Dear Mona Asadi Namin
Atoms in solid-state compounds cannot be generally described as the
same atoms in solution. When you simulate Mn(II) or Ni(II) in
solution, you actually find atomic charges fully compatible with a
description based on the concept of "oxidation state". However, in the
solid state the same atoms are generally involved in back-donation
processes which alter their totaal charge, so that their atomic charge
is not anymore compatible with their formal oxidation states. You may
want to read this article to have an idea:
H Raebiger, S Lany, A Zunger; Nature 2008, 453, 763.
When feasible, local magnetic moment is a better indicator of
oxidation state of atoms in solid-state systems (you may want to flick
through the supporting information of J. Am. Chem. Soc. 2015, 137,
10254).
HTH
Giuseppe
Quoting Mona Asadinamin <[email protected]>:
Dear all;
I am doing Hubbard calculations on lithium niobate; specifically, I
am simulating the Nb antisite (that sits in a Li site) with charge
state +5.
How do we simulate this situation in QE?
I have set tot_charge=5 and also set the initial occupations of the
Nb d orbitals to zero; however after a scf iteration, the antisite
Nb d orbitals will gain partial occupations, similar to other Nb
atoms.
So, how can I empty the d orbitals?
This procedure is what was mentioned in the paper:
PHYSICAL REVIEW MATERIALS 1, 054406 (2017)
But, I do not understand the underlying physics. When we put a Nb
atom in Li site, it should lose 4 electrons and in real world these
are shared with oxygens in the lattice. But if we empty the Nb d
orbitals initially, is that physical?
Best regards;
Mona Asadi Namin
Graduate student
Center for simulational physics
University of Georgia
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