Categories: Quantum mechanics.

# 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 both $$\psi(\vec{r})$$ and $$\psi(\vec{r} + \vec{a})$$ are eigenstates with the same energy:

\begin{aligned} \hat{H} \psi(\vec{r}) = E \psi(\vec{r}) \qquad \hat{H} \psi(\vec{r} + \vec{a}) = E \psi(\vec{r} + \vec{a}) \end{aligned}

Now define the unitary translation operator $$\hat{T}(\vec{a})$$ such that $$\psi(\vec{r} + \vec{a}) = \hat{T}(\vec{a}) \psi(\vec{r})$$. From the previous equation, we then know that:

\begin{aligned} \hat{H} \hat{T}(\vec{a}) \psi(\vec{r}) = E \hat{T}(\vec{a}) \psi(\vec{r}) = \hat{T}(\vec{a}) \big(E \psi(\vec{r})\big) = \hat{T}(\vec{a}) \hat{H} \psi(\vec{r}) \end{aligned}

In other words, if $$\hat{H}$$ is lattice-periodic, then it will commute with $$\hat{T}(\vec{a})$$, i.e. $$[\hat{H}, \hat{T}(\vec{a})] = 0$$. Consequently, $$\hat{H}$$ 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 $$\psi(\vec{r})$$.