[Re:09] plotting DOS

Posted on : September 11, 2015 (Fri) 22:26:13

by Jeremy

Hi Hitoshi,

'anclr' for Ti should be 22 indeed.

The 'natm' is the number of atoms of basis, and thus the 'atmicx' are corresponding to the atomic position of the atoms of basis.
Is this right?

The atomic positions of the atoms of basis in the unit cell are as fellow.

Zn(0,0,0); Ti (0.5,0.5,0); P(0.25,0.25,0.125)

And the fellowing are all the atoms in the unit cell
Zn
0.00000000 0.00000000 0.00000000
0.00000000 0.50000000 0.25000000
0.50000000 0.50000000 0.50000000
0.50000000 0.00000000 0.75000000
Ti
0.50000000 0.50000000 0.00000000
0.50000000 0.00000000 0.25000000
0.00000000 0.00000000 0.50000000
0.00000000 0.50000000 0.75000000
P
0.25000000 0.25000000 0.12500000
0.25000000 0.75000000 0.87500000
0.75000000 0.75000000 0.12500000
0.75000000 0.25000000 0.87500000
0.75000000 0.75000000 0.62500000
0.75000000 0.25000000 0.37500000
0.25000000 0.25000000 0.62500000
0.25000000 0.75000000 0.37500000

In the fact, the 2 P are belong to the same atom of basis.

So what do think?