[gpaw-users] hartree potential and LCAO coefficients

[Ask Hjorth Larsen]

Hello Ondřej

The LCAO coefficients are stored in calc.wfs.kpt_u[u].C_nM and
distributed over bands n, if you use that (probably not).

C_nM[n] is the set C_M of coefficients for band n.  As for which M
belong to which atom, calc.wfs.setups.M_a[a] gives the first index M
of the basis functions on atom a, and calc.wfs.setups[a].nao is the
number of indices M on that atom, i.e.,

if M1=calc.wfs.setups.M_a[a] and M2 = M1 + calc.wfs.setups[a].nao,
then C_nM[n, M1:M2] are the coefficients of state n on atom a.

As for which M correspond to which actual basis functions, the M are
ordered in the same way as in the basis set (as listed in the txt file
or shown by tools/analyse-basis), with different spherical harmonics
of the same basis function adjacent.  Typically s, py, pz, px, ...

I hope this makes sense. :)

We should have an object which knows this layout and allows accessing
basis functions and projectors in a sensible way.

As for the Hartree potential, there should be a way but it escapes me
for the moment.  It is available "raw" (in Hartree) as
calc.hamiltonian.vHt_g on the fine grid distributed over domains.  I
can further reveal that there's a method get_electrostatic_corrections
which is meant to account for the PAW effects to the pseudo Hartree
potential.  Maybe someone else remembers this better.

Best regards
Ask

2015-03-11 15:29 GMT+01:00 Ondřej Krejčí <Okrejcio at seznam.cz>:
> Dear GPAW developers,
>
> I would like to know, if there is some way in the gpaw, how to get Hartree
> potential or valence-Hartree potential from GPAW? I guess that
> get_effective_potential() gives you potential with included XC corrections.
>
> I have a second question, whether in the LCAO mode is there some possibility
> how to obtain LCAO coefficients for each band, atom, orbital momentum and
> spin? The projection to Gaussian as it is in get_projections() or real-space
> wave-function - get_pseudo_wave_function() - is not something I'm looking
> for.
>
> Thank you for your answers!
> Sincerely
> Ondrej Krejci
>
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