Physics 451 - II - Course Notes

David L. Feder

January 8, 2013 Contents

1 Boltzmann Statistics (aka The ) 3 1.1 Example: (1D) ...... 3 1.2 Example: Harmonic Oscillator (3D) ...... 4 1.3 Example: The rotor ...... 5 1.4 The (reprise) ...... 6 1.4.1 of States ...... 8 1.5 The Maxwell Speed Distribution ...... 10 1.5.1 Interlude on Averages ...... 12 1.5.2 Molecular Beams ...... 12

2 and the 14 2.1 Virial Theorem ...... 14 2.1.1 Example: ideal ...... 15 2.1.2 Example: Average of the ...... 15 2.2 ...... 16 2.2.1 Free revisited ...... 17 2.2.2 Example: Pauli Paramagnet ...... 18 2.3 Grand Partition Function ...... 19 2.3.1 Examples ...... 20 2.4 ...... 21

3 Quantum Counting 22 3.1 Gibbs’ Paradox ...... 22 3.2 Chemical Potential Again ...... 24 3.3 Arranging Indistinguishable ...... 25 3.3.1 ...... 25 3.3.2 ...... 26 3.3.3 ! ...... 28 3.4 Emergence of Classical Statistics ...... 29

4 Quantum Statistics 32 4.1 Bose and Fermi Distributions ...... 32 4.1.1 Fermions ...... 33 4.1.2 Bosons ...... 35 4.1.3 ...... 37 4.2 Quantum-Classical Transition ...... 39 4.3 Entropy and Equations of State ...... 40

1 PHYS 451 - Statistical Mechanics II - Course Notes 2

5 Fermions 43 5.1 3D Box at zero temperature ...... 43 5.2 3D Box at low temperature ...... 44 5.3 3D isotropic harmonic trap ...... 46 5.3.1 ...... 46 5.3.2 Low ...... 47 5.3.3 Spatial Profile ...... 48 5.4 A Few Examples ...... 50 5.4.1 in ...... 50 5.4.2 Electrons in the Sun ...... 50 5.4.3 Ultracold Fermionic in a Harmonic Trap ...... 51

6 Bosons 52 6.1 Quantum Oscillators ...... 52 6.2 Phonons ...... 53 6.3 Blackbody Radiation ...... 56 6.4 Bose-Einstein Condensation ...... 59 6.4.1 BEC in 3D ...... 59 6.4.2 BEC in Lower Dimensions ...... 60 6.4.3 BEC in Harmonic Traps ...... 62 Chapter 1

Boltzmann Statistics (aka The Canonical Ensemble)

This chapter covers the material in Ch. 6 of the PHYS 449 course notes that we didn’t get to last term. 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 the quantum levels. Because the system is known to exhibit periodic motion, we can again use Bohr-Sommerfeld quantization and avoid having to solve Schr¨odinger’sequation. The total energy is p2 kx2 p2 mω2x2 E = + = + , 2m 2 2m 2 √ where ω = pk/m is the classical frequency. Inverting this gives p = 2mE − m2ω2x2. Insert this into the Bohr-Sommerfeld quantization condition:

I I p pdx = 2mE − m2ω2x2dx = nh, where the integral is over one full period of oscillation. Let x = p2E/mω2 sin(θ) so that m2ω2x2 = 2mE sin2(θ). Then

I r Z 2π √ 2E 2 2E 1 2πE pdx = 2 2mE cos (θ)dθ = 2π = = nh. mω 0 ω 2 ω

So, again making the switch E → n, we obtain hω  = n = n¯hω. n 2π The full solution to Schr¨odinger’sequation (a lengthy process involving Hermite polynomials) gives 1 n = ¯hω(n + 2 ). Except for the constant factor, Bohr-Sommerfeld quantization has done a fine job of determining the energy states of the harmonic oscillator.

3 PHYS 451 - Statistical Mechanics II - Course Notes 4

Armed with the energy states, we can now obtain the partition function: X X Z = exp(−n/kBT ) = exp(−βn) = 1 + exp(−β¯hω) + exp(−2β¯hω) + .... n n

But this is just a : if I make the substitution x ≡ exp(−β¯hω), then Z = 1 + x + x2 + x3 + .... But I also know that xZ = x + x2 + x3 + .... Since both Z and xZ have an infinite number of terms, I can subtract them and all terms cancel except the first: Z − xZ = 1, which immediately yields Z = 1/(1 − x), or 1 Z = . (1.1) 1 − exp(−β¯hω) Now I can calculate the mean energy:

2 2 ∂ ln(Z) NkBT ∂Z 2 [1 − exp(−β¯hω)] ¯hω U = NkBT = = NkBT 2 (−1) 2 (−1) exp(−β¯hω) ∂T Z ∂T [1 − exp(−β¯hω)] kBT exp(−β¯hω) N¯hω = N¯hω = . 1 − exp(−β¯hω) exp(β¯hω) − 1 1 = N¯hωhn(T )i, where hn(T )i ≡ is the occupation factor. exp(¯hω/kBT ) − 1

At very high temperatures T  1, exp(¯hω/kBT ) ≈ 1 + (¯hω/kBT ), so hn(T )i → 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-dimensional 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

nx,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 can 3 be simply written as the product of three 1D partition functions, i.e. Z3D = (Z1D) . Because 3 almost all thermodynamic quantities are related to ln (Z3D) = ln (Z1D) = 3 ln (Z1D), almost all quantities will simply be mupltiplied by a factor of 3. For example, U3D = 3NkBT = 3U1D and CV (3D) = 3NkB = 3CV (1D).

One can think of atoms in a crystal as N point masses connected to each other with springs. To a 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 room temperature, accounting for 92% of this simple classical picture. It is interesting to consider the expression for the specific heat at low temperatures. At low temperature, the mean energy goes to U → 3N¯hω exp(−¯hω/kBT ), so that the specific heat approaches

3N¯hω  ¯hω   ¯hω 2  ¯hω  CV → − 2 (−¯hω) exp − = 3NkB exp − . kBT kBT kBT kBT PHYS 451 - Statistical Mechanics II - Course Notes 5

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

1.3 Example: The rotor

Now let’s consider the energies associated with . In , the rotational is 1 T = ~ω · I · ~ω, 2 where I is the tensor and ~ω is the angular vector. In the inertial ellipsoid, this can be rewritten L2 L2 L2 T = x + y + z , 2Ixx 2Iyy 2Izz where Lj is the angular momentum along direction ˆand Ijj is the corresponding moment of inertia. Suppose that we have a spherical top, so that Ixx = Iyy = Izz = I:

1 L2 T = L2 + L2 + L2 = . 2I x y z 2I In the quantum version, the kinetic energy is almost identical, except now the angular momentum is an operator, denoted by a little hat: Lˆ2 T = . 2I The eigenvalues of this operator are `(` + 1)¯h2/2I, where ` = −L, −L + 1, −L + 2,...,L − 1,L so that ` can take one of 2L + 1 possible values.

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

∞ L ∞ X X  `(` + 1)¯h2  X  L(L + 1)¯h2  Z = exp − ≈ (2L + 1) exp − , 2IkBT 2IkBT L=0 `=−L L=0 where the second term assumes that the contributions from the different ` values are more or less equal. This assumption should be pretty good at high temperatures where the argument of the exponential 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 can approximate the sum by an integral:

Z ∞  L(L + 1)¯h2  Z ≈ (2L + 1) exp − dL. 0 2IkBT PHYS 451 - Statistical Mechanics II - Course Notes 6

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

Z ∞  2  ¯h x 2IkBT Z = exp − dx = 2 . 0 2IkBT ¯h Again, I can calculate the mean energy

2 2 2 ∂ ln(Z) NkBT ∂Z 2 ¯h 2IkB U = NkBT = = NkBT 2 = NkBT. ∂T Z ∂T 2IkBT ¯h This is exactly the contribution that we expected from the equipartition theorem: there are two ways the linear top can rotate, so there should be two factors of (1/2)NkBT contributing to the energy.

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

∞ L  2  ∞  2  X X `(` + 1)¯h X 2 L(L + 1)¯h Z = (2L + 1) exp − ≈ (2L + 1) exp − , 2IkBT 2IkBT L=0 `=−L L=0 where the second term assumes that the contributions from the different ` values are more or less equal. This assumption should be pretty good at high temperatures where the argument of the exponential 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 ∞  L(L + 1)¯h2  Z ≈ (2L + 1)2 exp − dL. 0 2IkBT At high temperatures, one needs large values of L before the argument of the exponentials will be significant, so it is reasonable to make the substitution L(L + 1) → L2 and (2L + 1)2 → 4L2. This yields Z ∞  2 2   3/2 2 ¯h L √ 2IkBT Z = 4L exp − dL = π 2 . 0 2IkBT ¯h Again, I can calculate the mean energy

2  2 3/2  3/2 2 ∂ ln(Z) NkBT ∂Z 2√ ¯h 1 3 2IkB 1/2 3 U = NkBT = = NkBT π √ 2 T = NkBT. ∂T Z ∂T 2IkBT π 2 ¯h 2 For the speherical top, there are now three contributions to the total energy, which accounts for the additional 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 mean energy for the particle-in-the-box and harmonic oscillator problems were very similar: the mean energies were all some multiple of NkBT/2 and the specific heat some multiples of NkB/2. Notice PHYS 451 - Statistical Mechanics II - Course Notes 7 that for both the particle in the box and the harmonic oscillator, both quantities were three larger 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 freedom of the particles in a given system.

To make further progress with this idea, we need to revisit the hokey derivation of the for an presented in the PHYS 449 course notes. Recall that, in order to enumerate all the accessible states in a 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 the entropy was given by S = kB ln(V/∆V ). But tells us that we can’t know both the exact position and momentum of a particle at the same . The relationship between the uncertainty in the position ∆x and the momentum ∆p is quantified in the Heisenberg uncertainty principle ∆x∆p ≥ h, where h again is Planck’s constant. For our purposes, this means that Planck’s constant sets a fundamental limit on the size of the volumelets on can partition the system into. In a sense, the Bohr-Sommerfeld quantization condition already stated this: the integral of momentum over space is minimally Planck’s constant.

Mathematically, this means that we can write the general partition function in one dimension as a continuous function Z ∞ Z ∞   1 (x, px) Z1D → dx dpx exp − , h −∞ −∞ kBT where the accessible classical states  are explicitly assumed to be functions of both position x and momentum px. In three dimensions, it would be Z Z   1 3 3 (r, p) Z3D → 3 d r d p exp − . h kBT The six dimensions (r, p) together constitute , and the two three-dimensional integrals are denoted phase-space integrals.

To make these ideas more concrete, let’s again calculate the partition function for a particle in a 1D box of length L. The classical energy is entirely kinetic, so (p) = p2/2m, and Z L/2 Z ∞  2  r 1 p 1 √ p mkBT L Zbox = dx dp exp − = L π 2mkBT = L 2 = , h −L/2 −∞ 2mkBT h 2π¯h 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 (x, p) = p2/2m + mω2x2/2. Now the partition function is Z ∞ Z ∞ p2 mω2x2 ! 1 2m + 2 Zh.o. = dx dp exp − h −∞ −∞ kBT 1 Z ∞  mω2x2  Z ∞  p2  = dx exp − dp exp − . h −∞ 2kBT −∞ 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 √ r π 2kBT √ p kBT Z = π 2mk T = , h.o. 2π mω2 B ¯hω PHYS 451 - Statistical Mechanics II - Course Notes 8 which is identical to the high-temperature limit of the 1D harmonic oscillator partition function (6.1): 1 1 k T lim = ≈ B . T 0 1 − exp(−β¯hω) 1 − (1 − β¯hω) ¯hω The mean energy in this limit is

2 2 NkBT ∂Z NkBT ¯hωkB U(T  0) = ≈ = NkBT. Z ∂T kBT ¯hω

But we already know that the second of the two integrals contributed NkBT/2 to the mean energy U at high temperature, because this was the result for the particle in the 1D box. This means that the first integral also contributed NkBT/2 to the mean energy. This is very much analogous to each spatial integral contributing a factor of NkBT/2 to the mean energy going from 1D to 3D. If you think of the kinetic and terms in the classical expression for the energy as each standing for one degree of freedom, then one can finally state the D Equipartition Theorem: Each degree of freedom that contributes a term quadratic in position or momentum to the classical single-particle energy contributes an average energy of kBT/2 per particle. Many examples showing how one can predict the value of U at high temperatures will be covered in class.

1.4.1 Density of States

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

¯h2π2  = n2 + n2 + n2 . n 2mL2 x y z We can think of the energy levels as giving us the coordinates of objects on the surface of a sphere in Cartesian coordinates, except only in the first octant (because {nx, ny, nz} ≥ 0). So, instead, we can think of the energies as continuous,  = γn2, where γ =h ¯2π2/2mL2 and n2 is the length of the vector in ‘energy space.’ So, n = p/γ and therefore

1 d dn = √ √ . 2 γ 

In spherical coordinates, we can write

1 dn = d3n = dn dn dn = n2 sin(θ)dndθdφ → 4πn2dn (integrating over angles) x y z 8 √ π    1 d π d = √ √ = ≡ g()d. 2 γ 2 γ  4γ3/2

So, the density of states per unit energy g() for a particle in a 3D box is given by √ dn π √ m 2m√ V (2m)3/2 √ g() ≡ =  = V  = , d 4γ3/2 2¯h3π2 4¯h3π2 PHYS 451 - Statistical Mechanics II - Course Notes 9 where in the last line I have put back in the explicit form for γ. The most√ important thing to notice is that the density of states per unit energy for the 3D box goes like .

