosed in an energy surface of radius ε = εx + εy + εz. The result is

G(ε) =1

hωxhωyhωz

∫ ε

0

dεx

∫ ε−εx

0

dεy

∫ ε−εx−εy

0

dεz =ε3

6h3ωxωyωz.

The density of states is then

g(ε) =dG(ε)

dε=

ε2

2h3ωxωyωz=

ε2

2(hω)3,

in agreement with the guess above.

Here are another couple of ways to see this, if we assume that the trap is isotropic (ωx = ωy = ωz).We know that the partition function for the one-dimensional oscillator with energy levels given by


εn = hωn, n = 0, 1, 2, . . . (neglecting the zero-point energy) is

Z1D =1

1− x, x = exp(−βhω).

Because the partition function for the s-dimensional harmonic oscillator is the s-fold product of theone-dimensional partition functions, we can write

ZsD =1

(1− x)s=

(1 + sx+

s(s− 1)

2x2 +

s(s− 1)(s− 2)

6x3 + . . .

)=

∞∑n=0

(n+ s− 1

n

)xn =

∞∑n=0

g(s, n)xn. (5.12)

So, the degeneracy of the s-dimensional oscillator is

g(s, n) =

(n+ s− 1

n

)=

(n+ s− 1)!

n!(s− 1)!.

For three dimensions (s = 3), g(n) = (n + 1)(n + 2)/2. This expression becomes simpler g(n′) =n′(n′ + 1)/2 if we make the replacement n′ = n+ 1 and count from 1 instead of zero. For large n′,this becomes g(n′) = n′2/2, as found above.

Alternatively, the degeneracy for the three dimensional isotropic trap can be found by counting thenumber of ways we can distribute M distinct energy quanta into a common (degenerate) energylevel, just like in the Planck case which we’ll see in the next chapter. We have WN = (N + M −1)!/[M !(N − 1)!]. If M = 0 then WN = 1, i.e. there is only one way to fill up each level; this isequivalent to the one-dimensional case (s = 1). If M = 1 then WN = N !/(N − 1)! = N , i.e. thereare N ways to fill up each level indexed by N ; this is equivalent to s = 2. If M = 2 then there areWN = (N + 1)!/[2(N −1)!] = N(N + 1)/2, ways to fill up each level indexed by N ; this is equivalentto s = 3 and clearly s = M + 1 in this notation.

5.3.2 Low Temperatures

Armed with the density of states, we are in a position to calculate the finite-temperature expressionsfor the mean number, the mean energy, and the pressure for the 3D isotropic oscillator, as we didfor the 3D case above. The mean number is

N =

∫ ∞0

g(ε)n(ε)dε =g

2(hω)3

∫ ∞0

ε2n(ε)dε ≈ g

2(hω)3

[∫ µ

0

ε2dε+π2

3(kBT )2µ

]=

g

6

( µ

hω

)3[

1 + π2

(kBT

µ

)2]. (5.13)

At zero temperature this expression defines the Fermi energy:

EF =

(6N

g

)1/3

hω. (5.14)


Again, if we fix N at all temperatures then

N =g

6

(EFhω

)3

=g

6

( µ

hω

)3[

1 + π2

(kBT

µ

)2]

which can be inverted to yield

µ ≈ EF

[1− π2

3

(kBT

EF

)2]. (5.15)

Next let’s calculate the mean energy:

U =

∫ ∞0

g(ε)εn(ε)dε =g

2(hω)3

∫ ∞0

ε3n(ε)dε ≈ g

2(hω)3

[∫ µ

0

ε3dε+π2

2(kBT )2µ2

]=

gµ

8

( µ

hω

)3[

1 + 2π2

(kBT

µ

)2]. (5.16)

Substituting into Eq. (5.13) gives

U ≈ 3

4Nµ

1 + 2π2(kBTµ

)2

1 + π2(kBTµ

)2 ≈3

4Nµ

[1 + π2

(kBT

µ

)2].

Finally, inserting the expression (5.15) gives

U ≈ 3

4NEF

[1− π2

3

(kBT

EF

)2][

1 + π2

(kBT

µ

)2]≈ 3

4NEF

[1 +

2π2

3

(kBT

EF

)2].

The pressure follows directly from here as it did for the 3D case.

5.3.3 Spatial Profile

The above analysis doesn’t tell us anything about the spatial profile of the confined fermions. Inprinciple, one must put a gi fermions in each available energy state εi. So one would need to takeinto consideration each level and its associated degeneracy, then calculate the appropriate spatialwavefunction to build up the total density profile. This is a lot of work! Thankfully, there is a betterway to accomplish this task when the number of particles is large, so that a detailed knowledge ofthe single-particle wavefunctions is not necessary. We can use the local density approximation,which assumes that the particles at any given region of the trap behave locally as if there was noexternal potential at all. In other words, the Fermi occupation factor would be expressed as

n(r,k) =1

2π3

1

exp[β(h2k2/2m+mω2r2/2− µ

)]+ 1

.

