[Wien] Equivalent and inequivalent atoms in practice

[Atta-fynn, Raymond]

Dear Peter, Stefan, and WIEN users,
 
As far as WIEN is concerened, atoms are considered equivalent if they have the 
same atomic number, same number of nearest neighbors, and if there exist symmetry 
transformations (translations + rotations) between the atomic coordinates. 
 
Before I ask my questions, let us consider a (2 x 2) surface unit cell with a 
slab of 3 layers'  thickness for (111) surface of an FCC lattice (4 atoms per layer, total of 12).
 
Cell dimensions: a=b=12.27 Bohrs, c=70.0184131 Bohrs (vacuum = 60 Bohr)
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)
 
where d is the inter-layer separation = lattice const./sqrt(3.)
 
It can clearly be seen that atoms on each layer are connected by simple symmetry operations.
To create the case.struct file, I can EITHER enter all 12 atoms as inequivalent (Space group # 1 P1) 
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). Hence using the latter for calculations is faster than 
the former. 
 
My first question concerns the accuracy of the energies at the AFM level with spin-orbit interactions: 
Why are the energies obtained in the case for 3 inequiv. atoms lower than the energies obtained in 
the case 12 inequiv. using the same set conditions (Rmt, Kmax, energy convergence criterion, etc.)? 
I must state, however, that I did not use the same number of  irreducible k-points because of the 
different symmetries in the two cases;
12 inequivalent atoms --->16 irred. k-points,  
3 inequivalent atoms ---> 19 irred. k-points.
 
My second question:  In the case where I have 3 inequiv. atoms, how do  I obtain information 
like magnetic moments in the sphere for each of the original 12 atoms? Should I assume that  
MM for an inequivalent atom is the same for all four atoms? 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?
 
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? 
The reason why I ask is that for O or C say, the gamma point, large FCC box method outlined 
in the FAQ section yield energies that are significantly higher than the energies from lstart.  
 
 
Thanks
 
Ray Atta-Fynn
 
 
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