# [molpro-user] one question about calculating polarizability

CC XX xcjscu at gmail.com
Tue Sep 4 14:03:00 BST 2012

```Dear Ms/Mr,
Now, I am obsessed by one question, can you help me solve the
problem? Thank you very much!
It is about calculating the polarizability of one molecule by
differential method. Based on a case (H2O molecule) in Molpro manual, the
input file is displayed as following:
"

***,H2O finite field calculations

r=1.85,theta=104                   !set geometry parameters
geometry={O;                       !z-matrix input
H1,O,r;
H2,O,r,H1,theta}
basis=avtz                         !define default basis
field=[0,0.005,-0.005]             !define finite field strengths
\$method=[hf,mp4,ccsd(t),casscf,mrci]

k=0
do i=1,#field                      !loop over fields
dip,,,field(i)                   !add finite field to H
do m=1,#method                   !loop over methods
k=k+1
\$method(m)                     !calculate energy
e(k)=energy                    !save energy  enddo
enddo

k=0
n=#method
do m=1,#method
k=k+1
energ(m)=e(k)
dipmz(m)=(e(k+n)-e(k+2*n))/(field(2)-field(3))   !dipole moment as
first energy derivative
dpolz(m)=(e(k+n)+e(k+2*n)-2*e(k))/((field(2)-field(1))*(field(3)-field(1)))
!polarizability as second der.enddo

table,method,energ,dipmz,dpolz
title,results for H2O, r=\$R, theta=\$theta, basis=\$basis
---

"

The result is as following:
"
METHOD        ENERG        DIPMZ        DPOLZ
HF        -76.05828804   0.79028512   8.66914177
MP4       -76.34319951   0.72435361   9.84241227
CCSD(T)   -76.34178065   0.72957181   9.67032932
CASSCF    -76.11356760   0.72500885   8.89662369
MRCI      -76.32839170   0.70823783   9.20622885
"
As showing above, the value of polarizability in z direction is positive.
But when I calculate the z direction component of the polarizability by
adding a finite electronic field in z direction, and then compute the
derivative of dipole with respect to the field, the value of this
derivative is found be negative but have the same magnitude by and large.
The input file of normal calculation is:
"
***,H2O finite field calculations
r=1.85,theta=104               !-->set geometry parameters
geometry={O;                   !-->z-matrix input
H1,O,r;
H2,O,r,H1,theta}
basis=avtz                     !-->define default basis
{hf}
{mrci;dm;
natorb,2140.2}
"
The result is:
"
!RHF STATE 1.1 Dipole moment           0.00000000     0.00000000
0.79032209
!CI(SD) STATE 1.1 Dipole moment        0.00000000     0.00000000
0.74610325
"
"
***,H2O finite field calculations
r=1.85,theta=104                   !-->set geometry parameters
geometry={O;                       !-->z-matrix input
H1,O,r;
H2,O,r,H1,theta}
basis=avtz                         !-->define default basis
field=0.001                        !-->define finite field strengths
dip,,,field
{hf}
{mrci;dm;
natorb,2140.2
"
The result is:
"
!RHF STATE 1.1 Dipole moment           0.00000000 0.00000000 0.78165042
!CI(SD) STATE 1.1 Dipole moment        0.00000000 0.00000000 0.73678704

"
So, for the HF method, polzz=(0.78165042-0.79032209)/0.001=-8.67167
and, for the CD(SD),    polzz=(0.73678704-0.74610325)/0.001=-9.31621
Actually, this derivative is also the value of polarizability [I also found
this definition in some references pol_zz=d(dip_z)/d(E_z) ]. So why the
result of this method is negative? Or there're some details in Molpro
calculations which I neglected? Thanks a lot!

Best regards,
Changjian
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