Improve knowledge base

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 %}