8.3: Fermi-Dirac Distribution

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 \(\frac{g!}{(n!(g − n)!)}\) ways. The total number of distinct arrangements is thus given by

\[W(\{n_{\alpha}\}) = \prod_{\alpha} \frac{g_{\alpha}}{n_{\alpha}!(g_{\alpha} - n_{\alpha})} \]

The function to be maximized to identify the equilibrium distribution is therefore given by

\[\frac{S}{k} − βU + βµN = - n_α \log n_α - (g_α - n_α) \log (g_α - n_α) - \beta (\epsilon_{\alpha} - \mu)n_α + \text{constant} \]

The extremization condition reads

\[\log \left[ \frac{(g_α - n_α)}{n_α} \right] = \beta (\epsilon_{\alpha} - \mu) \]

with the solution

\[n_α = \frac{g_α}{e^{\beta(\epsilon_{\alpha} - \mu)} + 1} \]

So, for fermions in equilibrium, we can take the occupation number to be given by

\[n = \frac{1}{e^{\beta(\epsilon - \mu)} + 1} \]

with the degeneracy factor arising from summation over states of the same energy. This is the Fermi-Dirac distribution. The normalization conditions are again,

\[\sum \int \frac{d^3xd^3p}{(2 \pi ħ)^3} \frac{1}{e^{\beta(\epsilon - \mu)} + 1} = N \\ \sum \int \frac{d^3xd^3p}{(2 \pi ħ)^3} \frac{\epsilon }{e^{\beta(\epsilon - \mu)} + 1} = U \]

As in the case of the Bose-Einstein distribution, we can write down the partition function for free fermions as

\[ \begin{equation}
\begin{split}
\log Z & = \sum \log \left( 1 + e^{-\beta(\epsilon - \mu)} \right) \\[0.125in]
Z & = \prod \frac{1}{1 + e^{-\beta(\epsilon - \mu)}}
\end{split}
\end{equation} \]

Notice that, here too, the partition function for each state is \(\sum_n e^{-n \beta (\epsilon - \mu)}\) ; it is just that, in the present case, \(n\) can only be zero or 1.