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Dear VASP users

we have been struggling with the electronic ground state of a free Pb atom (POTCAR PBE_Pb_d) for a while now, especially when spin-orbit coupling is incorporated. The electron configuration that we end up with, is definitely not the one that is experimentally known (two 6p-electrons with parallel spin, 3P0 term symbol).

All our calculations yield a zero magnetization and zero orbital angular momentum, while the experimental 3P0 term symbol implies a spin moment of 2 bohr magneton and an orbital moment of 1 bohr magneton. We have been playing around with reduced symmetry of the unit cell (orthorhombic) and with MAGMOM, without success (regardless the value of MAGMOM, the magnetization converges to zero). The keywords FERWE and FERDO in the INCAR file provided much help in similar cases without spin-orbit coupling, but cannot be used for Pb as they are not consistent with spin-orbit coupling.

Any hints on how we can enforce this particular (experimental) electron configuration in a free heavy atom with spin-orbit coupling?

Thank you,
cmm - Ghent

Set NUPDOWN (IIRC) to 2, this fixes the magnetization to 2 (although you might consider this cheating ;-) )
(also check for different fixed magnetizations, maybe magnetization=1 blocks access to magnetization=2...or the obtained results might be degenerate?)

Danny

We tried this: NUPDOWN has no influence on the spins. Both the magnetization and the orbital momentum remain zero, when NUPDOWN is set to 2. We suspect NUPDOWN is not defined for spin-orbit coupling, since this part of the code makes use of non-collinear magnetization (and hence no real 'up' or 'down' is defined).

cmm