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Diffstat (limited to 'source/know/concept/fermi-dirac-distribution')
-rw-r--r-- | source/know/concept/fermi-dirac-distribution/index.md | 41 |
1 files changed, 21 insertions, 20 deletions
diff --git a/source/know/concept/fermi-dirac-distribution/index.md b/source/know/concept/fermi-dirac-distribution/index.md index 09a3e76..2a38eb3 100644 --- a/source/know/concept/fermi-dirac-distribution/index.md +++ b/source/know/concept/fermi-dirac-distribution/index.md @@ -11,67 +11,68 @@ layout: "concept" **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. +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 $$N$$ is variable, we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/), whose grand partition function $$\mathcal{Z}$$ is as follows, -where we sum over all microstates of $$s$$: +where $$\varepsilon$$ is the energy of $$s$$ +and $$\mu$$ is the chemical potential: $$\begin{aligned} \mathcal{Z} - = \sum_{N = 0}^1 \exp(- \beta N (\varepsilon - \mu)) - = 1 + \exp(- \beta (\varepsilon - \mu)) + = \sum_{N = 0}^1 \Big( e^{-\beta (\varepsilon - \mu)} \Big)^N + = 1 + e^{-\beta (\varepsilon - \mu)} \end{aligned}$$ -Where $$\mu$$ is the chemical potential, -and $$\varepsilon$$ is the energy contribution per particle in $$s$$, -i.e. the total energy of all particles $$E = \varepsilon N$$. - The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/) is the Landau potential $$\Omega$$, given by: $$\begin{aligned} \Omega = - k T \ln{\mathcal{Z}} - = - k T \ln\!\Big( 1 + \exp(- \beta (\varepsilon - \mu)) \Big) + = - k T \ln\!\Big( 1 + e^{-\beta (\varepsilon - \mu)} \Big) \end{aligned}$$ -The average number of particles $$\Expval{N}$$ -in state $$s$$ is then found to be as follows: +The average number of particles $$\expval{N}$$ +in $$s$$ is then found by taking a derivative of $$\Omega$$: $$\begin{aligned} - \Expval{N} + \expval{N} = - \pdv{\Omega}{\mu} = k T \pdv{\ln{\mathcal{Z}}}{\mu} - = \frac{\exp(- \beta (\varepsilon - \mu))}{1 + \exp(- \beta (\varepsilon - \mu))} + = \frac{e^{-\beta (\varepsilon - \mu)}}{1 + e^{-\beta (\varepsilon - \mu)}} \end{aligned}$$ -By multiplying both the numerator and the denominator by $$\exp(\beta (\varepsilon \!-\! \mu))$$, +By multiplying both the numerator and the denominator by $$e^{\beta (\varepsilon \!-\! \mu)}$$, we arrive at the standard form of the **Fermi-Dirac distribution** or **Fermi function** $$f_F$$: $$\begin{aligned} \boxed{ - \Expval{N} + \expval{N} = f_F(\varepsilon) - = \frac{1}{\exp(\beta (\varepsilon - \mu)) + 1} + = \frac{1}{e^{\beta (\varepsilon - \mu)} + 1} } \end{aligned}$$ -This tells the expected occupation number $$\Expval{N}$$ of state $$s$$, +This gives the expected occupation number $$\expval{N}$$ +of state $$s$$ with energy $$\varepsilon$$, given a temperature $$T$$ and chemical potential $$\mu$$. -The corresponding variance $$\sigma^2$$ of $$N$$ is found to be: + +{% comment %} +The corresponding variance $$\sigma^2 \equiv \expval{N^2} - \expval{N}^2$$ is found to be: $$\begin{aligned} \boxed{ \sigma^2 - = k T \pdv{\Expval{N}}{\mu} - = \Expval{N} \big(1 - \Expval{N}\big) + = k T \pdv{\expval{N}}{\mu} + = \expval{N} \big(1 - \expval{N}\big) } \end{aligned}$$ +{% endcomment %} |