Editing Bloch's theorem (section)

== Velocity and effective mass ==
If we apply the time-independent [[Schrödinger equation]] to the Bloch wave function we obtain
<math display="block">\hat{H}_\mathbf{k} u_\mathbf{k}(\mathbf{r}) = 
\left[ \frac{\hbar^2}{2m} \left( -i \nabla + \mathbf{k} \right)^2 + U(\mathbf{r}) \right] u_\mathbf{k}(\mathbf{r}) =
\varepsilon_\mathbf{k} u_\mathbf{k}(\mathbf{r})
</math>
with boundary conditions
<math display="block">u_\mathbf{k}(\mathbf{r}) = u_\mathbf{k}(\mathbf{r} + \mathbf{R})</math>
Given this is defined in a finite volume we expect an infinite family of eigenvalues; here <math>{\mathbf{k}}</math> is a parameter of the Hamiltonian and therefore we arrive at a "continuous family" of eigenvalues <math>\varepsilon_n(\mathbf{k})</math> dependent on the continuous parameter <math>{\mathbf{k}}</math> and thus at the basic concept of an electronic band structure.

{{math proof
| title = Proof<ref name=":1">{{Harvnb|Ashcroft|Mermin|1976|p=140}}</ref>
| proof =
<math display="block">
E_\mathbf{k} \left(e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x})\right) 
= 
\left[\frac{- \hbar^2}{2m} \nabla^2 + U(\mathbf{x} ) \right]
\left(e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x})\right)
</math>

We remain with
<math display="block">\begin{align}
E_\mathbf{k} e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x})
&=
\frac{- \hbar^2}{2m} \nabla \cdot 
\left( i \mathbf{k} e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x}) 
+ e^{i \mathbf{k} \cdot \mathbf{x} } \nabla u_\mathbf{k}(\mathbf{x}) \right) 
+ U(\mathbf{x}) e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x}) \\[1.2ex]

E_\mathbf{k} e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x})
&=
\frac{- \hbar^2}{2m} 
\left( i \mathbf{k} \cdot 
\left( i \mathbf{k} e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x}) 
+ e^{i \mathbf{k} \cdot \mathbf{x} } \nabla u_\mathbf{k}(\mathbf{x}) \right)
+ i \mathbf{k} \cdot e^{i \mathbf{k} \cdot \mathbf{x} } \nabla u_\mathbf{k}(\mathbf{x}) 
+ e^{i \mathbf{k} \cdot \mathbf{x} } \nabla^2 u_\mathbf{k}(\mathbf{x}) \right) 
+ U(\mathbf{x}) e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x}) \\[1.2ex]

E_\mathbf{k} e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x})
&=
\frac{ \hbar^2}{2m} 
\left(\mathbf{k}^2 e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x}) 
- 2i \mathbf{k} \cdot e^{i \mathbf{k} \cdot \mathbf{x} } \nabla u_\mathbf{k}(\mathbf{x}) 
- e^{i \mathbf{k} \cdot \mathbf{x} } \nabla^2 u_\mathbf{k}(\mathbf{x}) \right) 
+ U(\mathbf{x}) e^{i \mathbf{k} \cdot \mathbf{x} } u_\mathbf{k}(\mathbf{x}) \\[1.2ex]

E_\mathbf{k} u_\mathbf{k}(\mathbf{x})
&= 
\frac{ \hbar^2}{2m} 
\left(-i \nabla + \mathbf{k}\right)^2 u_\mathbf{k}(\mathbf{x}) 
+ U(\mathbf{x}) u_\mathbf{k}(\mathbf{x})
\end{align}</math>
}}

This shows how the effective momentum can be seen as composed of two parts,
<math display="block">\hat{\mathbf{p}}_\text{eff} =  -i \hbar \nabla + \hbar \mathbf{k} ,</math>
a standard momentum <math>-i \hbar \nabla</math> and a crystal momentum <math>\hbar \mathbf{k}</math>. More precisely the crystal momentum is not a momentum but it stands for the momentum in the same way as the electromagnetic momentum in the [[minimal coupling]], and as part of a [[canonical transformation]] of the momentum.

