1. Boltzmann equation

When using the Boltzmann equation to calculate currents, a function $f$ is defined which gives the probability that state $\vec{k}$ is occupied at position $\vec{r}$ and time $t$.

1. Give an expression for the electrical current density in terms of $f$ and the density of states $D(\vec{k)}$.
2. Write down the Boltzmann equation that must be solved to find $f$. (Hint: Take the total derivative of $f$.)
3. What is the meaning of the collision term in the Boltzmann equation?

Solution

1. Give an expression for the electrical current density in terms of $\boldsymbol{f}$ and the density of states $\boldsymbol{D(\vec{k})}$.

   The electrical current density is defined as

   \begin{equation*} \vec{j}_{elec}=-e \int \vec{v}(\vec{k})D(\vec{k})f(\vec{k})\,d\vec{k} = -\frac{e}{4\pi^3}\int\vec{v}(\vec{k})f(\vec{k})\,d\vec{k} \quad\text{with}\quad D(\vec{k})=\frac{2}{(2\pi)^3}. \end{equation*}

   $D(\vec{k})$ is the density of states in $\vec{k}$ space per unit volume where the factor of two takes the spin degeneracy into account. This general expression for the current density is basically the sum over all velocities times the occupation probability for each $\vec{k}$ state. The velocity $\vec{v}(\vec{k})$ can be obtained from the dispersion relation $E(\vec{k})$ and in particular for the free electron model $\vec{v}(\vec{k})$ is given by

   \begin{equation*} E(\vec{k})=\frac{\hbar^2\vec{k}^2}{2m}, \,\,\,\, \vec{v}(\vec{k}) =\frac{1}{\hbar} \nabla_k E(\vec{k}) = \frac{\hbar\vec{k}}{m}. \end{equation*}

2. Write down the Boltzmann equation that must be solved to find $\boldsymbol{f}$. (Hint: Take the total derivative of $\boldsymbol{f}$.)

   In a steady state there is no change in $f(\vec{k},\vec{r},t)$ and thus the total derivative of the probability distribution $f(\vec{k},\vec{r},t)$

   \begin{align*} \frac{df(\vec{k},\vec{r},t)}{dt} &= \frac{\partial f}{\partial t}+\frac{\partial f}{\partial k_x}\frac{dk_x}{dt}+\frac{\partial f}{\partial k_y}\frac{dk_y}{dt}+\frac{\partial f}{\partial k_z}\frac{dk_z}{dt}+\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z} \frac{dz}{dt}\\ &= \frac{\partial f}{\partial t} + \frac{d\vec{k}}{dt} \cdot \nabla_k f +\frac{d\vec{r}}{dt}\cdot \nabla f \overset{!}{=}0\\ \end{align*}

   has to be zero. Additionally using

   \begin{equation*} \vec{v}=\frac{d\vec{r}}{dt}\quad\text{and}\quad \hbar \frac{d\vec{k}}{dt}=\vec{F}_{ext} \end{equation*}

   and introducing a collision term (for its meaning see below) leads to the Boltzmann equation

   \begin{equation*} \frac{\partial f}{\partial t}= - \frac{\vec{F}_{ext}\cdot\nabla_k f}{\hbar}-\vec{v}\cdot\nabla f +\frac{\partial f}{\partial t }\bigg \vert_{coll}. \end{equation*}

   The first term on the right hand side describes the effect on f due to external forces and fields (e.g. electromagnetic fields). The process of diffusion is described by the second term and the last term is the heuristically introduced collision term.

3. What is the meaning of the collision term in the Boltzmann equation?

   This term is introduced to take the transition between different $\vec{k}$ states into account. The simplest model is the relaxation time approximation

   \begin{equation*} \frac{\partial f}{\partial t }\bigg \vert_{coll} = \frac{f_0(\vec{k})-f(\vec{k})}{\tau(\vec{k})} \end{equation*}

   where $f_0$ is the equilibrium distribution function (Fermi-Dirac distribution) and $\tau$ denotes the relaxation time. The physical interpretation of this relaxation time is the time associated with the rate of return to the equilibrium distribution when all external fields or thermal gradients are turned off. A more sophisticated model for the collision term might invoke Fermi’s golden rule to describe the transition rates between different states.

2. Seebeck Effect

1. What is the electrochemical potential?
2. What rank tensor is the Seebeck coefficient?
3. How can you calculate the Seebeck coefficient?
4. If you wanted to calculate the Seebeck coefficient for a free electron gas, what dispersion relation would you use?

Solution

1. What is the electrochemical potential?

   The electrochemical potential $\tilde{\mu}$ is the sum of the electrostatic potential and the chemical potential,

   \begin{equation*} \tilde{\mu} = -e\phi + \mu \end{equation*}

   and thus corresponds to the total potential energy of a charged particle if an electric field and a concentration gradient are present. It is the energy required to add a charged particle (in this case an electron with charge $-e$) to the system at constant pressure, temperature and amount of other species. This quantity is actually measured by a voltmeter (not the electrostatic potential alone).

   See also http://lampz.tugraz.at/~hadley/ss2/transport/ecp.php.

