Statistical Mechanics - Physics at Oregon State University

248 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
S(T, µ, V ) = NkB −
fM (ɛo; T, µ) log(fM (ɛo; T, µ))
where the sum is over orbitals.
There are several ways to do this. For example:
S = −
o
< n >o= kBT
∂ log Zo
∂µ T,V
< n >o= fM (ɛo; T, µ) = e 1
k B T (µ−ɛo)
∂ log Zo
∂µ T,V
= 1 1
k e B T
kBT (µ−ɛo)
log Zo = e 1
k B T (µ−ɛo) = fM (ɛo; T, µ)
Ω = −kBT
log(Zo) = −kBT
fM (ɛo; T, µ)
o
∂Ω
= kB fM (ɛo; T, µ) + kBT
∂T V,µ
∂
∂T fM (ɛo; T, µ)
o
fM (ɛo; T, µ) = N
o
∂
∂T fM (ɛo; T, µ) = − 1
1
k (µ−ɛo)e B T
kBT 2 (µ−ɛo) 1
= −
T log(fM (ɛo; T, µ))fM (ɛo; T, µ)
which gives the result we need
Problem 5.
Consider a system of independent particles. The number of orbitals with energy
between E and E + dE is given by N(E)dE. The function N(E) is called the
density of states. One measures the expectation value of a certain operator O.
For a particle in an orbital o the value of the operator depends only on the
energy of the orbital, or Oo = O(ɛo). Show that in the thermodynamic limit
the ensemble average of the operator is given by
∞
< O >= O(E)N(E)f(E; T, µ)dE
−∞
where f(E; T, µ) is the distribution function for these particles.
o
o