For the effective velocity we can derive
{{Equation box 1
|indent=:
|title='''mean velocity of a Bloch electron'''
|equation=<math>\frac{\partial \varepsilon_n}{\partial \mathbf{k}} = \frac {\hbar^2}{m} \int 
d\mathbf{r}\, \psi^{*}_{n\mathbf{k}} (-i \nabla)\psi_{n\mathbf{k}} = \frac {\hbar}{m}\langle\hat{\mathbf{p}}\rangle = \hbar \langle\hat{\mathbf{v}}\rangle</math>
|cellpadding
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|border colour = rgb(80,200,120)
|background colour = rgb(80,200,120,10%)}}
{{math proof
| title = Proof<ref name=":2">{{Harvnb|Ashcroft|Mermin|1976|p=765 Appendix E}}</ref>
| proof =
We evaluate the derivatives <math>\frac{\partial \varepsilon_n}{\partial \mathbf{k}}</math> and <math>\frac{\partial^2 \varepsilon_n(\mathbf{k})}{\partial k_i \partial k_j}</math>
given they are the coefficients of the following expansion in {{math|'''q'''}} where {{math|'''q'''}} is considered small with respect to {{math|'''k'''}}
<math display="block">
\varepsilon_n(\mathbf{k} + \mathbf{q}) = \varepsilon_n(\mathbf{k}) + 
\sum_i \frac{\partial \varepsilon_n}{\partial k_i} q_i + 
\frac{1}{2} \sum_{ij} \frac{\partial^2 \varepsilon_n}{\partial k_i \partial k_j} q_i q_j +
O(q^3)
</math>
Given <math>\varepsilon_n(\mathbf{k}+\mathbf{q})</math> are eigenvalues of <math>\hat{H}_{\mathbf{k}+\mathbf{q}}</math>
We can consider the following perturbation problem in q:
<math display="block">
\hat{H}_{\mathbf{k}+\mathbf{q}} = 
\hat{H}_\mathbf{k} + 
\frac{\hbar^2}{m} \mathbf{q} \cdot ( -i\nabla + \mathbf{k} ) +
\frac{\hbar^2}{2m} q^2
</math>
Perturbation theory of the second order states that
<math display="block">
E_n =E^0_n + \int d\mathbf{r}\, \psi^{*}_n \hat{V} \psi_n + 
\sum_{n' \neq n} \frac{|\int d\mathbf{r} \,\psi^{*}_n \hat{V} \psi_n|^2}{E^0_n - E^0_{n'}} + ...
</math>
To compute to linear order in {{math|'''q'''}}
<math display="block">
\sum_i \frac{\partial \varepsilon_n}{\partial k_i} q_i = 
\sum_i \int d\mathbf{r}\, u_{n\mathbf{k}}^{*} \frac{\hbar^2}{m} ( -i\nabla + \mathbf{k} )_i q_i u_{n\mathbf{k}}
</math>
where the integrations are over a primitive cell or the entire crystal, given if the integral 
<math display="block">\int d\mathbf{r}\, u_{n\mathbf{k}}^{*} u_{n\mathbf{k}}</math>
is normalized across the cell or the crystal.

We can simplify over {{math|'''q'''}} to obtain
<math display="block">
\frac{\partial \varepsilon_n}{\partial \mathbf{k}} = 
\frac{\hbar^2}{m} \int d\mathbf{r} \, u_{n\mathbf{k}}^{*}( -i\nabla + \mathbf{k} ) u_{n\mathbf{k}}
</math>
and we can reinsert the complete wave functions
<math display="block">
\frac{\partial \varepsilon_n}{\partial \mathbf{k}} = 
\frac{\hbar^2}{m} \int d\mathbf{r} \, \psi_{n\mathbf{k}}^{*}( -i\nabla) \psi_{n\mathbf{k}}
</math>
}}

