<div dir="ltr"><div dir="ltr">Dear <span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif">Fhokrul,</span><div><span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif"><br></span></div><div><span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif">In my experience with my own work (non spin-polarized CdS and CdSe calculations), the Hamiltonian matrix elements are technically complex due to small numerical imprecision. My strongest elements might be order 1 eV for the real part and order 10^-9 eV for the imaginary part. While technically complex, I think you would agree this value is basically real. My weakest elements are perhaps </span>10^-9 eV for both the real and imaginary parts, i.e. these elements are really basically 0. I can't say how this translates to your system, but in my experience for systems like mine (non spin-polarized semiconductors), if the Wannier orbitals you get out are mostly real, as they should be, then the Hamiltonian elements will be mostly real as well.</div><div><span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif"><br></span></div><div><span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif">You can check this by increasing the precision of the printed *hr.dat file. Or you can extract the unitary matrices U(k) from the *.chk file (they are double precision in this file, 16 or 17 digits for each element) and build your Hamiltonian from the DFT eigenvalues yourself in, say, Matlab. </span></div><div><span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif"><br></span></div><div><span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif">Best,</span></div><div><span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif">Peyton Cline</span></div><div><span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif">5th year PhD Student</span></div><div><span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif">Eaves Group</span></div><div><span style="color:rgb(0,0,0);font-family:Calibri,Helvetica,sans-serif">CU-Boulder</span></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Tue, May 26, 2020 at 2:44 AM Md. Fhokrul Islam <<a href="mailto:fislam@hotmail.com">fislam@hotmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">
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Dear developers and users,</div>
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I have recently started to use Wannier90 code and I am wondering if someone can help me clarify some confusions. </div>
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I am working with periodic systems. With spin-orbit coupling, the matrix elements of the Hamiltonian are, in general, expected to be complex except for the diagonal terms. But without spin-orbit, can the matrix element also be complex? Most of my calculations
without spin-orbit show that the matrix elements are real. But one of my most recent calculations shows the elements are complex. So, I am wondering if it is correct or the result of some numerical error? In this case the system doesn't have inversion symmetry.
Could it could be that since the DFT basis set are complex without inversion symmetry, and so the matrix elements of the Wannier90 Hamiltonian are also complex?</div>
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Also, is it possible to obtain complex matrix elements even with inversion symmetry and without spin-orbit coupling if disentanglement procedure is not done properly?</div>
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Regards,</div>
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Fhokrul</div>
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