[Wien] Fwd:
Anup Shakya
npshakya31 at gmail.com
Thu May 2 03:07:48 CEST 2019
Dear All,
I am doing calculations on a material Nd2PdSi3 which is of AlB2 hexagonal
structure (space group P6/mmm) with lattice constants a=b= 4.103 ang and c
= 4.204 ang. The positions of the atoms are as follows: R 1a site 0 0 0 and
Pd and Si statistically distributed on the 2d site 1/3 2/3 1/2 with 25 %
and 75% respectively. I started making the structure using R at 1a site and
Si at 2d site, so I get one Nd atom and two Si atoms.
Title
H LATTICE,NONEQUIV.ATOMS: 2 191_P6/mmm
MODE OF CALC=RELA unit=ang
7.754116 7.754116 7.944223 90.000000 90.000000120.000000
ATOM -1: X=0.00000000 Y=0.00000000 Z=0.00000000
MULT= 1 ISPLIT= 8
Nd NPT= 781 R0=0.00001000 RMT= 2.5000 Z: 60.000
LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
0.0000000 1.0000000 0.0000000
0.0000000 0.0000000 1.0000000
ATOM 2: X=0.33333333 Y=0.66666667 Z=0.50000000
MULT= 2 ISPLIT= 8
ATOM 2:X= 0.66666667 Y=0.33333333 Z=0.50000000
Si NPT= 781 R0=0.00010000 RMT= 2.2200 Z: 14.000
LOCAL ROT MATRIX: 0.0000000 0.0000000 0.0000000
0.0000000 0.0000000 0.0000000
0.0000000 0.0000000 0.0000000
0 NUMBER OF SYMMETRY OPERATIONS
~
In order to make the ratio of Pd:Si as 1:3, I have doubled the structure
along x axis using the super cell program, The structure is shown below:
Title
P LATTICE,NONEQUIV. ATOMS 4
MODE OF CALC=RELA unit=ang
15.508232 7.754116 7.944223 90.000000 90.000000120.000000
ATOM 1: X=0.00000000 Y=0.00000000 Z=0.00000000
MULT= 1 ISPLIT= 8
Nd NPT= 781 R0=0.00001000 RMT= 2.5000 Z: 60.0
LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
0.0000000 1.0000000 0.0000000
0.0000000 0.0000000 1.0000000
ATOM 2: X=0.50000000 Y=0.00000000 Z=0.00000000
MULT= 1 ISPLIT= 8
Nd NPT= 781 R0=0.00001000 RMT= 2.5000 Z: 60.0
LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
0.0000000 1.0000000 0.0000000
0.0000000 0.0000000 1.0000000
ATOM 3: X=0.16666666 Y=0.66666667 Z=0.50000000
MULT= 2 ISPLIT= 8
ATOM 3: X=0.33333334 Y=0.33333333 Z=0.50000000
Si NPT= 781 R0=0.00010000 RMT= 2.2200 Z: 14.0
LOCAL ROT MATRIX: 0.0000000 0.0000000 0.0000000
0.0000000 0.0000000 0.0000000
0.0000000 0.0000000 0.0000000
ATOM 4: X=0.66666666 Y=0.66666667 Z=0.50000000
MULT= 2 ISPLIT= 8
ATOM 4: X=0.83333334 Y=0.33333333 Z=0.50000000
Si NPT= 781 R0=0.00010000 RMT= 2.2200 Z: 14.0
LOCAL ROT MATRIX: 0.0000000 0.0000000 0.0000000
0.0000000 0.0000000 0.0000000
0.0000000 0.0000000 0.0000000
0 NUMBER OF SYMMETRY OPERATIONS
After this I splitted the 2nd atomic position of the 3rd Si atom and then
changed the atom from Si to Pd and the Z value to 46 i.e, for Pd and then
saved the structure.
When I initialized I got the following warning:
SUMS TO 4.44000 LT. NN-DIST= 4.47684
WARNING: Mult not equal. PLEASE CHECK outputnn-file
WARNING: ityp not equal. PLEASE CHECK outputnn-file
WARNING: ityp not equal. PLEASE CHECK outputnn-file
Then I accepted the structure given by x sgroup and I get the following
struture
Title
P LATTICE,NONEQUIV.ATOMS: 6 25 Pmm2
RELA
7.754116 7.944223 13.430523 90.000000 90.000000 90.000000
ATOM -1: X=0.00000000 Y=0.00000000 Z=0.00000000
MULT= 1 ISPLIT= 8
Nd1 NPT= 781 R0=0.00001000 RMT= 2.5000 Z: 60.0
LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
0.0000000 1.0000000 0.0000000
0.0000000 0.0000000 1.0000000
ATOM -2: X=0.50000000 Y=0.00000000 Z=0.50000000
MULT= 1 ISPLIT= 8
Nd2 NPT= 781 R0=0.00001000 RMT= 2.5000 Z: 60.0
LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
0.0000000 1.0000000 0.0000000
0.0000000 0.0000000 1.0000000
ATOM -3: X=0.49999999 Y=0.50000000 Z=0.83333334
MULT= 1 ISPLIT= 8
Pd1 NPT= 781 R0=0.00001000 RMT= 2.5000 Z: 46.0
LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
0.0000000 1.0000000 0.0000000
0.0000000 0.0000000 1.0000000
ATOM -4: X=0.00000000 Y=0.50000000 Z=0.33333334
MULT= 1 ISPLIT= 8
Si1 NPT= 781 R0=0.00010000 RMT= 1.8900 Z: 14.0
LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
0.0000000 1.0000000 0.0000000
0.0000000 0.0000000 1.0000000
ATOM -5: X=0.50000001 Y=0.50000000 Z=0.16666666
MULT= 1 ISPLIT= 8
Si2 NPT= 781 R0=0.00010000 RMT= 1.8900 Z: 14.0
LOCAL ROT MATRIX: 1.0000000 0.0000000 0.0000000
0.0000000 1.0000000 0.0000000
0.0000000 0.0000000 1.0000000
ATOM -6: X=0.00000000 Y=0.50000000 Z=0.66666666
MULT= 1 ISPLIT= 8
Si3 NPT= 781 R0=0.00010000 RMT= 1.8900 Z: 14.0
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
1 0 0 0.00000000
Now the question is my initial experimental structure was AlB2 hexagonal
type but since I could not have Pd and Si distribution in this structure in
1:3 ratio I made a supercell and the structure which I have generated now
is Pmm2 structure which is orthorhombic but has a distribution of Pd and Si
in 1:3 ratio. Is it ok to perform calculations on this structure?
Secondly, I want to perform Antiferromagnetic calculations with different
magnetic configurations like A, C, G type and for that I have to further
increase the supercell. That in itself will further change the symmetry.
Can I use this structure and make a super cell to define the magnetic
configuration for the Nd ion?
Please suggest.
Sincerely,
Anup Pradhan Sakhya (Ph.D.)
Visiting Post-Doctoral Fellow
DCMP&MS, TIFR, Mumbai
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