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diff --git a/content/know/concept/blochs-theorem/index.pdc b/content/know/concept/blochs-theorem/index.pdc new file mode 100644 index 0000000..1828d8a --- /dev/null +++ b/content/know/concept/blochs-theorem/index.pdc @@ -0,0 +1,115 @@ +--- +title: "Bloch's theorem" +firstLetter: "B" +publishDate: 2021-02-22 +categories: +- Quantum mechanics + +date: 2021-02-22T20:02:14+01:00 +draft: false +markup: pandoc +--- + +# 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 both $\psi(\vec{r})$ and $\psi(\vec{r} + \vec{a})$ +are eigenstates with the same energy: + +$$ +\begin{aligned} + \hat{H} \psi(\vec{r}) = E \psi(\vec{r}) + \qquad + \hat{H} \psi(\vec{r} + \vec{a}) = E \psi(\vec{r} + \vec{a}) +\end{aligned} +$$ + +Now define the unitary translation operator $\hat{T}(\vec{a})$ such that +$\psi(\vec{r} + \vec{a}) = \hat{T}(\vec{a}) \psi(\vec{r})$. +From the previous equation, we then know that: + +$$ +\begin{aligned} + \hat{H} \hat{T}(\vec{a}) \psi(\vec{r}) + = E \hat{T}(\vec{a}) \psi(\vec{r}) + = \hat{T}(\vec{a}) \big(E \psi(\vec{r})\big) + = \hat{T}(\vec{a}) \hat{H} \psi(\vec{r}) +\end{aligned} +$$ + +In other words, if $\hat{H}$ is lattice-periodic, +then it will commute with $\hat{T}(\vec{a})$, +i.e. $[\hat{H}, \hat{T}(\vec{a})] = 0$. +Consequently, $\hat{H}$ 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 $\psi(\vec{r})$. |