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2D lattice, primitive cell choice

  1. Sep 2, 2017 #1
    1. The problem statement, all variables and given/known data
    When calculating the fourier coefficients of the potential of the following lattice (the potential is a sum of deltas at the atom sites):
    Iy1Sw.jpg

    I get the wrong coefficients if I choose the following primitve cell, with primitve vectors a1,a2:
    YcEmZ.jpg

    And the right coefficients if I choose the following primitve cell:
    FRcXt.jpg

    Question: What's wrong with the first choice? Is the cell chosen in the second picture even a primitve cell?

    3. The attempt at a solution

    First choice
    :

    ## U_K=\int_{primitve cell} \textbf[\delta(r)+\delta(r-a1)+\delta(r-a1-\frac{1}{2}a2)+\delta(r-a1-a2)+\delta(r-a2) +\delta(r-\frac{1}{2})+\delta(r-\frac{1}{2}a1-\frac{1}{2}a2)\textbf]\space e^{-i K r}\space dr
    ##


    for ##K=\frac{2\pi}{a}(m,n)## I then get ##U_K= 1+(-1)^m+(-1)^{m+n}##

    Second choice (the right one):

    ##U_K=1+(-1)^m + (-1)^n##
     
    Last edited: Sep 2, 2017
  2. jcsd
  3. Sep 2, 2017 #2

    Orodruin

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    Why do you think you got the wrong coefficients in the first case?
     
  4. Sep 2, 2017 #3
    In the book they state the first case as a solution, so I suppose it's correct...
     
  5. Sep 2, 2017 #4

    Orodruin

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    Well, this does not really answer the question. If you have a different origin you will get different Fourier coefficients. Did you try making the translation?
     
  6. Sep 3, 2017 #5
    I don't understand why the choice of origin would make any difference. The coefficients depend only on K which is a reciprocal lattice vector, and for the same K both choices (if they're correct) should give the same value.
    If I take, for example the reciprocal vector ##K=\frac{2\pi}{a}(-1,-1)## (m=-1, n=-1) they dont give the same value.
     
  7. Sep 3, 2017 #6

    Orodruin

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    Changing the origin changes the function you are Fourier transforming according to ##f(\vec x) \to f(\vec x - \vec d)##, where ##\vec d## is the displacement of the origin. You should not expect to get the same Fourier coefficients for different functions. It should hold that
    $$
    U_{\vec k, \vec d} = \int f(\vec x - \vec d) e^{-i \vec k \cdot \vec x} d^2x = \int f(\vec x) e^{-i\vec k\cdot (\vec x + \vec d)} d^2x
    = e^{-i\vec k \cdot \vec d} \int f(\vec x) e^{-\vec k \cdot \vec x} d^2 x = e^{-i\vec k \cdot\vec d} U_{\vec k},
    $$
    which is not equal to ##U_{\vec k}## unless ##\vec k \cdot \vec d## is a multiple of ##2\pi##.

    Edit: Also note that you mixed up ##m## and ##n## when computing your first transform. Your result is not compatible with your definition of ##m## and ##n## so be careful with this.
     
    Last edited: Sep 3, 2017
  8. Sep 3, 2017 #7
    (You're correct about the mixup). The problem was that when I was calculating band gaps that aries due to the peridodic potential in the corner of the 1st Brillouin zone you get a matrix with different fourier coefficients of the potential and you calculate the gaps using the condition ##det(Matrix)=0## , I thought that if the coefficients for different ##K=\frac{2\pi}{a}(m,n)## change based on the choice of origin, so will the matrix change and I will get different energy gaps depending on what I choose as the origin (which would be pretty stupid :D). So I checked and the matrix does change but at the end (I'm not sure exactly why) the energies stay the same, which is only logical, since both choices are a primitive cell. So the main thing that I was concerned about was wheter or not I have made the wrong choice for the primitive cell(2nd image) .
     
  9. Sep 3, 2017 #8
    This is the matrix I was talking about :

    MvOKLLv.jpg
     
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