If you found this derivation confusing, here’s another one. One way to think about the energy levels for 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):

¯h2 πn2 ¯h2k2  = ≡ , k 2m L 2m which is just the usual expression for the kinetic energy if you recognize that p ≡ ¯hk. This is known as 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: L3  g()d = k2 sin(θ)dθdφdk . π3 Integrating over the angles gives

V V dk  g()d = k2dk ⇒ g () = k2 2π2 3D 2π2 d for the 3D box. Now,  =h ¯2k2/2m so k = p2m/¯h. Putting it together we obtain √ V V 2mr2m 1 V (2m)3/2  g()d = k2dk = √ d = d, 2π2 2π2 ¯h2 ¯h2 2  2π2 2¯h3 as we obtained above.

It is straightforward to generalize the 3D box to the 2D and 1D cases. Going through the same procedure gives Ak dk  L dk  g () = ; g () = . 2D 2π d 1D π d

Explicitly using the free-particle dispersion relation as above shows that the√ density of states per unit volume is a constant (independent of energy) for 2D, and goes like 1/  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() is the integral of the density of states up to energy state :

 3/2 3/2 2 3/2 Z V (2m) √ 0 V (2m) π 2mL  G() = g(0)d0 =  d0 = 3/2 =  . 2 3 2 3 2 2 0 2π 2¯h 6π ¯h 6 ¯h π

3 At room temperature,  = 2 kBT where T = 300 K. So this means that the total number of states accessible for in a 1 m box at room temperatures is approximately G() ≈ (π/6)(1020)3/2 ∼ 1030. But the density of air at sea level is about 1.25 kg/m3, or about 1025 per m3. So even with these rough estimates it is clear that there are far more states accessible to the molecules in the air at room temperature than there are particles. Alternatively, the probability that a given energy state has a particle in it is something like 10−5. PHYS 451 - Statistical Mechanics II - Course Notes 10

1.5 The Maxwell Speed Distribution

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

X Z Z N = ni ⇒ dn() = dg() exp(−β) i A(2m)3/2V Z ∞ √ = d  exp(−β), in three dimensions. 3 2 4¯h π 0

3 Because Z = N/A, this expression also quickly yields Z = V/λD in the usual way. Alternatively, we can set A = N/Z and obtain

3/2 √ N  2π¯h2  (2m)3/2V √ 2π N n() = Ag() exp(−β) =  exp(−β) = exp(−β). 3 2 3/2 V mkBT 4¯h π (πkBT )

2 √Now suppose that the energy levels were simply classical kinetic energy states:  = mv /2, so that  = pm/2v and d = mvdv. Then,

r 2  m 3/2  βmv2  n(v)dv = N v2dv exp − . π kBT 2 This is the Maxwell- of . The important thing to notice is that for small velocities, the distribution increases quadratically, n(v)dv ∝ v2, while for large velocities it decreases 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 consequences for the statistics.

Let’s obtain the mean velocity v first:

 2  R R ∞ 3 βmv vn(v)dv 0 v exp − 2 dv v = = , R  2  n(v)dv R ∞ 2 βmv 0 v exp − 2 dv where I’ve cancelled all the common constant terms. To turn these into integrals we can evaluate, 2 2 p let’s set x ≡ βmv /2, so that v = x 2kBT/m. Then

r ∞ 2k T R x3 exp −x2 dx v = B 0 . m R ∞ 2 2 0 x exp (−x ) dx Now it is very useful to know the following integrals: Z ∞ √ Z ∞ 2n 2 2 (2n)! π 2n+1 2 2 n! x exp(−a x )dx = 2n and x exp(−a x )dx = 2n+2 . 0 n!2(2a) a 0 2a In the current case, we have a = 1 and r r r 2k T 1 8 8k T k T v = B √ = B ≈ 1.596 B . m 2 2 π πm m PHYS 451 - Statistical Mechanics II - Course Notes 11

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

∞  βmv2  R 2 R 4 v n(v)dv 0 v exp − 2 dv v2 = = R  2  n(v)dv R ∞ 2 βmv 0 v exp − 2 dv ∞ √ 2k T R x4 exp −x2 dx 2k T 24 π 8 3k T = B 0 = B √ = B m R ∞ 2 2 m 64 2 π πm 0 x exp (−x ) dx r r 3k T k T ⇒ ∆v = B ≈ 1.73 B . πm m

The most probable speed v˜ corresponds to the point at which the distribution is maximum: " # ∂ r 2  m 3/2  βmv2  N v2dv exp − = 0 ∂v π kBT 2  βmv2   m   βmv2  2˜v exp − − v˜2 2˜v exp − = 0 2 2kBT 2 r r 2k T k T ⇒ v˜ = B ≈ 1.414 B . m m The amazing thing about the Maxwell-Boltzmann distribution of velocities is that all the quantities v, ∆v, andv ˜ 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. So the 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’s estimate the velocity a particle at the Earth’s surface would need to fully escape the gravitational field. The balance of kinetic and potential energies implies (1/2)mv2 = mgR, where R is the radius of the Earth. (Actually I think that this implies that the particle is at the√ Earth’s center, where all of it’s mass would be concentrated?). In any case, we obtain vescape = 2gR ≈ 11 000 m/s. The −27 p mean velocity for molecules (mass of 1.66×10 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 6 times the average velocity. So there are lots of hydrogen molecules escaping forever all the time. Thankfully, our supply is continually replenished by bombarding us from the sun. Much more serious is , which is being lost irretrievably, with no replenishing. In fact, the U.S. Government has been stockpiling huge reserves of for years in preparation of a world- wide shortage. But over the past few years the current administration has softened its policy on PHYS 451 - Statistical Mechanics II - Course Notes 12 helium conservation and these stockpiles are slowly dwindling. In any case, for and nitrogen molecules, whose mean velocities are about a factor of 4 lower than that of hydrogen, very few of these 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 the energy,  = mv2/2. Of course, if the energy levels are different, or the dimension of the problem is not 3D like it was above, then the way we take averages is going to be different. So how does one take averages in general using the canonical ensemble?

Recall that the general form for the average of something we can measure, call it B, is X B = piBi, i where the sum is over all the accessible states of the system, and pi are the probabilities of occupying those states. In the canonical ensemble, those probabilities are

g exp(−β ) p = i i , i Z where Z is the usual partition function, and I have explicitly inserted the degeneracy factor. So the average of B is defined as

∞ P g B exp(−β ) R dk g(k)B(k) exp[−β(k)] R dg()B() exp(−β) B = i i i i ≈ = 0 . P g exp(−β ) R dk g(k) exp[−β(k)] R ∞ i i i 0 dg() exp(−β) Thus, the dimensionality, the dependence of the energy  on the wavefector k or velocity v, and the inherent degeneracy of a given energy level are all buried in the density of states per unit energy, g(). 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, where (k) = αk3/2. This is an effective 2D system, so we use the expression for the 2D density of states,

Ak dk Ak d   2/3 A   2/3  2  A g() = = = = 1/3. 2π d 2π d α 2π α 3α2/31/3 3πα4/3

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 per particle for this problem would be

∞ R d 4/3 exp(−β) (k T )7/3Γ 7  4k T U = 0 = B 3 = B . R ∞ 1/3 4/3 4  3 0 d  exp(−β) (kBT ) Γ 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 about it! Suppose that we have an oven containing lots of hot molecules. There’s a small hole in one end, out of which shoot the molecules. On the same table in front of the hole is a series of walls PHYS 451 - Statistical Mechanics II - Course Notes 13 with small holes lined up horizontally with the exit hole of the oven. The idea here is that most of the molecules moving off the horizontal axis will hit the various walls, and only those molecules moving 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 is area 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

dθ sin(θ)dφ number of particles = Avt cos(θ)n(v) dv dΩ = Avt cos(θ)n(v) dv . 4π The flux of molecules f(v)dv is the number of molecules per unit area per unit time,

vn(v)dv Z π/2 Z 2π vn(v)dv f(v)dv = dθ cos(θ) sin(θ) dφ = . 4π 0 0 4 For the first integral, set x = sin(θ) so that dx = cos(θ). The integral is therefore x2/2 = 2 π/2 sin (θ)/2|0 = 1/2. And the second integral is 2π. Using the Maxwell-Boltzmann distribution of velocities, we obtain the flux density as

 3/2  2  3 3  2  Nπ 2m 3 mv NλD m 3 mv f(v) = v exp − = 2 v exp − . 8 πkBT 2kBT 8π ¯h 2kBT

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

Z Z π/2 Z 2π sin(θ)dθdφ P = dv dθ dφ2p cos(θ)v cos(θ)n(v)dv 0 0 4πV 1 Z ∞ m Z ∞ Nm Nm 3k T Nk T = mvn(v)v dv = v2n(v)dv = v2 = B = B . 3V 0 3V 0 3V 3V m V A slightly less hokey derivation (maybe) is to say that the pressure per unit volume is the mean force along the horizontal per unit area:

F Z N P = = dvf(v)mv2 = mv2. A x V x

2 2 2 2 2 But v = vx + vy + vz = 3vx. So Nm Nk T P = v2 = B . 3V V Chapter 2

Virial Theorem and the Grand Canonical Ensemble

2.1 Virial Theorem

Before launching into the theory of quantum , it’s useful to learn a powerful thing in statistical P mechanics, 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 X X X X = (p˙ · r + r˙ · p ) = p˙ · r + mv2 = 2K + p˙ · r = 2K + F · r , dt i i i i i i i i i i i i i i i 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 some fluctuating quantity (call it A(t)) is defined as hA(t)it: 1 Z τ hA(t)it ≡ A(t)dt. τ 0 We can use this to define the following time average:

dG 1 Z τ dG G(τ) − G(0) = dt = . dt t τ 0 dt τ Now here comes the crucial part. If G(t) is a periodic function, with period P , then clearly for τ = P the time average must be zero since G(τ) = G(0). Likewise, if G has finite values for all times, then as the elapsed time τ gets to be very large the average will also go to zero. This means that dG = 0 dt t as long as G is finite. Finally we obtain the virial theorem * + 1 X hKi = − F · r . (2.1) t 2 i i i i

14 PHYS 451 - Statistical Mechanics II - Course Notes 15

Apparently, the word ‘virial’ comes from the Latin word for ‘force’ though I remember from my high school Latin that force was ‘fortis’ meaning strength. Oh well. You see anyhow where the force comes 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 ideal gas (again! this probably won’t be the last time either). We’ve already seen from the equipartition theorem 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 the collide elastically, dFi = −P ndAˆ for all particles, wheren ˆ is the normal to the wall surface (this is nothing but a restatement of P = dF/dV ). So

X Z Z Fi · ri = −P nˆ · rdA = −P ∇ · rdV = −3PV i 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 is 30 7 M ≈ 3 · 10 kg and its radius is R ≈ 3 · 10 m. Assuming that the sun is made up entirely of hydrogen with mass 1.67 · 10−27 kg, then this gives N = 1.8 · 1057 atoms and therefore a density ρ = 1.6 × 1034 m−3. The force on a particle of mass m due to the of the sun is F = 2 −11 3 2 −(GM m/R )ˆr, where G ≈ 6.674 × 10 m /kg/s is the , and M and R are the mass and radius of the sun, respectively. The density of the sun (assuming that it’s constant 3 everywhere, which isn’t likely to be the case!) is ρ = M /V = M /(4πR /3). Now, we’d like to sum up all the gravitional forces from the sun’s center to the surface. The radial-dependent mass is therefore M 4πr3  r 3 M(r) = = M . 4πR3 3 R 3 Also, for the test mass we have dm = ρdV = ρr2 sin(θ)dr dθ dφ. Putting it all together, we have

1 X 1 Z G r3 K = − F · r = r2 sin(θ) dr dθ dφ M ρrˆ · r, 2 i i 2 r2 R3 i where K is the mean kinetic energy. Butr ˆ · r = (r/r) · r = r, so

Z R 5 2 GM ρ 4 2πGM ρR 3 GM K = 3 4π r dr = 3 = , 2R 0 5R 10 R where in the last part I used the expression for the density ρ. Because K = (3/2)NkBT by equipar- 2 6 tition, we obtain T ≈ (1/5)(GM /NkBR ) ≈ 5 × 10 K. In fact, except for the factor 5 I could have guessed this from , because there is only one way to combine G, M , and R in a way that gives units of energy. In fact, the temperature of the sun varies from about 10 million degrees at the center to about 6 000 degrees at the surface. So the virial theorem has done a pretty good job of estimating the average temperature. PHYS 451 - Statistical Mechanics II - Course Notes 16

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 ∂U dU = T dS − P dV + dN ≡ T dS − P dV + µdN. ∂N This can also be inverted to give dU + P dV − µdN dS = . T So the chemical potential for the system is defined as

 ∂U   ∂S  X µ ≡ = −T ,N = n . ∂N ∂N i S,V U,V i∈subsystem But remember to be careful when using the second definition of the chemical potential, because

T = (∂U/∂S)V,N .

This reminds me about general forces: here S and T are conjugate variables linked through the total 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 the chemical potential above shows that µ and N are also conjugate variables: µ is the generalized force associated 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 change of the of the reservoir SR and subsystem Ss:   ∂ ∂SR ∂Ss ∂SR ∂NR ∂Ss dSsystem = (SR + Ss) dNs = dNs + dNs = dNs + . ∂Ns ∂Ns ∂Ns ∂NR ∂Ns ∂Ns

But NR = N − Ns so ∂NR/∂Ns = −1 giving   ∂Ss ∂SR dNs dSsystem = dNs − = − (µs − µR) . ∂Ns ∂NR T If you didn’t like this sloppy math, then how about this:

∂Ss ∂SR dSsystem = dNs + dNR. ∂Ns ∂NR

Since dNR = −dNs this gives   ∂Ss ∂SR dSsystem = dNs − , ∂Ns ∂NR which is what we obtained above.

