# 2D lattice, primitive cell choice

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1. Sep 2, 2017

### luka77777778

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):

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

And the right coefficients if I choose the following primitve cell:

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 by a moderator: Sep 2, 2017
2. Sep 2, 2017

### Orodruin

Staff Emeritus
Why do you think you got the wrong coefficients in the first case?

3. Sep 2, 2017

### luka77777778

In the book they state the first case as a solution, so I suppose it's correct...

4. Sep 2, 2017

### Orodruin

Staff Emeritus
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?

5. Sep 3, 2017

### luka77777778

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.

6. Sep 3, 2017

### Orodruin

Staff Emeritus
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
7. Sep 3, 2017

### luka77777778

(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) .

8. Sep 3, 2017

### luka77777778

This is the matrix I was talking about :