[Pw_forum] Problems with phonones in Fe
Nicola Marzari
nicola.marzari at epfl.ch
Sat Oct 26 15:03:26 CEST 2013
Dear Ivan,
maybe you could trade emails with Daniele, cc'd here, that has
been working on iron. At first sight, though:
1) use the BCC unit cell, with only one atom per cell,
and use the minimum number of bands, so that you have no bands
with occupation less than 0.01 .
2) 8x8x8 sampling for q-points is very dense. I'd guess 4x4x4
is more than enough.
I guess that 1)+2) would give a 10-20 times speedup. Also, which
parallelization strategy you are using? For iron on 48 cores I would
probably parallelize on k-points using npools, if the memory allows.
Re the change of sign, at Gamma 3 frequencies should have been zero
to begin with. Again, Daniele can send you a standard setup for these
calculations.
Also, look carefully at the work of Tilmann Hickel and Joerg Neugebauer
on phonons in paramagnetic iron (a very difficult system), and Shobhana
Narasimhan and Stefano de Gironcoli for ferromagnetic bcc iron.
nicola
On 26/10/2013 12:06, Иван Булдашев wrote:
> Dear QE users,
> I studied the thermodynamical parameters of bcc Fe, so I need to obtain
> the phonon frequencies, but there are some problems with ph.x.
> I normally performed scf calculations after convergence tests and got
> electronic structure. But phonon calculations worked very long. For
> instance, look at this input (two atoms of Fe, simple cubic, nbnd=24):
>
> phonons of Fe
> &inputph
> tr2_ph=1.e-10,
> alpha_mix=0.1,
> ldisp=.true.,
> nq1=8, nq2=8, nq3=8
> amass(1)=55.845,
> outdir='/home/buldashev/temp9',
> fildyn='fe.dyn',
> /
>
> This calculation took more than five days on 48 processor cores.
> When I try to compute 16 atoms of Fe, it worked incredibly long and
> spent more than two days for one q-point.
> Why it work so long and how I can calculate phonones for big system (16
> or more atoms)?
>
> The second problem seems very strange. I got the positive frequencies
> (for system with two Fe) in reciprocal-space:
>
> Diagonalizing the dynamical matrix
> q = ( 0.000000000 0.000000000 0.000000000 )
> **************************************************************************
> omega( 1) = 7.991400 [THz] = 266.564426 [cm-1]
> omega( 2) = 7.991400 [THz] = 266.564426 [cm-1]
> omega( 3) = 7.991400 [THz] = 266.564426 [cm-1]
> omega( 4) = 16.792406 [THz] = 560.134373 [cm-1]
> omega( 5) = 16.792406 [THz] = 560.134373 [cm-1]
> omega( 6) = 16.792406 [THz] = 560.134373 [cm-1]
> **************************************************************************
>
> but after matdyn.x they become negative:
>
> q = 0.0000 0.0000 0.0000
> **************************************************************************
> omega( 1) = -14.768968 [THz] = -492.639743 [cm-1]
> ( 0.130412 0.000000 -0.554126 0.000000 0.419449 0.000000 )
> ( -0.130412 0.000000 0.554126 0.000000 -0.419449 0.000000 )
>
> Why they change their signs and how I can improve it?
>
> Thank you.
> Buldashev Ivan, student.
> South Ural State University.
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--
----------------------------------------------------------------------
Prof Nicola Marzari, Chair of Theory and Simulation of Materials, EPFL
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