Derivation of the Drude model conductivity from the Boltzmann transport equation

Question

The electric conductivity in the Drude model according to Wikipedia is \begin{align*} \sigma&=\frac{n q^2\tau}{m}. \end{align*} I'm trying to derive this from the Boltzmann equation in the relaxation time approximation, \begin{align*} \frac{\partial f}{\partial t}+\boldsymbol{v}\cdot\frac{\partial f_{}}{\partial \boldsymbol{r}}+\boldsymbol{F}\cdot\frac{\partial f}{\partial \boldsymbol{p}}&=-\frac{f-f_{\text{eq}}}{\tau}, \end{align*} where $f$ is the probability distribution, $f_{\text{eq}}$ is the equilibrium distribution, $\tau$ is the relaxation time, and $\boldsymbol{E}=\boldsymbol{F}/q$ is the electric field. Assuming a stationary situation and to first order in the electric field, this reduces to \begin{align*} f(\boldsymbol{p})=f_{\text{eq}}({p})-\tau q\boldsymbol{E}\cdot \boldsymbol{\nabla}_{\boldsymbol{p}} f_{\text{eq}}({p}). \end{align*} I use that the electrons obey Fermi-Dirac statistics in the low-temperature limit, \begin{align*} f_{\text{eq}}({p})=\frac{2}{h^3}H\left(\epsilon_F-p^2/2m\right), \end{align*} where $H$ is the Heaviside function, and the Fermi energy $\epsilon_F$ is given by the normalization condition using the density $n$, \begin{align*} n&=\int \text{d}^3p\ f_{\text{eq}}({p})=\frac{8\pi}{3h^3}\left(2m\epsilon_F\right)^{3/2} = \frac{8\pi}{3h^3}p_F^3 , \end{align*} where I let $p_F=\sqrt{2m\epsilon_F}$. This gives \begin{align*} f(\boldsymbol{p})=f_{\text{eq}}({p})+\frac{2\tau q}{mh^3}\boldsymbol{E}\cdot\boldsymbol{p} \ \delta\left(\epsilon_F-p^2/2m\right). \end{align*}

Now, the electric current is, using this property of the Dirac delta, \begin{align*} \boldsymbol{J}&=\frac{q}{m}\langle \boldsymbol{p}\rangle=\frac{2\tau q^2}{m^2h^3}\int\text{d}^3p\ (\boldsymbol{E}\cdot\boldsymbol{p}) \boldsymbol{p} \ \delta\left(\epsilon_F-p^2/2m\right)=\frac{2\tau q^2}{m^2h^3}\int\text{d}^3p\ (\boldsymbol{E}\cdot\boldsymbol{p}) \boldsymbol{p} \frac{\delta\left(p-p_F\right)}{\left|-p_F/m\right|}. \end{align*} The Dirac delta distribution further reduces this to the surface integral over a sphere, and using the expression for $n$ from before, I get \begin{align*} \boldsymbol{J}&=\frac{2\tau q^2}{mh^3}\frac{1}{p_F}\int_{p=p_F} p_F^2\boldsymbol{E}\text{d}S=\frac{2\tau q^2}{mh^3}p_F\boldsymbol{E}4\pi p_F^2=\frac{3nq^2\tau }{m}\boldsymbol{E}=\sigma\boldsymbol{E}, \end{align*} which is a factor $3$ off compared to the Drude model. I redid this calculation a few times and can't find anything off, so help is appreciated!

Answer 1

Choosing a coordinate system where $\vec{E}= \left|\vec{E}\right| \vec{e}_z$ and switching to spherical coordinates for the integration variable $\vec{p}= p \, (\sin \theta \cos \phi \, \vec{e}_x +\sin \theta \sin \phi \, \vec{e}_y + \cos \theta \, \vec{e}_z)$ with $0 \le p \le \infty$, $\, 0\le \theta \le \pi$ and $0 \le \phi <2\pi$, one finds \begin{align} & \int\limits_{\mathbb{R}^3} \! d^3p \, (\vec{E} \cdot \vec{p})\, \vec{p} \, \delta(\epsilon_F-\vec{p}^2/2m) \\=&\int\limits_0^\infty \!\!dp \, p^2 \!\!\int\limits_0^\pi \!\! d \theta \, \sin \theta \! \int\limits _0^{2\pi}\!\! d\phi \, (|\vec{E}|\, p \, \cos \theta) \, p \, (\sin \theta \cos \phi \, \vec{e}_x\!+\!\sin \theta \sin \phi \, \vec{e}_y\!+\! \cos \theta \, \vec{e}_z) \, \delta\!\left(\!\epsilon_F-\frac{p^2}{2m}\!\right)\\ =&\, 2 \pi|\vec{E}| \, \vec{e}_z \!\int\limits_0^\infty \! \! dp \, p^4 \, \delta(\epsilon_F-p^2/2m)\!\int\limits_0^\pi \! \! d\theta \, \sin \theta \, \cos^2 \theta \\ =& \,2\pi m \vec{E} \,\underbrace{\int\limits_0^\infty \! dp^2 \, p^3\, \delta(p^2-2m \epsilon_F)}_{(2m \epsilon_F)^{3/2}} \; \;\; \underbrace{\int\limits_{-1}^1 \!du \, u^2}_{2/3}\\ =& \, \frac{4\pi m p_F^3}{3} \vec{E} \end{align}

Answer 2

Let $\chi=-\cos\vartheta$, which has the convenient result that $\mathrm d\chi=\sin\vartheta\,\mathrm d\vartheta$, no need to reverse the order of integration.

The second last equation of yours ended with $$ \begin{align} \tag1\vec J&=\frac{2q^2\tau}{m^2h^3}\int_0^{2\pi}\int_{-1}^{+1}\int_0^\infty\left(\vec E\cdot\vec p\right)\vec p\frac{\delta(p-p_F)}{|-p_F/m|}p^2\,\mathrm dp\,\mathrm d\chi\,\mathrm d\varphi \end {align} $$ You moved too fast for the next step. Let the direction of the $\vec E$ field define the $z$-axis from which the angle $\vartheta$ is measured from. This means that $\vec E\cdot\vec p$ gives a factor of $-\chi$ and $\vec p$ itself also gives a factor of $-\chi$. This means that the next step is actually $$ \begin{align} \tag2\vec J&=\frac{2q^2\tau}{mh^3}\frac1{p_F}\cdot E\cdot\hat{\vec z}\cdot\int_0^{2\pi}\int_{-1}^{+1}p_F^2\,\chi^2\,p_F^2\,\mathrm d\chi\,\mathrm d\varphi\\ \tag3&=\frac{2q^2\tau}{mh^3}\cdot\vec E\cdot\frac{4\pi}3p_F^3=\frac{nq^2\tau}m\vec E=\sigma\vec E \end {align} $$ without the errant factor of 3.

Note that this is very much sensible. If you did not have the $\vec E\cdot\vec p$ term, the lack of the $\chi$ term from there would lead to be an odd integrand in even integral, leading to a spurious zero. If you did not have the $\vec p$ term, again, same spurious zero. In any direction perpendicular to the $\vec E$ field, not only would the $\varphi$ integral send the integral to zero, there would also be $\sqrt{1-\chi^2}=\sin\vartheta$ in the integration, and again you have odd integrand of even integral, guaranteeing a zero too. It is only when you get everything correct that you obtain the result that agrees with both theory and experiment.