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---
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 $\psi(\vb{r})$.
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