Homework 8 SolutionsProblem 1: Kittel 10-4 (a) The partition function of a single oscillator

that can move in three dimensions is given by:

Zs =∑

nx,ny,nz

e(ε−(nx+ny+nz)h̄ω)/τ) = eετ (∑

n

e−nh̄ω/τ )3 =eε/τ

(1 − e−h̄ω/τ )3

where −ε is the ground state energy of an atom in the solid lattice. Here, thezero of the energy is taken to be the energy of a free atom at rest (which isthe ground state energy of an atom in the gas). We may estimate the chemicalpotential of the atoms in the solid:

µs = Gs = Fs + pvs ≈ Fs

⇒ λs = eµs/τ ≈ eFs/τ = e− log Zs =1

Zs= e−ε/τ (1 − e−h̄ω/τ )3

Gs, Fs, and vs are the Gibbs energy, Helmholtz energy, and volume per particlefor the atoms in the solid. p is the vapor pressure of the gas-solid system. Wehave neglected the term pvs, which is presumably very small for a solid. Weknow the chemical potential for the gas phase, which we treat as an ideal gas:

µg = τ log(n

nQ)

⇒ λg = eµg/τ =n

nQ=

p

τnQ=

p

τ

(2πh̄2

Mτ

)3/2

Because the system is in diffusive equilibrium, we have that λg = λs. From this,we may obtain the vapor pressure:

p

τ

(2πh̄2

Mτ

)3/2

= e−ε/τ (1 − e−h̄ω/τ )3

p =(M

2π

)3/2 τ5/2

h̄3 (1 − e−h̄ω/τ )3e−ε/τ

≈(M

2π

)3/2 τ5/2

h̄3

( h̄3ω3

τ3

)

e−ε/τ

=(M

2π

)3/2 ω3

τ1/2e−ε/τ

where we have taken a high temperature limit (τ >> h̄ω)(b) We start with the Clausius-Clapeyron equation (K&K equation 10-15):

dp

dτ=

L

τ(vg − vs)≈

L

τvg=

Lp

τ2

where the last step involved the ideal gas law. Next, we differentiate our ex-pression for the vapor pressure:

dp

dτ=

(M

2π

)3/2

ω3e−ε/τ( −1

2τ3/2+

ε

τ5/2

)

1

=(M

2π

)3/2 ω3

τ1/2e−ε/τ 1

τ2

(

ε −τ

2

)

=p

τ2

(

ε −τ

2

)

Problem 2 – Kittel 10-5 (a) The total free energy of the system is:

F = Fs + Fg = Us + Ug + p(Vs + Vg) ≈ Us + (Ug + pV ) = Us + Fg

= −Nsεo + Ngτ [log(Ng/V nQ) − 1]

In this derivation, we have assumed Vg ≈ V and Vs ≈ 0.(b) If the total particle number N = Ns + Ng is constant, then we may

write:

F (Ng) = (Ng − N)εo + Ngτ [log(Ng/V nQ) − 1]

Minimizing this with respect to Ng gives the condition:

dF

dNg= 0 = εo + τ [log(Ng/V nQ) − 1] + τ

⇒ Ng = nQV exp(−εo/τ)

(c) The equilibrium vapor pressure follows from the ideal gas law:

p =Ngτ

V= nQτ exp(−εo/τ)

Problem 3 - Kittel 10-6 (a) On Kittel page 256, we have an expressionfor the difference in free energy between the normal and superconducting phasesof a material:

[FN (τ) − FS(τ)]/V = B2c (τ)/2µo

Using the relation: σ = −(∂F/∂T )V , we see immediately that:

[σS(τ) − σN (τ)]/V =∂

∂τ

( 1

2µoB2

c (τ))

=Bc

µo

dBc

dτ

As τ → 0, both entropies go to zero by the third law. This implies that at τ = 0,(dBc/dτ) = 0. We are also given that Bc decreases with increasing temperature.The figure shows a curve consistent with these facts (not the only possibility).

