Dear all, I got confused when I looked into the Mulliken population in the output file of water molecule calculation. This is the Mulliken population part of the output file.
******************************************************************************* mulliken: Atomic and Orbital Populations: Species: O Atom Qatom Qorb 2s 2s 2py 2pz 2px 2py 2pz 2px 2Pdxy 2Pdyz 2Pdz2 2Pdxz 2Pdx2-y2 1 5.730 0.802 0.581 1.742 1.306 0.561 0.051 0.164 0.502 0.000 0.003 0.003 0.013 0.002 Species: H Atom Qatom Qorb 1s 1s 1Ppy 1Ppz 1Ppx 2 1.135 0.440 0.452 0.102 0.100 0.040 3 1.135 0.440 0.452 0.102 0.100 0.040 mulliken: Qtot = 8.000 ******************************************************************************* O has 6 original valence electrons and H has 1 original valence electron. >From the output file, O in H2O has 5.730 electrons and H in H2O has 1.135 electrons. This means that O has lost electrons and H has gained electrons, which is not true in an actual H2O molecule. In fact, O should be negatively charged while H is positively charged. Does it mean that I should not use Mulliken population for investigating charge transfer or that I misinterpreted it? I checked out the dipole moment of the system too from the output file and its direction turned out to be right. (opposite to the Mulliken pop result). Here are the coordinate information and the dipole moment. ******************************************************************************* -0.00000074 -0.00000064 0.01696909 1 O 1 0.77007244 -0.00000006 0.60794331 2 H 2 -0.77007216 0.00000001 0.60794310 2 H 3 siesta: Electric dipole (a.u.) = 0.000001 0.000001 0.525608 siesta: Electric dipole (Debye) = 0.000002 0.000002 1.335964 ******************************************************************************* Please let me know if anyone has any idea about this problem. Thank you, Sungjong Woo [EMAIL PROTECTED] Postdoctoral Research Associate University of Massachusetts Lowell Department of Physics and Applied Physics
