Statistical Mechanics - Physics at Oregon State University

A.2. SOLUTIONS FOR CHAPTER 2. 241
For one particle we have:
Z(T, 1) =
e −βɛn1 When we have N particles they are in states n1, n2, · · · , nN and we have
Z(T, N) =
· · ·
e −β(ɛn1 +ɛn2 +···+ɛnN )
n1
n2
Which is equal to :
Z(T, N) =
e −βɛn
1 e −βɛn2 · · ·
e −βɛnN = (Z(T, 1)) N
Problem 9.
n1
n2
nN
The energy eigenvalues of a system are given by 0 and ɛ + n∆ for n = 0, 1, 2, · · ·.
We have both ɛ > 0 and ∆ > 0. Calculate the partition function for this system.
Calculate the internal energy and the heat capacity. Plot the heat capacity as
a function of temperature for 0 < kBT < ɛ for (a) ∆ ≫ ɛ, (b) ∆ = ɛ , and (c)
∆ ≪ ɛ.
Z = e β0 +
n1
nN
∞
e −β(ɛ+n∆)
−βɛ
= 1 + e e −nβ∆
n=0
The sum can be done using
n rn = 1
1−r
U = ɛ
We have three cases:
and gives:
Z = 1 + e−βɛ
1 − e−β∆ = 1 + e−βɛ − e−β∆ 1 − e−β∆ 2 ∂
∂
U = kBT log(Z) = −
∂T ∂β log(Z)
U = ɛe−βɛ − ∆e−β∆ 1 + e−βɛ ∆e−β∆
+
− e−β∆ 1 − e−β∆ e−βɛ −β∆ e
−
1 − e−β∆ 1 + e−βɛ + ∆
− e−β∆ e
∆ ≫ ɛ ⇒ U ≈ ɛ
−βɛ 1
= ɛ
1 + e−βɛ eβɛ + 1
e
∆ = ɛ ⇒ U = ɛ
−βɛ 1
= ɛ
1 − e−βɛ eβɛ − 1
∆ ≪ ɛ ⇒ U ≈ ɛ + kBT (for kBT ≫ ∆)
n
e −β∆
1 + e −βɛ − e −β∆