Re: [Wien] Querry in a resultant structure
Hello again I have repeated the calculation using the low symmetry structure and i found a close results to those of the monoclinic one. I have checked the wyckoof positions and the symmetry operations of the orthorhombic and the monoclinic structures and I have found that the later is a subgroup of the former and the only difference is the change of the lattice parameters and the atomic positions. Here is the low symmetry structure MnP-Pnma-afmIII2 P LATTICE,NONEQUIV.ATOMS: 4 MODE OF CALC=RELA unit=bohr 11.637501 6.731303 11.245005 90.00 90.00 90.00 ATOM -1: X=0.00032814 Y=0.2500 Z=0.24996705 MULT= 2 ISPLIT= 8 -1: X=0.99967186 Y=0.7500 Z=0.75003295 Ni1NPT= 781 R0=0.5000 RMT= 2.25Z: 28.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 ATOM -2: X=0.50030217 Y=0.2500 Z=0.25003393 MULT= 2 ISPLIT= 8 -2: X=0.49969783 Y=0.7500 Z=0.74996607 Ni2NPT= 781 R0=0.5000 RMT= 2.25Z: 28.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 ATOM -3: X=0.25012913 Y=0.2500 Z=0.58455708 MULT= 2 ISPLIT= 8 -3: X=0.74987087 Y=0.7500 Z=0.41544292 S 1NPT= 781 R0=0.0001 RMT= 1.84Z: 16.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 ATOM -4: X=0.75011125 Y=0.2500 Z=0.91545439 MULT= 2 ISPLIT= 8 -4: X=0.24988875 Y=0.7500 Z=0.08454561 S 2NPT= 781 R0=0.0001 RMT= 1.84Z: 16.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 4 NUMBER OF SYMMETRY OPERATIONS -1 0 0 0. 0-1 0 0. 0 0-1 0. 1 1 0 0 0. 0 1 0 0. 0 0 1 0. 2 -1 0 0 0. 0 1 0 0.5000 0 0-1 0. 3 1 0 0 0. 0-1 0 0.5000 0 0 1 0. 4 Here is the monoclinic low symmetry structure NiS-MnP-afmIII P LATTICE,NONEQUIV.ATOMS: 4 11_P21/m MODE OF CALC=RELA unit=bohr 11.262961 10.563150 7.029377 90.00 90.00 90.00 ATOM -1: X=0.25021227 Y=0.1424 Z=0.2500 MULT= 2 ISPLIT= 8 -1: X=0.74978773 Y=0.8576 Z=0.7500 Ni1NPT= 781 R0=0.5000 RMT= 2.13Z: 28.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -2: X=0.75024661 Y=0.50012576 Z=0.7500 MULT= 2 ISPLIT= 8 -2: X=0.24975339 Y=0.49987424 Z=0.2500 Ni2NPT= 781 R0=0.5000 RMT= 2.13Z: 28.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -3: X=0.08986359 Y=0.75012482 Z=0.7500 MULT= 2 ISPLIT= 8 -3: X=0.91013641 Y=0.24987518 Z=0.2500 S 1NPT= 781 R0=0.0001 RMT= 1.74Z: 16.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -4: X=0.41013993 Y=0.25012471 Z=0.7500 MULT= 2 ISPLIT= 8 -4: X=0.58986007 Y=0.74987529 Z=0.2500 S 2NPT= 781 R0=0.0001 RMT= 1.74Z: 16.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 4 NUMBER OF SYMMETRY OPERATIONS 1 0 0 0. 0 1 0 0. 0 0 1 0. 1 -1 0 0 0. 0-1 0 0. 0 0 1 0.5000 2 -1 0 0 0. 0-1 0 0. 0 0-1 0. 3 1 0 0 0. 0 1 0 0. 0 0-1 0.5000 4 The results are as follows respectively AFMIII2 Equation of state: Birch-Murnaghaninfo 2 E = E0 + 9/16*(B/14703.6)*V0*[(eta**2-1)**3*BP + (eta**2-1)**2*(6-4*eta**2)] --> eta = (V0/V)**(1/3) Pressure = 3/2*B*(eta**7 - eta**5)*(1 + 3/4*(BP-4)*[eta**2 - 1]) V0,B(GPa),BP,E0 760.8174 104.6494 4.1456 -15359.539382 vol energy de(Birch-Murnaghan) Pressure(GPa) 762.9081 -15359.539337-0.25 -0.286 739.3131 -15359.536947-0.0001633.184 715.7179 -15359.528892 0.747.259 880.8836 -15359.487014 0.40 -11.327 786.5033 -15359.536637 0.000170 -3.244 810.0984 -15359.529313 0.000130 -5.768 833.6935 -15359.517895-0.000258 -7.922
Re: [Wien] Querry in a resultant structure
