[Pw_forum] How to set nbnd value for berry phase calculations?

Lim Chiang Huay, Freda limch at ihpc.a-star.edu.sg
Wed Jun 30 09:43:52 CEST 2010


Hi all,

I'm doing Berry Phase calculations for the first time for BiFeO3. I met with some problems in setting nbnd values for the scf and nscf calculations. And I can't find an answer in previous forum post.

I used the pseudopotentials available in the website http://www.pwscf.org/pseudo.php
They are:
Bi.pbe-d-mt.UPF
Fe.pbe-nd-rrkjus.UPF
O.pbe-mt.UPF
and from the pseudopotentials, there should be 41 electrons in a cell. From the manual nbnd = nelect/2.

I've tried different combinations of nbnd values for scf and nscf jobs and non was able to give me the polarization value.
The following are the some selected cases and their error msg:
Case(1)
*.scf, nbnd=21; *.nscf, nbnd=21
Both jobs crash
Error for scf: too few bands
Error for nscf: starting and expected charges differ

Case(2)
*.scf, nbnd=43; *.nscf, nbnd=42
Scf converged. Nscf crashed.
Error for nscf: Polarization only for insulators and no empty bands

Case(3)
*.scf, nbnd=42; *.nscf, nbnd=42
Scf converged. Nscf crashed.
Error for scf: too few spin up bands
Error for nscf: starting and expected charges differ

My input files are as follows:
SCF JOB:
&control
    calculation  = 'scf',
    restart_mode = 'from_scratch'
    pseudo_dir   = '/Home6/cc/ihpclimc/BP_CAL/PPOT'
    outdir       = '/Scratch/ihpclimc/BERRY_PHASE_CALCULATIONS/Try_BP/OUTDIR',
 /
 &system
    ibrav=5
    celldm(1)=10.6523,
    celldm(4)=0.50988,
    nat=10
    ntyp=3
    nbnd=42
    ecutwfc=50.0
    occupations = 'fixed'
    degauss=0.00
    nspin=2
    multiplicity=5
    lda_plus_u=.true.
    Hubbard_U(2)=6
 /
 &electrons
    conv_thr = 1e-12,
    mixing_beta=0.3,
    diagonalization='cg',
/
ATOMIC_SPECIES
  Bi   209.0      Bi.pbe-d-mt.UPF
  Fe    55.85     Fe.pbe-nd-rrkjus.UPF
  O     16.00     O.pbe-mt.UPF

ATOMIC_POSITIONS crystal
  Bi    0.000    0.000    0.000         0       0       0
  Bi    0.500    0.500    0.500         0       0       0
  Fe    0.219    0.219    0.219
  Fe    0.719    0.719    0.719
  O     0.445    0.032    0.897
  O     0.897    0.445    0.032
  O     0.032    0.897    0.445
  O     1.397    0.532    0.945
  O     0.532    0.945    1.397
  O     0.945    1.397    0.532

K_POINTS {automatic}
  1 1 1 1 1 1

NSCF JOB:
&control
    calculation  = 'nscf',
    pseudo_dir   = '/Home6/cc/ihpclimc/BP_CAL/PPOT'
    outdir       = '/Scratch/ihpclimc/BERRY_PHASE_CALCULATIONS/Try_BP/OUTDIR',
    lberry       = .true.
    gdir         = 3
    nppstr       = 7
 /
 &system
    ibrav=5
    celldm(1)=10.6523,
    celldm(4)=0.50988,
    nat=10
    ntyp=3
    nbnd=42
    ecutwfc=50.0
    occupations = 'fixed'
    degauss=0.00
    nspin=2
    multiplicity=5
    lda_plus_u=.true.
    Hubbard_U(2)=6
 /
 &electrons
    conv_thr = 1e-12,
    mixing_beta=0.3,
    diagonalization='cg',
/
ATOMIC_SPECIES
  Bi   209.0      Bi.pbe-d-mt.UPF
  Fe    55.85     Fe.pbe-nd-rrkjus.UPF
  O     16.00     O.pbe-mt.UPF

ATOMIC_POSITIONS crystal
  Bi    0.000    0.000    0.000         0       0       0
  Bi    0.500    0.500    0.500         0       0       0
  Fe    0.219    0.219    0.219
  Fe    0.719    0.719    0.719
  O     0.445    0.032    0.897
  O     0.897    0.445    0.032
  O     0.032    0.897    0.445
  O     1.397    0.532    0.945
  O     0.532    0.945    1.397
  O     0.945    1.397    0.532

K_POINTS {automatic}
  4 4 4 1 1 1


Can anyone help?

Thanks,
Rgds

Freda Lim
Institute of High Performance Computing
1 Fusionopolis Way, #16-16 Connexis
Singapore 138632
Email: limch at ihpc.a-star.edu.sg<mailto:limch at ihpc.a-star.edu.sg>



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