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In the presence of a 2D crystal and spin-orbit coupling, the SU(2) spin axial vector (Sx,Sy,Sz) decouples into an in-plane (Sx,Sy) and out-of-plane part Sz. For Quantum ESPRESSO, is there a way to turn on only the out-of-plane component of spin-orbit coupling?

I happen to know that the effect of the in-plane spin-orbit is marginal for the bands near the Fermi level (as observed by photo-emission experiments). Therefore, I imagine turning off the in-plane part would not only be computationally easier, but also more accurate to my material.

I don't see why this shouldn't be possible, seeing that it amounts to a constraint that the up/down spin should be conserved (i.e not mixed at the level of the Hamiltonian). I believe we would say the spins should remain "collinear". However, I can't seem to find anything on this; and when I choose the following parameters

  lspinorb = .true.
  noncolin = .false.

then Quantum ESPRESSO (pw.x) immediately crashes with the following complaint (written in CRASH file):

spin orbit requires a non collinear calculation

🤔

Compared to a non-spin-orbit scf calculation, the only other thing I'm doing differently is use fully relativistic pseudopotentials:

Fe.rel-pbesol-spn-kjpaw_psl.1.0.0.UPF
Se.rel-pbesol-dn-rrkjus_psl.1.0.0.UPF

Any insight would be greatly appreciated!

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I believe this is just a polarized calculation, see here for the input flags: How to do spin polarization calculations using Quantum ESPRESSO?

I think in the case where they are separated, as you say, then spin-orbit does not play a role. Otherwise you are looking at spin-constrained calculations. I don't know if QE does this.

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  • $\begingroup$ Thanks for the response. I don't want to physically polarize the spins, so much as treat up/down spin as separately conserved quantum numbers, i.e the spin symmetry reduces from SU(2) to U(1). This way time-reversal symmetry is preserved, and the net magnetization is zero. This sort of spin-orbit can lead to gaps experimentally. (After some trial and error, I've come to the conclusion that vanilla QEsp doesn't allow for this. Though I suspect some clever hackers could make this work without too much difficulty, as I don't see anything about DFT itself preventing this type of calculation.) $\endgroup$
    – Paul Eugenio
    Mar 1, 2022 at 16:52
  • $\begingroup$ Ok, I don't think this will be computationally more easy since just zeroing non-z components may end up in non-physical results. This was why spin-constrained calculations originated, as far as I know. All-in-all, I think it will be safe to conduct them as implemented in QE. $\endgroup$
    – nickpapior
    Mar 2, 2022 at 8:04

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