(b) (1) Since Bc(τc) = 0, we have that FN (τc) = FS(τc). Because theentropies are also equal, we also know that F ′

N (τc) = F ′

S(τc). Therefore, thecurves are tangent at τ = τc. (2) By defintion,F = U − τσ. Because atτ = τc, FN = FS and σN = σS ,which together imply that UN = US. (3) Thelatent heat per volume of a transition between states N and S is defined asL = τ(σN − σS) = Bc

µ0

dBc

dτ . At τ = τc, Bc = 0 so therefore L = 0.

(c)

∆C = CS − CN = τ∂(σS/V )

∂τ− τ

∂(σN/V )

∂τ=

τ

2µ0

d2(B2c )

dτ2

2

Bc

T

Figure 1: Cartoon of Bc vs τ .

For τ << τc, we are told that ∆C is dominated by CN . Also, we are told thatCN = γτ + O(τ2) and C2 = O(τ2). Therefore,

∆C = −γτ + O(τ2) =τ

µ0

d

dτ

[

BcdBc

dτ

]

=τ

µ0

[(dBc

dτ

)2

+ Bcd2Bc

dτ2

]

−γ = O(τ) +1

µ0

[(dBc

dτ

)2

+ Bcd2Bc

dτ2

]

Since the above holds for any low temperature, we know that in particular, itshould hold for τ = 0. Using the fact that (dBc/dτ) = 0 at τ = 0, we have:

γ = −1

µ0Bc

(d2Bc

dτ2

)

τ=0

Problem 4 - Kittel 10-8 (a) We seek the free energy of a gas of phonons inthe Debye approximation. A fast way to obtain this is to note that for a photongas, the free energy density is Fphoton/V = −π2τ4/(45h̄3c3). The Debye theoryfollows the same derivation where the sound speed v is used instead of c and thenumber of polarizations is 3 instead of 2. Following the photon derivation (seeHomework 3, problem 5), we see that we need to multiply the photon expressionby 3/2:

Fphonon/V =3

2Fphoton/V = −

π2τ4

30h̄3v3

(b) Whether the α or β configuration is favored depends on which state hasthe lower free energy. When τ = τc, the two free energies are equal. The freeenergy of state α (per volume) is the sum of the free energy (per volume) of theground state crystal, Uα(τ = 0), plus the free energy density of the phonons,which are excited when the crystal experiences a finite temperature. Likewise

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for β.

Fα = Fβ ⇒ Uα(0) −π2τ4

c

30h̄3v3α

= Uβ(0) −π2τ4

c

30h̄3v3β

τ4c = (30h̄3/π2)[Uβ(0) − Uα(0)]/(v−3

β − v−3α )

(c) The latent heat (per volume) is:

L = τc(σα(τc) − σβ(τc)) = τc

(dFβ

dτ−

dFα

dτ

)

τ=τc

= τc

(

4π2τ3

c

30h̄3v3β

− 4π2τ3

c

30h̄3v3α

)

= 4( π2τ4

c

30h̄3v3β

−π2τ4

c

30h̄3v3α

)

= 4[Uβ(0) − Uα(0)]

Problem 5The equation for the isotherm is given on page 23-2 of the notes:

p(V ) =8τ

3V − 1−

3

V 2

where p is given in units of pc, τ in units of τc, and V in units of Vc. Forthe case of τ = 0.9, the isotherm is shown in the accompanying figure. Alsodrawn are the lines p = pmax and p = pmin which define the pressure range overwhich the volume is triple-valued. We find the pressure range using the “fsolve”function of Maple. The numbers turn out to be pmin ≈ 0.420 and pmax ≈ 0.724.The corresponding volumes are vmin,l = 0.719, vmin,g = 4.61, vmax,l = 0.591,and vmax,g = 1.53. In order to determine the pressure where we get liquid-gascoexistence, we use the procedure outlined in the hint.

