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+---
+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.