I have a small question about the final exam of 2018.
For the exercise about the crystal, we displace the atom in the middle of the plane by \delta in the z-direction. I don’t understand why the primitive unit cell in this case equals the conventional unit cell. I would say that it still contains 3 atoms, the ones from the cubic, and the ones from the x-z and y-z plane. As I understand it, only the atoms in the x-y plane are moved by \delta.
My guess was: a_1 = (a/2, a/2, \delta) a_2 = (a/2,0,a/2) a_3 = (0,a/2,a/2)
Could you explain me why in this cas, the conventional unit cell is a primitive unit cell.
I consider a primitive unit cell to be a unit cell that has only 1 atom in it. But when using the conventional unit cell, I count a total of 3 atoms. And that is why my proposal of a_1 = (a/2,a/2,\delta) has only 1 atom. So that is why I am a bit confused.
Thank you for your fast response
Note that there is a difference between atoms and lattice points. The definition of a primitive unit cell is that it contains at most 1 lattice point per basis. This means that there can be multiple atoms inside a primitive unit cell.
If there was a lattice vector a_1 = (a/2, a/2, \delta), then there would need to be for example an atom at 2a_1 = (a, a, 2\delta), but the problem doesn’t show the atom at (a, a, 0) moving. Does that help see what is going wrong with your reasoning?
I understand this reasoning, and it makes sense that this is not correct. I came up with the same reasoning as @Tomas. Although I can’t really figure out how the answer for \delta \neq 0 is correct. In my understanding you can only make integer combinations of the lattice vectors to arrive at a new point, and I don’t see how you would ever end up at the point moved by \delta .
