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

      

Relaxation time approximation

The Boltzmann equation that describes the probability density function for electrons in a solid is,

\begin{equation} \frac{\partial f}{\partial t} = - \frac{1}{\hbar}\vec{F}_{\text{ext}}\cdot\nabla_{\vec{k}}f-\vec{v}\cdot\nabla_{\vec{r}}f + \frac{\partial f}{\partial t} \bigg\rvert_{collisions}. \end{equation}

If an external force is applied, the system will move out of equilibrium. If this external force is removed, the collision term will cause scattering events that return the system to equilibrium. In the relaxation time approximation, the collision term is assumed to have the form,

\begin{equation} \frac{\partial f(\vec{k})}{\partial t} \bigg\rvert_{collision} = \frac{f_0(\vec{k}) - f(\vec{k})}{\tau(\vec{k})}. \end{equation}

Here $f_0$ is the Fermi function,

\begin{equation} f_0(\vec{k}) = \frac{1}{1+\exp\left(\frac{E(\vec{k})-\mu}{k_BT}\right)}. \end{equation}

This form for the collision term ensures that the probability density function returns to the Fermi function if all forces are removed. If there are no external forces and the electrons are uniformly distributed in space $\nabla_{\vec{r}}f = 0$, then the Boltzmann equation becomes,

\begin{equation} \frac{\partial f(\vec{k})}{\partial t} = \frac{f_0(\vec{k}) - f(\vec{k})}{\tau(\vec{k})}. \end{equation}

The solution to this equation is,

\begin{equation} f(\vec{k},t) = \left(f(\vec{k},t=0)-f_0(\vec{k})\right)\exp(-t/\tau(\vec{k})) +f_0(\vec{k}). \end{equation}

This equation describes a system that starts in some intitial non-equilibrium probability density function $f(\vec{k},t=0)$ at time $t=0$ and returns to equilibrium in a relaxation time $\tau(\vec{k})$. Some $\vec{k}$ states decay more quickly than others. Typically the states far from the Fermi energy decay the fastest. To include the possibility that there is a temperature gradient or a concentration gradient in the problem, the temperature and the chemical potential can depend on $\vec{r}$.

\begin{equation} \frac{\partial f(\vec{k},\vec{r},t)}{\partial t} = - \frac{1}{\hbar}\vec{F}_{\text{ext}}\cdot\nabla_{\vec{k}}f(\vec{k},\vec{r},t)-\vec{v}\cdot\nabla_{\vec{r}}f(\vec{k},\vec{r},t) + \frac{f_0(\vec{k},\vec{r}) - f(\vec{k},\vec{r},t)}{\tau(\vec{k})}. \end{equation}

Often we consider steady-state conditions where the external fields are held constant for a much longer time than the relaxation time. Under steady-state conditions, $\frac{\partial f}{\partial t}=0$ and we can solve for $f(\vec{r},\vec{k},t)$,

\begin{equation} f(\vec{k},\vec{r}) = f_0(\vec{k},\vec{r})- \tau (\vec{k}) \left( \frac{1}{\hbar}\vec{F}_{\text{ext}}\cdot\nabla_{\vec{k}}f(\vec{k},\vec{r})+\vec{v}\cdot\nabla_{\vec{r}}f(\vec{k},\vec{r}) \right) . \end{equation}

If the system is close to equilibrium then $f(\vec{k},\vec{r})$ will be nearly the same as $f_0(\vec{k},\vec{r})$. We can treat the relaxation time as a small parameter. To first order in $\tau$, $f(\vec{k},\vec{r})$ is,

\begin{equation} f(\vec{k},\vec{r}) \approx f_0(\vec{k},\vec{r})- \tau (\vec{k}) \left( \frac{1}{\hbar}\vec{F}_{\text{ext}}\cdot\nabla_{\vec{k}}f_0(\vec{k},\vec{r})+\vec{v}\cdot\nabla_{\vec{r}}f_0(\vec{k},\vec{r}) \right) . \end{equation}

If the system is spatially uniform so that it does not depend on $\vec{r}$, the probability density function function can be written,

\begin{equation} f(\vec{k}) \approx f_0(\vec{k})- \tau (\vec{k}) \left( \frac{1}{\hbar}\vec{F}_{\text{ext}}\cdot\nabla_{\vec{k}}f_0(\vec{k})\right) \approx f_0\left(\vec{k}-\frac{\tau\vec{F}_{\text{ext}}}{\hbar}\right), \end{equation}

and the external force shifts the Fermi function $f_0$ away from its equilibrium position.

Returning to the case where spatial dependence is allowed, when taking the derivatives of Fermi function, we assume that the $\vec{k}$ dependence arises only through the dispersion relation $E(\vec{k})$ and that the temperature and the chemical potential can depend on the position $\vec{r}$. Here we assume that the dispersion relation does not depend on position so that the material is uniform in space. The derivatives of the Fermi function are,

\begin{equation} \nabla_{\vec{k}}f_0(\vec{k},\vec{r}) = \frac{\partial f_0}{\partial E}\nabla_{\vec{k}} E(\vec{k}), \end{equation} \begin{equation} \nabla_{\vec{r}}f_0(\vec{k},\vec{r}) = \frac{\partial f_0}{\partial T}\nabla_{\vec{r}} T + \frac{\partial f_0}{\partial \mu}\nabla_{\vec{r}} \mu, \end{equation}

where

\begin{equation} \frac{\partial f_0}{\partial \mu} = -\frac{\partial f_0}{\partial E} = \frac{T}{E(\vec{k})-\mu}\frac{\partial f_0}{\partial T}=\frac{\exp\left(\frac{E(\vec{k})-\mu}{k_BT}\right)}{k_BT\left( \exp\left(\frac{E(\vec{k})-\mu}{k_BT}\right) +1\right)^2}. \end{equation}

The partial derivatives $\frac{\partial f_0}{\partial \mu}$, $ = \frac{\partial f_0}{\partial E}$ and $\frac{\partial f_0}{\partial T}$ are all functions that are only non zero for energies $E(\vec{k})$ close to the chemical potential. This means that only the states near the Fermi surface will contribute to the transport properties.

$k_BT \large \frac{\partial f_0}{\partial \mu}$

$\large \frac{E-\mu}{k_BT}$

The external force on an electron is the Lorentz force $\vec{F}_{\text{ext}} = -e\vec{E}-e\vec{v}_{\vec{k}}\times\vec{B} = -e\vec{E}-\frac{e}{\hbar}\nabla_{\vec{k}}E(\vec{k})\times\vec{B}$. Here $\vec{B}$ is the magnetic field and the electric field $\vec{E}$ should not be confused with the dispersion relation $E(\vec{k})$. 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 +\frac{e}{\hbar}\nabla_{\vec{k}}E(\vec{k})\times\vec{B}\right). \end{equation}

The term, $\nabla_{\vec{k}}E(\vec{k})\cdot \left( \nabla_{\vec{k}}E(\vec{k})\times\vec{B}\right)$ is zero. This is a statement that only applying a magnetic field will not change the distribution of $k$-states away from equilibrium. In order to describe magnetic effects like the Hall effect or the Nerst effect it is necessary to include higher orders of $\tau$ in the calculation. The form can be used to calculate the electrical and thermal current densities in the absence of a magnetic field 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}