This means that the spatial density is given by

n(r) =

∫d3k n(r,k) =

g(2m)3/2

4h3π2

∫ ∞0

√ε

exp [β (ε+mω2r2/2− µ)] + 1dε.


At sufficiently low temperatures we can approximate the Fermi occupation factor by a Θ-function,as discussed just below Eq. (4.3), and set µ ≈ EF . Then we can write

n(r) =g(2m)3/2

4h3π2

∫ EF−mω2r2/2

0

√εdε =

g(2m)3/2

6h3π2

(EF −

1

2mω2r2

)3/2

.

The Fermi energy can then be used to define the radius of the cloud R through EF = (1/2)mω2R2

so one obtains

n(r) =g

6π2

(mωh

)3

R3

[1−

( rR

)2]3/2

. (5.17)

Using the definition of the Fermi energy (5.14), the radius of the cloud is

R =

√2EFmω2

=

√2h

mω

(6N

g

)1/6

=√

2

(6N

g

)1/6

`, (5.18)

where I have defined the harmonic oscillator length ` ≡√h/mω (the meaning of this will be discussed

below). So finally the density is

n(r) =2

π2`3

√gN

3

[1−

( rR

)2]3/2

.

Where does the length ` come from? For a one-dimensional harmonic oscillator the force is Fi =−kriı = −mω2riı, and the virial theorem (2.1) states that on average

〈K〉 = −1

2

⟨∑i

Fi · ri

⟩=mω2

2

⟨∑i

r2i

⟩=mω2`2

2,

where `2 is a length scale describing the mean-square displacement of the oscillator. Meanwhile thekinetic energy term is ⟨∑

i

p2i

2m

⟩=

h2

2m

⟨∑i

k2i

⟩=

h2

2m`2,

where `−2 is the mean-square size of the oscillator in the wavevector representation.∗ Alternatively,you can think of the `−2 factor as simply ensuring the correct units. So we obtain `4 = h2/m2ω2 or` =

√h/mω as the characteristic size of the harmonic oscillator.

At high temperatures, the analysis above won’t apply because many of the fermions below the Fermienergy will be excited into levels above the Fermi energy (cf. Fig. 4.1). In this case, the fermionswill behave essentially as Boltzmannons (!). To find the density, we can either treat the potential asan external chemical potential, so that

n(r) =1

λ3D

exp

[µ− µext(r)

kBT

]=

1

λ3D

exp

(− µ

kBT

)exp

(− x

2

R2x

)exp

(− y

2

R2y

)exp

(− z

2

R2z

), (5.19)

∗The reason for this is that the particle distribution in the harmonic oscillator is a Gaussian. So the Fouriertransform of it (the momentum-space distribution) is also a Gaussian, whose mean-square size is the inverse of theoriginal Gaussian.


where R2i = 2kBT/mω

2i and ωi are the trapping frequencies in the x, y, and z directions. Alterna-

tively, we could use the definition of the total number in terms of the canonical partition functionN = AZ, and then use the equipartition form for Z,

N = AZ =1

λ3D

exp

(− µ

kBT

)∫d3r exp

(− x

2

R2x

)exp

(− y

2

R2y

)exp

(− z

2

R2z

),

and then use the normalization conditionN =∫d3rn(r) to obtain the chemical potential. In any case

we have the same thing. The effective volume of the cloud is V = (4π/3)R3i = (4π/3)(2kBT/mω)3/2

assuming all frequencies are equal.

5.4 A Few Examples

Let’s consider a few examples of fermions in familiar contexts. First and foremost, we need todetermine if we must treat them quantum mechanically or classically. In other words, we firstdetermine if the particles are quantum degenerate or not. If so, we can apply the results obtainedabove to obtain various properties if we are interested. Below are four examples of Fermi degeneratesystems where the temperatures span 14 orders of magnitude!

5.4.1 Electrons in Metals

Typical electron densities in metals are n = 1029 m−3. Using the mass of an electron as 9.109 ·10−31 kg, the de Broglie wavelength at room temperature (300 K) is found to be λD = 4.303·10−9 m.The mean interparticle separation is d = n−1/3 = 2 · 10−10 m. This means that λD d and theelectrons must be treated quantum mechanically. Another way to think about it is to use the Fermitemperature, TF ≡ EF /kB , which must be larger than the actual temperature of the system tobe quantum degenerate. Using the same parameters, I obtain EF = 2 · 10−18 J (or about 12 eV),and TF = 1.5 · 105 K. This is of course much larger than room temperature. So the electrons havesomething like 500 times the energy of atoms in the room.

Let’s calculate some other things while we’re at it. The Fermi velocity corresponds to the mean

velocity of fermions near the Fermi energy, and is defined as vF = hkF /m, where kF =√

2mEF /h2

is the Fermi wavevector. Putting in numbers gives vF ≈ 2.1 × 106 m/s, which is pretty fast!The electron pressure is given by Eq. (5.10) at low temperatures. Since at room temperaturekBT/EF ≈ 0.002 we are justified in neglecting the finite-temperature correction. The pressure isthen P ≈ 8 × 1010 N/m2. This is phenomenally high, almost a million atmospheres! It’s amazingthat metals don’t simply explode. . . . Or is it?

