[Pw_forum] Hello, why the calculated polarization through Berry Phase is so large
Xiaofang
lizhanfang at yahoo.com.cn
Tue Feb 19 16:00:43 CET 2008
Dear all,
I have calculated the polarization of KNbO3 using the Berry Phase. The atomic position and lattice constants of input are all same with the experimental values. The computed polarization is 1.1201929 C/m^2,but the experimental polarization is about 0.37 C/m^2. I even use the espressov.4.0cvs to calculate, but the result is similar. I have also tried to calculate the Born effective charge using ph.x, through the effective charge I get the polarization which is 0.44 C/m^2. I don’t know why there is so much difference between the two kinds of method and why the calculated polarization through Berry phase is so large. Any suggestion is appreciated.
This is the Scf’s input
&control
calculation = 'scf'
restart_mode = 'from_scratch'
pseudo_dir = '/home/pseudo/'
outdir = '/home/tmp/'
/
&system
ibrav=6
celldm(1)=7.5532353215993782928038366689241
celldm(3)=1.016512384288216162121591193395
nat=5
ntyp=3
nbnd=20
ecutwfc=30.0
ecutrho =300.0
occupations = 'fixed'
degauss=0.00
/
&electrons
diagonalization ='cg'
conv_thr = 1e-12,
mixing_beta=0.3,
/
ATOMIC_SPECIES
K 39.0983 019-K-ca-sp-vgrp.uspp.UPF
Nb 92.90638 041-Nb-ca-sp-vgrp.uspp.UPF
O 15.9994 008-O-ca--vgrp.uspp.UPF
ATOMIC_POSITIONS (crystal)
K 0.5 0.5 0.518
Nb 0.000000000 0.000000000 0.0
O 0.500000000 0.00000000 0.04
O 0.00000000 0.500000000 0.04
O 0.00000000 0.00000000 0.544
K_POINTS {automatic}
6 6 3 1 1 1
This is the BP’s input
&control
calculation = 'nscf'
pseudo_dir = '/home/pseudo/'
outdir = '/home/tmp/'
lberry = .true.
gdir = 3
nppstr = 9
/
&system
ibrav=6
celldm(1)=7.5532353215993782928038366689241
celldm(3)=1.016512384288216162121591193395
nat=5
ntyp=3
nbnd=20
ecutwfc=30.0
ecutrho =300.0
occupations = 'fixed'
degauss=0.00
/
&electrons
diagonalization ='cg'
conv_thr = 1e-5
mixing_beta = 0.3
/
ATOMIC_SPECIES
K 39.0983 019-K-ca-sp-vgrp.uspp.UPF
Nb 92.90638 041-Nb-ca-sp-vgrp.uspp.UPF
O 15.9994 008-O-ca--vgrp. uspp.UPF
ATOMIC_POSITIONS (crystal)
K 0.5 0.5 0.518
Nb 0.000000000 0.000000000 0.0
O 0.500000000 0.00000000 0.04
O 0.00000000 0.500000000 0.04
O 0.00000000 0.00000000 0.544
K_POINTS {automatic}
6 6 9 1 1 1
Thanks in advance.
Zhanfang Li
Joint training student, Penn State University, USA
---------------------------------
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