[Wien] Fwd:

Peter Blaha pblaha at theochem.tuwien.ac.at
Thu May 2 20:36:03 CEST 2019


Do you think the structure you generated is a "statistical" occupation ???

You need to generate bigger supercells (without a preferred direction as 
in your trial structure) and bring some "randomness" into the 
occupations. The smallest cell I would accept is a 2x2x2 supercell. And 
you can try different distributions. Maybe you find by E-tot comparisons 
some preferred neighborhood ....

PS: Due to the substitution you will of course break symmetry and no 
longer have P6/mmm. But usually you can use nn instead of sgroup to find 
equivalent atoms and keep the hexagonal unit cell (when the number of 
non-equivalent atoms and the number of symmetry operations are 
identical, there is no need to use the cell generated by sgroup).

Am 02.05.2019 um 03:07 schrieb Anup Shakya:
> 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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> 

-- 
Peter Blaha
Inst.Materials Chemistry
TU Vienna
Getreidemarkt 9
A-1060 Vienna
Austria
+43-1-5880115671


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