5.4.2 Electrons in the Sun

The sun was already considered in the context of the virial theorem, Sec. 2.1. Recall that the massof the sun is M = 3 · 1030 kg, and the average temperature is something like 5 × 106 K. Thenumber of hydrogen atoms gives a reasonable estimate for the number of electrons. With the massof hydrogen assumed to be 1.67 · 10−27 kg, I obtain Ne = 1.8 · 1057 electrons. The radius of the sunis 3 · 107 m, giving a density of 1.6 · 1034 m−3. The Fermi temperature is therefore TF = 2.7 · 108 K,which is something like 50 times larger than the mean temperature. So the electrons in the sunare also quantum degenerate! Note that one can also define the velocity of electrons at the Fermi


surface, vF ≡ hkF /m = 108 m/s for electrons in the sun. This is already one-third of the velocityof light, so if the sun were much hotter we’d have to treat the electrons relativistically. In fact, ifa star is heavier than about 1.4M then the electron degeneracy pressure will no longer be able tostabilize it against collapse. This so-called Chandrasekhar limit is quite easy to calculate when theelectrons are treated relativistically, as is needed for hot white dwarfs. There is a similar ‘Tolman-Oppenheimer-Volkoff’ limit for neutron stars that are stabilized by Fermi pressure associated withneutrons, in the range 1.4M <∼ M <∼ 3.5M; the actual values are hard to compute because thestrong force is really a pain to deal with.

5.4.3 Ultracold Fermionic Atoms in a Harmonic Trap

If we had a very good knowledge of the number of fermions in our trap, then we could use the Fermienergy (5.14) and compare it to kBT to test if the atoms were in fact quantum degenerate. In manyexperiments, the number of atoms is N ∼ 106 and trapping frequencies are of order ω/2π ∼ 100 Hz.Then with g = 1 (the atoms are spin-polarized which means that all the spins are pointing in thesame direction) one obtains EF ≈ 1.2× 10−29 J or a Fermi temperature TFEF /kB ≈ 1 µK. This isvery cold! But these days, experiments are operating around 100 nK or T = 0.1TF . So ultracoldfermions in traps are in fact strongly quantum degenerate.

Another way to estimate the temperature required to reach Fermi degeneracy, one can use theBoltzmann spatial profile (5.19). The density is n = 106/R3

i ≈ 3 · 1019 m−3 giving a mean spacingbetween particles of d ≈ 3 · 10−7 m. Fermi degeneracy is reached when λD = d which in this casecorresponds to T = TF ∼ 10−6 K, the same estimate we obtained above. Interestingly, the samedensity is found using the low-temperature Fermi distribution (5.17). For the parameters givenabove one obtains ` ≈ 1.6 µm and therefore R ≈ 22 µm. The volume is then 4πR3/3 ≈ 4.3× 10−14

which gives a density of n ≈ 2 × 1019 m−3. This is almost identical to the much rougher estimateabove, and so the qualitative results are the same.

Chapter 6

Bosons

That minus sign in the denominator of the Bose-Einstein distribution function really leads to totallydifferent behaviour, and we’ll now see.

6.1 Quantum Oscillators

Last term we obtained the partition function for the harmonic oscillator, and associated observables.To jog your memory, I’ll repeat some of this derivation here. The Bohr-Sommerfeld quantizationprocedure yields the energy eigenvalues εn = nhω, which is off by only a constant factor. The exactexpression is εn = hω(n + 1

2 ), but the additional factor makes no contribution to any statisticalproperties.

One can then obtain the (canonical) partition function:

Z =∑n

exp(−εn/kBT ) =∑n

exp(−βεn) = 1 + exp(−βhω) + exp(−2βhω) + . . . .

But this is just a geometric series: if I make the substitution x ≡ exp(−βhω), then Z = 1 +x+x2 +x3 + . . .. But I also know that xZ = x+x2 +x3 + . . .. Since both Z and xZ have an infinite numberof terms, I can subtract them and all terms cancel except the first: Z − xZ = 1, which immediatelyyields Z = 1/(1− x), or

Z =1

1− exp(−βhω). (6.1)

Now I can calculate the mean energy:

U = NkBT2 ∂ ln(Z)

∂T=NkBT

2

Z

∂Z

∂T= NkBT

2 [1− exp(−βhω)]

[1− exp(−βhω)]2 (−1)

hω

kBT 2(−1) exp(−βhω)

= Nhωexp(−βhω)

1− exp(−βhω)=

Nhω

exp(βhω)− 1(6.2)

= Nhω〈n(T )〉, where 〈n(T )〉 ≡ 1

exp(hω/kBT )− 1is the occupation factor.

Notice that the occupation factor is in fact identical to the Bose-Einstein distribution function,Eq. (4.4), with the identification εi = hω. There is in fact a very close connection between bosons

52


and oscillators, which you might already have anticipated. For example, photons are quantizedpackets of light, but light is also a wave (i.e. an oscillating field). Photons are also bosons, havinginteger (unit) spin.

