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Derivation of Kohn Effect

  1. May 31, 2015 #1
    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. jcsd
  3. Jun 5, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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