Editing Bloch's theorem (section)

=== Using operators ===
In this proof all the symmetries are encoded as commutation properties of the translation operators
{{math proof | title = Proof using operators | proof =
Source:<ref name=":4">{{Harvnb|Ashcroft|Mermin|1976|p=137}}</ref>

We define the translation operator
<math display="block">\begin{align}
\hat{\mathbf{T}}_{\mathbf{n}}\psi(\mathbf{r})&= \psi(\mathbf{r}+\mathbf{T}_{\mathbf{n}}) \\
&= \psi(\mathbf{r}+n_1\mathbf{a}_1+n_2\mathbf{a}_2+n_3\mathbf{a}_3) \\
&= \psi(\mathbf{r}+\mathbf{A}\mathbf{n})
\end{align}</math>
with
<math display="block">
\mathbf{A} = \begin{bmatrix} \mathbf{a}_1 & \mathbf{a}_2 & \mathbf{a}_3 \end{bmatrix}, \quad \mathbf{n} = \begin{pmatrix} n_1 \\ n_2 \\ n_3 \end{pmatrix}
</math>
We use the hypothesis of a mean periodic potential
<math display="block">U(\mathbf{x}+\mathbf{T}_{\mathbf{n}})= U(\mathbf{x})</math>
and the [[independent electron approximation]] with an Hamiltonian
<math display="block">\hat{H}=\frac{\hat{\mathbf{p}}^2}{2m}+U(\mathbf{x})</math>
Given the Hamiltonian is invariant for translations it shall commute with the translation operator
<math display="block">[\hat{H},\hat{\mathbf{T}}_{\mathbf{n}}] = 0</math>
and the two operators shall have a common set of eigenfunctions.
Therefore, we start to look at the eigen-functions of the translation operator:
<math display="block">\hat{\mathbf{T}}_{\mathbf{n}}\psi(\mathbf{x})=\lambda_{\mathbf{n}}\psi(\mathbf{x})</math>
Given <math>\hat{\mathbf{T}}_{\mathbf{n}}</math> is an additive operator 
<math display="block">
\hat{\mathbf{T}}_{\mathbf{n}_1} \hat{\mathbf{T}}_{\mathbf{n}_2}\psi(\mathbf{x}) =
\psi(\mathbf{x} + \mathbf{A} \mathbf{n}_1 + \mathbf{A} \mathbf{n}_2) = \hat{\mathbf{T}}_{\mathbf{n}_1 + \mathbf{n}_2} \psi(\mathbf{x})
</math>
If we substitute here the eigenvalue equation and dividing both sides for <math>\psi(\mathbf{x})</math> we have
<math display="block">
\lambda_{\mathbf{n}_1} \lambda_{\mathbf{n}_2} =
\lambda_{\mathbf{n}_1 + \mathbf{n}_2}
</math>

This is true for 
<math display="block">\lambda_{\mathbf{n}} = e^{s \mathbf{n} \cdot \mathbf{a} } </math>
where <math>s \in \Complex </math> if we use the normalization condition over a single primitive cell of volume V
<math display="block">
1 = \int_V |\psi(\mathbf{x})|^2 d \mathbf{x} = 
\int_V \left|\hat\mathbf{T}_\mathbf{n} \psi(\mathbf{x})\right|^2 d \mathbf{x} = 
|\lambda_{\mathbf{n}}|^2 \int_V |\psi(\mathbf{x})|^2 d \mathbf{x}
</math>
and therefore
<math display="block">1 = |\lambda_{\mathbf{n}}|^2</math> and <math display="block">s = i k </math> where <math>k \in \mathbb{R}</math>. Finally,
<math display="block">
\mathbf{\hat{T}_n}\psi(\mathbf{x})=
\psi(\mathbf{x} + \mathbf{n} \cdot \mathbf{a} ) = 
e^{i k \mathbf{n} \cdot \mathbf{a} }\psi(\mathbf{x})
,</math>
which is true for a Bloch wave i.e. for <math>\psi_{\mathbf{k}}(\mathbf{x}) = e^{i \mathbf{k} \cdot \mathbf{x} } u_{\mathbf{k}}(\mathbf{x})</math> with <math>u_{\mathbf{k}}(\mathbf{x}) = u_{\mathbf{k}}(\mathbf{x} + \mathbf{A}\mathbf{n})</math>
}}