[Pw_forum] Problems with phonones in Fe

[Krishnendu Mukherjee]

Respected Experts,

 We have installed Quantum Espresso 6.2 and interested in calculating the
Helmholtz free energy of bcc (iron) ferrite. I found out this discussion in
Pw_forum from 2013.

 There it was suggested by Dr. Nicola Marzari to use the BCC unit cell,
with only one atom per cell, and use the minimum number of bands, such that
there are no bands with occupation less than 0.01.

Can you please let me know where to set the number of bands in the set up?

Thanks,
Regards,
Krishnendu

====================================================================


Dear Ivan,

1) as Nicola already mentioned, you should use primitive cells if you
want to speed up your calculations.

2) The 4x4x4 suggested by Nicola should be a good reasonable grid even
for the primitive bcc cell. This corresponds roughly to a 3x3x3 grid
for the 2 atoms cubic cell if you want to keep the same
(approximately) linear density. You are using a 8x8x8 grid in the
cubic cell, do you really need it for thermodynamic quantities?

Here is a setup for the phonon calculation of the bcc ferromagnetic
iron (primitive bcc cell):

&INPUTPH
     outdir = "./scratch/",
     prefix = "--",
     fildyn = "--",
     amass(1)=55.845,
     ldisp  = .true.,
     nq1 = 4,
     nq2 = 4,
     nq3 = 4,
     tr2_ph = 1e-16,
/

As you see, even if it costs a bit more, I use a more strict
convergence threshold. Try to decrease your value.
Also, do not use poor/large scf thresholds in the pw calculation, you
gain very little.

If your pw calculation is converged properly, this should give you
frequencies that are zero in Gamma.

		Daniele


On Oct 26, 2013, at 3:03 PM, Nicola Marzari wrote:

> > >* 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.
*>>* _______________________________________________
*>>* Pw_forum mailing list
*>>* Pw_forum at pwscf.org <http://pwscf.org/mailman/listinfo/pw_forum>
*>>* http://pwscf.org/mailman/listinfo/pw_forum
<http://pwscf.org/mailman/listinfo/pw_forum>
*>> > > >* --
*> >* ----------------------------------------------------------------------
*>* Prof Nicola Marzari, Chair of Theory and Simulation of Materials, EPFL
*




-- 
Dr. Krishnendu Mukherjee,

Principal Scientist,
CSIR-NML,
Jamshedpur.
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.quantum-espresso.org/pipermail/users/attachments/20171201/b25e27ff/attachment.html>