PHYSICS 210A : STATISTICAL PHYSICS
HW ASSIGNMENT #2 SOLUTIONS
(1) A system consists of N ’molecules’. Each molecule consists of four ’spins’: σ, µ1 , µ2 ,
and µ3 , where each spin polarization can takes values ±1. The molecular Hamiltonian is
ĥ = Jσ(µ1 + µ2 + µ3 ) − H 3σ − µ1 − µ2 − µ3 .
(a) Enumerate all the molecular energy states along with their degeneracies.
(b) Find the molecular partition function ζ(T, H).
(c) Compute the magnetic susceptibility χ(T, H = 0).
Solution :
(a) The states and their degeneracies are listed in Tab. 1 below. Note that there ĥ exhibits
permutation symmetry among the (µ1 , µ2 , µ3 ) states.
σ
+
+
+
+
−
−
−
−
(µ1 , µ2 , µ3 )
(+, +, +)
(+, +, −)
(+, −, −)
(−, −, −)
(+, +, +)
(+, +, −)
(+, −, −)
(−, −, −)
g
1
3
3
1
1
3
3
1
E
+3J
+J − 2H
−J − 4H
−3J − 6H
−3J + 6H
−J + 4H
+J + 2H
+3J
Table 1: States and their degeneracies g.
(b) Accordingly,
ζ(T, H) = 2 e−3βJ + 2 e+3βJ cosh(6βH) + 6 e−βJ cosh(2βH) + 6 e+βJ cosh(4βH) .
(c) The molecular magnetization is
1 ∂ ln ζ
∂f
=
∂H
β ∂H
3βJ
6e
sinh(6βH) + 6 e−βJ sinh(2βH) + 12 eβJ sinh(4βH)
= −3βJ
e
+ e3βJ cosh(6βH) + 3 e−βJ cosh(2βH) + 3 eβJ cosh(4βH)
18 e3βJ + 6 e−βJ + 24 eβJ
· βH + O(H 3 ) .
=
cosh(3βJ) + 3 cosh(βJ)
m=−
1
Thus, the zero-field molecular susceptibility is
χ(T, H = 0) =
∂m
18 e3J/kB T + 6 e−J/kB T + 24 eJ/kB T
1
=
·
.
∂H
cosh(3J/kB T ) + 3 cosh(J/kB T )
kB T
Note that for J = 0 we obtain χ(T, H = 0) = 12/kB T . For a single spin with magnetic
moment p, i.e. ĥ = −pHσ, the susceptibility is p2 /kB T . Thus for our system, when J = 0
we have one spin (σ) with p = 3 and three (µ1,2,3 ) with p = 1, hence the total susceptibility
is χ = (32 + 12 + 12 + 12 )/kB T = 12/kB T .
(2) In §4.9.4 of the lecture notes, we considered a simple model for the elasticity of wool in
which each of N monomers was in one of two states A or B, with energies εA,B and lengths
ℓA,B . Consider now the case where the A state is doubly degenerate due to a magnetic
degree of freedom which does not affect the energy or the length of the A± monomers.
(a) Generalize the results from Eqs. 4.221 and 4.222 of the lecture notes and show that
you can write the Hamiltonian Ĥ and chain length L̂ in terms of spin variables Sj ∈
{−1, 0, 1}, where Sj = ±1 if monomer j is in state A± , and Sj = 0 if it is in state B.
Construct the appropriate generalization of Eqn. 4.223.
(b) Find the equilibrium length L(T, τ, N ) as a function of the temperature, tension, and
number of monomers.
(c) Now suppose an external magnetic field is present, so the energies of the A± states
are split, with εA± = εA ∓ µ0 H. Find an expression for L(T, τ, H, N ).
Solution :
(a) Take
Ĥ =
N h
X
εB + (εA − εB ) Sj2
j=1
i
,
L̂ =
N h
X
ℓB + (ℓA − ℓB ) Sj2
j=1
resulting in
K̂ = Ĥ − τ L̂ = N (εB − τ ℓB ) + ∆
N
X
Sj2 ,
j=1
where
∆ = (εA − εB ) − τ (ℓA − ℓB ) .
(b) The partition function is
Y (T, τ, N ) = e−G/kB T = Tr e−K̂/kB T
= e−N (εB −τ ℓB )/kB T 1 + 2 e−∆/kB T
2
N
.
i
,
Thus, the Gibbs free energy is
G(T, τ, N ) = −kB T ln Y (T, τ, N ) = N (εB − τ ℓB ) − N kB T ln 1 + 2 e−∆/kB T .
The equilibrium length is
L=−
2 e−∆/kB T
∂G
= N ℓB + N (ℓA − ℓB ) ·
.
∂τ
1 + 2 e−∆/kB T
Note that L = N ℓA for ∆ → −∞ and L = N ℓB for ∆ → +∞.
(c) Accounting for the splitting of the two A states,
L = N ℓB + N (ℓA − ℓB ) ·
2 e−∆/kB T cosh(µ0 H/kB T )
.
1 + 2 e−∆/kB T cosh(µ0 H/kB T )
(3) Consider a system of identical but distinguishable particles, each of which has a nondegenerate ground state with ε0 = 0, and a g−fold degenerate excited state with energy
ε > 0. Study carefully problems #1 and #2 in the example problems for chapter 4 of the
