[Wien] Comparing Total Energies

[David Tompsett]

Dear Fabien,

Thanks for the response. It is probably my k-mesh that is making it
inaccurate to compare.

David.

On Tue, Jul 10, 2012 at 3:48 PM, <tran at theochem.tuwien.ac.at> wrote:

> Hi,
>
> When comparing two total energies obtained from different unit cells, one
> has to be always very careful.
>
> First, if you did the calculation on the
> small unit cell with a k-mesh (n1,n2,n3), then the calculation on the
> large unit cell should be done with the corresponding k-mesh
> (n1/m1,n2/m2,n3/m3), where m1, m2, and m3 indicate by how many times
> (along the lattice vectors a, b and c) is the large unit cell larger than
> the small one. You can have problems if If n1/m1, n2/m2 or n3/m3 is not
> an integer. But this not a problem if you
> are using very good k-meshes such that the energies are well converged.
>
> It can also happen that the orientation of the lattice vectors are
> completely different in the small and large unit cells. In this case
> some numerical noise (e.g., from the FFTs) can introduce some errors.
>
> But actually, what do you mean by "I can't compare"?
>
> My recommendation: use the same unit cell.
>
> F. Tran
>
> On Tue, 10 Jul 2012, David Tompsett wrote:
>
> > Dear All,
> >
> > I have what is probably a very basic question about comparing total
> > energies. I have been considering a system that undergoes a charge
> density
> > wave transition from a high to low symmetry structure. I have been
> > comparing the total energy of the high and low symmetry structures. The
> > unit cell of the high symmetry structure has half as many atoms as does
> the
> > low symmetry one.
> >
> > In my calculations it seems I can't compare 2x(total energy of high
> > symmetry cell) with 1x(total energy of low symmetry cell). Instead I have
> > to perform the high symmetry calculation in the same size unit cell and
> > symmetry as for the low symmetry. Why is this the case? What is the
> effect
> > of having different sized unit cells?
> >
> > Thank you,
> > David Tompsett.
> >
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