[QE-users] Keep the symmetry during a calculation

[Giuseppe Mattioli]

Dear Carina
In the case of a simple lattice such as a cubic box full of water  
molecules plus one solvated ion (no symmetry), you can try to do it  
yourself: shrink isotropically the coordinates, step by step, and plot  
an E/V curve to find a minimum. A parabolic fit is often a quick and  
dirty way to have a reasonable (i.e. not so stressed) box.
I suppose that there is no simple way otherwise to do what you need.
HTH
Giuseppe

Carina Backtorp <carinabacktorp at hotmail.com> ha scritto:

> 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



GIUSEPPE MATTIOLI
CNR - ISTITUTO DI STRUTTURA DELLA MATERIA
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E-mail: <giuseppe.mattioli at ism.cnr.it>