From a8d31faecc733fa4d63fde58ab98a5e9d11029c2 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sun, 2 Apr 2023 16:57:12 +0200
Subject: Improve knowledge base

---
 source/know/concept/blochs-theorem/index.md | 46 ++++++++++-------------------
 1 file changed, 16 insertions(+), 30 deletions(-)

(limited to 'source/know/concept/blochs-theorem/index.md')

diff --git a/source/know/concept/blochs-theorem/index.md b/source/know/concept/blochs-theorem/index.md
index 6f445f1..d7fcf90 100644
--- a/source/know/concept/blochs-theorem/index.md
+++ b/source/know/concept/blochs-theorem/index.md
@@ -17,85 +17,72 @@ take the following form,
 where the function $$u(\vb{r})$$ is periodic on the same lattice,
 i.e. $$u(\vb{r}) = u(\vb{r} + \vb{a})$$:
 
-$$
-\begin{aligned}
+$$\begin{aligned}
 	\boxed{
 		\psi(\vb{r}) = u(\vb{r}) e^{i \vb{k} \cdot \vb{r}}
 	}
-\end{aligned}
-$$
+\end{aligned}$$
 
 In other words, in a periodic potential,
 the solutions are simply plane waves with a periodic modulation,
 known as **Bloch functions** or **Bloch states**.
 
-This is suprisingly easy to prove:
+This is surprisingly easy to prove:
 if the Hamiltonian $$\hat{H}$$ is lattice-periodic,
 then both $$\psi(\vb{r})$$ and $$\psi(\vb{r} + \vb{a})$$
 are eigenstates with the same energy:
 
-$$
-\begin{aligned}
+$$\begin{aligned}
 	\hat{H} \psi(\vb{r}) = E \psi(\vb{r})
 	\qquad
 	\hat{H} \psi(\vb{r} + \vb{a}) = E \psi(\vb{r} + \vb{a})
-\end{aligned}
-$$
+\end{aligned}$$
 
 Now define the unitary translation operator $$\hat{T}(\vb{a})$$ such that
 $$\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r})$$.
 From the previous equation, we then know that:
 
-$$
-\begin{aligned}
+$$\begin{aligned}
 	\hat{H} \hat{T}(\vb{a}) \psi(\vb{r})
 	= E \hat{T}(\vb{a}) \psi(\vb{r})
 	= \hat{T}(\vb{a}) \big(E \psi(\vb{r})\big)
 	= \hat{T}(\vb{a}) \hat{H} \psi(\vb{r})
-\end{aligned}
-$$
+\end{aligned}$$
 
 In other words, if $$\hat{H}$$ is lattice-periodic,
 then it will commute with $$\hat{T}(\vb{a})$$,
 i.e. $$[\hat{H}, \hat{T}(\vb{a})] = 0$$.
 Consequently, $$\hat{H}$$ and $$\hat{T}(\vb{a})$$ must share eigenstates $$\psi(\vb{r})$$:
 
-$$
-\begin{aligned}
+$$\begin{aligned}
 	\hat{H} \:\psi(\vb{r}) = E \:\psi(\vb{r})
 	\qquad \qquad
 	\hat{T}(\vb{a}) \:\psi(\vb{r}) = \tau \:\psi(\vb{r})
-\end{aligned}
-$$
+\end{aligned}$$
 
 Since $$\hat{T}$$ is unitary,
 its eigenvalues $$\tau$$ must have the form $$e^{i \theta}$$, with $$\theta$$ real.
 Therefore a translation by $$\vb{a}$$ causes a phase shift,
 for some vector $$\vb{k}$$:
 
-$$
-\begin{aligned}
+$$\begin{aligned}
 	\psi(\vb{r} + \vb{a})
 	= \hat{T}(\vb{a}) \:\psi(\vb{r})
 	= e^{i \theta} \:\psi(\vb{r})
 	= e^{i \vb{k} \cdot \vb{a}} \:\psi(\vb{r})
-\end{aligned}
-$$
+\end{aligned}$$
 
 Let us now define the following function,
 keeping our arbitrary choice of $$\vb{k}$$:
 
-$$
-\begin{aligned}
+$$\begin{aligned}
 	u(\vb{r})
-	= e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
-\end{aligned}
-$$
+	\equiv e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
+\end{aligned}$$
 
 As it turns out, this function is guaranteed to be lattice-periodic for any $$\vb{k}$$:
 
-$$
-\begin{aligned}
+$$\begin{aligned}
 	u(\vb{r} + \vb{a})
 	&= e^{- i \vb{k} \cdot (\vb{r} + \vb{a})} \:\psi(\vb{r} + \vb{a})
 	\\
@@ -104,8 +91,7 @@ $$
 	&= e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
 	\\
 	&= u(\vb{r})
-\end{aligned}
-$$
+\end{aligned}$$
 
 Then Bloch's theorem follows from
 isolating the definition of $$u(\vb{r})$$ for $$\psi(\vb{r})$$.
-- 
cgit v1.2.3