[Pw_forum] sawtooth method

hanghui chen chenhanghuipwscf at gmail.com
Sat Nov 15 19:27:53 CET 2008


Dear PWSCF users,
      I calculate a symmetric STO slab (separated by vacuum) and turn on a
symmetric sawtooth potential (the turning point is right at the center of
the STO slab). I fix all the atoms. Without the sawtooth potential, the
forces are symmetric and there are no forces in the central layer. However,
when I turn on the sawtooth potential which I set to be symmetric, the
forces on the atoms are NO longer symmetric and the forces in the layer are
NO longer zero. The input file and the forces (turning on sawtooth
potential) are given below.
     Why are the forces NOT symmetric but the configuration (including the
sawtooth potential) is symmetric? Does the turning point in the material
cause any anomaly?
     Thank you very much.

Hanghui Chen
Department of Physics, Yale University

Input file:
&CONTROL
  calculation='scf'
  wf_collect=.true.
  pseudo_dir = './psp'
  outdir='.'
  wfcdir='/scratch'
  prefix='test1'
  disk_io='low'
  verbosity='high'
  tprnfor = .true.
  tstress = .true.
  dt=80.D0
  tefield=.true.
/
&SYSTEM
  ibrav=  6
  celldm(1) = 7.27
  celldm(3) = 6.0
  nat=  12
  ntyp= 3
  ecutwfc = 30.0
  ecutrho = 180.0
  occupations='smearing'
  smearing='gauss'
  degauss=0.005
  edir=3
  emaxpos=0.5
  eopreg=0.5
  eamp=0.01
/
&ELECTRONS
  diagonalization='david'
  mixing_beta = 0.7D0
  diago_david_ndim = 4
  mixing_mode= 'local-TF'
  electron_maxstep = 300
  startingpot = 'file'
  conv_thr= 1.d-8
/
&IONS
  ion_dynamics = 'bfgs'
  phase_space = 'full'
  pot_extrapolation = 'first_order'
  wfc_extrapolation = 'first_order'
/
&CELL
  cell_dynamics = 'damp-w'
/
ATOMIC_SPECIES
 Sr  87.62  038-Sr-ca-sp-vgrp.uspp.format.UPF
 Ti  47.90  022-Ti-ca-sp-vgrp.uspp.format.UPF
 O   16.00  008-O-ca--vgrp.uspp.format.UPF
ATOMIC_POSITIONS alat
 Sr 0.00 0.00 2.00
 O  0.50 0.50 2.00
 Ti 0.50 0.50 2.50
 O  0.50 0.00 2.50
 O  0.00 0.50 2.50
 Sr 0.00 0.00 3.00
 O  0.50 0.50 3.00
 Ti 0.50 0.50 3.50
 O  0.50 0.00 3.50
 O  0.00 0.50 3.50
 Sr 0.00 0.00 4.00
 O  0.50 0.50 4.00
K_POINTS {automatic}
8 8 1 0 0 0
Forces:
 atom   1 type  1   force =     0.00000001   -0.00000001   -0.09670746
 atom   2 type  3   force =     0.00000000    0.00000000   -0.21698818
 atom   3 type  2   force =     0.00000000    0.00000001   -0.17490285
 atom   4 type  3   force =     0.00000000    0.00000000   -0.17341486
 atom   5 type  3   force =     0.00000001    0.00000000   -0.17341491
 atom   6 type  1   force =    -0.00000002    0.00000000    0.06540405
 atom   7 type  3   force =    -0.00000001   -0.00000001   -0.01253063
 atom   8 type  2   force =     0.00000001    0.00000001    0.32344891
 atom   9 type  3   force =     0.00000000    0.00000000    0.08311003
 atom  10 type  3   force =     0.00000000    0.00000000    0.08311005
 atom  11 type  1   force =     0.00000000    0.00000000    0.16600421
 atom  12 type  3   force =     0.00000001    0.00000000    0.12688164
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