Equilibrium corresponds to maximizing entropy, or dS = 0. This means that the condition for equilibrium between the subsystem and the reservoir is that the chemical potentials for each should PHYS 451 - Statistical Mechanics II - Course Notes 17

be equal, µs = µR. But even more important, as equilibrium is being approached, the entropy is changing with time like dS dN µ − µ  system = − s s R ≥ 0 dt dt T because the entropy must increase toward equilibrium (unless they are already at equilibrium). If initially µR > µs, then clearly dNs/dt ≥ 0 to satisfy the above inequality. This means that in order to reach equilibrium when the chemical potential for the reservoir is initially larger than that of the subsystem, particles must flow from the reservoir into the subsystem. So the chemical potential provides 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 chemical potential µ. Then the change in , and the changes in the , Helmholtz, and Gibbs free energies are given respectively by

dU = T dS − P dV + µdN;

dH = T dS + V dP + µdN; dF = −SdT − P dV + 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 from their cases seen above without allowing for changes to the total particle number! So we obtain the following rules:  ∂U  ∂H   ∂F   ∂G  µ = = = = . (2.2) ∂N S,V ∂N S,P ∂N T,V ∂N T,P 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

 ∂F  µ = . 1 ∂N 1 T,V,N2,N3,...

Let’s explore this last expression a bit more:

 ∂F  ∂N ln(Z) ∂ ln(Z) µ = = −kBT = −kBT ln(Z) − NkBT . ∂N T,V ∂N T,V ∂N T,V

Because Z = N/A, this last relation can be solved directly: ∂ ln(N/A)/∂N = ∂ ln(N)/∂N = 1/N so that µ = −kBT [ln(Z) + 1]. At high temperatures, the arguments of the exponentials in Z will all be small, which implies that Z > 0. This means that the chemical potential for classical particles is always negative. At low temperatures it might approach zero or get positive. This possibility will be explored a bit later in the term when we discuss quantum statistics. PHYS 451 - Statistical Mechanics II - Course Notes 18

2.2.2 Example: Pauli Paramagnet To clarify the various meanings of the chemical potential, let’s return first to the Pauli paramagnet. 1 Recall that in the , the entropy for the - 2 case was hn  n   n   n i S = −Nk 1 ln 1 + 1 − 1 ln 1 − 1 . B N N N N Assuming that we have some subsystem with number N in contact with a reservoir at temperature T , one might na¨ıvely assume that the chemical potential is defined as   ∂S N  n1  n2  µ = −T = −kBT ln = kBT ln 1 − = kBT ln ∂N N − n1 N 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-down atoms. But we don’t know from the entropy expression above which spin direction corresponds to the lower energy. This means that the answer should have been symmetric in n1 and n2, which it is definitely not. In fact, there is a fundamental error. As N varies, so also do the values of n1 and n2. In other words, n1 = n1(N) and n2 = n2(N). When taking this derivative we have to remember that the items kept constant are the volume (in this case the external magnetic field) and the total energy U.

So to calculate µ properly we need to convert n1 and n2 into explicit functions of U and N. Recally that n1 + n2 = N and −n1 + n2 = U. The second of these gives n2 − n1 = U/ and combining with the first gives 2n2 = N + U/e or n2/N = (1 + U/N)/2. We immediately then obtain n1/N = (1 + U/N)/2. The expression for the entropy is then:

1  U  1  U  1  U  1  U  S = −Nk 1 − ln 1 − + 1 + ln 1 + . B 2 N 2 N 2 N 2 N

Now we can calculate the chemical potential using the formula above. After some straightforward algebra one obtains n n  µ = k T ln 1 2 . B N 2

This result looks better, because it is symmetric under the exchange of labels n1 ↔ n2. At high temperatures n1 = n2 = N/2, and the chemical potential becomes large and negative, just as in the example for the 3D box. At low temperatures, n1 → 0 or n2 → 0, which makes the ln diverge logarithmically, but it’s o.k. because the temperature is going to zero linearly (i.e. fundamentally faster 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 zero chemical potential means that the change in the number of particles in the reservoir is not related to the change in the number of particles in the subsystem; alternatively, the entropy is invariant under changes in the number of particles. This implies that a zero chemical potential means that the system doesn’t conserve the number of particles. For the Pauli paramagnet, I can keep 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. PHYS 451 - Statistical Mechanics II - Course Notes 19

2.3 Grand Partition Function

Recall that in the derivation of the Boltzmann distribution in the canonical ensemble, we maximized P the entropy (or the number of microstates) subject to the two constraints N = ni and U = P i i nii. So we had ! ! X X d ln(Ω) + α dN − dni + β dU − idni = 0, i i where α and β were the Lagrange multipliers fixing the two constraints (alternatively, they are unknown multiplicative constants in front of terms that are zero). So we immediately have the following relations: ∂ ln(Ω) ∂ ln(Ω) α = ; β = . ∂N ∂U

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

∂ ln(Ω) ∂ ln(Ω) 1 µ = −kBT ; = . ∂N ∂U 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 find Ω 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 obtain Ωs = A exp [−β (Es − µNs)] . The physical system is actually comprised of a very large number of these subsystems, all of which have the same temperature and chemical potential at equilibrium, but all of which have different energies Ei and number of particles Ni. The total number of ways of distributing the particles P Ω 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,

Ωi exp [−β (Ei − µNi)] exp [−β (Ei − µNi)] pi = = P ≡ , Ω i exp [−β (Ei − µNi)] Ξ where Ξ is the grand partition function, X Ξ ≡ exp [−β (Ei − µNi)] . i PHYS 451 - Statistical Mechanics II - Course Notes 20

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

P (s ) Ω (s ) eSR(s2)/kB  1  2 = R 2 = = exp [S (s ) − S (s )] . S (s )/k R 2 R 1 P (s1) ΩR(s1) e R 1 B kB We know that 1 1 dS = (dU + P dV − µdN ) = − (dU + P dV − µdN ) R T R R R T s s s 1 = − {E(s ) − E(s ) − µ [N(s ) − N(s )]} . T 2 1 2 1 So we obtain n 1 o exp − [Es(s2) − µNs(s2)] P (s2) kB T = n o. P (s1) 1 exp − [Es(s1) − µNs(s1)] kB T The probability of occupying any state s is then

n 1 o exp − k T [Es(s) − µNs(s)] P (s) = B , 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 the way it was done in the canonical ensemble. Suppose that the subsystem contains three accessible energy levels 0, , and 2. Now suppose that there is only one in the total system. How many ways 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  (Es = , Ns = 1), or I might have one atom in state 2 (Es = 2, Ns = 1). So my grand partition function is:

Ξ = exp[−β(0 − µ0)] + exp[−β(0 − µ1)] + exp[−β( − µ1)] + exp[−β(2 − µ1)] = 1 + exp(βµ) [1 + exp(−β) + exp(−2β)] .

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

Ξ = 1 + exp(βµ) [1 + exp(−β) + exp[−2β)] + exp(2βµ) [1 + exp(−β) + 2 exp(−2β) + exp(−3β) + exp(−4β)] , where in the second line I have recognized that with two particles we can have Es = 0 (both in state 0), Es =  (one in state 0, the other in state ), Es = 2 (either both in state , or one in state 0 while the other is in state 2, thus the factor of two out front of this term), Es = 3 (one in state , the other in state 2), and Es = 4 (both in state 2). PHYS 451 - Statistical Mechanics II - Course Notes 21

Now suppose that the number of particles in our system is totally unknown, and also there is no restriction on the number of particles that can exist in a particular energy level. Then the grand partition function is

Ξ = 1 + exp(βµ) [1 + exp(−β) + exp[−2β)] + exp(2βµ) [1 + exp(−β) + exp[−2β)]2 = 1 + eβµZ + e2βµZ2 + ... ∞ X n 1 = eβµZ = , 1 − eβµZ n=0 where I have used Euler’s solution in the last line. In other words, the grand canonical ensemble for the fully unrestricted case is simply a linear combination of the canonical partition functions for no particles, one particle, two particles, etc., suitably weighted by their fugacities. In fact, this classical grand partition function leads to slightly paradoxical results, which will be discussed at length next term. 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 grand partition function instead of the regular partition function. The entropy is defined as

X X U µN S = −k p ln(p ) = −k p [−β(E − µN ) − ln(Ξ)] = − + k ln(Ξ), B i i B i i i T T B i i P P 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 mechanics books. Recall that in the canonical ensemble we had U − TS = F = −kBT ln(ZN ), where F is the . So the grand potential is simply related to the free energy by ΦG = F − µN. Anyhow, the following thermodynamic relations immediately follow:

∂Φ ∂Φ ∂ ln(Ξ) S = − G ; N = − G = k T . ∂T ∂µ B ∂µ Two other relations that follow in analogy with the results for the canonical ensemble are:

∂Φ ∂ ln(Ξ) ∂ ln(Ξ) P = − G ; U = − = k T 2 . ∂V ∂β B ∂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 to realize this was Gibbs (of Free Energy fame!), so it is called Gibbs’ Paradox. Basically, he showed that there was a problem with the definition of entropy that we have been using so far. The only way to 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 that particles can occupy. But so far, we haven’t been concerned with the particles themselves. In this section 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

U ∂ ln(Z) S = Nk ln(Z) + = Nk ln(Z) + Nk T . B T B B ∂T

3 As we saw last term, the partition function for a monatomic ideal gas in a 3D box is Z = V/λD, q 2 where V is the volume and λD = 2π¯h /mkBT is the de Broglie wavelength. Plugging this in, we obtain 3 2π¯h2  3 ln(Z) = ln(V ) − ln + ln(T ). 2 mkB 2 Evidently, ∂ ln(Z)/∂T = (1/Z)∂Z/∂T = 3/2T so

 3 2π¯h2  3 3  3  S = NkB ln(V ) − ln + ln(T ) + ≡ NkB ln(V ) + ln(T ) + σ , 2 mkB 2 2 2 where 3  2π¯h2  σ = 1 − ln 2 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 that divides the box into two regions of equal volume V/2, each containing N/2 atoms. Now the total

22 PHYS 451 - Statistical Mechanics II - Course Notes 23 entropy is the sum of the entropies for each region,

N  V  3  N  V  3  S = k ln + ln(T ) + σ + k ln + ln(T ) + σ 2 B 2 2 2 B 2 2  V  3   3  = Nk ln + ln(T ) + σ 6= Nk ln(V ) + ln(T ) + σ . B 2 2 B 2

In other words, simply putting in a barrier seems to have reduced the entropy of the particles in the box by a factor of NkB ln(2). Recall that this is the same as the high-temperature entropy for the two-level Pauli paramagnet (a.k.a. the coin). So it seems to suggest that the two sides of the partition once the barrier is up take the place of ‘heads’ or ‘tails’ in that there are two kinds of states atoms can occupy: the left or right partitions. But here’s the paradox: putting in the barrier hasn’t done any work on the system, or added any heat, so the entropy should be invariant! And simply removing the barrier brings the entropy back where it was before. Simply reducing the capability of 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 atoms on the left or on the right.

Clearly, we are treating particles on the left and right as distinguishable objects, which has given rise to this paradox. How can we fix things? The cheap fix is to realize that because all of the particles are fundamentally indistinguishable, the N! permutations of the particles among themselves shouldn’t lead to physically distinct realizations of the system. So, our calculation of Z must be too large by a factor of N!: ZN Z ≡ ZN = old , N correct N! where I explicitly make use of the fact that the N-particle partition function is N-fold product of the 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

 3    V  3  S = Nk ln(V ) + ln(T ) + σ − ln(N) + 1 = Nk ln + ln(T ) + σ + 1 . B 2 B N 2

Now, if we partition the box into two regions of volume V/2, each with N/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 function ZN are now ∂ ln(Z ) ∂ ln(Z ) U = k T 2 N ; S + k ln(Z ) + k T N ; F = −k T ln(Z ). B ∂T B N B ∂T B N That is, the expression are identical to the ones we’ve seen before, except the N factors in front have disappeared, and the Z is replaced by a ZN . PHYS 451 - Statistical Mechanics II - Course Notes 24

3.2 Chemical Potential Again

Now let’s return to the particle in the 3D box. Including the Gibbs’ correction, the entropy is       V 3 mkBT 5 S = NkB ln + ln + . N 2 2π¯h2 2

3 Now, U = 2 NkBT , so kBT = 2U/3N. Inserting this into the above expression for the entropy gives   V  3  mU  5 S = NkB ln + ln + . N 2 3π¯h2N 2 Now we can take the derivative with respect to N to obtain the chemical potential:   V  3  mU  5  3 µ = −kBT ln + ln + + kBT 1 + . N 2 3π¯h2N 2 2      V 3 mkBT 3  µ = −kBT ln + ln = kBT ln nλ , N 2 2π¯h2 D where n = N/V is the density. Note that the subsystem stops conserving the number of particles 3 when nλD = 1, or when the mean interparticle separation ` becomes comparable to the de Broglie −1/3 wavelength, ` ≡ n ∼ λD. But we already know that something quantum happens on the length scale 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 the 3D box.

How can you calculate the chemical potential from the free energy? I’m sure that you were burning to find this out. So let’s do it. The free energy is F = U − TS so dF = dU − T dS − SdT = T dS − P dV + µdN − T dS − SdT , so ∂F ∂ ln (Z ) µ = = −k T N . ∂N B ∂N

Let’s check that this gives the right answer for the particle in the 3D box. In that case, ZN = 3 N 3 3 (V/λD) /N! so ln(ZN ) = N ln(V/λD)−N ln(N)+N. So ∂ ln(ZN )/∂N = ln(V/λD)−ln(N)−1+1, 3 or µ = kBT ln(nλD). Suppose that the energies of the particles were larger by a constant factor of 2 2 ∆, so that k = (¯h k /2m) + ∆. Then

X  ¯h2k2  X  β¯h2k2  V Z = exp −β + ∆ = exp(−β∆) exp − = exp(−β∆). 2m 2m λ3 k k D So, the chemical potential is now 3 µ = kBT ln(nλD) + ∆. Clearly, the shift in the energies has led to exactly the same shift in the chemical potential. This additional 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 the equipartition 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 would have increased as well. These would be ‘internal chemical potential’ contributions. Would these tend to increase or decrease the chemical potential? PHYS 451 - Statistical Mechanics II - Course Notes 25

3.3 Arranging Indistinguishable Particles

The way we have derived statistical mechanics so far was to count the ways of arranging various outcomes into bins, and call the most-occupied bin the equilibrium state of the system. In the microcanonical ensemble, these outcomes tended to be things like ‘headness’ or ‘tailness’, and the bins were the number of heads and tails after N trials. In the canonical ensemble, the outcomes are usually designated as the accessible (quantum) energy levels of a given system, and the bins are the various ways of N particles occupying these levels. But both of these approaches assumed that we 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 particle that can be in the states ↑, ◦, and ↓, with energies 1, 0, and −1, respectively. Suppose we have three of these particles. Then we obtain the following table, where the ‘ways’ column denotes the number of ways we could have obtained the number of ↑, ◦, and ↓ in the state (i.e. the number of microstates in the macrostate).

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

We started with 10 distinguishable macrostates. But if the macrostates are the various values of the total energy of the macrostates, then we actually have only 7 macrostates, labeled by energies {3, 2, 1, 0, −1, −2, −3}, with {1, 3, 6, 7, 6, 3, 1} microstates in each, respectively. In fact, we’ve briefly run 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 g into account?

3.3.1 Bosons

Suppose that we want to enumerate the number of ways to arrange two particles that you can’t distinguish 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 many particles are in that level: In this example, there are 10 combinations. If you remember your binomial coefficients, this number  5  corresponds to . Why would this be? Let’s do another example to find some kind of trend. 2 How about three particles also with g = 4: PHYS 451 - Statistical Mechanics II - Course Notes 26

0 0 0 2 0 0 1 1 0 1 0 1 1 0 0 1 0 0 2 0 0 1 1 0 1 0 1 0 0 2 0 0 1 1 0 0 2 0 0 0

 6  This gives me 20 combinations, or . O.k., this is enough for a trend. The lower number of the 3 ‘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 ni particles into gi degenerate levels of energy state i is   (b) ni + gi − 1 (ni + gi − 1)! Ωi = = . ni ni!(gi − 1)! I haven’t imposed any kind of restrictions about how many particles can exist in one of the degenerate levels. These kinds of particles are called ‘bosons,’ because they obey ‘Bose-Einstein statistics,’ after the 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 total number of ways of distributing all the bosons is

Y (b) Y (ni + gi − 1)! Ω(b) = Ω = . (3.1) i n !(g − 1)! i i i i

(b) Note that if gi = 1 for a given level i, so that there is only one state per energy level, then Ωi = 1, i.e. that there is only one way to arrange the ni particles. How do we reconcile this with the counting method we used before? If all we get is one microstate in every macrostate, how do we maximize the entropy and obtain our equilibrium state? Let’s postpone this quandary for a moment, and first introduce fermions.

3.3.2 Fermions

Suppose that there is another kind of particle that refuses to share its with another particle; that is, each state within a degenerate energy level can only hold one particle. Particles with this restrictions are called ‘fermions,’ because the satisfy ‘Fermi-Dirac statistics.’ How many ways of arranging the particles are there now? Let’s consider the same examples as was done above for bosons. First, N = 2, and g = 4:  4  This gives Ω = 6 = . And the case with N = 3 and g = 2 is: 2 PHYS 451 - Statistical Mechanics II - Course Notes 27

0 0 0 3 0 0 1 2 0 1 0 2 1 0 0 2 0 0 2 1 0 1 1 1 1 0 1 1 0 2 0 1 1 1 0 1 2 0 0 1 0 0 3 0 0 1 2 0 1 0 2 0 0 2 1 0 1 1 1 0 2 0 1 0 0 3 0 0 1 2 0 0 2 1 0 0 3 0 0 0

0 0 1 1 0 1 0 1 1 0 0 1 0 1 1 0 1 0 1 0 1 1 0 0

 4  Now we only have four combinations, or . Again, this is enough to see the trend. The top 3 number is clearly the degeneracy, because it hasn’t changed. The bottom number looks like the number of particles. So for fermions we obtain

Y (f) Y gi! Ω(f) = Ω = . (3.2) i n !(g − n )! i i i i i

(f) For fermions, Ωi = 1 when gi = ni, i.e. when there is exactly one particle in each of the degenerate states within the energy level i. PHYS 451 - Statistical Mechanics II - Course Notes 28

0 1 1 1 1 0 1 1 1 1 0 1 1 1 1 0

3.3.3 Anyons!

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

(b) (ni + gi − 1)! (f) gi! Ωi = ;Ωi = . ni!(gi − 1)! ni!(gi − ni)!

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

(a) [gi + (ni − 1)(1 − α)]! Ωi = , ni![gi − 1 − α(ni − 1)]! where clearly for α = 0 we recover the expression for bosons, and for α = 1 we obtain that for fermions. But what about values of α in between? As discussed above, an unlimited number of bosons can occupy any state, independently of whether that state is already occupied by another particle or not; whereas only a single can occupy a given state. So, intuitively, there might max exist particles where only a finite number of them could occupy a given state, i.e. 1 < ni < ∞. 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. In two 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 the ones responsible for the fractional quantum Hall effect, for example. The ‘’ is the reason for (a) the ‘a’ superscript in the Ωi expression above.

(b) (f) Recall that for bosons, Ωi = 1 only if gi = 1, and for fermions, Ωi = 1 only if gi = ni; in other words, the distribution saturates for these values of the degeneracy gi. For what values of gi do (a) particles obeying F.E.S. saturate? For Ωi = 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:

gi − 1 αc = , 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 into the state with degeneracy gi, so that anyons with F.E.S. approach Fermi and Bose statistics in the two limits. PHYS 451 - Statistical Mechanics II - Course Notes 29

α gi ni gi ni gi ni gi ni 1 1 1 2 2 3 3 4 4 1 2 1 0 2 3 3 5 4 7 1 3 1 0 2 4 3 7 4 10 1 4 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 spans something like 105 quantum energy states. Then there would be on average 1 atoms per region, but there would be a huge degeneracy for this region, because the atom could be sitting in any of the 105 nearby energy levels. (I’m assuming that they are so closely-spaced as to be degenerate for our purposes). Then, we’re in the situation where gi  1. Now let’s go back to our expressions for the number of ways of distributing ni particles in gi states:

(a) [gi + (ni − 1)(1 − α)]! Ωi = , ni![gi − 1 − α(ni − 1)]!

If gi  1, then

(a) [gi + ni(1 − α)]! [gi + ni − niα)]! Ωi ≈ ≈ ni!(gi − αni)! ni!(gi − αni)! (g + n − αn )(g + n − αn − 1) ··· (g − αn + 1)(g − αn )(g − αn − 1) ··· (1) = i i i i i i i i i i i i ni!(gi − αni)(gi − αni − 1) ··· (1) gni ≈ i , ni! because not only is gi  1 but also gi  ni. Notice that this final result doesn’t depend on alpha and therefore on the kind of statistics. So the quantum way of counting at high temperatures reduces to the simple universal result Y gni Q gni Ω = i = i i . n ! Q n ! i i i i

In fact, this isn’t very different from the microcanonical result for a situation where there are various Q ni outcomes of N trials, Ωmicro = N!/ i ni! except that the factor of N! is replaced by gi . But as we will see now, this new factor resolves Gibbs’ paradox: ! !     Y Y X X gi ln(Ω) = ln gni − ln n ! = [n ln(g ) − n ln(n ) + n ] = n ln + n . i i i i i i i i n i i i i i i This means that        X ∂ gi X gi ni d ln(Ω) = 0 = n ln dn = ln − ) dn . ∂n i n i n n i i i i i i i PHYS 451 - Statistical Mechanics II - Course Notes 30

  X gi ⇒ ln dn = 0. n i i i This equation isn’t much different from the maximization of entropy condition used previously to P derive the Boltzmann distribution! Again, we have the supplementary conditions N = ni or P P i i dni = 0 and that of constant mean energy dU = i idni = 0. Again, I use the method of Lagrange multipliers to obtain the single equation     X gi ln − α − β dn = 0 n i i i i from which I get gi = exp [α + βi] ni or ni = gi exp[−βi − α] = giA exp(−βi). 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() = Ag() exp(−β), where g() is just the density of states per unit energy, rather than the degeneracy. The total particle number for a 3D box is then A(2m)3/2V Z √ N =  exp(−β)d 4¯h3π2 which determines the unknown constant A in terms of N and the box volume V .

Now the entropy is written "   # X gi X S = k ln(Ω) = k n ln + N = k n [ln(g ) − ln(A) − ln(g ) + β ] + k N B B i n B i i i i B i i i

= kB [βU − N ln(A) + N] . But N = AZ which means that ln(A) = ln(N) − ln(Z) and the entropy becomes  ∂  S = k [βU − N ln(N) + N ln(Z) + N] = k βNk T 2 ln(Z) + N ln(Z) − ln(N!) B B B ∂T U ZN  U = Nk ln(Z) + − k ln(N!) = k ln + , B T B B N! T which is different from the expression assuming distinguishable particles by a factor of ln(N!). This is exactly the term that we needed to resolve Gibbs’ paradox! Note that the expression for U is the same: P ( " #) X gii exp(−βi) ∂ X U = g A exp(−β ) = N i = Nk T 2 ln g exp(−β ) i i i P g exp(−β ) B ∂T i i i i i i i ∂ ∂ ZN  = Nk T 2 ln(Z) = k T 2 ln B ∂T B ∂T N! PHYS 451 - Statistical Mechanics II - Course Notes 31 where I got away with adding the N! term in the log because the derivative with respect to T would make this vanish! And of course I also have ZN  F = −k T ln . B N! Chapter 4

Quantum Statistics

Recall that the way one counts the number of ways quantum particles can be distributed in accessible quantum states is very different from the way classical particles are distributed in their quantum states. But we more or less dropped that story for a while in order to resolve Gibbs’ paradox. At large temperatures, when the number of states that are occupied is much larger than the number of particles, 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 closely spaced singly-degenerate quantum states into another state, which we called the highly degenerate quantum state. But we also found that at high temperatures, the mean number of levels occupied is also much larger than the number of particles in the system. In this chapter, we’ll relax the second assumption about high temperatures, so that the number of particles can begin to approach the degeneracy of one of these compound levels. In this case, we shouldn’t recover the classical Boltzmann 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 energy level i with degeracy gi is given by

(a) [gi + (ni − 1)(1 − α)]! Ωi = , ni![gi − 1 − α(ni − 1)]! where α = 0 for bosons and α = 1 for fermions, and any other α corresponds to particles with fractional exclusion statistics. The total number of ways of arranging the particles is then

Y [gi + (ni − 1)(1 − α)]! Ω = . n ![g − 1 − α(n − 1)]! i i i i Let’s follow the same procedure as we did previously in deriving the Boltzmann distribution, by P P maximing the entropy (or ln(Ω)) subject to the constraints N = i ni and U = i ini. Now, in the grand canonical ensemble we know that the Lagrange multiplier fixing the first condition is nothing but the chemical potential µ. Putting this together we have X {(1−α) ln[gi +(ni −1)(1−α)]+(1−α)−ln(ni)−1+α ln[gi −1−α(ni −1)]+α+βµ−βi}dni = 0 i

32 PHYS 451 - Statistical Mechanics II - Course Notes 33

⇒ (1 − α) ln[gi + (ni − 1)(1 − α)] − ln(ni) + α ln[gi − 1 − α(ni − 1)] + βµ − βi = 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 − βi = 0   gi ⇒ ln − 1 = β(i − µ). ni This immediately gives the Fermi-Dirac (FD) distribution function

FD gi ni = , (4.1) exp [β(i − µ)] + 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. Because each state within a degenerate level can hold either no particles (with energy zero), or one particle (with energy Ei = i and particle number Ni = 1), the FD partition function is

gi Y gi Ξi = {1 + exp[−β(i − µ)]} , Ξ = {1 + exp[−β(i − µ)]} . i

The single-particle partition function is taken to the power of gi, because each of the gi states is assumed to be independent. The grand potential is therefore X ΦG = −kBT ln(Ξ) = −kBT gi ln {1 + exp[−β(i − µ)]} . (4.2) i Now, the mean number of particles in the system is

∂ΦG ∂ ln(Ξ) X giβ exp[−β(i − µ)] X gi N = − = k T = k T = . ∂µ B ∂µ B 1 + exp[−β( − µ)] exp[β( − µ)] + 1 i i i i P Because N = i ni, we immediately obtain

FD gi ni = exp [β(i − µ)] + 1 in agreement with the expression above.

