Free electron model: Thermoelectric coefficient, Seebeck coefficient

Here $m^*$ is the effective mass. The thermoelectric coefficient $\kappa$ relates describes the current density that flows when a temperature gradient is applied, $\vec{j}_{elec}=\kappa\nabla T$. For an isotropic system, the thermoelectric coefficient is,

$$\kappa=\frac{e}{3\pi^2T}\int\limits_0^{\infty} \tau(\vec{k}) \frac{\partial f_0}{\partial \mu} \left( E(\vec{k}) - \mu \right) \left|\vec{v}_{\vec{k}}\right|^2 k^2 dk.$$

The derivative of the Fermi function is,

\begin{equation} \frac{\partial f_0}{\partial \mu}= \frac{\exp(x)}{k_BT\left(\exp(x)+1\right)^2} \end{equation}

where

\begin{equation} x= \frac{E-\mu}{k_BT}=\frac{\frac{\hbar^2k^2}{2m^*}-\mu}{k_BT}. \end{equation}

Differentiating to find $dk$

\begin{equation} dk= \frac{m^*k_BT}{\hbar^2k}dx \end{equation}

The thermoelectric coefficient can be written as,

$$\kappa=\frac{e}{3\pi^2T}\int\limits_0^{\infty} \tau(\vec{k}) \frac{\partial f_0}{\partial \mu} \left( E(\vec{k}) - \mu \right) \left|\vec{v}_{\vec{k}}\right|^2 k^2 dk.$$ $$\kappa=\frac{ek_BT}{3\pi^2T}\int\limits_0^{\infty} \tau(\vec{k}) \frac{\exp(x)}{k_BT\left(\exp(x)+1\right)^2} x \left(\frac{\hbar k}{m^*}\right)^2 k^2 \frac{m^*k_BT}{\hbar^2k}dx.$$ $$\kappa=\frac{ek_B}{3\pi^2m^*}\int\limits_0^{\infty} \tau(\vec{k}) \frac{x\exp(x)}{\left(\exp(x)+1\right)^2} k^3 dx.$$

At low temperatures, the derivative of the Fermi function is sharply peaked around the Fermi energy so $k$ is approximately $k_F$ and it can be pulled out the the integral. The remaining integral over $x$ evaluates to $\ln(2)$.

$$\kappa=\frac{\ln(2) ek_B\tau k_F^3 }{3\pi^2m^*}.$$

For free electrons,

\begin{equation} n =\frac{k_F^3}{3\pi^2}. \end{equation}

The thermoelectric coefficient in the free electron model is,

$$\kappa=\frac{\ln(2) ek_B\tau n }{m^*}.$$

Since $\sigma S = -\kappa$, the Seebeck coefficient in the free electron model is,

$$S = -\frac{\ln(2)k_B}{e} \approx -60\,\mu\text{V/K}.$$