Thank you Lyudmila for you comment As a last step I need from all of you to show me how to get the AFMIII structure from the these two links and the following non magnetic structure of NiS-MnP (orthorhombic): https://ibb.co/nt2nFm https://ibb.co/mqySFm NiS-MnP P LATTICE,NONEQUIV.ATOMS: 2 62_Pnma MODE OF CALC=RELA unit=ang 10.056514 6.692230 10.722761 90.00 90.00 90.00 ATOM -1: X=0.0050 Y=0.2500 Z=0.2000 MULT= 4 ISPLIT= 8 -1: X=0.9950 Y=0.7500 Z=0.8000 -1: X=0.5050 Y=0.2500 Z=0.3000 -1: X=0.4950 Y=0.7500 Z=0.7000 Ni NPT= 781 R0=0.5000 RMT= 2.08Z: 28.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 ATOM -2: X=0.2000 Y=0.2500 Z=0.5700 MULT= 4 ISPLIT= 8 -2: X=0.8000 Y=0.7500 Z=0.4300 -2: X=0.7000 Y=0.2500 Z=0.9300 -2: X=0.3000 Y=0.7500 Z=0.0700 S NPT= 781 R0=0.0001 RMT= 1.70Z: 16.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 8 NUMBER OF SYMMETRY OPERATIONS Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
> 29.10.2017, 1:20 +04:00 from Gavin Abo : > The problem I see with this is that WIEN2k only allows the up and dn to > be defined in the +z direction and -z direction (or along c axis), > respectively. A small comment to the exhaustive answer of Gavin: as far as I understand, there are no dependence on direction in the equations of a usual sp calculation. So, the solution is the same for any direction of magnetic field, We say z only for convinience. Only when we switch on the spin-orbit interaction we obtain a dependence on the direction (relationship with crystal field), though imagnetisation is still collinear. WIENncm gives us a possibility to calculate noncollinear magnetism. Best wishes Lyudmila Dobysheva ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
The spacegroup 11 struct in the post https://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/msg16619.html is likely slightly wrong for what you want to do. Take the AFM III picture from your link in the post https://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/msg16630.html and compare it to the struct viewed in VESTA (attached file NiS-MnP-afmIII.vesta). If you rotate the view of the structure so that the a axis points down (in the -z direction), then the AFM III and struct file pictures look the same with the up and dn aligned along the a axis (or -/+ x directions). The problem I see with this is that WIEN2k only allows the up and dn to be defined in the +z direction and -z direction (or along c axis), respectively. WIENncm most likely could handle the up and dn along the x axis. However, usually experience in both electronic structure theory and programming (e.g., Fortran and C) is need to use WIENncm. Your current question suggests you might currently lack the skills that would be needed to use it. This is because WIENncm has very limited support [ http://susi.theochem.tuwien.ac.at/reg_user/ncm/ ]. So a user of it needs to be able to do so with little to no help. One solution to use WIEN2k in your case, may be to use the rotateall function of the structeditor (refer to section 9.28 in the WIEN2k 17.1 usersguide). Or another possible solution should be to calculate by hand what the structure parameters would need to be after rotating the structure so that the up and dn are aligned along the z axis. #VESTA_FORMAT_VERSION 3.3.0 CRYSTAL TITLE NiS-MnP-afmIII GROUP 1 1 P 1 SYMOP 0.00 0.00 0.00 1 0 0 0 1 0 0 0 1 1 -1.0 -1.0 -1.0 0 0 0 0 0 0 0 0 0 TRANM 0 0.00 0.00 0.00 1 0 0 0 1 0 0 0 1 LTRANSL -1 0.00 0.00 0.00 0.00 0.00 0.00 LORIENT -1 0 0 0 0 1.