So this question is linked to the other question I asked about the electronic patition function and CASSCF. When calculating thermochemical data (e.g. Gibbs Free energy) the partition functions are calculated. This includes the vibrational and rotational partition functions along with the electronic and translational terms.

Now when we use CASSCF, we are using multiple determinants, which can be considered as low lying excited states of the molecule. My first thought was that considering excited states would mean the electronic partition function term is now included in the SCF energy (which is what my earlier question is about).

But what happens to the other degrees of freedom? For example, the vibrational modes are obvioulsy calculated based on the CASSCF solution, so it contains the contribution of the vibrational modes of the excited states. Does this type of calculation actually make sense?

In my lectures, what my professor told us is that we take the ground state, and then we calculate all the partition functions independently based on the ground state. So what happens when the contribution of excited determinants are included in the vibrational and rotational terms?


2 Answers 2


So what happens when the contribution of excited determinants are included in the vibrational and rotational terms?

Nothing, except that the Hessian and/or gradient are more difficult to evaluate, for example see coupled-perturbed Hartree-Fock.

I think maybe your confusion is arising because the excited determinants are not exactly excited-states. A multi-electron wave function can be solved exactly within the basis of determinants:

$$ \tag{1} \Psi(r_1,r_2, ...,r_N)= \sum_i c_i \psi_i $$

where $\psi_i$ are the Slater determinants of the ground, and "excited" electronic configuration.

These excited electronic configurations should not be confused for excited-states however, because the real excited electronic state is also composed of the same type of sum.

  • $\begingroup$ +1 Yes, I thought until now that excited states were the same to excited determinants. But I am confused now, because determinants represent one configuration, so it has energy levels. If we "excite" the determinant, then surely the configuration can match the actual excited electronic state? $\endgroup$
    – S R Maiti
    Apr 12, 2021 at 21:53
  • 2
    $\begingroup$ The true ground-state is usually similar to the ground determinant, this is why we are able to use Hartree-Fock theory. Likewise, the excited-state can sometimes resemble more closely one of the excited-determinants but typically requires a bit more excited configurations like singles, doubles, etc. It really depends on your level of approximation, technically the full wave function is equation (1) but if you truncate to only singles it's called configuration interaction singles (CIS), doubles it's CISD, etc. $\endgroup$
    – Cody Aldaz
    Apr 12, 2021 at 22:19
  • $\begingroup$ Sorry, I am still confused. Does CIS give the ground state or the excited state, for instance for a molecule that can undergo an one electron excitation, like a pi->pi* system? I thought CIS is only used for excited states? $\endgroup$
    – S R Maiti
    Apr 13, 2021 at 9:13
  • $\begingroup$ see me in chat.stackexchange.com/rooms/119831/… $\endgroup$
    – Cody Aldaz
    Apr 13, 2021 at 9:19
  • $\begingroup$ Basically all CI / CC calculations sit on the Fock space and when one diagonalize the Hamiltonian represented in basis of slater determinate, all eigen states (regardless of ground state or excited states) are linear combinations of Slater determinate. There is a gap between fundamental quantum mechanics and concepts is chemistry (e.g. orbitals, chemical bond, etc). It can be misleading when combine these concepts to fundamental theory. $\endgroup$
    – Paulie Bao
    Apr 15, 2021 at 9:38

In some cases, we need to consider vibrational - rotational coupling. But this effect is normally weak since the rotational energy level (~1e-4 eV)is much smaller comparing with vibrational energy level (~1e-1 eV). Rotational energy level is normally serves as fine structures in a high resolution photo-electron spectra and this effect is normally so weak that can be ignored for the photo-electron spectra of a poly-atomic system.

  • $\begingroup$ -1 This does not answer my question. I was asking about excited determinants, rotational-vibrational coupling is not related to that. $\endgroup$
    – S R Maiti
    Apr 15, 2021 at 9:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .