[Wien] Help on Core hole calculations
delamora
delamora at unam.mx
Mon May 20 06:25:44 CEST 2019
See the end
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So, you are saying that from this file
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 splitting and symmetrizing, I will arrive at this
R LATTICE,NONEQUIV.ATOMS: 3 161_R3c
9.519303 9.519303 26.005203 90.000000 90.000000120.000000
ATOM -1: X=0.14500000 Y=0.14500000 Z=0.14500000
Fe1 NPT= 781 R0=0.00005000 RMT= 1.97 Z: 26.000
ATOM -2: X=0.05500001 Y=0.75000000 Z=0.44500000
O NPT= 781 R0=0.00010000 RMT= 1.69 Z: 8.000
ATOM -3: X=0.35500000 Y=0.35500000 Z=0.35500000
Fe2 NPT= 781 R0=0.00005000 RMT= 1.97 Z: 26.000
Now, oxygen had multiplicity 6 and now it has 1. Is this correct? Also
The original cell has 10 atoms and this one has much less and it is quite different from the original.
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The crystal is SG 161, I only put the fist position.
There are two Fe1 (up), two Fe2 (dn) and 6 O
I started with the 'P' lattice (above) and from the 4 Fe I took the 1st and 3th as Fe2 and 2nd and 4th as Fe1
You can play other combinations, but this seems the best; each O has two Fe1and two Fe2 forming a cross.
Pablo
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