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bader value of H

Posted: Mon Apr 24, 2006 7:32 pm
by hunterdai
# X Y Z VORONOI BADER % MIN DIST
------------------------------------------------------------------------------------------

61 9.5265 2.2454 6.7469 1.9106 1.9567 0.4880 1.1065

This atom is a H atom, why the bader value is almost 2, does it mean it gets about a whole electron?

Thanks

Posted: Mon Apr 24, 2006 7:56 pm
by graeme
Yes, although it seems a little strange. Is the H atom is in a very electro-rich environment? Also, there are several versions of the H pseudopotential which have different charges, for terminating bonds. Make sure you have the standard 1 electron pseudopotential.

Posted: Wed Apr 26, 2006 8:36 pm
by hunterdai
I used H_200eV pseudopotential, is that ok?
It only said : VRHFIN =H: ultrasoft test in the POTCAR file of H_200eV

Posted: Wed Apr 26, 2006 8:43 pm
by graeme
Yes, that should be fine. If you post the configuration (POSCAR file) or send us a .tar.gz file of the run, we could take a closer look to make sure things are working properly.

Posted: Fri Apr 28, 2006 4:26 pm
by hunterdai
you can download my runs at :
http://puccini.che.pitt.edu/~bdai/withH.tar
http://puccini.che.pitt.edu/~bdai/withoutH.tar

in withoutH.tar the charge in atom 1,2 are 7.267 and 6.439 respectively. It should not be that different, right?

in withH.tar atom 3,4 are same position with atom 1,2 in withoutH.tar. This time the charge are 6.397 and 6.305 in withH.tar. It looks reasonable.

Am I right?

By the way, I calculate this because I want to figure out why H have different adsorption energies in different surfaces by comparing the local charge distrubition. Is that reasonable? Thanks

Posted: Fri Apr 28, 2006 9:41 pm
by graeme
Your results look just fine to me. The H atom is next to Mg atoms, so it makes sense for the H to accept an extra electron from the Mg.

You can certainly try to correlate charge transfer with binding energy.

It's not clear to me if the asymmetric charges are a problem. Your without H is a much smaller cell. Wouldn't the issue be clearer if you simply removed the H from your withH cell and compared those charges more directly?

Posted: Mon May 01, 2006 7:40 pm
by hunterdai
well, I did another calculation with 2X2 supercell of without H structure, which is pretty much similar with your suggestion. But the asymmetric charges are still there, same with 1X1 case. So....

Is that a problem?

Thanks