Re: [Wien] Optimization of an spinel structure
Hi Pablo, You repeat about the structure what I already told (maybe too short), if you check the intenational tables of crystallography you will find that there are two set-ups of the origin possible (0,0,0) or (1/8,1/8,1/8) in space group 227. Therfore, it is not unique if you give only the numbers (x,y,z) but you need to give the Wyckoff position and/or which origin setup you use. This you do now. x is already a free parameter of the structure in your case with origin at (1/8,1/8,1/8), therefore the -min should find whether x is 1/4 or 1/4+delta, so far the non magnetic case. My main concern, however, was not the structure but the antiferromagnetic set up and its symmetry assignment. a) if the spin arrangement is known from experiment then you should use it, (Neutron diffraction, XMCD, etc.) b) Is it true that only one species carries a magnetic moment or contibute both, Fe and Mn (and eventually with small part also O) ? c) was the measurement for the ground state or any excited state, if it was with any magnetic fields at non Zero temperature then it might be very different from your calculation. d) for magnetic systems one cannot just use the "non-magnetic" space groups but one has to use the magnetic ones. ... as Laurence already told "If you have constructed the AFM correctly the structure will have the symmetry of the Coulomb field, not the magnetism." The important thing is the end of the sentence: "not the magnetism" !!! You assume a very fixed symmetry which may be far off from reality, in your case, the Mn atoms can not beome antiferromagnetically alligned as you have only 1 distinguished side. One thing that you can do is to create the cubic cell using supercell, then you fill find that there are much mor possibillities to have an antiferromagnetic arrangement. Note, the direction of the magnetic moments must not be a long a cube axis [001] but may be along the space diagonal [111], which will result in different symmetry, too. You tell the problem how it has to look like (tetragonal) and then you wonder that it does not agree with you (cubic) because it takes you serious. ... are you sure it is cubic or was this just an interpretation of experimental data that used a cubic model to explain the data where the atoms (not the magnetic moments) are sitting ? again: ... as Laurence already told "If you have constructed the AFM correctly the structure will have the symmetry of the Coulomb field, not the magnetism." !!! Anyway, if the actual magnetic order is different from that what you use in the calculation, what do you like to learn ? That was my question. My other doubt is, that you select different x parameters but I would say if you have differnt results, then it's numerics or something is wrong with Wien2k (I did many structure optimizations starting from different position parameters but the result for the relaxed structure was always the same within numerics, if I did not make the mistake). ... or do you try to use the result that you like most and forget about which is the relaxed one. Youre misunderstanding is probably that you try to tell the system what it has to be, instead of asking what it is. Ciao Gerhard DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy: "I think the problem, to be quite honest with you, is that you have never actually known what the question is." Dr. Gerhard H. Fecher Institut of Physics Johannes Gutenberg - University 55099 Mainz Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora [delam...@unam.mx] Gesendet: Mittwoch, 3. November 2021 04:40 An: A Mailing list for WIEN2k users Betreff: Re: [Wien] Optimization of an spinel structure Trying to answer your comments; SG227 is FCC, that is, the conventional cell has 4 primitive cells the first wyckoff positions are; 8a 1/8,1/8,1/8 8b 3/8,3/8,3/8 16c 0,0,0 16d 1/2,1/2,1/2 32e x,x,x 8a and 8b have 2 atoms in the primitive cell 16c and 16d have 4 32e has 8 MnFe2O4 is a spinel Mn; 8a Fe; 16d O; 32e x=0.2639 The primitive cell has 2 formula units while the conventional cel has 8 formula units then the atomic positions in MnFe2O4.struct are; F LATTICE,NONEQUIV.ATOMS: 3 227_Fd-3m 16.059466 16.059466 16.059466 90.00 90.00 90.00 ATOM 1: X=0.1250 Y=0.1250 Z=0.1250 ATOM 1:X= 0.8750 Y=0.8750 Z=0.8750 Mn NPT= 781 R0=0.5000 RMT=1.8600 Z: 25.000 ATOM -2: X=0.5000 Y=0.5000 Z=0.5000 ATOM -2:X= 0.5000 Y=0.7500 Z=0.7500 ATOM -2:X= 0.7500 Y=0.7500 Z=0.5000 ATOM -2:X= 0.7500 Y=0.5000 Z=0.7500 Fe NPT= 781 R0=0.5000 RMT=2.1500 Z: 26.000 ATOM -3: X=0.2639 Y=0.2639 Z=0.2639 ATOM -3:X= 0.7361 Y=0.7361 Z=0.7361 ATOM -3:X= 0.7361 Y=0.5139 Z=0.5139 ATOM -3:X= 0.2639 Y=0.9861 Z=0.98610
