A-PDF Merger DEMO : Purchase from www.A-PDF.com to remove the watermark

1

Solutions to sample quiz problems and assigned problems

Sample Quiz Problems

Quiz Problem 1. Prove the expression for the Carnot efficiency for a perfectly reversible Carnot cycle using anideal gas.Solution: The ideal Carnot cycle consists of four segments as follows (1) An isothermal expansion during whichheat QH is added to the system at temperature TH ; (2) an adiabatic expansion during which the gas cools fromtemperature TH to TC ; (3) An isothermal compression during which heat QC is extracted from the system to a coldreservoir at temperature TC ; (4) Adiabatic compression during which the temperature of the gas rises from TC to TH .The efficiency of the engine is given by η = W/QH , where W is the useful work done by the system and QH is theheat added to the system. This is a thermodynamic cycle provided it is carried out reversibly (i.e. kept very closeto equilibrium), and then the internal energy is the same at the beginning and the end of the cycle. In that case,dU = 0, and Wtotal = QH −QL, so we can also write the efficiency as

η = 1− QLQH

= 1− TL∆S

TH∆S= 1− TL

TH(1)

The introduction of Entropy by Clausius greatly simplified understanding of engines.

—————–

Quiz Problem 2. Show that in the eigenfunction basis, the von Neuman entropy S = −kBtr(ρln(ρ)) reduces tothe Gibbs form S = −kB

∑i piln(pi) where the sum is over all eigenstates.

Solution. See Eq. (20) of Lecture notes.

—————–

Quiz Problem 3. Use Stirling’s approximation to show that,

ln(

(N

n

)) ≈ −N [pln(p) + (1− p)ln(1− p)], where p = n/N (2)

Solution. Using Stirling’s approximation we have,

ln(

(N

n

)) = ln(N !)− ln(n!)− ln(N − n)! = Nln(N)−N − nln(n) + n− (N − n)ln(N − n) + (N − n) (3)

which reduces to

ln(

(N

n

)) = Nln(N)− nln(n)− (N − n)ln(N − n). (4)

Defining p = n/N gives,

ln(

(N

n

)) = Nln(N)− pNln(pN)−N(1− p)ln(N(1− p)) = −N [pln(p) + (1− p)ln(1− p)] (5)

—————–

Quiz Problem 4. If S = −kBN [pln(p) + (1 − p)ln(1 − p)], by doing a variation with respect to p find the valueof p that gives the maximum entropy. Demonstrate that it is a maximum by showing that the second derivative withrespect to p is negative.

Solution. Taking a derivative with respect to p gives,

∂S

∂p= −kBN [ln(p)− ln(1− p)] (6)

2

The extrema occur at values of p = p∗ where this expression is zero, which yields p∗ = 1/2. To determine whetherthis is a minimum of maximum, we take the second derivative to find that,

∂2S

∂p2= −kBN [

1

p+

1

1− p]p=p∗ = −kBN (7)

Since the second derivative is negative the extremum at p∗ = 1/2 is a maximum. We have thus found the value ofthe probability at which the entropy is a maximum.

—————–

Quiz Problem 5. Use the Master equation to prove the second law of thermodynamics, i.e. in a closed systemdS/dt ≥ 0.

Solution. see lecture notes page 10.

—————–

Quiz Problem 6. Give three examples of systems where the ergodic hypothesis fails. Explain why it fails in thesecases.

Solution. Non-interacting classical gas, as in the absence of interactions the systems is non-chaotic. Coupledharmonic oscillators with no other interactions, as the normal modes do not share energy. Any system with spon-taneous symmetry breaking at low temperature, e.g. magnetic systems. Once in the up magnetized state at lowenough temperature the system stays in that state. Similarly once a crystal has formed from the gas phase at lowtemperature it almost never transforms to another crystal phase.

—————–

Quiz Problem 7. Why is the ideal gas law still a good starting point for high temperature gases, even though anon-interacting gas is non-ergodic?

Solution. Even very small interactions can make the system chaotic. These interactions have little effect onthe equilibrium thermodynamic behavior though they are critical to the transport properties.

—————–

Quiz Problem 8. Using pi = e−βEi/Z in the Gibbs form for the entropy, show that F = −kBT ln(Z), whereF = U − TS is the Helmholtz free energy. Here Z =

∑i e−βEi is the canonical partition function.

Solution. The Gibbs formula for the entropy is

S = −kB∑i

piln(pi). (8)

Using the Boltzmann probability in the canonical ensemble pi = exp(−βEi)/Z, we have,

S = −kB∑i

pi[−EikBT

− ln(Z)] =U

T+ kBln(Z) so U − TS = −kBT ln(Z), (9)

where we used∑i pi = 1, U =

∑i piEi

—————–

Quiz Problem 9. If x = e−β(Ei−Ej), so that wji → f(x). Show that the Metropolis algorithm is given byf(x) = min(x, 1).

Solution With this notation, the Metropolis algorithm is: If x > 1, wji = 1, and if x < 1, wji = x. This can beput in one equation as wji = f(x) = min(1, x).

3

—————–

Quiz Problem 10. If x = e−β(Ei−Ej), so that wji → f(x). Show that detailed balance is satisfied for any f(x)satisfying f(x)/f(1/x) = x. Moreover, show that f(x) = x/(1 + x), the heat bath algorithm, is one such solution.

Solution Since wij = 1/wji, if f(x) = wji, and if f(1/x) = wij , then detailed balance is satisfied providedf(x)/f(1/x) = x. For the heat bath case,

f(x)/f(1/x) = (x/1 + x)/(1/x/(1 + 1/x)) = x (10)

—————–

Quiz Problem 11. For a monatomic interacting classical gas, with interactions that only depend on the particleco-ordinates, derive the Maxwell Boltzmann distribution of velocities and show that the average kinetic energy isgiven by < KE >= 3NkBT/2.

Solution. See page 23 of the notes.

—————–

Quiz Problem 12. Using the fact that δE2 =< E2 > − < E >2= kBT2CV show that δE/E is proportional 1/N1/2.

Solution. See page 28 of the notes.

—————–

Quiz Problem 13. Write down the central difference form of the second derivative. Using this expression for theacceleration, write down the basic form of the Verlet algorithm for Molecular dynamics.

Solution. See. page 21-22 of the notes.

—————–

Quiz Problem 14. Given that the virial is defined to be G =∑k ~pk · ~rk. Explain why the average of its time

derivative is expected to be zero at long times.

Solution. The average of the time derivative of the virial gives,

<dG

dt>=

1

τ

∫ τ

0

dG

dtdt = (G(τ)−G(0))/τ

where the average is taken over the time interval τ . Since G is bounded, as τ goes to infinity, the average of the timederivative of G goes to zero.

—————–

Quiz Problem 15. The ground state structure of Argon is a fcc crystal. If a MD simulation is carried out wherea high temperature structure is quenched to a low temperature where the fcc structure is the equilibrium state. Ifthe starting high temperature state in your MD simulation is a liquid phase, do you expect the simulation to find thecorrect fcc state? Explain.

Solution. For most particle systems it is difficult to find the low temperature ground state, so the fcc single crystalstate is usually not found. In experiments a polycrystal is quit often found but it is harder to make single crystals.In MD simulations, the timescale is very short so unless special conditions are applied, the low temperature state isa frozen glass state.

4

—————–

Quiz Problem 16. In a MC calculation of the ferromagnetic Ising model on a square lattice, you calculate themagnetization as a function of temperature as accurately as you can. However your simulations remain roundedat the critical point, instead of showing the behavior |Tc − T |β expected in the thermodynamic limit. Explain thephysical origins of the rounding that you observe. Your experimental colleague does a measurement of a model Isingmagnet that consists of quantum dots with around 1000 spins in each quantum dot. She also observes a roundingof the behavior of the magnetization. Provide some reasons why her magnetization experiment also gives a roundedbehavior near the critical point.

Solution. Three reasons for rounding of the transition are finite size effects, sample inhomogeneity and non-equilibrium effects. Non-equilibrium effects occur when the system is not fully equilibrated. In experiments all threeeffects occur, while in simulations the sample inhomogeneity effect is avoided.

—————–

Quiz Problem 17. Derive the relation,

PV = NkBT +1

3

internal∑k

~rk · ~Fk (11)

Explain the physical origin of each of the three terms. This relation is not always true for periodic boundary condi-tions, though rather fortuitiously it is true for periodic boundaries provided the interaction potential is a pair potential.

Solution. See page 23-24 of the notes.

—————–

Quiz Problem 18. Show that

(δE)2 = kBT2CV (12)

Solution. See page 28 of the notes.

Solutions to assigned problems

Problem 1. Find the relation between pressure and volume for an ideal gas under going an adiabatic process.

Solution:. In an adiabatic process no heat is added to the system, so dU+dW = 0, where dW = PdV . Combiningthe ideal gas law and the equipartion theorem, we have

dU = αNkBdT = αd(PV ) = α(V dP + PdV ) (13)

where α = DOF/2. Using the first law, we then have,

−PdV = α(V dP + PdV ) or αdP

P= −(α+ 1)

dV

V(14)

Integrating gives,

P

P0= (

V0

V)1+1/α (15)

which may be written in its most common form,

PV γ = constant where γ =α+ 1

α(16)

For a mono-atomic ideal gas in three dimensions there are three translational degrees of freedom, so DOF = 3,α = 3/2, and γ = 5/3.

5

—————–

Problem 2: Derive the ideal monatomic gas law using kinetic theory. You may use the equipartition resultU = 3

2NkBT .

Solution: The internal energy of an ideal gas is purely kinetic energy, so that,

U =3

2NkBT =

1

2

∑i

m < [(vix)2 + (viy)2 + (viz)2] >=

1

2Nm < ~v2 > (17)

The pressure is calculated by considering a particle incident normally on a perfectly reflecting wall,

Fx = max = m∆pxδt

=2mvxδt

(18)

The time taken for the particle to strike the wall again is δt = 2L/vx, so that

Fwall =2mv2

x

2Lso that P =

∑i F

ix

A=m∑i(v

ix

2)

AL(19)

Noting that V = AL, with∑i(v

ix)2 = 1

3 < ~v2 > and combining (12) and (14) yields,

PV = m∑i

(vix)2 = NkBT (20)

—————–

Problem 3. Find the degeneracy Ω(E,N) for N a spin 1/2 Ising system with Hamiltonian H = −h∑i Si.

Solution: The energy of a spin state depends only on the number of up spins Nu, and the number of down spinsNd, but not their arrangement. We then have,

E = −h2

(Nu −Nd), with degeneracy Ω(E) =N !

Nu!Nd!(21)

We also have N = Nu +Nd, so that

E = −h2

(2Nu −N) or Nu =N

2− E

h, Nd =

E

h− N

2(22)

Since entropy is the kBln(Ω(E)), we use Stirling’s approximation to write,

ln(Ω(E)) = NlnN −N −NulnNu +Nu −NdlnNd +Nd = NlnN −NulnNu −NdlnNd (23)

It is useful to introduce the fraction f = Nu/N , so that Nd/N = 1− f and

ln(Ω) = NlnN − fNln(fN)− (1− f)Nln[(1− f)N ] = N [fln(f) + (1− f)ln(1− f)] (24)

which shows that the entropy in this model is extensive. In terms of f , E = (Nh/2)(2f − 1). We define ε = E/N , sothat f = 1/2− ε/h. Finally the entropy is,

S = −kBln(Ω(E)) = −kBN [(1/2− ε/h)ln(1/2− ε/h) + (ε/h− 1/2)ln(ε/h− 1/2)] (25)

—————–

6

Problem 4. Consider a random walk on a one dimensional lattice where the probability of stepping to the rightor to the left is 1/2. Find the probability that after N steps the walker is at position x lattice units from the startingposition. Use Stirling’s approximation to show that for large N the probability distribution is a Gaussian in x. Findthe average mean square distance the walker is from the origin as a function of N .

Solution: Let us define the number of steps that the student takes to be N = t/δt. The number of steps taken inthe positive x-direction, n, and the number taken in the negative x-direction, m, must add up to N , so that,

N = n+m (26)

The x position of the student at time t = Nδt is given by x = dδx, where

d = (n−m). (27)

But what is the probability that the student reaches this position at time t?

P (d,N) =1

2NN !

m!n!=

1

2NN !

((N + d)/2)!((N − d)/2)!(28)

Stirling’s approximation:

Log(n!) ≈ nln(n)− n (29)

We then have,

Ln(P (d,N)) ≈ −NLn(2) +NLn(N)−N

−N + d

2Ln((N + d)/2) + (N + d)/2

−(N − d

2)Log((N − d)/2) + (N − d)/2 (30)

This reduces to,

Ln(P (d,N)) ≈ −NLn(2) +NLn(N)

−N + d

2[Ln(N) + Ln(1 + d/N)− Ln(2)]

−(N − d

2)[Ln(N) + Ln(1− d/N)− Ln(2)] (31)

Defining y = d/N and expanding for small y yields,

ln(P (d,N)) = −N2

[(1 + y)ln(1 + y) + (1− y)ln(1− y)] ≈ −N2

[(1 + y)(y− 1

2y2) + ...) + (1− y)(−y− 1

2y2) + ...)] (32)

To leading order this yields

ln(P (d,N)) ≈ −Ny2 = −d2

N(33)

or,

P (d,N) ≈ e−d2/N → e−x

2δt/(δx)2t. (34)

Comparing to the solutions to the continuum diffusion equation in one dimension, this shows that the diffusionconstant is related to the stochastic parameters δx and δt through,

D ≈ (δx)2/δt. (35)

From the probability distribution (28), it is easy to see that the average displacement of an ink particle is < x >= 0,but the typical distance from the origin is (< x2 > − < x >2)1/2 ≈ N1/2δx.

7

—————–

Problem 5. Prove the central limit theorem for a set of N independent subsystems that each have an energy Eidrawn from a Gaussian distribution with standard deviation σ. You will need to use the integral,∫ ∞

−∞e−ax

2+bxdx = (π

a)1/2e

b2

4a (36)

Solution: Consider any intensive thermodynamic variable Xi (e.g. E/N) that has a different value in each of Nsubsystems. Consider each subsystem to have the same volume and that each subsystem is statistically equivalent.Each subsystem is assumed to have a statistical behavior that is Gaussian distributed, so that,

P1(Xi) =(2πσ2

)−1/2e−

X2i

2σ2 (37)

where we assumed that each subsystem has the same statistical variations, quantified by σ. We would like to calculatethe statistical behavior of the quantity

XN =1

N

N∑i=1

Xi (38)

The statistical behavior of XN is given by,

PN (XN ) =(2πσ2

)−N/2 ∫ ∞−∞

∏i

dXie− X2

i2σ2 δ(

∑i

Xi −NXN ) (39)

Using the delta function representation,

δ(α) =1

2π

∫ ∞−∞

eiyαdy (40)

we find,

PN (XN ) =(2πσ2

)−N/2 ∫ ∞−∞

1

2π

∫ ∞−∞

dy∏i

dXie− X2

i2σ2 eiy(

∑iXi−NXN ) (41)

which is equivalent to

PN (XN ) =(2πσ2

)−N/2 ∫ ∞−∞

dy

2πe−iyNXN [I]N (42)

where

I =

∫ ∞−∞

dXe−X2

2σ2 eiyX =(2πσ2

)1/2e−

σ2y2

2 (43)

Finally,

PN (XN ) =

∫ ∞−∞

dy

2πe−iyNXN e−

Nσ2y2

2 =1

(2πN)1/2σe−X

2N/2σ

2N (44)

where σN = σ/N1/2

—————–

Problem 6. Find the entropy of mixing for a system consisting of m different atom types, with ni of each atomtype and with the total of atoms, N =

∑mi ni.

Solution: The number of ways of arranging the system is given by the multinomial,

Ω(ni) =N !

n1!n2!...nm!(45)

8

Using Stirling’s approximation we find,

ln(Ω) = NlnN −N − [∑i

(niln(ni)− ni)] = NlnN −N − [∑i

(Npiln(Npi)−Npi)] (46)

where pi = ni/N . Using∑i pi = 1 this simplifies to,

S = kBln(Ω) = −kBN∑i

piln(pi) (47)

which is the same form as the Gibbs entropy, though the interpretation of pi is different here as it is not the probabilityof an individual many body state.

—————–

Problem 7. Derive Stirling’s approximation ln(N !) = Nln(N)−N .

Solution. Expanding the logarithm of N ! as a sum we write,

ln(N !) =

N∑i=1

ln(i) ≈∫ N

1

ln(x)dx = [xln(x)− x]|N1 = Nln(N)−N + 1 (48)

In the large N limit the constant “one” is negligible.

—————–

Problem 8. Find the entropy of a system of N non-interacting gas particles in a box of volume V = Ld, where dis the dimension. Consider two cases: Where the particles are distinguishable; and where they are indistinguishable.Is the entropy extensive in both cases? Assume that all configuration of the particles have the same energy, so wecan ignore the kinetic energy term.

SolutionIn the case where all of the particles are distinguishable, the number of available configurations is V N as each

particle can be at any position in the volume. The entropy is then S = kBNln(V ), which is not extensive. Gibb’srealized that identical particles are not distinguishable so the naive calculation of the number of states V N is notvalid. Instead, we must order the co-orindates to eliminate overcounting of indistinguishable particles.

Consider a set of particles with co-ordinates ~r1...~rN . If we order one component of these co-ordinates, say thex-component so that x1 ≥ x2 ≥ x3... ≥ xN then the co-ordinates of the particles are always different hence ensuringthat there is no overcounting. The integration over the spatial co-ordinates then becomes,

Ω = Ld−1

∫ L

0

dx1

∫ x1

0

dx2...

∫ xN−1

0

dxN = L(d−1)N LN

N !=V N

N !(49)

The entropy of indistinguishable particles is then,

S = kBN [ln(V )− ln(N) + 1]→ NkBln(V/N) (50)

which is extensive for V/N finite.

—————–

Problem 9. Prove the central limit theorem holds for any distribution provided it has a finite standard deviation.i.e. extend the solution to problem 5 to arbitrary distributions that have a finite second moment.

