[QE-users] Symmetry - Centered cell reduction

[Fnukal Frantisek]

Hello Paulo and thank you for your answer,


I will clarify what I meant with an example. Let's say I have a structure with space group 15 -> C 2/c, so monoclinic and base centered. If I create an input file specifying the space group, the code reduces the information to this structure in primitive cell, which I can see in the output file by examining the cell volume (or cell vectors).

(I'm somehow failing to create the input specifying ibrav, I'm clearly doing something wrong because the code requires smearing for a non-metallic structure, but I still can see that the volume is 2x smaller)

With ibrav = 0, the reduction is not done and on the output, I get exactly what I gave on the input (with optimized geometry). So up until this point I've been using ibrav=0, because it is more convenient for me.


What you're saying about k-points and computational time is interesting. I've done some small testing a while back for conventional vs primitive cell while keeping the same k-point spacing and it turned out some structures in primitive cell took longer to compute than in conventional cell. However, I'm very much a beginner at this, so there is a solid chance that, again, I may have done something wrong. I will have to educate myself on this matter.


Thank you,


Best regards,

Frantisek Fnukal

UCT Prague



________________________________
Od: Paolo Giannozzi <paolo.giannozzi at uniud.it>
Odesláno: sobota 17. prosince 2022 10:23:14
Komu: Quantum ESPRESSO users Forum
Kopie: Fnukal Frantisek
Předmět: Re: [QE-users] Symmetry - Centered cell reduction

I am not sure I understand your problem. For face-centered and
body-centered lattice, one may use a minimum-size "primitive" cell or a
larger but simpler  "conventional" cell, having a simple cubic (or
simple tetragonal, or whatever applies) lattice. Is this what you are
referring to?

QE uses primitive cells because it is _always_ convenient to use the
smallest cell: the computation increases as (cell volume)^3, while the
number of k-points decreases as 1/(cell volume).

If however you really want to use the conventional cell for whatever
reason, it is easy to do that using "way 3)" you describe below. Just
specify a simple cubic (simple tetragonal, whatever) lattice and all the
atomic positions in the conventional cell.

It wouldn't be difficult to force "way 2)" to do the same automatically,
but this requires a) adding an additional variable (not a big deal), and
b) checking all 230 space groups and changing accordingly many of them
(also simple but looong and error-prone)

Paolo

On 17/12/2022 04:39, Fnukal Frantisek via users wrote:
> Dear QE users,
>
>
> I've studied the possible ways of inputing structural data and I found there basically were 3 ways. 1) Using ibrav = 0, so in essence saying: "Find it yourself" 2) Specifying the space group 3) Specifying the Bravais lattice with ibrav.
>
> So far I've been using ibrav = 0, however output files keep telling me this way is not recommended as it may not correctly recognize symmetry in some cases.
>
> However, I have a very important problem with the other 2 ways of inputing data. I've noticed they both automatically perform transformation to primitive cell if I input a structure in centered cell. My problem is with k-points. If I want to have an equivalent amount of k-points in the primitive cell, then I have a smaller system, but more k-points and in some cases this may lead to completely erasing any profit gained by reducing the system. I would very much like to control whether the transformation is applied or not.
>
> My question is this: Is there any way of inputing structural data, preferably with space group, and turn off automatic centered cell to primitive cell conversion?
>
>
> Thank you for your answers.
>
>
> Best regards,
>
> Frantisek Fnukal
>
> UCT Prague
>
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--
Paolo Giannozzi, Dip. Scienze Matematiche Informatiche e Fisiche,
Univ. Udine, via delle Scienze 206, 33100 Udine Italy, +39-0432-558216