Re: [Wien] Optimization of an spinel structure
Technically, you seem to do everything correct. When you split the 4 Fe into 2 sets, you break cubic symmetry. You also see it from the resulting Sg #74. Therefore there will be a relaxation ... (even without relaxation, you already broke cubic symmetry). Your approximation of an AFM ordering in this specific way, simply breaks symmetry. And since this is a Kagome lattice, there will probably be no simple (collinnear) way to introduce AFM without breaking symmetry. Regards Peter Blaha Am 03.11.2021 um 04:40 schrieb delamora: Trying to answer your comments; SG227 is FCC, that is, the conventional cell has 4 primitive cells the first wyckoff positions are; 8a 1/8,1/8,1/8 8b 3/8,3/8,3/8 16c 0,0,0 16d 1/2,1/2,1/2 32e x,x,x 8a and 8b have 2 atoms in the primitive cell 16c and 16d have 4 32e has 8 MnFe2O4 is a spinel Mn; 8a Fe; 16d O; 32e x=0.2639 The primitive cell has 2 formula units while the conventional cel has 8 formula units then the atomic positions in MnFe2O4.struct are; F LATTICE,NONEQUIV.ATOMS: 3 227_Fd-3m 16.059466 16.059466 16.059466 90.00 90.00 90.00 ATOM 1: X=0.1250 Y=0.1250 Z=0.1250 ATOM 1:X= 0.8750 Y=0.8750 Z=0.8750 Mn NPT= 781 R0=0.5000 RMT= 1.8600 Z: 25.000 ATOM -2: X=0.5000 Y=0.5000 Z=0.5000 ATOM -2:X= 0.5000 Y=0.7500 Z=0.7500 ATOM -2:X= 0.7500 Y=0.7500 Z=0.5000 ATOM -2:X= 0.7500 Y=0.5000 Z=0.7500 Fe NPT= 781 R0=0.5000 RMT= 2.1500 Z: 26.000 ATOM -3: X=0.2639 Y=0.2639 Z=0.2639 ATOM -3:X= 0.7361 Y=0.7361 Z=0.7361 ATOM -3:X= 0.7361 Y=0.5139 Z=0.5139 ATOM -3:X= 0.2639 Y=0.9861 Z=0.9861 ATOM -3:X= 0.9861 Y=0.9861 Z=0.2639 ATOM -3:X= 0.5139 Y=0.5139 Z=0.7361 ATOM -3:X= 0.5139 Y=0.7361 Z=0.5139 ATOM -3:X= 0.9861 Y=0.2639 Z=0.9861 O NPT= 781 R0=0.0001 RMT= 1.6000 Z: 8.000 Now, if one wants to make it antiferro then one posibility is to take two Fe 'up' and two 'dn', and the system becomes; B LATTICE,NONEQUIV.ATOMS: 5 74_Imma 11.355757 11.355757 16.059466 90.00 90.00 90.00 ATOM 1: X=0. Y=0.2500 Z=0.1250 ATOM 1:X= 0. Y=0.7500 Z=0.8750 Mn NPT= 781 R0=0.5000 RMT= 1.8600 Z: 25.000 ATOM 2: X=0.2500 Y=0.7500 Z=0.2500 ATOM 2:X= 0.2500 Y=0.2500 Z=0.7500 Fe1 NPT= 781 R0=0.5000 RMT= 2.1500 Z: 26.000 ATOM 3: X=0. Y=0.5278 Z=0.2639 ATOM 3:X= 0. Y=0.4722 Z=0.7361 ATOM 3:X= 0. Y=0.0278 Z=0.7361 ATOM 3:X= 0. Y=0.9722 Z=0.2639 O 1 NPT= 781 R0=0.0001 RMT= 1.6000 Z: 8.000 ATOM 4: X=0.7778 Y=0.7500 Z=0.4861 ATOM 4:X= 0. Y=0.2500 Z=0.5139 ATOM 4:X= 0.7778 Y=0.2500 Z=0.5139 ATOM 4:X= 0. Y=0.7500 Z=0.4861 O 2 NPT= 781 R0=0.0001 RMT= 1.6000 Z: 8.000 ATOM 5: X=0. Y=0. Z=0.5000 ATOM 5:X= 0. Y=0.5000 Z=0.5000 Fe2 NPT= 781 R0=0.5000 RMT= 2.1500 Z: 26.000 As it can be seen the 8 O atoms in 32e position are now in two groups of 4 O atoms and as I mentioned before the 'runsp -min' will move them in a way that it would break the cubic FCC crystal symmetry. The problem is that Fe in this system has the kagome arangement and antiferro ordering is just a simplification, but the system should be cubic with the SG 227. Saludos Pablo *De:* Wien en nombre de Fecher, Gerhard *Enviado:* martes, 2 de noviembre de 2021 05:07 a. m. *Para:* A Mailing list for WIEN2k users *Asunto:* Re: [Wien] Optimization of an spinel structure Sorry, it depends on the setting for space group 227 whether X=1/4 results in a 8- or 32-fold degenerate position, I did not check which one Wien2k uses (unfortunately one can not use crystallographic Wykoff positions directly). if you have already the 32-fold degenerate site (I guess wien2k reports a smaller cell with 8 sites) then min should find the correct positions by minimizing the forces. The remarks on the magnetic order are not touched by that. Ciao Gerhard DEEP THOUGHT in D. Adams; Hitchhikers Guide to the Galaxy: "I think the problem, to be quite honest with you, is that you have never actually known what the question is." Dr. Gerhard H. Fecher Institut of Physics Johannes Gutenberg - University 55099 Mainz Von: Wien [wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora [delam...@unam.mx] Gesendet: Dienstag, 2. November 2021 04:49 An: wien@zeus.theochem.tuwien.ac.at Betreff: [Wien] Optimization of an spinel structure Dear WIEN community, I am trying to optimize a spinel structure. The n
Re: [Wien] Mixer error: NEC01 charge leakage too large
Your post is too large, but I'll answer here:
Your struct and in1 file for TiC is ok.
