PHYSICS 140A : STATISTICAL PHYSICS
HW ASSIGNMENT #6 SOLUTIONS
(1) Consider a monatomic ideal gas, represented within the grand canonical ensemble.
Show that the probability of finding the system to have N atoms is given by the Poisson
distribution,
1 −hN i
e
hN iN .
PN =
N!
P N
Solution : The grand partition function is Ξ =
N z ZN , where z = exp(βµ) is the
fugacity.
The
ordinary
canonical
partition
function
is ZN = (V /λ3T )N /N !, where λT =
p
2π~2 /mkB T is the thermal wavelength. Thus, Ξ = exp(zV /λ3T ), and the average number
of particles is
∂
zV
hN i = z
ln Ξ = 3 .
∂z
λT
The probability that there are N particles in the system is clearly
PN =
z N ZN
1 −hN i
=
e
hN iN .
Ξ
N!
(2) Derive the grand canonical distribution when there are several types of particles present.
Solution : Let Z(N1 , N2 , . . . , NK ) be the OCE partition function for K species of particles
when there are Nj particles of species j. Then introduce K chemical potentials µj (or K
fugacities zj = exp(βµj )) and write
X
X
Ξ(T, V, µ1 , . . . , µK ) =
···
eN1 βµ1 · · · eNK βµK Z(N1 , . . . , NK ) .
N1
NK
(3) An ideal paramagnet is described by the model in §3.11 of the notes, i.e.
Ĥ = −µ0 H
N
X
σj ,
j=1
where each σj = ±1. Suppose the system starts off at a temperature T = 10 mK and a field
H = 20 T. The field is then lowered adiabatically to H = 1 T. What is the final temperature
of the system?
Solution : This problem may be solved by dimensional analysis. Clearly the entropy S
is a function of T and H, and the only dimensionally correct possibility is S(T, H, N ) =
N kB f (µ0 H/kB T ). We conclude that during an adiabatic process that the ration H/T is
constant. Thus,
T
H
Tf = Ti · f = i = 500 nK .
Hi
20
1
Explicitly, we have
X
Z=
µo Hσ/kB T
e
σ=±1
!N
N
= 2 cosh
N
µ0 H
kB T
.
(1)
The free energy is then
µ H
F = −kB T ln Z = −N kB T ln 2 − N kB T ln cosh 0
kB T
.
The entropy is
∂F
S=−
∂T
H,N
= N kB
(
µ H
ln 2 + ln cosh 0
kB T
+
µ0 H
kB T
)
µ0 H
.
sinh
kB T
This is of the required form, S(T, H, N ) = N kB f (µ0 H/kB T ), with f (x) = ln(2 cosh x) +
x sinh x.
(4) Consider a nonrelativistic ideal gas. From dimensional analysis, we conclude that
|p|k = Ck (mkB T )k/2 .
Find the constants Ck . Use the OCE.
Solution : Canceling factors common to both numerator and denominator, we have
R∞ 2+k −p2 /2mk T
B
dp p
e
0
|p|k =
R∞
2
dp p2 e−p /2mkB T
.
0
Thus, after writing p = x
√
mkB T , we have
Ck =
R∞
2
dx x2+k e−x /2
0
R∞
dx x2 e−x2 /2
.
0
We now change variables, writing x =
Ck =
√
√
2y. Then dx = dy/ 2y and
R∞
1
k
2k/2 dy y 2 + 2 e−y
0
R∞
dx y
1
2
= 2k/2
e−y
0
Note C0 = 1, as required by normalization, C1 =
result hp2 /2mi =
3
2 kB T .
2
q
8
π,
Γ 3+k
2
.
Γ 32
and C2 = 3, which yields the familiar
(5) Show that
CV = −kB β 2
∂2
βF .
2
∂β
Solution : We have
CV = −T
∂ 2F
.
∂T 2
We now write β = 1/kB T , hence
∂
∂
= −kB β 2
.
∂T
∂β
(2)
Then
2
1
2 ∂
F
CV = −
−kB β
kB β
∂β
∂F
∂ 2F
∂F
∂
β2
= −kB β 3 2 + 2β 2
= kBβ
∂β
∂β
∂β
∂β
= −kB β 2
2
∂2
2 ∂ ln Z
βF
=
k
β
.
B
∂β 2
∂β 2
(6) Consider a three state system with energy levels at ε = 0, ε = ∆, and ε = W , with
0 ≤ ∆ ≤ W . Compute the free energy for such a system, f (T ). Derive an expression for
the heat capacity c(T ). You may find the results from problem (5) useful. Plot the specific
heat c(T ) versus kB T /∆ for W = ∆, W = 2∆, and W = 6∆.
Solution : The partition function is
ζ = Tr e−β Ĥ = 1 + e−β∆ + e−βW .
Thus,
f (T ) = −kB T ln ζ = −kB T ln 1 + e−∆/kB T + e−W/kB T .
The heat capacity is
∂2
ln ζ
∂β 2
∆2 e−β∆ + W 2 e−βW + (W − ∆)2 e−β(∆+W )
.
= kB β 2
2
1 + e−β∆ + e−βW
c(T ) = kB β 2
3
(3)
Figure 1: Specific heat c(T ) for W = ∆, W = 2∆, and W = 6∆.
4