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diff --git a/source/know/concept/bose-einstein-distribution/index.md b/source/know/concept/bose-einstein-distribution/index.md new file mode 100644 index 0000000..8397a8f --- /dev/null +++ b/source/know/concept/bose-einstein-distribution/index.md @@ -0,0 +1,77 @@ +--- +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/), +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 $N$ is variable, +we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/), +whose grand partition function $\mathcal{Z}$ is as follows, +where $\varepsilon$ is the energy per particle, +and $\mu$ is the chemical potential: + +$$\begin{aligned} + \mathcal{Z} + = \sum_{N = 0}^\infty \Big( \exp(- \beta (\varepsilon - \mu)) \Big)^{N} + = \frac{1}{1 - \exp(- \beta (\varepsilon - \mu))} +\end{aligned}$$ + +The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/) +is the Landau potential $\Omega$, given by: + +$$\begin{aligned} + \Omega + = - k T \ln{\mathcal{Z}} + = k T \ln\!\Big( 1 - \exp(- \beta (\varepsilon - \mu)) \Big) +\end{aligned}$$ + +The average number of particles $\Expval{N}$ +is found by taking a derivative of $\Omega$: + +$$\begin{aligned} + \Expval{N} + = - \pdv{\Omega}{\mu} + = k T \pdv{\ln{\mathcal{Z}}}{\mu} + = \frac{\exp(- \beta (\varepsilon - \mu))}{1 - \exp(- \beta (\varepsilon - \mu))} +\end{aligned}$$ + +By multitplying both the numerator and the denominator by $\exp(\beta(\varepsilon \!-\! \mu))$, +we arrive at the standard form of the **Bose-Einstein distribution** $f_B$: + +$$\begin{aligned} + \boxed{ + \Expval{N} + = f_B(\varepsilon) + = \frac{1}{\exp(\beta (\varepsilon - \mu)) - 1} + } +\end{aligned}$$ + +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. |