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Bloch’s theorem

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In quantum mechanics, Bloch’s theorem states that, given a potential V(r⃗)V(\vec{r}) which is periodic on a lattice, i.e. V(r⃗)=V(r⃗+a⃗)V(\vec{r}) = V(\vec{r} + \vec{a}) for a primitive lattice vector a⃗\vec{a}, then it follows that the solutions ψ(r⃗)\psi(\vec{r}) to the time-independent Schrödinger equation take the following form, where the function u(r⃗)u(\vec{r}) is periodic on the same lattice, i.e. u(r⃗)=u(r⃗+a⃗)u(\vec{r}) = u(\vec{r} + \vec{a}):

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ψ(r⃗)=u(r⃗)eik⃗⋅r⃗ -\begin{aligned} - \boxed{ - \psi(\vec{r}) = u(\vec{r}) e^{i \vec{k} \cdot \vec{r}} - } -\end{aligned} -

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In other words, in a periodic potential, the solutions are simply plane waves with a periodic modulation, known as Bloch functions or Bloch states.

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This is suprisingly easy to prove: if the Hamiltonian Ĥ\hat{H} is lattice-periodic, then it will commute with the unitary translation operator T̂(a⃗)\hat{T}(\vec{a}), i.e. [Ĥ,T̂(a⃗)]=0[\hat{H}, \hat{T}(\vec{a})] = 0. Therefore Ĥ\hat{H} and T̂(a⃗)\hat{T}(\vec{a}) must share eigenstates ψ(r⃗)\psi(\vec{r}):

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Ĥψ(r⃗)=Eψ(r⃗)T̂(a⃗)ψ(r⃗)=τψ(r⃗) -\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} -

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Since T̂\hat{T} is unitary, its eigenvalues τ\tau must have the form eiθe^{i \theta}, with θ\theta real. Therefore a translation by a⃗\vec{a} causes a phase shift, for some vector k⃗\vec{k}:

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ψ(r⃗+a⃗)=T̂(a⃗)ψ(r⃗)=eiθψ(r⃗)=eik⃗⋅a⃗ψ(r⃗) -\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} -

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Let us now define the following function, keeping our arbitrary choice of k⃗\vec{k}:

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u(r⃗)=e−ik⃗⋅r⃗ψ(r⃗) -\begin{aligned} - u(\vec{r}) - = e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r}) -\end{aligned} -

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As it turns out, this function is guaranteed to be lattice-periodic for any k⃗\vec{k}:

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u(r⃗+a⃗)=e−ik⃗⋅(r⃗+a⃗)ψ(r⃗+a⃗)=e−ik⃗⋅r⃗e−ik⃗⋅a⃗eik⃗⋅a⃗ψ(r⃗)=e−ik⃗⋅r⃗ψ(r⃗)=u(r⃗) -\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} -

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Then Bloch’s theorem follows from isolating the definition of u(r⃗)u(\vec{r}) for ψ(r⃗)\psi(\vec{r}).

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