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 <[email protected]> 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 <[email protected]>
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
<[email protected]<mailto:[email protected]>> 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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I-00015 - Monterotondo Scalo (RM)
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E-mail: <[email protected]>
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