8.3: Fermi-Dirac Distribution
( \newcommand{\kernel}{\mathrm{null}\,}\)
The counting of distinct arrangements for fermions is even simpler than for the Bose-Einstein case, since each state can have an occupation number of either zero or 1. Thus consider g states with n particles to be distributed among them. There are n states which are singly occupied and these can be chosen in g!(n!(g−n)!) ways. The total number of distinct arrangements is thus given by
W({nα})=∏αgαnα!(gα−nα)
The function to be maximized to identify the equilibrium distribution is therefore given by
Sk−βU+βµN=−nαlognα−(gα−nα)log(gα−nα)−β(ϵα−μ)nα+constant
The extremization condition reads
log[(gα−nα)nα]=β(ϵα−μ)
with the solution
nα=gαeβ(ϵα−μ)+1
So, for fermions in equilibrium, we can take the occupation number to be given by
n=1eβ(ϵ−μ)+1
with the degeneracy factor arising from summation over states of the same energy. This is the Fermi-Dirac distribution. The normalization conditions are again,
∑∫d3xd3p(2πħ)31eβ(ϵ−μ)+1=N∑∫d3xd3p(2πħ)3ϵeβ(ϵ−μ)+1=U
As in the case of the Bose-Einstein distribution, we can write down the partition function for free fermions as
logZ=∑log(1+e−β(ϵ−μ))Z=∏11+e−β(ϵ−μ)
Notice that, here too, the partition function for each state is ∑ne−nβ(ϵ−μ) ; it is just that, in the present case, n can only be zero or 1.