
Advanced Solid State Physics  

Thermoelectric effectsThermoelectric effects describe the electrical currents and heat currents that flow when electric fields, magnetic fields and thermal gradients are applied to a sample. These effects are used in thermoelectric generators to convert heat into electricity, they are used in Peltier coolers to pump heat using an electrical current, and they are used in thermocouples to measure temperature. We will use the Boltzmann equation to describe the electrical and heat currents that arise when electric fields, magnetic fields, or thermal gradients are present. This includes phenomena such as the electrical conductivity, the thermal conductivity, the Hall effect, the Seebeck effect, the Peltier effect, and the Nernst effect. We will see that all of these effects are related to the band structure of the material. In a band structure calculation, for every state $\vec{k}$ there is a state $\vec{k}$ with the same energy. In thermal equilibrum, the occupation of these states is given by the Fermi function. Since the energy of states $\vec{k}$ and $\vec{k}$ are the same, they both have the same ocupation probability. In other words, there are as many right moving states as left moving states and the net current is zero. If an electric field is applied, the occupation of the left and right moving states can be different and the electrical current density is given by an integral 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}Here $e$ is the charge of an electron, $\vec{v}_{\vec{k}}$ is the group velocity of an electron in state $\vec{k}$, $D(\vec{k})$ is the density of states in $\vec{k}$ per unit volume, and $f(\vec{k})$ is the probability density function which gives the probability that state $\vec{k}$ is occupied. Similarly, the particle current density is, \begin{equation} \vec{j}_n= \int \vec{v}_{\vec{k}}D(\vec{k})f(\vec{k})d^3k, \end{equation}and 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}The first law of thermodynamics states that the change in internal energy is, \begin{equation} dU= dQdW+\mu dN. \end{equation}Here $dU$ is change in the internal energy, $dQ$ is the change in the heat energy, $dW$ is the work performed by the system, $\mu dN$ is the chemical potential times the change in particle number. Under steadystate conditions no work is performed and $dQ = dU  \mu dN$. The electrical contribution to the heat current density is thus $j_Q = j_U \mu j_n$, \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}To perform these integrals we must determine $\vec{v}_{\vec{k}}$, $D(\vec{k})$, and $f(\vec{k})$. The allowed $\vec{k}$ states are uniformly distributed in reciprocal space do the density of states is simply, \begin{equation} D(\vec{k}) = \frac{2}{(2\pi)^3}. \end{equation}The factor of 2 in the numerator is for spin. The group velocity can be calculated from the electron dispersion relation, \begin{equation} \vec{v}_{\vec{k}} = \frac{\nabla_{\vec{k}}E(\vec{k})}{\hbar}. \end{equation}The probability density function $f(\vec{k})$ that describes the occupation of the $\vec{k}$ states is the more difficult to calculate and will be explained in the following sections. 