[QE-users] Consequence of incorrect setup of electron configuration.

[Eduardo Menendez]

Dear Evgeny,

1. Let's consider a supercell with odd number of electron with
nspin=1 (non-polarized calculation). The result should probably be
incorrect, but how would the error manifest? I'm specifically
interested in errors for electron state structure and total energy.
A/ The calculation may be good if the system is not magnetic.
There are some other important parameters: k-points and smearing.
If  you have a nk1 x nk2 x nk3 k-point mesh, the calculation if equivalent to
a calculation of the supercell of size nk1 x nk2 xnk3 with only the
gamma point.  Hence, if the number of k-points is even, you have an
equivalent even number of electrons.
Even if you choose an odd k-points grid, the smearing may cause the
HOMO to be half occupied.
This is the case for a simple non-magnetic metal, e.g. Al, Cu, where
you need a dense k-points grid,
and there wide conduction band, and the lowest half of the conduction
band gets occupied.

In my experience, the incorrect solutions appear when the Fermi level
intersects a narrow, non-dispersive conduction band,
or have an isolated HOMO level half-occupied. In these cases, applying
nspin=2, and breaking the symmetry
with starting_magnetization, the half-occupied
level or band can split into an occupied and an empty band. This the
case of defects in a big supercell (usually gamma point),
or the case of a dissociating H2 molecule. The same happens for some
crystals where the Fermi level is within a narrow d-band,
nspin=1 gives a metallic system, and nspin=2 gives an insulating state.
Well, if d-levels are involved the system may still be incorrectly metallic due
to the self-interaction error, see the LDA+U method.

In case of doubt, and I would say, always, do a test spin-polarized
calculation and see if you get a smaller total energy.



2. Let's consider a supercell A which is a half of supercell B. A
calculation with supercell B reveals a band with electron population
of 1 electron /per B (the system is paramagnetic and the calculation
uses nspin=2). Would a band with close energy manifest for a run with
supercell A ?

A/ If supercell B is the double of A, then you have an even number of
electrons, unless you charge the system.
I think that you cannot have such band with just one electron in supercell B,
assuming the supercell B is not relaxed after duplication of A.

Also, note that to make A and B equivalent, for supercell A you need
to duplicate the number of k-points, therefore you end
with the same number of occupied energy levels and the same energies,
just distributed in a different fashion.

Best regards

Eduardo Menendez

www.gnm.cl/emenendez
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