# Fourier transform of periodic potential in crystal lattice

Tags:
1. Mar 30, 2017

### vbrasic

1. The problem statement, all variables and given/known data

2. Relevant equations
I'm not sure.

3. 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, $N\gg 1$.

2. Mar 30, 2017

### Dr Transport

write $\cos(\frac{2\pi x}{a})$ in terms of complex exponentials and the first part will pop out immediately.

3. Mar 30, 2017

### vbrasic

So for the very first part I have, $$\cos(\frac{2\pi x}{a})=\frac{e^{i2\pi x/a}+e^{-i2\pi x/a}}{2}.$$

I'm not sure how this will pull out a term.

4. Mar 30, 2017

### Dr Transport

When you do the integral, you'll get two values of $k$.

5. Apr 1, 2017

### vbrasic

Okay... so I'll take the integral, $$\frac{\beta}{2L}\int_0^L (e^{i2\pi x/a}+e^{-i2\pi x/a})e^{-ikx}.$$ I get, $$\frac{\beta}{2L}\int_0^Le^{ix(2\pi/a-k)}+e^{-ix(2\pi/a+k)}.$$ This resolves to $$\frac{\beta}{2L}(\frac{2a\sin(L(2\pi/a+k))}{ak+2\pi}).$$ I'm still not sure how I'd manipulate this to pull out two values of $k$.

6. Apr 1, 2017

### Dr Transport

Leave it as complex exponentials, the integral is the definition of the Kronecker $\delta_{ij}$

7. Apr 1, 2017

### vbrasic

Ah. Alright. So by definition, because we are integrating on the bound, $[0,L]$ we will have the following for the integral, $$\beta(\frac{1}{L}\int_0^L e^{-ix(k-2\pi/a)}\,dx+\frac{1}{L}\int_0^Le^{-ix(k+2\pi/a)}\,dx).$$ This evaluates to $\beta(\delta_{k,\,2\pi/a}+\delta_{k,\,-2\pi/a})$, such that $k=\pm \frac{2\pi}{a}$?

8. Apr 1, 2017

### Dr Transport

exactly......

9. Apr 1, 2017

### vbrasic

For the second part, we are given $$\psi(x)=\frac{1}{\sqrt{L}}\Sigma_G\tilde{u_Q}(G)e^{i(Q+G)x}.$$ This is the same as $$e^{iQx}\frac{1}{\sqrt{L}}\Sigma_G\tilde{u_Q}(G)e^{iGx}.$$ This is exactly, Bloch's theorem, $\psi(x)=e^{ipx}u(x),$ where $u(x)$ is periodic. We can then write the Schrodinger equation as, $$-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}(e^{iQx}u(x))+e^{iQx}V(x)u(x)=e^{iQx}Eu(x).$$ This becomes, $$-\frac{\hbar^2}{2m}(-Q^2u(x)+2iQu'(x)+u''(x))+V(x)u(x)=Eu(x).$$ We rewrite this as, $$-\frac{\hbar^2}{2m}u''(x)-\frac{\hbar^2}{2m}(2iQ)u'(x)+\frac{\hbar^2}{2m}Q^2u(x)=Eu(x).$$ My professor claims that this is equivalent to, $$\frac{\hbar^2}{2m}(G^2+2GQ+Q^2)\tilde{u_Q}(G)+\Sigma_{G'}V(G-G')\tilde{u_Q}(G')=E\tilde{u_Q}(G)\to\frac{\hbar^2}{2m}(G+Q)^2\tilde{u_Q}(G)+\Sigma_{G'}V(G-G')\tilde{u_Q}(G')=E\tilde{u_Q}(G).$$ The goal is naturally then to somehow subsitute in the expression we got from part (i) for potential. However, I'm confused on how to do this. I have that $G=\frac{2n\pi}{a}$ from translational invariance arguments. Are the $G'$s simply the values pulled out by the Kronecker delta from part (i)? And if so, how would one find $\tilde{u_Q}(G)$?

(Sorry, the reason I'm asking is because I'm not too familiar with Fourier expansions. I'm a visiting student at my current university, and never learned Fourier series at my home university.)

10. Apr 2, 2017

### Dr Transport

Yes, the $G'$'s are just the $\pm\frac{2\pi n}{a}$ from the first part. Your resulting equation should be a set of equations for $\tilde{u}_Q(G)$

11. Apr 2, 2017

### vbrasic

So we have, $$\frac{\hbar^2}{2m}(G+Q)^2+\tilde{V}(G-\frac{2\pi}{a})\tilde{u_Q}(\frac{2\pi}{a})+\tilde{V}(G+\frac{2\pi}{a})\tilde{u_Q}(-\frac{2\pi}{a})=E\tilde{u_Q}(G).$$ We can subsitute $G=\frac{2n\pi}{a}$ in to get, $$\frac{\hbar^2}{2m}(G+Q)^2+\tilde{V}(\frac{2\pi(n-1)}{a})\tilde{u_Q}(\frac{2\pi}{a})+\tilde{V}(\frac{2\pi(n+1)}{a})\tilde{u_Q}(-\frac{2\pi}{a})=E\tilde{u_Q}(G).$$ But isn't $\tilde{V}=0$ for all values besides $\pm\frac{2\pi}{a}$?

12. Apr 2, 2017

### Dr Transport

your missing a factor of $\tilde{u}_Q$ in the first part of your expressions. Yes, $\tilde{V}$ is zero for everything except $\pm \frac{2\pi n}{a}$. you can write this as a matrix then find the eigenvalues for the $\tilde{u}$.

13. Apr 2, 2017

### vbrasic

Alright, so then, $$\frac{\hbar^2}{2m}(G+Q)^2\tilde{u_Q}(G)+\frac{\beta}{2}\tilde{u_Q}(\frac{2\pi}{a})+\frac{\beta}{2}\tilde{u_Q}(-\frac{2\pi}{a})=E\tilde{u_Q}(G)?$$

14. Apr 2, 2017

### Dr Transport

$G = \pm\frac{2\pi n}{a}$ gives you two equations.....

15. Apr 2, 2017

### vbrasic

So for $G=\frac{2n\pi}{a},$ it's, $$\frac{\hbar^2}{2m}(\frac{2n\pi}{a}+Q)^2\tilde{u_Q}(\frac{2n\pi}{a})+\tilde{V}(\frac{2n\pi}{a}-\frac{2\pi}{a})\tilde{u_Q}(\frac{2\pi}{a})+\tilde{V}(\frac{2n\pi}{a}+\frac{2\pi}{a})\tilde{u_Q}(-\frac{2\pi}{a})=E\tilde{u_Q}(G),$$ and for $G=-\frac{2n\pi}{a}$ it's, $$\frac{\hbar^2}{2m}(-\frac{2n\pi}{a}+Q)^2\tilde{u_Q}(\frac{2n\pi}{a})+\tilde{V}(-\frac{2n\pi}{a}-\frac{2\pi}{a})\tilde{u_Q}(\frac{2\pi}{a})+\tilde{V}(-\frac{2n\pi}{a}+\frac{2\pi}{a})\tilde{u_Q}(-\frac{2\pi}{a})=E\tilde{u_Q}(G)?$$ Sorry, I kind of don't understand why there are two separate values of $G$; isn't the summation over $G'$?

16. Apr 2, 2017

### Dr Transport

I have to sit down and think about it a little more.....

17. Apr 2, 2017

### vbrasic

Thanks for all the help. Expanding that sum over $G'$ is proving very challenging for me because I'm not sure exactly what I'm summing over...