[molpro-user] Fwd: quadrupole moment

Evangelos Miliordos emiliord at chem.uoa.gr
Fri Oct 8 17:53:05 BST 2010

Dear molpro users,

I am trying to calculate the quadrupole moment of H2. I know that the
first way is calculating the<wfc|QMZZ|wfc>   integral, which gives 0.48
a.u. at MRCI, exactly the experimental one. But, I would also like to
calculate the QMZZ applying a gradient field fgzz
(or Vzz) at the origin. According to A. J. Stone in "The theory of
intermolecular forces" the perturbation in the Hamiltonian is
H'=1/3*Vzz*Θzz, where Θzz operator is
Θzz=3/2*sum{i}{qi*(zi^2-1/3*ri^2)}, and the perturbed energy is (if
field is zero at origin) E' = E0 + 1/3*Θzz*Vzz - 1/6*Czz,zz*Vzz^2 + ...
So, I would expect that
Θzz=3*dE/dVzz. However, using the finite field approach, using a
possitive and a negative Vzz, I get 29.33 a.u. What am I missing? Is the
perturbation in molpro2006.1 the aforementioned one?

Thanks a lot,
Evangelos Miliordos

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