[Pw_forum] convergence of phonon

Eduardo Ariel Menendez P emenendez at macul.ciencias.uchile.cl
Thu May 26 01:23:05 CEST 2005

Thank you Stefano and Nicola for your advice.
I just want to close this issue, after
almost three months calculating the phonons.
I added random displacements to the atoms and relaxed the atomic
position and the cell shape. In this way, the calculations were done
without symmetry. The atomic positions and the cell shape changed
a little bit, but the structure was not unstable.

Then, for each cutoff (30,40, and 50 Ry), I relaxed the positions and
followed a Gamma-point phonon calculation. A summary is

Cutoff                 30 Ry       40 Ry            50 Ry
smallest omegas   -63,-45,14   -57,-27,-12    -44,-32,-9
largest omegas    637,691,692   640,691,692   639,692,692
(omegas in cm^-1)

Hence, I consider that I have reasonable convergence, and
even the awful  negative frecuencies go towards 0.

Was I think that is to be learnt, is that before a phonon calculation one
must relax the positions after breaking the symmetry. The bad news is
that the calculation becomes much more expensive.

Best regards

Message: 3
Date: Thu, 20 Jan 2005 12:43:33 +0100
From: Stefano de Gironcoli <degironc at sissa.it>
To: pw_forum at pwscf.org
Subject: Re: [Pw_forum] Re: convergence of phonon
Reply-To: pw_forum at pwscf.org

There can be many reasons for appearance of negative (actually
immaginary) frequencies
in a stable system (insufficient convergence, poor k-point sampling, bad
bug in the code...).
But it is also possible that the calculation is telling you something
about your system.
Your forces are zero, most of them for symmetry reasons (judging from
the fact that they are exacly zero!!!).
It is possible that your system is not in stable equilibrium but in a
saddle point (that can be
physically relevant or due to poor k-point sampling, insufficient
cut-off, bad xc-functional,
bad pseudopotential, ...)
Your negative fequencies are two-fold degenerate that means for sure
that they do not belong
to the totally symmetric irreducible representation of your crystal
symmtry (the one that is
preserved during relaxation).
Try breaking the symmetry of your system moving the atoms along one ove
the unstable eigenvectors
or just add some random displacements to your coordinates and relax again .
Does the system go back to the original position ? if this happens there
is a problem in the phonon calculation.
Or the system goes somewhere else ? If this happens your original
positions actually  correspond
to a saddle point and the phonon code spotted it.

I'm suspicious about your degauss, it looks very small to me.
I would rather use a larger degauss and m-p or m-v smearing
Make a plot of your DOS broadening it with  FD with degauss=0.002
If your sampling is sufficiently dense the DOS will be smooth, otherwise
it will  wildy oscillate.

best regards,

   Stefano de Gironcoli

Eduardo Ariel Menendez P wrote:

>Hello phonon community,
>Thanks to nicola marzari for his advice on the parameters for phonon
>However, I have tested the convergence of the phonon
>frequencies against the wavefunction cutoff, and I do not find convegence.
>I am surprised that I obtain very large negative frequencies like
>     omega( 1) =      -4.995406 [THz] =    -166.629919 [cm-1]
>     omega( 2) =      -4.995406 [THz] =    -166.629919 [cm-1]
>     omega( 3) =      -4.474172 [THz] =    -149.243320 [cm-1]
>     omega( 4) =      -4.474172 [THz] =    -149.243320 [cm-1]
>     omega( 5) =      -2.511886 [THz] =     -83.788069 [cm-1]
>     omega( 6) =      -2.511886 [THz] =     -83.788069 [cm-1]
>     omega( 7) =       0.734339 [THz] =      24.495069 [cm-1]
>     omega( 8) =       1.858145 [THz] =      61.981460 [cm-1]
>  etc, up to omega(36)

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