Solution We consider that the statistics of an extensive quantity X obeys a distribution g(X) that has mean X∞and standard deviation σ. Without loss of generality, we can set X∞ = 0 by a simple shift of X. The statisticalbehavior of XN is given by,

PN (XN ) =(2πσ2

)−N/2 ∫ ∞−∞

∏i

dXig(xi)δ(∑i

Xi −NXN ) (51)

9

Using the delta function representation,

δ(α) =1

2π

∫ ∞−∞

eiyαdy (52)

we find,

PN (XN ) =(2πσ2

)−N/2 ∫ ∞−∞

1

2π

∫ ∞−∞

dy∏i

dXig(Xi)eiy(

∑iXi−NXN ) (53)

which is equivalent to

PN (XN ) =(2πσ2

)−N/2 ∫ ∞−∞

dy

2πe−iyNXN [I]N (54)

where

I =

∫ ∞−∞

dXg(X)eiyX (55)

Now X is extensive, so the exponential is a rapidly oscillating function and the integral is dominated by the regionnear y = 0. We then expand the exponential to find,

I =

∫ ∞−∞

dXg(X)eiyX ≈∫ ∞−∞

dXg(X)(1− iyX − 1

2y2X2 + ...) (56)

Assuming that g(X) is symmetric, this reduces to,

I =

∫ ∞−∞

dXg(X)eiyX ≈ (1− 1

2y2σ2 + ...) ≈ e− 1

2y2σ2

(57)

Finally, this is of the same form as that of Problem 5 and hence the result there applies,

PN (XN ) =

∫ ∞−∞

dy

2πe−iyNXN e−

Nσ2y2

2 =1

(2πN)1/2σe−X

2N/2σ

2N (58)

where σN = σ/N1/2

—————–

Problem 10. Which of the following is an exact differential? a) dU(x, y) = 2xdx + 3xydy; b) dU(x, y) =(−1/y)dx + (x/y2)dy; c) dU(x, y) = −ydx + xdy ? For these three examples, set dU = 0 and solve the differentialequations to find y(x). In cases where the differential is exact, find U(x, y).

Solution:a) We have

∂U

∂x= 2x; and

∂U

∂y= 3xy (59)

so that

∂2U

∂x∂y= 1/y2 6= ∂2U

∂y∂x= 3y (60)

Therefore dU IS NOT an exact differential.b) We have

∂U

∂x=−1

y; and

∂U

∂y=

x

y2(61)

10

so that

∂2U

∂x∂y=

1

y2=

∂2U

∂y∂x=

1

y2(62)

Therefore dU IS an exact differential.c) We have

∂U

∂x= −y; and

∂U

∂y= x (63)

so that

∂2U

∂x∂y= −1 6= ∂2U

∂y∂x= 1 (64)

Therefore dU IS NOT an exact differential.Now we solve the three cases and if possible find U .a) We find,

dy

dx=−2

3y, with solution y =

4

3(c− x)1/2; or y = −4

3(c+ x)1/2 (65)

b) In this case,

dy

dx=y

x, with solution y = cx (66)

This is an exact differential, where U(x, y) = −x/y.c) The solution for dU = 0 is the same as b), but for finite U , it is not an exact differential so we cannot find

U(x, y) to satisfy the equation.

—————–

Problem 11. By using the fact that S is extensive show that

N

(∂S

∂N

)V,U

+ V

(∂S

∂V

)N,U

+ U

(∂S

∂U

)N,V

= S (67)

and hence that Nµ = U + PV − TS.

Solution. Taking a derivative with respect to lambda of the homogeneous equation,

λS(U, V,N) = S(λU, λV, λN) (68)

and then setting λ = 1 yields Eq. (68). From the fundamental thermodynamic relation we have,

dS =1

TdU +

P

TdV − µ

TdN =

(∂S

∂U

)N,V

dU +

(∂S

∂V

)N,U

dV +

(∂S

∂N

)V,U

dN (69)

Using the relations inferred from this equation in Eq. (68) yields Nµ = U + PV − TS.

—————–

Problem 12. Using the fundamental thermodynamic relation and Nµ = U + PV − TS prove the Gibbs-Duhemrelation.

Solution. From the fundamental thermodynamic relation we have,

dU = TdS − PdV + µdN = d(TS − PV + µN) (70)

From the equality of the last two expressions we deduce

SdT − V dP +Ndµ = 0 (71)

11

—————–

Problem 13. From the fundamental thermodynamic relation show that,(∂µ

∂P

)S,N

=

(∂V

∂N

)S,P

. (72)

Solution: Since the independent variables are S, P,N , we choose the enthalpy and write,

dH = TdS + V dP + µdN =

(∂H

∂S

)P,N

dS +

(∂H

∂P

)S,N

dP +

(∂H

∂N

)S,P

dN (73)

hence, (∂H

∂P

)S,N

= V ;

(∂H

∂N

)S,P

= µ. (74)

By doing a derivative with respect to N of the first equation in (75) we find,(∂

∂N

(∂H

∂P

)S,N

)S,P

=

(∂V

∂N

)S,P

. (75)

while a derivative with respect to P of the second equation in (75) yields(∂

∂P

(∂H

∂N

)S,P

)S,N

=

(∂µ

∂P

)S,N

. (76)

Because dH is an exact differential, the order of the derivatives of H in Eq. (76) and (77) does not matter so theseexpressions are equivalent hence proving the Maxwell relation (73).

—————–

Problem 14. From the fundamental thermodynamic relation show that,(∂CP∂P

)T,N

= −T(∂2V

∂T 2

)P,N

. (77)

Solution: Since the independent variables in the expression are T, P , we consider the Gibb’s free energy G(T, P,N),so that,

dG = −SdT + V dP + µdN =

(∂G

∂T

)P,N

dT +

(∂G

∂P

)T,N

dP +

(∂G

∂N

)T,P

dN (78)

hence, (∂G

∂T

)P,N

= −S;

(∂G

∂P

)T,N

= V. (79)

This leads to the Maxwell relation,

−(∂S

∂P

)T,N

=

(∂V

∂T

)P,N

(80)

Now take a temperature derivative at constant pressure of this equation, to find,

−

(∂

∂T

(∂S

∂P

)T,N

)P,N

=

(∂2V

∂T 2

)P,N

(81)

12

We can change the order of the derivatives of the expression on the left hand side, and use the fact that CP =T (∂S/∂T )P to write,

−(∂(CP /T )

∂P

)T,N

=

(∂2V

∂T 2

)P,N

; or

(∂CP∂P

)T,N

= −T(∂2V

∂T 2

)P,N

(82)

which solves the problem.

—————–

Problem 15. The equipartition theorem states that for a classical particle system of volume V with N particles,described by Hamiltonian H, the following relations hold,

< si∂H

∂sj>= kTδij (83)

where si can be either the spatial or momentum co-ordinate. Using the canonical partition function, carry out theintegrals explicitly to verify this theorm for a system described by the quadratic Hamiltonian,

H =∑i

p2i

2m+∑i

1

2kq2i . (84)

Find the internal energy of this system as a function of temperature. Is this consistent with the equipartion theorem?

Solution We need to evaluate the following quantities explicitly,

< qi∂H

∂qj>= k < qiqj >; < pi

∂H

∂pj>=

1

m< pipj > (85)

It is convenient to define,

I(l)i =

∫(qi)

le−βkq2i /2dqi; J

(l)i =

∫(pi)

le−βp2i /(2m)dpi (86)

It is then evident that,

< qiqj >=

∫dωqiqje

−βH∫dωe−βH

(87)

Reduces to,

< qiqj >=I

(1)i I

(1)j

I(0)i I

(0)j

, for i 6= j; < qiqj >=I

(2)i

I(0)i

, for i = j; (88)

Now note that I(i l) and J

(i l) are zero for l odd, and also that they are independent on the index i, we thus find,

< qiqj >= δij

∫∞−∞ q2e−βkq

2/2dq∫∞−∞ e−βkq2/2dq

(89)

Now we use the standard integrals, ∫ ∞−∞

q2ne−αq2

dq =(2n− 1)!!

(2α)n(π

α)1/2 (90)

where (2n−1)!! = (2n−1)(2n−3)... and is one for n = 0 or n = 1. Using this result verifies the equipartition formula.The calculation for the momentum is essentially the same.

The internal energy is < H > and since we have found that k < q2i >=< p2

i > /m = kBT , the internal energyof this problem is 3NkBT . This is consistent with the equipartition formula that assigns kBT/2 for every quadraticdegree of freedom. Here there are six degrees of freedom of that type.

13

—————–

Problem 16. You discover a new state of matter that has equation of state and internal energy given by,

P = AT 3/V, U = BTnLn(V/V0) + f(T ) (91)

where B,n, V0 are constants while f(T ) only depends on the temperature. Find a numerical value for n, and find arelation between the constants B and A.

Solution It is convenient to define an entropy that is a Legendre transform of the conventional entropy so that,

SL = S − U

T(92)

We then have,

dSL =P

TdV +

U

T 2dT (93)

where we used dS = dU/T +PdV/T . The Maxwell relation derived from dSL in combination with the functions givenin the problem definition yield,

∂

∂T(AT 2

V) =

∂

∂V(BTn−2ln(V/V0) +

f(T )

T 2) (94)

This equation is satisfied provided n = 3 and B = 2A.

—————–

Problem 17. From the fundamental thermodynamic relation show that (here we use β = 1/T )

T

(∂N

∂T

)V,βµ

=

(∂E

∂µ

)T,V

(95)

Solution: Since the independent variables are β, V, βµ, we start with the entropy form of the fundamental relation,

dS = βdU + βPdV − βµdN (96)

where β = 1/T . Since the independent variables in the desired expression are β, V, βµ, we define, Y = S−βU+(βµ)N ,so that,

dY (β, V, βµ) = −Udβ + βPdV +Nd(βµ) =

(∂Y

∂β

)V,βµ

dβ +

(∂Y

∂V

)β,βµ

dV +

(∂Y

∂βµ

)β,V

d(βµ) (97)

hence, (∂Y

∂β

)V,βµ

= −U ;

(∂Y

∂(βµ)

)β,V

= N, (98)

that leads to the Maxwell relation,

−(

∂U

∂(βµ)

)β,V

=

(∂N

∂β

)V,βµ

(99)

Since β is constant for the left hand side derivative, we can take it outside. For the right hand derivative, we use thechain rule to find

−1

β

(∂U

∂µ

)β,V

=

(∂N

∂T

)βµ,V

(∂T

∂β

)βµ,V

(100)

which reduces to the equation posed in the problem.

14

—————–

Problem 18. From the fundamental thermodynamic relation show that

1

T 2

(∂T

∂βµ

)E,V

= −(∂N

∂E

)βµ,V

and −ρ2

(∂(S/N)

∂ρ

)T,N

=

(∂P

∂T

)ρ,N

(101)

where ρ = N/V is the number density.

Solution: For the first equation, the variables held constant are βµ and E, so we use the entropy form of thefundamental relation, dS = βdU + βPdV − βµdN . We also define Y = S + (βµ)N to find a relation that has βµ asan independent variable. We then have,

dY (U, V, βµ) = βdU + βPdV +Nd(βµ) =

(∂Y

∂U

)V,βµ

dV +

(∂Y

∂V

)U,βµ

dV +

(∂Y

∂βµ

)U,V

d(βµ) (102)

which leads to the Maxwell relation, (∂β

∂(βµ)

)U,V

=

(∂N

∂U

)V,βµ

(103)

Using the chain rule on the left hand side yields,(∂β

∂(βµ)

)U,V

=

(∂β

∂T

)U,V

(∂T

∂(βµ)

)U,V

= − 1

T 2

(∂T

∂(βµ)

)U,V

(104)

which is the first relation of Eq. (101). The second relation of equation (101) has T, V,N as independent variables,so we use the Helmholtz free energy,

dF = −SdT − PdV + µdN =

(∂F

∂T

)V,N

dT +

(∂F

∂V

)T,N

dV +

(∂F

∂N

)T,V

dN (105)

leading to the Maxwell relation, (∂S

∂V

)T,N

=

(∂P

∂T

)V,N

(106)

The expression on the left hand side may be rewritten using the chain rule,(∂S

∂ρ

)T,N

=

(∂S

∂V

)T,N

(∂V

∂ρ

)T,N

= −V2

N

(∂S

∂V

)T,N

(107)

Equation (106) is then,

−N2

V 2

(∂(S/N)

∂V

)T,N

= −ρ2

(∂(S/N)

∂V

)T,N

=

(∂P

∂T

)V,N

=

(∂P

∂T

)ρ,N

(108)

which is the second relation in Eq. (101)

—————–

Problem 19. From the second result of problem 18 show that(∂P

∂ρ

)S/N,N

=

(∂P

∂ρ

)T,N

+ [

(∂P

∂T

)ρ,N

]2NT

ρ2CV(109)

Solution: We use the relation for a non-natural derivative,(∂P

∂ρ

)S/N,N

=

(∂P

∂ρ

)T,N

+

(∂P

∂T

)ρ,N

(∂T

∂ρ

)S/N,N

(110)

15

Using the triple product rule, we have,(∂ρ

∂T

)S/N,N

(∂(S/N)

∂T

)ρ,N

(∂ρ

∂(S/N)

)T,N

= −1 (111)

This yields, (∂ρ

∂T

)S/N,N

= −(

∂T

∂(S/N)

)ρ,N

(∂(S/N)

∂ρ

)T,N

=NT

CV

1

ρ2

(∂P

∂T

)ρ,N

(112)

Using the second expression in (101) and (112) in (110) yields the desired result.

—————–

Problem 20. Consider a system with an order parameter, x. If the volume(V ), energy(E) and particle number(N)are fixed, then the system will choose the value of x with the maximum entropy. Now consider that the same systemis connected to a reservoir that has temperature T and chemical potential µ, so it can exchange energy and particleswith the reservoir. Show that the total entropy, that is the sum of the entropy of the reservoir and the system, ismaximized when (

∂P

∂x

)T,µ,V

= 0 (113)

Hint: You can assume the system and reservoir are at the same temperature and chemical potential as required byequilibrium. Moreover you may use PV = TS − E + µN .

Solution: Since the variables that are kept fixed are T, µ, V , we use the grand potential ΦG = U − TS − µN , sothat,

dΦG(T, V, µ) = −SdT + PdV −Ndµ = −d(PV ) (114)

Now we take a derivative of this expression with respect to x at fixed T, V, µ. Since the entropy is maximized, theGrand potential is minimized, so that (

∂ΦG∂x

)T,V,µ

= 0 = −V(∂P

∂x

)T,V,µ

(115)

—————–

Problem 21. Consider a system where the thermodynamics are a function of temperature and another variable xwhich can be varied by an experimentalist. Assuming that(

∂E

∂S

)x

= T ; and using F = E − TS (116)

show that, (∂E

∂x

)S

=

(∂F

∂x

)T

(117)

Solution: We take a derivative of the equation F = E − TS with respect to x at constant T , to find,(∂F

∂x

)T

=

(∂E

∂x

)T

− T(∂S

∂x

)T

(118)

Using the formula for a non-natural derivative, we have,(∂E

∂x

)T

=

(∂E

∂S

)x

(∂S

∂x

)T

+

(∂E

∂x

)S

(119)

Using this result, along with the first of Eq. (116), in Eq. (118) proves Eq. (117).

16

—————–

Problem 22. Show that

CV = −T(∂2F

∂T 2

)N,V

(120)

Solution: The specific heat at constant volume is given by,

CV = T

(∂S

∂T

)V,N

(121)

The Helmholtz free energy is given by,

F = U − TS; so dF = −SdT − PdV + µdN =

(∂F

∂T

)V,N

dT +

(∂F

∂V

)T,N

dV +

(∂F

∂N

)T,V

dN (122)

so that,

S = −(∂F

∂T

)V,N

; and

(∂S

∂T

)V,N

= −(∂2F

∂T 2

)V,N

(123)

Multiplying by T and using Eq. (121) proves the relation (120).

—————–

Problem 23. Show that

U =

(∂(F/T )

∂(1/T )

)N,V

(124)

Solution: Defining β = 1/T , we have,(∂(βF )

∂β

)N,V

= F + β

(∂F

∂β

)N,V

= F − T(∂F

∂T

)N,V

(125)

where the last expression on the right hand side is found using the chain rule. We also have,

dF = −SdT − PdV + µdN =

(∂F

∂T

)V,N

dT +

(∂F

∂V

)T,N

dV +

(∂F

∂N

)T,V

dN (126)

so that (∂F

∂T

)V,N

= −S. (127)

Therefore, (∂(βF )

∂β

)N,V

= F + TS = U (128)

as posed in Eq. (124).

—————–

Problem 24. Show that

V

(∂P

∂T

)µ,N

= S; and V

(∂P

∂µ

)T,N

= N (129)

17

Now express the pressure of the ideal classical gas in terms of the variables µ and T and using this expression verifythe equations above.

Solution: We use the fact that the Gibbs free energy is µN to write(∂G

∂P

)T,N

= N

(∂µ

∂P

)T,N

(130)

We also have,

dG =

(∂G

∂T

)P,N

dT +

(∂G

∂P

)T,N

dP +

(∂G

∂N

)T,P

dN = −SdT + V dP + µdN (131)

so that (∂G

∂P

)T,N

= V. (132)

Using this relation in Eq. (130) proves the second identity in (129). To prove the first relation in (129), we use thetriple product identity, (

∂P

∂T

)µ,N

(∂µ

∂P

)T,N

(∂T

∂µ

)P,N

= −1. (133)

From Eq. (134) and G = µN , we have,

−S =

(∂G

∂T

)P,N

= N

(∂µ

∂T

)P,N

(134)

Using the second relation in Eq. (129) for the second term in the product (133) and using (134) leads to the firstrelation in (129).

—————–

Problem 25. From the fundmental thermodynamic relation show that(∂S

∂P

)T,µ

= −(∂V

∂T

)P,µ

(135)

Solution: The fundamental thermodynamic relation is dU = TdS − PdV + µdN , so the independent variables areS, V,N . To find the desired expression using a Maxwell relation approach, we need independent variables T, P, µ, sowe define Y = U − TS + PV − µN , so that,

dY = −SdT + V dP −Ndµ =

(∂Y

∂T

)P,µ

dT +

(∂Y

∂P

)T,µ

dP +

(∂Y

∂µ

)T,P

dµ (136)

so that, (∂Y

∂T

)P,µ

= −S;

(∂Y

∂P

)T,µ

= V ;

(∂Y

∂µ

)T,P

= −N (137)

and hence, (∂

∂P

(∂Y

∂T

)P,µ

)T,µ

= −(∂S

∂P

)T,µ

(∂

∂T

(∂Y

∂P

)T,µ

)P,µ

=

(∂V

∂T

)P,µ

(138)

Since the order of differentiation does not matter, as dY is an exact derivative, the relation (135) is proven.

—————–

18

Problem 26. Consider a single harmonic oscillator with energies εn = hω(n+1/2). Find the average energy of thesystem as a function of temperature. Find the average occupancy of each energy level as a function of temperature.Is there are regime where your result for the average energy is in agreement with the equipartition theorem?

