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

[Peter Blaha]

> >                 Energy to separate semicore and valencestates:  -0.23419
> >
> >
> >         :NOE  : NUMBER OF ELECTRONS          =  50.000
> >
> >         :FER  : F E R M I - ENERGY(TETRAH.M.)=   0.70230
>
> Thanks a lot, Kevin, that was what I've been looking for. At least the
> TiC-example shows a cutoff energy of 0.26299 eV and a Fermi energy of
> 0.74115 - sounds reasonable.

Again you must "think" !!

C has 1s, 2s,2p
Ti    1s,2s,2p, 3s,3p, 3d,4s


Now check the scf file:
        1.ATOM      Ti                    4 CORE STATES
:1S 01: 1S                   -360.968068 Ry
:2S 01: 2S                    -40.563324 Ry
:2PP01: 2P*                   -34.247583 Ry
:2P 01: 2P                    -33.809461 Ry

        2.ATOM      C                     1 CORE STATES
:1S 02: 1S                    -18.600064 Ry

Ah!!! C 1s and Ti 1s,2s,2p,2p* (the is p1/2 and 3/2) are    core    in WIEN.


> TiC-example shows a cutoff energy of 0.26299 eV and a Fermi energy of

??? Let's check again the scf file:  We need to identify Ti 3s, 3p (because
they should be semicore) and (maybe) C 2s (and now it comes: it is your
responsibility to say: C 2s is "semicore" or it belongs to valence (the
more common interpretaation)


          ATOMIC SPHERE DEPENDENT PARAMETERS FOR ATOM  Ti
          OVERALL ENERGY PARAMETER IS    0.3000
          OVERALL BASIS SET ON ATOM IS LAPW
          E( 0)=    0.3000
             APW+lo
->>>      E( 0)=   -3.4675   E(BOTTOM)=   -3.770   E(TOP)=   -3.165  ->>> Ti 3s
             LOCAL ORBITAL
->>>      E( 1)=   -1.7250   E(BOTTOM)=   -2.180   E(TOP)=   -1.270  ->>> Ti 3p
             APW+lo
          E( 1)=    0.3000
             LOCAL ORBITAL
          E( 2)=    0.3000   E(BOTTOM)=    0.250   E(TOP)= -200.000
             APW+lo

          ATOMIC SPHERE DEPENDENT PARAMETERS FOR ATOM  C
          OVERALL ENERGY PARAMETER IS    0.3000
          OVERALL BASIS SET ON ATOM IS LAPW
 ->>>     E( 0)=   -0.7800   E(BOTTOM)=   -0.910   E(TOP)= -200.000 ->>> C 2s
             APW+lo
          E( 0)=    0.3000
             LOCAL ORBITAL
          E( 1)=    0.3000
             APW+lo

       K=   0.00000   0.00000   0.00000            1
:RKM  : MATRIX SIZE   83LOs:  18  RKM= 5.84  WEIGHT= 1.00  PGR:
       EIGENVALUES ARE:
        -3.4182681   -1.6543690   -1.6543690   -1.6543690   -0.1729034
         0.8000601    0.8000601    0.8000601    0.8649445    0.8649445
         0.8649445    0.8752640    0.8752640    1.5130949    1.7755933

--->>> So the Ti 3s orbitals are around -3.4 Ry,
                 3p                     -1.65
              C  2s                     -0.15

higher energies are definitely "valence" (and furthermore cannot be
identified just by "looking" on the eigenvalue. Of course, these states form
bands, thus will cover a certain energy range, not just a sharp number as
seen from 1 k-point only.

SO now it is up to you to decide a good energy to "separate" valence and
semicore. Either you could choose  eg. -1.  (and cutoff just Ti 3s,3p, but
keep C 2s), or you "follow" the suggestion of the scf file:

> TiC-example shows a cutoff energy of 0.26299 eV and a Fermi energy of

which cuts-off also the C 2s states.

Please read the UG (-in1new switch) how the program finds this energy.

Regards


                                      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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