Advanced Solid State Physics

Outline

Electrons

Magnetic effects and
Fermi surfaces

Magnetism

Linear response

Transport

Crystal Physics

Electron-electron
interactions

Quasiparticles

Structural phase
transitions

Landau theory
of second order
phase transitions

Superconductivity

Quantization

Photons

Exam questions

Appendices

Lectures

Books

Course notes

TUG students

      

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,

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

The group velocity is given by,

\begin{equation} \vec{v}_{\vec{k}} = \frac{\nabla_{\vec{k}}E(\vec{k})}{\hbar}. \end{equation}

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

\begin{equation} D(\vec{k}) = \frac{2}{(2\pi)^3}. \end{equation}

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

\begin{equation} 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). \end{equation}

The electrical current density is thus,

\begin{equation} \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. \end{equation}

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 $-\vec{k}$.

\begin{equation} \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. \end{equation}

Thus the current density is,

\begin{equation} \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. \end{equation}

Similarly, the particle current is,

\begin{equation} \vec{j}_n= \int \vec{v}_{\vec{k}}D(\vec{k})f(\vec{k})d^3k, \end{equation} \begin{equation} \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. \end{equation}

The energy current density is,

\begin{equation} \vec{j}_U= \int \vec{v}_{\vec{k}}E(\vec{k})D(\vec{k})f(\vec{k})d^3k. \end{equation} \begin{equation} \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. \end{equation}

Finally, the electronic component to the thermal current is,

\begin{equation} \vec{j}_Q= \int \vec{v}_{\vec{k}}\left( E(\vec{k}) - \mu \right) D(\vec{k})f(\vec{k})d^3k. \end{equation} \begin{equation} \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. \end{equation}

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.