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 1.00 LMATRIX 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 CELLP 5.820074 5.294696 3.632391 90.00 90.00 90.00 0.00 0.00 0.00 0.00 0.00 0.00 STRUC 1 NiNi1 1. 0.250212 0.14 0.251a 1 0.00 0.00 0.00 0.00 2 NiNi1 1. 0.749788 0.86 0.751a 1 0.00 0.00 0.00 0.00 3 NiNi2 1. 0.750247 0.500126 0.751a 1 0.00 0.00 0.00 0.00 4 NiNi2 1. 0.249753 0.499874 0.251a 1 0.00 0.00 0.00 0.00 5 S S1 1. 0.089864 0.750125 0.751a 1 0.00 0.00 0.00 0.00 6 S S1 1. 0.910136 0.249875 0.251a 1 0.00 0.00 0.00 0.00 7 S S2 1. 0.410140 0.250125 0.751a 1 0.00 0.00 0.00 0.00 8 S S2 1. 0.589860 0.749875 0.251a 1 0.00 0.00 0.00 0.00 0 0 0 0 0 0 0 THERI 0 1Ni1 1.00 2Ni1 1.00 3Ni2 1.00 4Ni2 1.00 5 S1 1.00 6 S1 1.00 7 S2 1.00 8 S2 1.00 0 0 0 SHAPE 0 0 0 0 0.00 0 192 192 192 192 BOUND 01 01 01 0 0 0 0 0 SBOND 0 0 0 0 SITET 1Ni1 1.2500 183 187 189 183 187 189 204 1 2Ni1 1.2500 183 187 189 183 187 189 204 1 3Ni2 1.2500 183 187 189 183 187 189 204 1 4Ni2 1.2500 183 187 189 183 187 189 204 1 5 S1 1.0400 255 250 0 255 250 0 204 1 6 S1 1.0400 255 250 0 255 250 0 204 1 7 S2 1.0400 255 250 0 255 250 0 204 1 8 S2 1.0400 255 250 0 255 250 0 204 1 0 0 0 0 0 0 VECTR 1 -1.00.00.0 0 1 0000 2 0000 0 0 0 0 0 21.00.00.0 0 3 0000 4 0000 0 0 0 0 0 0 0 0 0 0 VECTT 1 0.250 255 0 0 1 2 0.250 255 0 0 1 0 0 0 0 0 SPLAN 0 0 0 0 LBLAT 0 1 2 3 4 5 6 7 -1 LBLSP -1 DLATM -1 DLBND -1 DLPLY -1 PLN2D 0 0 0 0 ATOMT 1 Ni 1.2500 183 187 189 183 187 189 204 2 S 1.0400 255 250 0 255 250 0 204 0 0 0 0 0 0 SCENE -0.026724 0.999626 -0.005734 0.00 -0.996986 -0.027070 -0.072699 0.00 -0.072827 0.003774 0.997337 0.00 0.00
Re: [Wien] Querry in a resultant structure
Thank you very Delamora Now I still waiting for your answer on my question in this link about the magnetic structure of the zinc blende structure Re: [Wien] The magnetic structure for zinc blende non magnetic structure Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
I do not know if it is correct, but yes, the AFM has lower symmetry due to the separation of Ni up and Ni dn De: Wien en nombre de Abderrahmane Reggad Enviado: sábado, 28 de octubre de 2017 08:18 a. m. Para: wien@zeus.theochem.tuwien.ac.at Asunto: Re: [Wien] Querry in a resultant structure Sorry for the missed word "correct" According to your example , you mean that the AFM structure may be correct Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
Sorry for the missed word "correct" According to your example , you mean that the AFM structure may be correct Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
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 en nombre de Abderrahmane Reggad 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 90.00 90.00 90.00 ATOM -1: X=0.0050 Y=0.2500 Z=0.2000 MULT= 4 ISPLIT= 8 -1: X=0.9950 Y=0.7500 Z=0.8000 -1: X=0.5050 Y=0.2500 Z=0.3000 -1: X=0.4950 Y=0.7500 Z=0.7000 Ni NPT= 781 R0=0.5000 RMT= 2.08Z: 28.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 ATOM -2: X=0.2000 Y=0.2500 Z=0.5700 MULT= 4 ISPLIT= 8 -2: X=0.8000 Y=0.7500 Z=0.4300 -2: X=0.7000 Y=0.2500 Z=0.9300 -2: X=0.3000 Y=0.7500 Z=0.0700 S NPT= 781 R0=0.0001 RMT= 1.70Z: 16.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 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=0.0050 Y=0.2500 Z=0.2000) and (X=0.9950 Y=0.7500 Z=0.8000) to UP and the atoms (X=0.5050 Y=0.2500 Z=0.3000) and (X=0.4950 Y=0.7500 Z=0.7000) 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 90.00 90.00 90.00 ATOM -1: X=0.25021227 Y=0.1424 Z=0.2500 MULT= 2 ISPLIT= 8 -1: X=0.74978773 Y=0.8576 Z=0.7500 Ni1NPT= 781 R0=0.5000 RMT= 2.21Z: 28.