Polarizability and hyperpolarizability as Hartree-Fock energy derivatives

Photo Credit: Vincent van Gough - The Starry Night - Museum of Modern Art, New York

Polarizability and hyperpolarizability as Hartree-Fock energy derivatives

In the previous post I was interested in how the force constants change with an applied field. In order to deal with that question, the dipole moment and polarizability were defined as electric field derivatives.

In this post I expand on where these definitions come from, and how such polarizabilities may be calculated – at least, within Hartree-Fock theory.

Helmann-Feynman theorem

The dipole moment and polarizability definitions ultimately arise from the Helmann-Feynman theorem.

The Helmann-Feynman theorem states that if \( \lambda \) is a parameter on which the Hamiltonian depends then

The parameter \( \lambda \) may be

  • The coordinate of a particular nucleus (in which case the left-hand-side represents the force on that nucleus along that coordinate), or

  • It may represent a component of an applied electric field (in which case the left-hand side represents a component of the dipole moment).

Proof

The proof is trivial:

The last term is zero because of the normalization condition \( \braket{\Psi}{\Psi} = 1 \).

Dipole and polarizabilities as energy derivatives

In order to understand how the dipole moment arises from the Helmann-Feynman theorem we need to know what the Hamiltonian for a system looks like in the presence of an electric field.