Einstein constructed a model of a solid in 1907, where he assumed that the atoms making up thesolid were all able to oscillate independently around their equilibrium positions. (This is the ‘Ein-stein solid’ model discussed in Schroeder’s textbook in Chapter 2.2). The point was to understandexperimental results obtained early in the 20th century that showed that the heat capacity decreasesexponentially at low temperatures. Assuming each oscillator could vibrate in three directions inde-pendently, the mean energy becomes

U =3Nhω

exp(hω/kBT )− 1.

The specific heat is then

CV =∂U

∂T= 3NkB

(hω

kBT

)2exp(hω/kBT )

(exp(hω/kBT )− 1)2.

At low temperature, the exponential term becomes large, so we obtain

CV (T → 0) ≈ 3NkB

(hω

kBT

)2

exp

(− hω

kBT

),

which is indeed exponentially small at low temperature. This was a great early success of thequantum theory, because it was the only theory to work, and helped vindicate the idea of quantummechanics. Unfortunately, better experiments subsequently showed that the heat capacity at lowtemperatures is in fact not exponential, but rather goes like T 3. A better theory is evidently needed,which is discussed in the next section.

6.2 Phonons

In fact the atoms in a crystal are not able to oscillate completely independently, because the bondsfrom site to site are by definition strong. A better model, originally proposed by Debye in 1912, isto imagine a regular array of N masses m connected to each other by springs of length a with springconstant K. Suppose that the coordinate ηj describes the displacement of the mass at site j. TheHamiltonian (energy function) for the system is then

H =1

2

∑j

[mη2

j +K (ηj+1 − ηj)2].

The force can be obtained by using

Fi = −∂V∂ηi

= − ∂

∂ηi

K

2

∑j

(ηj+1 − ηj)2= −K

∑j

[(ηj+1 − ηj) δi,j+1 − (ηj+1 − ηj) δi,j ]

= −K (2ηj − ηj+1 − ηj−1)

so that the equations of motion become

mηj −K (ηj+1 + ηj−1 − 2ηj) = 0.


With equilibrium positions identified as x = ja, one can posit that the solutions are given bytravelling waves, ηj ∝ exp[i(kx − ωt)]. Imposing periodic boundary conditions η0 ≡ ηN requires1 = exp(ikNa) which implies kNA = 2πn or k = (2πn/N)(1/a). These crystal momenta are in factidentical to the quantized momenta seen for the particle in a box k = 2πn/L, except that now weidentify the total length L = Na with the product of crystal lattice spacings. Inserting the travelingwave solution into the equation of motion gives

−mω2 −K(eika + e−ika − 2

)= 0.

This can be simplified by noting that

sin2

(ka

2

)= −1

4

(eika/2 − e−ika/2

)2

= −1

4

(eika + e−ika − 2

).

We then obtain the dispersion relation for phonons

ω2(k) =4K

msin2

(ka

2

). (6.3)

For small wavevector k (momenta) or long wavelength, the frequency is linear: ω(k → 0) ≈2√K/m|k|a. This is the same as the dispersion relation for a regular wave, ω = ck, where

c = a√K/m is the wave velocity. It is quite different from the free-particle dispersion εk ∝ k2

encountered in the context of the 3D box.

In the canonical ensemble, the mean energy is given by

U =∑k

hωknk =V

2π2

∫ ∞0

k2dkhω(k)

exp(hω(k)kBT

)− 1

.

You might wonder why I blithely used the Bose distribution factor here. In principle, there are onlytwo kinds of distributions that I am allowed to use and be consistent with quantum mechanics, theBose or Fermi. Because there are no restriction about how many vibrations I am allowed to have ina given energy, it is natural to use the Bose one.

Nevertheless, it will be hard to evaluate this expression using the full dispersion relation (6.3)obtained above. So instead let’s assume that at low temperatures only the lowest-energy states willbe excited, so that we can make the identification ω(k) ≈ ck. The energy then becomes

U ≈ 3V

2π2

∫ ∞0

hck3

exp(hckkBT

)− 1

dk =3V

2π2

(kBT

hc

)4

hc

∫ ∞0

x3

ex − 1dx, (6.4)

where the factor of 3 comes from the three different polarizations: two transverse ones like thephoton, and one longitudinal. There are 3N phonon modes in total; a one-dimensional chain withN sites has N modes, in 3D there are 3N . The last integral is related to a couple of special functions:∫ ∞

0

xn

ex − 1dx = ζ(n+ 1)Γ(n+ 1),

where Γ(n) is Euler’s gamma function and the zeta function was defined earlier (5.5). For n = 3 oneobtains Γ(4)ζ(4) = π4/15 so that the heat capacity becomes at low temperature

CV (T → 0) ≈ ∂

∂T

3V

30π2

(πkBT

hc

)4

hc =2π2

5V kB

(kBT

hc

)3

.


This now has the desired T 3-dependence, due to the phonons. Because nothing was assumed in thisderivation except for the fact that the dispersion was linear, the same result is true for any systemcharacterized by travelling waves.

