Hello,
I have a question about example 10, which does calculations on PbTiO_3. The
number of valence electrons specified is 25, nbnd=25, corresponding to a Ti
cation (4s2 3d1) and the Ti pseudopotential , Ti.vdb.UPF also states that it is
generated for the cation,
Wavefunctions nl l occ
3S 0 2.00
3P 1 6.00
4S 0 2.00
3D 2 1.00
Giving 11 valence electrons corresponding to 4s2 3d1.
THough in the same file it states
12.00000000000 Z valence
Is the calculation really being carried out for PbTiO_3 with a titanium cation
and the rest neutral atoms? IF so why? Or is there something I am
misunderstanding about how the pseudopotential and run file are made?
Thank you,
Dr. Helen Eisenberg,
Fritz Haber Center,
Hebrew University
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