A Inequivalent K and K' points in graphene

Good evening to everybody!
I have a question concerning monolayer graphene.
In all the papers I read it is well specified that K and K' in graphene are not equivalent points, but I didn't find anywhere where is the difference between them. Can anybody tell me where this difference is coming from theoretically and how can be detected?
 

TeethWhitener

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The K and K’ points are inequivalent for the same reason the A and B sublattices of graphene are inequivalent. Here’s a picture:
https://images.app.goo.gl/1PnPU5HPaW4FZawD8
In the momentum space picture, remember that only the first (hexagonal) Brillouin zone is shown; the Brillouin zones actually tile the plane. So with that in mind, it’s clear that (e.g.) the K point occurs at a vertex with two hexagons below it and one hexagon above it, and the K’ point occurs at the mirror image vertex, with one hexagon below and two above. In other words, both K and K’ have threefold rotational symmetry, but they’re inversions of each other.
 
Ok but if this is the case, why when in a DFT calculation we use the Irreducible Brillouin Zone, we only consider K? In principle it is not enough to consider only 1/12 of the hexagon as everybody do, right?
 

TeethWhitener

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The electronic structure (edit: the band energy structure, not the wavefunctions) is going to be the same, but the wavefunction at the K point is related to the wavefunction at the K' point by time reversal symmetry. It's kind of a momentum space analogy to different enantiomers of a molecule: they have the same energy, but they are mirror images of each other. Another way you can think about it is that near the K (K') points, the Hamiltonian ends up having the same form as that of a massless spin 1/2 particle. But the phases of the wavefunction components differ by a sign change. So at the K point, you get a "left-handed"spinor and at the K' point you get a "right-handed" spinor. Here's a pretty good review:
https://cdn.journals.aps.org/files/sample-rmp-revtex4.pdf
 

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