The Fermi-Dirac distribution function for various inverse temperatures β is shown in Fig. 4.1. As temperature increases (lower β), the distribution becomes flatter and shows a longer exponential tail, which begins to look like the Boltzmann distribution. At lower temperature (larger β), the distribution approaches a step-like function, signifying full occupation for levels  < µ and zero occupation for levels larger than the chemical potential. PHYS 451 - Statistical Mechanics II - Course Notes 34

1

0.8

) 0.6 ε ( FD

n 0.4

0.2

0 0 10 20 5 ε 15

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

In order to get a better feeling for the Fermi distribution (4.1), let’s modify the way it looks a little. gi Consider ni − 2 :

1 1 gi 1 − 2 exp [β(i − µ)] − 2 gi  1 − exp [β(i − µ)] ni − = gi = . 2 exp [β(i − µ)] + 1 2 exp [β(i − µ)] + 1

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

gi gi  exp [−β(i − µ)/2] − exp [β(i − µ)/2] ni − = 2 2 exp [β(i − µ)/2] + exp [−β(i − µ)/2] g  β  = − i tanh ( − µ) . 2 2 i

So we are finally left with a slightly more intuitive expression (I hope!) for the Fermi distribution function that now involves a tanh: g  β  n = i 1 − tanh ( − µ) . (4.3) i 2 2 i

Recall that the tanh function has the following property: ( −1 x  0 tanh(ax) = +1 x  0 0 x = 0 Also, tanh(ax) ∼ ax when x ∼ 0. So the value of a determines how quickly the function changes from its value of −1 to +1 as x passes through zero. In our case, x = i − µ and a = β. This means that the Fermi distribution is equal to gi for i  µ and is exactly zero when i  µ. Also, as the PHYS 451 - Statistical Mechanics II - Course Notes 35

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

n gi i < µ ni(T = 0) = giΘ(i − µ) ≡ . 0 i > µ

This means that exactly gi particles occupy each energy level i 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 − βi = 0.

Now, the degeneracy of a given level is still assumed to be large, so that gi  1. In this case we obtain   gi ln + 1 = β(i − µ). ni This can be inverted to give gi = exp [β(i − µ)] − 1, ni or finally the Bose-Einstein (BE) distribution:

BE gi ni = , (4.4) exp [β(i − µ)] − 1 otherwise known as the Bose-Einstein occupation factor. Note that the BE and FD distributions differ only in the sign of the ‘1’ in the denominator! As we’ll see later, this small difference leads to profoundly different kinds of behaviour. But one thing is immediately clear: the chemical potential for bosons cannot ever exceed the value of the lowest energy level 0. If it did, the value of the distribution function for this level n0 would be a negative number, which is unphysical. So for bosons we must have µ < 0. Again, we can derive the BE distribution within the grand canonical ensemble instead. There is no restriction to how many bosons can occupy a single state within a degenerate level. So the grand partition function for a given level is now a sum over contributions with no particles, one particle (with energy Ei = i and number Ni = 1), two particles (with energy Ei = 2i and Ni = 2), etc.:

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

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

 1 gi Ξi = . 1 − exp[−β(i − µ)] The grand potential is X ΦG = −kBT ln(Ξ) = kBT gi ln {1 − exp[−β(i − µ)]} . (4.5) i PHYS 451 - Statistical Mechanics II - Course Notes 36

10

8

) 6 ε ( FD

n 4

2

0 10 12 14 18 20 ε 16

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

The mean number of particles is

∂ΦG X giβ exp[−β(i − µ)] X gi N = − = k T = . ∂µ B 1 − exp[−β( − µ)] exp[β( − µ)] − 1 i i i i P Again, we know that N = i ni so

BE gi ni = , exp[β(i − 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. As temperature increases (lower β), the distribution shows a longer exponential tail, which again begins to look like the Boltzmann distribution. At lower temperature (larger β), the distribution tightens up, with more probability near the chemical potential.

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

FES gi ni = , w {exp[β(i − µ)]} + α where w(x) is a function of x = exp[β(i − 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, again reproducing the expected result. For rational fractions α = p/q with p and q mutually prime, this PHYS 451 - Statistical Mechanics II - Course Notes 37

2

1.5 ) ε (

1 (Boltz,FD,BE) n 0.5

0 0 10 20 5 ε 15

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 chemical potential (µ = 10). The degeneracy factor is g = 1. implies a polynomial of order q. The distribution for semions, which satisfy α = 1/2, is

semion gi ni = . q 1 exp[2β(i − µ)] + 4

The Boltzmann, Fermi-Dirac, and Bose-Einstein distributions are plotted together in Fig. 4.3. Notice that they all coincide for large energies levels , 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 obtained in terms of the grand potential using ∂Φ S = − G . ∂T The grand potential for bosons and fermions, derived above, are X ΦG = ∓kBT gi ln {1 ± exp[−β(i − µ)]} , i where the upper sign corresponds to fermions, and the lower sign to bosons. The derivative is carried out more easily in terms of β, so we’ll define the entropy instead as

∂ΦG ∂ 1 X S = k β2 = ∓k β2 g ln {1 ± exp[−β( − µ)]} B ∂β B ∂β β i i i   X ∓(i − µ) exp[−β(i − µ)] = g ±k ln {1 ± exp[−β( − µ)]} ∓ k β i B i B 1 ± exp[−β( − µ)] i i PHYS 451 - Statistical Mechanics II - Course Notes 38

  X β(i − µ) = k g ± ln {1 ± exp[−β( − µ)]} + . B i i exp[β( − µ)] ± 1 i i This is a pretty ugly expression, but it can be made to look nicer using the expressions for the FD and BE distributions gi ni = . exp[β(i − µ)] ± 1 This can be inverted to give

gi exp[β(i − µ)] ± gi ∓ gi gi gi ∓ ni = = exp[β(i − µ)] ± 1 1 ± exp[−β(i − µ)] which yields   gi ∓ ni ln {1 ± exp[−β(i − µ)]} = − ln . gi We’re getting there! We also know that

gi gi ∓ ni exp[β(i − µ)] = ∓ 1 = ni ni

⇒ β(i − µ)] = ln(gi ∓ ni) − ln(ni). Putting all of these into the expression for the entropy gives   X ni ni S = k g ∓ ln(g ∓ n ) ± ln(g ) + ln(g ∓ n ) − ln(n ) B i i i i g i i g i i i i X = kB {(ni ∓ gi) ln(gi ∓ ni) − ni ln(ni) ± gi ln(gi)} . i For fermions, the resulting entropy takes the form:

FD X S = −kB {ni ln(ni) + (gi − ni) ln(gi − ni) + gi ln(gi)} , (4.6) i which reduces so something intuitive when gi = 1:

FD X S (gi = 1) = −kB {ni ln(ni) + (1 − ni) ln(1 − ni)} . i In other words, it is the sum of two contributions, corresponding to each level either empty, or containing one particle. This is very reminiscent of the entropy for a two-state system (coins or spins)! Note that the entropy is zero if each state is occupied by one particle (ni = 1). This is the ground state of the system at zero temperature, and it reflects the Pauli exclusion principle. For bosons, the form of the entropy is less intuitive looking, but still nice:

BE X S = kB {(gi + ni) ln(gi + ni) − ni ln(ni) − gi ln(gi)} . (4.7) i

Note that for bosons, the entropy is zero only if all states are unoccupied, i.e. ni = 0 ∀i. But this must be nonsense, because we have a finite number of particles in our system! Where have all the particles gone? This is the origin of Bose-Einstein condensation, which we’ll get to below. Notice also that there is a complete in the Bose-Einstein case between ni and gi. PHYS 451 - Statistical Mechanics II - Course Notes 39

4.2 Quantum-Classical Transition

Last term we found that the chemical potential at high temperatures is large and negative. This means that one might expect exp[β(i −µ)]  1 at high temperatures, but only if |µ|  kBT at high temperatures as well. Supposing that the magnitude of the chemical potential gets large faster than kBT gets large (checked below), then both the Bose and Fermi distributions become the Boltzmann distribution: BE,FD ni (T  0) ≈ gi exp(βµ) exp(−βi), where A = exp(βµ) is called the fugacity. But remember that we defined A = N/Z in the canonical ensemble. So this immediately gives N  µ = k T ln , B Z where Z is understood as the high-temperature limit of the canonical partition function. This is consistent with the results of the examples (Pauli paramagnet and particles in a 3D box) discussed last term. If µ/kBT  −1, then this implies that N  Z or Z  N. For a 3D box, this means V 1 3  1 or n  3 NλD λD at high temperatures. But the mean distance between particles d = (1/n)1/3 so this criterion is equivalent to saying that the classical limit is equivalent to stating that the mean distance between particles 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 equipartition 2 theorem, which√ states that the mean kinetic energy of each particle is U/N = p /2m = (3/2)kBT . Then p = 3mkBT  ¯h/d, whereh/d ¯ is a measure of the largest quantum wavevector in the system. The condition is equivalent to

s 2 ¯h λD d  = √ or d  λD. 3mkBT 6π So, the transition from the classical to the quantum regimes occurs when the particles get sufficiently close 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. Consider the mean particle (fermion or ) density for a 3D box:

N 1 Z ∞ k2dk n = = 2 , V 2π 0 exp[β( − µ)] ± 1

2 2 2 2 2 where k =h ¯ k /2m. Making the change of variables k = (2mkBT/¯h )x , we obtain

 3/2 Z ∞ 2 1 2mkBT x dx n = 2 2 2 , (4.8) 2π ¯h 0 exp(x − η) ± 1 where η ≡ µ/kBT . We assume that the mean density in the box does not change as a function of temperature. So let’s suppose that we start increasing the temperature. Clearly, the integral PHYS 451 - Statistical Mechanics II - Course Notes 40 has to decrease like 1/T if n is conserved; at very large temperatures, the integral must approach zero. But the only thing in the integral that knows about the temperature is η = µ/kBT . For this to be true, we clearly require η → −∞ at high temperatures, or |µ|/kBT  1. This justifies the connection between the grand canonical function for indistinguishable quantum particles and the canonical distribution function for classical particles discussed above. At large (but not infinite) temperature, the integral becomes

 3/2 Z ∞  3/2   √ 1 2mkBT 2 2 1 2mkBT µ π n ≈ 2 2 dx x exp[−(x − η)] = 2 2 exp . 2π ¯h 0 2π ¯h kBT 4 Inverting this gives  2  µ 3 2π¯h 3  = ln(n) + ln ⇒ µ = kBT ln nλD kBT 2 mkBT 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 a 3D box at high temperature we’re expecting the result to be P = nkBT , where n is the density. What about 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 ∂Φ P = − G , ∂V and the grand potential for fermions and bosons is Z ∞ X kBTV Φ = ∓k T g ln {1 ± exp[−β( − µ)]} ≈ ∓ k2dk ln {1 ± exp[−β( − µ)]} . G B i i 2π2 k i 0

2 2 In the second line I have explicitly assumed a 3D box, and that the energy levels k = ¯h k /2m are sufficiently closely spaced relative to the temperature that the energy levels are more or less continuously distributed. The pressure is therefore Z ∞ kBT 2 P = ± 2 k dk ln {1 ± exp[−β(k − µ)]} . 2π 0 To make progress, let’s integrate this expression by parts once. Recall to integrate by parts we use the rule Z b Z b b udv = uv a − vdu. a a The first term on the right hand side is called the surface contribution. In the pressure expression, 2 3 we’ll let u = ln {1 ± exp[−β(k − µ)]} and dv = k dk. Now, v = k /3 which is zero for k = 0, and u = 0 when k = ∞ because the argument of the ln becomes one. So the surface term vanishes. The pressure expression becomes

Z ∞ 2 4 Z ∞ 2 4 kBT ¯h k exp[−β(k − µ)] 1 ¯h k P = ∓ 2 (∓β) dk = 2 nkdk. 2π 0 3m 1 ± exp[−β( − µ)] 2π 0 3m PHYS 451 - Statistical Mechanics II - Course Notes 41

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

We can divide the factor of exp[−β(k − µ)] in the numerator into the denominator, which makes the integrand explicitly depend on the distribution functions nk. Making the same substitution 2 2 2 x = (¯h /2mkBT )k as was done in the evaluation of the mean particle density, and simplifying a bit, one obtains  3/2 Z ∞ 4 kBT 2mkBT x P = 2 2 2 dx, 3π ¯h 0 exp(x − η) ± 1 where as before η = µ/kBT . Note that the prefactor is almost exactly the same as the right hand side of the expression for the particle density, Eq. (4.8). So finally we have

R ∞ x4 2k T n 2 dx P = B 0 exp(x −η)±1 . (4.9) 3 R ∞ x2 0 exp(x2−η)±1 dx

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

∞ 2 √ √ 2nk T R x4e−x dx 2nk T 3 π  π −1 P (T  1) ≈ B 0 = B = nk T, 3 R ∞ 2 −x2 3 8 4 B 0 x e dx 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 can write 1  1  1 = ≈ exp(−x2 + η) 1 ∓ exp(−x2 + η) . exp(x2 − η) ± 1 exp(x2 − η) 1 ± exp(−x2 + η)

Inserting this into the expression for the pressure gives

√ √ R ∞ 4  −x2 η −2x2  3 π 3 π x e ∓ e e dx ∓ eη √ 1 ∓ eη √1 2kBT n 0 2kBT n 8 32 2 4 2 P ≈ = √ √ ≈ nk T . R ∞ 2 −x2 η −2x2  B η 1 3 x e ∓ e e dx 3 π ∓ eη √π 1 ∓ e √ 0 4 8 2 2 2

η βµ 3 The results below Eq. (4.8) suggest that we simply make the identification e = e = nλD. The pressure then becomes  3  nλD P ≈ nkBT 1 ± √ . 4 2 The two terms in this expression are called the first and second virial coefficients, respectively. The first virial coefficient simply reflects the , as we found above anyhow. The second is a density-dependent correction that arises strictly from the . The contribution to the pressure is positive for fermions, but negative for bosons. For fermions, the pressure increases 3 as the temperature is lowered (i.e. nλD increases) because the Pauli exclusion principle effectively pushes 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 together at lower temperature; this presages the idea of Bose-Einstein condensation, discussed in detail below. PHYS 451 - Statistical Mechanics II - Course Notes 42

In general, the deviation of the ideal gas law for non-interacting particles is one important signature for 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

 2α − 1  P = nk T 1 + nλ2 . B 4 D

For fermions (α = 1), the second virial coefficient is clearly positive, while for bosons (α = 0), it is 1 negative. Interestingly, it is exactly zero for semions (α = 2 ), which are halfway between the two limits.