lecture notes, where this system is treated in the microcanonical and ordinary canonical
ensembles. Here you are invited to work out the results for the grand canonical ensemble.
(a) Find the grand partition function Ξ(T, z) and the grand potential Ω(T, z). Express
your answers in terms of the temperature T and the fugacity z = eµ/kB T .
(b) Find the entropy S(T, z).
(c) Find the number of particles, N (T, z).
(d) Show how, in the thermodynamic limit, the entropy agrees with the results from the
microcanonical and ordinary canonical ensembles.
Solution :
(a) There ordinary canonical partition function is clearly
Z(T, N ) = 1 + ge−ε/kB T
hence the grand partition function is
Ξ=
∞
X
z N Z(T, N ) =
N =0
N
,
1
,
1 − z(1 + g e−ε/kB T )
where z = exp(µ/kB T ) is the fugacity. The grand potential is
Ω(T, z) = −kB T ln Ξ = kB T ln 1 − z(1 + g e−ε/kB T ) .
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(b) The entropy is S = − ∂Ω
∂T µ , so we must take care to allow z = exp(µ/kB T ) to vary. The
result is
µ
z(1 + g e−ε/kB T )
zg e−ε/kB T
ε
·
S(T, µ) = −kB ln 1−z(1+g e−ε/kB T ) − ·
+
.
T 1 − z(1 + g e−ε/kB T ) T 1 − z(1 + g e−ε/kB T )
(c) The particle number is
N =−
∂Ω
∂µ
T
z
=−
kB T
Thus,
z=
∂Ω
∂z
T
=
z(1 + g e−ε/kB T )
.
1 − z(1 + g e−ε/kB T )
1
1
·
.
−1
1+N
1 + g e−ε/kB T
(d) Expressing the entropy S(T, z) in terms of T and N , we find
N ε g e−ε/kB T
−1
S(T, N ) = N kB ln 1 + g e−ε/kB T +
+
k
ln(N
+
1)
+
N
k
ln
1
+
N
.
B
B
T 1 + g e−ε/kB T
The first two terms are extensive, i.e. of order N 1 . They agree with the results in problem
(2c) of Example HW #4. The penultimate term is of order ln N and the last term is of order
N 0 , hence they are subleading and negligible in the thermodynamic limit.
(4) The grand partition function for a system is given by the expression
Ξ = (1 + z)V /v0 1 + z αV /v0 ,
where α > 0. In this problem, you are to work in the thermodynamic limit. You will also
need to be careful to distinguish the cases |z| < 1 and |z| > 1.
(a) Find an expression for the pressure p(T, z).
(b) Find an expression for the number density n(T, z).
(c) Plot v(p, T ) as a function of p for different temperatures and show there is a first
order phase transition, i.e. a discontinuity in v(p), which occurs for |z| = 1. What is
the change in volume at the transition? .
Solution :
(a) The grand potential is
Ω(T, z) = −kB T ln Ξ = −
kB T V
ln(1 + z) − kB T ln 1 + z αV /v0 .
v0
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Now take the thermodynamic limit V /v0 → ∞. One then has
(
0
if
kB T V
Ω(T, z) = −
ln(1 + z) − αkB T V
v0
ln z if
v
|z| < 1
|z| > 1 .
0
From this we compute the pressure,
αk T z αV /v0 ln z
kB T
ln(1 + z) + B ·
v0
v0
1 + z αV /v0
T,µ
(
0
if |z| < 1
kB T
ln(1 + z) + αkB T
=
v0
ln z if |z| > 1 .
v
p=−
∂Ω
∂V
=
0
(b) For the density, we have
∂Ω
=
∂z T,V
(
0
1
z
=
·
+
v0 1 + z
α/v0
n=−
z
V kB T
z
z αV /v0
α
1
·
·
+
v0 1 + z v0 1 + z αV /v0
if
if
|z| < 1
|z| > 1 .
(c) We eliminate z from the above equations, and we write v = 1/n as the volume per
particle. The fugacity z(v) satisfies

v0

if v > 2v0

 v−v0







2v0

if 1+2α

< v < 2v0
1
z(v) =


2v0
v0
v0 −αv

< v < 1+2α
if 1+α


(1+α)v−v0







v0
∞
if v < 1+α
We then have
 v

ln

v−v0










ln 2
if
if
pv0
=

kB T

v0 −αv
v


ln (1+α)v−v
+ α ln (1+α)v−v
if


0
0







∞
if
v > 2v0
2v0
1+2α
< v < 2v0
v0
1+α
<v<
v<
v0
1+α
2v0
1+2α
Plots of z(v) and p(v) for three representative α values are shown in Fig. 1.
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Figure 1: z(v) and p(v) for α = 0.2, 1.0, and 3.0.
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