[Pw_forum] Getting correct phonon frequencies

Stefano de Gironcoli degironc at sissa.it
Thu Jun 16 11:31:14 CEST 2005


I think that there is nothing really wrong in the frequencies you get.
The isolated molecule should have 1 non zero frequency, corresponding
to the bond-stretching mode and 5 zero frequencies (3 due to global 
translational
symmetry and 2 due to rotational symmetry of a diatomic molecule).

In the periodic calculation you are performing the two rotational mode 
never
really vanish due to weak interaction with the periodic images of the 
molecule.

The three translational modes would be zero for a perfect integration of the
exchange-correlation energy on the real-space grid used to reperesent 
the density.
This is never achieved in practice (except for exceedingly dense real 
space grids)
thus some violation of the order of a few percent of the stiffer 
frequency is common
expecially when using gradient corrected XC functionals and/or US 
pseudopotentials.

Your  calculation is reasonably accurate, I think, and the violation of 
the zero-frequency
result that you correctly expect is maybe smaller than you think.

Think in terms of interatomic force constants:

omega_bend = sqrt (PHI/reduced_mass) where PHI is the on site force 
constant of Hydrogen(1)
that by translational symmetry should be equal and opposite to the 
 Hydorgen(1)-Hydrogen(2)
interatomic force constant.
The fact that this sum rule (Acoustic Sum Rule says \sum_I 
PHI_{I\alpha,J\beta} = 0 for each J,\alpha,\beta)
is violated gives a NON-ZERO translational frequency of the order of
omega_trasl = sqrt(Delta PHI/ total_mass)

Therefore the relative error in the determination of the interatomic 
force constants in your calculation
is of the order of

Delta PHI/PHI = (omega_trals/omega_bend)^2 *total_mass/reduced_mas = 
approx (150/4300)^2 *4 = 0.5 %

which is not that bad, although not wonderful.

Translational Acoustic Sum Rule can be enforced (with no noticable 
effect on the real non-zero frequencies)
in the auxiliary code dynmat.x that you can find in the pwtools 
 subdirectory of the espresso distribution.

A more refined treatment of Translation and Rotational Acoustic Sum 
Rules can be enforced in the CVS version
of the distribution (and in future web distributions) thanks to a 
contribution by Nicolas Mounet .

best regards,

Stefano de Gironcoli



Mousumi Upadhyay Kahaly wrote:

>Dear PWscf devepolers and users,
>
>
>       I am trying to get phonon frequencies of H2 molecule(gas phase)
>with PWSCF v.2.0.1.
>
>
>The results for H2 from PWSCF with all other parameters remanining same
>except different energy cutoffs are given below. I had done for dual= 4, 6
>and 8, where "ecutrho= dual* ecutwfc".
>
>Results- with box-size 20 bohr, relaxed atomic positions:
>================================
>        ATOMIC_POSITIONS (angstrom)
>H        1.325979436   0.000000000   0.000000000
>H        0.574020564   0.000000000   0.000000000
>
>Phonon frequencies:
>==================
>A1.  ecutwfc = 25 Ry,  ecutrho = 100 Ry :
>
> **************************************************************************
>     omega( 1) =       2.829060 [THz] =      94.367920 [cm-1]
>     omega( 2) =       2.829060 [THz] =      94.367920 [cm-1]
>     omega( 3) =       3.861495 [THz] =     128.806463 [cm-1]
>     omega( 4) =       4.643165 [THz] =     154.880338 [cm-1]
>     omega( 5) =       4.643165 [THz] =     154.880338 [cm-1]
>     omega( 6) =     129.812814 [THz] =    4330.118153 [cm-1]
>**************************************************************************
>
>
>A2.   ecutwfc = 25 Ry,  ecutrho = 150 Ry :
>
> **************************************************************************
>     omega( 1) =       2.153650 [THz] =      71.838519 [cm-1]
>     omega( 2) =       2.153650 [THz] =      71.838519 [cm-1]
>     omega( 3) =       4.145972 [THz] =     138.295665 [cm-1]
>     omega( 4) =       4.145972 [THz] =     138.295665 [cm-1]
>     omega( 5) =       4.817966 [THz] =     160.711103 [cm-1]
>     omega( 6) =     129.800590 [THz] =    4329.710425 [cm-1]
> **************************************************************************
>
>A3.    ecutwfc = 25 Ry,  ecutrho = 200 Ry :
>
> **************************************************************************
>     omega( 1) =      -1.588702 [THz] =     -52.993762 [cm-1]
>     omega( 2) =      -1.588702 [THz] =     -52.993762 [cm-1]
>     omega( 3) =       3.878442 [THz] =     129.371746 [cm-1]
>     omega( 4) =       4.580893 [THz] =     152.803167 [cm-1]
>     omega( 5) =       4.580893 [THz] =     152.803167 [cm-1]
>     omega( 6) =     129.718516 [THz] =    4326.972705 [cm-1]
> **************************************************************************
>
>
>Thus I got negative and positive non-zero frequencies, where I am
>expecting 4 zero-frequencies, one frequency close to zero(due to bending)
>and one frequency in the order of 4560cm-1.
>
>Changing the atomic positions and with the same three dual values, the
>frequencies and their sign vary. Results are given below-
>
>Results- in box-size = 20 bohr, relaxed atomic positions:
>================================
>        ATOMIC_POSITIONS (angstrom)
>H        0.000000000   0.000000000   0.000000000
>H        0.434143657   0.434143657   0.434143657
>
>Phonon frequencies:
>==================
>B1.  ecutwfc = 25 Ry,  ecutrho = 100 Ry :
>
> **************************************************************************
>     omega( 1) =      -4.811247 [THz] =    -160.486990 [cm-1]
>     omega( 2) =      -4.811247 [THz] =    -160.486990 [cm-1]
>     omega( 3) =      -2.203041 [THz] =     -73.486037 [cm-1]
>     omega( 4) =       0.500005 [THz] =      16.678469 [cm-1]
>     omega( 5) =       0.500005 [THz] =      16.678469 [cm-1]
>     omega( 6) =     129.696261 [THz] =    4326.230348 [cm-1]
> **************************************************************************
>
>                                                                                B2.
>
>ecutwfc
>=
>25
>Ry,
>
>ecutrho
>=
>150
>Ry
>:
>
> **************************************************************************
>     omega( 1) =       2.357793 [THz] =      78.648043 [cm-1]
>     omega( 2) =       2.357793 [THz] =      78.648043 [cm-1]
>     omega( 3) =       2.816217 [THz] =      93.939507 [cm-1]
>     omega( 4) =       2.816217 [THz] =      93.939507 [cm-1]
>     omega( 5) =       4.476603 [THz] =     149.324395 [cm-1]
>     omega( 6) =     129.796002 [THz] =    4329.557366 [cm-1]
> **************************************************************************
>
>B3.  ecutwfc = 25 Ry,  ecutrho = 200 Ry :
>
> **************************************************************************
>     omega( 1) =       2.005189 [THz] =      66.886334 [cm-1]
>     omega( 2) =       2.005189 [THz] =      66.886334 [cm-1]
>     omega( 3) =       3.773415 [THz] =     125.868397 [cm-1]
>     omega( 4) =       4.558315 [THz] =     152.050044 [cm-1]
>     omega( 5) =       4.558315 [THz] =     152.050044 [cm-1]
>     omega( 6) =     129.748872 [THz] =    4327.985258 [cm-1]
> **************************************************************************
>
>
>         H2 molecule system is so simple and still phonon frequencies
>obtained are so off! Will you please help me to understand what
>is the origin of these non-zero(both negative and positive<200
>cm-1 )frequencies?
>How should I get rid of them and have correct frequencies?
>
>         Best regards,    mousumi.
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>







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