[Wien] valence charge

[Yundi Quan]

Hi, All,
I encountered a a problem when calculating BaBiO3. Actually, I want to
reproduce results that other people have already published. Bismuth has [Xe]
4f14 5d10 6s2 6p3configuration. I want to know the how many 6s and 6p
electrons are inside the muffin-tin sphere. I set the cutoff energy for
separating core and valence charge to be -9 Ry. Therefore, 6s and 6p are
both valence electrons. The hybrid functional calculation is carried out for
both s and p states. (I know that they are not quite localized. I just want
to compare the results with hybrid functional to other results). The QTL in
the scf file is as follows:
:PCS002: PARTIAL CHARGES SPHERE =  2
S,P,D,F,PX,PY,PZ,D-Z2,D-X2Y2,D-XY,D-XZ,D-YZ
:QTL002: 0.0000 3.3196 0.0000 0.0000 1.1098 1.1006 1.1091 0.0000 0.0000
0.0000 0.0000 0.0000
:PCS003: PARTIAL CHARGES SPHERE =  3
S,P,D,F,PX,PY,PZ,D-Z2,D-X2Y2,D-XY,D-XZ,D-YZ
:QTL003: 0.0000 3.3871 0.0000 0.0000 1.1242 1.1364 1.1261 0.0000 0.0000
0.0000 0.0000 0.0000
Why are there no s electrons inside the Muffin-Tin sphere? I did include 6s
electrons as valence electrons. Is there a way of finding out how many 6s
electrons are inside the muffin-tin sphere? Thanks a lot.



Yundi
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