## Current densities

The electrical current density is the charge $-e$ associated with an electron state $\vec{k}$, times the group velocity of that state $\vec{v}_{\vec{k}}$, times the density of states per unit volume $D(\vec{k})$, times the probability density function $f(\vec{k})$ which gives the probability that state $\vec{k}$ is occupied, summed over all $\vec{k}$ states,

$$\vec{j}_{\text{elec}}= -e\int \vec{v}_{\vec{k}}D(\vec{k})f(\vec{k})d^3k.$$

The group velocity is given by,

$$\vec{v}_{\vec{k}} = \frac{\nabla_{\vec{k}}E(\vec{k})}{\hbar}.$$

The density of states tell us how many states with wave number $\vec{k}$ there are per unit volume,

$$D(\vec{k}) = \frac{2}{(2\pi)^3}.$$

In the relaxation time approximation in the absence of a magnetic field, the probability density function is,

$$f(\vec{k},\vec{r}) \approx f_0(\vec{k},\vec{r})- \frac{\tau (\vec{k})}{\hbar} \frac{\partial f_0}{\partial \mu} \nabla_{\vec{k}}E(\vec{k})\cdot\left(e\vec{E}+ \nabla_{\vec{r}}\mu+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T\right).$$

The electrical current density is thus,

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

The term that integrates over the Fermi function vanishes because in thermal equilibrium the occupation of a state $\vec{k}$ is the same as the occupation of state $-\vec{k}$.

$$\int \frac{1}{\hbar}\nabla_{\vec{k}}E(\vec{k}) f_0(\vec{k},\vec{r}) d^3k =\int \vec{v}(\vec{k})f_0(\vec{k},\vec{r}) d^3k =0.$$

Thus the current density is,

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

Similarly, the particle current is,

$$\vec{j}_n= \int \vec{v}_{\vec{k}}D(\vec{k})f(\vec{k})d^3k,$$ $$\vec{j}_n= -\frac{1}{4\pi^3\hbar^2}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}\nabla_{\vec{k}}E(\vec{k})\left( \nabla_{\vec{k}}E(\vec{k})\cdot\left(e\vec{E}+ \nabla_{\vec{r}}\mu+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T \right) \right) d^3k.$$

The energy current density is,

$$\vec{j}_U= \int \vec{v}_{\vec{k}}E(\vec{k})D(\vec{k})f(\vec{k})d^3k.$$ $$\vec{j}_U= -\frac{1}{4\pi^3\hbar^2}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}E(\vec{k})\nabla_{\vec{k}}E(\vec{k})\left( \nabla_{\vec{k}}E(\vec{k})\cdot\left(e\vec{E}+ \nabla_{\vec{r}}\mu+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T \right) \right) d^3k.$$

Finally, the electronic component to the thermal current is,

$$\vec{j}_Q= \int \vec{v}_{\vec{k}}\left( E(\vec{k}) - \mu \right) D(\vec{k})f(\vec{k})d^3k.$$ $$\vec{j}_Q= -\frac{1}{4\pi^3\hbar^2}\int \tau (\vec{k}) \frac{\partial f_0}{\partial \mu}\left( E(\vec{k}) - \mu \right)\nabla_{\vec{k}}E(\vec{k})\left( \nabla_{\vec{k}}E(\vec{k})\cdot\left(e\vec{E}+ \nabla_{\vec{r}}\mu+\frac{E(\vec{k})-\mu}{T}\nabla_{\vec{r}}T\right) \right) d^3k.$$

The relaxation time $\tau(\vec{k})$ can be calculated using Fermi's golden rule of time dependent perturbation theory. However, often it is assumed that the same relaxation time can be used for all $\vec{k}$ states and $\tau$ is pulled out of the integral.