[Wien] Help on Core hole calculations

[delamora]

Professor Blaha, Israel,
The following suggestion does not work;
-----------------------------------
No. You should not run a supercell for a R structure, since this will
even with 1x1x1 create a 3 times (conventional hexagonal) cell.
Instead,
i) remove the spacegroup and change to "R"-lattice.
ii) split the 4 Fe positions into 2 and 2 (Fe1 and Fe2). This splitting
can be done in different ways (up-up-dn-dn along z; or in other ways.
Please check literature which one is correct). After nn (or sgroup) you
should have only 2 Fe positions.
-----------------------------------
The original cell is;

Title
R   LATTICE,NONEQUIV.ATOMS:  2 167_R-3c
MODE OF CALC=RELA unit=ang
  9.519310  9.519310 26.005287 90.000000 90.000000120.000000
ATOM  -1: X=0.35500000 Y=0.35500000 Z=0.35500000
ATOM  -1:X= 0.64500000 Y=0.64500000 Z=0.64500000
ATOM  -1:X= 0.85500000 Y=0.85500000 Z=0.85500000
ATOM  -1:X= 0.14500000 Y=0.14500000 Z=0.14500000
Fe         NPT=  781  R0=0.00005000 RMT=    2.0000   Z: 26.000

ATOM   2: X=0.94499999 Y=0.55500000 Z=0.25000000
ATOM   2:X= 0.05500001 Y=0.44500000 Z=0.75000000
ATOM   2:X= 0.55500000 Y=0.25000000 Z=0.94499999
ATOM   2:X= 0.44500000 Y=0.75000000 Z=0.05500001
ATOM   2:X= 0.25000000 Y=0.94499999 Z=0.55500000
ATOM   2:X= 0.75000000 Y=0.05500001 Z=0.44500000
O          NPT=  781  R0=0.00010000 RMT=    2.0000   Z:  8.000
.................................
but if space group 167 is changed to R the cell is changed to
.................................
Title
R   LATTICE,NONEQUIV.ATOMS:  2
MODE OF CALC=RELA unit=ang
  9.519310  9.519310 26.005287 90.000000 90.000000120.000000
ATOM  -1: X=0.00000000 Y=0.00000000 Z=0.35500000
ATOM  -1:X= 0.64500000 Y=0.64500000 Z=0.64500000
ATOM  -1:X= 0.85500000 Y=0.85500000 Z=0.85500000
ATOM  -1:X= 0.14500000 Y=0.14500000 Z=0.14500000
Fe         NPT=  781  R0=0.00005000 RMT=    2.0000   Z: 26.000

ATOM   2: X=0.97166667 Y=0.33333333 Z=0.58333333
ATOM   2:X= 0.05500001 Y=0.44500000 Z=0.75000000
ATOM   2:X= 0.55500000 Y=0.25000000 Z=0.94499999
ATOM   2:X= 0.44500000 Y=0.75000000 Z=0.05500001
ATOM   2:X= 0.25000000 Y=0.94499999 Z=0.55500000
ATOM   2:X= 0.75000000 Y=0.05500001 Z=0.44500000
O          NPT=  781  R0=0.00010000 RMT=    2.0000   Z:  8.000
.................................
but if instead one changes 167 to P, but the cell parameters are changed to
 10.263863 10.263863 10.263863 55.256000 55.256000 55.256000
then the cell has all the characteristics of the original but the Fe can be split into 'up' and 'dn'
.................................
Title
P   LATTICE,NONEQUIV.ATOMS:  2
MODE OF CALC=RELA unit=ang
 10.263863 10.263863 10.263863 55.256000 55.256000 55.256000
ATOM  -1: X=0.35500000 Y=0.35500000 Z=0.35500000
ATOM  -1:X= 0.64500000 Y=0.64500000 Z=0.64500000
ATOM  -1:X= 0.85500000 Y=0.85500000 Z=0.85500000
ATOM  -1:X= 0.14500000 Y=0.14500000 Z=0.14500000
Fe         NPT=  781  R0=0.00005000 RMT=    2.0000   Z: 26.000

ATOM   2: X=0.94499999 Y=0.55500000 Z=0.25000000
ATOM   2:X= 0.05500001 Y=0.44500000 Z=0.75000000
ATOM   2:X= 0.55500000 Y=0.25000000 Z=0.94499999
ATOM   2:X= 0.44500000 Y=0.75000000 Z=0.05500001
ATOM   2:X= 0.25000000 Y=0.94499999 Z=0.55500000
ATOM   2:X= 0.75000000 Y=0.05500001 Z=0.44500000
O          NPT=  781  R0=0.00010000 RMT=    2.0000   Z:  8.000
...............................................
After this one can split Fe, with two Fe 'up' and two 'dn' then 'sgroup' will modify the parameters and the SG=161
the cell will look different, since one Fe will be moved to 0,0,0 since it does not have inversion symmetry
but you can symmetrize it with
****
Fe2O3.outputsgroup;
.....
Note that shift vectors for this space group are defined
only up to the vector { Z, Z, Z }.
Here Z can take any value.
****
and obtain the cell that I suggested last friday

Pablo

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