% 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. $[\hat{H}, \hat{T}(\vec{a})] = 0$.
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 $\psi(\vec{r})$.