In metals at very low temperatures the T 3-dependence becomes very small and the heat capacityinstead goes like T due to the electrons in the crystal, as discussed in Section 5.2. In insulators theelectrons are not mobile and this contribution doesn’t exist.

At higher temperatures, one needs to worry about higher-energy modes in the phonon distributionfunction. In principle, we need to include the full k-dependence of the phonon dispersion rela-tion (6.3), but this is a problem because the integrand diverges at high frequencies. But we haveoverlooked an important fact: phonons with wavelengths shorter than the lattice spacing cannotexist. So there is in fact a natural cut-off at high frequency (short wavelength) that will remove thedivergence in the integrals in practice.

This cut-off is called the Debye frequency ωD, and is determined at zero temperature where thedispersion relation is linear. With ω = ck the 3D density of states becomes

V k2dk

2π2=

3V ω2dω

2π2c3,

where the factor of 3 again comes from the polarizations. So

3N =

∫ ωD

0

3V ω2

2π2c3dω =

V ω3D

2π2c3,

which defines the Debye frequency

ωD ≡(

6π2c3N

V

)1/3

= c(6π2n

)1/3. (6.5)

The Debye temperature is then defined as θD ≡ hωD/kB = (hc/kB)(6π2n)1/3.

Let’s use this to re-derive the zero-temperature heat capacity of the phonon gas, but this time let’suse the grand partition function, just for fun. When the chemical potential is zero, the free energycoincides with the grand potential (4.5):

F = −kBT ln(Ξ) = kBT∑i

gi ln 1− exp[−β(εi − µ)] =3kBTV

2π2c3

∫ ωD

0

ω2 ln[1− e−hω/kBT

].

Setting x = hω/kBT gives

F =3V kBT

2π2

(kBT

hc

)3 ∫ θD/T

0

x2 ln[1− e−x

]= −9NkBT

(T

θD

)3(π4

45

),

where the value of the integral is obtained under the assumption θD/T >∼ 20; at higher temperaturesit will be smaller, but the linear approximation to the dispersion relation will not be applicable. Thespecific heat is then found to be

CV = −T ∂2F

∂T 2=

12π4

5NkB

(T

θD

)3

,

which is easily shown to be identical to the expression found earlier.


0 5 10 15 20

ν (arb. units)

0

1

2

3

4

5

u(ν

) (a

rb. unit

s)

Figure 6.1: The blackbody distribution u(ν) is plotted as a function of the frequency ν.

6.3 Blackbody Radiation

A blackbody is a body that looks completely black when it is cold, i.e. it absorbs light perfectly atall wavelengths. Conversely, these same bodies when heated emit light at all wavelengths, whichmight make their ‘blackbody’ moniker somewhat confusing. Of course, there is no real material thatis a perfect blackbody, but there are several systems that come very close. One obvious one is coal.Perhaps the best current one is actually the cosmic background radiation, believed to be a relic ofthe Big Bang. As the universe has expanded (effectively adiabatically), the temperature has cooledsuch that the current temperature is on the order of a few degrees Kelvin. Alternatively, the fabricof spacetime has stretched to such an extent that the radiation emitted from the Big Bang (more orless isotropically) has become severely red-shifted.

The distribution of wavelengths of light λ (or frequencies ν = c/λ) from a blackbody was investi-gated extensively in the 19th century, and was found to be a universal curve characterized by thetemperature. This is plotted in Fig. 6.1. As a model of a natural blackbody, experimentalists studieda small 3D box (cavity) with a hole in it. The heat radiation absorbed by the box is transformedinto light inside, which was considered to rattle around inside, coming to equilibrium, and eventu-ally escaping out of the small hole. It was found that the amount of radiation escaping from thehole was proportional to the area of the hole A and was related to the temperature through theStefan-Boltzmann law:

dQ

dt= AσT 4,

with σ ≈ 5.67× 10−8 W/m2K4. There was no microscopic explanation for this result, however, nordid anyone know the expression for the distribution of wavelengths. In 1896 Wilhelm Wien suggestedu(λ) = c1e

−c2/λT /λ5, which worked well at short wavelengths but terribly at long wavelengths.

The Rayleigh-Jeans theory (1905) attempted to do a better job in deriving the blackbody distri-


bution. They started by assuming that the cavity walls contained charged particles that oscil-lated about their equilibrium positions. Using the equipartition theorem, they proposed that eachoscillator has kBT of energy. Assuming a 3D cubic cavity of length L, the density of states isg(k)dk = (L3/π2)k2dk (the missing factor of two comes from the fact that light has two polarizationdirections). Using the relationship between wavevector and wavelength for photons k = 2π/λ onehas k2 = 4π2/λ2 and dk = −(2π/λ2)dλ so that k2dk = −(8π3/λ4)dλ. The density of states is theng(λ)dλ = (8πL3/λ4)dλ. The distribution is then

u(λ) =8πkBT

λ4.

Unfortunately this does not agree well with the observed distribution, because there is too muchradiation implied at short wavelengths (λ→ 0): you would be microwaved looking at a burning log!Even worse, the total energy density is infinite:

u =

∫ ∞0

u(λ)dλ =∞,

whereas it should go like T 4. This was called the ‘ultraviolet catastrophe’ by Ehrenfest in 1911, andit is because the integral diverges in the ultraviolet (short wavelength limit).

