[ase-users] Space group #68 and Sr3Ru2O7

[Jesper Friis]

What you are doing looks correct. The warning says that the symmetry operations imposed by the spacegroup will map atom 1 and 3 to the same sites. I tested to enter your structure in VESTA and got the same (wrong) structure as shown by ASE, so I guess that this is due to an error in the description of the structure and not in the spacegroup module. I found the attached cif-file for the Sr3Ru2O7 structure from Physical Review, Serie 3. B - Condensed Matter (18,1978-) (1998) 58, p8515-p8521, which seems to display correctly.

Regards
/Jesper


________________________________________
From: ase-users-bounces at listserv.fysik.dtu.dk [ase-users-bounces at listserv.fysik.dtu.dk] On Behalf Of Jens Jørgen Mortensen [jensj at fysik.dtu.dk]
Sent: Tuesday, September 20, 2011 2:28 PM
To: ase-users at listserv.fysik.dtu.dk
Subject: [ase-users] Space group #68 and Sr3Ru2O7

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




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