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Date: Sat, 20 Feb 2021 13:46:18 +0100
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+% Bloch's theorem
+
+
+# Bloch's theorem
+In quantum mechanics, *Bloch's theorem* states that,
+given a potential $V(\vec{r})$ which is periodic on a lattice,
+i.e. $V(\vec{r}) = V(\vec{r} + \vec{a})$
+for a primitive lattice vector $\vec{a}$,
+then it follows that the solutions $\psi(\vec{r})$
+to the time-independent Schrödinger equation
+take the following form,
+where the function $u(\vec{r})$ is periodic on the same lattice,
+i.e. $u(\vec{r}) = u(\vec{r} + \vec{a})$:
+
+$$
+\begin{aligned}
+	\boxed{
+		\psi(\vec{r}) = u(\vec{r}) e^{i \vec{k} \cdot \vec{r}}
+	}
+\end{aligned}
+$$
+
+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 it will commute with the unitary translation operator $\hat{T}(\vec{a})$,
+i.e. $\comm{\hat{H}}{\hat{T}(\vec{a})} = 0$.
+Therefore $\hat{H}$ and $\hat{T}(\vec{a})$ must share eigenstates $\psi(\vec{r})$:
+
+$$
+\begin{aligned}
+	\hat{H} \:\psi(\vec{r}) = E \:\psi(\vec{r})
+	\qquad
+	\hat{T}(\vec{a}) \:\psi(\vec{r}) = \tau \:\psi(\vec{r})
+\end{aligned}
+$$
+
+Since $\hat{T}$ is unitary,
+its eigenvalues $\tau$ must have the form $e^{i \theta}$, with $\theta$ real.
+Therefore a translation by $\vec{a}$ causes a phase shift,
+for some vector $\vec{k}$:
+
+$$
+\begin{aligned}
+	\psi(\vec{r} + \vec{a})
+	= \hat{T}(\vec{a}) \:\psi(\vec{r})
+	= e^{i \theta} \:\psi(\vec{r})
+	= e^{i \vec{k} \cdot \vec{a}} \:\psi(\vec{r})
+\end{aligned}
+$$
+
+Let us now define the following function,
+keeping our arbitrary choice of $\vec{k}$:
+
+$$
+\begin{aligned}
+	u(\vec{r})
+	= e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r})
+\end{aligned}
+$$
+
+As it turns out, this function is guaranteed to be lattice-periodic for any $\vec{k}$:
+
+$$
+\begin{aligned}
+	u(\vec{r} + \vec{a})
+	&= e^{- i \vec{k} \cdot (\vec{r} + \vec{a})} \:\psi(\vec{r} + \vec{a})
+	\\
+	&= e^{- i \vec{k} \cdot \vec{r}} e^{- i \vec{k} \cdot \vec{a}} e^{i \vec{k} \cdot \vec{a}} \:\psi(\vec{r})
+	\\
+	&= e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r})
+	\\
+	&= u(\vec{r})
+\end{aligned}
+$$
+
+Then Bloch's theorem follows from
+isolating the definition of $u(\vec{r})$ for $\psi(\vec{r})$.
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