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Advanced Solid State Physics | |
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Free-electron model: electrical conductivityThe dispersion relation in the free electron model is, \begin{equation} E(\vec{k})= \frac{\hbar^2 k^2}{2m^*}. \end{equation}Here $m^*$ is the effective mass. The gradient in $k$ of the dispersion relation is, \begin{equation} \nabla_{\vec{k}} E(\vec{k})= \frac{\hbar^2}{m^*}\left(k_x\hat{x}+k_y\hat{y}+k_z\hat{z}\right). \end{equation}For an isotropic system, the electrical conductivity is, \begin{equation} \sigma =\frac{e^2}{4\pi^3\hbar^2}\int \tau(\vec{k}) \frac{\partial f_0}{\partial \mu}\left( \nabla_{\vec{k}}E(\vec{k})\cdot \hat{z}\right)^2 d^3k. \end{equation}Using the expression for the gradient of the dispersion relation, \begin{equation} \sigma =\frac{\hbar^2e^2}{4\pi^3m^{*2}}\int \tau(\vec{k}) \frac{\partial f_0}{\partial \mu} k_z^2 d^3k. \end{equation}The differential volume is, \begin{equation} d^3k =k^2\sin\theta dk d\theta d\varphi \qquad k_z = k\cos\theta, \end{equation} \begin{equation} \sigma =\frac{\hbar^2e^2}{4\pi^3m^{*2}}\int \tau(k) \frac{\partial f_0}{\partial \mu} k^4\cos^2\theta \sin\theta dk d\theta d\varphi. \end{equation}The integral over $\varphi$ contributes a factor of $2\pi$. \begin{equation} \sigma =\frac{\hbar^2e^2}{2\pi^2m^{*2}}\int \tau(k) \frac{\partial f_0}{\partial \mu} k^4\cos^2\theta \sin\theta dk d\theta . \end{equation}The integral over $\theta$ contributes a factor of $2/3$. \begin{equation} \sigma =\frac{\hbar^2e^2}{3\pi^2m^{*2}}\int \tau(k) \frac{\partial f_0}{\partial \mu} k^4 dk . \end{equation}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^2}{2m^*}(k^2-k_F^2)}{k_BT}, \end{equation}and $\mu = \hbar^2k_F^2/2m^*$. Differentiating to find $dk$ \begin{equation} dk= \frac{m^*k_BT}{\hbar^2k}dx \end{equation}The conductivity can be written as, \begin{equation} \sigma =\frac{e^2}{3\pi^2m^{*}}\int \limits_{\frac{-\mu}{k_BT}}^{\infty}\tau(k) \frac{\exp(x)}{\left(\exp(x)+1\right)^2} k^3 dx. \end{equation}At low temperatures, the derivative of the Fermi function is sharply peaked around the Fermi energy so $k$ is approximately $k_F$ and $\tau (k_F) k_F^3$ can be pulled out of the integral. The remaining integral over $x$ evaluates to 1. \begin{equation} \sigma =\frac{e^2\tau (k_F) k_F^3}{3\pi^2m^{*}}. \end{equation}For free electrons, \begin{equation} n =\frac{k_F^3}{3\pi^2}. \end{equation}In terms of the electron density, the electrical conductivity is, \begin{equation} \sigma =\frac{ne^2\tau}{m^{*}}. \end{equation} |