Homework Help: 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 :