PHYSICS 140A : STATISTICAL PHYSICS
Examples from Nov. 25 problem session
(1) The single particle energy levels of a system are ε = 0 and ε = ∆. Find the OCE partition function Z(T, N ) for the case N = 2 under the following conditions: (i) distinguishable
particles, (ii) Maxwell-Boltzmann statistics, (iii) Bose-Einstein statistics, (iv) Fermi-Dirac
statistics.
Solution : The single particle partition function is ζ = Tr e−β Ĥ = 1 + e−β∆ . For two
distinguishable particles, we have
ZD (T, N ) = ζ N
ZD (T, N = 2) = 1 + 2 e−β∆ + e−2β∆ .
(1)
Note that we can identify by the coefficients that there is one state of energy 0, two states
of energy ∆ (corresponding to one particle being in the state ε = 0 and the other in ε = ∆),
and one state of energy ε = 2∆.
For Maxwell-Boltzmann statistics, we have
ZMB (T, N ) =
ZMB (T, N = 2) =
ζN
N!
1
2
+ e−β∆ + 12 e−2β∆ .
(2)
(3)
Note that the coefficients are no longer integers. Maxwell-Boltzmann statistics therefore
does not correspond to any sensible state counting. Rather, it makes sense only in a certain
limit – one in which there are a large number of possible states and the average occupancy
of each state is very small.
For quantum statistics, we must count occupation number eigenstates of the many-body
Hamiltonian.
If α is used to label
P the single particle basis, and nα is the occupancy of
α , then we must have N =
α nα . The individual occupancies nα range over the set
of nonnegative integers in the case of BE statistics, and over the set {0,1} in the
case of
FD statistics. Thus, with N = 2 bosons, and two single particle levels 0 and 1 , with
energies ε0 = 0 and ε1 = ∆, we can have (n0 , n1 ) = (2, 0) or (1, 1) or (0, 2). The resulting
partition function is
ZBE (T, N = 2) = 1 + e−β∆ + e−2β∆ .
(4)
For Fermi-Dirac statistics, we can have at most one particle in any single particle level,
hence the only possibility consistent with N = 2 is (n0 , n1 ) = (1, 1), and
ZFD (T, N = 2) = e−β∆ .
(5)
(2) Consider two particles each with three internal states labeled by S ∈ {−1, 0, +1},
interacting according to the Hamiltonian
Ĥ = −J S1 S2 − µ0 B(S1 + S2 ) .
1
(6)
Find (i) the partition function Z, (ii) the magnetization M , and (iii) the magnetic suscep
tibility at zero field χ = ∂M
∂B B=0 .
Solution : To find the partition function in cases with a tractable number of energy states,
one can simply enumerate all the states and their energies. We do this in table 1 below.
This allows us to write
Z(T, B) = Tr e−β Ĥ
1
X
=
1
X
e−βE(S1 ,S2 )
S1 =−1 S2 =−1
= 2eβJ cosh(2βµ0 B) + 4 cosh(βµ0 B) + 2e−βJ + 1 .
(7)
Note that Z(T → ∞, B) = 9, which is the total number of states.
E
+
0
−
+
−J − 2µ0 B
−µ0 B
J
0
−µ0 B
0
µ0 B
−
J
µ0 B
−J + 2µ0 B
Table 1: Energy E(S1 , S2 ) for the Hamiltonian of eqn. 6.
To find the magnetization, we use
M =−
=
∂F
1 1 ∂Z
=−
∂B
β Z ∂B
(8)
4µ0 eβJ sinh(2βµ0 B) + 4µ0 sinh(βµ0 B)
2eβJ cosh(2βµ0 B) + 4 cosh(βµ0 B) + 2e−βJ + 1
(9)
To find the zero field susceptibility, simply expand M (T, B) in powers of B and keep the
lowest (linear) term. This gives
2 eβJ + 1
· 4βµ20 B
4 cosh(βJ) + 5
(10)
∂M 2 eJ/kB T + 1
4µ20
=
·
.
∂B B=0 4 cosh(J/kB T ) + 5 kB T
(11)
M=
Therefore, the susceptibility is
χ=
Note that as T → 0 we have χ ∼ 4µ20 /kB T . This is because the two states | + + i and
| − − i are preferred by the J term, and at sufficiently low temperatures the two spins
become ‘locked’ into a single state of magnetic moment 2µ0 .
2
(3) Consider a gas of noninteracting classical indistinguishable particles with dispersion
ε(p) = |p|4 /2L, where L is a dimensionful constant. Evaluate the OCE partition function
Z(T, N ) and find the statistical entropy S(T, N ), the heat capacity CV , and the equation
of state p = p(T, V, N ).
Solution : The single particle partition function is
Z∞
d3p −p4 /2Lk T
4πV
4
B
e
= 3 dp p2 e−p /2LkB T .
ζ=V
3
h
h
Z
(12)
0
Change variables to s = p4 /2LkB T . Then
p = (2LkB T )1/4 s1/4
dp = (2LkB T )1/4 · 14 s−3/4 ds
,
(13)
and
Z∞
V
ζ = 3 3 · 4π · (2LkB T )3/4 · 14 · ds s−1/4 e−s
8π ~
0
3/4
3
Γ 4
2LkB T
=
·V ·
.
2
8π
~4
(14)
The N -particle partition function is Z(T, V, N ) = ζ N /N ! and using Stirling’s formula
ln(N !) = N ln N − N + O(ln N ) we have
F (T, V, N ) = −kB T ln Z
= −N kB T ln ζ − N kB T ln N + N kB T
"
#
Γ 43
V
2LkB T 3/4
= −N kB T ln
+ N kB T − N kB T ln
.
N
8π 2
~4
(15)
The entropy is
∂F
S(T, V, N ) = −
∂T V,N
"
#
Γ 34
2LkB T 3/4
= N kB ln
+ 43 N kB .
8π 2
~4
(16)
The heat capacity is
CV = T
∂S
∂T
= 34 N kB .
(17)
V,N
Thus, the heat capacity differs from the familiar result 32 N kB for the ideal monatomic gas
with ballistic ε = p2 /2m dispersion. Nevertheless, this system still obeys the ideal gas law:
∂F
N kB T
p=−
=
.
(18)
∂V T,N
V
3