# Derivation of Kohn Effect

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1. May 31, 2015

### LucidLunatic

I'm working through a derivation of the [Kohn Effect](http://en.wikipedia.org/wiki/Kohn_effect), as presented in Ziman's 'Principles of the Theory of Solids.' However, I find myself getting a somewhat different result. The book states that calculating

$$\epsilon(\mathbf{q}, \omega) = 1+ \frac{4\pi e^2}{q^2}\sum_{\mathbf{k}} \frac{f_0(\mathbf{k})-f_0(\mathbf{k+q})}{E_{k+q}-E_{k}-\hbar\omega + i\hbar\alpha}$$

at zero temperature (and $\omega = 0$) grants

$$\epsilon(\mathbf{q}, 0) = 1+\frac{4\pi e^2}{q^2}\frac{n}{\frac{2}{3}\mathcal{E}_F}[\frac{1}{2}+\frac{4k_F^2-q^2}{8k_Fq}\ln|{\frac{2k_F+q}{2k_F-q}}|]$$

Here is my work:
$$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}\sum_{\mathbf{k}} \frac{f_0(\mathbf{k})-f_0(\mathbf{k+q})}{2\mathbf{k\cdot q}+q^2}$$

$$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}\sum_{\mathbf{k}}[ \frac{f_0(\mathbf{k}))}{2\mathbf{k\cdot q}+q^2}- \frac{f_0(\mathbf{k+q})}{2\mathbf{k\cdot q}+q^2} ]$$

We may exchange $\sum_\mathbf{k} = \frac{1}{(2\pi)^3}\int\,d^3\mathbf{k}$

$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}[ \int\,d^3k\frac{f_0(\mathbf{k}))}{2\mathbf{k\cdot q}+q^2}- \int\,d^3k\frac{f_0(\mathbf{k+q})}{2\mathbf{k\cdot q}+q^2} ]$
Given the fact that $f_0(\mathbf{k})$ is zero outside the Fermi sphere, and one inside it allows us to select bounds on the integrals. For the second integral, a change of variables to $u = k+q$ allows us to center the Fermi sphere.
$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}[\int_0^{2\pi}\int_0^\pi\int_0^{k_F}\,k^2\sin\theta\,d\phi\,d\theta\,dk\frac{f_0(\mathbf{k}))}{2\mathbf{k\cdot q}+q^2}-\int_0^{2\pi}\int_0^\pi\int_0^{k_F}\,u^2\sin\theta\,d\phi\,d\theta\,du\frac{f_0(\mathbf{u})}{2\mathbf{(u-q)\cdot q}+q^2} ]$

$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}[ \int_0^{2\pi}\int_0^\pi\int_0^{k_F}\,k^2\sin\theta\,d\phi\,d\theta\,dk\frac{1}{2\mathbf{k\cdot q}+q^2}-\int_0^{2\pi}\int_0^\pi\int_0^{k_F}\,u^2\sin\theta\,d\phi\,d\theta\,du\frac{1}{2\mathbf{(u\cdot q}-q^2} ]$

If we relabel our dummy integration variable $u$ as $k$, we may combine back into a single integral
$$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_0^{2\pi}\int_0^\pi\int_0^{k_F}\,k^2\sin\theta\,d\phi\,d\theta\,dk[\frac{1}{2\mathbf{k\cdot q}+q^2}-\frac{1}{2\mathbf{k\cdot q}-q^2} ]$$

We may select our coordinate system so that $\mathbf{q}$ lies along the $z$ axis.

$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}A\int_0^{2\pi}\int_0^\pi\int_0^{k_F}\,k^2\sin\theta\,d\phi\,d\theta\,dk[\frac{1}{2kq\cos\theta+q^2}-\frac{1}{2kq\cos\theta-q^2} ]$
$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_0^{2\pi}\int_0^\pi\int_0^{k_F}\,k^2\sin\theta\,d\phi\,d\theta\,dk[\frac{2kq\cos\theta-q^2}{(2kq\cos\theta)^2-q^4}-\frac{2kq\cos\theta+q^2}{(2kq\cos\theta)^2)-q^4} ]$
$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_0^{2\pi}\int_0^\pi\int_0^{k_F}\,k^2\sin\theta\,d\phi\,d\theta\,dk[\frac{-2q^2}{(2kq\cos\theta)^2-q^4} ]$
$\epsilon(\mathbf{q}, 0) = 1- \frac{8\pi e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_0^{2\pi}\int_0^\pi\int_0^{k_F}\,\,d\phi\,d\theta\,dk[\frac{k^2\sin\theta}{(2k\cos\theta)^2-q^2} ]$
$\epsilon(\mathbf{q}, 0) = 1+ \frac{16\pi^2 e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_1^{-1}\int_0^{k_F}\,d\cos\theta\,dk[\frac{k^2}{(2k\cos\theta)^2-q^2} ]$
$\epsilon(\mathbf{q}, 0) = 1+ \frac{16\pi^2 e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_1^{-1}\int_0^{k_F}\,d\cos\theta\,dk[\frac{k^2}{(2k\cos\theta)^2-q^2} ]$
$\epsilon(\mathbf{q}, 0) = 1+ \frac{16\pi^2 e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_0^{k_F}\,dk[\frac{k(\tanh^{-1}(\frac{2k}{q})- \tanh^{-1}(\frac{-2k}{q})}{2q} ]$
Since $\tanh^{-1}(x) = \frac{1}{2}[\ln(1+x)-\ln(1-x)]$
$\epsilon(\mathbf{q}, 0) = 1+ \frac{8\pi^2 e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_0^{k_F}\,dk[\frac{k(\ln(\frac{q+2k}{q-2k}- \ln(\frac{q-2k}{q+2k})}{q} ]$
$\epsilon(\mathbf{q}, 0) = 1+ \frac{8\pi^2 e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_0^{k_F}\,dk[\frac{k\ln(\frac{q+2k}{q-2k})^2}{q} ]$
$\epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi^2 e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}\int_0^{k_F}\,dk[\frac{k\ln(\frac{q+2k}{q-2k})}{q} ]\\ \epsilon(\mathbf{q}, 0) = 1+ \frac{4\pi^2 e^2}{\frac{\hbar^2q^2}{2m}}\frac{1}{(2\pi)^3}[\frac{2k_Fq+(4k_F^2-q^2)\ln|\frac{q+2k}{q-2k}|}{8q} ]$

Which holds similarities to the desired answer, but both the factor in front of and the first term of the integral are incorrect.

2. Jun 5, 2015