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+---
+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.