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -2: X=0.75024661 Y=0.50012576 Z=0.7500 MULT= 2 ISPLIT= 8 -2: X=0.24975339 Y=0.49987424 Z=0.2500 Ni2NPT= 781 R0=0.5000 RMT= 2.21Z: 28.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -3: X=0.08986359 Y=0.75012482 Z=0.7500 MULT= 2 ISPLIT= 8 -3: X=0.91013641 Y=0.24987518 Z=0.2500 S 1NPT= 781 R0=0.0001 RMT= 1.81Z: 16.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -4: X=0.41013993 Y=0.25012471 Z=0.7500 MULT= 2 ISPLIT= 8 -4: X=0.58986007 Y=0.74987529 Z=0.2500 S 2NPT= 781 R0=0.0001 RMT= 1.81Z: 16.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 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 ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
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 90.00 90.00 90.00 ATOM -1: X=0.0050 Y=0.2500 Z=0.2000 MULT= 4 ISPLIT= 8 -1: X=0.9950 Y=0.7500 Z=0.8000 -1: X=0.5050 Y=0.2500 Z=0.3000 -1: X=0.4950 Y=0.7500 Z=0.7000 Ni NPT= 781 R0=0.5000 RMT= 2.08Z: 28.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 ATOM -2: X=0.2000 Y=0.2500 Z=0.5700 MULT= 4 ISPLIT= 8 -2: X=0.8000 Y=0.7500 Z=0.4300 -2: X=0.7000 Y=0.2500 Z=0.9300 -2: X=0.3000 Y=0.7500 Z=0.0700 S NPT= 781 R0=0.0001 RMT= 1.70Z: 16.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 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=0.0050 Y=0.2500 Z=0.2000) and (X=0.9950 Y=0.7500 Z=0.8000) to *UP* and the atoms (X=0.5050 Y=0.2500 Z=0.3000) and (X=0.4950 Y=0.7500 Z=0.7000) 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 90.00 90.00 90.00 ATOM -1: X=0.25021227 Y=0.1424 Z=0.2500 MULT= 2 ISPLIT= 8 -1: X=0.74978773 Y=0.8576 Z=0.7500 Ni1NPT= 781 R0=0.5000 RMT= 2.21Z: 28.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -2: X=0.75024661 Y=0.50012576 Z=0.7500 MULT= 2 ISPLIT= 8 -2: X=0.24975339 Y=0.49987424 Z=0.2500 Ni2NPT= 781 R0=0.5000 RMT= 2.21Z: 28.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -3: X=0.08986359 Y=0.75012482 Z=0.7500 MULT= 2 ISPLIT= 8 -3: X=0.91013641 Y=0.24987518 Z=0.2500 S 1NPT= 781 R0=0.0001 RMT= 1.81Z: 16.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -4: X=0.41013993 Y=0.25012471 Z=0.7500 MULT= 2 ISPLIT= 8 -4: X=0.58986007 Y=0.74987529 Z=0.2500 S 2NPT= 781 R0=0.0001 RMT= 1.81Z: 16.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 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* ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
Very simple, you need to break the rectangle character of one face, but without changing angles; In an orthorhombic cell Put atoms in the face; 0 0 0 a b 0 but a and b not equal to 0.5 If you put a=b=1/2 then it changes to face center C m m m De: Wien en nombre de Abderrahmane Reggad Enviado: viernes, 27 de octubre de 2017 04:41:28 p. m. Para: wien@zeus.theochem.tuwien.ac.at Asunto: Re: [Wien] Querry in a resultant structure I see that you haven''t understood me this time my question is as follows: is it possible to get a monoclinic structure with orthorhombic lattice parameters ? Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
I see that you haven''t understood me this time my question is as follows: is it possible to get a monoclinic structure with orthorhombic lattice parameters ? Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
