What's Wrong with the First Primitive Cell Choice?

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SUMMARY

The discussion centers on the calculation of Fourier coefficients for a lattice potential represented by delta functions at atomic sites. The first choice of the primitive cell, defined by vectors a1 and a2, yields incorrect coefficients, while the second choice produces the correct results. The discrepancy arises from the choice of origin, which affects the Fourier transform, leading to different functions and coefficients. The participants clarify that while both choices can be valid primitive cells, the origin's selection is crucial for accurate coefficient calculations.

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Homework Statement


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?

The Attempt at a Solution



First choice[/B]:

## 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##
 
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Why do you think you got the wrong coefficients in the first case?
 
In the book they state the first case as a solution, so I suppose it's correct...
 
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?
 
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 don't give the same value.
 
luka77777778 said:
I don't understand why the choice of origin would make any difference.

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:
(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) .
 
This is the matrix I was talking about :

MvOKLLv.jpg
 

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