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Particle interference and DOS

  1. May 3, 2010 #1

    In this paper, http://people.ccmr.cornell.edu/~jcdavis/publicationPDF/NaturePhys_3_865.pdf, page 2 figure 1c, there is a plot of the Fourier transformation of the local density of states for some energy E. There is a thing I don't quite understand: The LDOS is discrete, i.e. it depends on spatial position (e.g. (0,0), (0,1), .....), so how is it possible to talk about its Fourier transform it?
  2. jcsd
  3. May 3, 2010 #2
    You can Fourier transform a function defined on a lattice g(R) with this transformation:

    f(\mathbf{k}) = \sum_{\mathbf{R} } e^{i \mathbf{k} \cdot \mathbf{R} } g(\mathbf{R})

    where R runs over the vectors in the lattice. This results in a continuous function f(k) which is periodic in k (momentum) space, which perodicity given by the reciprocal lattice vectors.
  4. May 3, 2010 #3
    Hi kanato

    Thanks for replying so fast. I will try it out.
  5. May 4, 2010 #4
    I don't know how much experience you have in this subject kanato, but I have another question regarding the article (figure 1).

    How is it that we know that the Fermi surface intersects the two nodes in the bananas?
  6. May 4, 2010 #5
    I think this is following from the fact that the Bogoliubov transformation is particle-hole symmetric, since it mixes particle states with hole states. So the quasiparticle dispersion is the same for +E as for -E, which means the Fermi surface has to be at the node. Well, I don't have much experience in high Tc superconductivity, and I'm not sure if that really answers your question.
  7. May 7, 2010 #6
    Can anyone tell me how to calculate the energy difference of a "proton and a electron independent and without influence of any forces" from those "who combines to form a hydrogen atom"?
  8. May 11, 2010 #7
    Ok kanato, I hope there is a final issue you can help me with. The function f(k) in post #2 is the local DOS of my lattice. I am looking at a homogeneous case, so it should be constant everywhere.

    Now, I have found f(k), and it is given by something like

    f(\mathbf k) = 0.1+0.1e^{i\left( {2k_x + k_y } \right)} + 0.1e^{i\left( {k_x + 2k_y } \right)} + 0.1e^{i\left( {2k_x + 0k_y } \right)} + \ldots ,

    where the argument of the exponential is determined by R. Now, what I have done is to divide my Brillouin zone (ranging from -Pi..Pi in both x and y) into cells with width 2Pi/N, where N2 is the total number of atoms.

    When I find f(k) for these different wavevectors and make a contour-plot, it does not show up as a constant. I think it is because of the division of the BZ.

    How should I partition my BZ in order for it to be constant for all kx and ky? Your help will be greatly appreciated.
  9. May 11, 2010 #8
    Hmm I don't think it should show up as a constant. If you mean your local DOS g(R) is constant, then its Fourier transform should be strongly peaked at k = 0 and nearly zero at other k. You should look at your expansion and see if you can combine terms, for instance if you have [tex]e^{i(2k_x+k_y)}[/tex] and [tex]e^{-i(2k_x+k_y)}[/tex] then you will wind up with [tex]2\cos(2k_x+k_y)[/tex] (this should definitely be the case if you have inversion symmetry in your lattice).

    Don't forget the way the inverse Fourier transform works:
    [tex]g(R) = \sum_{k} e^{-ikR} f(k)[/tex]

    For the term with k = 0, e^(-ikR) = 1, so that term is constant, independent of R. If g(R) is a constant function, then the only term which can have a coefficient will be the one with k = 0.

    prakash kumar: You should make a new post and not try to hijack a thread which is not relevant to your question.
  10. May 11, 2010 #9
    Regarding the LDOS being constant: You are right. In my LDOS(k) all terms are there (not just the constant one), which I really don't understand. I am doing this numerically, and LDOS(k)) is of the form written in post #7.

    I have inversion symmetry, and I do see the "complementary terms" (i.e. complex conjugate of eachother). But I really don't understand why there are other terms than the k=0 term. You have any ideas?
    Last edited: May 11, 2010
  11. May 12, 2010 #10
    You say that I should combine terms from the exponential (I am still talking about the homogeneous case). Shouldn't they equal zero all of them, such that LDOS(k) is constant?

    And besides this specific problem, there really is a problem for the general case. The lattice FT is given in post #2, and in the homogeneous case LDOS(k) is constant (for some energy). When we FT, we expect only the k=0 term to be there, which we can see from the inverse lattice FT (post #8). But that is generally not the case, since

    f(\mathbf{k}) = \sum_{\mathbf{R} } e^{i \mathbf{k} \cdot \mathbf{R} } g(\mathbf{R})

    Am I the only one who sees the problem? By the way, did you see my PM?

    Best regards,
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