To get around these problems, Planck in 1905 suggested a different form for the distribution function.His reasoning went as follows. Suppose that there are N oscillators (electrons) in the walls of thecavity vibrating at frequency ν. The total energy is UN = NU and the total entropy is SN = NS =kB ln(Ω) (we are clearly in the microcanonical ensemble here!). How can one distribute the UNenergy amont the N oscillators? Suppose the energy is made up of discrete elements, UN = Mε,where M 0. (This is the quantum hypothesis). Then the number of ways of distributing Mindistinguishable energy states among N distinguishable oscillators is

Ω =(N − 1 +M)!

M !(N − 1)!,

which we also derived in Eq. (5.12). You will now also recognize this as the number of ways ofarranging indistinguishable bosons, Eq. (3.1). The Boltzmann entropy is then

SN = kB ln(Ω) = kB

[(1 +

M

N

)ln

(1 +

M

N

)− M

Nln

(M

N

)]= kB

[(1 +

U

ε

)ln

(1 +

U

ε

)− U

εln

(U

ε

)]This is strongly reminiscent of the Bose-Einstein entropy (4.7). Now, using the microcanonicaldefinition of the temperature gives

1

T=

(∂S

∂U

)V

= kB

[1

εln

(1 +

U

ε

)+

1

ε− 1

εln

(U

ε

)− 1

ε

]=kBε

ln( εU

+ 1).

Inverting gives

U =ε

exp(

εkBT

)− 1

,

which is the just the same result I obtained previously for a set of N oscillators, Eq. (6.2), exceptin that case I used the canonical ensemble.


Planck then connected the frequencies of oscillation to the energies of the (light) standing modes inthe cavity. He used the classical wave equation for the light modes

1

c2∂2y

∂t2= ∇2y

and assumed that the light had to vanish at the walls, so that y(x, y, z) = A sin(kxx) sin(kyy) sin(kzz)with kn = nπ/Ln. Inserting into the wave equation gives ∂2y/∂t2 = −c2k2y = −ω2(k)y or ω(k) =ck, a linear spectrum just as found for phonons. Then with the final assumption that ε(k) ∝ ck orω = hck, one obtains

U(k) =hck

exp(hckkBT

)− 1

⇒ U(λ) =hc

λ

1

exp(

hcλkBT

)− 1

with the identification k = 2π/λ. Combining this with the density of states found in the Rayleigh-Jeans theory gives the full Planck distribution function

u(λ) =8πhc

λ5

1

exp(

hcλkBT

)− 1

.

This fit the experimental data perfectly! For short wavelengths λ→ 0 it reproduces Wien’s predic-tion, u(λ→ 0) ≈ (8πhc/λ5)e−hc/kBTλ. At long wavelengths one obtains u(λ→∞) ≈ 8πkBT/λ

4 →0 which is the same as the Rayleigh-Jeans result.

It is straightforward to calculate the energy of radiation:

dQ

dt≡ u =

∫ ∞0

8πhc

λ5

1

exp(

hcλkBT

)− 1

dλ.

With the replacement z = hc/λkBT , then dλ = −(hc/kBTz2)dz and

u = 8π(kBT )4

(hc)3

∫ ∞0

z3

ez − 1dz.

The integral we have seen before, in Eq. (6.4), so the result is

u =8π5

15

(kBT )4

(hc)3.

This is both finite (good thing!) and also proportional to T 4, in accordance with the Stefan-Boltzmann law. The constant is then

σ =8π5k4

B

15(hc)3

which allows one to obtain the ratio h3/k4B experimentally. Furthermore, the maximum of the dis-

tribution occurs when hc/λkBT ≈ 4.965, which allows one to obtain the ratio h/kB experimentally.These two facts together uniquely determine both Planck’s and Boltzmann’s constants, neither ofwhich were known previous to this work.

This is the end of the story, more or less. It is interesting to note, though, that while Planckquantized the vibrational levels in the walls of the cavity, he used this only as a ‘trick’ to get a


sensible answer. At the time he didn’t believe that the energies were really discrete. He certainlydidn’t seem to realize that his theory also implied the discreteness of the light energies. The theorywas developed earlier (and therefore independently) of Einstein’s theory on the photoelectric effect,where the quantization of light was key. In fact, the relationship between oscillators and bosonswas really only tied together properly much later (in 1924) by Bose. But this is another interestingstory, discussed in the next section.

6.4 Bose-Einstein Condensation

The prediction of Bose-Einstein condensation (BEC) in 1924 remained more-or-less a curiosity inphysics for about 70 years, until gaseous BECs were first produced in the laboratory. In the mean-time, it formed the theoretical foundation for phenomena such as superfluidity in liquid helium andsuperconductivity, and led to the development of the laser; but in reality the BEC of non-interactingparticles envisaged by Bose and Einstein is very different from the phenomena in these condensedmatter systems. BEC is essentially what happens when the temperature is lowered sufficiently thatthe bosons reach quantum degeneracy (that is, the de Broglie wavelength approaches the meaninterparticle distance).