The high-temperature correction to the total energy can be calculated in a similar fashion. We have   2 ∂ ln(Ξ) 2 ∂ ΦG ∂ΦG U = kBT = kBT − = ΦG − T . ∂T ∂T kBT ∂T We could go through the same procedure as above, integrating by parts, etc. But we can save lots P of time by observing that U = i nii, so that

V Z ∞ ¯h2k4 3 U = 2 nkdk = PV 2π 0 2m 2 when comparing to the expression for the pressure above. So one obtains

3  1  U ≈ Nk T 1 ± nλ3 . 2 B 25/2 D Chapter 5

Fermions

5.1 3D Box at zero temperature

2 2 For a three-dimensional box, the highest energy level is EF =h ¯ kF /2m, where kF is the Fermi wavevector, or the radius of the Fermi sphere in momentum space (recally p = ¯hk). What is the Fermi energy at zero temperature in this case? We know that the mean number of particles is therefore i XF gV Z kF N = n = k2dk, i 2π2 i=0 0 where I’ve kept a factor of g out front in case each energy level has any intrinsic degeneracy over 1 and above the usual density of states. For example, if I assume that each fermion has spin- 2 , then I could place exactly two particles (g = 2) in each energy level (one with spin up, one with spin down) and still satisfy the Pauli exclusion principle. Integrating, and making use of the fact that 2 2 EF =h ¯ kF /2m, I obtain gV 2mE 3/2 N = F , (5.1) 6π2 ¯h2 which can be inverted to give ¯h2 6π2n2/3 E = . (5.2) F 2m g Here, n = N/V is the mean density. This expression immediately gives the value for the Fermi wavector at zero temperature: 6π2n1/3 k = . F g

Let’s also calculate the mean energy for the fermions at zero temperature: i XF gV Z kF ¯h2k2  gV ¯h2k5 6 gV k3  ¯h2k2  3 U = g  n = k2 dk = F = F F = NE . (5.3) i i i 2π2 2m 10π22m 10 6π2 2m 5 F i=0 0 The pressure for a is meanwhile

Z ∞ 2 4 2 Z kF  3   2 2  g ¯h k g¯h 4 gV kF ¯h kF 2 2nEF P = 2 nkdk = 2 k dk = 2 = . 2π 0 3m 6mπ 0 6π 2m 5V 5

43 PHYS 451 - Statistical Mechanics II - Course Notes 44

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

5.2 3D Box at low temperature

At low but finite temperature, we need to evaluate expressions involving the full Fermi occupation factors, i.e. Z Z ∞ Z ∞ X 3 X = Xini = d k g(k)n(k)X(k) = g()n()X()d = n()Y ()d, (5.4) i −∞ −∞ where Y () ≡ g()X(). Notice that I have explicitly inserted the density of states into the integral expressions, but that the inherent degeneracy of states g (for example originating in the spin degree of freedom) is already included in the ni. Here it is also assumed that Y () vanishes when  → −∞ and is at most a power of  for  → ∞. If we define K() in terms of Y () as Z  K() ≡ Y (0)d0, −∞ so that Y () = dK()/d, then one can integrate Eq. (5.4) by parts to obtain Z ∞ Z ∞ Z ∞  ∂n n()Y ()d = n()dK() = K() − d. −∞ −∞ −∞ ∂ The surface term is zero because K( = −∞) = 0 and n( = ∞) is exponentially suppressed compared with K( = ∞). Now, n() at low temperatures is constant for most , and varies rapidly only near  = µ. So it is reasonable to expand K() in a around  = µ, so that we have

Z ∞ Z ∞ " ∞ n n #   X ( − µ) ∂ K() ∂n n()Y ()d = K(µ) + − d. n! ∂n ∂ −∞ −∞ n=1 =µ The leading term is simply gK(µ), and only even terms will appear at higher-order because ∂n/∂ is an even function of  − µ. So we have

Z ∞ Z µ ∞ Z ∞ 2n   2n−1 0 0 X ( − µ) ∂n ∂ Y () n()Y ()d = g Y ( )d + − d . (2n)! ∂ ∂2n−1 −∞ −∞ n=1 −∞ =µ

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

Z ∞ Z µ ∞ 2n−1 X 2n ∂ n()Y ()d = g Y ()d + g an(kBT ) Y () , ∂2n−1 −∞ −∞ n=1 =µ where Z ∞ x2n  d 1   1  a ≡ − dx = 2 − ζ(2n), n x 2(n−1) −∞ (2n)! dx e + 1 2 with the Riemann zeta functions defined as 1 1 1 ζ(n) ≡ 1 + + + + .... (5.5) 2n 3n 4n PHYS 451 - Statistical Mechanics II - Course Notes 45

In practice we don’t care about anything except the first correction term, n = 1, for which a1 = ζ(2) = π2/6. So Z ∞ Z µ 2 π 2 0 n()Y ()d ≈ g Y ()d + g (kBT ) Y (µ). (5.6) −∞ −∞ 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 mean energy, and the pressure. Using the 3D density of states seen last term, the mean number is Z ∞ 3/2 Z ∞ 3/2 Z µ 2  gV (2m) 1/2 gV (2m) 1/2 π 2 1 −1/2 N = g()n()d = 2 3  n()d ≈ 2 3  d + (kBT ) µ 0 2π 2¯h 0 2π 2¯h 0 6 2 " # gV (2m)3/2 π2 k T 2 = µ3/2 1 + B . (5.7) 6π2 ¯h3 8 µ This means that as we raise the temperature from zero at constant chemical potential, the mean particle number will increase. Alternatively, one can assume that the mean number of particles N remains constant as temperature is increased from zero, so that the Fermi energy EF (5.2) remains well-defined at all temperatures, then the chemical potential must also vary: " # gV 2mE 3/2 gV (2m)3/2 π2 k T 2 N = F = µ3/2 1 + B . 6π2 ¯h2 6π2 ¯h3 8 µ

" 2  2# π kBT ⇒ µ ≈ EF 1 − . (5.8) 12 EF

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

Next let’s calculate the mean energy: Z ∞ 3/2 Z ∞ 3/2 Z µ 2  gV (2m) 3/2 gV (2m) 3/2 π 2 3√ U = g()n()d = 2 3  n()d ≈ 2 3  d + (kBT ) µ 0 2π 2¯h 0 2π 2¯h 0 6 2 " # gV (2m)3/2 5π2 k T 2 = µ5/2 1 + B . (5.9) 10π2 2¯h3 8 µ While correct, this expression isn’t so nice because it is difficult to compare it to the zero-temperature result (5.3). Substituting into Eq. (5.7) gives 2 5π2  k T  1 + B " 2  2# 3 8 µ 3 π kBT U ≈ Nµ 2 ≈ Nµ 1 + . 2   5 π kB T 5 2 µ 1 + 8 µ Finally, inserting the expression (5.8) gives

" 2  2#" 2  2# " 2  2# 3 π kBT π kBT 3 5π kBT U ≈ NEF 1 − 1 + = NEF 1 + . 5 12 EF 2 µ 5 12 EF PHYS 451 - Statistical Mechanics II - Course Notes 46

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 low temperature was zero. It’s also the reason for the linear-temperature specific heat observed in clean crystals at very low temperatures, because electrons are fermions. Again, using the relation P = 2U/3V gives the correction to the pressure: " # 2nE 5π2 k T 2 P ≈ F 1 + B . (5.10) 5 12 EF

The finite-temperature correction term is exactly the same as for the mean energy above. This is nothing but the second virial coefficient, which is clearly positive for all finite temperatures. Recall that at very high temperatures, the pressure expression is instead

 1  P (T → ∞) = nk T 1 + nλ3 . B 25/2 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 temperatures have been trapped by magnetic fields or focused laser beams. The confining potentials are generally 3D harmonic traps. So let’s consider this case in more detail. You might be interested to note that 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 (accessible energy states) are given by (nx, ny, nz) = nx¯hωx + ny¯hωy + nz¯hωz. In order to evaluate the various integrals, we first need to obtain the density of states per unit energy. A rough way to do this is to 2 2 2 2 2 2 2 2 2 1/3 simply set ki = ni, so that  = kx(¯hωx) + ky(¯hωy) + kz (¯hωz) ≡ k (¯hω) , where ω = (ωxωyωz) is the mean frequency, and dki/i = 1/¯hω. Because ki = ni now rather than ki = πni/L, the 3D density of states is given by k2 dk 2 g() = = . (5.11) 2 d 2(¯hω)3

Another (more rigorous) way to obtain the density of states is to ask how many states G() are enclosed in an energy surface of radius  = x + y + z. The result is

1 Z  Z −x Z −x−y 3 G() = dx dy dz = 3 . ¯hωx¯hωy¯hωz 0 0 0 6¯h ωxωyωz The density of states is then

dG() 2 2 g() = = 3 = 3 , d 2¯h ωxωyωz 2(¯hω) 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 PHYS 451 - Statistical Mechanics II - Course Notes 47

n =hωn, ¯ n = 0, 1, 2,... (neglecting the zero-point energy) is 1 Z = , x = exp(−β¯hω). 1D 1 − x Because the partition function for the s-dimensional harmonic oscillator is the s-fold product of the one-dimensional partition functions, we can write

1  s(s − 1) s(s − 1)(s − 2)  Z = = 1 + sx + x2 + x3 + ... sD (1 − x)s 2 6 ∞ ∞ X  n + s − 1  X = xn = g(s, n)xn. (5.12) n n=0 n=0 So, the degeneracy of the s-dimensional oscillator is

 n + s − 1  (n + s − 1)! g(s, n) = = . n n!(s − 1)!

For three dimensions (s = 3), g(n) = (n + 1)(n + 2)/2. This expression becomes simpler g(n0) = n0(n0 + 1)/2 if we make the replacement n0 = n + 1 and count from 1 instead of zero. For large n0, this becomes g(n0) = n02/2, as found above.

Alternatively, the degeneracy for the three dimensional isotropic trap can be found by counting the number of ways we can distribute M distinct energy quanta into a common (degenerate) energy level, 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 is equivalent to the one-dimensional case (s = 1). If M = 1 then WN = N!/(N − 1)! = N, i.e. there are N ways to fill up each level indexed by N; this is equivalent to s = 2. If M = 2 then there are WN = (N + 1)!/[2(N − 1)!] = N(N + 1)/2, ways to fill up each level indexed by N; this is equivalent to 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 expressions for the mean number, the mean energy, and the pressure for the 3D isotropic oscillator, as we did for the 3D case above. The mean number is Z ∞ Z ∞ Z µ 2  g 2 g 2 π 2 N = g()n()d = 3  n()d ≈ 3  d + (kBT ) µ 0 2(¯hω) 0 2(¯hω) 0 3 " # g  µ 3 k T 2 = 1 + π2 B . (5.13) 6 ¯hω µ

At zero temperature this expression defines the Fermi energy:

1/3 6N  E = ¯hω. (5.14) F g PHYS 451 - Statistical Mechanics II - Course Notes 48

Again, if we fix N at all temperatures then " # g E 3 g  µ 3 k T 2 N = F = 1 + π2 B 6 ¯hω 6 ¯hω µ which can be inverted to yield " 2  2# π kBT µ ≈ EF 1 − . (5.15) 3 EF

Next let’s calculate the mean energy:

Z ∞ Z ∞ Z µ 2  g 3 g 3 π 2 2 U = g()n()d = 3  n()d ≈ 3  d + (kBT ) µ 0 2(¯hω) 0 2(¯hω) 0 2 " # gµ  µ 3 k T 2 = 1 + 2π2 B . (5.16) 8 ¯hω µ

Substituting into Eq. (5.13) gives  2 2 kB T " 2# 3 1 + 2π µ 3 k T  U ≈ Nµ ≈ Nµ 1 + π2 B .  2 4 2 kB T 4 µ 1 + π µ

Finally, inserting the expression (5.15) gives

" 2  2#"  2# " 2  2# 3 π kBT 2 kBT 3 2π kBT U ≈ NEF 1 − 1 + π ≈ NEF 1 + . 4 3 EF µ 4 3 EF

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. In principle, one must put a gi fermions in each available energy state i. So one would need to take into consideration each level and its associated degeneracy, then calculate the appropriate spatial wavefunction to build up the total density profile. This is a lot of work! Thankfully, there is a better way to accomplish this task when the number of particles is large, so that a detailed knowledge of the 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 no external potential at all. In other words, the Fermi occupation factor would be expressed as 1 1 n(r, k) = . 2π3 exp β ¯h2k2/2m + mω2r2/2 − µ + 1

This means that the spatial density is given by √ Z g(2m)3/2 Z ∞  n(r) = d3k n(r, k) = d. 3 2 2 2 4¯h π 0 exp [β ( + mω r /2 − µ)] + 1 PHYS 451 - Statistical Mechanics II - Course Notes 49

