3
$\begingroup$

When we employ Hartree-Fock or coupled cluster to calculate an atom's energy, can we directly obtain the valence electron energy? Is it as simple as looking at the orbital energy?

I know that summing up orbital energies won't give us the total energy because of double-counting electron-electron interactions. However, this should not be a problem when dealing with a single orbital energy.

$\endgroup$
2
  • 1
    $\begingroup$ Just to clarify, do you mean a hydrogenic atom (i.e. a single electron around a nucleus). Or do you mean the energy of an arbitrary orbital of a multielectron atom/molecule? $\endgroup$
    – Tyberius
    Jun 27, 2023 at 23:07
  • $\begingroup$ @Tyberius I am interested in an answer for a hydrogenic atom as well as for an arbitrary multielectron atom. $\endgroup$
    – Peter234
    Jun 28, 2023 at 5:37

2 Answers 2

6
$\begingroup$

Normally $\textit{ab initio}$ methods should give approximately the orbital energies that the electrons will occupy after applying the exclusion principle. The accuracy of the total energy is related to the accuracy of these orbital energies and how the method has treated $V_\text{ee}$. Most of the used methods solve an electron in a mean field potential with a more or less advanced description of the orbitals.

The highest occupied atomic orbital energy is important as it can be related to the experimental ionization energy to compare the accuracy of these eigenvalues.

A mean field approximation is quite similar to the solution of a hydrogenic atom with an effective potential. The eigenvalues of a hydrogenic atom are analytical: $-Z^2/2n^2$ and very inaccurate for a multi-electronic system. Because of $+V_\text{ee}$ the eigenvalues should be upshifted. A crude approximation will be to determine an effective potential $Z_\text{eff}$ acting on one electron using the previous expression. This parameter gives a more efficient description but is not accurate enough. The effective potential should have a radial dependency, but all the methods with a local potential only dependent on $r$ especially a local exchange potential do not lead to accurate eigenvalues. The effective potential should be non-local this means should contain two or more spatial variables: $V_\text{eff}(r,r')$ this is the main reason why Hartree-Fock and related methods have the most accurate eigenvalues (orbital energies). You should rely on these methods for what you are looking for: valence electron energies that are merely the eigenvalues.

$\endgroup$
4
$\begingroup$

In Hartree-Fock theory, orbital energies are indeed good estimates for the energies of single electrons. This is because of Koopmans' theorem: if you assume that the other electrons do not relax when you remove an electron, you will find that the difference in the total energies is given exactly by the orbital energy.

However, here you already have two approximations: Hartree-Fock, and that the other orbitals do not relax. Hartree-Fock neglects electron correlation, and the relaxation effects can be significant, especially in the case of core orbitals.

The better alternative is to study Dyson orbitals, which can also be computed from accurate many-body methods.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks for the hint regarding Dyson orbitals! It's the first time I hear about them and I will have to do some research. $\endgroup$
    – Peter234
    Jul 3, 2023 at 5:40

You must log in to answer this question.

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