[Wien] Equivalent and inequivalent atoms in practice

Lyudmila Dobysheva lyu at otf.pti.udm.ru
Tue Feb 21 09:52:10 CET 2006


On Tuesday 21 February 2006 02:15, Atta-fynn, Raymond wrote:
> Angles: alpha = beta = 90, gamma = 120  Lattice type: H
> (111) surface coordinates in (x, y, z) format:
>      Layer 1                 Layer 2                        Layer 3 
> (0.0, 0.0, 0.0) (0.33333,0.166666, d/c) (0.166666, 0.33333, 2*d/c)
> (0.0, 0.5, 0.0) (0.33333,0.666666, d/c) (0.166666, 0.83333, 2*d/c)
> (0.5, 0.0, 0.0) (0.83333,0.166666, d/c) (0.666666, 0.33333, 2*d/c)
> (0.5, 0.5, 0.0) (0.83333,0.666666, d/c) (0.666666, 0.83333, 2*d/c)
> To create the case.struct file, I can EITHER enter all
> 12 atoms as inequivalent (Space group # 1 P1)

This is clear.

> OR I can enter 4 atoms on each layer as equivalent, i.e.  total of 3
> inequivalent atoms each with 4 atomic positions (Space group # 156 P3m1). 

If I take this group it gives me:
H   LATTICE,NONEQUIV.ATOMS:  3156_P3m1
 12.270000 12.270000 70.018413 90.000000 90.000000120.000000
ATOM  -1: X=0.00000000 Y=0.00000000 Z=0.00000000
ATOM   2: X=0.33333333 Y=0.16666667 Z=0.10000000
ATOM   2:X= 0.83333333 Y=0.16666666 Z=0.10000000
ATOM   2:X= 0.83333333 Y=0.66666667 Z=0.10000000
ATOM   3: X=0.16666667 Y=0.33333333 Z=0.20000000
ATOM   3:X= 0.66666667 Y=0.83333334 Z=0.20000000
ATOM   3:X= 0.16666666 Y=0.83333333 Z=0.20000000

And not your 12 atoms. How did you receive such positions?

> Why are the energies obtained in the case for
> 3 inequiv. atoms lower than the energies obtained in the case 12 inequiv.

I suspect there is somthing wrong in the consistency of the struct files

> My second question:  Should I assume that MM for an inequivalent atom is the
> same for all four atoms? 

Yes.

> Also, for the case of 3 inequiv. atoms, I flipped
> the spins for the atom on layer 2 for AFM calculations. Is this equivalent
> to flipping the spins for the 4 individual atoms on layer 2 in the case
> corresponding to 12 inequiv. atoms?

Yes.

> Third question (unrelated): In the user guide, it states that the energy
> obtained using lstart (which is fully relativistic) can be used as the
> energy of an isolated atom if the element is light (section 6.4, pp. 67).
> This therefore applies to elements such as C, N, O, Si, etcetera. Is this
> always true? 

I don't know

Best regards
  Lyudmila Dobysheva 
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