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-boun...@zeus.theochem.tuwien.ac.at> on behalf of delamora <delam...@unam.mx> Sent: Sunday, May 19, 2019 10:25:44 PM To: wien@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. =============== 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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