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Fourier transform of periodic potential in crystal lattice

  1. Mar 30, 2017 #1
    1. The problem statement, all variables and given/known data
    by58f4p.png

    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##.
     
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  3. Mar 30, 2017 #2

    Dr Transport

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    write [itex] \cos(\frac{2\pi x}{a})[/itex] in terms of complex exponentials and the first part will pop out immediately.
     
  4. Mar 30, 2017 #3
    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.
     
  5. Mar 30, 2017 #4

    Dr Transport

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    When you do the integral, you'll get two values of [itex] k [/itex].
     
  6. Apr 1, 2017 #5
    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##.
     
  7. Apr 1, 2017 #6

    Dr Transport

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    Leave it as complex exponentials, the integral is the definition of the Kronecker [itex] \delta_{ij}[/itex]
     
  8. Apr 1, 2017 #7
    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}##?
     
  9. Apr 1, 2017 #8

    Dr Transport

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    exactly......
     
  10. Apr 1, 2017 #9
    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.)
     
  11. Apr 2, 2017 #10

    Dr Transport

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    Yes, the [itex] G'[/itex]'s are just the [itex] \pm\frac{2\pi n}{a}[/itex] from the first part. Your resulting equation should be a set of equations for [itex] \tilde{u}_Q(G)[/itex]
     
  12. Apr 2, 2017 #11
    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}##?
     
  13. Apr 2, 2017 #12

    Dr Transport

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    your missing a factor of [itex] \tilde{u}_Q[/itex] in the first part of your expressions. Yes, [itex] \tilde{V}[/itex] is zero for everything except [itex] \pm \frac{2\pi n}{a}[/itex]. you can write this as a matrix then find the eigenvalues for the [itex] \tilde{u}[/itex].
     
  14. Apr 2, 2017 #13
    Alright, so then, $$\frac{\hbar^2}{2m}(G+Q)^2\tilde{u_Q}(G)+\frac{\beta}{2}\tilde{u_Q}(\frac{2\pi}{a})+\frac{\beta}{2}\tilde{u_Q}(-\frac{2\pi}{a})=E\tilde{u_Q}(G)?$$
     
  15. Apr 2, 2017 #14

    Dr Transport

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    [itex] G = \pm\frac{2\pi n}{a}[/itex] gives you two equations.....
     
  16. Apr 2, 2017 #15
    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'##?
     
  17. Apr 2, 2017 #16

    Dr Transport

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    I have to sit down and think about it a little more.....
     
  18. Apr 2, 2017 #17
    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...
     
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