For the [[Effective mass (solid-state physics)|effective mass]]
{{Equation box 1
|indent=:
|title='''effective mass theorem'''
|equation=<math>
\frac{\partial^2 \varepsilon_n(\mathbf{k})}{\partial k_i \partial k_j} = 
\frac {\hbar^2}{m} \delta_{ij} + 
\left(
\frac {\hbar^2}{m}
\right)^2
\sum_{n' \neq n}
\frac{
\langle n\mathbf{k} | -i \nabla_i | n'\mathbf{k} \rangle 
\langle n'\mathbf{k} | -i \nabla_j | n\mathbf{k} \rangle 
+
\langle n\mathbf{k} | -i \nabla_j | n'\mathbf{k} \rangle 
\langle n'\mathbf{k} | -i \nabla_i | n\mathbf{k} \rangle
}{
\varepsilon_n(\mathbf{k}) - \varepsilon_{n'}(\mathbf{k})
}
</math>
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|background colour = rgb(80,200,120,10%)}}
{{math proof
| title = Proof<ref name=":2"/>
| proof = 
The second order term
<math display="block">
\frac{1}{2} \sum_{ij} \frac{\partial^2 \varepsilon_n}{\partial k_i \partial k_j} q_i q_j =
\frac {\hbar^2}{2m} q^2 + 
\sum_{n' \neq n} 
\frac{|
\int d\mathbf{r} \,
u_{n\mathbf{k}}^{*} 
\frac{\hbar^2}{m} \mathbf{q} \cdot (-i\nabla + \mathbf{k}) 
u_{n'\mathbf{k}}
|^2} 
{\varepsilon_{n\mathbf{k}} - \varepsilon_{n'\mathbf{k}}}
</math>
Again with <math> \psi_{n\mathbf{k}} =| n\mathbf{k}\rangle = e^{i\mathbf{k}\mathbf{x}} u_{n\mathbf{k}}</math> 
<math display="block">
\frac{1}{2} \sum_{ij} \frac{\partial^2 \varepsilon_n}{\partial k_i \partial k_j} q_i q_j =
\frac {\hbar^2}{2m} q^2 + 
\sum_{n' \neq n} 
\frac{|
\langle n\mathbf{k} | 
\frac{\hbar^2}{m} \mathbf{q} \cdot (-i\nabla) 
| n'\mathbf{k}\rangle
|^2} 
{\varepsilon_{n\mathbf{k}} - \varepsilon_{n'\mathbf{k}}}
</math>
Eliminating <math>q_i</math> and <math>q_j</math> we have the theorem
<math display="block">
\frac{\partial^2 \varepsilon_n(\mathbf{k})}{\partial k_i \partial k_j} = 
\frac {\hbar^2}{m} \delta_{ij} + 
\left(
\frac {\hbar^2}{m}
\right)^2
\sum_{n' \neq n}
\frac{
\langle n\mathbf{k} | -i \nabla_i | n'\mathbf{k} \rangle 
\langle n'\mathbf{k} | -i \nabla_j | n\mathbf{k} \rangle 
+
\langle n\mathbf{k} | -i \nabla_j | n'\mathbf{k} \rangle 
\langle n'\mathbf{k} | -i \nabla_i | n\mathbf{k} \rangle
}{
\varepsilon_n(\mathbf{k}) - \varepsilon_{n'}(\mathbf{k})
}
</math>
}}
The quantity on the right multiplied by a factor<math>\frac{1}{\hbar^2}</math> is called effective mass tensor <math>\mathbf{M}(\mathbf{k})</math><ref name=":5">{{Harvnb|Ashcroft|Mermin|1976|p=228}}</ref> and we can use it to write a semi-classical equation for a [[charge carrier]] in a band<ref name=":6">{{Harvnb|Ashcroft|Mermin|1976|p=229}}</ref>
{{Equation box 1
|indent=:
|title='''Second order semi-classical equation of motion for a [[charge carrier]] in a band'''
|equation=<math>
\mathbf{M}(\mathbf{k}) \mathbf{a} = \mp e \left(\mathbf {E} + \mathbf{v}(\mathbf{k}) \times \mathbf{B}\right)
</math>
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where <math>\mathbf{a}</math> is an [[acceleration]]. This equation is analogous to the [[Matter wave|de Broglie wave]] type of approximation<ref name=":7">{{Harvnb|Ashcroft|Mermin|1976|p=227}}</ref>
{{Equation box 1
|indent=:
|title='''First order semi-classical equation of motion for electron in a band'''
|equation=<math>
\hbar \dot{k} = - e \left(\mathbf {E} +  \mathbf{v} \times \mathbf{B}\right)
</math>
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As an intuitive interpretation, both of the previous two equations resemble formally and are in a semi-classical analogy with [[Newton's laws of motion#Newton's second law|Newton's second law]] for an electron in an external [[Lorentz force]].