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There is a very similar question here Reciprocal Lattice of a non-bravais lattice, but I don't fully understand the answer, and the question is now obsolete so I feel that I should ask it again.

How can I derive the reciprocal lattice of the HCP lattice? It has two sets of nonequivalent sites, and can not be comprehensively described by its three primitive vectors. @delete000 indicated in the question above that HCP can be treated as "a hexagonal lattice with a base", so how can I derive the reciprocal lattice with that method? Thanks!

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  • $\begingroup$ Yes, HCP is a simple hexagonal lattice with a basis. HCP is commonly referenced as an ABAB stacking of hexagonal close packed planes. So, the full simple hexagonal cell consists of an AB pair. The basis is a pair of atoms, one in A, and one in B. So, the vectors a in the close packed plane, and the vector c being the distance from one A plane to the next, constitute the simple hexagonal lattice. From there it is left as an exercise to crank through the vector algebra to get the reciprocal lattice... $\endgroup$ – Jon Custer Oct 21 '14 at 16:06
  • $\begingroup$ @JonCuster In my textbook the process of deriving the reciprocal lattice is as such: find the primitive vectors of the original lattice; calculate the primitive vectors of the reciprocal lattice (with all the vector algebra); find all the integral linear combination of the reciprocal primitive vectors to get the reciprocal lattice. Where should I plug the basis in the process? $\endgroup$ – arax Oct 21 '14 at 16:10
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    $\begingroup$ Any point in real space has a corresponding point in reciprocal space (and vice versa of course). So, the real space position of the basis atoms, as expressed by the real space lattice vectors, can be converted into the reciprocal space position of the basis atoms, as expressed by the reciprocal space lattice vectors. $\endgroup$ – Jon Custer Oct 21 '14 at 17:31
  • $\begingroup$ @JonCuster I did the calculation but the basis atoms in the reciprocal space looked a bit awkward. Their projection in xOy plane is still at the geometric center of the hexagonal grid. but as the reciprocal operation turns $\gamma=120^{\circ}$ to $\gamma '=60^{\circ}$, these atoms are in the geometric center of a 120-degree isosceles triangle, instead of the equilateral triangle in the real space. Did I do it wrong? Thanks! $\endgroup$ – arax Oct 22 '14 at 8:08
  • $\begingroup$ So, remember 2 things: First, don't worry where the 'atoms' are - reciprocal space is all about scattering vectors for x-ray or electron diffraction, or about momentum space for crystals (phonon dispersion and whatnot). Second, don't forget that you can always shift things about by adding/subtracting lattice vectors (now in reciprocal space) to an equivalent position that fits your aesthetic sense better. $\endgroup$ – Jon Custer Oct 22 '14 at 13:12

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