For crystal structures we sometimes need to determine how many lattice points are in a unit cell. If these lattice points are at the corner or edge of a unit cell, they only count for a certain fraction.
In the lecture this morning, this fraction was determined by how many neighbours share an a lattice point: the corners of a 2d square each have four neighbours, so they count for 1/4 of a lattice point.
In the lecture notes, on the other hand, this fraction is determined by the angle that the edges of the unit cell make at the lattice point: for the parallellogram in the first figure, the edges make a 45\degree angle at the top right corner, so it counts for 45/360 = 1/8. In the end you get the result that 1 lattice point occupies the unit cell.
By coincidence (or not?) you can get the same result by noting that each vertex is shared between four unit cells, and so each lattice point in the corner contributes 1/4, to get 4\cdot1/4=1 lattice point in the unit cell.
Can you always use this reasoning to get correct results? Or are there some shapes where you truly need to consider the angle, and can’t just look at how many neighbours share a lattice point.
If the angles are 90 degrees, the fractions become easy to calculate. In particular in 2D: 90/360=1/4. In general the angles may be less simple, and one needs to take these into account when calculating the fraction of a corner or vertex point that sits in a particular unit cell.
Indeed, but does this ever give physically different results?
different from what? If you have a point that sits at the edge of a unit cell, you always need to consider the angles to calculate what fraction of the point sits in a unit cell.
Yes, so the correct method is to consider the angles.
The thing I was confused by is that just looking at the amount of neighbours always seems to give the correct answer as well.
For the parallellogram in the first figure (spanned by \tilde{a_1} and \tilde{a_2}) you get the correct result if you (wrongly) assume that all the angles are 90\degree. In that case, each vertex counts 1/4 so you get 4\cdot1/4=1 lattice points inside the unit cell.
I don’t see any reason why this wouldn’t work in general. After all, all the unit cells are exactly the same shape, so if there is a sharp angle at one vertex, the neighbouring unit cells must have a less sharp angle to compensate for it. Thus the original unit cell also has a less sharp angle.
Is this actually the case, or am I missing something?
not sure about the general proof, but it sounds reasonable