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I did this in VASP but I guess this would be the case for any program. I have a structure that has a cubic conventional cell (a = 9.86 A). I found the primitive cell (a=b=c=6.97, alpha=90, beta=60, gamma=120) and used that as the input structure.

I relaxed the structure and that results in changes in the cell (a=7.23,b=6.97,c=7.2,alpha=90.99,beta=58.1,gamma=121).

Isn't the cubic symmetry now broken because the lattice parameters and angles have all changed? How do I go from the relaxed primitive cell back to the conventional cubic cell??

Edit: Initial CIF file (conventional cell), as well as the input POSCAR (primitive cell) and output CONTCAR files are here: https://github.com/DoubleKx/arg_example

Edit2: I'm trying to replicate the examples from this paper. I'm on Part 1. I've found the primitive cell, enumerated the structure, and run a relaxation (on just one of the enumerated structures).

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    $\begingroup$ Welcome, +1. What do you mean go from relaxed back to conventional cubic cell? $\endgroup$
    – Cody Aldaz
    Jul 1, 2020 at 4:14
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    $\begingroup$ +1. Thanks for asking your question here, and Welcome to the site!!! We hope to see a lot more of you !!! $\endgroup$
    – Nike Dattani
    Jul 1, 2020 at 5:01
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    $\begingroup$ +1. Perhpas you could provide the initial and relaxed POSCAR files? I think this would help understand your question better. $\endgroup$
    – ProfM
    Jul 1, 2020 at 5:54
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    $\begingroup$ Yes, the cubic symmetry is broken. If that happens make sure that the result is converged with respect to energy cutoff and k-point grid. If you want to enforce the symmetry set the ISIF parameter appropriately. $\endgroup$
    – Fabian
    Jul 1, 2020 at 8:45
  • $\begingroup$ @CodyAldaz my input was a primitive cell. So the output is still primitive - I need to go back to the conventional cell. The issue is the symmetry is broken after relaxation $\endgroup$
    – DoubleKx
    Jul 1, 2020 at 21:19

1 Answer 1

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From what I understand in your question you have you started with a conventional cell with bases $B_{conv}$ and you found a transformation matrix $C$ to the unit cell such that

$B_{unit} = C B_{conv}$.

Then you relaxed the unit cell and you received a "new" unit cell $\tilde{B}_{unit}$. The closest approximation to your old conventional cell should thus be.

$\tilde{B}_{conv} = C^{-1} \tilde{B}_{unit}$.

Naturally, it will not be perfectly cubic again as the first unit cell was relaxed.

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  • $\begingroup$ That's exactly right. Where Bunit is the primitive cell. The example I'm trying to follow (please see the link I added to my post) has somehow gone from the relaxed primitive cell to the conventional cubic cell. Perhaps I've missed a step somewhere. $\endgroup$
    – DoubleKx
    Jul 1, 2020 at 21:33
  • $\begingroup$ In the supplementary of the paper they provide a quite extensive jupyter notebook and also the results of their relaxation run in a file (vasprun.xml.relax2.gz). Have you checked that out? With Pymatgen it should be possible to see what they have done exactly. $\endgroup$
    – CKl
    Jul 2, 2020 at 9:23
  • $\begingroup$ +1. You are sooooo close, you just need to post 3 more questions or answers within the next 24 hours to fulfill commitment: area51.stackexchange.com/users/208946/ckl do you think you can do it please? :) $\endgroup$
    – Nike Dattani
    Oct 24, 2020 at 21:40

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