From 6ce0bb9a8f9fd7d169cbb414a9537d68c5290aae Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Fri, 14 Oct 2022 23:25:28 +0200
Subject: Initial commit after migration from Hugo

---
 source/know/concept/blochs-theorem/index.md | 110 ++++++++++++++++++++++++++++
 1 file changed, 110 insertions(+)
 create mode 100644 source/know/concept/blochs-theorem/index.md

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

diff --git a/source/know/concept/blochs-theorem/index.md b/source/know/concept/blochs-theorem/index.md
new file mode 100644
index 0000000..496d8d3
--- /dev/null
+++ b/source/know/concept/blochs-theorem/index.md
@@ -0,0 +1,110 @@
+---
+title: "Bloch's theorem"
+date: 2021-02-22
+categories:
+- Quantum mechanics
+layout: "concept"
+---
+
+In quantum mechanics, **Bloch's theorem** states that,
+given a potential $V(\vb{r})$ which is periodic on a lattice,
+i.e. $V(\vb{r}) = V(\vb{r} + \vb{a})$
+for a primitive lattice vector $\vb{a}$,
+then it follows that the solutions $\psi(\vb{r})$
+to the time-independent Schrödinger equation
+take the following form,
+where the function $u(\vb{r})$ is periodic on the same lattice,
+i.e. $u(\vb{r}) = u(\vb{r} + \vb{a})$:
+
+$$
+\begin{aligned}
+	\boxed{
+		\psi(\vb{r}) = u(\vb{r}) e^{i \vb{k} \cdot \vb{r}}
+	}
+\end{aligned}
+$$
+
+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:
+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:
+
+$$
+\begin{aligned}
+	\hat{H} \psi(\vb{r}) = E \psi(\vb{r})
+	\qquad
+	\hat{H} \psi(\vb{r} + \vb{a}) = E \psi(\vb{r} + \vb{a})
+\end{aligned}
+$$
+
+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:
+
+$$
+\begin{aligned}
+	\hat{H} \hat{T}(\vb{a}) \psi(\vb{r})
+	= E \hat{T}(\vb{a}) \psi(\vb{r})
+	= \hat{T}(\vb{a}) \big(E \psi(\vb{r})\big)
+	= \hat{T}(\vb{a}) \hat{H} \psi(\vb{r})
+\end{aligned}
+$$
+
+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 $\psi(\vb{r})$:
+
+$$
+\begin{aligned}
+	\hat{H} \:\psi(\vb{r}) = E \:\psi(\vb{r})
+	\qquad \qquad
+	\hat{T}(\vb{a}) \:\psi(\vb{r}) = \tau \:\psi(\vb{r})
+\end{aligned}
+$$
+
+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 $\vb{k}$:
+
+$$
+\begin{aligned}
+	\psi(\vb{r} + \vb{a})
+	= \hat{T}(\vb{a}) \:\psi(\vb{r})
+	= e^{i \theta} \:\psi(\vb{r})
+	= e^{i \vb{k} \cdot \vb{a}} \:\psi(\vb{r})
+\end{aligned}
+$$
+
+Let us now define the following function,
+keeping our arbitrary choice of $\vb{k}$:
+
+$$
+\begin{aligned}
+	u(\vb{r})
+	= e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
+\end{aligned}
+$$
+
+As it turns out, this function is guaranteed to be lattice-periodic for any $\vb{k}$:
+
+$$
+\begin{aligned}
+	u(\vb{r} + \vb{a})
+	&= e^{- i \vb{k} \cdot (\vb{r} + \vb{a})} \:\psi(\vb{r} + \vb{a})
+	\\
+	&= e^{- i \vb{k} \cdot \vb{r}} e^{- i \vb{k} \cdot \vb{a}} e^{i \vb{k} \cdot \vb{a}} \:\psi(\vb{r})
+	\\
+	&= e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
+	\\
+	&= u(\vb{r})
+\end{aligned}
+$$
+
+Then Bloch's theorem follows from
+isolating the definition of $u(\vb{r})$ for $\psi(\vb{r})$.
-- 
cgit v1.2.3