---
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/),
will distribute themselves across the available states in a system at equilibrium.

Consider one single-particle state $$s$$,
which can contain $$0$$ or $$1$$ fermions.
Because the occupation number $$N$$ is variable,
we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
whose grand partition function $$\mathcal{Z}$$ is as follows,
where we sum over all microstates of $$s$$:

$$\begin{aligned}
    \mathcal{Z}
    = \sum_{N = 0}^1 \exp(- \beta N (\varepsilon - \mu))
    = 1 + \exp(- \beta (\varepsilon - \mu))
\end{aligned}$$

Where $$\mu$$ is the chemical potential,
and $$\varepsilon$$ is the energy contribution per particle in $$s$$,
i.e. the total energy of all particles $$E = \varepsilon N$$.

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

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

The average number of particles $$\Expval{N}$$
in state $$s$$ is then found to be as follows:

$$\begin{aligned}
    \Expval{N}
    = - \pdv{\Omega}{\mu}
    = k T \pdv{\ln{\mathcal{Z}}}{\mu}
    = \frac{\exp(- \beta (\varepsilon - \mu))}{1 + \exp(- \beta (\varepsilon - \mu))}
\end{aligned}$$

By multiplying both the numerator and the denominator by $$\exp(\beta (\varepsilon \!-\! \mu))$$,
we arrive at the standard form of
the **Fermi-Dirac distribution** or **Fermi function** $$f_F$$:

$$\begin{aligned}
    \boxed{
        \Expval{N}
        = f_F(\varepsilon)
        = \frac{1}{\exp(\beta (\varepsilon - \mu)) + 1}
    }
\end{aligned}$$

This tells the expected occupation number $$\Expval{N}$$ of state $$s$$,
given a temperature $$T$$ and chemical potential $$\mu$$.
The corresponding variance $$\sigma^2$$ of $$N$$ is found to be:

$$\begin{aligned}
    \boxed{
        \sigma^2
        = k T \pdv{\Expval{N}}{\mu}
        = \Expval{N} \big(1 - \Expval{N}\big)
    }
\end{aligned}$$



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