∫

V dp =

∫ V2

V1

Vdp

dVdV

= −2.4(ln(3V2 − 1

3V1 − 1

)

+1

3V1 − 1−

1

3V2 − 1) + 6(

1

V1−

1

V2)

= f(V1, V2)

Here V1 and V2 are the liquid and gas volumes respectively corresponding toa particular value of the pressure. Coexistence occurs at the pressure whichcauses f(V1, V2) = 0, for reasons discussed more fully in the lecture notes. Bynumerical iteration, we find this happens for a pressure p ≈ 0.65.

Problem 6 On page 302 of Kittel, the model of ferromagnetism is discussed.The energy per unit volume of this system is given by U/V = −λM2/2 =−nτcm

2/2, where the variable are defined in the problem set. We also knowthat m = tanh(m/t) where t = τ/τc. Armed with this information, we maycalculate the heat capacity (per volume). First, we need to find dm/dτ :

dm

dt= sech2(m/t)

(

−m

t2+

1

t

dm

dt

)

⇒dm

dτ=

1

τc

msech2(m/t)

tsech2(m/t) − t2

4

0.5

1

1.5

2

1 2 3 4 5 6

V

Figure 2: Plot of P/Pc vs V/Vc where p = pmax and p = pmin are also shown.

Then:

C =d(U/V )

dτ= nτcm

dm

dτ=

nm2sech2(m/t)

t2 − tsech2(m/t)

The following figure is a plot of C vs t for t < 1 as requested. For t > 1, m = 0is the only solution to the self-consistent equation m = tanh(m/t). This impliesthat C = 0 above the critical temperature. The jump discontinuity is finite.

Problem 7 (a) Suppose a system can be in a number of states and theprobability of it being in state i is given by pi. We want to calculate the averagevalue of some property A of the system, < A >=

∑

i Aipi. Formally, thisinvolves summing an infinite number of terms. Monte-Carlo simulation is a wayto get an approximate answer by summing a finite number of terms. The basicidea is to randomly choose some states {i} and calculate the average of the {Ai}.The Metropolis algorithm is a way to ensure that, if we take enough states, thenthe probability that a given state i will appear in our random sequence is pi.Therefore, our Monte-Carlo average will well-approximate the actual average.

Here is the idea of Metropolis: Suppose our system begins in some state i.We randomly choose another state j. If pi < pj , then we accept the choice andour system is in state j. If pi > pj , then we accept the choice with probabilitypj/pi. Depending on the outcome of the random number generator, our systemwill stay in state i or move to state j. The important point is that either waywe update our running average for A.

(b) Suppose we have two states 0 and 1 which occur with probabilities pand 1 − p. Then if our random number generator gives a number x < p, then

5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

0 0.2 0.4 0.6 0.8 1 1.2 1.4

’h8_p6.txt’

Figure 3: Plot of C vs t = τ/τc.

we say our system is in state 0. Otherwise, it is in state 1.(c) If our system is in state 0 after n steps, then after the (n + 1)st step,

we know that it is either in state 0 (with probability t00) or in state 1 (withprobability t10). Thus, t00 + t10 = 1. Likewise, t01 + t11 = 1.

(d) Suppose we impose the condition that if our inputs are canonical , thenthe outputs must be canonical. This means:

( 1e−E/τ

)

=(

t00 1 − t111 − t00 t11

)( 1e−E/τ

)

which means that:

1 = t00 + (1 − t11)e−E/τ

⇒ t00 = 1 − e−E/τ (1 − t11)

(e) The transition matrix for Metropolis is given by:

TMetro =(

1 − e−E/τ 1e−E/τ 0

)

To see this is Metropolis, we note that if we start in state 1, then it musttransition to the lower state 0, which implies that t01 = 1 and t00 = 0. On theother hand, if the system is in state 0, then it will transition to state 1 with atransition probability which is the ratio of the probabilities p1/p0 = e−E/τ .

6

(f) If we want canonical outputs regardless of the inputs, then we need thefollowing matrix. You may explicitly verify that it behaves as required:

TCanon =1

1 + e−E/τ

( 1 1e−E/τ e−E/τ

)

7