Is the AFMIII structure or not? **** I do not understand this question
NB: What's mysterious is this AFM structure that the space group is monoclinic and the lattice parameters are orthorhombic. **** My answer fits in this question; you have an orthorhombic cell, but it has atoms inside that break this symmetry, just as the example that I gave you, now the new symmetry is reduced. Imagine a 2D square with atoms at a,a -a,a a,-a -a,-a The symmetry is still square Now if you associate, due to magnetic ordering, that is you put two atoms up and two down Up; a,a -a,-a Dn; -a,a a,-a Then the structure is no longer square, it does not have 90 degrees rotations ________________________________ De: Wien <wien-boun...@zeus.theochem.tuwien.ac.at> en nombre de Abderrahmane Reggad <jazai...@gmail.com> Enviado: sábado, 28 de octubre de 2017 06:22 a. m. Para: wien@zeus.theochem.tuwien.ac.at Asunto: Re: [Wien] Querry in a resultant structure Thanks Delamora for your answer and this is not what i am looking for and I am sorry that I couldn't communicate what I want. My problem is as follows: 1- I am considering this non magnetic structure (orthorhombic ) * the non-magnetic structure is in this link https://ibb.co/nt2nFm * The struct file is: NiS-MnP P LATTICE,NONEQUIV.ATOMS: 2 62_Pnma MODE OF CALC=RELA unit=ang 10.056514 6.692230 10.722761 90.000000 90.000000 90.000000 ATOM -1: X=0.00500000 Y=0.25000000 Z=0.20000000 MULT= 4 ISPLIT= 8 -1: X=0.99500000 Y=0.75000000 Z=0.80000000 -1: X=0.50500000 Y=0.25000000 Z=0.30000000 -1: X=0.49500000 Y=0.75000000 Z=0.70000000 Ni NPT= 781 R0=0.00005000 RMT= 2.08 Z: 28.0 LOCAL ROT MATRIX: 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 1.0000000 0.0000000 0.0000000 ATOM -2: X=0.20000000 Y=0.25000000 Z=0.57000000 MULT= 4 ISPLIT= 8 -2: X=0.80000000 Y=0.75000000 Z=0.43000000 -2: X=0.70000000 Y=0.25000000 Z=0.93000000 -2: X=0.30000000 Y=0.75000000 Z=0.07000000 S NPT= 781 R0=0.00010000 RMT= 1.70 Z: 16.0 LOCAL ROT MATRIX: 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 1.0000000 0.0000000 0.0000000 8 NUMBER OF SYMMETRY OPERATIONS 2- I want to construct the AFM structure of type III as in the following link https://ibb.co/mqySFm * I have used the supercell 1x1x1 P to make the four Ni atoms independent * I have put the atoms (X=0.00500000 Y=0.25000000 Z=0.20000000) and (X=0.99500000 Y=0.75000000 Z=0.80000000) to UP and the atoms (X=0.50500000 Y=0.25000000 Z=0.30000000) and (X=0.49500000 Y=0.75000000 Z=0.70000000) to DOWN * I have used the sgroup - I get the following AFM structure NiS-MnP-afmIII P LATTICE,NONEQUIV.ATOMS: 4 11_P21/m MODE OF CALC=RELA unit=bohr 10.998349 10.005529 6.864228 90.000000 90.000000 90.000000 ATOM -1: X=0.25021227 Y=0.99991424 Z=0.25000000 MULT= 2 ISPLIT= 8 -1: X=0.74978773 Y=0.00008576 Z=0.75000000 Ni1 NPT= 781 R0=0.00005000 RMT= 2.21 Z: 28.0 LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 ATOM -2: X=0.75024661 Y=0.50012576 Z=0.75000000 MULT= 2 ISPLIT= 8 -2: X=0.24975339 Y=0.49987424 Z=0.25000000 Ni2 NPT= 781 R0=0.00005000 RMT= 2.21 Z: 28.0 LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 ATOM -3: X=0.08986359 Y=0.75012482 Z=0.75000000 MULT= 2 ISPLIT= 8 -3: X=0.91013641 Y=0.24987518 Z=0.25000000 S 1 NPT= 781 R0=0.00010000 RMT= 1.81 Z: 16.0 LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 ATOM -4: X=0.41013993 Y=0.25012471 Z=0.75000000 MULT= 2 ISPLIT= 8 -4: X=0.58986007 Y=0.74987529 Z=0.25000000 S 2 NPT= 781 R0=0.00010000 RMT= 1.81 Z: 16.0 LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 0.0000000 0.0000000 0.0000000 1.0000000 4 NUMBER OF SYMMETRY OPERATIONS My question is: Is the AFMIII structure or not? NB: What's mysterious is this AFM structure that the space group is monoclinic and the lattice parameters are orthorhombic. Best regards
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