[Pw_forum] symmetry breaking in QE and VASP

Hanghui Chen chenhanghuipwscf at gmail.com
Tue Jan 29 03:43:36 CET 2013

Dear QE developers,
     This is a comment rather than a question.
     In QE, if the system has some symmetry (say cubic symmetry), and even
if you intentionally turn off the symmetry (set nosym = .TRUE.), the
resulting self-consistent charge density in fact still has the original
cubic symmetry (i.e. three 3 t2g orbitals have the occupancy and so do the
two eg orbitals). Interestingly, if you do the same procedure in VASP, even
the atomic structure has the cubic symmetry, the symmetry could be
spontaneously broken in the self-consistent charge density. This is
particular the case when the Hubbard U is turned on. Because of this, it is
extremely difficult to stabilize an insulating LaTiO3, since the three t2g
orbitals always have the same occupancy in QE. But in VASP, the symmetry is
spontaneously broken and one t2g orbital is largely occupied and an
insulating ground state (d^1 configuration) can be stabilized with a
reasonable U (~7 eV).
     I am wondering whether there is some automatic symmetrization in
charge density after each consistent iteration? And sometimes, when the
physical system does have an orbital ordering, how can we correctly do the
calculations (i.e. avoid missing the orbital-ordered state) with the
rotationally invariant LDA+U in QE?
     Thank you very much.

Dr. Hanghui Chen
Department of Physics
Columbia University
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