[QE-users] Ground state of DFT+U

[Malte Sachs]

Dear DFT+U experts,

I am wondering how to know what is the ground state of a DFT+U 
calculation from an ab initio point of view. As approximated DFT is not 
variational I think I cannot use the total energy as an argument.

In my case I have calculated the electronic structure of NdCo in the 
CsCl structure type using QE-6.6. The f-orbitals splits into three 
representations: au, t2u and t2g. For Nd, I expect that it has a valency 
of three and that the t2g orbitals are fully occupied for one spin. 
However, I find that occupying the three t2g orbitals by 2/3 and the au 
orbital by one electron yields a lower energy than the expected 
occupation. So I wonder if I can trust in this "ground state" that is 
against my intuition or if this is a well known effect of DFT+U.

All hints are welcome, best regards,

Malte

-- 
Malte Sachs
Anorganische Chemie, Fluorchemie
Philipps-Universität Marburg
Hans-Meerwein-Straße 4
35032 Marburg (Paketpost: 35043 Marburg)
Tel.: +49 (0)6421 28 - 25 68 0
http://www.uni-marburg.de/fb15/ag-kraus/


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