[Wien] Help on Core hole calculations

[Israel Omar Perez Lopez]

Hi Pablo,

Can you please share with me the whole structure file? What is the multiplicity of each Fe atom? According to Prof Blaha, for core hole calculations I should have MULT=1 for each Fe atom!

Thanks
________________________________
From: Wien <wien-bounces at zeus.theochem.tuwien.ac.at> on behalf of delamora <delamora at unam.mx>
Sent: Sunday, May 19, 2019 10:25:44 PM
To: wien at zeus.theochem.tuwien.ac.at
Subject: Re: [Wien] Help on Core hole calculations

See the end
===============

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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