Why Does My Kohn Effect Derivation Differ from Ziman's Model?

[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.

 

[eljose79]

Thank you for sharing your work on the derivation of the Kohn Effect. I have reviewed your calculations and I believe I have found the error in your derivation.

In your third step, you have incorrectly exchanged the sum over k with the integral. This is not valid because the sum is over a discrete set of k values, while the integral is over a continuous range of k values. This means that the sum and integral cannot be simply interchanged.

To correct this, you will need to use the Riemann sum approximation to convert the sum into an integral. This will result in a different factor in front of the integral, which should lead to the correct result.

I hope this helps in resolving the discrepancy in your derivation. Keep up the good work!