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I just had my second solid state physics lecture and we were talking about bravais lattices. As far as I understand a Bravais lattice is an infinite network of points that looks the same from each point in the network. For example:

enter image description here

would be a Bravais lattice. On the other hand, this:

enter image description here

is not a bravais lattice because the network looks different for different points in the network. However, in lecture it was briefly mentioned that we could make this into a Bravais lattice by choosing a suitable basis:

enter image description here

The problem is, I don't really see how that changes anything. The positions of the atoms/points didn't change relative to each other.

1) Do I have to imagine the two atoms "combined" into one? If I do that, where is the new "2-in-1" atom located?

2) How can I construct a primitive vector that will go to this point?

3) Is there an infinite amount of points/atoms I can combine? Are there an infinite amount of basis I can choose?

4) Would the Wigner-Seitz cell have to be over two points if I choose a two atom basis?

Edit:

enter image description here

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  • $\begingroup$ Yes, the two atoms are the 'basis' of the space group. The Bravais lattice vectors go between, say, the middle of the lines connecting the basis atoms to equivalent points of the other atom pairs on other Bravais lattice sites. $\endgroup$ – Jon Custer Apr 24 '16 at 17:06
  • $\begingroup$ @JonCuster Thanks for the quick reply. So the vectors $a_1, a_2$ I have drawn are not viable basis vectors? $\endgroup$ – qmd Apr 24 '16 at 17:10
  • $\begingroup$ No, they absolutely are just fine. Moving along those vectors gives the same 'scenery' wherever you are on the lattice. Placing the vertex on one of the basis atoms yields every other equivalent basis atom. $\endgroup$ – Jon Custer Apr 24 '16 at 17:13
  • $\begingroup$ the cell and the vectors in your drawing are good $\endgroup$ – Ilja Apr 24 '16 at 17:14
  • $\begingroup$ @JonCuster So you are saying a better choice of grid would be to put the "origin" of the grid on top of one of the atoms? $\endgroup$ – qmd Apr 24 '16 at 17:19
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The answer to nearly everything is: yes :) your intuition about it is quite right, and your picture is good, too.
You have two different kinds of points, and any pair with one point from each kind would be a suitable basis. You will of course take adjacent ones in practice.
You could also take more than two points as primitive cell, but it will not be a good choice, it will be not primitive. You are interested in the smallest cell, because then the symmetry is better seen.
Then the neighborhood "looks the same" from any cell. Or to be more precise, you can get the whole network by translating your cell by integer multiples of the two vectors. So it's in essence a rhombic lattice.

The Wigner-Seitz cell has to contain two atoms, yes, you can take one hexagon (which will contain three thirds of each atom)

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  • $\begingroup$ Ok I see. In my second picture I have a set of primitive vectors. If I draw the grid like I did in the third picture, is it not going to be impossible to find the new basis vectors? $\endgroup$ – qmd Apr 24 '16 at 17:18
  • $\begingroup$ Your grid in the third picture is fine. What do you mean by "impossible to find", you have drawn it well (you mean $a_1$ and $a_2$, right?); you can also draw them from one atom to the neighbouring atoms of the same type, this is the same. $\endgroup$ – Ilja Apr 24 '16 at 17:22
  • $\begingroup$ Let me draw another picture. I will edit my opening post. $\endgroup$ – qmd Apr 24 '16 at 17:30
  • $\begingroup$ I added another diagramm to my opening post. My problem is, how would I express the new red basis vectors by using the old unit vectors $z_1,z_2$. There seems to be no connection $\endgroup$ – qmd Apr 24 '16 at 17:36
  • $\begingroup$ But what is the meaning of $z_1$ and $z_2$? Why do you want to express the basis vectors that are appropriate for the problem through others that are not? :) Anyway: it seems, that the basis vectors are $2z_2$ and $3/2*z_1 + z_2$, if I understand correctly what you mean by the $z_{1,2}$ $\endgroup$ – Ilja Apr 24 '16 at 17:39

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