Solution The energy levels of the Harmonic oscillator are given by εn = hω(n + 1/2), so the canonical partitionfunction is given by,

Z1 =∑n=0

e−βhω(n+1/2) =e−βhω/2

1− e−βhω(139)

so the average energy is,

< E >= −∂lnZ∂β

=1

2hω +

hω

eβhω − 1(140)

This can be written as,

< E >= hω(< n > +1/2); where < n >=1

eβhω − 1(141)

In the high temperature limit, this reduces to < E >= 12 hω + kBT which has the equipartition term (2 degrees of

freedom) plus a quantum correction that is negligible at high temperature.

—————–

Problem 27. For a pure state show that

ρ2 = ρ (142)

Also show that for a mixed state this identity fails.

Solution For a pure state,

(ρ)nm = anka∗mk (143)

and

(ρ2)nm =∑l

ana∗l ala

∗m = anka

∗mk (144)

where we used∑l a∗l al = 1 for an orthonormal wavefunction. For a mixed state the elements of the density matrix

are given by,

(ρ)nm =∑k

pkanka∗mk (145)

so the elements of the square of the density matrix are,

(ρ2)nm =∑kk′

∑l

pkpk′anka∗lkalk′a

∗mk′ (146)

which is clearly not the same. The purity of the mixed state is,

Pu = tr(ρ2) =∑kk′

∑ln

pkpk′anka∗lkalk′a

∗nk′ (147)

—————–

19

Problem 28. Consider two single particle pure states with wavefunctions

|ψ >±=1

21/2(|0 > ±|1 >) (148)

Write down the density matrix for each of these pure states and verify that they have purity equal to one. Nowconsider the density matrix of a mixed state consisting of ρ = p+ρ+ + p−ρ−, where p+ + p− = 1. Find the purity ofthis state, and the value of the mixing probability p+ at which the purity is smallest.

ρ+ =1

2

(1 11 1

)ρ− =

1

2

(1 −1−1 1

)(149)

(ρ+)2 =1

4

(2 22 2

)(ρ−)2 =

1

4

(2 −2−2 2

)(150)

which confirms the identity. The purity is clearly Pu = 1. For the mixed case we have,

ρ =p+

2ρ+ +

p−2ρ− (151)

From which we find,

(ρ)2 = (p+

2ρ+ +

p−2ρ−)2 = (p+)2ρ+ + (p−)2ρ− = (p+)2 + (1− p+)2 = 1− 2p+ + 2p2

+ ≤ 1 (152)

the minimum occurs at p+ = 1/2 and the maximum at p+ = 1.

Problem 29. Show that Eqs. (179) of the lecture notes reduce to Eqs. (178) in the eigenvalue basis.

Solution We use the relation

UHU† = λE (153)

where U is a unitary matrix and λE is the diagonal matrix with the eigenvalues of the hamiltonian H on the diagonal.We then have,

Z =∑i

e−βEi = tr(e−βλE ) = tr(U†e−βλE U) = tr(e−βU†λEU ) = tr(e−βH) (154)

where we used the trace property tr(ABC) = Tr(BCA) and the property of unitary matrices U†U = U U† = 1, where

1 is the identity matrix. The other expressions are transformed in a similar way.

—————–

1

Solutions to problems for Part 2

Sample Quiz Problems

Quiz Problem 1. Write down the equation for the thermal de Broglie wavelength. Explain its importance in thestudy of classical and quantum gases.

Solution

λ =h√

2πmkBT(1)

This is of the form h/pT , where pT = (2πmkBT )1/2 is an average thermal momentum. Define the average interparticlespacing of a gas Lc = (V/N)1/3. If λ > Lc quantum effects become important in the thermodynamics.

—————–

Quiz Problem 2. Why are the factors 1/N ! and 1/h3N introduced into the derivation of the partition function ofthe ideal classical gas?

SolutionThe factor 1/N ! is needed to account for the fact that when an intergration is carried out over all phase space for

N particles, all permutations of the particle identities is included. For indentical particles this must be removed. Thefactor 1/h3N takes account of the Heisenberg uncertainty principle which states that the smallest phase space volumethat makes sense is (h/2)3. The fact that it is 1/h3 instead of 1/(h/2)3 for each particle is to reproduce the hightemperature behavior of quantum gases.

—————–

Quiz Problem 3. Explain why the heat capacity at constant volume of a nitrogen molecule is roughly 1.5NkB at1K but 2.5NkB at room temperature. What are the expressions for the heat capacity at constant pressure for thesetwo cases?

SolutionAt low temperatures only the translational degrees of freedom are active so we have kBT/2 of energy for each

of three degrees of freedom. The specific heat is then 3kBTN/2. Once the temperature is larger than the spacingbetween rotational energy levels h2/2I where I is the moment of inertia, then the rotational degrees of freedom arealso active as we get two more degrees of freedom. Only two rotational degrees of freedom contribute as the momentof inertia about the molecule axis is very small, so the level spacing is large and this mode is not active at roomtemperature. The heat capacity at constant pressure is CP = CV +NkBT .

—————–

Quiz Problem 4. In modeling the earth’s atmosphere using the equilibrium isothermal model, the gas pressuredecreases exponentially with altitude. Explain how this is reconciled with the equilibrium condition of constantpressure found in our discussion of thermodynamics in Part 1 of the course.

SolutionThe condition that the pressure of a thermodynamic system such as a gas must be a constant applies when there

are no externally applied potentials in the system. An external potential such as a gravitational field or a harmonictrapping potential as occurs in atom traps leads to a force and this force is balanced by a pressure gradient in thegas. This leads to the condition of hydrostatic equilibrium like that as used in the analysis earth’s atmosphere.

—————–

2

Quiz Problem 5. Discuss the assumptions used in the isothermal and adiabatic models of the atmosphere.

SolutionThe isothermal model assumes that the atmostphere is at constant temperature which requires that thermal equi-

librium has been established in all regions of the atmosphere. The adiabatic model assumes that no heat flow occursin the gas and is justified by the argument that the thermal conductivity of air is very low. These models are clearlylimiting cases with the truth being somewhere between the two. Surprisingly the adiabatic model seems to fit thebehavior of the troposphere quite well.

—————–

Quiz Problem 6. What is the Curie law? Derive this law for the spin half Ising paramagnet.

SolutionThe spin half Ising paramagnet has Hamiltonian,

H = −µsh∑i

Si (2)

where Si = ±1. The partition function and magnetization are then,

ZN = (2cosh(βµsh))N ; m =∂(ln(ZN ))

∂(βh)= Nµstanh(βµsh) (3)

The magnetic susceptibility is then,

χ =∂m

∂h= Nβµ2

ssech2(βµsh)) ≈ Nµ2

s

kBT(4)

which shows that in the limit h→ 0, the susceptibility behaves as C/T which is the Curie law.

—————–

Quiz Problem 7. Paramagnets and also the Ising model in one dimension exhibit a peak in their specific heat,with the low temperature behavior being an approach to zero exponentially and at high temperature an approach tozero as 1/T 2. Why is there a peak in the specific heat in these models. What can’t we use the rule kBT/2 per degreeof freedom to find the specific heat?

SolutionA peak in the heat capacity usually occurs in systems where there is a gap between the ground state energy and

the first excited state, though this is not the only mechanism for a peak in CV . If the gap is ∆, the the peak inthe specific heat occurs at roughly kBTc ≈ ∆, though the precise location depends on the system. The peak occursbecause at the temperature the first excited state begins to be populated leading to a rapid change in the internalenergy with temperature and this leads to an increase in the heat capacity

—————–

Quiz Problem 8. Explain the meaning of negative temperature in the Ising paramagnetic and how it can be usedto make a magnetic cooling device.

Solution Negative temperature occurs when ∂S/∂E is negative. For the Ising paramagnet this condition appliesin the energy regime E > 0 as the peak in the many body density of states is at E = 0. Magnetic cooling canbe produced by preparing a state in a polarized spin configuration followed by removing the field. Once the fieldis removed, the system becomes more disordered and in doing so absorbs heat from its surroundings. This leads tocooling, just as gas cooling is produced when a gas expands as in conventional gas cycle refridgeration.

—————–

3

Quiz Problem 9. What is the meaning of the terms “spontaneous symmetry breaking” (or spontaneous magne-tization) and “breaking of ergodicity” in the context of the two dimensional Ising model.

Solution In a ferromagnet there are two ground states: all up spins; or all down spins. At low temperatures, thesetwo states are separated by a large energy barrier that is impossible to overcome at zero applied field. This systemis non-ergodic as phase space is separated into to regions that are not connnected by any trajectories. Spontaneoussymmetry breaking refers to the fact that the system chooses between to states (either all up or all down) and so theup down symmetry of the system is broken without the application of a symmetry breaking field.

—————–

Quiz Problem 10. State and give a physical explanation of the behavior of the chemical potential µ and thefugacity z = eβµ as temperature T → 0, for both the non-relativistic and ultra-relativistic Bose gas.

Solution.For the Bose gas as temperature goes to zero, the internal energy contribution dominates. As temperature goes to

zero all of the particles that are added go into the ground state, so the chemical potential goes to the ground stateenergy. For the ideal gas case the ground state energy is zero, so the chemical potential goes to zero. The fugacitytherefore goes to one. This is true for both non-relativistic and ultra-relativistic gases. In the Bose condensed phasethe fugacity remains at one up to the BEC critical temperature.

—————–

Quiz Problem 11. Write down the starting expression in the derivation of the grand partition function, ΞB forthe ideal Bose gas, for a general set of energy levels εl, with degeneracy gl. Carry out the sums over the energy leveloccupancies, nl and hence write down an expression for ln(ΞB).

SolutionFor the case of Bose statistics the possibilities are nl = 0, 1, 2...∞ so we find

ΞB =∏l

(∑nl

e−β(εl−µ)nl

)gl=

M∏l=1

(1

1− e−β(εl−µ)

)gl=

M∏l=1

(1

1− ze−βεl

)gl(5)

where the sums are carried out by using the formula for a geometric progression. We thus find,

ln(ΞB) = −M∑l=1

glln(1− ze−βεl

)(6)

< nl >= − 1

β

∂

∂εl[ln(ΞB)] =

ze−βεl

1− ze−βεl(7)

—————–

Quiz Problem 12. In the condensed phase, superfluids are often described in terms of a two fluid model. Basedon the analysis of the ideal Bose gas, explain the physical basis of the two fluid model.

SolutionThe two fluid model considers that the Bose condensed phase is a superfluid while the particles in the excited states

behave as a normal fluid. The normal fluid exhibits dissipation and viscosity, while the superfluid has very low valuesof viscosity and other remarkable properties such as phase coherence.

Quiz Problem 13. Why is the chemical potential of photons in a box, and also acoustic phonons in a crystal,taken to be zero?

Solution.

4

The lowest energy state of these systems is zero so any additional photons or phonons may be placed in this state.This is explained by the fact that photons and phonons are massless bosons that can be created and destroyed, orby arguing that we in essence have a Bose condensed phase of zero energy particles so µ = 0. A more subtle andultimately the full explanation is through an understanding of the interactions with the reservoir. In the case ofmassive particles the reservoir contains a very large number of the same massive particles so the exchange with thereservoir is through exchange of the same type of particle. In a photon or phonon gas, the reservoir is a systemof atoms where the photons or phonons may be absorbed and re-emitted as combinations of different photons orphonons. For this reason the same amount of total free energy in the phonon or photon gas may be divided amongstan arbitrary number of particles, so the chemical potential to add another particle must be zero.

—————–

Quiz Problem 14. Derive or write down the blackbody energy density spectrum in three dimensions.

Solution. The blackbody energy density spectrum follows from the equation for the energy of the photon gas inthree dimensions,

U = 2(L

2π)3

∫ ∞0

dk 4πk2(hkce−βhkc

1− e−βhkc= 2(

L

2πc)3

∫ ∞0

dω 4πω2(hω)e−βhω

1− e−βhω(8)

where we used ω = kc. we then write,

U

V=

∫dω u(ω), where, u(ω) =

h

π2c3ω3

eβhω − 1

—————–

Quiz Problem 15. Write down and explain the relationship between the intensity of radiation emitted by ablackbody (Stefan-Boltzmann law) and the energy density of a photon gas in the blackbody.

Solution. The relationship between the intensity and the energy density of blackbody radiation is,

I =c

4

U

V= σT 4 (10)

The factor c/4 is explained as follows: The factor of c converts the energy density of an EM wave into the intensityof radiation crossing a surface whose surface normal is in the same direction as the direction of wave propagation.The factor of 1/4 has two pieces. First we note that emission from the surface of a blackbody is isotropic so half ofthe radiation is emitted back into the blackbody. Moreover, the amount of radiation emitted to the exterior is also inall directions on a hemisphere. To find the radiation emitted in the normal direction, we take the component of theelectric field in the normal direction, leading to a factor of cos(θ). However the intensity is the square of the electricfield, so it comes with a factor of cos2(θ). The average of cos2(θ) leads to the second factor of 1/2.

—————–

Quiz Problem 16. Explain the physical origin of the cosmic microwave background (CMB) blackbody spectrumof the universe. It is currently at a temperature of TCMB = 2.713K. If the universe is expanding at a constant rateL(t) = H0t, where H0 is a constant what is the expected behavior of the temperature TCMB(t).

Solution. During the “photon epoque” of the early universe that is believed to have existed during the period from10 seconds after the big bang to 377 thousand years after the big bang (that is believed to have occured roughly 13.7billions years ago), the universe consisted of a gas of charged particles and photons that was equilibrated. At around380 thousand years after the big bang, Hydrogen and Helium began to form, reducing the scattering of photons andthe universe became “transparent”. The cosmic microwave background is a remnant of the photon gas that existed380 thousand years ago. Assuming that the photon gas making up the CMB has not changed significantly due toscattering since that time, we can relate the temperature of the CMB to the size of the universe by assuming that theenergy in the photon gas is conserved, so that,

U = constant = L(t)3 π2k4B

15h3c3T 4 (11)

where L(t) is the size of the universe. We then find T ∝ L−3/4.

5

—————–

Quiz Problem 17. Derive or write down the spectral energy density for blackbody radiation in a universe withtwo spatial dimensions.

Solution. The blackbody energy density spectrum follows from the equation for the energy of the photon gas intwo dimensions,

U = 2(L

2πc)2

∫ ∞0

dω 2πω(hω)e−βhω

1− e−βhω= L2

∫dω u(ω) (12)

Note that here I kept the two polarizations of light even though one of them is along the third direction. We thenhave,

u(ω) =h

πc2ω2

eβhω − 1

—————–

Quiz Problem 18. Derive or write down the Debye theory for the internal energy for phonons in a square lattice.Derive the low and high temperature limits of the internal energy and specific heat for this system.

Solution. The energy density for the Debye model for the case of a square lattice comes from assuming thatphonons are an ideal Bose gas where there is one acoustic mode per atom. The chemical potential is taken to be zero,so we have,

U = 2(L

2π)2

∫ kD

0

dk 2πk(hkvs)e−βhkvs

1− e−βhkvs(14)

where,

NA = (L

2π)2

∫ kD

0

2πk dk, so that kD =

(4πNAL2

)1/2

(15)

The factor of two in the front of the energy equation takes into account the fact that there are two phonon modesfor the square lattice. This is a rough approximation as only one of the the two modes has the dispersion relationεp = pvs.

We define x = βpvs, and kBTD = hkDc leading to,

U

L2= 4T (

T

TD)2

∫ TD/T

0

dxx2

ex − 1(16)

In the low temperature and high temperature limits this reduces to,

CVNkB

∼ (T

TD)2, T << TD; CV = 2NkB T >> TD (17)

—————–

Quiz Problem 19. On a graph, illustrate the behavior of the chemical potential µ and the fugacity z = eβµ ofBose, Fermi and classical gases as a function of temperature. What is the chemical potential and fugacity of the theFermi and Bose gases as temperature T →∞ and as T → 0.

SolutionAt low temperatures the chemical potential of quantum gases approaches the mean of the energy of the highest

occupied and lowest unoccupied energy levels. For Fermions with a negligible gap at the Fermi energy the chemicalpotential at T = 0 is the same as the Fermi energy. For Bosons, all particles may be in the lowest energy leveland for gases this is close to zero for large volumes, so the chemical potential of the Bose gas is zero at T = 0,V∞. At high temperatures the Bose and Fermi gases act like a classical gas so their chemical potential is given by,µ = kBT ln(Nλ3/V ) to leading order. The fugacity of all three gases approaches zero at sufficiently high T , while atlow T the fugacity of the Bose gas approaches one while that of the Fermi gas diverges.

6

—————–

Quiz Problem 20. Write down the starting expression in the derivation of the grand partition function, ΞF forthe ideal Fermi gas, for a general set of energy levels εl. Carry out the sums over the energy level occupancies, nl andhence write down an expression for ln(ΞF ).

Solution

ΞF =∑n1

...∑nM

e−β∑M

l=1(εl−µ)nl =

M∏l=1

(1 + e−β(εl−µ)

)=

M∏l=1

(1 + ze−βεl

)(18)

where z = eβµ and each sum is over the possiblities nl = 0, 1 as required for Fermi statistics. We thus find,

ln(ΞF ) =

M∑l=1

ln(1 + ze−βεl

)(19)

—————–

Quiz Problem 21. White dwarf stars are stable due to electron degeneracy pressure. Explain the physical originof this pressure.

SolutionEven in the ground state, the internal energy of the Fermi gas is positive. This is due to the fact that only one

Fermion can be in each energy level so high energy states are occupied at zero temperature. As the density increase,the Fermi energy or energy of the highest occupied state, increases. The pressure is the rate of change of the energywith volume so the pressure increases with the density. This “degeneracy pressure” opposes gravitational collapseand stabilizes white dwarf stars.

—————–

Quiz Problem 15. Explain the physical origins of the paramagnetic and diamagnetic contributions to the mag-netization of the free electron gas.

Solution. The paramagnetic contribution to the magnetization of the free electron gas is the change in the spinpolarization due to the application of a magnetic field. The diamagnetic contribution to the magnetization is due tochanges in the electron orbitals due to the application of a magnetic field. The diamagnetic contribution can occureven if there is no net spin. To a first approximation, we can add the paramagnetic and diamagnetic contributions.When a paramagnetic contribution occurs, these two contributions are usually of opposite sign.

—————–

Assigned problems

Assigned Problem 1. From the density of states for an ideal monatomic gas Ω(E) given in Eq. (26) of the notes,find the Sackur-Tetrode equation for the entropy, Eq. (27) of the notes.

Solution.

Ω(E) =2π1/2V N

N !h3N

(2πmE)3N/2−1/2

( 3N2 − 1)!

