PHYS 449 - Statistical Mechanics I

Assignment 7 Solutions

1. (a) Return now to problem #2 in Assignment 5, where only three energy levels of a particle in a one-dimensionalbox are accessible to a particle: ε = 0, 1, 4ε1, where ε1 = h2π2/2mL2. The partition function is

Z = 1 + e−βε1 + e−4βε1 .

The probabilities of being in the energy states are therefore p0 = 1/Z, p1 = e−βε1/Z, and p4 = e−4βε1/Z. In themicrocanonical ensemble, we found that the maximum entropy condition required the following relationshipamong the probabilities:

p2p30

p41

= 1.

Clearly, our definition of the probabilities is consistent with this:

p4p30

p41

=e−4βε1

e−4βε1= 1.

We also automatically satisfy that the sum of probabilities is unity. The last equation is p1 + 4p4 = U/N ,which is also satisfied. So these probabilities must be identical to those found in the microcanonical ensemble.This means that all quantities involving those probabilities must be the same, and the numerical results in theattached Mathematica notebook bear this out.

The mean energy per particle can be found directly using the definition of the mean:

U

N= p1ε1 + p44ε4 =

e−βε1 + 4e−4βε1

1 + e−βε1 + e−4βε1ε1.

It is useful to check that this expression is consistent with the formal one tied to the partition function:

U

N= − ∂

∂βln(Z) = −−ε1e

−βε1 − 4ε1e−4βε1

1 + e−βε1 + e−4βε1,

which is exactly the same. The entropy per particle is therefore

S

N= kB ln(Z) +

U

NT= kB ln(1 + e−βε1 + e−4βε1) + kB(βε1)

e−βε1 + 4e−4βε1

1 + e−βε1 + e−4βε1.

The remainder of this question is done on the attached Mathematica notebook.

(b) This is also done on the attached Mathematica notebook.

2. (a) Recall that for a particle in a box, the energy levels depend entirely on the volume. This means that changesin volume actually change the energy levels (rather than the distribution of particles in the energy levels); thisdoes work in the form of dW = −PdV . Likewise, in a magnetic system, the energy levels depend only on theexternal magnetic field B; changing the magnetic field therefore does work on the spin direction, and this willchange the magnetization. The work for a magnetic system is therefore dW = −MdB. Thus, volume andpressure for an ideal gas become magnetic field and magnetization, respectively. One simply needs to recopythe formulas in Chapter 5, replacing all instances of V with B and P with M .

Consider first the total energy. dU = Q+W = TdS −MdB. Clearly then U is a simultaneous function of thetwo parameters S and B, i.e. U = U(S,B). So we can write:

dU =

(∂U

∂S

)B

dS +

(∂U

∂B

)S

dB ≡ TdS −MdB,

PHYS 449 - Assignment 7 Solutions 2

which immediately yields the following two equations

T =

(∂U

∂S

)B

, M = −(∂U

∂B

)S

.

The second of these equations unfortunately isn’t very handy, because it is difficult to keep entropy fixed inpractise.

The fact that dU is an exact differential of the quantity U allows us to derive another useful relation. If wetook a second derivative of U , the result can’t depend on the order, i.e.

∂2U

∂B∂S≡ ∂2U

∂S∂B(∂

∂B

)S

(∂U

∂S

)B

≡(∂

∂S

)B

(∂U

∂B

)S

⇒(∂T

∂B

)S

= −(∂M

∂S

)B

.

For the Helmholtz free energy U−TS, instead consider the quantity TS, for which we have d(TS) = TdS+SdT .Then

dU = TdS −MdB = d(TS)− SdT −MdB ⇒ d(U − TS) ≡ dF = −SdT −MdB.

Now, F = F (T,B) so that

dF =

(∂F

∂T

)B

dT +

(∂F

∂B

)T

dB ≡ −SdT −MdB,

which immediately yields the following two equations

S = −(∂F

∂T

)B

, M = −(∂F

∂B

)T

.

These are very useful equations! Now either the magnetic field or the temperature have to be kept constant,which is much easier to do. And again, using the second-derivative trick above, one finds:(

∂S

∂B

)T

=

(∂M

∂T

)B

.

Thus, knowing how the magnetization varies with temperature (at constant external magnetic field) is enoughto tell you how the entropy varies with magnetic field (at constant temperature)! This is very handy.

