- #1
vbrasic
- 73
- 3
Homework Statement
Homework Equations
I'm not sure.
The Attempt at a Solution
I started on (i) -- this is where I've gotten so far.
I am asked to compute the Fourier transform of a periodic potential, ##V(x)=\beta \cos(\frac{2\pi x}{a})## such that, $$\tilde{V}(k)=\frac{1}{L}\int_0^LV(x)e^{-ikx}\,dx$$ and hence show that it is non-zero for only two values of ##k##. I have, $$\tilde{V}(k)=\frac{\beta}{L}\int_0^L\cos(\frac{2\pi x}{a})e^{-ikx}\,dx.$$ According to Wolfram, this gives, $$\tilde{V}(k)=\frac{ia\beta e^{-ikL}(-ake^{ikL}+ak\cos(\frac{2\pi L}{a})+2i\pi\sin(\frac{2\pi L}{a}))}{L(a^2k^2-4\pi^2)}.$$ I am also given that ##L=Na##, ##N\in\mathbb{Z}^+##, such that the expression simplifies to, $$\frac{i\beta e^{-ikL}(-ake^{ikL}+ak)}{N(a^2k^2-4\pi^2)}.$$ From this we have, $$\tilde{V}(k)=\frac{ia\beta(ke^{-ikNa}-1)}{N(a^2k^2-4\pi^2)}.$$ However, given this, I have no idea how to show that this is not equal to ##0## for only two values of ##k##. I tried setting ##ke^{-ikNa}-1## to ##0##, such that, $$e^{-ikNa}=\frac{1}{k},$$ and solving for ##k##, however, I'm not sure if this is the correct approach.
Also, ##N\gg 1##.