(20)

Using Stirling’s approximation and dropping constants, we have

kBln(Ω(E)) = kBN [ln(V

h3)− ln(N) + 1− 3

2ln(

3N

2) +

3

2+

3

2ln(2πmU)] = NkB

[ln[

V

N

(4πmU

3Nh2

)3/2

] +5

2

](21)

7

—————–

Assigned Problem 2. Using the canonical partition function for the ideal gas, show that,

(δE)2 = kBT2Cv (22)

Solution.From Part 1 page 28, and using the ideal classical gas expression ZN = V N/(N !λ3N ), we have,

δE2 =∂2ln(Z)

∂β2= δE2 =

∂2

∂β2[ln(

V N (2πm)3N/2

N !h3N− 3N

2ln(β)] =

3

2N(kBT )2 = kBT

2CV (23)

where CV = 3NkB/2 for the classical monatomic non-relativistic ideal gas in three dimensions.

—————–

Assigned Problem 3. Using the grand partition function of the ideal classical gas show that,

(δN)2 = NkBTρκT (24)

Solution. From Part 1 page 28, and using the expression for the grand partition function for the classical gas wehave,

(δN)2 = (kBT )2 ∂2ln(Ξ)

∂µ2= (kBT )2 V

λ3β2eβµ =

V

λ3eβµ =

PV

kBT= N (25)

where we used (II.25) to write αz = ln(Ξ) = PV/kBT . Also the right hand side of Eq. (17) for the classical ideal gasis,

NkBTρκT = kBTN2

VκT = kBT

N2

PV= N (26)

where we used Eq. (II.17) for κT for the ideal gas

—————–

Assigned Problem 4Consider a gas of N atoms in volume V . Each atom has one unpaired electron in its outer energy level that is a zero

angular momentum state, i.e. the ground state is l = 0 and has an unpaired electron. A magnetic field is applied tothis gas. Write down the canonical and grand canonical partition functions of this system taking into account center ofmass and spin degrees of freedom, assuming that the magnetic interaction is paramagnetic with no spin-orbit coupling.

Solution The translational and spin degrees of freedom do not interact, so the canonical partition function issimply a product of the two subsystem canonical partition functions,

ZN =V N

N !λ3N(2cosh(βµsh))N (27)

The grand partition function is,

Ξ =∑N

zNZN = eαz where α =V

λ32cosh(βµsh) (28)

—————–

Assigned Problem 5.Consider a gas where each atom can have one of two energy levels ε0 and ε1. Find the average energy per particle

taking into account the ideal monatomic gas behavior and the two internal energy levels. Find the specific heat ofthe system in the limits T → 0 and T →∞.

8

Solution The canonical partition function is again a product of the two subsystem partitions functions so that,

ZN =V N

N !λ3N(e−βε0 + e−βε1)N (29)

so the internal energy is,

U = Un + Ui =3

2NkBT +N

ε0e−βε0 + ε1e

−βε1

e−βε0 + e−βε1(30)

where Un is the contribution from the nuclear or translational degrees of freedom while Ui is the contribution fromthe internal degrees of freedom.

—————–

Assigned Problem 6.Consider a dissociating system at thermal equilibrium, where,

A ⇐⇒ B +B (31)

Ignore the internal degrees of freedom of both species and take the mass of an A particle to be twice that of a Bparticle. Given that N = 2NA +NB is fixed, that the volume is fixed, and that the binding energy of A is ε, find thetemperature dependence of NA and NB and find a simple expression for the ratio < NB >2 / < NA >.

SolutionIn this system there is an exchange of particles between the subsystem consisting of B atoms and the subsystem A

consisting of diatomic molecules with 2 B atoms. If there are NA molecules and NB free atoms the canonical partitionfunctions for the two subsystems is,

ZA =V NA

NA!λ3NA

A

eβεNA ; ZB =V NB

NB !λ3NB

B

(32)

where ε is the binding energy of the molecule. The equilibrium condition is that the chemical potential of an atom ofB must be the same whether it is bound in a molecule or free. The chemical potential of a free B atom is,

µB =∂

∂NB(−kBT lnZB) = −kBT ln(

V

NBλ3B

) (33)

while the chemical potential of an A dimer consisting of two bound B atoms is,

µA =∂

∂NA(−kBT lnZA) = −kBT ln(

V

NAλ3A

)− ε (34)

At equilibrium a B atom has the same chemical potential whether it is free or bound into an A dimer. We then havethe condition,

µB =1

2µA (35)

and using this condition we find the relation,

N2B

NA=

V

8λ3A

e−βε (36)

where we used mA = 2mB so that λ2B = 2λ2

A. Using this relation in combination with the condition N = 2NA +NB ,we can find the behavior of NB as a function of temperature by solving,

2N2B

N −NB=

V

8λ3A

e−βε = x (37)

so that,

2N2B + xNB − xN = 0; so NB =

1

4[−x±

√x2 + 8Nx] =

x

4[−1±

√1 + 8N/x] (38)

9

The grand partition function may also be used to find the ratio < NB >2 / < NA >, by realizing that Ξ = ΞAΞBand using,

ΞB =∑NB

zNB

B ZB(N) = Exp[zBV/λ3B ]; ΞA =

∑NA

zNA

A ZA(N) = Exp[zAV/λ3B ] (39)

The average number of atoms of each type is given by,

< NB >=∂ln(ΞB)

∂(βµB)= eβµB

V

λ3B

(40)

and

< NA >=∂ln(ΞA)

∂(βµA)= eβµAeβε

V

λ3A

(41)

Now use the fact that µA = 2µB and take the ratio < NB >2 / < NA > to reproduce the result found earlier.

—————–

Assigned Problem 7.Calculate the specific heat of the model defined by the Hamiltonian Eq. (51) of the text. Is there a peak in the

specific heat? Why or why not?

Solution The internal energy per spin of this model is −hm so that,

U = −µshN [coth(βµsh)− 1

βµsh] (42)

Defining x = βµsh, we find the specific heat to be

Ch = µshkBN [1

x2− 1

sinh2(x)] (43)

This increases monotonically with temperature and has no peak.

—————–

Assigned Problem 8.Generalize the transfer matrix method for the one dimensional Ising model to the case of an applied field (i.e.

combine the Hamiltonians in Eq. (41) and (55)) and calculate the magnetic susceptibility at low field. Does it obeya Curie law?

Solution It is convenient to write the Hamiltonian in the form,

H = −J∑i

SiSi+1 − µsh∑i

1

2(Si + Si+1) (44)

as this makes the transfer matrix more symmetric. We then find,

T++ = eβJ+βµsh; T+− = e−βJ = T−+; T−− = eβJ−βµsh (45)

and we seek the largest eigenvalue of this matrix. Defining α = eβJ ; γ = eβµsh the characteristic equation for T is,

λ2 − α(γ +1

γ) + α2 − 1

α2= λ2 − 2eβJcosh(βµsh)λ+ 2sinh(2βJ) (46)

which has solutions,

λ± =1

2[2eβJcosh(βµsh)±

√4e2βJcosh2(βµsh)− 8sinh(2βJ)] (47)

10

which simplifies to,

λ± = eβJ [cosh(βµsh)±√sinh2(βµsh) + e−4βJ ] (48)

The eigenvalue with the plus sign dominates, so we find the magnetization using,

m =M

N=

1

N

∂

∂(βh)ln(Z) =

∂

∂(βh)ln

(eβJ [cosh(βµsh) +

√sinh2(βµsh) + e−4βJ ]

)(49)

This simplifies to,

m(h, T ) = µsSinh(βhµs)

[Sinh2(βµsh) + e−4βJ ]1/2→ βµ2

she2βJ ; 1d Ising. (50)

where the latter from is correct in the small h limit. The magnetic susceptibility at small field is then

χ =∂m

∂h→ µ2

s

kBTe2βJ as h→ 0 (51)

This diverges exponentially at low T due to the incipient phase transition in the 1D Ising model.

—————–

Assigned Problem 9.Consider the Peierls argument for the two dimensional spin half Ising model on a square lattice, where the topological

excitation is a flat domain wall dividing a domain of up spins and down spins. Does this argument suggest that finitetemperature spontaneous symmetry breaking is possible in the two dimensional Ising model?

SolutionIn the two dimensional Ising model the Peierls estimate of the difference in free energy between the ground state

and a state with one domain wall in it is,

δF = 2JL− kBT ln(L). (52)

Clearly the energy cost is larger than the entropy gain so the ordered state is stable.

—————–

Assigned Problem 10. Using the energy levels of a particle in a box and MB statistics to fill the energy levels,derive the classical gas thermodynamics.

SolutionThe grand partition function is given by,

ΞMB =∏i

Exp[gie−β(εl−µ)] so ln(ΞMB) =

∑i

gie−β(εl−µ) (53)

Using the energy levels of a particle in a box and taking the continuum limit and setting gi = 1, this becomes,

ln(ΞMB) = (L

2π)3

∫ ∞0

4πk2ze−βp2/(2m) (54)

Using cartesian co-ordinates we find,

ln(ΞMB) = z(L

h)3

(∫ ∞−∞

e−βp2/(2m)

)3

= zV

(2mπ

β

)3/2

(55)

so that,

ln(ΞMB) = zV

λ3(56)

as for the classical gas.

11

Assigned Problem 11. In lectures we showed that at high temperatures the equation of state of the Bose gasreduces to the ideal classical gases. Derive the next term in the expansion of the equation of state of the ideal Boseat high temperatures.

Solution. For the Bose case, expanding to second order gives

Nλ3

V= g3/2(z) = z +

z2

2√

2≈ z1 +

z20

2√

2(57)

where z0 = Nλ3/V is the leading order solution, we then find,

z1 = z0 −z2

0

2√

2(58)

We substitute this into the second order expansion of the equation of state,

P

kBT=

1

λ3[z +

z2

4√

2+ ...] ≈ 1

λ3[z1 +

z21

4√

2+ ...] ≈ 1

λ3[z0 −

z20

2√

2+ (

z0 − z202√

2)2

4√

2+ ...] (59)

keeping terms to order z20 gives,

P

kBT=z0

λ3[1− z0(

1

2√

2− 1

4√

2) + ...] (60)

substituting z0 = Nλ3/V gives,

PV

NkBT= 1− 1

4√

2

λ3N

V+ .... Bose gas (61)

The pressure is lower than the classical gas due to the effective attraction of Bosons.

—————–

Assigned Problem 12. Derive expressions for Ξ, PV , N/V and U for the non-relativistic Bose gas in one andtwo dimensions. White general expressions that are valid in any dimension. Find the leading order terms in the hightemperature expansions for these quantities. Discuss the behavior of the non-relativistic Bose gas at low temperaturesin one and two dimensions. Is a finite temperature Bose condensation predicted ? Explain your reasoning.

Solution. For a Bose gas with dispersion relation εp = p2/2m in d dimensions

P

kBT=

1

λdgd/2+1(z)− 1

Ldln(1− z); N

Ld=

1

λdgd/2(z) +

1

Ldz

1− z;

U

Ld=d

2

kBT

λdgd/2+1(z) (62)

These results are found from the integral forms in one and two dimensions below along with the three dimensionalresult derived in lectures.

ln(ΞB) = − L

2πh

∫ ∞0

2dpln(1− ze−βp2/2m)− ln(1− z); 1− dimension (63)

and

ln(ΞB) = −(L

2πh)2

∫ ∞0

2πpdpln(1− ze−βp2/2m)− ln(1− z) 2− dimension (64)

A series expansion as carried out for these integrals and the similar forms for N/V and U lead to the results above.To leading order in the fugacity of the equation for N/V , we find the chemical potential to be the same as that of theideal classical gas in d dimensions, i.e. z = N(λ/L)d. The dimensional dependence comes from the different powersof the factor (L/h)d, and the factors of p in the integral. The integrals that are needed are,∫ ∞

0

xe−x2ldx =

1

2l;

∫ ∞0

e−x2ldx =

1

2(π

l)1/2 (65)

12

We then have,

PL

kBT= ln(ΞB) =

L

2πh(2m

β)1/2

∞∑l=1

zl

l

∫ ∞0

2dxe−x2l − ln(1− z); 1− dimension (66)

and

PL2

kBT= ln(ΞB) = (

L

2πh)2(

2m

β)

∞∑l=1

zl

l

∫ ∞0

2πxdxe−x2l − ln(1− z); 2− dimensions (67)

which reduce to the expression given above for P/(kBT ).At at any temperature, the chemical potential potential of the ideal non-relativistic Bose gas in dimensions less

than 2 + δ cannot be one in the thermodynamic limit, as,

gn(z) =∑l

zl

ln(68)

diverges for z = 1 and n ≤ 1. Since z cannot approach one, the term z/(V (1− z)) approaches zero in the thermody-namic limit, indicating that it is impossible for a finite fraction of the particles to be in the ground state. There istherefore no BEC phase transition at finite temperature in one and two dimensional ideal non-relativistic Bose gases.

—————–

Assigned Problem 13. For the 3-D non-relativistic case, find the entropy of the ideal Bose gas in the condensedphase T < Tc. Compare to the classical gas.

Solution. Using the thermodynamic relation, U = TS − PV + µN , we find,

TS = U + PV − µN =5

2PV − µN (69)

For the Bose gas at T < Tc where µ = 0, we have P/kBT = ζ(5/2)/λ3 and using PV = 2U/3 we find,

TS =5

2PV =

5

2

kBTV

λ3ζ(5/2) (70)

The entropy of the Bose gas in the condensed phase is thus proportional to T 3/2, which goes to zero at low temperature.The entropy of the classical gas is given by Eq. (16) of lecture notes part 2, so that,

S = NkB [ln(V

Nλ3) +

5

2] (71)

The classical gas has higher entropy than the Bose gas in the condensed phase. In the condensed phase the number ofexcited state particles grows at T 3/2 and only these particles contribute to the entropy of the Bose gas. The groundstate particles do not contribute to the entropy and as the temperature is reduced almost all particles are in theground state.

—————–

Assigned Problem 14. Show that a d − dimensional Bose gas with dispersion relation εp = cps obeys therelation,

P =s

d

U

Ld(72)

Solution. The equations for a Bose gas with this dispersion relation in d dimensions is written as,

P

kBT= − cd

hd

∫ ∞0

dp pd−1ln(1− ze−βcps

);U

Ld=cdhd

∫ ∞0

dp pd−1(cps)ze−βcp

s

1− ze−βcps(73)

Integrating the pressure equation by parts gives,

I1 = −∫ ∞

0

dp pd−1ln(1− ze−βcps

) = −1

dpdln(1− ze−βcp

s

)|∞0 +sβ

d

∫ ∞0

dp pd−1(cps)ze−βcp

s

1− ze−βcps(74)

The first term on the right hand side is zero and comparison of the remaining integral with the energy equation provesrelation (63).

13

—————–

Assigned Problem 15. Find the thermodynamic properties, PV , U , S, CV , N of a photon gas in d dimensions.Show that the entropy per photon is independent of temperature.

Solution. We use the relations,

N = 2

(L

2πh

)d ∫ ∞0

cdpd−1dp

e−βpc

1− e−βpc= 2cd

(L

h

)d(

1

βc)d∫ ∞

0

dxxd−1

ex − 1(75)

and

U = 2

(L

2πh

)d ∫ ∞0

cdpd−1dp (pc)

e−βpc

1− e−βpc= 2cdc

(L

h

)d(

1

βc)d+1

∫ ∞0

dxxd

ex − 1(76)

along with PV = sU/d, with s = 1 and the integral,∫ ∞0

xs−1dx

ex − 1= Γ(s)ζ(s), (77)

to find,

U = 2cdcd!ζ(d+ 1)

(L

h

)d(kBc

)d+1T d+1; N = 2cd(d− 1)!ζ(d)

(L

h

)d(kBc

)dT d (78)

We also have,

TS = U + PV − µN = (d+ 1)U/d ∝ T d+1 (79)

From this it is evident that both S and N are proportional to T d, so S/N is temperature independent. Also,

CV =∂U

∂T= 2(d+ 1)!ζ(d+ 1)cdc

(L

h

)d(kBc

)d+1T d (80)

—————–

Assigned Problem 16. Find the thermodynamic properties, U and CV for the Debye phonon model in d dimen-sions.

Solution. We use the relation,

U = d

(L

2πh

)d ∫ pd

0

cdpd−1dp (pvs)

e−βpvs

1− e−βpvs= dcdvs

(L

h

)d(kBvs

)d+1T d+1

∫ xD

0

dxxd

ex − 1(81)

where the factor of d in front ensures that we recover the high temperature equipartition result. xD = βpDvs and wedefine the Debye temperature through kBTD = pDvs, so that xD = βkBTD = TD/T . We also use the integral,∫ ∞

0

xs−1dx

ex − 1= Γ(s)ζ(s), (82)

to find that at low temperatures TD/T → ∞, so the behavior is like that of the photon gas in d dimensions, withc→ vs, and multiplied by d/2 due to the difference in degeneracy (2 for photons, d for phonons). We then have ,

U = dcdvsd!ζ(d+ 1)

(L

h

)d(kBvs

)d+1T d+1; CV =∂U

∂T= d(d+ 1)!ζ(d+ 1)cdvs

(L

h

)d(kBvs

)d+1T d (83)

At high temperatures, the behavior is like that of a classical gas in a harmonic potential so that U = dNkBT,CV =NdkB , and PV = 2U/d.

14

—————–

Assigned Problem 17. Carry through the analysis of the Bose gas for the case of massive ultrarelativistic particleswhere µ 6= 0. What is the lower critical dimension for a BEC transition in this case?

Solution. For the ultrarelativistic Bose gas, we use, εk = hkc, so that,

N =

(L

2πh

)3 ∫ ∞0

4πk2dkze−βhkc

1− ze−βhkc+

z

1− z(84)

which reduces to,

N =V

2π2

1

(βhc)3

∫ ∞0

dxx2ze−x∞∑l=0

zle−xl +z

1− z. (85)

where we used the expansion 1/(1− x) = 1 + x+ x2 + x3.... Using the integral,∫ ∞0

dxxs−1e−lx =1

ls

∫ ∞0

ys−1e−ydy =(s− 1)!

ls→ (s = 3)

2

l3(86)

N =V

2π2

1

(βhc)3

∫ ∞0

dx

∞∑l=1

2zl

l3+

z

1− z. (87)

Since for z = 1, the series is convergent, there is the possibility of Bose condensation at finite temperature and bysetting z = 1 we find the critical condition,

N =V

π2

1

(βhc)3ζ(3) (88)

A similar calculation shows that in two dimensions the ultrarelativistic ideal Bose gas has a phase transition at finitetemperature while in one dimension a finite temperature BEC transition does not occur.

