[Wannier] Kramers degeneracy broken by Wannier interpolation
Aguilera, Irene
i.aguilera at fz-juelich.de
Tue Jul 25 16:14:09 CEST 2017
Dear all,
somebody suggested me to use "use_ws_distance = .true.". This seems to have solved the problem with the broken Kramers degeneracy.
Thanks to everybody who gave me suggestions.
Best,
Irene
________________________________
From: Aguilera, Irene
Sent: Wednesday, June 21, 2017 2:32 PM
To: wannier at quantum-espresso.org
Subject: Kramers degeneracy broken by Wannier interpolation
Dear all,
this is a general question about the Wannier interpolation technique. I'm using wannier90 to perform a Wannier interpolation for a system with inversion symmetry (IS), time-reversal (TRS) symmetry, and spin-orbit coupling. Therefore, all states must be doubly degenerate (Kramers pairs). The interpolated band structure is in excellent agreement with the explicitly calculated one (DFT or GW). The construction of the Wannier functions seems correct and the bands at the k points which are present in the explicit DFT or GW calculation are indeed doubly degenerate. But this is not the case for the arbitrary k points (q) at which the bands are interpolated. For those q points there is a small splitting between the Kramers partners. I do understand the following:
In order for the Kramers degneracy to be preserved, the Hamiltonian has to be invariant with respect to a certain symmetry operation (IS+TRS in this case):
H(k) = S^T(k)*H(k)*S(k)
where S(k) is the corresponding transformation matrix. This transformation matrix is k-dependent. For the explicit k points, the above condition is fulfilled. For an interpolated q point, we would have to demand
H(q) = S^T(q)*H(q)*S(q) (*)
However, in the Wannier interpolation, H(q) is a linear combination of the H(k):
H(q) = SUM(k) c(q,k) H(k)
with some coefficients c(q,k).
This probably leads to the fact that, because of the k dependence of S(k), the requirement (*) is not fulfilled (in general), and therefore the degeneracy is lifted.
So, I think I understand why this happens and I understand how to improve it and make the splitting negligible (increasing the k points in the DFT calculation does it). But I cannot find a solution to obtain real degeneracies and I could not find discussions about this issue in the literature or in the forum.
Are there approaches to solve this problem? How can one symmetrize H(q)? Can one impose some conditions on H(R) (the Hamiltonian in real space) such that H(q) presents the correct symmetries?
Thank you very much.
Best regards,
Irene
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