[QE-developers] failing to calculating matrix representations of symmetry operators

[Gerson J. Ferreira]

Dear Paolo,

I'm sorry I wasn't precise in that sentence. What I should have said is
that I've read a message in the mailing list about a possible issue with
magnetic symmetries:

https://lists.quantum-espresso.org/pipermail/developers/2020-May/002220.html

After I wrote the previous email, I realized that I could have written the
question to the mailing list in simple terms. So let me try again here:

SIMPLIFYING THE QUESTION
-----------------------------------------
Graphene lattice has the symmetry D6h + time-reversal (T) [and ignoring
translations for simplicity]. On the K point we lose inversion (I) and
time-reversal, since these take K to K', and the point group becomes D3h at
K. But the combined symmetry T.I is preserved, as TRS goes from K to K' and
inversion takes it back to K.

I use the method of invariants to develop effective k.p models and for
graphene this T.I symmetry is important, as it tells us that a k-dependent
SOC term is zero near the Dirac cone. If I don't use the T.I symmetry to
constrain my effective H, I get an extra term that shouldn't exist.

Now, running QE for graphene at the K point, the XML file does not show
this symmetry in the list of symmetries. My numerical analysis seems to
show that this symmetry is weakly broken in the QE results.

So I wonder if this might be a bug. Should QE have found this symmetry? Is
there a way to help it find it?
------------------------------------------


For the QE results this might be a very subtle issue that won't change the
results significantly, but for my project it is actually important. If I
don't apply all constraints in my code I end up with an ambiguity in the
definition of the unitary matrix for the change of basis between the
numerical QE results and the optimized basis based on the expected IRREPS.
Essentially, the unitary matrix becomes a linear combination of two
matrices, as in U = a.U1 + b.U2, and the resulting effective H will depend
on these parameters (a,b). Any choice of (a,b) coefficients will result in
a valid effective model H, but to "make it look good" we need to find an
optimal (a,b) and the T.I constraint would do this for us automatically in
this case.

If I don't get rid of this ambiguity, my code will require an input from
the user to choose the coefficients (a,b) in this case, and maybe even more
coefficients in more complex systems.

Thanks for the attention,
--
Gerson J. Ferreira
Prof. Dr. @ InFis - UFU
----------------------------------------------
gjferreira.wordpress.com
Institute of Physics
Federal University of Uberlândia, Brazil
----------------------------------------------


On Thu, Jul 29, 2021 at 3:30 AM Paolo Giannozzi <p.giannozzi at gmail.com>
wrote:

> On Thu, Jul 29, 2021 at 6:55 AM Gerson J. Ferreira <
> gersonjferreira at gmail.com> wrote:
>
>>
>> I know that QE can handle the magnetic symmetries, but it still has some
>> issues there
>>
>
> which issues are you referring to?
>
> Paolo
> --
> Paolo Giannozzi, Dip. Scienze Matematiche Informatiche e Fisiche,
> Univ. Udine, via delle Scienze 206, 33100 Udine, Italy
> Phone +39-0432-558216, fax +39-0432-558222
>
>
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