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Dear all,
When I do a spin-polarization BSE calculation by using VASP, is it possible to calculate the transition between different spin channels?
When I check my BSEFATBAND file, it seems that I can only obtain the transition between the same spin channel, probably because of the spin selection rule.
Thanks a lot.

Dear all,

When the BSE equation is solved, one makes an ansatz for the excited state. In VASP, and all BSE codes that I know, this ansatz involves a hole and an electron pair prepared within each spin channel (for instance an electron with up spin is removed from the ground state determinant and placed with the same spin into a previously unoccupied orbital). If the ground state is not spin polarized, the excited state is hence corresponding to a singlet.

Since the Hamiltonian commutes with the spin operator, this ansatz is- as far as I recall -exact. In order to obtain excitations between different spin channels, your Hamiltonian must involve terms that do not commute with the spin operator. This means that you have to include spin-orbit coupling (LSORBIT = .TRUE.) and perform BSE calculations also using LSORBIT = .TRUE. to obtain excitations between the different spin channels. However, the transition probabilities between different spin channels will be small (selection rules still apply to some extent).

Alternatively, if your groundstate is not spinpolarized (ISPIN=1), you can specify LHARTREE=.FALSE or the undocumented flag LTRIPLET = .TRUE. in the INCAR file (for the BSE calculation). This assumes that you remove an up electron from the ground state determinant and place a down electron in a previously unoccupied orbital (so the resulting ansatz corresponds to a triplet state). This will yield many transitions with zero transition probabilities. However, I am not sure whether the BSEFATBAND approach "interprets" this case correctly. Also, I do not recall whether we have tested this approach for spin polarized ground states i.e. ISPIN=2 (I believe that I tested this only for anti-ferromagnetic systems).

Georg

The following is a related post: LTRIPLET in BSE calculation!

Dear Georg,

As for a none-spinpolarized system, if I include spin-orbit coupling (LSORBIT = .TRUE.) and perform BSE calculations, the lowest exciton I obtained is singlet or triplet?
Thanks a lot!

Best,
Yuanchang Li

Dear Georg,
You have said that "If you start from an antiferromagnetic ground state, and perform standard BSE calculations (LHARTREE=.TRUE.; LTRIPLET = .FALSE), you will obtain one set of singlet solutions and one set of triplet solutions. If you set LTRIPLET=.TRUE., which does the same as setting LHARTREE=.FALSE., you will obtain the remaining two sets of triplet solutions, with many (or maybe even all) transition probabilities being zero (i.e. these correspond to dark, potentially long lived triplet excitons)." in the topic forum/viewtopic.php?f=4&t=18153
However, I do a test in an AFM system and find that the "triplet solution" in the calculation with "LTRIPLET=.TRUE." and without "LTRIPLET=.TRUE." is not match. I don't know what happened, could you please help me?
Thank you so much!

Dear yuanchang_li,

In spin-polarized systems, the triplet solutions with LTRIPLET=.TRUE. and LTRIPLET=.FALSE. are not the same. In case of LTRIPLET=.FALSE., the triplet solutions correspond to (S=1,Sz=0) and for LTRIPLET=.TRUE. the solutions correspond to (S=1, Sz=1) and (S=1,Sz=-1), where S is the spin and Sz its projection.

You can find a detailed discussion in M. Rohlfing, S.G. Louie, Phys. Rev. B 62, 4927-4944 (2000).