(b) Next let’s see if these equations apply to the Pauli paramagnet, the ‘canonical’ example of a magneticsystem (pun intended). Let’s keep things simple, and assume that we have spin-1/2 particles. The two energylevels are ε and −ε, where ε = gµBB/2 ≡ αB. The partition function is then

Z = eβε + e−βε = 2 cosh(βε) = 2 cosh(βαB).

The mean energy is

U = −N ∂

∂βln(Z) = −NαB tanh(βαB),

and the entropy is

S = NkB ln(Z) +U

T= NkB ln [2 cosh(βαB)]− NαB

Ttanh(βαB)

= NkB ln [2 cosh(βαB)]− βαB tanh(βαB) .

The (Helmholtz) free energy is therefore

F = U − TS = −NkBT ln(Z) = −NkBT ln [2 cosh(βαB)] .

PHYS 449 - Assignment 7 Solutions 3

The easiest equations to check are those associated with the free energy, so let’s do those first:

S = −(∂F

∂T

)B

= NkB∂

∂T

T ln

[2 cosh

(αB

kBT

)]= NkB ln

[2 cosh

(αB

kBT

)]− NkBαB

kBTtanh

(αB

kBT

),

which is exactly the expression above. Now for the magnetization:

M = −(∂F

∂B

)T

= NkBT∂

∂B

ln

[2 cosh

(αB

kBT

)]=NkBα

kBtanh

(αB

kBT

),

which is just the expression in the notes for the magnetization. Last let’s check the Maxwell relation: UsingMathematica (see the attached notebook) I obtain(

∂S

∂B

)T

= −NBα2

kBT 2sech2

(αB

kBT

)=

(∂M

∂T

)B

.

So it all hangs together!

In order to use the expressions involving the mean energy, we would need to figure out how to express themean energy as a function of S and B only. The problem is that the expression for the entropy above isvery complicated! We would need to invert the equation for entropy to solve for β, and then plug it into theexpression for the energy. I don’t think that there is any way to do this! So I will give up on the idea.

3. The variance of the energy about the mean is (∆U)2 = U2 − U2. Consider the mean energy per particle; in

the canonical ensemble, these are:

u =∑i

piεi =∑i

εie−βε1

Z= − 1

Z

∂Z

∂β= − ∂

∂βln(Z),

where Z ≡∑i e

−βεi . Let’s work out u2:

u2 =∑i

piε2i =

∑i

ε2i e−βε1

Z.

We know that∂2Z

∂β2=∑i

ε2i e−βεi

so that

u2 =1

Z

∂2Z

∂β2.

The variance is therefore

(∆u)2 =1

Z

∂2Z

∂β2−(

1

Z

∂Z

∂β

)2

=1

Z

∂2Z

∂β2−(

1

Z

∂Z

∂β

)2

.

Notice that this expression is consistent with:

(∆u)2 =1

Z

∂2Z

∂β2−(

1

Z

∂Z

∂β

)2

=∂

∂β

(1

Z

∂Z

∂β

)=

∂2

∂β2ln(Z).

But ∂ ln(Z)/∂β = −u = −U/N , which means that

(∆u)2 = − 1

N

∂U

∂β.

(b) For the particle in the 3D box, the partition function was found to be (at high temperatures) Z = V/λ3D =

V (mkBT/2πh2)3/2 = α/β3/2, where α includes all of the constants. Using Mathematica, I obtain gives (∆u)2 =

PHYS 449 - Assignment 7 Solutions 4

(3/2)(kBT )2. This is consistent with the famous ‘fluctuation-dissipation theorem’ N(∆u)2 = kBT2CV . Thus,

the standard deviation ∆u ∼ N−1/2, just as we saw at the beginning of the term when talking about randomwalks.

For the 3D harmonic oscillator, the full partition functon is more complicated (though the high-temperatureversion is also simple):

Z =1

(1− e−βhω)3 .

The variance comes out to be

(∆u)2 =3

4

[hωcsch

(βhω

2

)]2

.

This isn’t very informative. At high temperatures, the mean energy and the variance become u = u = 3kBTand 3(kBT )2, respectively. These results are again consistent with the fluctuation-dissipation theorem.