—————–

Assigned Problem 18. At high temperatures we found that the ideal quantum gases reduce to the ideal classicalgases. Derive the next term in the expansion of the equation of state of the ideal Fermi gas at high temperatures,and verify that,

PV

NkBT= 1 +

1

4√

2

λ3N

V+ .... Fermi gas (89)

The pressure of the ideal Fermi gas is higher than that of the classical gas at the same temperature and volume.Why? Carry out a similar expansion for the Bose gas. Is the pressure higher or lower than the ideal classical gas atthe same values of T, V ? Why?

Solution. For the Bose case, expanding to second order gives

Nλ3

V= g3/2(z) = z +

z2

2√

2≈ z1 +

z20

2√

2(90)

where z0 = Nλ3/V is the leading order solution, we then find,

z1 = z0 −z2

0

2√

2(91)

We substitute this into the second order expansion of the equation of state,

P

kBT=

1

λ3[z +

z2

4√

2+ ...] ≈ 1

λ3[z1 +

z21

4√

2+ ...] ≈ 1

λ3[z0 −

z20

2√

2+ (

z0 − z202√

2)2

4√

2+ ...] (92)

15

keeping terms to order z20 gives,

P

kBT=z0

λ3[1− z0(

1

2√

2− 1

4√

2) + ...] (93)

substituting z0 = Nλ3/V gives,

PV

NkBT= 1− 1

4√

2

λ3N

V+ .... Bose gas (94)

Analysis for the Fermi gas is the same, except that the sign on the correction term is positive. The Bose gas hasreduced pressure as compared to the classical gas at the same temperature, while the Fermi gas has higher pressurethan the classical case. This expansion can be extended to higher order and in general is written as,

PV

NkBT=

∞∑l=1

alαl−1; where α =

λ3N

V(95)

This expansion is valid when α is small, which means low density and/or high temperatures. In general expansionsof this type are called virial expansions and have played an important role in characterizing interactions in gases.In classical gases the second virial coefficient a2 is determined by the strength of the pair interactions, as we shallsee in Part 3 of the course. Here the terms l > 1 are due to quantum effects. In real quantum gases, both quantumeffects and interactions can be important. Recall than our condition for quantum effects to be important was that theinterparticle spacing Lc < λ. When this is true α is significant and more terms are required in the virial expansion.

—————–

Assigned Problem 19. By expanding the denominator of the integral, 1/(1 + y) for small y = ze−x2

show that,

f3/2(z) =4

π1/2

∫ ∞0

dx x2 ze−x2

1 + ze−x2 =

∞∑l=1

(−1)l+1zl

l3/2(96)

Solution.We use the expansion

1

1 + y= (1− y + y2 − y3...); and

∫ ∞0

x2e−ax2

dx =

√π

4a3/2(97)

so that,

f3/2(z) =4

π1/2

∫ ∞0

dx x2 ze−x2

1 + ze−x2 =

∞∑l=1

(−1)l+1zl∫ ∞

0

dx x2e−lx2

=

∞∑l=1

(−1)l+1zl

l3/2(98)

—————–

Assigned Problem 20a. Derive expressions for Z, Ξ, PV , µ and U for the classical gas in one and two dimen-sions. Are the results what you expect? How do they compare with the result in three dimensions. White generalexpressions that are valid in any dimension.

Solution In d-dimensions, the partition functions are,

Z =LdN

λdNN !, Ξ = eαz α = (

L

λ)d (99)

F = −kBT ln(Z) = −kBT ln(LdN

λdNN !) (100)

S = −(∂F

∂T

)Ld,N

= kBln(LdN

λdNN !) +

d

2NkB (101)

16

The internal energy is found by combining (31) and (32), so that,

U = F + TS =d

2NkBT (102)

The pressure is given by,

P = −(∂F

∂Ld

)T,N

= kBTN

Ld=kBNT

Ld, (103)

which is the ideal gas law, while the chemical potential is,

µ =

(∂F

∂N

)T,Ld

= kBT ln(λdN/Ld) (104)

—————–

Assigned Problem 20b. Derive expressions for Ξ, PV , N/V and U for the Fermi gas in one and two dimensions.Write general expressions that are valid in any dimension. Find the leading order terms in the high temperatureexpansions for these quantities. Are the results what you expect? How do they compare with the result in threedimensions, and with the classical gas.

Solution The relations for the non-relativistic Fermi gas are,

P

kBT=

1

λdfd/2+1(z);

N

Ld=

1

λdfd/2(z);

U

Ld=d

2

kBT

λdfd/2+1(z) (105)

These results are found from the integrals,

ln(ΞF ) =L

2πh

∫ ∞0

2dpln(1 + ze−βp2/2m); 1− dimension (106)

and

ln(ΞF ) = (L

2πh)2

∫ ∞0

2πpdpln(1 + ze−βp2/2m) 2− dimension (107)

These integrals and the analogous equations for N and V are expanded as in the three dimensional case. Followingthe procedure given in the solution to problem 7, we have,

PL

kBT= ln(ΞF ) =

L

2πh(2m

β)1/2

∞∑l=1

(−1)l+1zl

l

∫ ∞0

2dxe−x2l; 1− dimension (108)

and

PL2

kBT= ln(ΞF ) = (

L

2πh)2(

2m

β)

∞∑l=1

(−1)l+1zl

l

∫ ∞0

2πxdxe−x2l; 2− dimensions (109)

which reduce to the expression for P/kBT given in Eq. (42). The leading order expansion of fd/2(z) at hightemperature gives the chemical potential of the ideal classical gas in d dimensions so we recover the classical gas in ddimensions at sufficiently high temperatures.

—————–

Assigned Problem 21. Discuss the behavior of the Fermi gas at zero temperature in one and two dimensions. Isthere different behavior as a function of dimension? Explain your reasoning.

Solution. We calculate the energy and degeneracy pressure to see if there is a dependence on dimension. We onlycarry out the ground state calculation. The Fermi wavevector in one, two and three dimensions is given by,

kF1 =πN

L; kF2 = (

4πN

L2)1/2; kF3 = (

6π2N

V)1/3 (110)

17

so the Fermi energy is given by,

εF1 =h2

2mL2(πN)2; εF2 =

h2

2mL2(4πN); εF2 =

h2

2mL2(6π2N)2/3; (111)

The internal energy is given by,

U = (L

2π)d∫ddk

h2k2

2m(112)

In one two and three dimensions we find,

U1 =1

3NEF1; U2 =

1

2NEF2; U3 =

3

5NEF3 (113)

The degeneracy pressure is given by,

P = −(∂U

∂Ld

)N,T

(114)

Since EF is proportional to 1/L2 doing the derivative with respect to L, L2 and L3 in one two and three dimensions,leads to the following expressions for the degeneracy pressure,

P1 =2

3LNEF1; P2 =

1

2L2NEF2; U3 =

2

5L3NEF3. (115)

For fixed number of particles, the degeneracy pressure and internal energy are much higher for the one and twodimensional cases, first because EF grows much more rapidly with N as the dimension is reduced and second becausethe prefactor grows more slowly with L as the dimension is reduced. This results are expected as particles are moreconfined in one and two dimensions, so the effect of Pauli exchange is stronger, so the total energy is expected togrow more rapidly in lower dimension and the degeneracy pressure should be higher.

—————–

Assigned Problem 22. Show that a d − dimensional Fermi gas with dispersion relation εp = cps obeys therelation,

P =s

d

U

Ld(116)

Solution. The equations for a Fermi gas with this dispersion relation in d dimensions is written as,

P

kBT=cdhd

∫ ∞0

dp pd−1ln(1 + ze−βcps

);U

Ld=cdhd

∫ ∞0

dp pd−1(cps)ze−βcp

s

1 + ze−βcps(117)

Integrating the pressure equation by parts gives,

I1 =

∫ ∞0

dp pd−1ln(1 + ze−βcps

) =1

dpdln(1 + ze−βcp

s

)|∞0 +sβ

d

∫ ∞0

dp pd−1(cps)ze−βcp

s

1 + ze−βcps(118)

The first term on the right hand side is zero and comparison of the remaining integral with the energy equation provesthe relation.

—————–

Quiz Problem 23. Find the leading order term in the temperature dependence of the internal energy and specificheat of an three dimensional ultrarelativistic Fermi gas at low temperature.

Solution. The equations for U and N for the three-dimensional ultra-relativistic Fermi gas are,

N = 4π(L

h)3

∫ ∞0

dp p2 ze−βpc

1 + ze−βpc=

4π

(βc)3(L

h)3

∫ ∞0

dx x2 ze−x

1 + ze−x(119)

18

and

U = 4πc(L

h)3

∫ ∞0

dp p3 ze−βpc

1 + ze−βpc=

4πc

(βc)4(L

h)3

∫ ∞0

dx x3 ze−x

1 + ze−x(120)

We may expand the integral at small z, but this is not useful at low temperature. Instead we carry out the Sommerfeldexpansion. Here we write it in more general form, generalizing Eq. (II.73) to,

Is =

∫ ∞0

dx xs−1 ze−x

1 + ze−x=

∫ ∞0

dx xs−1 1

ex−ν + 1=

1

s

∫ ∞0

dx xsex−ν

(ex−ν + 1)2(121)

Expanding xs about ν we have,

xs = (ν + (x− ν))s = f(0) + (x− ν)f ′(0) +1

2(x− ν)2f”(0) + .... (122)

where f(y) = (ν + y)s, so that f(0) = νs, f ′(0) = sνs−1, ′′(0) = s(s− 1)νs−2. so that

xs = (ν + (x− ν))s = νs + (x− ν)sνs−1 +1

2(x− ν)2s(s− 1)νs−2 + .... (123)

Following the procedure of Eq. (II.77) and (II.78), we then have,

Is =1

s[νsI0 + sνs−1I1 +

1

2s(s− 1)νs−2I2 + ...); where In =

∫ ∞−∞

dttnet

(et + 1)2(124)

I0 = 1, while by symmetry In is zero for odd n. For even n > 0, In is related to the Reimann zeta function, through,

In = 2n(1− 21−n)(n− 1)!ζ(n), with ζ(2) =π2

6, ζ(4) =

π4

90, ζ(6) =

π6

945(125)

Since the odd integral I0 = 1, I1 = 0, I2 = π2/3, we find,

Is =1

s[νs +

π2

6s(s− 1)νs−2 + ...] (126)

Up to a prefactor that is defined differently here, the expansion above is consistent with Eqs. (II.78) and (II.88) asthey must be. The expansions we need are then,

N

V=

4π

(hβc)3

1

3[ν3 + π2ν + ...] (127)

and

U

V=

4πc

h3(βc)4

1

4[ν4 + 2π2ν2 + ...] (128)

The leading order term in the expansion of the chemical potential is found using,

N

V=

4π

(hβc)3

1

3(βµ0)3 so βµ0 = hβc

(3N

4πV

)1/3

(129)

The next correction is found using,

ν30 = (ν1)3 + π2ν0; so that ν1 = ν0(1− π2

3(ν0)2) = βεF [1− π2

3(kBT

εF)2 + ...]. (130)

where εF = µ0 = ν0/β. The internal energy expansion is,

U

V=

πc

h3(βc)4ν4[1 +

2

3π2ν−2 + ...] ≈ πc

h3(βc)4βε4F [1− 4π2

3(kBT

εF)2 + ...][1 + 2π2βε−2

F + ...] (131)

—————–

1

Solutions to problems for Part 3

Assigned problems and sample quiz problems

Sample Quiz Problems

Quiz Problem 1. Draw the phase diagram of the Ising Ferromagnet in an applied magnetic field. Indicate thecritical point. Plot the magnetization as a function of the applied field for three temperatures T < Tc, T = Tc, T > Tc.

Quiz Problem 2. Plot the behavior of the magnetization of the Ising ferromagnet as a function of the temperature,for three applied field cases: h < 0, h = 0, h > 0. Indicate the critical point.

Quiz Problem 3. Write down the definition of the critical exponents α, βe, γ, δ, η and ν. What values do theseexponents take within mean field theory.

Solution.

CV ∼ t−α; m ∼ tβ ; χ ∼ t−γ ; m(Tc) ∼ h1/δ; c(r) ∼ e−r/ξ/rd−2+η (1)

where ξ = t−ν , and t = |T − Tc|. Within mean field theory α = 0, β = 1/2, γ = 1, δ = 3, η = 0, ν = 1/2

—————–

Quiz Problem 4. Write down the mean field equation for the Ising ferromagnet in an applied field, on a latticewith co-ordination number z and exchange constant J . From this equation find the critical exponent δ for the Isingferromagnet within mean field theory.

Solution.

m = Tanh(βJzm+ βh) ∼ (βJzm+ βh)− 1

3(βJzm+ βh)3 (2)

at the critical point βJz = 1, so m ∼ h1/3 and hence δ = 3.

—————–

Quiz Problem 5. Write down the scaling hypothesis for the magnetization, susceptibility, free energy and corre-lation function. From these relations, find the Fisher, Widom and Rushbrooke critical exponent relations. Also writedown the hyperscaling relation.

SolutionWe assume that the correlation length is the key quantity in the scaling theory so that the scaling behavior is of

the form,

F (T, h) = t2−αFs(hξy); M(T, h) = tβMs(hξ

y); χ(T, h) = t−γχs(hξy); C(r) = r−pCs(r/ξ, hξ

y) (3)

where t = |T − Tc|, and y > 0. We also define ξy = t−∆, so that νy = ∆, where ∆ is the gap exponent. We also havep = d− 2 + η, and ξ = t−ν . The scaling functions have the property that as their argument x = hξy = h/t∆ goes tozero, the scaling functions must approach a constant. Moreover the scaling assumption states that for h < ξ−y thescaling functions are constant. Moreover, as x → ∞, the scaling functions go to zero. First consider the behavior ofthe magnetization when we are at the critical point, so that,

M(t = 0, h 6= 0) ∼ tβMs(x→∞) ∼ h1/δ; so that Ms(x) ∼ xk (4)

where,

tβxk = tβ(h

t∆)k = h1/δ; so that k = 1/δ; and ∆ = βδ (5)

2

Now consider the relation between the magnetization and the susceptibility,

M ∼∫ t∆

0

χdh ∼ t−γt∆ ∼ tβ ; so that β = ∆− γ (6)

In a similar manner,

F ∼∫ t∆

0

Mdh ∼ tβt∆ ∼ t2−α; so that β + ∆ = 2− α (7)

Finally, consider the scaling of the correlation function in the case where hξy is zero, so that Cs is a constant for r < ξand zero otherwise. We then have,

χ ∼∫d3rC(r) ∼

∫ ξ

a

drrd−1r−pCs(r/ξ, hξy) ∼ ξd−(d−2+η) ∼ t−γ ; so that γ = ν(2− η) (8)

These exponent relations are usually written in the form,

∆ = β + γ; γ = ν(2− η) (Fisher); α+ 2β + γ = 2 (Rushbrooke); γ = β(δ − 1) (Widom) (9)

Since we have added the “gap” exponent ∆, there are seven exponents in the problem. We have four exponent relationsso that only three exponents are independent. Josephson introduced another relation, called the hyperscaling relation.He introduced the hypothesis that the singular part of the free energy scales as 1/ξd. This implies that,

fsing ≈ ξ−d ≈ t2−α; so that dν = 2− α (Josephson, or hyperscaling relation) (10)

The hyperscaling relation is considered the most likely of the scaling relations to fail and for example is known to failin some heterogeneous models such as the Spin glass model.

—————–

Quiz Problem 6. Find the domain wall energy for the Ising (O(1) model) and for the O(2) model. From theseexpressions find the lower critical dimension for these two problems.

SolutionWe consider the ground state energy of a domain wall. The O(1) model is the Ising case so the spin is either

up or down, while O(2) corresponds to a spin vector that has unit length and can take any angle 0 < θ < 2π. Aferromagnetic interaction favors all the spins pointing in the same direction for both models. For both cases, a domainwall created by setting the spins on one side of a hypercubic system in the up direction, while the spins on the oppositesize are in the down direction. The interface between the up and down domains that this boundary condition createsis called the domain wall. For an Ising system we found that in the ground state,

EIsingDW = 2JLd−1 (11)

For the O(2) case, the domain wall energy is,

EcontinuousDW ≈ π

2JLd−2 (12)

The lower energy of domain walls in systems with an order parameter that can take a continuum of values, like theO(2) model is due to the possibility of a wide domain wall that changes smoothly between the boundary states. Inthe Ising case the domain wall is abrupt and this costs more energy. For the O(2) case, the Hamiltonian is,

H = −J∑<ij>

~Si · ~Sj = −J∑<ij>

|S|2cos(θij) (13)

where ~Si is a unit vector with angle θi to the z axis, and θij = θj − θi. To find the domain wall energy we considerthe direction perpendicular to the domain wall and notice that the lowest energy domain wall is formed by making awide domain wall with small differences in energy between spins. Since the domain wall corresponds to changing the

3

spin orientation from up to down, the angle must rotate by π. If the width of the domain wall is L, the angle betweenadjacent spins is π/L. The energy cost of the domain wall is then,

EO(2)DW = Ld−1J

L∑i=1

(1− cos(πL

)) = Ld−1JL1

2(π

L)2 =

π

2JLd−2 (14)

The lower critical dimension is the dimension below which the domain wall energy is finite, while above the lowercritical dimension the energy of a domain grows with the length of the domain wall. From the above we find that forthe Ising system dlc = 1 + ε, while for the O(2) model dlc = 2 + ε, where ε is small and positive.

—————–

Quiz Problem 7. Write down the van der Waals equation of state. Draw the P, v phase diagram of the van derWaals gas and indicate the critical point.

Solution.

P =kbT

v − b− a

v2(15)

—————–

Quiz Problem 8. Make plots of the van der Waals equation of state isotherms, for T > Tc, T < Tc and forT = Tc. For the case T < Tc explain why the non-convex part of the curve cannot occur at equilibrium and theMaxwell construction to obtain a physical P, v isotherm.

Quiz Problem 9. Write down the Landau free energy for the Ising and fluid-gas phase transitions. Explain thecorrespondences between the quantities in the magnetic and classical gas problems.

Solution.

F = a(T − Tc)y2 + by4 + cy (16)

For the Ising model y = m, c = h, for the van der Waals gas, y = vg − vl, c = P − Pc.

—————–

Quiz Problem 10. Derive the Helmholtz free energy of the van der Waals gas and explain the physical meaningof the parameters a and b. Using your free energy explain the Maxwell construction.

SolutionSee lecture notes.

—————–

Quiz Problem 11. Write down the Gibb’s free energy of the van der Waals gas. Explain the conditions underwhich co-existence is expected to occur.

