[Pw_forum] Getting correct phonon frequencies

[Mousumi Upadhyay Kahaly]

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.