Assignment 7 Solutions(2) Here is the definition of the partition function, where Α = Ε1/kB:

Z = 1 + Exp@-Α TD + Exp@-4 Α TD

1 + ã-

4 Α

T + ã-

Α

T

Here is the definition of the mean energy per particle, noting that kB = Ε1/Α:

u = FullSimplify@Ε1 Α * T^2 * D@Log@ZD, TDD

J4 + ã

3 Α

T N Ε1

1 + ã

3 Α

T + ã

4 Α

T

Let’s express the total energy per particle in units of the energy scale Ε1, i.e. up = U/NΕ1:

up =

J4 + ã

3 Α

T N

1 + ã

3 Α

T + ã

4 Α

T

4 + ã

3 Α

T

1 + ã

3 Α

T + ã

4 Α

T

s = FullSimplify@kB * Log@ZD + u TD

J4 + ã

3 Α

T N Ε1

J1 + ã

3 Α

T + ã

4 Α

T N T

+ kB LogB1 + ã-

4 Α

T + ã-

Α

T F

noting that Ε1 = kB*Α, we can rewrite this as sp, where sp = S/N/kB

sp =

J4 + ã

3 Α

T N Α

J1 + ã

3 Α

T + ã

4 Α

T N T

+ LogB1 + ã-

4 Α

T + ã-

Α

T F

J4 + ã

3 Α

T N Α

J1 + ã

3 Α

T + ã

4 Α

T N T

+ LogB1 + ã-

4 Α

T + ã-

Α

T F

Here are the probabilities:

p0 = 1 Z

p1 = Exp@-Α TD Z

p4 = Exp@-4 * Α TD Z

1

1 + ã-

4 Α

T + ã-

Α

T

ã-

Α

T

1 + ã-

4 Α

T + ã-

Α

T

ã-

4 Α

T

1 + ã-

4 Α

T + ã-

Α

T

Now plot the probabilities as a function of the total energy, as was done in Assignment 5:

Α = 1; Show@8ParametricPlot@8up, p0<, 8T, 0.01, 100<, PlotRange ® 880, 5 3<, 80, 1<<,

PlotStyle ® RedD, ParametricPlot@8up, p1<, 8T, 0.01, 100<,

PlotRange ® 880, 5 3<, 80, 1<<, PlotStyle ® BlueD, ParametricPlot@8up, p4<,

8T, 0.01, 100<, PlotRange ® 880, 5 3<, 80, 1<<, PlotStyle ® GreenD<D

0.5 1.0 1.5

0.0

0.2

0.4

0.6

0.8

1.0

Unsurprisingly, these curves are identical to those found in Assignment 5 for the range of validity. Now

for the energy as a function of temperature:

Plot@up, 8T, 0.01, 50<, PlotRange ® 880, 50<, 80, 5 3<<D

10 20 30 40 50

0.0

0.5

1.0

1.5

It is clear that at low temperature the mean energy goes to zero, while at high temperatures it asymp-

totes as 5/3, as expected. Now for the entropy:

2 ass7sol.nb

It is clear that at low temperature the mean energy goes to zero, while at high temperatures it asymp-

totes as 5/3, as expected. Now for the entropy:

Plot@sp, 8T, 0.01, 5<, PlotRange ® 880, 5<, 80, Log@3D<<D

1 2 3 4 5

0.0

0.2

0.4

0.6

0.8

1.0

The entropy also saturates (to Log[3]), but much more quickly than the energy.

(b) Now imagine that there are four energy levels. Let’s repeat the calculations above:

Clear@ΑD; Z = 1 + Exp@-Α TD + Exp@-4 Α TD + Exp@-9 Α TD

1 + ã-

9 Α

T + ã-

4 Α

T + ã-

Α

T

u = FullSimplify@Ε1 Α * T^2 * D@Log@ZD, TDD

J9 + 4 ã

5 Α

T + ã

8 Α

T N Ε1

1 + ã

5 Α

T + ã

8 Α

T + ã

9 Α

T

up =

J9 + 4 ã

5 Α

T + ã

8 Α

T N

1 + ã

5 Α

T + ã

8 Α

T + ã

9 Α

T

9 + 4 ã

5 Α

T + ã

8 Α

T

1 + ã

5 Α

T + ã

8 Α

T + ã

9 Α

T

s = FullSimplify@kB * Log@ZD + u TD

J9 + 4 ã

5 Α

T + ã

8 Α

T N Ε1

J1 + ã

5 Α

T + ã

8 Α

T + ã

9 Α

T N T

+ kB LogB1 + ã-

9 Α

T + ã-

4 Α

T + ã-

Α

T F

sp =

J9 + 4 ã

5 Α

T + ã

8 Α

T N Α

J1 + ã

5 Α

T + ã

8 Α

T + ã

9 Α

T N T

+ LogB1 + ã-

9 Α

T + ã-

4 Α

T + ã-

Α

T F

J9 + 4 ã

5 Α

T + ã

8 Α

T N Α

J1 + ã

5 Α

T + ã

8 Α

T + ã

9 Α

T N T

+ LogB1 + ã-

9 Α

T + ã-

4 Α

T + ã-

Α

T F

ass7sol.nb 3

p0 = 1 Z

p1 = Exp@-Α TD Z

p4 = Exp@-4 * Α TD Z

p9 = Exp@-9 Α TD Z

1

1 + ã-

9 Α

T + ã-

4 Α

T + ã-

Α

T

ã-

Α

T

1 + ã-

9 Α

T + ã-

4 Α

T + ã-

Α

T

ã-

4 Α

T

1 + ã-

9 Α

T + ã-

4 Α

T + ã-

Α

T

ã-

9 Α

T

1 + ã-

9 Α

T + ã-

4 Α

T + ã-

Α

T

The highest energy corresponds to all probabilies equal to 1/4, which gives

umax = 1 * 1 4 + 4 * 1 4 + 9 * 1 4

7

2

Α = 1; Show@8ParametricPlot@8up, p0<, 8T, 0.01, 100<,

PlotRange ® 880, 7 2<, 80, 1<<, PlotStyle ® RedD, ParametricPlot@8up, p1<,

8T, 0.01, 100<, PlotRange ® 880, 7 2<, 80, 1<<, PlotStyle ® BlueD,

ParametricPlot@8up, p4<, 8T, 0.01, 100<, PlotRange ® 880, 7 2<, 80, 1<<,

PlotStyle ® GreenD, ParametricPlot@8up, p9<, 8T, 0.01, 100<,

PlotRange ® 880, 7 2<, 80, 1<<, PlotStyle ® OrangeD<D

0.5 1.0 1.5 2.0 2.5 3.0 3.5

0.0

0.2

0.4

0.6

0.8

1.0

Not much to say about these, they look much like the case with three levels! They all meet at the same

place at the right energy.

4 ass7sol.nb

Plot@up, 8T, 0.01, 50<, PlotRange ® 880, 50<, 80, 7 2<<D

10 20 30 40 50

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

This seems to approach the upper limit only later, at higher temperatures.

Plot@sp, 8T, 0.01, 10<, PlotRange ® 880, 10<, 80, Log@4D<<D

2 4 6 8 10

0.0

0.2

0.4

0.6

0.8

1.0

1.2

The entropy saturates (to Log[4]), but now only does so at higher temperatures. The inference is that as

the number of accessible states increases, one will need much higher temperatures to guarantee that

all of the probabilities will become close to equal.

(3) First let’s define the Helmholtz free energy (well, its negative actually):

Clear@ΑD; func = Num * k * T * Log@2 * Cosh@Α * B k TDD

k Num T LogB2 CoshBB Α

k T

FF

The entropy is therefore:

S = FullSimplify@D@func, TDD

k Num LogB2 CoshBB Α

k T

FF -

B Num Α TanhA B Α

k TE

T

ass7sol.nb 5

FullSimplify@D@S, BDD

-

B Num Α2SechA B Α

k TE

2

k T2

Likewise, the magnetization is:

M = FullSimplify@D@func, BDD

Num Α TanhBB Α

k T

F

FullSimplify@D@M, TDD

-

B Num Α2SechA B Α

k TE

2

k T2

Notice that the derivative of the entropy with respect to the magnetic field indeed coincides with the

derivative of the magnetization with respect to the temperature, as predicted by the Maxwell relation.

(4) Let’s plug in the variance formula for the 3D box first:

Z = Α Β^H3 2L;

mean = -D@Z, ΒD Z

variance = D@Z, 8Β, 2<D Z - HD@Z, ΒD ZL^2

3

2 Β

3

2 Β2

Next let’s do the 3D harmonic oscillator:

Z = 1 H1 - Exp@-Β * hbar * ΩDL^3

mean = FullSimplify@-D@Z, ΒD ZDvariance = FullSimplify@D@Z, 8Β, 2<D Z - HD@Z, ΒD ZL^2D

1

I1 - ã-hbar Β ΩM3

3 hbar Ω

-1 + ãhbar Β Ω

3

4

hbar2

Ω2

CschBhbar Β Ω

2

F2

At high temperatures one obtains:

6 ass7sol.nb

Series@mean, 8Β, 0, 1<DSeries@variance, 8Β, 0, 1<D

3

Β

-3 Hhbar ΩL

2

+1

4

hbar2

Ω2

Β + O@ΒD2

3

Β2

-hbar

2 Ω2

4

+ O@ΒD2

ass7sol.nb 7