--- 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.