[Wien] Vacancy problem with Wien2k

[Peter Blaha]

> Now comes my question, when I'm not wrong in this case I should have 18
> atoms per layer(or 17 + vacancy on the surface). This gives the total
> number of atoms 88 + 2 vacancies. The symetry group I have in this case
> is Pmmm and I have 23 inequivalent atoms.
> Maybe this is a silly question but in "case.outputs" file I found line
> which says that the number of atom is 352 (also 4 times bigger).
>
> After the list of all 88 atoms with positions comes section:
>
> DETERMINATION OF POINTGROUP FOR ALL POSITIONS
> and there i see lines like this
> ...
> Fe        :  16 Atoms, Index  337 to  352
> number of atoms: 352
> ...
>
> I would like to know what is Wien2k doing here? It tries to generate all
> possible positions of atoms, and then reduce the total number to only
> inequivalent ones ? I hope it is calculating _only_ the cell with 88 atoms.

Yes, WIEN does the calculation for your 88 atoms.

Only for the determination of symmetry I need more atoms:

Consider a simple tetragonal lattice with just one atom at the origin.

With a single atom at (0,0,0) you cannot determine any symmetry!!!
You must add the lattice vectors and "create" 3 additional atoms like
(a,0,0), (0,a,0) and (0,0,c)

With these "4" atoms you can no go ahead and determine the symmetry....

                                      P.Blaha
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Peter BLAHA, Inst.f. Materials Chemistry, TU Vienna, A-1060 Vienna
Phone: +43-1-58801-15671             FAX: +43-1-58801-15698
Email: blaha at theochem.tuwien.ac.at    WWW: http://info.tuwien.ac.at/theochem/
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