From f04a80d30c6f4bd7a0c17fc5ec26ed7968621edf Mon Sep 17 00:00:00 2001 From: Prefetch Date: Fri, 19 Feb 2021 20:25:37 +0100 Subject: Proof-of-concept for knowledge base --- static/know/blochs-theorem/index.html | 103 ++++++++++++++++++++++++++++++++++ 1 file changed, 103 insertions(+) create mode 100644 static/know/blochs-theorem/index.html (limited to 'static/know/blochs-theorem') diff --git a/static/know/blochs-theorem/index.html b/static/know/blochs-theorem/index.html new file mode 100644 index 0000000..26e3480 --- /dev/null +++ b/static/know/blochs-theorem/index.html @@ -0,0 +1,103 @@ + + + + + + + Prefetch | Bloch’s theorem + + + +

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

$+\begin{aligned} + \boxed{ + \psi(\vec{r}) = u(\vec{r}) e^{i \vec{k} \cdot \vec{r}} + } +\end{aligned} +$

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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