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. \(u(\vec{r}) = u(\vec{r} + \vec{a})\):
\[ @@ -53,7 +59,7 @@ \end{aligned} \]
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 \(\psi(\vec{r})\):
+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. \(\comm{\hat{H}}{\hat{T}(\vec{a})} = 0\). Therefore \(\hat{H}\) and \(\hat{T}(\vec{a})\) must share eigenstates \(\psi(\vec{r})\):
\[ \begin{aligned} \hat{H} \:\psi(\vec{r}) = E \:\psi(\vec{r}) @@ -91,7 +97,6 @@ \end{aligned} \]
Then Bloch’s theorem follows from isolating the definition of \(u(\vec{r})\) for \(\psi(\vec{r})\).
-