What's Wrong with the First Primitive Cell Choice?

In summary, the conversation discusses two potential choices for a lattice, with one resulting in incorrect Fourier coefficients and the other in correct coefficients. The question arises about the difference in choice of origin and whether or not it affects the coefficients. The response explains how changing the origin affects the function being Fourier transformed and the resulting coefficients. The conversation concludes with the discussion of a matrix used to calculate energy gaps, with the main concern being whether or not the choice of primitive cell (second image) was correct.
  • #1
luka77777778

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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  • #2
Why do you think you got the wrong coefficients in the first case?
 
  • #3
In the book they state the first case as a solution, so I suppose it's correct...
 
  • #4
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
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.
 
  • #6
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.
 
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  • #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) .
 
  • #8
This is the matrix I was talking about :

MvOKLLv.jpg
 

FAQ: What's Wrong with the First Primitive Cell Choice?

What is a 2D lattice in science?

A 2D lattice is a regular geometric arrangement of points or atoms in a two-dimensional space. It is used to describe the crystal structure of materials and is an important concept in materials science and solid state physics.

How is a 2D lattice different from a 3D lattice?

A 2D lattice only has two dimensions, while a 3D lattice has three dimensions. This means that a 2D lattice can be represented on a flat surface, while a 3D lattice requires a three-dimensional representation. Additionally, 2D lattices have different types of symmetries compared to 3D lattices.

What is a primitive cell in a 2D lattice?

A primitive cell is the smallest repeating unit of a 2D lattice. It contains all the information about the arrangement of atoms or points in the lattice and can be used to construct the entire lattice.

How do scientists choose the primitive cell in a 2D lattice?

The choice of primitive cell in a 2D lattice depends on the symmetries of the lattice. The primitive cell should have the same symmetries as the entire lattice and should be the smallest possible cell that can be repeated to form the entire lattice.

Can a 2D lattice have more than one type of primitive cell?

Yes, a 2D lattice can have multiple types of primitive cells. This is because different primitive cells can have the same symmetries as the entire lattice. However, the primitive cell that is chosen should be the most convenient for describing the lattice and predicting its properties.

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