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Graphene is two-dimensional honeycomb crystal lattice with the two Carbon atoms in its unit cell. Clearly, it is inversion symmetric. But, suppose we have two different atoms in the hexagonal system like in case of SiGe, then can we call it an inversion-symmetric system? If yes, then how?

Thanks in advance!

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Perhaps it would help to draw a picture! Once you have a picture, pick a convenient point upon which to carry out the inversion, and see if the structure is different. If it is the same it has inversion symmetry, if not then no inversion symmetry.

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  • $\begingroup$ Thank you for your valuable comments. Overall structure remains the same, but the atoms positions get changed. So, will it be inversion symmetric? Kindly find the attached link, pubs.acs.org/doi/pdf/10.1021/acsomega.7b01957 In this paper in SiGe, there are two different atoms in the unit cell and they are considering it as the inversion symmetric. Thanks again! $\endgroup$ – Payal Wadhwa Feb 28 at 3:15
  • $\begingroup$ Because the identities of the atoms swapped, it should not be considered to have inversion symmetry, no. $\endgroup$ – speeze Mar 1 at 4:16
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No, such a lattice is not inversion symmetric. For graphene, the inversion needs to be done at the center of a bond (because otherwise the lattice sites move) and the only thing that happens is that you swap carbon for carbon, so nothing changes, but if you put different atoms on the two sites, then that is no longer true.

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  • $\begingroup$ Thank you for your valuable comments. Kindly find the attached link. pubs.acs.org/doi/pdf/10.1021/acsomega.7b01957 In this paper, there are two different atoms in the unit cell, still, they are using Kane Mele criterion which can be applied to inversion symmetric systems only. Please have a look. Thank again! $\endgroup$ – Payal Wadhwa Feb 28 at 3:11
  • $\begingroup$ @PayalWadhwa I'm unfamiliar with that criterion and its range of applicability. That's probably worth asking separately, providing a detailed question about what exactly it is you want to know, along with as much background as you can give. $\endgroup$ – Emilio Pisanty Mar 4 at 18:47

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