---
title: "Bose-Einstein distribution"
firstLetter: "B"
publishDate: 2021-07-11
categories:
- Physics
- Statistics
- Quantum mechanics

date: 2021-07-11T18:22:44+02:00
draft: false
markup: pandoc
---

# Bose-Einstein statistics

**Bose-Einstein statistics** describe how bosons,
which do not obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/),
will distribute themselves across the available states
in a system at equilibrium.

Consider a single-particle state $s$,
which can contain any number of bosons.
Since 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 $\varepsilon_s$ is the energy per particle,
and $\mu$ is the chemical potential:

$$\begin{aligned}
    \mathcal{Z}_s
    = \sum_{N_s = 0}^\infty \Big( \exp\!(- \beta (\varepsilon_s - \mu)) \Big)^{N_s}
    = \frac{1}{1 - \exp\!(- \beta (\varepsilon_s - \mu))}
\end{aligned}$$

The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/)
is the Landau potential $\Omega$, 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 $\expval{N_s}$
is found by taking a derivative of $\Omega$:

$$\begin{aligned}
    \expval{N_s}
    = - \pdv{\Omega_s}{\mu}
    = k T \pdv{\ln{\mathcal{Z_s}}}{\mu}
    = \frac{\exp\!(- \beta (\varepsilon_s - \mu))}{1 - \exp\!(- \beta (\varepsilon_s - \mu))}
\end{aligned}$$

By multitplying both the numerator and the denominator by $\exp\!(\beta(\epsilon_s \!-\! \mu))$,
we arrive at the standard form of the **Bose-Einstein distribution** $f_B$:

$$\begin{aligned}
    \boxed{
        \expval{N_s}
        = f_B(\varepsilon_s)
        = \frac{1}{\exp\!(\beta (\varepsilon_s - \mu)) - 1}
    }
\end{aligned}$$

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.