[QE-users] Keep the symmetry during a calculation

[Carina Backtorp]

Hi!

Thank you for the explanation.
I understand it as although I have given ibrav=1 as an input for the vc-relax calculation, Quantum Espresso may find that the crystal does not have ibrav=1 symmetry. Is that correct?
As an example, if one places ONE  water molecule in a large cubic box and runs a vc-relax calculation in QE with ibrav = 1, i.e. cubic symmetry, then
QE does a symmetry analysis concluding that the crystal symmetry is not ibrav 1 and therefore does the calculation with the correct symmetry.

 If so, is there a key word in QE that I can include into the input file so that QE gives the obtained corrected symmetry in the output file?

Also, is it possible to give QE a constraint in the input file so that it keeps the symmetry that was given? For example: even if the crystal structure does not have the cubic symmetry according to QE, can one still keep all three box lengths equal during a vc-relax calculation?

Thank you for all help!
Kind regards,
Carina

________________________________
From: Paolo Giannozzi <p.giannozzi at gmail.com>
Sent: Wednesday, September 19, 2018 3:44 PM
To: Quantum Espresso users Forum
Cc: Carina Backtorp
Subject: Re: [QE-users] Keep the symmetry during a calculation

On Sat, Sep 15, 2018 at 10:09 AM, Carina Backtorp <carinabacktorp at hotmail.com<mailto:carinabacktorp at hotmail.com>> wrote:

1) When doing an vc-relax calculation in espresso, I expected that once the symmetry was given (ibrav=1) espresso should keep the symmetry during all the calculation

it does. Note however that the symmetry of the lattice is just part of the story.  QE finds the starting symmetry of the crystal, and this is what is kept during the run (although occasionally, numerical noise and poor convergence may lead to the loss of the original symmetry). If the starting symmetry is not cubic, the final cell may lose its original cubic aspect.  Also note that QE uses its own criteria for symmetry, that may differ from criteria used by other codes, so something that QE deems cubic may be classified as orthorhombic by a pickier code

Paolo


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--
Paolo Giannozzi, Dip. Scienze Matematiche Informatiche e Fisiche,
Univ. Udine, via delle Scienze 208, 33100 Udine, Italy
Phone +39-0432-558216, fax +39-0432-558222

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