SolutionSee lecture notes

—————–

Quiz Problem 12. Explain the meaning of the upper critical dimension and lower critical dimension in the theoryof critical phenomena.

SolutionBelow the lower critical dimension, there is no traditional phase transition to an ordered phase at finite temperature.

It should be noted however that at the critical dimension special transitions may occur, for example in the case of theO(2) model the Kosterlitz-Thouless transition occurs in two dimensions and this is a special phase transition that isdifferent that the phase transitions of the model above the lower critical dimensions.

Above the upper critical dimension, the critical phenomena is correctly described by mean field critical exponents.Between the upper and lower critical dimensions, critical exponents and the nature of the transition (continuous ordiscontinuous) may change.

4

—————–

Quiz Problem 13. State the universality hypothesis in the theory of critical phenomena and using it explain whythe liquid gas phase transition is in the same universality class as the Ising mdoel.

SolutionThe universality hypothesis refers to continuous phase transitions where the correlation length diverges so that

fluctuations occur on all length scales. The universality hypothesis states that the critical exponents describingcontinuous phase transitions depend on only three things:

(i) The spatial dimension(ii) The order parameter symmetry(iii) The range of the interactionsThis predicts for example that systems described by very different physics, such as the liquid-gas transition and

Ising magnetic phase transition, should be described by the same critical exponents.

—————–

Quiz Problem 14. Explain the importance of the “linked-cluster” theorems in perturbation theory of manyparticle systems.

Solution.Linked cluster theorems state that averages of observables can be described by sums of linked diagrams, and that

non-extensive contributions from disconnected diagrams must cancel when the diagrams are summed to all orders.

—————–

Quiz Problem 15. Draw the high temperature series expansion diagrams to order t8 (where t = tanh(βJ)) forthe square lattice, nearest neighbor, spin half Ising ferromagnet partition function. What is the degeneracy of eachof these diagrams? Write down the expansion for the Helmholtz free energy and give a physical reason why only theterms of order N are kept.

Solution.See Lecture Notes

—————–

Quiz Problem 16. Draw the low temperature series expansion diagrams to order s8 (where s = Exp[−2βJ) forthe square lattice, nearest neighbor, spin half Ising ferromagnet partition function. What is the degeneracy of eachof these diagrams? Write down the expansion for the Helmholtz free energy and give a physical reason why only theterms of order N are kept.

Solution.See Lecture Notes

—————–

Quiz Problem 17. Write down the mathematical form of the virial expansion for many particle systems andexplain why it is important. What physical properties can be extracted from the second virial coefficient?

SolutionThe virial expansion is given by,

P

kBT= ρ+B2ρ

2 +B3ρ3 + .... (17)

where ρ = N/V is small. Keeping only the leading order term on the right hand side of this equation leads to theideal gas law. The first correction to the ideal gas law is due to the prefactor B2, which is called the second virialcoefficient. The second virial coefficient for a particle system described by central force pair potential is given by,

B2 = −∫dR R2(e−βu(r) − 1) (18)

From this expression we can see that the second virial coefficient is related to the interatomic potential. If we measureB2 as a function of temperature it is possible to extract a lot of information about u(r).

5

—————–

Quiz Problem 18. Explain the meaning of second quantization. Discuss the way that it can be used in positionspace and in the basis of single particle wavefunctions. Write down the commutation relations for Bose and Fermisecond quantized creation and annihilation operators.

Solution.Second quantization is a formulation of quantum mechanics and of quantum field theory that is expressed in

terms of creation and annihilation operators. In many body quantum physics creation and annihilation operatorscreate and destroy particles in many body basis sets constructed from single particle wave functions. In the caseof Fermions a many body basis function is a determinant, while for Bosons it is a permanent. The commutationrelations for Fermions and Bosons are similar, except that for fermions we have anticommutators and for Bosons wehave commutators. In many body quantum mechanics we have,

[ai, a†j ] = δij ; for bosons; and ai, a†j = δij ; for fermions; (19)

while for quantum fields, we have,

[ψ(x), ψ†(x′)] = δ(x− x′); for bosons; and ψ(x), ψ†(x′) = δ(x− x′); for fermions; (20)

These days, new theories are often formulated using creation and annihilation operators rather than the Heisenbergor Schrodinger formulations of quantum theory.

—————–

Quiz Problem 19. Write down the Hamiltonian for BCS theory, and the decoupling scheme used to reduce it toa solvable form. Explain the physical reasoning for the decoupling scheme that is chosen.

Solution.In the s-wave BCS theory a singlet state is assumed, so that,

Hpair − µN =∑~kσ

(ε~kσ − µ) a†~kσa~kσ +

∑~k~l

V~k~l a†~k↑a†−~k,↓

a−~l↓a~l↑, (21)

where N =∑~kσ n~kσ is the number of electrons in the Fermi sea. We carry out an expansion in the fluctuations,

a−~l↓a~l↑ = bl + (a−~l↓a~l↑ − bl); a†~k↑a†−~k↓

= b∗k + (a†~k↑a†−~k↓− b∗k) (22)

The mean field Hamiltonian keeps only the leading order term in the fluctuations so that,

HMF − µN =∑~kσ

(ε~kσ − µ) a†~kσa~kσ +

∑~k~l

V~k~l (a†~k↑a†−~k,↓

b~l + b∗~ka−~l↓a~l↑ − b∗~kb~l) (23)

This is the Hamiltonian that leads to the BSC solution.

—————–

Quiz Problem 20. Consider the inverse Bogoliubov-Valatin transformation,

γ~kσ = u∗~ka~kσ − σv~ka†−~k−σ

. (24)

Show that if the operators a, a† obey standard fermion anti-commutator relations, then the operators γ, γ† also obeythese relations, provided,

|uk|2 + |vk|2 = 1 (25)

Solution The anticommutator,

γ~kσ, γ~lα = u∗~ka~kσ − σv~ka†−~k−σ

, u∗~l a~lα − αv~la†−~l−α

. (26)

6

Expanding the anticommutator gives,

γ~kσ, γ~lα = (u∗~ka~kσ − σv~ka†−~k−σ

)(u∗~l a~lα − αv~la†−~l−α

) + (u∗~l a~lα − αv~la†−~l−α

)(u∗~ka~kσ − σv~ka†−~k−σ

) (27)

which reduces to,

γ~kσ, γ~lα = u∗~ku∗~la~kσ, a~lα+ σαv~kv~la

†~−k−σ

, a†−~l−α

− αu∗~kv~la~kσ, a†−~l−α

− σv~ku∗~la ~−k−σ, a

†~lα (28)

The first two anticommutators are zero. The second two anticommutators are finite when the conditions δ(k,−l)δ(σ,−α)hold. However when this condition holds, the two commutators are equal and opposite, so they sum to zero. Takingthe adjoint of Eq. (15) shows that γ~kσ, γ~lα = 0. Now consider,

γ~kσ, γ†~lα = (u∗~ka~kσ − σv~ka

†−~k−σ

)(u~la†~lα− αv∗~l a−~l−α) + (u~la

†~lα− αv∗~l a−~l−α)(u∗~ka~kσ − σv~ka

†−~k−σ

) (29)

γ~kσ, γ†~lα = u∗~ku~la~kσ, a

†~lα+ σαv~kv

∗~la†~−k−σ, a−~l−α − αu

∗~kv∗~l a~kσ, a−~l−α − σv~ku~la

†~−k−σ

, a†~lα = 0 (30)

which reduces to

γ~kσ, γ†~lα = δkl|u~k|

2a~kσ, a†~lα+ σα|v~k|

2a†~−k−σ, a−~l−α = δklδσα(|u~k|2 + |v~k|

2). (31)

This anticommutator thus requires Eq. (12) in order for the γ operators to obey Fermion anti commutator relations.

—————–

Quiz Problem 21. Given that the energy of quasiparticle excitations from the BCS ground state have thespectrum,

E = [(ε− εF )2 + |∆|2]1/2, (32)

where ∆ is the superconducting gap and EF is the Fermi energy, show that the quasiparticle density of states if givenby,

D(E) =N(εF )E

(E2 −∆2)1/2(33)

Solution We use the relation,

N(εF )dε = D(E)dE; and dE =ε− εF

[(ε− εF )2 + ∆2]1/2dε (34)

to find,

D(E) =N(εF )[(ε− εF )2 + ∆2]1/2

ε− εF(35)

and using,

(ε− εF )2 = E2 − |∆|2 (36)

yields,

D(E) =N(εF )E

[E2 − |∆|2]1/2(37)

—————–

Quiz Problem 22. Describe the physical meaning of the superconducting gap, and the way in which BCS theorydescribes it.

SolutionThe superconducting gap is the energy required to generate a quasiparticle excitation from the superconducting

ground state. In BCS theory, the quasiparticles behave like non-interacting fermions and the energy required togenerate a quasiparticle is at least 2∆(T ).

7

—————–

Quiz Problem 23. Given the general solutions to the BCS mean field theory

∆~k = −∑~l

V~k~l∆~l

2E~l, E~k = ((ε~k − µ)2 + |∆~k|

2)1/2 (38)

Describe the assumptions that are made in deducing that,

1 =N(εF )V

2

∫ εF +hωc

εF−hωc

dε

((ε− εF )2 + |∆|2)1/2=

N(εF )V

∫ hωc/∆

0

dx

(1 + x2)1/2= N(εF )V Sinh−1(

hωc∆

) (39)

and hence,

∆ = 2hωcExp[−1

N(εF )V] (40)

SolutionWe assume an isotropic gap, and that the attractive coupling between electrons is constant −V , over the range

εF − hωc < ε < εF + hωc. The density of states is assumed constant with value N(εF ).

—————–

Assigned problems

Assigned Problem 1. From Eq. (18) of the notes, derive the Landau form Eq. (44). Explain the approximationsthat are made. Plot FL as a function if m for T > Tc and T < Tc for h = 0 and for h 6= 0. Explain the concept ofspontaneous symmetry breaking (SSB) using your graphs.

Solution.Equation (18) is,

−fR =−βFMF

N+ ln(2) = −1

2Jzm2 + ln(Cosh(βJzm+ βh))]

≈ −1

2βJzm2 +

1

2(β(Jzm+ h))2 − 1

12(β(Jzm+ h))4 +O(m6) (41)

where we used the expansion of log(cosh(x)) for small x. Now we use the fact that kBTc = Jz and expanding in h tolinear order we find,

fL = a(T − Tc)m2 + bm4 + hm (42)

where constant prefactors in the term hm are absorbed into fL, a and b.

—————–

Assigned Problem 2. Consider the Ising ferromagnet in zero field, in the case where the spin can take three valuesSi = 0,±1. a) Find equations for the mean field free energy and magnetization. b) Find the critical temperature andthe behavior near the critical point. Are the critical exponents (β, γ, α, δ) the same as for the case S = ±1? Is thecritical point at higher or lower temperature than the spin ±1 case? c) Is the free energy for the the spin 0,±1 casehigher or lower than the free energy of the ±1 case? Why? d) Carry out an expansion of the free energy to fourthorder in the magnetization. Does this free energy have the Landau form expected for an Ising ferromagnet?

Solution. The partition function and Helmholtz free energy are,

Z = [1 + 2Cosh(βJzm)]N ; F = −kBTNln[1 + 2Cosh(βJzm)] (43)

8

The mean field equation is,

m =2Sinh(βJzm)

1 + 2Cosh(βJzm)≈ 2

3βJzm− 1

3(βJzm)3 + ... (44)

The critical point is at βJz = 3/2, so the critical temperature is at kBTc = 2Jz/3 which is lower than that for spin1/2 due to the additional entropy of the spin one system. The critical exponents β and δ are clearly the same as forthe spin 1/2 case. The free energy is lower for the spin 1 case due to the higher entropy.

—————–

Assigned Problem 3. By using b = ξ show that Eqs. (58) reduces to Eqs. (50).

Assigned Problem 4. Consider the Landau free energy,

F = a(T − Tc)m2 + bm4 + cm6 (45)

where c(T ) > 0 as required for stability. Sketch the possible behaviors for a(T ), b(T ) positive and negative, and showthat the system undergoes a first order transition at some value Tc. Find the value of a(Tc) and the discontinuity inm at the transition.

Solution.The mean field equation is,

δF

δm= am+ bm3 + cm5 = 0; (46)

Note that in general we do a variation with respect to m, so when we add fluctuations later, we need to use theEuler-Lagrange equation. Here the variation is the same as a partial derivative with respect to m. Solving the meanfield equation, we find five solutions.

m = 0, m = ±m±; m2± =

−b± (b2 − 4ac)1/2

2c(47)

Though there are always five solutions, only the real solutions are physical. Analysis of the behavior of the modelreduces to identifying the real solutions, and finding which real solution has the lowest free energy. We can understandthe nature of the solutions by looking at the second derivative,

∂2F

∂m2= a+ 3bm2 + 5cm4, (48)

which enables us to distinguish between maxima and minima. We also use the fact that F is symmetric in m andthat at large m, because c is positive, F is large and positive for large |m|. Finally, without loss of generality, we candivide through by c, or equivalently set c = 1. Now consider the four cases for a, b.

(i) a > 0, b > 0. In this case b2 − 4ac < b, so m2± is always negative. Therefore the solutions m± are always

imaginary. The only real solution is m = 0, which is a minimum having F (0) = 0.(ii) a < 0, b > 0. In this regime, b2 + 4|a|c > b2, so m2

+ > 0, so that m+ is real. However −b − (b2 + 4|a|c)1/2

remains negative, so m− remains imaginary. The real solutions are thus a maximum at m = 0 and two symmetricminima at m2

+ = ± 12 ([b2 + 4|a|]1/2 − b).

There is a phase transition between states (i) and (ii) that occurs at a = 0 where two new solutions emerge and theextremum at m = 0 changes from a maximum for a < 0 to a minimum for a > 0. The nature of the transition isfound by making a small |a| expansion of the solutions m+, which leads to m ≈ |a|1/2. This is the Ising/Van derWaals univesality class we have studied using mean field theory, where we found |a| ≈ |T − Tc|

(iii) a < 0, b < 0. In this regime, b2 + 4|a|c > b2, so m+ is real but m− remains imaginary. Therefore, as in case(ii), there is a maximum at m = 0, and minima at ±m+.

(iv) a > 0, b < 0. In this regime there are several things going on. First, the discriminant b2 − 4a is negative forb2 < 2a, so in this regime there is only one real solution, a minimum at m = 0. For b2 > 2a, there are five realsolutions because |b| ± (b2 − 4a)1/2 > 0. Moreover, we know that the solution at m = 0 is a minimum, so we knowthat ±m− are maxima, while ±m+ are minima.

The final issue we have to resolve is the behavior of the minima m+ as a function of a, b, in particular we need toknow if F (m+) is greater than or less than F (0). If the lowest free energy state changes it corresponds to a phase

9

transition. We can solve this problem by evaluating F (m+), or we can find solutions where F (m0) = 0 and then solvem0 = m+. The later leads to,

m20 =

3

2[|b|2± (

b2

4− 4a

3)1/2] (49)

and setting m20 = m2

+, we find,

3

2[|b|2± (

b2

4− 4a

3)1/2] =

1

2[|b|+ (b2 − 4a)1/2] (50)

which has solution

|b|∗ = 4(a

3)1/2; with m2

∗ = m20(|b|∗, a) = 2(

a

3)1/2 (51)

m∗ is the magnetization on the phase boundary defined by |b|∗. There is then a first order phase transition frommagnetization m∗ for |b| < |b|∗, to magnetization m = 0 for |b| > |b|∗. The behavior of the magnetization on thephase boundary is m∗ ≈ a1/4 ≈ |T − Tc|1/4, which is the mean field result for the order parameter near a tricriticalpoint where a line of second order phase transitions meets a line of first order phase transitions.

—————–

Assigned Problem 5. Consider a Landau theory with a cubic term,

F = a(T − Tc)m2 + bm3 + cm4. (52)

Analyse the behavior of this model, particularly the nature of the different phases and phase transition(s) that occur.

SolutionsIt is convenient to write this in the form,

F =α

2m2 − β

3m3 +

γ

4m4, (53)

We require that γ > 0 to ensure that the magnetization at the free energy minimum is bounded, and we can takeγ = 1 without loss of generality. We also note that the transformation β → −β, m → −m leaves F invariant, so weonly consider β > 0. First consider the case where β = 0. In that case we have the Landau theory that we studiedbefore and we know that there is a continuous mean field phase transition at α = 0 from a spontaneously orderedphase for α < 0(T < Tc) to a disordered m = 0 phase for α > 0(T > Tc). For β > 0, the positive magnetization stateis further stabilized, so we look for a transition at α > 0. To determine the behavior for β > 0, we find the extremeof F ,

∂F

∂m= 0 = αm− βm2 +m3,

where has solutions,

m = 0;m± =1

2[β ±

√β2 − 4α] (56)

when the disriminant is negative there is only one real solution (m = 0), while if the discriminant is positive, thereare three real solutions. The critical condition is called the spinodal line and it is given by,

βs = 2√α α > 0 (57)

for β < βs there is no ordered phase, while for β > βs there are three extrema, but we don’t yet know if there is astable ordered phase. To find out if a positive magnetization state is stable, we compare the free energies, f(0) andf(m+) and find the critical condition by equating them,

f(0) = 0 =α

2m2

+ −β

3m3

+ +1

4m4

+ (58)

10

or,

α

2− β

3m+ +

1

4m2

+ = 0 (59)

We also have,

m2+ = βm+ − α (60)

so that,

α

2− β

3m+ +

1

4(βm+ − α) = 0 (61)

or

m+ +3α

β= 0. (62)

Using the solution for m+, we find, √β2 − 4α = −6α

β− β (63)

Solving yields,

βc =3√2

√α. (64)

For β < βc, the model predicts an ordered phase with finite magnetization, and at βc there is a first order transitioninto a phase with finite magnetization. In the regime βs < β < βc the system has zero magnetization but there is ametastable state with finite magnetization. For β < βs the zero magnetization state is the only local minimum in themodel.

—————–

Assigned Problem 6.A spin half Ising model with four spin interactions on a square lattice has Hamiltonian,

H = −∑

ijkl in square

JSiSjSkSl (65)

where the sum is over the smallest squares on an infinite square lattice, the interaction is ferromagnetic J > 0 andSi = ±1. Each elementary square is counted only once in the sum.

Using a leading order expansion in the fluctuations (i.e. write Si = mi + (Si −mi) and expand to leading order inthe fluctuations), find the mean field Hamiltonian for this problem (Here mi =< Si > is the magnetization at site i).

Using the mean field Hamiltonian and assuming a homogeneous state where mi = m, find an expression for themean field Helmholtz free energy, and the mean field equation for this problem.

Taking J = 1, sketch the behavior of the solutions to the mean field equation as a function of temperature. Doesthe non-trivial solution move continuously toward the m = 0 solution as the temperature increases? Is the behaviorof the order parameter at the critical temperature discontinuous or continuous? Do you expect the correlation lengthto diverge at the critical point in this problem?

SolutionsWe start by writing a three spin term in terms of an expansion in the fluctuations,

SiSjSkSl = [mi + (Si −mi)][mj + (Sj −mj)][mk + (Sk −mk)][ml + (Sl −ml)] (66)

Expanding to leading order in the fluctuations, we then have,

SiSjSkSl ≈ mimjmkml + (Si −mi)mjmkml +mi(Sj −mj)mkml +mimj(Sk −mk)ml +mimjmk(Sl −ml) (67)

11

Since the interaction is ferromagnetic we can make the uniform assumption mi = m, which yields,

HMF = −∑

ijkl in square

[−3Jm4 +m3(Si + Sj + Sk + Sl)] = 3JNm4 − 4Jm3∑i

Si (68)

The mean field partition function is then,

ZMF = e−3βJNm4

[2cosh(4βJm3)]N (69)

which leads to the free energy,

FMF = −kBT lnZMF = 3JNm4 − kBTNln[2cosh(4βJm3)] (70)

and the mean field equation is,

m = tanh(4βJm3) (71)

The mean field equation can be found in several ways, for example by doing the variation δF/δm = 0, or by findingthe expectation < Si > using the mean field Hamiltonian.

By plotting the graphs of m and tanh(4βJm3), it is evident that at large values of K = βJ (i.e. low temperature),there are two positive m solutions as well as the solution at m = 0. However as K decreases (i.e. T increases) there isa critical point at which the two non-trivial positive m solutions merge. This defines the critical point. At this pointm is finite. For T > Tc the positive m solution disappears so there is a discontinuous jump in the magnetization froma finite value to zero at Tc. The transition in this model is then first order, in contrast to the pair interaction casewhere the transition is continuous. The correlation length at the transition found here is then remains finite at thecritial point.

Assigned Problem 7. The Dieterici equation of state for a gas is,

P =kBT

v − be−a/(kBTv) (72)

where v = V/N . Find the critical point and the values of the exponents β, δ, γ for this model.

Solution. The critical point is found by solving,

Pc =kBTcvc − b

e−a/(kBTvc); (73)

∂P

∂v= 0 = − kBTc

(vc − b)2e−a/(kBTvc) +

kBTcvc − b

a

kBTv2c

e−a/(kBTvc) (74)

Which simplify to,

− 1

vc − b+

a

kBTcv2c

= 0; soa

kBTc=

v2c

vc − b(75)

The second derivative yields,

∂2P

∂v2= 0 = 2

kBTc(vc − b)3

e−a/(kBTvc) − 2kBT

(vc − b)2

a

kBTcv2c

e−a/(kBTcvc)

−2kBTcvc − b

a

kBTcv3c

e−a/(kBTcvc) +kBTcvc − b

(a

kBTcv2c

)2e−a/(kBTcvc) (76)

so that,

21

(vc − b)2− 2

1

vc − ba

kBTcv2c

− 2a

kBTcv3c

+ (a

kBTcv2c

)2 = 0 (77)

12

and using Eq. (),

−21

vc

1

vc − b+ (

1

vc − b)2 = 0; so vc = 2b (78)

so we find that,

vc = 2b; kBTc =a

4b; Pc =

a

4e2b2(79)

To find the critical exponents we write v = vc + δv, T = Tc + δT, P = Pc + δP , so that,

Pc + δP =kB(Tc + δT )

vc + δv − be−a/[kB(Tc+δT )(vc+δv)] =

kBTcb

(1 + δTTc

)

1 + δvb

e−a/[kBTcvc(1+δT/Tc)(1+δv/vc)] (80)

This reduces to,

1 + p = e2 1 + t

1 + 2xExp[− 2

(1 + t)(1 + x)] (81)

where p = δP/Pc, t = δT/Tc, x = δv/vc. To third order this expansion gives,

p = 3t+ 2t2 − 2

3t3 + (−2t− 4t2)x+ 2tx2 − 2

3x3 (82)

Taking a derivative with respect to v at setting x = 0 leads to,

κT = (−V (∂P

∂V)T )−1 ≈ |T − Tc|−1 (83)

so γ = 1. The exponent δ is found by setting t to zero so that p ≈ x3, so that δ = 3. To find β we assume thatpl = pg, xl = −xg, so that,

pl = 3t+ 2t2 − 2

3t3 + (−2t− 4t2)xl + 2tx2

l −2

3x3l (84)

p− g = 3t+ 2t2 − 2

3t3 − (−2t− 4t2)xl + 2tx2

l +2

3x3l (85)

Setting these equations to be equal yields,

2(−2t− 4t2)xl =2

3x3l , so that xl ≈ |T − Tc|1/2 (86)

where we dropped the t2 term as it is higher order. In the above analysis the signs of the t and x are consistent buthave to be checked each time.

—————–

Assigned Problem 8. Consider a phase co-existence curve in a P − T phase diagram, separating two phases“A” and “B”. Consider two points on the phase coexistence curve at P, T and P + ∆P, T + ∆T . Since the chemicalpotential of the phases A and B are the same at any given point on the co-existence curve, we have,

∆gA = gA(P + ∆P, T + ∆T )− gA(P, T ) = gB(P + ∆P, T + ∆T )− gB(P, T ) = ∆gB (87)

From this relation, prove the Clausius-Clapeyron relation,

∂P

∂T=

L

T (VB − VA)(88)

where L is the latent heat. Find the form of this relation for the van der Waals equation of state. What is thedependence of the latent heat as T → Tc. Is this exponent related to any of the other exponents in the problem?

13

SolutionThe Gibb’s energies G(B), G(A) are the same at temperature T and at temperature T +∆T , so dGA = dGB , where

dG = G(T + dT )−G(T ). We also have,

dG = −SdT + V dP ; sodG

dT= −S + V

dP

dT(89)

Using the latter expression for both phases we have,

−SA + VAdP

dT= −SB + VB

dP

dT(90)

where we used the fact that the pressure is the same in the A and B phases. Now we use the Clausius relationdS = dQ/T , and the fact that dQ = L to find,

dP

dT=SB − SAVB − VA

=L

T (VB − VA)(91)

—————–

Assigned Problem 9. Recent work on black hole thermodynamics has suggested that a black hole with chargeQ obeys the equation of state,

P =T

v− 1

2πv2+

2Q2

πv4(92)

where the physical pressure and temperature are given by,

Press =hc

l2P; Temp =

hc

kBT ; v = 2l2P r+ (93)

where lP = hGN/c3 is the Planck length and r+ is the black hole event horizon. Yes, the ”Temp” temperature

expression looks strange, but that is the expression in the paper I got this from and I have not had time to figureout if there is a typo in the paper. In any case you don’t need to use that equation. Find the critical point of thisequation of state and find the critical exponents δ and β.

SolutionThe conditions for the critical point are,

∂P

∂v=∂2P

∂v2= 0 (94)

which lead to the two equations,

− Tv2

+1

πv3− 8Q2

πv5= 0;

2T

v3− 3

πv4+

40Q2

πv6= 0 (95)

which reduce to,

−Tv3 +v2

π− 8Q2

π= 0; 2Tv3 − 3v2

π+

40Q2

π= 0 (96)

Multiplying the first equation by two and adding it to the second equation gives,

−v2

π+

24Q2

π= 0; so vc =

√24Q (97)

and hence,

Tc =1

3√

6πQ; Pc =

1

96πQ2(98)

Defining p = P/Pc, ν = v/vc, t = T/Tc, we find,

8t = 3ν(p+2

ν2)− 1

ν3(99)

14

In these variables the critical point p, t, v is at 1, 1, 1. To find the critical exponents we expand as in the van der Waalscase, so we define, t = 1− δt, ν = 1 + δν; p = 1 + δp.

8

3

(1− δt)1 + δν

= 1 + δp+2

(1 + δν)2− 1

3(1 + ν)4(100)

Using the Taylor expansion,

1

(1 + x)l= 1− lx+

1

2l(l + 1)x2 − 1

3!l(l + 1)(l + 2) + ... (101)

we find,

8

3(1− δt)(1− δv + δv2 − δv3 + ...) = 1 + δp+ 2(1− 2δv + 3δv2 − 4δv3 + ...)− 1

3(1− 4δv + 10δv2 − 20δv3 + ...) (102)

Expanding gives,

8

3(1− δv + δv2 − δv3 − δt+ δtδv − δtδv2 + ..) =

8

3+ δp− 8

3δv +

8

3δv2 − 4

3δv3 (103)

which reduces to,

δp = −δt+ δtδv − δtδv2 − 4

3δv3 (104)

Using the fact that δvg = −δvl and subtracting equations for both the gas and liquid phases leads to,

δvg =√

0.75δt1/2. (105)

Finally setting δt = 0, we find δvg ∼ δp1/3. We thus find the exponents β = 1/2 and δ = 3 as expected for mean fieldsystems.

—————–

Assigned Problem 10. Consider a ferromagnetic nearest neighbor, spin 1/2, square lattice Ising model where theinteractions along the x-axis, Jx, are different than those along the y-axis, Jy. Extend the low and high temperatureexpansions Eq. (150) and Eq. (152) to this case. Does duality still hold? From your expansions, find the internalenergy and the specific heat.

Solution

Z =∑

Si=±1

∏<ij>

eKijSiSj = (Cosh(Kx))N (Cosh(Ky))N∑

Si=±1

∏<ij>

(1 + tijSiSj). (106)

tij = tx for horizontal bonds and tij = ty for vertical bonds, where tx = tanh(Kx), ty = tanh(Ky). The diagramsused are similar, but now we have to treat subclasses with different numbers of hirizontal and vertical bonds, so that,

−βF = ln(Z) =zN

2ln(Cosh(K)) +Nln(2) + ln[1 +Nt4 + 2Nt6 +N(N +

9

2)t8 + 0(t10)] (107)

becomes

−βF = N [ln(2) + ln(Cosh(Kx)) + ln(Cosh(Kx)) + t2xt2y + t2xt

2y(t2x + t2y) +

5

2(t4xt

4y) + t2xt

6y + t6xt

2y + ....] (108)

Similarly, the extension of the low temperature expansion,

Z =∑

Si=±1

∏<ij>

eKSiSj = eKzN [1 +Ns4 + 2Ns6 +N(N +9

2)s8 +O(s10)] (109)

to the anisotropic case leads to,

−βF = N [Kx +Ky + s2xs

2y + s2

xs2y(s2

x + s2y) +

5

2s4xs

4y + s2

xs6y + s6

xs2y...] (110)

where sx = e−2Kx , sy = e−2Ky . Duality holds for both x and y directions.

15

—————–

Assigned Problem 11. Find the second virial coefficient for four cases: (i) the classical hard sphere gas; (ii)Non-interacting Fermions; (iii) Non-interacting Bosons; (iv) The van der Waals gas.

Solution

Pv

kBT=

∞∑l=1

al(T )(λ3

v)l−1 (111)

where the first virial coefficient a1(T ) = 1, and the second virial coefficient is a2(T ). The virial expansion is mostoften carried out in the grand canonical ensemble, where we may write,

P

kBT=

1

λ3

∞∑l=1

blzl;

N

V=

1

λ3

∑l=1

lblzl (112)

so that,

Pv

kBT=

∑∞l=1 blz

l∑∞l=1 lblz

l=

∞∑l=1

al(T )(λ3

v)l−1 (113)

which gives relations between the quantities al(T ) and bl(T ). For the second virial coefficient the relationship isa2(T ) = −b2(T ). We have already calculated the coefficients bl for the ideal Bose and Fermi gases, with the results,

bl =(−1)l+1

l5/2; (Ideal Fermi) bl =

1

l5/2; (Ideal Bose) (114)

so we have,

Pv

kBT= 1− 1

25/2(λ3

v) + ....; (Ideal Bose) (115)

and

Pv

kBT= 1 +

1

25/2(λ3

v) + ....; (Ideal Fermi) (116)

The van der Waals equation of state may be expanded to find the second virial coefficient,

P =kBT

v − b− a

v

2, so that

Pv

kBT=

1

(1− b/v)− a

kBTv≈ 1 +

b− a/kBTv

+ ... (117)

For the classical interacting gas, b2 is given by,

b2 =1

2V λ3

∫dr1d

3r2[e−βu(~r1−~r2 − 1] =1

2λ3

∫d3r[e−βu(~r) − 1] (118)

For the hard sphere problem, with a hard core radius of R, we then find that b2 = −2πR3/(3λ3), so the virialexpansion for this case is,

Pv

kBT= 1 +

2πR3

3v+ .... (119)

The behavior of the second virial coefficient as a function of temperature can be used to deduce the interactionpotential, and the importance of quantum effects as they have different temperature dependences.

Assigned Problem 12. The BCS pairing Hamiltonian is a simplified model in which only pairs with zero center ofmass momentum are included in the analysis. We also assume that the fermion pairing that leads to superconductivityoccurs in the singlet channel. The BCS Hamiltonian is then,

Hpair − µN =∑~kσ

(ε~kσ − µ) a†~kσa~kσ +

∑~k~l

V~k~l a†~k↑a†−~k,↓

a−~l↓a~l↑, (120)

16

where N =∑~kσ n~kσ is the number of electrons in the Fermi sea. By making an expansion in the fluctuations and

defining,

b~k =< a−~k↓a~k↑ >, and b∗~k =< a†~k↑a†−~k↓

> . (121)

where b∗~k is the average number of pairs in the system at wavevector ~k, show that the mean field BCS Hamiltonian is

given by,

HMF − µN =∑~kσ

(ε~kσ − µ) a†~kσa~kσ +

∑~k~l

V~k~l (a†~k↑a†−~k,↓

b~l + b∗~ka−~l↓a~l↑ − b∗~kb~l) (122)

This is the Hamiltonian that we will solve to find the thermodynamic behavior of superconductors, using an atomisticmodel.

—————–

Assigned Problem 13. The BCS pairing Hamiltonian is a simplified model in which only pairs with zero center ofmass momentum are included in the analysis. We also assume that the fermion pairing that leads to superconductivityoccurs in the singlet channel. The BCS Hamiltonian is then,

Hpair − µN =∑~kσ

(ε~kσ − µ) a†~kσa~kσ +

∑~k~l

V~k~l a†~k↑a†−~k,↓

a−~l↓a~l↑, (123)

where N =∑~kσ n~kσ is the number of electrons in the Fermi sea. Defining,

b~k =< a−~k↓a~k↑ >, and b∗~k =< a†~k↑a†−~k↓

> . (124)

carry out a leading order expansion in fluctuations, leading to,

HMF − µN =∑~kσ

(ε~kσ − µ) a†~kσa~kσ +

∑~k~l

V~k~l (a†~k↑a†−~k,↓

b~l + b∗~ka−~l↓a~l↑ − b∗~kb~l) (125)

This is the Hamiltonian that we will solve to find the thermodynamic behavior of superconductors, using an atomisticmodel.

Solution. The mean-field Hamiltonian can be considered as a first order expansion in the fluctuations i.e.

a−~k↓a~k↑ =< a−~k↓a~k↑ > +[a−~k↓a~k↑− < a−~k↓a~k↑ >] (126)

Substition of this into the Hamilonian, along with the definitions above lead to HMF .

—————–

Assigned Problem 14. Using the Bogoliubov-Valatin transformation (Eq. 163), show that the mean field BCSHamiltonian (Eq. (162)) reduces to Eq. (166), provided Equations (167) and (168) are true.

Solution. With this transformation, the mean field Hamiltonian looks messy,

HMF − µN =

∑~k

(ε~k − µ) (a†~k↑a~k↑ + a†~k↓

a~k↓)−∑~k

(∆~ka†~k↑a†−~k,↓

+ ∆∗~ka−~k↓a~k↑ − b∗~k∆~k) =

∑~k

(ε~k − µ) ([u∗~kγ†~k↑

+ v~kγ−~k↓][u~kγ~k↑ + v∗~kγ†−~k↓

] + [u∗~kγ†~k↓− v~kγ−~k↑][u~kγ~k↓ − v

∗~kγ†−~k↑

])

−∆~k[u∗~kγ†~k↑

+ v~kγ−~k↓][u∗~kγ†−~k↓− v~kγ~k↑]−∆∗~k[u~kγ−~k↓ − v

∗~kγ†~k↑

][u~kγ~k↑ + v∗~kγ†−~k↓

] + b∗~k∆~k

17

Expanding this yields,∑~k

(ε~k − µ)[u∗~ku~kγ†~k↑γ~k↑ + u∗~kv

∗~kγ†~k↑

γ†−~k↓

+ v~ku~kγ−~k↓γ~k↑ + v~kv∗~kγ−~k↓γ

†−~k↓

]

+(ε~k − µ)[u∗~ku~kγ†~k↓γ~k↓ − u

∗~kv∗~kγ†~k↓γ†−~k↑− v~ku~kγ−~k↑γ~k↓ + v~kv

∗~kγ−~k↑γ

†−~k↑

]

−∆~k[u∗~ku∗~kγ†~k↑

γ†−~k↓− u∗~kv~kγ

†~k↑γ~k↑ + v~ku

∗~kγ−~k↓γ

†−~k↓− v~kv~kγ−~k↓γ~k↑]

−∆∗~k[u~ku~kγ−~k↓γ~k↑ + u~kv∗~kγ−~k↓γ

†−~k↓− v∗~ku~kγ

†~k↑γ~k↑ − v

∗~kv∗~kγ†~k↑γ†−~k↓

] + b∗~k∆~k

We now collect the terms in this expression into three catagories: Those which have no operators in them; those which

can be reduced to diagonal form ie. those which are of the form γ†kγk and; those that are off diagonal (e.g. γ†γ†).The first stage is to collect together the terms which look to be in these three catagories. First the constant term,

HMF − µN =∑~k

b∗~k∆~k

The following terms can be converted to diagonal form,

+(ε~k − µ) (|u~k|2γ†~k↑

γ~k↑ + |v~k|2γ−~k↓γ

†−~k↓

+ |u~k|2γ†~k↓

γ~k↓ + |v~k|2γ−~k↑γ

†−~k↑

)

−∆~k[−u∗~kv~kγ†~k↑γ~k↑ + v~ku

∗~kγ−~k↓γ

†−~k↓

]−∆∗~k[u~kv∗~kγ−~k↓γ

†−~k↓− v∗~ku~kγ

†~k↑γ~k↑]

Finally the off-diagonal terms are,

+(ε~k − µ) (u∗~kv∗~kγ†~k↑

γ†−~k↓− u∗~kv

∗~kγ†~k↓

γ†−~k↑

+ v~ku~kγ−~k↓γ~k↑ − v~ku~kγ−~k↑γ~k↓)

−∆~k[(u∗~k)2γ†~k↑γ†−~k↓− v2

~kγ−~k↓γ~k↑]−∆∗~k[u2

~kγ−~k↓γ~k↑ − (v∗~k)2γ†~k↑

γ†−~k↓

]

We have to rearrange the terms catagorized as diagonal above, as we need them in the form γ†γ. We do this usingthe commutation relation, i.e. γγ† = 1− γ†γ. This yields,

HMF − µN =∑~k

b∗~k∆~k + 2(ε~k − µ)|v~k|2 −∆~ku

∗~kv~k −∆∗~ku~kv

∗~k

+(ε~k − µ) (|u~k|2γ†~k↑

γ~k↑ − |v~k|2γ†−~k↓

γ−~k↓ + |u~k|2γ†~k↓

γ~k↓ − |v~k|2γ†−~k↑

γ−~k↑)

+∆~k[u∗~kv~kγ†~k↑γ~k↑ + v~ku

∗~kγ†−~k↓

γ−~k↓] + ∆∗~k[u~kv∗~kγ†−~k↓

γ−~k↓ + v∗~ku~kγ†~k↑γ~k↑]

Finally the off-diagonal terms are (as before),

+(ε~k − µ) (u∗~kv∗~kγ†~k↑

γ†−~k↓− u∗~kv

∗~kγ†~k↓

γ†−~k↑

+ v~ku~kγ−~k↓γ~k↑ − v~ku~kγ−~k↑γ~k↓)

−∆~k[(u∗~k)2γ†~k↑γ†−~k↓− v2

~kγ−~k↓γ~k↑]−∆∗~k[u2

~kγ−~k↓γ~k↑ − (v∗~k)2γ†~k↑

γ†−~k↓

]

We note that n~kσ = n−~kσ, and that the expectation of other terms that transform into one another through the

transformation ~k → −~k, are equivalent. Collecting terms then leads to Eqs. (166)-(168) of the lecture notes.

