古いバージョンのBBSは閲覧のみ可能です。
The old BBS is read only.

6654

[Re:10] plotting DOS

Posted on : September 12, 2015 (Sat) 18:37:11

by Hitoshi

Hello Jeremy,

One possible solution is to calculate ZnTiP2 as a simple tetragonal (st) crystal considering with all atomic positions within the (conventional) unit cell. But, it may requires large computational resources.

c------------------------------------------------------------
go data/ZnTiP2-st
c------------------------------------------------------------
c brvtyp a c/a b/a alpha beta gamm
st 10.4636 2 1 90 90 90
c------------------------------------------------------------
c edelt ewidth reltyp sdftyp magtyp record
0.001 1.5 sra mjw nmag 2nd
c------------------------------------------------------------
c outtyp bzqlty maxitr pmix
update 8 200 0.02
c------------------------------------------------------------
c ntyp
3
c------------------------------------------------------------
c type ncmp rmt field mxl anclr conc
Zn 1 1 0 2 30 100
Ti 1 1 0 2 22 100
P 1 1 0 2 15 100
c------------------------------------------------------------
c natm
16
c------------------------------------------------------------
c atmicx atmtyp
0.00000000 0.00000000 0.00000000 Zn
0.00000000 0.50000000 0.25000000 Zn
0.50000000 0.50000000 0.50000000 Zn
0.50000000 0.00000000 0.75000000 Zn
0.50000000 0.50000000 0.00000000 Ti
0.50000000 0.00000000 0.25000000 Ti
0.00000000 0.00000000 0.50000000 Ti
0.00000000 0.50000000 0.75000000 Ti
0.25000000 0.25000000 0.12500000 P
0.25000000 0.75000000 0.87500000 P
0.75000000 0.75000000 0.12500000 P
0.75000000 0.25000000 0.87500000 P
0.75000000 0.75000000 0.62500000 P
0.75000000 0.25000000 0.37500000 P
0.25000000 0.25000000 0.62500000 P
0.25000000 0.75000000 0.37500000 P
c------------------------------------------------------------




Another possibility is to use body-centered tetoragonal (bct) Bravais lattice with basis.
In other words, bct have 2 lattice point in the conventional unit cell, whereas st have 1. Thus, bct lattice and st lattice require 8 and 16 atoms in the basis in order to represent ZnTiP2 structure, respectively.

I have calculated following input file. The result shows very small (about 0.02 Ry) bandgap.

c------------------------------------------------------------
go data/ZnTiP2-bct
c------------------------------------------------------------
c brvtyp a c/a b/a alpha beta gamm
bct 10.4636 2 1 90 90 90
c------------------------------------------------------------
c edelt ewidth reltyp sdftyp magtyp record
0.0001 1.5 sra mjw nmag 2nd
c------------------------------------------------------------
c outtyp bzqlty maxitr pmix
update 8 200 0.02
c------------------------------------------------------------
c ntyp
3
c------------------------------------------------------------
c type ncmp rmt field mxl anclr conc
Zn 1 1 0 2 30 100
Ti 1 1 0 2 22 100
P 1 1 0 2 15 100
c------------------------------------------------------------
c natm
8
c------------------------------------------------------------
c atmicx atmtyp
0.0 0.0 0.0 Zn
0.0 1/2 1/4 Zn
1/2 1/2 0.0 Ti
1/2 0.0 1/4 Ti
1/4 1/4 1/8 P
3/4 3/4 1/8 P
3/4 1/4 3/8 P
1/4 3/4 3/8 P
c------------------------------------------------------------

Actually I have little confidence in my input files, because I have limited experience in such a complicated system... So please be careful to check the atomic positions.

# And also, lattice constant "a" should be given in Bohr (not Angstrom). Is it ok?

Hitoshi