Example on shc_alpha, beta, gamma is confusing

[jaemolihm]

https://github.com/wannier-berri/wannier-berri-org/blob/master/source/exampleFe.rst#spin-hall-conductivity says that "However, specification of shc_alpha, shc_beta, and shc_gamma does not work with symmetry.", but the code block below uses symmetry with shc_alpha.

It would be better to add two examples, one with symmetry and without shc_alpha, the other without symmetry and with shc_alpha.

Or am I misunderstanding something?

@manxkim

[stepan-tsirkin]

We had a discussion, and @manxkim was convincing me that when shc_* are used, the result works as a scalar, and use of symmetry is justified. However I was still not convienced, because if you take a forbidden component, and integrate over an irreducible wedge, you will get smth nonzzero. Then, multiplication by a number of irreducible wedges gives also non-vanishing result, which is incorrect. I think shc_* works only for some specific cases, like the xyz component for a cubic system. But I am still not sure. @manxkim, @jaemolihm , do you have something to add to this discussion?

[manxkim]

As I remember, for some space groups a single component of shc can be calculated using both symmetry and shc*, but in general it is not true. I found that there were also inconsistency between applying both and applying only shc* for some cases.

For now I guess a single only shc component for cubic I Laue class can be calculated using both, since (shc tensor) = (scalar) * (Levi-Civita tensor), and use of symmetry is justified even one calculates only one component. I am going to investigate for other Laue classes(where shc is not expressed as above) and let you know.

2021년 5월 30일 (일) 오전 12:18, Stepan Tsirkin @.***>님이 작성:

We had a discussion, and @manxkim https://github.com/manxkim was convincing me that when shc* are used, the result works as a scalar, and use of symmetry is justified. However I was still not convienced, because if you take a forbidden component, and integrate over an irreducible wedge, you will get smth nonzzero. Then, multiplication by a number of irreducible wedges gives also non-vanishing result, which is incorrect. I think shc* works only for some specific cases, like the xyz component for a cubic system. But I am still not sure. @manxkim https://github.com/manxkim, @jaemolihm https://github.com/jaemolihm , do you have something to add to this discussion?

— You are receiving this because you were mentioned. Reply to this email directly, view it on GitHub https://github.com/wannier-berri/wannier-berri-org/issues/3#issuecomment-850850100, or unsubscribe https://github.com/notifications/unsubscribe-auth/ALBFO63GZZIDOX2E3FS6XNTTQEAVTANCNFSM45YJUL7A .