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

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In quantum mechanics, Bloch’s theorem states that, given a potential V(r⃗)V(\vec{r}) which is periodic on a lattice, i.e. V(r⃗)=V(r⃗+a⃗)V(\vec{r}) = V(\vec{r} + \vec{a}) for a primitive lattice vector a⃗\vec{a}, then it follows that the solutions ψ(r⃗)\psi(\vec{r}) to the time-independent Schrödinger equation take the following form, where the function u(r⃗)u(\vec{r}) is periodic on the same lattice, i.e. u(r⃗)=u(r⃗+a⃗)u(\vec{r}) = u(\vec{r} + \vec{a}):

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

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In other words, in a periodic potential, the solutions are simply plane waves with a periodic modulation, known as Bloch functions or Bloch states.

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This is suprisingly easy to prove: if the Hamiltonian Ĥ\hat{H} is lattice-periodic, then it will commute with the unitary translation operator T̂(a⃗)\hat{T}(\vec{a}), i.e. [Ĥ,T̂(a⃗)]=0[\hat{H}, \hat{T}(\vec{a})] = 0. Therefore Ĥ\hat{H} and T̂(a⃗)\hat{T}(\vec{a}) must share eigenstates ψ(r⃗)\psi(\vec{r}):

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Ĥψ(r⃗)=Eψ(r⃗)T̂(a⃗)ψ(r⃗)=τψ(r⃗) +\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} +

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Since T̂\hat{T} is unitary, its eigenvalues τ\tau must have the form eiθe^{i \theta}, with θ\theta real. Therefore a translation by a⃗\vec{a} causes a phase shift, for some vector k⃗\vec{k}:

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ψ(r⃗+a⃗)=T̂(a⃗)ψ(r⃗)=eiθψ(r⃗)=eik⃗⋅a⃗ψ(r⃗) +\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} +

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Let us now define the following function, keeping our arbitrary choice of k⃗\vec{k}:

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u(r⃗)=e−ik⃗⋅r⃗ψ(r⃗) +\begin{aligned} + u(\vec{r}) + = e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r}) +\end{aligned} +

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As it turns out, this function is guaranteed to be lattice-periodic for any k⃗\vec{k}:

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u(r⃗+a⃗)=e−ik⃗⋅(r⃗+a⃗)ψ(r⃗+a⃗)=e−ik⃗⋅r⃗e−ik⃗⋅a⃗eik⃗⋅a⃗ψ(r⃗)=e−ik⃗⋅r⃗ψ(r⃗)=u(r⃗) +\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} +

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

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