[ase-users] Space group #68 and Sr3Ru2O7

[Jens Jørgen Mortensen]

Can anyone here help me build the Sr3Ru2O7 structure described here:

http://www.sciencedirect.com/science/article/pii/S0022459600987966

Table 2 says something like this:

Atom  x       y      z       fraction
Ru    1/4     1/4    0.40298 1
Sr1   1/4     1/4    0       1
Sr2   1/4     1/4    0.18631 1
O1    1/4     1/4    1/2     1
O2    1/4     1/4    0.30448 1
O3    0.5293 -0.0293 0.09710 0.959
O4    0.4707  0.0293 0.09710 0.041

a=5.5006 Å, b=5.5006 Å, c=20.725 Å
Space group: Bbcb (#68)

I tried:

from ase.lattice.spacegroup import crystal
a = 5.5006
c = 20.725
x = crystal('RuSr2O3',
            [(0.0, 0.0, 0.40298),
             (0.25, 0.25, 0.0),
             (0.25, 0.25, 0.18631),
             (0.25, 0.25, 0.5),
             (0.25, 0.25, 0.30448),
             #(0.5293, -0.0293, 0.09710),
             (0.4707, 0.0293, 0.09710)],
            spacegroup=68,
            #setting=1,
            cellpar=[a, a, c, 90, 90, 90])
x.edit()

But I must admit, I have no idea what I am doing.  I get this warning:

/home/jensj/ase/ase/lattice/spacegroup/spacegroup.py:375: UserWarning:
scaled_positions 1 and 3 are equivalent
  'are equivalent'%(kinds[ind], kind))

and the structure I get has too many atoms.  Any idea what I am doing
wrong?

Jens Jørgen