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I'm trying to calculate dipole moments from Wannier centers for the first time and there are basics I don't understand and can't figure out from online resources.

The software I'm using has a simple "output Wannier centers" option. So I now have output, which is a file with Wannier centers and atomic positions. My system contains many water molecules but here's a simple example for a system that has only one H2O. The "X" are Wannier centers:

X          6.50000026       6.18706578       6.50000000
X          6.50000000       6.71337672       6.50000002
X          6.50000008       6.95339960       6.49999998
X          6.49999971       6.54636884       6.50000000
H          6.50000000       7.09480000       7.26880000
H          6.50000000       7.09480000       5.73120000
O          6.50000000       6.50000000       6.50000000
  • Why are there four X for one H2O? I vaguely assumed that there would be one X, corresponding to one H2O. Do the four X correspond somehow to different molecular configuration possibilities? I have tried to look this up but probably don't even know the right search terms.

I included "Wannier90" as a tag because it seems relevant and generates output formatted like my example but I am not using Wannier90.

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1 Answer 1

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H2O has 10 electrons, two of which are oxygen 1s. This leaves you eight electrons. These eight electrons fit onto 4 spatial orbitals in a spin-restricted calculation for the singlet.

Determining localized orbitals [1] (for periodic boundary conditions these are called Wannier functions), the 4 orbitals localize into two covalent OH bonds, and two lone pairs on the oxygen atom. You have obtained the centroids of these orbitals, which are determined by the expectation value $\langle {\bf r} \rangle$.

[1]: Note that there is more than one way to determine localized orbitals; for instance, we have proposed Pipek-Mezey Wannier functions in J. Chem. Theory Comput. 2017, 13, 2, 460–474

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  • $\begingroup$ +1 for another very quick answer, backed with a good reference, as always! I'm sure the OP appreciates your efficiency! $\endgroup$
    – Nike Dattani - No Free Time
    Jun 17, 2021 at 23:42
  • $\begingroup$ Yes, I certainly do appreciate it! Thank you for the comment about periodic BC too. That clears up some confusion. May I please make sure I'm clear about something? Does it matter if the H2O is H2O or if it dissociates? I am dealing with high-ish temperature and this may have happened. $\endgroup$
    – NTS
    Jun 18, 2021 at 5:20
  • $\begingroup$ @NTS no; what matters is only the number of electrons. Of course, the dissociation is not captured by the calculation if you use spin-restricted orbitals; you need a broken-symmetry unrestricted calculation to be able to dissociate H2O $\endgroup$
    – Susi Lehtola
    Jun 18, 2021 at 23:19
  • $\begingroup$ @NTS does this answer your question? please accept the answer if yes $\endgroup$
    – Susi Lehtola
    Jun 18, 2021 at 23:20
  • $\begingroup$ @SusiLehtola, I was thinking over your comments about dissociation and orbitals. I don't understand them but will ask another question if my own efforts to understand are not successful. You have certainly answered my main question and thank you very much for taking the time and providing such a straightforward answer. $\endgroup$
    – NTS
    Jun 19, 2021 at 6:33

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