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 \(\psi(\vec{r})\).
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