Linear response derivatives for CI

When using a configuration interaction expansion of the wave function, the energy is given by: \[\begin{equation} \require{physics} E(\vb{c}, \boldsymbol{\eta}) = \ev{\hat{\mathcal{H}}(\boldsymbol{\eta})}{\vb{c}} \thinspace , \end{equation}\] where \(\ket{\vb{c}}\) is a normalized linear expansion inside a Fock subspace. The response force constant is then calculated as \[\begin{equation} \qty( \vb{k}_{\vb{c}} )_{ij} = \eval{ \pdv{ E(\boldsymbol{\eta}, \vb{c}) }{c_i}{c_j} }_{\boldsymbol{\eta}_0, \vb{c}^\star} = 2 \matrixel{i}{ \hat{\mathcal{H}}_{\text{elec}}(\boldsymbol{\eta}_0) }{j} \end{equation}\] and the response force as \[\begin{equation} \qty( \vb{F}_{\vb{c}} )_{im} = \eval{ \pdv[2]{ E(\boldsymbol{\eta}, \vb{c}) }{c_i}{\eta_m} }_{\boldsymbol{\eta}_0, \vb{c}^\star} = 2 \matrixel{i}{ \eval{ \pdv{ \hat{\mathcal{H}}_{\text{elec}}(\boldsymbol{\eta}) }{\eta_m} }_{\boldsymbol{\eta}_0} }{ \Psi(\vb{c}^\star) } \thinspace . \end{equation}\]

For an electric field, this becomes: \[\begin{equation} \qty( \vb{F}_{\vb{c}} )_{im} = \eval{ \pdv[2]{ E(\vb{F}, \vb{c}) }{c_i}{F_m} }_{\vb{F}_0, \vb{c}^\star} = -2 \sum_j^{\text{dim}} \matrixel{i}{ \hat{\mu}_m }{j} c^\star_j \thinspace , \end{equation}\] where we have used the theory on perturbation-dependent Fock spaces and the Schrödinger Hamiltonian for a uniform electric field.

We see that the nuclei do not contribute to these response quantities. I would like to thank Trygve Helgaker to point out a previous mistake that I had made.