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---
title: "Bloch's theorem"
sort_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 surprisingly 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 $$\vb{k}$$:
$$\begin{aligned}
u(\vb{r})
\equiv e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
\end{aligned}$$
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 $$\psi(\vb{r})$$.
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