2. What rank tensor is the Seebeck coefficient?

   The Seebeck coefficient relates the gradient of the electrochemical potential $\nabla\tilde{\mu}$ to the temperature gradient $\nabla T$:

   \begin{equation*} \nabla\tilde{\mu} = -S\nabla T \ . \end{equation*}

   Since both are vectors, the Seebeck coefficient must be a rank 2 tensor,

   \begin{equation*} \frac{\partial\tilde{\mu}}{\partial x_i} = -S_{ij}\frac{\partial T}{\partial x_j} \ . \end{equation*}

3. How can you calculate the Seebeck coefficient?

   The Boltzmann equation in the relaxation time approximation reads

   \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\mu+\frac{E(\vec{k})-\mu}{T}\nabla T +\frac{e}{\hbar}\nabla_{\vec{k}}E(\vec{k})\times\vec{B}\right) \end{equation*}

   and in terms of the distribution function $f$ the electrical current density is given by

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

   With

   \begin{equation*} \vec{v}_{\vec{k}} = \frac{1}{\hbar}\nabla_{\vec{k}}E(\vec{k}), \quad D(\vec{k}) = \frac{2}{(2\pi)^3}, \quad \vec{B} = 0, \quad \nabla\tilde{\mu} = \nabla(-e\phi + \mu) = e\vec{E} + \nabla\mu \end{equation*}

   and the distribution function from above we obtain

   \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(\nabla\tilde{\mu}+\frac{E(\vec{k})-\mu}{T}\nabla T \right) \right)d^3k \ . \end{equation*}

   Now consider a sample with finite length and open ends, then $\vec{j}_{\text{elec}}=0$ and assuming $\nabla\tilde{\mu} = -S\nabla T$ we arrive at the equation

   \begin{equation*} \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(-S\nabla T+\frac{E(\vec{k})-\mu}{T}\nabla T \right) \right)d^3k = 0 \ . \end{equation*}

   Now the elements of the Seebeck tensor $S$ can be calculated by inserting unit vectors for $\nabla T$ (the equation must be valid for any $\nabla T$) and numerically solving the implicit equations for $S$. There are 9 equations (1 vector equation = 3 scalar equations for each unit vector) for the 9 elements of $S$. All other quantities in the equation are known or can be calculated with other methods: $E(\vec{k})$ is the dispersion relation, $f_0$ the Fermi function and $\tau(\vec{k})$ the relaxation time.

   See also http://lampz.tugraz.at/~hadley/ss2/transport/seebeck.php

4. If you wanted to calculate the Seebeck coefficient for a free electron gas, what dispersion relation would you use?

   The dispersion relation for a free electron gas is

   \begin{equation*} E(\vec{k}) = \frac{\hbar^2 k^2}{2m^*} \ , \end{equation*}

   where $m^*$ is the effective mass.

3. Nernst effect

1. In the Nernst effect, a sample is placed in a magnetic field and a temperature gradient that are perpendicular to each other. The electric field in the direction perpendicular to both the magnetic field and the temperature gradient is measured. How could the Boltzmann equation be used to calculate this electric field?
2. How does the Nernst effect depend on the electron and phonon dispersion relations?
3. What is the relaxation time approximation? What changes in a relaxation time?

Solution

1. In the Nernst effect, a sample is placed in a magnetic field and a temperature gradient that are perpendicular to each other. The electric field in the direction perpendicular to both the magnetic field and the temperature gradient is measured. How could the Boltzmann equation be used to calculate this electric field?

   Starting from the Boltzmann equation, in principle the derivation is the same as for the Seebeck effect in question 2, but to include magnetic effects in the relaxation time approximation higher orders in $\tau$ must be taken into account. Again setting $\vec{j}_{elec}=0$ in

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

   the electric field (or more precisely the gradient of the electrochemical potential $\nabla\tilde{\mu}$) is given implicitly in terms of the temperature gradient and magnetic field (these quantities enter through the distribution function $f(\vec{k})$).

2. How does the Nernst effect depend on the electron and phonon dispersion relations?

   The electron dispersion relation appears in the group velocity $\vec{v}_k = \nabla_kE(\vec{k})$ as well as in the distribution function $f(\vec{k})$ obtained from the Boltzmann equation. The phonon dispersion does not directly appear in this description of the Nernst effect, but it may have an influence on the scattering time $\tau$.

3. What is the relaxation time approximation? What changes in a relaxation time?

   Transport phenomena are described by the Boltzmann equation

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

   The collision term contains information about all scattering events and an exact treatment would be very complicated. In the relaxation time approximation the simple expression

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

   is used for the collision term, where $f_0$ is the Fermi function.

   Without external forces and for a uniform distribution function $f$ the Boltzmann equation then reads

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

   and for the initial condition $f(\vec{k},t=0) = f_0(\vec{k}) + \delta f(\vec{k})$ has the solution

   \begin{equation*} f(\vec{k},t) = f_0(\vec{k}) + \delta f(\vec{k}) \mathrm{e}^{-t/\tau(\vec{k})}. \end{equation*}

   Thus, $\tau$ is the time scale on which the system returns to equilibrium after a perturbation.