[Wien] where can i find the energy to eliminate semicore states

[Peter Blaha]

> > ->>>      E( 0)=   -3.4675   E(BOTTOM)=   -3.770   E(TOP)=   -3.165 ->>>Ti 3s
> > ->>>      E( 1)=   -1.7250   E(BOTTOM)=   -2.180   E(TOP)= -1.270  ->>>Ti 3p
> > ->>>     E( 0)=   -0.7800   E(BOTTOM)=   -0.910  E(TOP)= -200.000 ->>> C 2s
>
> > > > --->>> So the Ti 3s orbitals are around -3.4 Ry,
> > > >                  3p                     -1.65
> > > >               C  2s                     -0.15
> belonging to the different states, I was just wondering how you got to "-0.15 Ry" for the
> C 2s-state if the line in the scf-file states:  E( 0)=   -0.7800.


Agreed: for TiC it is "trivial", what else could it be, the "next"
eigenvalue must be C-2s. However, in more complicated cases with several
different elements,... it might be a little more complicated. The fastest
way during (after) scf is inspection of case.help* files. These files
contain the partial charges for each atom (XXX) and each eigenvalue and
allow a quick and unique determination which character is present for a
particular eigenvalue.

                                      P.Blaha
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Peter BLAHA, Inst.f. Materials Chemistry, TU Vienna, A-1060 Vienna
Phone: +43-1-58801-15671             FAX: +43-1-58801-15698
Email: blaha at theochem.tuwien.ac.at    WWW: http://info.tuwien.ac.at/theochem/
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