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<h1 id="blochs-theorem">Bloch’s theorem</h1>
<p>In quantum mechanics, <em>Bloch’s theorem</em> states that, given a potential <span class="math inline">\(V(\vec{r})\)</span> which is periodic on a lattice, i.e. <span class="math inline">\(V(\vec{r}) = V(\vec{r} + \vec{a})\)</span> for a primitive lattice vector <span class="math inline">\(\vec{a}\)</span>, then it follows that the solutions <span class="math inline">\(\psi(\vec{r})\)</span> to the time-independent Schrödinger equation take the following form, where the function <span class="math inline">\(u(\vec{r})\)</span> is periodic on the same lattice, i.e. <span class="math inline">\(u(\vec{r}) = u(\vec{r} + \vec{a})\)</span>:</p>
<p><span class="math display">\[
\begin{aligned}
\boxed{
\psi(\vec{r}) = u(\vec{r}) e^{i \vec{k} \cdot \vec{r}}
}
\end{aligned}
\]</span></p>
<p>In other words, in a periodic potential, the solutions are simply plane waves with a periodic modulation, known as <em>Bloch functions</em> or <em>Bloch states</em>.</p>
<p>This is suprisingly easy to prove: if the Hamiltonian <span class="math inline">\(\hat{H}\)</span> is lattice-periodic, then it will commute with the unitary translation operator <span class="math inline">\(\hat{T}(\vec{a})\)</span>, i.e. <span class="math inline">\([\hat{H}, \hat{T}(\vec{a})] = 0\)</span>. Therefore <span class="math inline">\(\hat{H}\)</span> and <span class="math inline">\(\hat{T}(\vec{a})\)</span> must share eigenstates <span class="math inline">\(\psi(\vec{r})\)</span>:</p>
<p><span class="math display">\[
\begin{aligned}
\hat{H} \:\psi(\vec{r}) = E \:\psi(\vec{r})
\qquad
\hat{T}(\vec{a}) \:\psi(\vec{r}) = \tau \:\psi(\vec{r})
\end{aligned}
\]</span></p>
<p>Since <span class="math inline">\(\hat{T}\)</span> is unitary, its eigenvalues <span class="math inline">\(\tau\)</span> must have the form <span class="math inline">\(e^{i \theta}\)</span>, with <span class="math inline">\(\theta\)</span> real. Therefore a translation by <span class="math inline">\(\vec{a}\)</span> causes a phase shift, for some vector <span class="math inline">\(\vec{k}\)</span>:</p>
<p><span class="math display">\[
\begin{aligned}
\psi(\vec{r} + \vec{a})
= \hat{T}(\vec{a}) \:\psi(\vec{r})
= e^{i \theta} \:\psi(\vec{r})
= e^{i \vec{k} \cdot \vec{a}} \:\psi(\vec{r})
\end{aligned}
\]</span></p>
<p>Let us now define the following function, keeping our arbitrary choice of <span class="math inline">\(\vec{k}\)</span>:</p>
<p><span class="math display">\[
\begin{aligned}
u(\vec{r})
= e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r})
\end{aligned}
\]</span></p>
<p>As it turns out, this function is guaranteed to be lattice-periodic for any <span class="math inline">\(\vec{k}\)</span>:</p>
<p><span class="math display">\[
\begin{aligned}
u(\vec{r} + \vec{a})
&= e^{- i \vec{k} \cdot (\vec{r} + \vec{a})} \:\psi(\vec{r} + \vec{a})
\\
&= e^{- i \vec{k} \cdot \vec{r}} e^{- i \vec{k} \cdot \vec{a}} e^{i \vec{k} \cdot \vec{a}} \:\psi(\vec{r})
\\
&= e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r})
\\
&= u(\vec{r})
\end{aligned}
\]</span></p>
<p>Then Bloch’s theorem follows from isolating the definition of <span class="math inline">\(u(\vec{r})\)</span> for <span class="math inline">\(\psi(\vec{r})\)</span>.</p>
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