I'm not quite sure, why you have at the beginning the *broyd* present
message ?? These files are produced by mixer. Did you run this before,
with the same error or did it go through ?
You had a save_lapw command, which failed because of an already saved
calculation.
save_lapw will not overwrite previous "saves", unless you specify -f
Did you run in mpi-parallel mode ? For lapw0_mpi you did, but we don't
know for
lapw1/2 (What is your .machines file).
It could be that these small examples are too small for mpi mode. Try to
run them without parallelization or in (moderate) k-parallel mode.
Please read the UG (Parallelization) for more details. Running a small
case like TiC in highly parallel mode will actually run MUCH SLOWER than
in sequential mode.
Also: what gives:
grep :NEC01 *scf *.scfm
grep ':CHA ' *scf
Regards
Am 03.11.2021 um 00:43 schrieb ChenJie:
Hi,
I am new to Wien2k. I am trying to run the TiC example and MoS2
example from the user guide using Wien2k_21.1, and I was able to go
through all the steps in init_lapw without any issue, but when I tried
to run the scf (via run_lapw –p –ec 0.0001), I encountered the issue
at the mixer step (after lcore), and I have copy-pasted the STDOUT
error message below (for mos2):
/kpoint///
/mos2.broyd* files present ! You did not save_lapw a previous
clculation.///
/You have 60 seconds to kill this job ( ^C or kill 56323 )///
/or the script will rm *.broyd* and continue (use -NI to avoid
automatic rm)///
/ LAPW0 END///
/[1] Done srun -N1 -n4
/public1/home/sca2456/soft/lapw0_mpi lapw0.def >> .time00///
/ LAPW1 END///
/[1] + Done ( ( $remote $machine[$p] "cd
$PWD;$set_OMP_NUM_THREADS;$t $taskset0 $exe ${def}_$loop.def
;fixerror_lapw ${def}_$loop"; rm -f .lock_$lockfile[$p] ) >& .///
/stdout1_$loop; if ( -f .stdout1_$loop ) bashtime2csh.pl_lapw
.stdout1_$loop > .temp1_$loop; grep \% .temp1_$loop >> .time1_$loop;
grep -v \% .temp1_$loop | perl -e "print stderr " )///
/LAPW2 - FERMI; weights written///
/ LAPW2 END///
/[1] Done ( ( $remote $machine[$p] "cd
$PWD;$set_OMP_NUM_THREADS;$t $taskset0 $exe ${def}_${loop}.def
$loop;fixerror_lapw ${def}_$loop"; rm -f .lock_$lockfile[$p]///
/ ) >& .stdout2_$loop; if ( -f .stdout2_$loop ) bashtime2csh.pl_lapw
.stdout2_$loop > .temp2_$loop; grep \% .temp2_$loop >> .time2_$loop;
grep -v \% .temp2_$loop | perl -e "print stderr
/IN>" )///
/ SUMPARA END///
/ CORE END///
*/ERROR: NEC01 charge leakage too large/**//*
//
/> stop error///
I read from the manual that this issue could be caused by improper
setting of RMT, or the energy to separate core from valence states
(which defaults to -6 Ry), so I played with these two values as well
the kpoint mesh size during the init_lapw step, but I kept on getting
the above error message, for both TiC and mos2 system.
I am wondering what could be causing this error, for example, could it
be due to some other settings?
Thank you very much, and please let me know if any more information is
needed.
Regards,
Jie//
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Peter Blaha, Inst. f. Materials Chemistry, TU Vienna, A-1060 Vienna
Phone: +43-158801165300
Email: peter.bl...@tuwien.ac.at
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