[Octopus-users] Spin factor in OEP

[Miguel Marques]

  Hi,

On Tue, 2005-04-05 at 18:00 -0400, Nam A. Nguyen wrote:
> I'm doing TDKLI on the excited state of H2 (2 electrons). The relevant 
> part of the input is:
> 
> SpinComponents=1
> ExtraStates=1
> %Occupations
>  1.0|1.0
> %
> 
> So this is a spin-UNpolarized calculation with 2 orbitals, and Octopus 
> assigns
> oep%socc=M_HALF
> oep%sfact= M_TWO
> 
> In the  xc_OEP_SIC.F90 file, the ORBITAL density is calculated as follows:
> 
> st%rho(:, 1) = oep%socc*st%occ(i, is)*R_ABS(st%X(psi)(:, 1, i, is))**2
> 
> This means that:
> oep%socc*st%occ(i, is)=1/2
> 
> Is it not an error here ?

No, I don't think so. If you excite in this manner (singlet state), you
effectively put 0.5 electrons in a spin-up and 0.5 electrons in a spin-
down orbital. You may get other combinations by doing a spin-polarized
calculation. Note, however, than it is sometimes not trivial to get some
spin states with a single Kohn-Sham determinant of purely orbital parts.
(I hope I am not saying something completely wrong ;))

  cheers,
  miguel