[molpro-user] one question about calculating polarizability
Tatiana Korona
tania at tiger.chem.uw.edu.pl
Wed Sep 5 09:30:32 BST 2012
Hi,
A dipole polarizability is defined as minus one times polarization propagator,
i.e.
alpha_yy= - << Y;Y>>
This minus in included in the example you cited (see the second part of the
denominator: field(3)=-0.005):
dpolz(m)=(e(k+n)+e(k+2*n)-2*e(k))/((field(2)-field(1))*(field(3)-field(1)))
/ 0.005 * (-0.005)
Best wishes,
Tatiana
On Tue, 4 Sep 2012, CC XX wrote:
> 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
> "
> Add the field (0.001):
> "
> ***,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
>
Dr. Tatiana Korona http://tiger.chem.uw.edu.pl/staff/tania/index.html
Quantum Chemistry Laboratory
University of Warsaw
Pasteura 1, PL-02-093 Warsaw, POLAND
`The man who makes no mistakes does not usually make anything.'
Edward John Phelps (1822-1900)
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