The basic problem with attaining a gaseous BEC is that when gases get cold enough, they condenseinto very ordinary liquids, and not into BECs. The trick that took 70 years is to try to keep theparticles from ordinary condensation before reaching BEC. It turned out that spin-polarized gaseswere ideal candidates, because the spin-spin interaction is van der Waals-like, which means onlyvery weakly attractive when the gases are sufficiently dilute. But if they are very dilute, it meansthat the mean separation is very large, which means ridiculously cold temperatures are needed toachieve BEC (on the order of nanoKelvins). The achievement of BEC followed years of experimentalprogress on trapping and cooling of atoms with lasers, which took place mostly in the ’70s. This isa fascinating story, and if these notes ever become a book, I’ll tell it in detail!

6.4.1 BEC in 3D

The mean number of atoms in a 3D box is

N =∑i

1

exp[β(εi − µ)]− 1=

V

2π2

∫ ∞0

k2dk

exp[β(εk − µ)]− 1.

Making the substitution x = (h2/2mkBT )k2 gives the mean density N/V :

n =1

4π2

(2mkBT

h2

)3/2 ∫ ∞0

√xdx

z−1ex − 1,

where z ≡ exp(βµ) is the fugacity, as seen previously. Now, the mean density is fixed. So as thetemperature decreases, clearly the integral also needs to increase in order to keep up. The only thingdepending on the temperature is the fugacity, which appears in the denominator: exp(−βµ) needsto keep decreasing which means that µ needs to keep increasing (i.e. approach zero from below).But recall that the chemical potential for the Bose case can never exceed the lowest accessibleenergy level ε0, otherwise the occupation of a given level could be negative. At some point as thetemperature drops, µ will hit zero, and the integral will no longer be able to properly count particles.


The particles will suddenly start disappearing! This is the phenomenon of Bose-Einsteincondensation, and the temperature at which this occurs is called the Bose-Einstein condensation(BEC) transition temperature, Tc.

Where have they gone? Notice that the integrand is proportional to√x ∝ k. The lowest energy

accessible energy level εk=0 = 0 isn’t strictly included in the integral! The particles that appearto be disappearing are simply piling up in the ground state of the system, the zero-energy k = 0state. The density of atoms in the ground state is denoted by n0. If the total density n is known(or assumed), then the number of atoms in the ground state can only be inferred from the numberthat we can count: n0 = n− n′, where n′ is the value of the integral above.

It’s easy to calculate the value of Tc: simply set µ = 0 or z = 1. Then we can use the handy integraldefining the Bessel functions

Jν−1 ≡∫ ∞

0

xν−1dx

ex − 1= Γ(ν)ζ(ν).

In our case, ν = 32 and Γ

(32

)≈ 2.612 and ζ

(32

)= 1

2Γ(

12

)=√π

2 . So for a 3D box, the BEC transitiontemperature occurs when

n =1

4π2

(2mkBTc

h2

)3/2

Γ

(3

2

)ζ

(3

2

).

Inverting this gives a simple relationship between the transition temperature and the particle density:

kBTc ≈ 3.31h2

mn2/3. (6.6)

How does the BEC temperature relate to the quantum degeneracy temperature, when the de Broglie

wavelength becomes comparable to the interparticle separation, λD = d? We have√

2πh2/mkBTd =

(1/n)1/3 or h2n2/3/mkBTd = 1/2π. Thus Td/Tc = 2π/3.31, which means that the degeneracyand BEC transition temperatures are basically the same, apart from some factors of order unity.Alternatively, the BEC condition is that the phase-space density be larger than 2.612:

nλ3D ≥ 2.612.

What is the fraction of atoms in the condensate for temperatures T < Tc?

n′ =1

4π2

(2mkBT

h2

)3/2

Γ

(3

2

)ζ

(3

2

)= n

(T

Tc

)3/2

.

Together with n0 = n− n′, we obtain

n0

n= 1− n′

n= 1−

(T

Tc

)3/2

. (6.7)

6.4.2 BEC in Lower Dimensions

Is there a BEC transition in 2D or 1D systems? The issue is somewhat interesting, and relevant tomodern physics, so let’s explore it briefly. The mean particle number is

N =

∫ ∞0

g(ε)dε

exp[β(ε− µ)]− 1,


where in two dimensions the density of states g2D(ε) = Am/2πh2 is independent of energy and in

one dimension it is given by g1D(ε) =√m/8π2h2(L/

√ε). In two dimensions, we obtain

N2D =mA

2πh2

∫ ∞0

dε

z−1 exp(βε)− 1=mAkBT

2πh2

∫ ∞0

dx

z−1ex − 1(substituting x = βε)

=A

λ2D

ln(1− ze−x

)∣∣∣∣∞0

= − A

λ2D

ln (1− z) .

