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Example on shc_alpha, beta, gamma is confusing #3
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We had a discussion, and @manxkim was convincing me that when |
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?
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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 withshc_alpha
.Or am I misunderstanding something?
@manxkim
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