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

2 2 3/2 g(2m)3/2 Z EF −mω r /2 √ g(2m)3/2  1  n(r) = d = E − mω2r2 . 3 2 3 2 F 4¯h π 0 6¯h π 2

2 2 The Fermi energy can then be used to define the radius of the cloud R through EF = (1/2)mω R so one obtains g mω 3   r 23/2 n(r) = R3 1 − . (5.17) 6π2 ¯h R Using the definition of the Fermi energy (5.14), the radius of the cloud is

r r 1/6 1/6 2E 2¯h 6N  √ 6N  R = F = = 2 `, (5.18) mω2 mω g g where I have defined the harmonic oscillator length ` ≡ p¯h/mω (the meaning of this will be discussed below). So finally the density is s 2 gN   r 23/2 n(r) = 1 − . π2`3 3 R

Where does the length ` come from? For a one-dimensional harmonic oscillator the force is Fi = 2 −kriˆı= −mω riˆı,and the virial theorem (2.1) states that on average * + * + 1 X mω2 X mω2`2 hKi = − F · r = r2 = , 2 i i 2 i 2 i i where `2 is a length scale describing the mean-square displacement of the oscillator. Meanwhile the kinetic energy term is * + * + X p2 ¯h2 X ¯h2 i = k2 = , 2m 2m i 2m`2 i i 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 =h ¯2/m2ω2 or ` = p¯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 Fermi energy will be excited into levels above the Fermi energy (cf. Fig. 4.1). In this case, the fermions will behave essentially as Boltzmannons (!). To find the density, we can either treat the potential as an external chemical potential, so that

     2   2   2  1 µ − µext(r) 1 µ x y z n(r) = 3 exp = 3 exp − exp − 2 exp − 2 exp − 2 , (5.19) λD kBT λD kBT Rx Ry Rz

∗The reason for this is that the particle distribution in the harmonic oscillator is a Gaussian. So the Fourier transform of it (the momentum-space distribution) is also a Gaussian, whose mean-square size is the inverse of the original Gaussian. PHYS 451 - Statistical Mechanics II - Course Notes 50

2 2 where Ri = 2kBT/mωi 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 function N = AZ, and then use the equipartition form for Z,

  Z  2   2   2  1 µ 3 x y z N = AZ = 3 exp − d r exp − 2 exp − 2 exp − 2 , λD kBT Rx Ry Rz and then use the normalization condition N = R d3rn(r) to obtain the chemical potential. In any case 3 3/2 we have the same thing. The effective volume of the cloud is V = (4π/3)Ri = (4π/3)(2kBT/mω) 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 to determine if we must treat them quantum mechanically or classically. In other words, we first determine if the particles are quantum degenerate or not. If so, we can apply the results obtained above to obtain various properties if we are interested. Below are four examples of Fermi degenerate systems where the temperatures span 14 orders of magnitude!

5.4.1 Electrons in Metals Typical in metals are n = 1029 m−3. Using the mass of an electron as 9.109 · −31 −9 10 kg, the de Broglie wavelength at room temperature (300 K) is found to be λD = 4.303·10 m. −1/3 −10 The mean interparticle separation is d = n = 2 · 10 m. This means that λD  d and the electrons must be treated quantum mechanically. Another way to think about it is to use the Fermi temperature, TF ≡ EF /kB, which must be larger than the actual temperature of the system to −18 be quantum degenerate. Using the same parameters, I obtain EF = 2 · 10 J (or about 12 eV), 5 and TF = 1.5 · 10 K. This is of course much larger than room temperature. So the electrons have something 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 q 2 velocity of fermions near the Fermi energy, and is defined as vF =hk ¯ F /m, where kF = 2mEF /¯h 6 is the Fermi wavevector. Putting in numbers gives vF ≈ 2.1 × 10 m/s, which is pretty fast! The electron pressure is given by Eq. (5.10) at low temperatures. Since at room temperature kBT/EF ≈ 0.002 we are justified in neglecting the finite-temperature correction. The pressure is then P ≈ 8 × 1010 N/m2. This is phenomenally high, almost a million atmospheres! It’s amazing that 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 mass 30 6 of the sun is M = 3 · 10 kg, and the average temperature is something like 5 × 10 K. The number of hydrogen atoms gives a reasonable estimate for the number of electrons. With the mass −27 57 of hydrogen assumed to be 1.67 · 10 kg, I obtain Ne = 1.8 · 10 electrons. The radius of the sun 7 34 −3 8 is 3 · 10 m, giving a density of 1.6 · 10 m . The Fermi temperature is therefore TF = 2.7 · 10 K, which is something like 50 times larger than the mean temperature. So the electrons in the sun are also quantum degenerate! Note that one can also define the velocity of electrons at the Fermi PHYS 451 - Statistical Mechanics II - Course Notes 51

8 surface, vF ≡ ¯hkF /m = 10 m/s for electrons in the sun. This is already one-third of the velocity of light, so if the sun were much hotter we’d have to treat the electrons relativistically. In fact, if a is heavier than about 1.4M then the electron degeneracy pressure will no longer be able to stabilize it against collapse. This so-called is quite easy to calculate when the electrons are treated relativistically, as is needed for hot white dwarfs. There is a similar ‘Tolman- Oppenheimer-Volkoff’ limit for that are stabilized by Fermi pressure associated with < < neutrons, in the range 1.4M ∼ M ∼ 3.5M ; the actual values are hard to compute because the strong 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 Fermi energy (5.14) and compare it to kBT to test if the atoms were in fact quantum degenerate. In many experiments, 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 the −29 same direction) one obtains EF ≈ 1.2 × 10 J or a Fermi temperature TF EF /kB ≈ 1 µK. This is very cold! But these days, experiments are operating around 100 nK or T = 0.1TF . So ultracold fermions in traps are in fact strongly quantum degenerate.

Another way to estimate the temperature required to reach Fermi degeneracy, one can use the 6 3 19 −3 Boltzmann spatial profile (5.19). The density is n = 10 /Ri ≈ 3 · 10 m giving a mean spacing −7 between particles of d ≈ 3 · 10 m. Fermi degeneracy is reached when λD = d which in this case −6 corresponds to T = TF ∼ 10 K, the same estimate we obtained above. Interestingly, the same density is found using the low-temperature Fermi distribution (5.17). For the parameters given above 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 estimate above, 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 totally different 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 quantization procedure yields the energy eigenvalues n = n¯hω, which is off by only a constant factor. The exact 1 expression is n = ¯hω(n + 2 ), but the additional factor makes no contribution to any statistical properties.

One can then obtain the (canonical) partition function: X X Z = exp(−n/kBT ) = exp(−βn) = 1 + exp(−β¯hω) + exp(−2β¯hω) + .... n n But this is just a geometric series: if I make the substitution x ≡ exp(−β¯hω), then Z = 1 + x + x2 + x3 + .... But I also know that xZ = x + x2 + x3 + .... Since both Z and xZ have an infinite number of terms, I can subtract them and all terms cancel except the first: Z − xZ = 1, which immediately yields Z = 1/(1 − x), or 1 Z = . (6.1) 1 − exp(−β¯hω) Now I can calculate the mean energy: 2 2 ∂ ln(Z) NkBT ∂Z 2 [1 − exp(−β¯hω)] ¯hω U = NkBT = = NkBT 2 (−1) 2 (−1) exp(−β¯hω) ∂T Z ∂T [1 − exp(−β¯hω)] kBT exp(−β¯hω) N¯hω = N¯hω = (6.2) 1 − exp(−β¯hω) exp(β¯hω) − 1 1 = N¯hωhn(T )i, where hn(T )i ≡ is the occupation factor. exp(¯hω/kBT ) − 1

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

52 PHYS 451 - Statistical Mechanics II - Course Notes 53 and oscillators, which you might already have anticipated. For example, are quantized packets of light, but light is also a wave (i.e. an oscillating field). Photons are also bosons, having integer (unit) spin.

Einstein constructed a model of a solid in 1907, where he assumed that the atoms making up the solid 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 understand experimental results obtained early in the 20th century that showed that the decreases exponentially at low temperatures. Assuming each oscillator could vibrate in three directions inde- pendently, the mean energy becomes 3N¯hω U = . exp(¯hω/kBT ) − 1 The specific heat is then  2 ∂U ¯hω exp(¯hω/kBT ) CV = = 3NkB 2 . ∂T kBT (exp(¯hω/kBT ) − 1) At low temperature, the exponential term becomes large, so we obtain

 ¯hω 2  ¯hω  CV (T → 0) ≈ 3NkB exp − , kBT kBT which is indeed exponentially small at low temperature. This was a great early success of the quantum theory, because it was the only theory to work, and helped vindicate the idea of quantum mechanics. Unfortunately, better experiments subsequently showed that the heat capacity at low temperatures 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 bonds from site to site are by definition strong. A better model, originally proposed by Debye in 1912, is to imagine a regular array of N masses m connected to each other by springs of length a with constant K. Suppose that the coordinate ηj describes the displacement of the mass at site j. The Hamiltonian (energy function) for the system is then

1 X h 2i H = mη˙2 + K (η − η ) . 2 j j+1 j j The force can be obtained by using

∂V ∂ K X 2 X F = − = − (η − η ) = −K [(η − η ) δ − (η − η ) δ ] i ∂η ∂η 2 j+1 j j+1 j i,j+1 j+1 j i,j i i j j

= −K (2ηj − ηj+1 − ηj−1) so that the equations of motion become

mη¨j − K (ηj+1 + ηj−1 − 2ηj) = 0. PHYS 451 - Statistical Mechanics II - Course Notes 54

With equilibrium positions identified as x = ja, one can posit that the solutions are given by travelling waves, ηj ∝ exp[i(kx − ωt)]. Imposing periodic boundary conditions η0 ≡ ηN requires 1 = exp(ikNa) which implies kNA = 2πn or k = (2πn/N)(1/a). These crystal momenta are in fact identical to the quantized momenta seen for the particle in a box k = 2πn/L, except that now we identify the total length L = Na with the product of crystal lattice spacings. Inserting the traveling wave solution into the equation of motion gives −mω2 − K eika + e−ika − 2 = 0. This can be simplified by noting that ka 1  2 1 sin2 = − eika/2 − e−ika/2 = − eika + e−ika − 2 . 2 4 4 We then obtain the dispersion relation for phonons 4K ka ω2(k) = sin2 . (6.3) m 2 For small wavevector k (momenta) or long wavelength, the frequency is linear: ω(k → 0) ≈ 2pK/m|k|a. This is the same as the dispersion relation for a regular wave, ω = ck, where p 2 c = a K/m is the wave velocity. It is quite different from the free-particle dispersion k ∝ k encountered in the context of the 3D box.

In the canonical ensemble, the mean energy is given by Z ∞ X V 2 ¯hω(k) U = ¯hωknk = k dk . 2  hω¯ (k)  2π 0 exp − 1 k kB T You might wonder why I blithely used the Bose distribution factor here. In principle, there are only two kinds of distributions that I am allowed to use and be consistent with quantum mechanics, the Bose or Fermi. Because there are no restriction about how many vibrations I am allowed to have in a 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 will be excited, so that we can make the identification ω(k) ≈ ck. The energy then becomes Z ∞ 3  4 Z ∞ 3 3V ¯hck 3V kBT x U ≈ 2   dk = 2 ¯hc x dx, (6.4) 2π 0 exp hck¯ − 1 2π ¯hc 0 e − 1 kB T where the factor of 3 comes from the three different polarizations: two transverse ones like the , and one longitudinal. There are 3N phonon modes in total; a one-dimensional chain with N sites has N modes, in 3D there are 3N. The last integral is related to a couple of special functions: Z ∞ xn x dx = ζ(n + 1)Γ(n + 1), 0 e − 1 where Γ(n) is Euler’s and the zeta function was defined earlier (5.5). For n = 3 one obtains Γ(4)ζ(4) = π4/15 so that the heat capacity becomes at low temperature ∂ 3V πk T 4 2π2 k T 3 C (T → 0) ≈ B ¯hc = V k B . V ∂T 30π2 ¯hc 5 B ¯hc PHYS 451 - Statistical Mechanics II - Course Notes 55

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

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

At higher temperatures, one needs to worry about higher-energy modes in the phonon distribution function. 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 have overlooked an important fact: phonons with wavelengths shorter than the lattice spacing cannot exist. So there is in fact a natural cut-off at high frequency (short wavelength) that will remove the in the integrals in practice.

This cut-off is called the Debye frequency ωD, and is determined at zero temperature where the dispersion relation is linear. With ω = ck the 3D density of states becomes V k2dk 3V ω2dω = , 2π2 2π2c3 where the factor of 3 again comes from the polarizations. So

Z ωD 2 3 3V ω V ωD 3N = 2 3 dω = 2 3 , 0 2π c 2π c which defines the Debye frequency

2 3 1/3 6π c N  1/3 ω ≡ = c 6π2n . (6.5) D V

2 1/3 The Debye temperature is then defined as θD ≡ ¯hωD/kB = (¯hc/kB)(6π n) .

