[Pw_forum] Relaxation of low symmetry lattices

[Gabriele Sclauzero]

Dear Manoj,

Manoj Srivastava wrote:

> Result of relaxation : The covergence is achieved with follwing forces on
> the atoms. 
> Forces acting on atoms (Ry/au):
> 
>    Forces acting on atoms (Ry/au):
> 
>      atom   1 type  1   force =     0.00000000    0.00000000    0.00000000
>      atom   2 type  1   force =     0.00000567    0.00000185   -0.00000172
>      atom   3 type  1   force =     0.00001155    0.00000377   -0.00000372
>      atom   4 type  1   force =    -0.00001155   -0.00000377    0.00000372
>      atom   5 type  1   force =    -0.00000567   -0.00000185    0.00000172
> 
> 
> I realized that I should not have chosen cubic symmetry group, as my unit
> cell is not cubic. So, as described in the input file parameters, for the

in my understanding, the option cubic does not mean that you use a cubic symmetry group, 
but that you search for symmetry operations starting from the cubic ones (and removing 
those which do not hold in your system).
If you try with hexagonal it will find only 2 symmetries (identity and inversion), if you 
use cubic it should find 2 more, but they are not compatible with your FFT grid (see at 
the beginning of the output).

> low symmetry lattices, I specifed nosym=.TRUE., and again the relaxation
> stopped after 1 scf cycle, acheiving convergence. But the forces on the
> atoms are -  
> 
> Forces acting on atoms (Ry/au):
> 
>      atom   1 type  1   force =     0.00001355    0.00000524   -0.00000516
>      atom   2 type  1   force =    -0.00001493   -0.00000395    0.00000395
>      atom   3 type  1   force =     0.00000253   -0.00000043    0.00000045
>      atom   4 type  1   force =     0.00000820    0.00000230   -0.00000239
>      atom   5 type  1   force =    -0.00000935   -0.00000316    0.00000315
> 
> So, still forces are .001 ev/a.u, which i think is higher for a perfect

 From your data I see that all components of the forces are lower than 0.0002 ev/au, which 
is pretty small for practical purposes.

> FCC lattice. I can specify stricter convergence of force, but that would
> move atoms around and break symmetry. So, as long as the lattice parameter

Typically, a good convergence threshold for forces in relaxations is around 0.01 ev/au, 
which is 2 orders of magnitude higher than your residual forces.

> is correct, I think for perfect FCC lattice, forces on atoms should be
> smaller than what they are presently. I have checked the lattice constant

You have to take into account that you are using several approximations in the code (as in 
any other numeric solver), and you are using a finite basis set to represent wavefunctions 
and charge density, a finite number of k-point for sampling the BZ,...

> for bulk FCC with a different calculation, so lattice constant is not an
> issue. Another possibility is symmetry group, as you can see with the same

sure, lattice constant cannot be an issue in this case, since it simply rescales 
proportionally the lattice vectors lengths

> input variables, changing symmetry group changes results.I was wondering
> if there is a way i can choose correct symmetry group for this low
> symmetry lattice which might give me smaller forces? Are there any other

The code gives you the correct symmetry group of your system, which depends on the 
symmetry of the Bravais lattice and can be lowered if you have more than one atom per cell 
(sometimes it gives you a subgroup for the FFT incompatibility issues I mentioned). I 
don't see any other group that you could use other than the correct one... for sure not 
one of higher order: that makes no sense to me.

> parameters I need to specify? On a broad view I am curious to know how to
> do relaxation of low symmetry lattices.

Usually you lower the symmetry by purpose in order to let atoms explore non-symmetric 
configurations (surfaces, defects, nanocontacts, ...), so I don't understand you aim of 
chosing a low symmetry lattice to describe a very high symmetry system.

> 
> Regards, 
> Manoj Srivastava
> University of Florida
> Gainesville, USA
> 
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