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---
title: "Fermi-Dirac distribution"
firstLetter: "F"
publishDate: 2021-07-11
categories:
- Physics
- Statistics
- Quantum mechanics
date: 2021-07-11T18:22:37+02:00
draft: false
markup: pandoc
---
# Fermi-Dirac distribution
**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 $N_s$ is variable,
we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
whose grand partition function $\mathcal{Z}_s$ is as follows,
where we sum over all microstates of $s$:
$$\begin{aligned}
\mathcal{Z}_s
= \sum_{N_s = 0}^1 \exp\!(- \beta N_s (\varepsilon_s - \mu))
= 1 + \exp\!(- \beta (\varepsilon_s - \mu))
\end{aligned}$$
Where $\mu$ is the chemical potential,
and $\varepsilon_s$ is the energy contribution per particle in $s$,
i.e. the total energy of all particles $E_s = \varepsilon_s N_s$.
The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/)
is the Landau potential $\Omega_s$, given by:
$$\begin{aligned}
\Omega_s
= - k T \ln{\mathcal{Z}_s}
= - k T \ln\!\Big( 1 + \exp\!(- \beta (\varepsilon_s - \mu)) \Big)
\end{aligned}$$
The average number of particles $\expval{N_s}$
in state $s$ is then found to be as follows:
$$\begin{aligned}
\expval{N_s}
= - \pdv{\Omega_s}{\mu}
= k T \pdv{\ln{\mathcal{Z}_s}}{\mu}
= \frac{\exp\!(- \beta (\varepsilon_s - \mu))}{1 + \exp\!(- \beta (\varepsilon_s - \mu))}
\end{aligned}$$
By multiplying both the numerator and the denominator by $\exp\!(\beta (\varepsilon_s \!-\! \mu))$,
we arrive at the standard form of
the **Fermi-Dirac distribution** or **Fermi function** $f_F$:
$$\begin{aligned}
\boxed{
\expval{N_s}
= f_F(\varepsilon_s)
= \frac{1}{\exp\!(\beta (\varepsilon_s - \mu)) + 1}
}
\end{aligned}$$
This tells the expected occupation number $\expval{N_s}$ of state $s$,
given a temperature $T$ and chemical potential $\mu$.
The corresponding variance $\sigma_s^2$ of $N_s$ is found to be:
$$\begin{aligned}
\boxed{
\sigma_s^2
= k T \pdv{\expval{N_s}}{\mu}
= \expval{N_s} \big(1 - \expval{N_s}\big)
}
\end{aligned}$$
## References
1. H. Gould, J. Tobochnik,
*Statistical and thermal physics*, 2nd edition,
Princeton.
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