What happens when z → 1−, i.e. when µ→ 0−? Then

limµ→0

n2D =1

λ2D

ln

(kBT

|µ|

)which diverges logarithmically! So it would seem that the density in two dimensions is undefined asT → 0, implying that BEC is impossible. But there is a loophole: BEC can happen at exactly zerotemperature. When µ approaches zero as temperature approaches zero, the ratio µ/kBT actuallyremains well-defined and finite. The transition is called the Kosterlitz-Thouless transition. The BECis pure at T = 0; as temperature increases, vortex-antivortex pairs are spontaneously produced,destroying the nice properties of the BEC.

In one dimension, the story is more or less the same. The mean density is given by

n1D =

√m

8π2h2

∫ ∞0

dε√ε [z−1 exp(βε)− 1]

.

The same substitution x = βε, and also z ≡ e−α, give

n1D =

√mkBT

2πh2

1√4π

∫ ∞0

dx√x [exp(x+ α)− 1]

≈ 1

λD√

4π

∫ ∞0

x−1/2dx

x+ α

where in the last line I have recognized that the dominant contribution to the integral comes fromthe singular portion in the vicinity of x = 0. This last integral is each to evaluate, giving

n =1

λD√

4π

[Γ(

12

)]2√α

=

√π

4α

1

λD.

Clearly, this diverges for α → 0+ or µ → 0−, and so there is no BEC in 1D at finite temperatureeither. In fact, µ/kBT also diverges so there is no BEC at any temperature in one dimension.

Actually, the absence of BEC in one and two dimensions at finite temperature is only true in infinitesystems. I know that I’ve been considering particles in various boxes and the length has been well-defined, but really it isn’t. The lowest energy state always has k = 0, or k = πn/L where n = 0.This means that the lowest energy state has a characteristic length scale proportional to k−1 =∞.In 1D and 2D, this gives rise to a so-called infrared divergence in the perturbation theory expansionof the gas, and therefore the singular behaviour. This is the same ‘infrared catastrophe’ that youmight have heard of in the context of quantum electrodynamics (QED), and it’s no surprise: themediators of the electromagnetic field are photons which have Bose-Einstein statistics. The problemgoes away when the bosons are truly confined in finite volumes, such as in harmonic oscillator traps,discussed in the next subsection. In this case, BEC persists at finite temperatures.


6.4.3 BEC in Harmonic Traps

The spin-polarized atomic gases that have been coaxed into BEC have all been trapped 3D harmonicpotentials, just like the ultracold fermions. So we can use some of the results found in that case,most importantly the density of states (5.11). Assuming that the energy level spacing is muchsmaller than the temperature (which isn’t really true for current experiments, but you have to startsomewhere!), the mean number of atoms is

N =

∫ ∞0

ε2dε

2(hω)3

1

exp[β(ε− µ)]− 1.

The BEC critical temperature is found by setting µ = 0 as for the box (if we had kept the zero-pointmotion, the we would set µ = h(ωx + ωz + ωz)/2 rather than zero). With the substitution x = βε,the mean number becomes:

N =1

2

(kBTchω

)3 ∫ ∞0

x2dx

ex − 1=

1

2

(kBTchω

)3

Γ(3)ζ(3),

where Γ(3) = 2 and ζ(3) ≈ 1.202. Inverting gives the transition temperature as a function of thenumber of atoms:

kBTc ≈ 0.94hωN1/3.

Notice the different power law than that for a BEC in a 3D box, Eq. (6.6). The temperature-dependence of the condensate number is then

N0

N= 1−

(T

Tc

)3

,

which again differs from the 3D box case, Eq. (6.7).

The first test that the atoms truly had formed a BEC was that the atomic cloud showed a bimodaldistribution. That is, the noncondensed atoms formed a cloud that looked more-or-less classical, andwas identical to the semiclassical (equipartition) profile of the Fermi gas (5.19):

n′(r) =1

λ3D

exp

[µ− µext(r)

kBT

]=

1

λ3D

exp

(− µ

kBT

)exp

(− x

2

R2x

)exp

(− y

2

R2y

)exp

(− z

2

R2z

),

where R2i = 2kBT/mω

2i . On the other hand, the atoms in the BEC all occupy the lowest energy

state of the 3D harmonic potential. This is the solution of the 3D Schrodinger equation, and I won’tbore you with the details, but give you the answer instead:

n0(r) =N0

π3/2axayazexp

(−x

2

a2x

− y2

a2y

− z2

a2z

),

where ai =√h/mωi. The prefactor guarantees that

∫d3rn0(r) = N0. Now, it is clear that both

the BEC and noncondensate densities are Gaussian, but the length scales over which the Gaussiansvary are very different. The length scales for the condensate depend only on the trap parameters,and are therefore quite small (of order microns); the length scales for the noncondensate depend ontemperature, and for T ∼ Tc are on the order of tens of microns. To sum up: because of the bigN0 factor in the BEC density, the total density looks like a sharp, small spike in the centre of anextended noncondensate density. This bimodial distribution can easily be seen simply by taking apicture of the atomic cloud with a CCD!

Physics 451 - Statistical Mechanics II - Course Notespeople.ucalgary.ca/~dfeder/451/notes.pdfPHYS 451 - Statistical Mechanics II - Course Notes 5 This expression was rst derived by

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