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

Z ωD X 3kBTV h i F = −k T ln(Ξ) = k T g ln {1 − exp[−β( − µ)]} = ω2 ln 1 − e−hω/k¯ B T . B B i i 2π2c3 i 0

Setting x =hω/k ¯ BT gives

 3 Z θD /T  3  4  3V kBT kBT 2  −x T π F = 2 x ln 1 − e = −9NkBT , 2π ¯hc 0 θD 45 > where the value of the integral is obtained under the assumption θD/T ∼ 20; at higher temperatures it will be smaller, but the linear approximation to the dispersion relation will not be applicable. The specific heat is then found to be ∂2F 12π4  T 3 CV = −T 2 = NkB , ∂T 5 θD which is easily shown to be identical to the expression found earlier. PHYS 451 - Statistical Mechanics II - Course Notes 56

5

4

3

2 ) (arb. units) ν u( 1

0 0 5 10 15 20 ν (arb. units)

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 at all wavelengths. Conversely, these same bodies when heated emit light at all wavelengths, which might make their ‘blackbody’ moniker somewhat confusing. Of course, there is no real material that is 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 of the Big Bang. As the universe has expanded (effectively adiabatically), the temperature has cooled such that the current temperature is on the order of a few degrees . Alternatively, the fabric of spacetime has stretched to such an extent that the radiation emitted from the Big Bang (more or less 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 the temperature. This is plotted in Fig. 6.1. As a model of a natural blackbody, experimentalists studied a small 3D box (cavity) with a hole in it. The heat radiation absorbed by the box is transformed into 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 the hole was proportional to the area of the hole A and was related to the temperature through the Stefan-Boltzmann law: dQ = AσT 4, dt with σ ≈ 5.67 × 10−8 W/m2K4. There was no microscopic explanation for this result, however, nor did anyone know the expression for the distribution of wavelengths. In 1896 Wilhelm Wien suggested −c2/λT 5 u(λ) = c1e /λ , 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- PHYS 451 - Statistical Mechanics II - Course Notes 57 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 each oscillator has kBT of energy. Assuming a 3D cubic cavity of length L, the density of states is g(k)dk = (L3/π2)k2dk (the missing factor of two comes from the fact that light has two polarization directions). Using the relationship between wavevector and wavelength for photons k = 2π/λ one has k2 = 4π2/λ2 and dk = −(2π/λ2)dλ so that k2dk = −(8π3/λ4)dλ. The density of states is then g(λ)dλ = (8πL3/λ4)dλ. The distribution is then 8πk T u(λ) = B . λ4 Unfortunately this does not agree well with the observed distribution, because there is too much radiation implied at short wavelengths (λ → 0): you would be microwaved looking at a burning log! Even worse, the total energy density is infinite: Z ∞ u = u(λ)dλ = ∞, 0 whereas it should go like T 4. This was called the ‘’ by Ehrenfest in 1911, and it 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 the cavity 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 UN energy amont the N oscillators? Suppose the energy is made up of discrete elements, UN = M, where M  0. (This is the quantum hypothesis). Then the number of ways of distributing M indistinguishable energy states among N distinguishable oscillators is (N − 1 + M)! Ω = , M!(N − 1)! which we also derived in Eq. (5.12). You will now also recognize this as the number of ways of arranging indistinguishable bosons, Eq. (3.1). The Boltzmann entropy is then  M   M  M M  S = k ln(Ω) = k 1 + ln 1 + − ln N B B N N N N  U   U  U U  = k 1 + ln 1 + − ln B     This is strongly reminiscent of the Bose-Einstein entropy (4.7). Now, using the microcanonical definition of the temperature gives         1 ∂S 1 U 1 1 U 1 kB    = = kB ln 1 + + − ln − = ln + 1 . T ∂U V        U Inverting gives  U =   , exp  − 1 kB T which is the just the same result I obtained previously for a set of N oscillators, Eq. (6.2), except in that case I used the canonical ensemble. PHYS 451 - Statistical Mechanics II - Course Notes 58

Planck then connected the frequencies of oscillation to the energies of the (light) standing modes in the cavity. He used the classical wave equation for the light modes 1 ∂2y = ∇2y c2 ∂t2 and assumed that the light had to vanish at the walls, so that y(x, y, z) = A sin(kxx) sin(kyy) sin(kzz) 2 2 2 2 2 with kn = nπ/Ln. Inserting into the wave equation gives ∂ y/∂t = −c k y = −ω (k)y or ω(k) = ck, a linear spectrum just as found for phonons. Then with the final assumption that (k) ∝ ck or ω =hck ¯ , one obtains ¯hck hc 1 U(k) =   ⇒ U(λ) =   exp hck¯ − 1 λ exp hc − 1 kB T λkB T 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 8πhc 1 u(λ) = 5   . λ exp hc − 1 λkB T This fit the experimental data perfectly! For short wavelengths λ → 0 it reproduces Wien’s predic- 5 −hc/kB T λ 4 tion, u(λ → 0) ≈ (8πhc/λ )e . At long wavelengths one obtains u(λ → ∞) ≈ 8πkBT/λ → 0 which is the same as the Rayleigh-Jeans result.

It is straightforward to calculate the energy of radiation: dQ Z ∞ 8πhc 1 ≡ u = 5   dλ. dt 0 λ exp hc − 1 λkB T

2 With the replacement z = hc/λkBT , then dλ = −(hc/kBT z )dz and 4 Z ∞ 3 (kBT ) z u = 8π 3 z dz. (hc) 0 e − 1 The integral we have seen before, in Eq. (6.4), so the result is 8π5 (k T )4 u = B . 15 (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 8π5k4 σ = B 15(hc)3 3 4 which allows one to obtain the ratio h /kB 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 of which were known previous to this work.

This is the end of the story, more or less. It is interesting to note, though, that while Planck quantized the vibrational levels in the walls of the cavity, he used this only as a ‘trick’ to get a PHYS 451 - Statistical Mechanics II - Course Notes 59 sensible answer. At the time he didn’t believe that the energies were really discrete. He certainly didn’t seem to realize that his theory also implied the discreteness of the light energies. The theory was 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 bosons was really only tied together properly much later (in 1924) by Bose. But this is another interesting story, 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 in physics 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 and , and led to the development of the laser; but in reality the BEC of non-interacting particles envisaged by Bose and Einstein is very different from the phenomena in these condensed matter systems. BEC is essentially what happens when the temperature is lowered sufficiently that the bosons reach quantum degeneracy (that is, the de Broglie wavelength approaches the mean interparticle distance).

The basic problem with attaining a gaseous BEC is that when gases get cold enough, they condense into very ordinary liquids, and not into BECs. The trick that took 70 years is to try to keep the particles from ordinary condensation before reaching BEC. It turned out that spin-polarized gases were ideal candidates, because the spin-spin interaction is van der Waals-like, which means only very weakly attractive when the gases are sufficiently dilute. But if they are very dilute, it means that the mean separation is very large, which means ridiculously cold temperatures are needed to achieve BEC (on the order of nanoKelvins). The achievement of BEC followed years of experimental progress on trapping and cooling of atoms with lasers, which took place mostly in the ’70s. This is a 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 X 1 V Z ∞ k2dk N = = . exp[β( − µ)] − 1 2π2 exp[β( − µ)] − 1 i i 0 k

2 2 Making the substitution x = (¯h /2mkBT )k gives the mean density N/V :

 3/2 Z ∞ √ 1 2mkBT xdx n = 2 2 −1 x , 4π ¯h 0 z e − 1 where z ≡ exp(βµ) is the fugacity, as seen previously. Now, the mean density is fixed. So as the temperature decreases, clearly the integral also needs to increase in order to keep up. The only thing depending on the temperature is the fugacity, which appears in the denominator: exp(−βµ) needs to 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 accessible energy level 0, otherwise the occupation of a given level could be negative. At some point as the temperature drops, µ will hit zero, and the integral will no longer be able to properly count particles. PHYS 451 - Statistical Mechanics II - Course Notes 60

The particles will suddenly start disappearing! This is the phenomenon of Bose-Einstein condensation, 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 k=0 = 0 isn’t strictly included in the integral! The particles that appear to be disappearing are simply piling up in the ground state of the system, the zero-energy k = 0 state. 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 number 0 0 that 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 integral defining the Bessel functions Z ∞ xν−1dx Jν−1 ≡ x = Γ(ν)ζ(ν). 0 e − 1 √ 3 3  3  1 1  π In our case, ν = 2 and Γ 2 ≈ 2.612 and ζ 2 = 2 Γ 2 = 2 . So for a 3D box, the BEC transition temperature occurs when 1 2mk T 3/2 3 3 n = B c Γ ζ . 4π2 ¯h2 2 2 Inverting this gives a simple relationship between the transition temperature and the particle density:

¯h2 k T ≈ 3.31 n2/3. (6.6) B c m How does the BEC temperature relate to the quantum degeneracy temperature, when the de Broglie q 2 wavelength becomes comparable to the interparticle separation, λD = d? We have 2π¯h /mkBTd = 1/3 2 2/3 (1/n) orh ¯ n /mkBTd = 1/2π. Thus Td/Tc = 2π/3.31, which means that the degeneracy and 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:

3 nλD ≥ 2.612.

What is the fraction of atoms in the condensate for temperatures T < Tc?  3/2      3/2 0 1 2mkBT 3 3 T n = 2 2 Γ ζ = n . 4π ¯h 2 2 Tc

0 Together with n0 = n − n , we obtain

n n0  T 3/2 0 = 1 − = 1 − . (6.7) n n Tc

6.4.2 BEC in Lower Dimensions Is there a BEC transition in 2D or 1D systems? The issue is somewhat interesting, and relevant to modern physics, so let’s explore it briefly. The mean particle number is Z ∞ g()d N = , 0 exp[β( − µ)] − 1 PHYS 451 - Statistical Mechanics II - Course Notes 61

2 where in two dimensions the density of states g2D() = Am/2π¯h is independent of energy and in q √ 2 2 one dimension it is given by g1D() = m/8π ¯h (L/ ). In two dimensions, we obtain

Z ∞ Z ∞ mA d mAkBT dx N 2D = 2 −1 = 2 −1 x (substituting x = β) 2π¯h 0 z exp(β) − 1 2π¯h 0 z e − 1 ∞ A −x A = 2 ln 1 − ze = − 2 ln (1 − z) . λD 0 λD What happens when z → 1−, i.e. when µ → 0−? Then   1 kBT lim n2D = ln µ→0 2 λD |µ| which diverges logarithmically! So it would seem that the density in two dimensions is undefined as T → 0, implying that BEC is impossible. But there is a loophole: BEC can happen at exactly zero temperature. When µ approaches zero as temperature approaches zero, the ratio µ/kBT actually remains well-defined and finite. The transition is called the Kosterlitz-Thouless transition. The BEC is 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

r m Z ∞ d n = √ . 1D 2 2 −1 8π ¯h 0  [z exp(β) − 1] The same substitution x = β, and also z ≡ e−α, give

r Z ∞ Z ∞ −1/2 mkBT 1 dx 1 x dx n1D = 2 √ √ ≈ √ 2π¯h 4π 0 x [exp(x + α) − 1] λD 4π 0 x + α where in the last line I have recognized that the dominant contribution to the integral comes from the singular portion in the vicinity of x = 0. This last integral is each to evaluate, giving

2 1 Γ 1  r π 1 n = √ √2 = . λD 4π α 4α λD Clearly, this diverges for α → 0+ or µ → 0−, and so there is no BEC in 1D at finite temperature either. 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 infinite systems. 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 expansion of the gas, and therefore the singular behaviour. This is the same ‘infrared catastrophe’ that you might have heard of in the context of quantum electrodynamics (QED), and it’s no surprise: the mediators of the electromagnetic field are photons which have Bose-Einstein statistics. The problem goes away when the bosons are truly confined in finite , such as in harmonic oscillator traps, discussed in the next subsection. In this case, BEC persists at finite temperatures. PHYS 451 - Statistical Mechanics II - Course Notes 62

6.4.3 BEC in Harmonic Traps The spin-polarized atomic gases that have been coaxed into BEC have all been trapped 3D harmonic potentials, 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 much smaller than the temperature (which isn’t really true for current experiments, but you have to start somewhere!), the mean number of atoms is Z ∞ 2d 1 N = 3 . 0 2(¯hω) exp[β( − µ)] − 1 The BEC critical temperature is found by setting µ = 0 as for the box (if we had kept the zero-point motion, the we would set µ =h ¯(ωx + ωz + ωz)/2 rather than zero). With the substitution x = β, the mean number becomes:  3 Z ∞ 2  3 1 kBTc x dx 1 kBTc N = x = Γ(3)ζ(3), 2 ¯hω 0 e − 1 2 ¯hω where Γ(3) = 2 and ζ(3) ≈ 1.202. Inverting gives the transition temperature as a function of the number of atoms: 1/3 kBTc ≈ 0.94¯hωN . 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

N  T 3 0 = 1 − , N Tc 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 bimodal distribution. That is, the noncondensed atoms formed a cloud that looked more-or-less classical, and was identical to the semiclassical (equipartition) profile of the Fermi gas (5.19):

     2   2   2  0 1 µ − µext(r) 1 µ x y z n (r) = 3 exp = 3 exp − exp − 2 exp − 2 exp − 2 , λD kBT λD kBT Rx Ry Rz

2 2 where Ri = 2kBT/mωi . 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 Schr¨odingerequation, and I won’t bore you with the details, but give you the answer instead: N  x2 y2 z2  n (r) = 0 exp − − − , 0 3/2 2 2 2 π axayaz ax ay az

p R 3 where ai = ¯h/mωi. The prefactor guarantees that d rn0(r) = N0. Now, it is clear that both the BEC and noncondensate densities are Gaussian, but the length scales over which the Gaussians vary 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 on temperature, and for T ∼ Tc are on the order of tens of microns. To sum up: because of the big N0 factor in the BEC density, the total density looks like a sharp, small spike in the centre of an extended noncondensate density. This bimodial distribution can easily be seen simply by taking a picture of the atomic cloud with a CCD!