Bloch Function Recursion Relation of Fourier Components

[MisterX]

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.

 

[anathema]

Thank you for posting this problem for me to help you with. it is my pleasure to assist you in understanding this concept.

To 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)##, we will need to solve the infinite system of coupled equations you have written down. This system can be solved by using the method of Fourier series.

First, let's define the Fourier coefficients ##c_n(k)## as
$$c_n(k) = \int_{-\frac{a}{2}}^{\frac{a}{2}} \psi_{nk}(x)e^{-ikx} dx$$
where ##\psi_{nk}(x)## is the eigenfunction corresponding to the eigenvalue ##E_n(k)##.

Next, we can express the potential ##U(x) = 2A\cos(\frac{2\pi}{a} x)## as a Fourier series:
$$U(x) = \sum_{n=-\infty}^{\infty} u_n e^{i\frac{2\pi}{a}nx}$$
where ##u_n## are the Fourier coefficients of the potential.

Using the orthogonality property of Fourier series, we can write the infinite system of equations as
$$\left(\frac{\hbar^2k^2}{2m} - E_n(k)\right)c_n(k) + \sum_{n'} u_{n-n'}c_{n'}(k) = 0$$
where ##n'## runs over all integers.

Solving this system of equations, we can obtain the eigenvalues ##E_n(k)## and the corresponding eigenfunctions ##\psi_{nk}(x)##, which will give us the dispersion relations for the three lowest electron bands.

I hope this helps you understand how to approach this problem. Please let me know if you have any further questions or need clarification.