In my humble opinion YOU have to decide wether your structure looks like what you want. I only can/will expand somewhat on my previous advice to look at case.outputnn and to compare structures in xcrysden: Since xcrysden is difficult about arrows, do this in colors: You have 4 purple marbles in your unit cell. There are 3 ways to replace two of them by identical green marbles, depicted as AFM I - III in your figure. So, to avoid the fuss of talking xcrysden into plotting arrows, take the supercell generated .struct file and produce 3 different .struct files by replacing two Ni with, say, Co. Look at these with xcrysden and decide which is what. Rename all Ni-colored atoms Ni1, all Co-colored atoms Ni2 (and perhaps make all S identical) and ask sgroup what it thinks of the idea. As for the mysterious downgrading from rhombohedral to monoclinic: Look at your structures along the (1,1,1) direction. Is there a 3-fold symmetry? I hope this helps, --- Dr. Martin Pieper Karl-Franzens University Institute of Physics Universitätsplatz 5 A-8010 Graz Austria Tel.: +43-(0)316-380-8564 Am 25.10.2017 22:39, schrieb Abderrahmane Reggad: Hello again Here is the 3 different AFM configurations https://ibb.co/mqySFm Here are the 4 independent Ni atoms https://ibb.co/nt2nFm Now the big problem lies in if is it possible tp get a monoclinic structure (space group #11) with orthorhombic lattice parameters as we know that the rhombohedral structure in wien2k is represented bt a hexagonal lattice parameters. Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
Hello again Here is the 3 different AFM configurations https://ibb.co/mqySFm Here are the 4 independent Ni atoms https://ibb.co/nt2nFm Now the big problem lies in if is it possible tp get a monoclinic structure (space group #11) with orthorhombic lattice parameters as we know that the rhombohedral structure in wien2k is represented bt a hexagonal lattice parameters. Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
It is a valid procedure to generate a unit cell that supports placing antiparallel spins on the two Ni sublattices containing Ni1 and Ni2 atoms. I cannot tell you wether the new structure is 'true' or not. Concerning the crystal structure you simply should check youerself with xcrysden (or some similar program): There should be no visible change in the positions when viewing the two structures from the same angle. You also should compare the distances to nearest and next nearest neighbors for the old and new structure - they are listed in the respective case.outputnn and should be identical. Concerning the magnetic structure I don't know what exactly you mean with 'type III' af. If you are uneasy with the concept of an antiferromagnetic wave vector you might try to visualize what the difference to type I and type II is. Then look into outputnn again and decide if you really want these neighbors being antiparallel, or if it's the neighbours at some other distance. Best regards, Martin Pieper --- Dr. Martin Pieper Karl-Franzens University Institute of Physics Universitätsplatz 5 A-8010 Graz Austria Tel.: +43-(0)316-380-8564 Am 24.10.2017 14:04, schrieb Abderrahmane Reggad: Thank you pieper for your answer I am studying the magnetic order of type III with Ni1,Ni2 up and Ni3, Ni4 down according the figure included. This is the procedure that I adopted: -supercell 1x1x1 