Categories: Quantum mechanics.

# Bloch’s theorem

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