[Pw_forum] PDOS in the presence of spin-orbit coupling

[Sclauzero Gabriele]

In general, you cannot, because in presence of spin-orbit coupling the spherical harmonics are not eigenstates of the atomic Hamiltonian. Therefore the code projects onto spin-angle functions (see PRB 71, 115106), which are eigenstates of total angular momentum J=L+S, and of its projections along z.
As you can see from eqs. 3 and 4 of that paper, spin-angle functions of, say, j=1/2 and m_j=1/2 contains contributions from both m=0 and m=1, while the one with  j=1/2,m_j=-1/2 from both m=0 and m=-1.
The only spin-angle functions that have contributions from a unique spherical harmonic are the j=3/2,m_j=3/2 (only m=1 contributions), and the j=3/2,m_j=-3/2 (only m=-1).

HTH

GS

Dear all,

I face with a problem in the pdos calculations. In the absence of spin-orbit coupling the for pdos has a form like this "X.pdos_atm#1(Y)_wfc#2(p)" in the data file one can find the Pz, Py and Px contribution.
But by inclusion of SOC the I have "X.pdos_atm#1(Y)_wfc#2(p_j0.5)" and "X.pdos_atm#1(Y)_wfc#3(p_j1.5)". The first one has 2 column (2*0.5+1) and the second one has 4 column (2*1.5+1).
How can I recognize the contribution from different component of P (Px, Py and Pz)?

Best,

--
Mohsen Modarresi,
PhD student of Solid State Physics, Ferdowsi University of Mashhad, Iran.
Phone +98-9133452131
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Dr. Gabriele Sclauzero
Materials Theory (D_MATL)
ETH Zurich, HIT G 43.2
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8093 Zürich, Switzerland

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