# Can I distinguish hexagonal close packing (HCP) from face centred cubic (FCC) arrangement based on Fourier transform

First of all I would say that I'm not a physicist, but I have recently been given the task of distinguishing a hexagonal close packing (HCP) from a face centred cubic (FCC) arrangement in a set of 3D data. The data essentially form 3D probability maps and may contain different arrangements of spherical densities.

This seems to be very close to what others do in crystallography, so I wondered if this problem may already have been solved.

My first ides was to compute the 3D Fourier transform (FFT) of the data. With simulated HCP and FCC arrangements I can see that there is a difference in the FFTs of these data. But I'm a bit stuck as to how to proceed. This problem is maybe related to reciprocal lattices? Does the shape in the FFT match the reciprocal lattice of the lattice in the data?

If anyone has other ideas on how to distinguish HCP from FCC stacking I would also love to hear them.

• Are you familiar with systematic absences at all? That might help. Sep 12, 2020 at 23:43
• Hmmm, I didn't find that very useful, sorry. I looked around on that website and it seems like lattice planes & miller indices are maybe more linked to what I am doing. When I take the Fourier transform it tells me the 3D wavefunctions (or planes) that can be used to make the lattice arrangement. This seems very similar to lattice planes. What I don't understand is how people find these in real life, like in this paper (figure 2): pubs.acs.org/doi/10.1021/acs.cgd.7b01489 They give the plane indices but I don't know how they got them. I'm not even sure if their FFT was 3D or just 2D... Sep 13, 2020 at 10:37

The coefficients of the discrete fourier transform of FCC are all the same, may be taken equal to 1, whereas the coefficients of the HCP fourier transform are not all the same. That is because FCC is a lattice, but the HCP packing is not a lattice.

In other words, because FCC is a lattice the structure factors of FCC are trivially all equal to one another, whereas HCP has nontrivial structure factors that will appear in the coefficients of your fourier transform.

If you plot |FT(FCC)|^2, you should see peaks of uniform height. If you plot |FT(diamond)|^2, you should see peaks with heights in ratio 1:2. If you plot |FT(HCP)|^2, according to https://en.wikipedia.org/wiki/Structure_factor#Hexagonal_close-packed_(HCP), you should see peaks with heights in ratio 4:3:1.

A lattice or packing may be represented by a distribution of dirac delta functions with equal coefficients. A lattice has one of the delta functions located at the origin. The fourier transform of a lattice is a lattice, called the reciprocal lattice.

FCC is a lattice and its reciprocal lattice is BCC. The fourier transform of FCC is the BCC lattice.

A general regular non-lattice packing like diamond cubic or HCP will usually be a sum of N lattices that are offset from one another. For instance, diamond cubic is the sum of two offset FCC lattices, so its fourier transform is the BCC lattice times structure factors. I think that HCP is the sum of two offset (A2 x Z1) lattices.

In general, the fourier transform of such a regular non-lattice packing that is a sum of N offset lattices will be the reciprocal lattice times phase factors, "structure factors" per wikipedia. You can see a discussion of diamond cubic https://en.wikipedia.org/wiki/Structure_factor#Diamond_crystal_structure. They define the structure factors to be complex, (1+(-i)^(h+k+l)), but if you use symmetric offsets of +/-(1/8,1/8,1/8) instead, that allows the structure factors of diamond cubic to be real-valued, 2 cos(pi/2*(h+k+l)).

Regardless, the absolute values squared of the diamond cubic structure factors, which correspond to nodes of the BCC lattice, occur in ratio 0:1:2; and those of HCP, which correspond to nodes of A2xZ1, occur in ratio 0:1:3:4. Again, you should see this ratio 1:3:4 when you plot the peaks of |FT(HCP)|^2, whereas the peaks of |FT(FCC)|^2 should all be the same height.