Quantum Chemistry – Is Hartree-Fock a Ground State Theory?

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To calculate the orbitals in Hartree-Fock (HF) theory we imply the variations principle, so we try to find the wavefunction which minimises the energy $\left<\psi|H|\psi\right>$.
This variational principle does only apply to the ground state of a system, so it is said that also HF theory is a ground state theory and can not describe excites states.

But is this really true? If we do a HF calculation we usually get a set of virtual unoccupied orbitals. We could take an electron from an occupied orbital and put it in a virtual orbital. Wouldn’t this correspond to an excited state (at least approximately)? Something like this is often done in introductory courses when discussing the H2-molecule: One says that if both electrons are in the lowest orbital (sigma bonding orbital) it is the ground state. But one could put one electron in the higher orbital (sigma antibodies orbital) and this is often referred to as excited state.

My question in short:

Can HF describe excited states i.e. do the virtual unoccupied orbitals correspond to excited states? If yes, why is this so (the variational principle does not apply to excited states)? If no, what is the interpretation of these virtual orbitals?

Comment: Is there some connection to Koopmans Theorem which says that the energy of a Hartree Fock molecular orbital corresponds (approximately) to the ionisation energy /electron affinity.

Best Answer

Hartree-Fock is a groundstate method. The equations are constructed to optimize the groundstate energy and you will not obtain reasonable excited state energies by replacing an occupied orbital with an unoccupied one in your Slater determinant. This yields a very poor approximation for an excited state. Virtual orbitals do not correspond to excited states in such a simple manner.

But the results from a Hartree-Fock calculation can be used in Post-Hartree-Fock Methods to obtain excited states.

The virtual orbital eigenvalues can be interpreted as electron attachment energy/electron affinity. By looking at the energy of $E(N) - E(N+1)=-\epsilon_v$ where N stands for the number of electrons. $E(N)$ stands for the energy of your groundstate calculated within HF with $N$ occupied orbitals. $E(N+1)$ stands for the energy of a determinant that you obtain by occupying one additional virtual orbital with index $v$ while keeping the former $N$ occupied orbitals as they were. The difference of the energies of these two determinants corresponds to the electron affinity of the unoccupied orbital and is given by its negative orbital energy eigenvalue $\epsilon_v$. This is basically the same as Koopmans theorem but now for electron affinity instead of ionization potential.

But the energies calculated using Koopmans theorem are usually not good approximations. The determinant energies with $N\pm 1$ are bad approximations, when we keep using orbitals that were optimized for a $N$ electron system. If you did a HF calculation starting witn $N+1$ electrons, you would get different orbitals since your effective potential that is part of the Fock Operator changes with the number of electrons. The orbitals of a $N$ and a $N\pm1$ calculation are often similar but not the same. This difference is also often called orbital relaxation when we take the $N$ to $N-1$ case. Since the electrons feel the repulsion of one electron less and thus can "relax" compared to the system with one electron more. And to get these "relaxed" orbitals you would need to do new HF calculation with just $N-1$ electrons, instead of using the orbitals from the $N$ calculation.

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