>From your attachment I would guess you are using a DZP basis set for the
6s orbitals, and a DZ basis set for the d orbitals (this can be checked in
the output file, a bit before the calculation starts), right? So, after
the two lines listing the atomic orbitals:
6s 6s 6py 6pz 6px 5dxy 5dyz 5dz2
5dxz 5dx2-y2 5dxy 5dyz 5dz2 5dxz 5dx2-y2
you have the following entry:
1 4.729 0.268 0.145 0.053 0.043 0.095 0.638 0.894 0.562
0.609 0.534 0.205 0.040 0.204 0.204 0.235
The first number refers to the atom number - let's call it iat - in the
list of atoms (in the line above, iat=1). The second is the total Mulliken
charge on atom iat (4.729). Then, in the first line, you have the Mulliken
charge due to the first zeta of orbital 6s, the (Mulliken) charge due to
the second zeta of orbital 6s, and the charge due to the polarization
orbitals, which are 6p orbitals. Following, the charge due to the 1st zeta
of 5dxy, 5dyz and 5dz2, respectively. On the second line, you have the
charge due to the first zetas of 5dxz and 5dx2-y2, and finally the last
four numbers are the charges due to the second zetas of the 5dxy, 5dyz,
5dz2, 5dxz, 5dx2-y2.
So, to have the total charge on a given orbital, just sum the charges due
to the zetas. To get the charge due to d-electrons only, you would sum the
charges over all d-orbitals and zetas. To get the charge due to the
s-states, you will have to sum also over the 6p-orbitals. This is because,
as stated in the siesta manual, although you have no 6p orbitals
explicitly, they are the higher angular-momentum solution to the problem
of a 6s orbital in a weak electric field (that is, they are the
polarization orbitals as defined in siesta). Therefore, they also
contribute to the s-orbitals' charge. (Please someone correct me if I'm
wrong...)
Hope this helps you. Cheers,
Marcos
> Dear Siester users,
>
> I got the results of Mulliken charge (see my attachment), but how to
> evaluate the contribution of different atomic orbitals? My system is
> Platinum (Pt, 5d96s1). Is the main attributor to its stability 5dx2-y2,
> 5dxy, or 5dz2?
>
> Thanks in advance!
>
> Best regards,
> C. H. Hu
>
--
Dr. Marcos Verissimo Alves
Post-Doctoral Fellow
Condensed Matter and Statistical Physics Sector
International Centre for Theoretical Physics
Trieste, Italy
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