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Bloch Function Recursion Relation of Fourier Components

  1. Nov 1, 2015 #1
    1. The problem statement, all variables and given/known data
    This is just a problem to help me understand. Determine the dispersion relations for the three lowest electron bands for a 1-D potential of the form
    ##U(x) = 2A\cos(\frac{2\pi}{a} x)##

    2. Relevant equations
    I will notate ##G, \,G'## as reciprocal lattice vectors.
    $$\psi_{nk}(x) = \sum_{G} c_n(k-G)e^{i(k-G)x}$$
    $$U(x) = \sum_{G'} u(G')e^{iG'x} $$
    We arrive at coupled equations of the form
    $$ \left(\frac{\hbar^2\left(k-G\right)^2}{2m} - E\right)c_n\left(k-G\right) + \sum_{G'}u(G')c_n\left(k-G-G'\right) =0 $$

    3. The attempt at a solution
    with ##G_\eta = \frac{2\pi\eta}{a} ##.
    $$U(x) = Ae^{iG_1x} + Ae^{-iG_1x}$$
    So only the terms in the sum with ##G' = \pm G_1## will be nonzero. This leads to an infinite system of equations for each ##k## in the first BZ, such as
    $$\dots$$
    $$ \left(\frac{\hbar^2\left(k+G_1\right)^2}{2m} - E\right)c_n\left(k+G_1\right) + Ac_n(k+ G_2)+ Ac_n(k) =0$$
    $$ \left(\frac{\hbar^2k^2}{2m} - E \right)c_n\left(k\right) + Ac_n(k + G_1)+ Ac_n(k-G_1) =0$$
    $$ \left(\frac{\hbar^2\left(k-G_1\right)^2}{2m} - E\right)c_n\left(k-G_1\right) + Ac_n(k)+ Ac_n(k-G_2) =0$$
    $$ \left(\frac{\hbar^2\left(k-G_2\right)^2}{2m} - E\right)c_n\left(k-G_2\right) + Ac_n(k-G_1)+ Ac_n(k-G_3) =0$$
    and so on. I'm not sure how to go about solving this infinite system of coupled equations, or how to go about getting the dispersion relation. This recurrence relation has a form that goes both ways so I don't think the generating function/characteristic polynomial method applies. I tried to Z transform it but I got a nasty looking 2nd order diff eq with an irregular singular point.
     
    Last edited: Nov 1, 2015
  2. jcsd
  3. Nov 7, 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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