—————–

18

Assigned Problem 15. We define

v~k =g~k

(1 + |g~k|2)1/2(127)

show that Eq. (167) reduces to (173).

Solution Starting with,

2(ε~k − µ)(1− |v~k|2)1/2v~k + ∆~kv

2~k−∆∗~k(1− |v~k|

2) = 0, (128)

We have,

|u~k|2 = 1− |v~k|

2 =1

1 + |g~k|2; uk =

1

(1 + |g~k|2)1/2. (129)

Substitution into Eq. () leads to,

2(ε~k − µ)g~k + ∆~kg2~k−∆∗~k = 0. (130)

—————–

Assigned Problem 16. Show that E~k as defined in Eq. (174) is in agreement with Eq. (168).

Solution We need to show that the definitions,

E~k = (ε~k − µ)(|u~k|2 − |v~k|

2) + ∆~ku∗~kv~k + ∆∗~kv

∗~ku~k. (131)

and

E~k = [(ε~k − µ)2 + |∆~k|2]1/2 (132)

are the same. To do this we use the results of Problem 9 in Eq. (53) to write,

E~k = (ε~k − µ)([E~k + (ε~k − µ)]2

4E2~k

−[E~k − (ε~k − µ)]2

4E2~k

) + ∆~k

∆∗~k2E~k

+ ∆∗~k∆~k

2E~k. (133)

which reduces to Eq. (54) as required.

—————–

Assigned Problem 17. Prove the relations Eq. (175-177).

Solution We have,

g~k =E~k − (ε~k − µ)

∆~k

; E~k = [(ε~k − µ)2 + |∆~k|2]1/2 (134)

so that,

|∆~k|2 = E2

~k− (ε~k − µ)2; (135)

so that,

|g~k|2 =

E~k − (ε~k − µ)

∆~k

=[E~k − (ε~k − µ)]2

E2~k− (ε~k − µ)2

=E~k − (ε~k − µ)

E~k + (ε~k − µ)(136)

We then find,

|v~k|2 =

|g~k|2

1 + |g~k|2=E~k − (ε~k − µ)

2E~k; |u~k|

2 =E~k + (ε~k − µ)

2E~k(137)

and hence,

u~kv∗~k

=g~k

1 + |g~k|2= g~k|u~k|

2 =∆∗~k2E~k

(138)

—————–

1

Solutions to assigned problems and sample quiz problems for Part 4

Sample Quiz Problems

Quiz Problem 1. Write down the BCS gap equation at finite temperature and compare it to the zero temperatureexpression. Explain the origin of the differences between the two expressions.

Quiz Problem 2. Make a plot of the BCS gap as a function of temperature. Write down the scaling behavior ofthe gap on approach to the critical temperature.

Quiz Problem 3. Starting from the quantum mechanical expression for the current, show that in neutral super-fluids circulation is quantized. Give an expression of the quantum of circulation.

Quiz Problem 4. Starting from the quantum mechanical expression for the current in the presence of a vectorpotential, show that flux quantized. Give an expression for the flux quantum.

Quiz Problem 5. Describe the physical meaning of the healing length ( ξ ) in superfluids and superconductors.By considering the linearized Ginzburg-Landau equation in zero field find a solution describing the attenuation ofsuperconducting pair density near the surface of a superconductor.

Quiz Problem 6. Using either London’s original argument or starting with the Ginzburg-Landau equation, derivethe London differential equation describing the penetration of parallel magnetic field into a superconducting surface.Show that it has the solution B(x) = B0e

−x/λ, where λ is the London penetration depth.

Quiz Problem 7. What is the mixed phase of a type II superconductor? Give a physical reasoning to explainwhy the mixed phase of a type II superconductor can have, at sufficienty high external field, a lower free energy thanthe Meissner state.

Quiz Problem 8. Using London’s theory, find an expression for the lower critical field of a type II superconductor.

Quiz Problem 9. Using G-L theory find an expression for the upper critical field of a type II superconductor.

Quiz Problem 10. Write down the Langevin equation for motion of a particle in the presence of random forces.Describe the physical meaning of the random forces and the meaning of Gaussian white noise.

Quiz problem 11. Write down the non-conserved order parameter relaxation equation for the Ising model withinLandau theory and show that near the critical point the relaxation time diverges.

Quiz problem 12. Explain the difference between conserved dynamics and non-conserved dynamics. Give anexample of the two cases.

Quiz problem 13. Sketch the phase diagrams of the van der Waals gas and the ferromagnetic Ising model,indicating the spinodal lines. How are the spinodal lines defined in the two cases.

Quiz problem 14. Explain the physical basis for the concept of a critical droplet in the homogeneous nucleationof a new stable phase from a metastable phase. Find an expression for the critical droplet size for an Ising model ina magnetic field at low temperature.

Quiz problem 15. Derive the Vlasov equation for the time evolution of the single particle distribution functionf(t, ~r,~v) for a collisionless system.

Assigned problems

Assigned Problem 1. By doing a variation with respect to ψ∗ of the Ginzburg-Landau free energy (Eq. (64)) ofthe notes, derive the Ginzburg-Landau equation (65).

SolutionWithin Ginzburg-Landau theory, the Gibbs’ free energy for a superconductor is given by,

2

gGL =

∫d3r

(1

2m|(ih∇+ qA)ψ(~r))|2 + a(T )|ψ(~r)|2 +

b(T )

2|ψ(~r)|4 +

B2

2µ0+µ0H

2

2−B ·H

). (1)

Expanding this expression(and dropping function arguments), we have,

gGL =

∫d3r

(1

2m[(−ih∇+ qA)ψ∗ · (ih∇+ qA)ψ] + aψ∗ψ +

b

2(ψ∗)2ψ2 +

(∇∧A)2

2µ0+µ0H

2

2− (∇∧A) ·H

). (2)

Expanding the first term yields,

gGL =

∫d3r

(1

2m

3∑α=1

[h2∂ψ∗

∂xα

∂ψ

∂xα+ ihqAα(ψ∗

∂ψ

∂xα− ψ∂ψ

∗

∂xα) + q2A2

αψ∗ψ]

+ aψ∗ψ +b

2(ψ∗)2ψ2 +

(∇∧A)2

2µ0+µ0H

2

2− (∇∧A) ·H

)(3)

Using ~r = (x1, x2, x3), the Euler-Lagrange equation for a variation with respect to ψ∗ is,

δgGLδψ∗

=∂gGL∂ψ∗

−∑α

∂

∂xα(∂gGL

∂(∂ψ∗

∂xα)) (4)

Carring out this variation yields,

δgGLδψ∗

=1

2m

∑α

(ihqAα∂ψ

∂xα+ q2A2

αψ) + aψ + b|ψ|2ψ − 1

2m

∑α

∂

∂xα(h2

∂ψ

∂xα− ihqAαψ). (5)

This is equivalent to,

δgGLδψ∗

=1

2m(ih∇+ qA)2ψ + aψ + b|ψ|2ψ (6)

which is the GL equation in an applied field, as required. Another useful form is found by using the London gaugewhere ∇ ·A = 0 to find,

δgGLδψ∗

=1

2m

∑α

(−h2 ∂2ψ

∂x2α+ 2ihqAα

∂ψ

∂xα+ q2A2

αψ) + aψ + b|ψ|2ψ = 0 (7)

—————–

Assigned Problem 2. By doing a variation with respect to the vector potential A of the Ginzburg-Landau freeenergy (Eq. (64)) of the notes, derive the expression for the current (66).

SolutionThe Euler-Lagrange equation for a variation with respect to one component Ai of the vector potential is,

δgGLδAi

=∂gGL∂Ai

−∑α

∂

∂xα(∂gGL

∂( ∂Ai∂xα)) (8)

We then have,

δgGLδAi

= ihq

2m(ψ∗

∂ψ

∂xi− ψ∂ψ

∗

∂xi) +

q2

mAiψ

∗ψ −∑α

1

2µ0

∂

∂xα(∂(∇∧A)2

∂( ∂Ai∂xα)

) (9)

The term B ·H does not contribute as it is linear in the derivatives of A and hence is zero after the application ofthe second term in the Euler-Lagrange equation. The expansion of (∇∧A)2 is,

(∇∧A) · (∇∧A) = (∂A3

∂x2− ∂A2

∂x3)2 + (

∂A1

∂x3− ∂A3

∂x1)2 + (

∂A2

∂x1− ∂A1

∂x2)2 (10)

3

Using this expression to evaluate the last term in the case Ai = A1, yields∑α

∂

∂xα(∂(∇∧A)2

∂(∂A1

∂xα)

) = 2(− ∂2A2

∂x1∂x2+∂2A1

∂x22+∂2A1

∂x23− ∂2A3

∂x1∂x3) (11)

Similar expressions hold for A2 and A3. Now note that ∇ ∧ (∇ ∧ A) = ∇(∇ · A) − ∇2A. The first element of thisexpression is,

(∇(∇ ·A)−∇2A)1 =∂2A1

∂x21+

∂2A2

∂x1x2+

∂2A3

∂x1x3− ∂2A1

∂x21− ∂2A1

∂x22− ∂2A1

∂x23

We thus find the identity,

δ(∇∧A)2

δA= 2∇∧ (∇∧A) = 2[∇(∇ ·A)−∇2A] (12)

We also have,

∇∧ (∇∧A) = ∇∧B = µ0J (13)

Substituting these results into the equation above yields the simple result,

−∑α

1

2µ0

∂

∂xα(∂(∇∧A)2

∂( ∂Ai∂xα)

) = js

Using this in equation (9) for we find,

js = −i hq2m

(ψ∗∇ψ − ψ∇ψ∗)− q2

mAiψ

∗ψ (14)

as required.

—————–

Assigned Problem 3. Show that minimizing the Helmholtz free energy within London theory,

f1 =1

2µ0

∫ ∞0

(B2 + λ2µ20J

2)2πrdr (15)

gives the London equation.

SolutionWe apply the Euler-Lagrange equations to,

fLondon =1

2µ0

∫dV [λ2(∇∧B)2 +B2]. (16)

Using the Euler-Lagrange equation with the identities

δ(∇∧B)2

δB= 2∇∧ (∇∧B) = 2[∇(∇ ·B)−∇2B] (17)

and using ∇ ·B = 0, yields the London equation λ2∇2B = B.

—————–

Assigned Problem 4. Using London theory, find the Helmholtz free energy per unit length of a vortex in asuperconductor in an external field H.

Solution. The energy(per unit length) of an isolated vortex is given by,

ε1 =

∫ ∞0

(B2

2µ0+

1

2ρsv

2s)2πrdr (18)

4

The first term is the field energy and the second is the kinetic energy of the superconducting electrons (see Eq. (4)).The supercurrent is related to the velocity by,

J = nbqvs (19)

and using Maxwells equation, we then write,

ε1 =1

2µ0

∫ ∞0

(B2 + λ2µ20J

2)2πrdr (20)

which explicitly shows the contributions of the field and the current. In the large λ/ξ limit, this is dominated by theregime ξ ≤ r ≤ λ, so we find an approximate value of the vortex energy by using,

ε1 ≈1

2µ0(φ0

2πλ2)22π

∫ λ

ξ

[r(Ln(λ

r))2 + r(

λ

r)2]dr

=φ20

4πµ0λ2

∫ 1

ξ/λ

[x(Ln(x))2 +1

x]dx

=φ20

4πµ0λ2[x2

2(1

2− Ln(x) + Ln(x)2) + Ln(x)]|1ξ

λ

≈ φ204πµ0λ2

[Ln(λ

ξ)] (21)

Notice that the energy cost of forming the vortex is dominated by the kinetic energy of the superconducting electrons(the logarithmic term). The energy cost due to the magnetic field is relatively small.

—————–

Assigned Problem 5. Using the Gibb’s free energy within London theory, demonstrate that the triangular latticevortex array has lower energy than the square lattice array, for a fixed applied field, H > Hc1, which is close to Hc1.Solution

The energy to add the N th vortex to a vortex lattice is given by the energy of the added vortex plus the vortex-vortex interactions. Within London theory, the Gibbs free energy (or chemical potential) to add the N th vortex to avortex array is given by,

δg = (ε1 − φ0H) +∑j

B0φ0µ0

K0(|~r − ~rj |/λ) (22)

Using Hc1 = ε1/φ0, the relation between the applied field and the lattice spacing is found by setting δg = 0, whichyields,

H −Hc1 =B0

µ0N

∑ij

K0(|~ri − ~rj |/λ) (23)

This fixes the lattice spacing, a, of a flux array. This spacing is different for different lattice structures. Associatedwith the lattice spacing there is a total number of vortices inside the superconductor. This total number of vortices,at fixed a, is larger for the triangular lattice, Ntr than it is for the square lattice, Ntr > Nsq. In fact the highestpacking possible is for the triangular lattice case.

In setting up a vortex lattice, the gain in Gibb’s free energy is given by,

δG = N(ε1 − φ0H) +∑ij

B0φ02µ0

K0(|~ri − ~rj |/λ) (24)

Notice that there is a factor of two in the second term(compared to Eq. (33)), which is the energy cost of vortex-vortexrepulsion, due to the fact that the energy cost in setting up the flux lattice is small for the first vortices added to thelattice. Using Eq. (34), this can be rewritten as,

φ0δG

N= (Hc1 −H) +

1

2(H −Hc1); so that

δG

N=

1

2φ0(Hc1 −H) (25)

5

This is the energy gain per vortex in setting up the lowest energy vortex lattice. Notice that if we consider onlynearest neighbor interactions, the energy per vortex is the same no matter what the vortex lattice structure. Howeverthe number of vortices which can be added to the superconductor does depend on the structure of the final vortexlattice. Since the number of vortices in the case of a square lattice is smaller than the number added for the triangularlattice, the total energy is lowest for the case of a triangular lattice. Notice that this result is general, the triangularlattice is the true ground state because it has the highest possible vortex density.

—————–

Assigned Problem 6. Go through the calculations to show that the kinetic energy and angular momentum perunit length of a quantum of circulation in a neutral superfluid are given by equations (40) and (41) of the notes.

Solution We have,

εkin =

∫ b

ξ

1

2dm(r)vs(r)

2 =

∫ b

ξ

1

2ρs(2πrdr)

(κ0)2

(2πr)2=κ20ρs4π

ln(b/ξ) (26)

and

lkin =

∫ b

ξ

r2 dm(r)vs(r)

r=

∫ b

ξ

ρs(2πr3dr)

κ02πr

=κ0ρs

2(b2 − ξ2) (27)

—————–

Assigned Problem 7. Considering only the kinetic energy of the supercurrent and using arguments like thoseof Problem 6, show that the supercurrents circulating around a flux quantum in a superconductor have energyapproximately given by Eq. (55) of the notes.

Solution Using the same procedure as for the neutral superfluid, with vs = mjs/(qρs) with current given by,

js(r) = j0λ

r; j0 =

φ02πµ0λ3

(28)

we find,

εkin =

∫ λ

ξ

1

2dm(r)vs(r)

2 =

∫ b

ξ

1

2ρs(2πrdr)(

mφ02πqρsµ0λ3

)2λ2

r2=

φ204πµ0λ2

ln(λ/ξ) (29)

where we used,

λ = (m2

ρsq2µ0)1/2 (30)

—————–

Assigned Problem 8. Derive the drift diffusion equation

~Jn = enµn ~E + eDn∂n

∂x; ~Jp = epµp ~E − eDp

∂p

∂x(31)

from the Boltzmann equation using the relaxation time approximation.

SolutionSame as the solution for ~Jn with a change in sign in the force and in the definition of the current.

—————–

6

Assigned Problem 9. The collision integral (Eq. (113) of the notes) is zero when,

f(~p1)f(~p2) = f(~p3)f(~p4). (32)

Show that the Maxwell Boltzmann distribution

f(t, ~r,~v) = Ce−α(~v−~V )2 (33)

satisfies this equation, for the case of elastic scattering.SolutionSubsituting into the equation, cancelling C factors and taking a logarithm of both sizes, then cancelling α factors

gives,

(~v1 − ~V )2 + (~v1 − ~V )2 = (~v3 − ~V )2 + (~v4 − ~V )2 (34)

Expanding and rearranging gives,

|~v1|2 + |~v2|2 − (|~v3|2 + |~v4|2) = 2[(~v1 + ~v2)− (~v3 + ~v4)] · ~V (35)

The left hand side is zero for elastic scattering, while the right hand side is zero by momentum conservation, hencethe Maxwell-Boltzmann form leads to zero for the collision integral in this case.

—————–