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<span style="font-variant-ligatures: no-common-ligatures" class="">Dear all,</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">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:</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">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):</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">H(k) = S^T(k)*H(k)*S(k)</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">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</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">H(q) = S^T(q)*H(q)*S(q) (*)</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">However, in the Wannier interpolation, H(q) is a linear combination of the H(k):</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">H(q) = SUM(k) c(q,k) H(k)</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">with some coefficients c(q,k).</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">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.</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">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.</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">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?</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">Thank you very much.</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">Best regards,</span></div>
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<span style="font-variant-ligatures: no-common-ligatures" class="">Irene</span></div>
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