@@ -49,41 +44,41 @@

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. $u(\vec{r}) = u(\vec{r} + \vec{a})$ :

$+In quantum mechanics,$ 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 $\psi(\vec{r})$ :

$+This is suprisingly easy to prove: if the Hamiltonian$ $\hat{H}$ 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 $\vec{k}$ :

$+\] +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 $\vec{k}$ :

$+\] +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 $\vec{k}$ :

$+\] +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}) @@ -94,8 +89,8 @@ \\ &= u(\vec{r}) \end{aligned} -

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

+\]

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