Categories: Quantum mechanics.

Bloch’s theorem

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

ψ(r)=u(r)eik⋅r\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 H^\hat{H} is lattice-periodic, then both ψ(r)\psi(\vb{r}) and ψ(r+a)\psi(\vb{r} + \vb{a}) are eigenstates with the same energy:

H^ψ(r)=Eψ(r)H^ψ(r+a)=Eψ(r+a)\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 T^(a)\hat{T}(\vb{a}) such that ψ(r+a)=T^(a)ψ(r)\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r}). From the previous equation, we then know that:

H^T^(a)ψ(r)=ET^(a)ψ(r)=T^(a)(Eψ(r))=T^(a)H^ψ(r)\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 H^\hat{H} is lattice-periodic, then it will commute with T^(a)\hat{T}(\vb{a}), i.e. [H^,T^(a)]=0[\hat{H}, \hat{T}(\vb{a})] = 0. Consequently, H^\hat{H} and T^(a)\hat{T}(\vb{a}) must share eigenstates ψ(r)\psi(\vb{r}):

H^ ψ(r)=E ψ(r)T^(a) ψ(r)=τ ψ(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 T^\hat{T} is unitary, its eigenvalues τ\tau must have the form eiθe^{i \theta}, with θ\theta real. Therefore a translation by a\vb{a} causes a phase shift, for some vector k\vb{k}:

ψ(r+a)=T^(a) ψ(r)=eiθ ψ(r)=eik⋅a ψ(r)\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 k\vb{k}:

u(r)≡e−ik⋅r ψ(r)\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 k\vb{k}:

u(r+a)=e−ik⋅(r+a) ψ(r+a)=e−ik⋅re−ik⋅aeik⋅a ψ(r)=e−ik⋅r ψ(r)=u(r)\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(r)u(\vb{r}) for ψ(r)\psi(\vb{r}).