2
$\begingroup$

Would like to find if given two crystal structures whether they are symmetry distinct or symmetry equivalent in fastest (in terms of computation) possible way. The structure may have same lattice symmetry but the atom distribution could be different.

As an specific example: on FCC (face-centred cubic) lattice containing 18 A atoms and 14 B atoms are distributed randomly. Need to find any two random configurations, say conf-1 and conf-2, are symmetry equivalent or distinct?

Trivial way to do this would be: reduce conf-1 and conf-2 to their primitive lattices, then perform all symmetry-operations on conf-1 and check if resulting structure is same as conf-2 (I am interested in 'equivalent' or 'not equivalent' binary output). Unfortunately, this would be computationally expensive.

Is there any alternate way to achieve this?

Thanks to All.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

I guess you want to solve many such problems, otherwise you wouldn't care so much about the efficiency. The approach I would take is to first apply a sequence of necessity tests, of increasing cost. For example, first the number of atoms of each type, then the profile of distances between the atoms, etc.. The idea is that most pairs are eliminated quickly and you only need the expensive test for a few remaining cases.

Also, if your purpose is to eliminate equivalent cases for a long list, or if you want to look up a new case in a database of known cases, you are far better off devising a canonical form than using pairwise comparisons.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Yes, I want to create a database of symmetry-distinct structures, hence, emphasis on computational cost. $\endgroup$
    – MC70
    Commented Oct 21, 2015 at 14:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .