From 6ce0bb9a8f9fd7d169cbb414a9537d68c5290aae Mon Sep 17 00:00:00 2001 From: Prefetch Date: Fri, 14 Oct 2022 23:25:28 +0200 Subject: Initial commit after migration from Hugo --- source/know/concept/blochs-theorem/index.md | 110 ++++++++++++++++++++++++++++ 1 file changed, 110 insertions(+) create mode 100644 source/know/concept/blochs-theorem/index.md (limited to 'source/know/concept/blochs-theorem') diff --git a/source/know/concept/blochs-theorem/index.md b/source/know/concept/blochs-theorem/index.md new file mode 100644 index 0000000..496d8d3 --- /dev/null +++ b/source/know/concept/blochs-theorem/index.md @@ -0,0 +1,110 @@ +--- +title: "Bloch's theorem" +date: 2021-02-22 +categories: +- Quantum mechanics +layout: "concept" +--- + +In quantum mechanics, **Bloch's theorem** states that, +given a potential $V(\vb{r})$ which is periodic on a lattice, +i.e. $V(\vb{r}) = V(\vb{r} + \vb{a})$ +for a primitive lattice vector $\vb{a}$, +then it follows that the solutions $\psi(\vb{r})$ +to the time-independent Schrödinger equation +take the following form, +where the function $u(\vb{r})$ is periodic on the same lattice, +i.e. $u(\vb{r}) = u(\vb{r} + \vb{a})$: + +$$ +\begin{aligned} + \boxed{ + \psi(\vb{r}) = u(\vb{r}) e^{i \vb{k} \cdot \vb{r}} + } +\end{aligned} +$$ + +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 suprisingly easy to prove: +if the Hamiltonian $\hat{H}$ is lattice-periodic, +then both $\psi(\vb{r})$ and $\psi(\vb{r} + \vb{a})$ +are eigenstates with the same energy: + +$$ +\begin{aligned} + \hat{H} \psi(\vb{r}) = E \psi(\vb{r}) + \qquad + \hat{H} \psi(\vb{r} + \vb{a}) = E \psi(\vb{r} + \vb{a}) +\end{aligned} +$$ + +Now define the unitary translation operator $\hat{T}(\vb{a})$ such that +$\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r})$. +From the previous equation, we then know that: + +$$ +\begin{aligned} + \hat{H} \hat{T}(\vb{a}) \psi(\vb{r}) + = E \hat{T}(\vb{a}) \psi(\vb{r}) + = \hat{T}(\vb{a}) \big(E \psi(\vb{r})\big) + = \hat{T}(\vb{a}) \hat{H} \psi(\vb{r}) +\end{aligned} +$$ + +In other words, if $\hat{H}$ is lattice-periodic, +then it will commute with $\hat{T}(\vb{a})$, +i.e. $[\hat{H}, \hat{T}(\vb{a})] = 0$. +Consequently, $\hat{H}$ and $\hat{T}(\vb{a})$ must share eigenstates $\psi(\vb{r})$: + +$$ +\begin{aligned} + \hat{H} \:\psi(\vb{r}) = E \:\psi(\vb{r}) + \qquad \qquad + \hat{T}(\vb{a}) \:\psi(\vb{r}) = \tau \:\psi(\vb{r}) +\end{aligned} +$$ + +Since $\hat{T}$ is unitary, +its eigenvalues $\tau$ must have the form $e^{i \theta}$, with $\theta$ real. +Therefore a translation by $\vb{a}$ causes a phase shift, +for some vector $\vb{k}$: + +$$ +\begin{aligned} + \psi(\vb{r} + \vb{a}) + = \hat{T}(\vb{a}) \:\psi(\vb{r}) + = e^{i \theta} \:\psi(\vb{r}) + = e^{i \vb{k} \cdot \vb{a}} \:\psi(\vb{r}) +\end{aligned} +$$ + +Let us now define the following function, +keeping our arbitrary choice of $\vb{k}$: + +$$ +\begin{aligned} + u(\vb{r}) + = e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r}) +\end{aligned} +$$ + +As it turns out, this function is guaranteed to be lattice-periodic for any $\vb{k}$: + +$$ +\begin{aligned} + u(\vb{r} + \vb{a}) + &= e^{- i \vb{k} \cdot (\vb{r} + \vb{a})} \:\psi(\vb{r} + \vb{a}) + \\ + &= e^{- i \vb{k} \cdot \vb{r}} e^{- i \vb{k} \cdot \vb{a}} e^{i \vb{k} \cdot \vb{a}} \:\psi(\vb{r}) + \\ + &= e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r}) + \\ + &= u(\vb{r}) +\end{aligned} +$$ + +Then Bloch's theorem follows from +isolating the definition of $u(\vb{r})$ for $\psi(\vb{r})$. -- cgit v1.2.3