---
title: "Bose-Einstein distribution"
sort_title: "Bose-Einstein distribution"
date: 2021-07-11
categories:
- Physics
- Statistics
- Quantum mechanics
layout: "concept"
---

**Bose-Einstein statistics** describe how bosons,
which do not obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/),
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 $$N$$ is variable,
we use the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
whose grand partition function $$\mathcal{Z}$$ is as shown below,
where $$\varepsilon$$ is the energy per particle,
and $$\mu$$ is the chemical potential.
We evaluate the sum in $$\mathcal{Z}$$ as a geometric series:

$$\begin{aligned}
    \mathcal{Z}
    = \sum_{N = 0}^\infty \Big( e^{-\beta (\varepsilon - \mu)} \Big)^{N}
    = \frac{1}{1 - e^{-\beta (\varepsilon - \mu)}}
\end{aligned}$$

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

$$\begin{aligned}
    \Omega
    = - k T \ln{\mathcal{Z}}
    = k T \ln\!\big( 1 - e^{-\beta (\varepsilon - \mu)} \big)
\end{aligned}$$

The average number of particles $$\expval{N}$$ in $$s$$
is then found by taking a derivative of $$\Omega$$:

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

By multiplying both the numerator and the denominator by $$e^{\beta(\varepsilon \!-\! \mu)}$$,
we arrive at the standard form of the **Bose-Einstein distribution** $$f_B$$:

$$\begin{aligned}
    \boxed{
        \expval{N}
        = f_B(\varepsilon)
        = \frac{1}{e^{\beta (\varepsilon - \mu)} - 1}
    }
\end{aligned}$$

This gives the expected occupation number $$\expval{N}$$
of state $$s$$ with energy $$\varepsilon$$,
given a temperature $$T$$ and chemical potential $$\mu$$.

{% comment %}
The corresponding variance $$\sigma^2 \equiv \expval{N^2} - \expval{N}^2$$ is found to be:

$$\begin{aligned}
    \boxed{
        \sigma^2
        = k T \pdv{\expval{N}}{\mu}
        = \expval{N} \big(1 + \expval{N}\!\big)
    }
\end{aligned}$$
{% endcomment %}



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