Re: [Spglib-users] get_symmetry basis

[Anders C. Riis-J. <ac...@fy...>]

Here is my code:


import sys
import spglib
import numpy as np
from numpy import pi, sin, cos, arccos, sqrt, dot
from ase import Atoms

# Input: Structure
a = 2.46
cell = np.array([[a, 0, 0,], [-a/2, a*sin(60*pi/180), 0], [0, 0, 5*a]])

structure = Atoms('C2', cell=cell,
             scaled_positions=np.array([[0, 0, 0],
                        [2./3, 1./3, 0]]),
             )

structure.pbc = (1, 1, 0)

cell = structure.get_cell()

#Define rotation matrix:
rot60 = np.array([[1/2, -sqrt(3)/2, 0], [sqrt(3)/2, 1/2, 0], [0, 0, 1]])

# Get all symmetry operations
symop = spglib.get_symmetry(structure, symprec=1e-5)
rotaa = symop['rotations']
trlaa = symop['translations']

rotxy = []
trlxy = []

#Transform matrices to cartesian coordinates:
for j in range(len(rotaa)):
    rot_t = np.dot(np.linalg.inv(cell), np.dot(rotaa[j], cell))
    trl_t = np.dot(np.dot(np.linalg.inv(cell), trlaa[j]), cell)
    rotxy.append(rot_t)
    trlxy.append(trl_t)




Anders

________________________________
Fra: Atsushi Togo <atz...@gm...>
Sendt: 11. oktober 2017 02:14:17
Til: Anders Christian Riis-Jensen
Cc: spg...@li...
Emne: Re: [Spglib-users] get_symmetry basis

Hi,

Could you give your script?

Togo

On Tue, Oct 10, 2017 at 7:18 PM, Anders Christian Riis-Jensen
<ac...@fy...> wrote:
> I am using the built-in function get_symmetry from spglib, and I seem to
> have some issues with this. In this context I have a question.
> I will here give an example showing my problem:
>
> If I define the unit cell for graphene:
> cell = np.array([[a, 0, 0,], [-a/2, a*sin(60*pi/180), 0], [0, 0, 5*a]])
> and put the carbon atoms in the (scaled) positions [0,0,0] and [0, 2/3, 1/3]
> I then run spglib.get_symmetry on the system (with the default tolerance). I
> get a set of symmetry transformations (I understand each transformation is a
> combination of a matrix multiplication and a translation). I want to have
> the symmetry matrices (M) in cartesian coordinates and therefore make the
> following basis transformation:
> inverse(cell).M.cell
> And here I would for example expect to find the standard 60 degrees rotation
> matrix, however this I dot not get. I therefore realize that the original
> symmetry matrices might not be written in the lattice basis as I expected.
>
> Which basis is used and what is the proper transformation to get to
> cartesian coordinates?
>
> Any answer will be greatly appreciated.
>
> Best regards
> Anders Riis-Jensen
>
>
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--
Atsushi Togo
Elements Strategy Initiative for Structural Materials, Kyoto university
E-mail: atz...@gm...