Statistical Mechanics - Physics at Oregon State University

250 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
1
1 − k e B T (N∆−µ) + 1
1
e 1
k B T ((N+1)∆−µ) + 1
N
i=1
∞
1 −
e
i=N+1
k B T (N∆−µ) + 1
1 − k e B T (i∆−µ) + 1
e 1
k B T ((N+1)∆−µ)
e 1
k B T (i∆−µ) + 1
When the temperature is small, the sums go to one and we have:
1
1 − k e B T (N∆−µ) + 1 =
1
e 1
k B T ((N+1)∆−µ) + 1
1 − k e B T (N∆−µ) + 1 = e 1
kB T ((N+1)∆−µ) + 1
− 1
1
(N∆ − µ) = ((N + 1)∆ − µ)
kBT kBT
−(N∆ − µ) = ((N + 1)∆ − µ)
which gives the desired result, since at T = 0 µ = ɛF . In words, for low temperature
the amount on electrons leaving the state i = N is equal to the amount
gained in i = N + 1, and fF D(µ + x; T, µ) + fF D(µ − x; T, µ) = 1
Problem 7.
The entropy for a system of independent Fermions is given by
S = −kB (fF D log(fF D) + (1 − fF D) log(1 − fF D))
o
Calculate lim
T →0 fF D(ɛ, T, µ) for ɛ < µ ,ɛ = µ , and ɛ > µ.
fF D(ɛ, T, µ) =
1
e β(ɛ−µ) + 1
fF D(µ, T, µ) = 1
2
Therefore, the limits are 1, 1
2 , and 0, respectively.
The number of orbitals with energy ɛo equal to the Fermi energy ɛF is M.
Calculate the entropy at T = 0 in this case.
1
S = −kBM
2 log(1
1
) +
2 2 log(1
2 )
=
= kB log(2 M )
Explain your answer in terms of a multiplicity function. Pay close attention to
the issue of dependent and independent variables.