- #1
Telemachus
- 835
- 30
I was trying to determine the bandgap in the nearly free electron model. I'm having trouble to determine the band gap bewteen the second and the third band. Its a one dimensional problem.
So, the central equation reads:
##\displaystyle \left [ \frac{\hbar}{2m} (k-G)^2-E \right ]c_{k-G}+ \sum_{K'} U_{G-K'} c_{k-K'}##
So, I want to determine the band gap which would be at the point ##k=2{\pi}{a}## in the extended diagram (the band gap between the second and the third band).
And my ##K'## I think that should be: ##K'=\frac{2\pi}{a},\frac{4\pi}{a}##. When I carry on the whole calculation I get something that don't make any sense to me, because I don't find any degeneracy on the energy, and I think that's wrong.
For ##G=\frac{2\pi}{a}## I get:
##\displaystyle \left [ \frac{\hbar}{2m} (k-\frac{2\pi}{a})^2-E-U_0 \right ]c_{k-\frac{2\pi}{a}}+ U_{\frac{2\pi}{a}}c_{k-\frac{4\pi}{a}}##
And for ##G=\frac{4\pi}{a}##:
##\displaystyle \left [ \frac{\hbar}{2m} (k-\frac{4\pi}{a})^2-E-U_0 \right ]c_{k-\frac{4\pi}{a}}+ U_{-\frac{2\pi}{a}}c_{k-\frac{2\pi}{a}}##
I don't know what I'm doing wrong.
Thanks in advance.
So, the central equation reads:
##\displaystyle \left [ \frac{\hbar}{2m} (k-G)^2-E \right ]c_{k-G}+ \sum_{K'} U_{G-K'} c_{k-K'}##
So, I want to determine the band gap which would be at the point ##k=2{\pi}{a}## in the extended diagram (the band gap between the second and the third band).
And my ##K'## I think that should be: ##K'=\frac{2\pi}{a},\frac{4\pi}{a}##. When I carry on the whole calculation I get something that don't make any sense to me, because I don't find any degeneracy on the energy, and I think that's wrong.
For ##G=\frac{2\pi}{a}## I get:
##\displaystyle \left [ \frac{\hbar}{2m} (k-\frac{2\pi}{a})^2-E-U_0 \right ]c_{k-\frac{2\pi}{a}}+ U_{\frac{2\pi}{a}}c_{k-\frac{4\pi}{a}}##
And for ##G=\frac{4\pi}{a}##:
##\displaystyle \left [ \frac{\hbar}{2m} (k-\frac{4\pi}{a})^2-E-U_0 \right ]c_{k-\frac{4\pi}{a}}+ U_{-\frac{2\pi}{a}}c_{k-\frac{2\pi}{a}}##
I don't know what I'm doing wrong.
Thanks in advance.