# [molpro-user] Computation of the dipole moment with CCSD(T)

Omololu prayerz at yahoo.com
Fri Sep 8 22:15:45 BST 2006

If H0 is the hamiltonian of the system and we are
interested in calculating the expectation value
<0|dip|0> of the operator "dip" where |0> is the
wavefunction for H0, then we can do this by
finite-differencing, irrespective of Hellman-Feynman's
theorem (HFT).

Basically, form the Hamiltonian H= H0 + L*dip. Then,
assume the eigen-energy (E) of the new hamiltonian H
can be expanded in powers of L (like is done in
perturbation theory). Then the derivative dE/dL at L=0
would give <0|dip|0>. This has nothing to do with HFT;
it only relies on the fact that E (and the eigenvector
of H) can be expanded in powers of L.

Where does HFT come in? If |L> is the eigenvector of H
and you compute <L|dip|L>, HFT tells you that this
expectation value is equal to dE/dL. Thus, HFT
guarantees that your calculated value of <L|dip|L> at
L=0 (this is the orbitally-relaxed dipole) would be
equal to <0|dip|0>.

If your wavefunction does not satisfy the conditions
you listed, then you are correct that
orbitally-relaxed dipole may not be equal to <0|dip|0>
as you found for CCSD. So, if you need <0|dip|0>, then
finite-differencing would work if you use a small
enough field. Ditto for CCSD(T) or any other methods.

o.

--- Lorenzo Lodi <l.lodi at ucl.ac.uk> wrote:

> I am computing the dipole moment of water with
> various methods and I'd
> now like to use the CCSD(T) method.
>
> At the CCSD level I can compute the dipole with
> ---
> RHF
> CCSD ; CORE, 0,0,0,0
> EXPEC, DM
> ---
> and I understand that the value given as "orbitally
> relaxed CCSD dipole
> moment" is the same value as the expectation value
> of \mu on the CCSD
> wavefunction. (Incidentally, the GEXPEC directive
> does not work for me,
> is this behaviour normal?)
>
> Now, for the CCSD(T) dipole.
> Looking at some old posts it was suggested to
> calculate it using the DIP
> keyword with a small field and then taking the
> derivative of the energy
> at zero field. Now, my objection is as follows: the
> Hellmann-Feynmann
> theorem, on which this finite-field approach is
> based, holds for
> 1) the exact wavefunction
> and
> 2) a wavefunction which is variationally optimised
> in all its parameters
> The coupled cluster method is, of course,
> non-variational so I don't
> expect the finite-field approach to work well in
> this case.
> In fact, I verified this comparing the CCSD dipole
> obtained by
> finite-field and as given by molpro (equilibrium
> geometry, cc-pV6Z
> basis) and I got a difference of ~0.003 a.u., which
> is too large for the
> level of accuracy I am looking for.
>
> My conclusion would be that there is no accurate way
> to compute the
> dipole with the CCSD(T) method (in fact, as there is
> no CCSD(T)
> wavefunction, this may not be anything new...).
>
> Could anyone confirm these comments and/or give any
> to proceed in this direction?
>
> Thank you.
>
> Lorenzo Lodi
>
>

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