bader charge calculation

[dwyuan]

Dear all,

I want to calculate the charge transfer between metal cluster and reduced TiO2 surface with bader charge analysis with the CHGCAR from
VASP calculation.
I put a Au dimer on a O-vacancy of TiO2(110) surface. However, I didn't
find any charge transfer with bader charge.
the calculation details:
4x2 cell, 3 triple-layers, only gamma-point,
400 eV for ecutt off.

As reported in theoretical or experimental studies, some charge transfer was expected in these systems. I will be very appreciated to have your suggestions to improve my results!

Thanks a lot in advance.

[graeme]

The charge transfer could be small, but it can't be precisely zero.

What were the charges on the Au atoms and the neighboring Ti and O atoms? Is the defect charged or neutral?

Did you include core charges, using the LAECHG=.TRUE. flag?

[schretien]

We wrote a paper about charging of small Au clusters when supported on a partially reduced surface (S. Chretien and H. Metiu J. Chem. Phys. 126, 104701 (2007). In contrast to the general belief, charging does not automatically take place when Au clusters are sitting on a vacancy site. Charge transfer depends on the shape of the LUMO for even clusters and SOMO for odd clusters: charge transfer from the reduced surface to the cluster takes place only when the LUMO (or SOMO) has a lobe pointing in the direction of a fivefold coordinated atom. Otherwise, clusters with even "n" stay neutral whil odd "n" become positively charged. In the specific case of the dimer, we found a charge of +0.03 electron for the equilibrium structure in which the dimer is bounded to the bridging oxygen atoms located on each side of the missing oxygen. However, a charge of -0.70 electron is obtained for the second lowest energy structure in which one Au is sitting at the vacancy site while the other one is adsorbed over a fivefold corrdinated Ti atom.

These charges were obtained using the soft-pseudopotential (ENCUT =396 eV), the PW91 functional, a [4x2] supercell, a 12 layer slab.

All the best,

Steeve