Bloch Function Recursion Relation of Fourier Components

  • #1
MisterX
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Homework Statement


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)##

Homework 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
$$ \left(\frac{\hbar^2\left(k-G\right)^2}{2m} - E\right)c_n\left(k-G\right) + \sum_{G'}u(G')c_n\left(k-G-G'\right) =0 $$

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.
 
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