Statistical Mechanics - Physics at Oregon State University

260 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
Split in odd/even terms:
∞ 1
Z = T r(
(2k)! β2k −
k=0
∞
k=0
1
(2k + 1)! β2k+1 H)
Z = T r(cosh(β) − H sinh(β))
The internal energy is U = − ∂
∂β log(Z), which gives:
In the limit T → ∞ we find
T r(sinh(β) − H cosh(β))
U = −
T r(cosh(β) − H sinh(β))
T rH
U =
T r1
which makes sense, it is the average of all energies.
In the other limit T → 0 we use
U = − T r(eβ (1 − H) + e −β (−1 − H))
T r(e β (1 − H) + e −β (1 + H))
U = − eβ T r(1 − H) − e −β T r(1 + H)
e β T r(1 − H) + e −β T r(1 + H)
which gives U = −1 unless T rH = T r1 , in which case all energy values are
+1 and only the second terms survive to give U = 1.
Solution to Problem 7.
As stated the problem is easy. Since the sum of the eigenvalues is 1 there is
exactly one eigenvalue equal to 1, the rest is zero. The system is in a specific
state and the entropy is zero.
The question should have been that the eigenvalues are either 0 or 1
N
in which
case we have exactly N eigenvalues equal to 1
N and the entropy is kB log(N).
Part C is simple: ρ = ρ1ρ2 · · · where ρ1 only acts on the part of the wave
function pertaining to particle 1, and gives 1 for everything else.