---
title: "Fermi-Dirac distribution"
sort_title: "Fermi-Dirac distribution"
date: 2021-07-11
categories:
- Physics
- Statistics
- Quantum mechanics
layout: "concept"
---

**Fermi-Dirac statistics** describe how identical **fermions**,
which obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/),
distribute themselves across the available states in a system at equilibrium.

Consider one single-particle state $$\ket{i}$$,
which can contain $$0$$ or $$1$$ fermions.
Because the occupation number $$n_i$$ is variable,
we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
whose grand partition function $$\mathcal{Z}$$ is as follows,
where $$\varepsilon_i$$ is the energy of $$\ket{i}$$
and $$\mu$$ is the chemical potential:

$$\begin{aligned}
    \mathcal{Z}
    = \sum_{m = 0}^1 \Big( e^{-\beta (\varepsilon_i - \mu)} \Big)^m
    = 1 + e^{-\beta (\varepsilon_i - \mu)}
\end{aligned}$$

The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/)
is the Landau potential $$\Omega$$, given by:

$$\begin{aligned}
    \Omega
    = - k T \ln{\mathcal{Z}}
    = - k T \ln\!\Big( 1 + e^{-\beta (\varepsilon_i - \mu)} \Big)
\end{aligned}$$

The average number of particles $$\expval{n_i}$$
in $$\ket{i}$$ is then found by taking a derivative of $$\Omega$$:

$$\begin{aligned}
    \expval{n_i}
    = - \pdv{\Omega}{\mu}
    = k T \pdv{\ln{\mathcal{Z}}}{\mu}
    = \frac{e^{-\beta (\varepsilon_i - \mu)}}{1 + e^{-\beta (\varepsilon_i - \mu)}}
\end{aligned}$$

By multiplying both the numerator and the denominator by $$e^{\beta (\varepsilon_i - \mu)}$$,
we arrive at the standard form of
the **Fermi-Dirac distribution** or **Fermi function** $$f_F$$:

$$\begin{aligned}
    \boxed{
        \expval{n_i}
        = f_F(\varepsilon_i)
        = \frac{1}{e^{\beta (\varepsilon_i - \mu)} + 1}
    }
\end{aligned}$$

This gives the expected occupation number $$\expval{n_i}$$
of state $$\ket{i}$$ with energy $$\varepsilon_i$$,
given a temperature $$T$$ and chemical potential $$\mu$$.

{% comment %}
The corresponding variance $$\sigma^2 \equiv \expval{n_i^2} - \expval{n_i}^2$$ is found to be:

$$\begin{aligned}
    \boxed{
        \sigma^2
        = k T \pdv{\expval{n_i}}{\mu}
        = \expval{n_i} \big(1 - \expval{n_i}\big)
    }
\end{aligned}$$
{% endcomment %}



## References
1.  H. Gould, J. Tobochnik,
    *Statistical and thermal physics*, 2nd edition,
    Princeton.