[DFTB-Plus-User] Spin-polarized md simulation using DFTB+

[Bálint Aradi]

Dear Kamal,

> 
> 1) The basis set for Sulfur atom in DFTB3 includes d-orbitals. I
> have read in the literature (
> 
> "Zoltan Bodrog* and Balint Aradi Phys. Status Solidi B 249, No. 2, 
> 259–269 (2012) / DOI 10.1002/pssb.201100524) the following: "The 
> simplistic, ‘semidiagonal’ U matrix is a good approximation for
> elements up to the third period, but it is significantly wrong above
> (subshell hardness errors going well beyond 10%, see Table 1) if d
> orbitals are present in the atomic basis."
> 
> I understand that S is not third period but certainly polarizable. Is
> it OK to therefore use an atomic based Hubbard matrix in this 
> particular case for open shell situations for example including an 
> excess electron? In other words, since I would like to use DFTB3
> and the 3ob-3-1 S-K parameters and orbital resolved Hubbard
> derivatives are not available for most of the atoms is it reasonable
> to use the atomically-resolved methodology with the option
> ShellResolvedSpin=Yes in DFTB+? I notice that the spin parameters 
> exist for all the orbitals in my system.

First of all, you have to use the same settings (orbital resolved,
non-orbital resolved, etc.) as was used during the parameterisation.
Otherwise, your electronic Hamiltonian won't be compatible with the
repulsive. So, if the 3ob set contains only atomic Hubbard derivatives,
use only atomic resolved SCC.

> How should I understand this (beyond the fact that d orbitals are in 
> principle empty but are there for polarization effects)? How should
> one decide about using Wpp or Wdd in this case, particularly
> considering that the condensed phase system includes an excess
> electron (not necessarily localized on S but who knows)?

Spin coupling constants (and also Hubbard U values) for empty orbitals
can be very artificial (and physicall wrong) sometimes, due to the
problems of empty orbitals in DFT in general. So, take the highest
occupied one (pp) instead. Actually, for the chemical hardness (Hubbard
U), the highest occupied orbital was taken for the entire atom, so
taking the same for the spin would make sense.

Best regards,

Bálint

-- 
Dr. Bálint Aradi
Bremen Center for Computational Materials Science, University of Bremen
http://www.bccms.uni-bremen.de/cms/people/b-aradi/


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