## Electrical conductivity

The electrical current density is,

$$\vec{j}_{\text{elec}}= \frac{e}{4\pi^3\hbar^2}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}\nabla_{\vec{k}}E(\vec{k})\left( \nabla_{\vec{k}}E(\vec{k})\cdot\left(\nabla_{\vec{r}}\tilde{\mu}+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T \right) \right) d^3k.$$

If there is no temperature gradient, the current density is,

$$\vec{j}_{\text{elec}}= \frac{e}{4\pi^3\hbar^2}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}\nabla_{\vec{k}}E(\vec{k})\left( \nabla_{\vec{k}}E(\vec{k})\cdot\nabla_{\vec{r}}\tilde{\mu} \right) d^3k.$$

Here $-\nabla_{\vec{r}}\tilde{\mu}$ is the force driving the electrons. It has an electrostatic component $-e\vec{E}$ and concentration component $-\nabla_{\vec{r}}\mu$. Usually the conductivity matrix is defined as the relationship between the current density and the electric field,

$$$\left[ \begin{array}{cccc} j_x \\ j_y \\ j_z \\ \end{array} \right]= \left[ \begin{array}{cccc} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \\ \end{array} \right]\left[ \begin{array}{cccc} E_x \\ E_y \\ E_z \\ \end{array} \right].$$$

An equivalent definition would be,

$$$\left[ \begin{array}{cccc} j_x \\ j_y \\ j_z \\ \end{array} \right]=\frac{1}{e} \left[ \begin{array}{cccc} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \\ \end{array} \right]\left[ \begin{array}{cccc} \frac{\partial \tilde{\mu}}{\partial x} \\ \frac{\partial \tilde{\mu}}{\partial y} \\ \frac{\partial \tilde{\mu}}{\partial z} \\ \end{array} \right].$$$

This second form emphasizes the fact that currents flow not only because of electric fields but also because of concentration gradients. The elements of the conductivity matrix can be determined by choosing the driving force to be the unit vectors $\hat{x}$, $\hat{y}$, and $\hat{z}$. We see that the symmetry of the conductivity matrix depends on the band structure $E(\vec{k})$ of the material.

$$$\sigma_{ij}=\frac{e^2}{4\pi^3\hbar^2}\int \tau(\vec{k}) \frac{\partial f_0}{\partial \mu}\nabla_{\vec{k}}E(\vec{k})\cdot \hat{e}_i\left( \nabla_{\vec{k}}E(\vec{k})\cdot \hat{e}_j\right) d^3k.$$$

Here $\hat{e}_i$ are the unit vectors $i = [x,y,z]$. For cubic crystals the electrical conductivity is a constant,

$$$\sigma =\frac{e^2}{4\pi^3\hbar^2}\int \tau(\vec{k}) \frac{\partial f_0}{\partial \mu}\left( \nabla_{\vec{k}}E(\vec{k})\cdot \hat{z}\right)^2 d^3k.$$$

This can be written in terms of the velocity using $\vec{v}_{\vec{k}} = \frac{\nabla_{\vec{k}}E(\vec{k})}{\hbar}$.

$$\sigma =\frac{e^2}{4\pi^3}\int \tau(\vec{k}) \frac{\partial f_0}{\partial \mu}\left( \vec{v}_{\vec{k}}\cdot \hat{z}\right)^2 d^3k.$$

The differnential volume is $d^3k =k^2\sin\theta dk d\theta d\varphi$.

$$\sigma =\frac{e^2}{4\pi^3}\int \tau(\vec{k}) \frac{\partial f_0}{\partial \mu}\left| \vec{v}_{\vec{k}}\right|^2 \cos^2\theta k^2\sin \theta dk d\theta d\phi.$$

The integral over $\phi$ contributes a factor of $2\pi$ and the integral over $\theta$ contributes a factor of $2/3$.

$$\sigma =\frac{e^2}{3\pi^2}\int\limits_0^{\infty} \tau(\vec{k}) \frac{\partial f_0}{\partial \mu}\left| \vec{v}_{\vec{k}}\right|^2 k^2 dk.$$