# Bloch Function Recursion Relation of Fourier Components

Tags:
1. Nov 1, 2015

### MisterX

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. Nov 7, 2015