P to make the 4 atoms Ni indepedent a - I make the Ni1 and Ni2 to be the atom Ni1 and the atoms Ni3 and Ni4 to be the atom Ni2. - sgroup to search a new space group - I got the new strycture with monoclinic space group but with orthorhombic lattice parameters. I want to know if this new structure is true or not. The non magnetic structure is as follows: NiS-MnP P LATTICE,NONEQUIV.ATOMS: 2 62_Pnma MODE OF CALC=RELA unit=ang 10.056514 6.692230 10.722761 90.00 90.00 90.00 ATOM -1: X=0.0050 Y=0.2500 Z=0.2000 MULT= 4 ISPLIT= 8 -1: X=0.9950 Y=0.7500 Z=0.8000 -1: X=0.5050 Y=0.2500 Z=0.3000 -1: X=0.4950 Y=0.7500 Z=0.7000 Ni NPT= 781 R0=0.5000 RMT= 2.08Z: 28.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 ATOM -2: X=0.2000 Y=0.2500 Z=0.5700 MULT= 4 ISPLIT= 8 -2: X=0.8000 Y=0.7500 Z=0.4300 -2: X=0.7000 Y=0.2500 Z=0.9300 -2: X=0.3000 Y=0.7500 Z=0.0700 S NPT= 781 R0=0.0001 RMT= 1.70Z: 16.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 8 NUMBER OF SYMMETRY OPERATIONS Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
Thank you pieper for your answer I am studying the magnetic order of type III with Ni1,Ni2 up and Ni3, Ni4 down according the figure included. This is the procedure that I adopted: -supercell 1x1x1 P to make the 4 atoms Ni indepedent a - I make the Ni1 and Ni2 to be the atom Ni1 and the atoms Ni3 and Ni4 to be the atom Ni2. - sgroup to search a new space group - I got the new strycture with monoclinic space group but with orthorhombic lattice parameters. I want to know if this new structure is true or not. The non magnetic structure is as follows: NiS-MnP P LATTICE,NONEQUIV.ATOMS: 2 62_Pnma MODE OF CALC=RELA unit=ang 10.056514 6.692230 10.722761 90.00 90.00 90.00 ATOM -1: X=0.0050 Y=0.2500 Z=0.2000 MULT= 4 ISPLIT= 8 -1: X=0.9950 Y=0.7500 Z=0.8000 -1: X=0.5050 Y=0.2500 Z=0.3000 -1: X=0.4950 Y=0.7500 Z=0.7000 Ni NPT= 781 R0=0.5000 RMT= 2.08Z: 28.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 ATOM -2: X=0.2000 Y=0.2500 Z=0.5700 MULT= 4 ISPLIT= 8 -2: X=0.8000 Y=0.7500 Z=0.4300 -2: X=0.7000 Y=0.2500 Z=0.9300 -2: X=0.3000 Y=0.7500 Z=0.0700 S NPT= 781 R0=0.0001 RMT= 1.70Z: 16.0 LOCAL ROT MATRIX:0.000 1.000 0.000 0.000 0.000 1.000 1.000 0.000 0.000 8 NUMBER OF SYMMETRY OPERATIONS Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
Re: [Wien] Querry in a resultant structure
I don't know what your question is, but hope that a comment might help. Apparently you did not change size and shape of the unit cell (using a 1*1*1 supercell), so the lattice constants and angles are the same. However, you told sgroup that there is not one type of Ni in the unit cell but two. This change in the decoration of the bricks your crystal is built of broke some symmetries in their arrangement - now you are down to monoclinic. I don't know exactly what af wave vector you want to establish in your structure (+-+-, or ++--++--, or ... counting along which direction?). Is your problem that your Ni1 and Ni2 are not the two you actually wanted to have antiparallel moments? If 'afIII' fits into this unit cell (Ni1 and Ni2 do have antiparallel moments), and if the atoms are where you want them, run symmetry, set spin directions in case.inst, run lstart ... Type III sounds comlicated enough to not fit into this unit cell - meaning that you cannot complete one period of the +/- count in your chosen direction within the unit cell. In that case you have to change shape and/or size of the unit cell. Maybe the options provided by supercell (double the cell along basis directions ...) are sufficient. There is no guarantee for this to be the case. You may have to work out the fitting new shape and the positions of the atoms in the new unit cell using additional tools. I hope this helps, Martin Pieper --- Dr. Martin Pieper Karl-Franzens University Institute of Physics Universitätsplatz 5 A-8010 Graz Austria Tel.: +43-(0)316-380-8564 Am 22.10.2017 20:57, schrieb Abderrahmane Reggad: Dear wien users I am studing the magnetic structure of type III for a orthorhombic structure. After doing the supercell and labelling the independent atoms for spin-up and spin-dn atoms and using the sgroup command I got the following structure: The space group of a monoclinic structure and the lattice parameters are those of orthorhombic structure. NiS-MnP-afmIII P LATTICE,NONEQUIV.ATOMS: 4 11_P21/m MODE OF CALC=RELA unit=bohr 10.998349 10.005529 6.864228 90.00 90.00 90.00 ATOM -1: X=0.25021227 Y=0.1424 Z=0.2500 MULT= 2 ISPLIT= 8 -1: X=0.74978773 Y=0.8576 Z=0.7500 Ni1NPT= 781 R0=0.5000 RMT= 2.21Z: 28.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -2: X=0.75024661 Y=0.50012576 Z=0.7500 MULT= 2 ISPLIT= 8 -2: X=0.24975339 Y=0.49987424 Z=0.2500 Ni2NPT= 781 R0=0.5000 RMT= 2.21Z: 28.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -3: X=0.08986359 Y=0.75012482 Z=0.7500 MULT= 2 ISPLIT= 8 -3: X=0.91013641 Y=0.24987518 Z=0.2500 S 1NPT= 781 R0=0.0001 RMT= 1.81Z: 16.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -4: X=0.41013993 Y=0.25012471 Z=0.7500 MULT= 2 ISPLIT= 8 -4: X=0.58986007 Y=0.74987529 Z=0.2500 S 2NPT= 781 R0=0.0001 RMT= 1.81Z: 16.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 4 NUMBER OF SYMMETRY OPERATIONS Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
[Wien] Querry in a resultant structure
Dear wien users I am studing the magnetic structure of type III for a orthorhombic structure. After doing the supercell and labelling the independent atoms for spin-up and spin-dn atoms and using the sgroup command I got the following structure: The space group of a monoclinic structure and the lattice parameters are those of orthorhombic structure. NiS-MnP-afmIII P LATTICE,NONEQUIV.ATOMS: 4 11_P21/m MODE OF CALC=RELA unit=bohr 10.998349 10.005529 6.864228 90.00 90.00 90.00 ATOM -1: X=0.25021227 Y=0.1424 Z=0.2500 MULT= 2 ISPLIT= 8 -1: X=0.74978773 Y=0.8576 Z=0.7500 Ni1NPT= 781 R0=0.5000 RMT= 2.21Z: 28.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -2: X=0.75024661 Y=0.50012576 Z=0.7500 MULT= 2 ISPLIT= 8 -2: X=0.24975339 Y=0.49987424 Z=0.2500 Ni2NPT= 781 R0=0.5000 RMT= 2.21Z: 28.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -3: X=0.08986359 Y=0.75012482 Z=0.7500 MULT= 2 ISPLIT= 8 -3: X=0.91013641 Y=0.24987518 Z=0.2500 S 1NPT= 781 R0=0.0001 RMT= 1.81Z: 16.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 ATOM -4: X=0.41013993 Y=0.25012471 Z=0.7500 MULT= 2 ISPLIT= 8 -4: X=0.58986007 Y=0.74987529 Z=0.2500 S 2NPT= 781 R0=0.0001 RMT= 1.81Z: 16.0 LOCAL ROT MATRIX:1.000 0.000 0.000 0.000 1.000 0.000 0.000 0.000 1.000 4 NUMBER OF SYMMETRY OPERATIONS Best regards ___ Wien mailing list Wien@zeus.theochem.tuwien.ac.at http://zeus.theochem.tuwien.ac.at/mailman/listinfo/wien SEARCH the MAILING-LIST at: http://www.mail-archive.com/wien@zeus.theochem.tuwien.ac.at/index.html
