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N$_2$ is a canonical example of this aspect of the failure of HF method, for instance, see table 4.9 in "Modern Quantum Chemistry" by Szabo and Ostlund. Here it is shown that in HF, the HOMO of N$_2$ is of $\pi$ symmetry, whereas correlated method and experiments show that its HOMO should have $\sigma$ symmetry, which becomes HOMO-1 in HF.

user1271772
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nougako
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  • Maybe this answer on a related question helps: https://chemistry.stackexchange.com/questions/154162/is-strength-of-pi-bonds-greater-than-sigma-bond-of-nitrogen-molecule/155418#155418 – Libavius Dec 22 '21 at 12:59
  • The linked question discusses N$_2$ only. – nougako Dec 23 '21 at 02:16
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    Yes, but the linked answer discusses the notion of "wrong" orbital ordering and the limits of the interpretability of orbitals in general. – Libavius Dec 23 '21 at 21:51
  • The idea of the question is to address the symmetry of the highest occupied orbital in N$_2$ in HF method. The fact that there is a "experiment" column in the cited table above indicates that experiments can actually identify the symmetry of the lowest ionization potential of N$_2$, which is associated with the HOMO of the neutral. The next ionization potential is associated with ionization that brings the cation to its first excited state, and in HF this is approximated by the removal of HOMO-1 of the neutral. – nougako Dec 24 '21 at 01:00
  • Therefore, despite the idea of orbital being unphysical in the exact sense, it still bear certain degree of information in the approximate single determinant sense. – nougako Dec 24 '21 at 01:01
  • Yes, that is Koopman's theorem. So by "experiments show that its HOMO should have \sig symmetry" you mean, that the ion has \sig symmetry? – Libavius Dec 27 '21 at 12:19
  • @libavius it is called Koopmans' Theorem as the s is part of the name. – Martin - マーチン Dec 27 '21 at 16:15
  • @Libavius I should have been more specific. The kind of ionization I was referring to is tunneling ionization, not single-photon ionization. The former requires no resonance between ionizing photon energy and neutral's energy levels but intensity must be high. In tunnel ionization, you are most likely ionizing the valence orbitals. Hence the $\sigma$ symmetry encoded in the experimental ion yield as a function of polarization angle of the laser comes from the symmetry of the HOMO of the neutral N$_2$. – nougako Dec 27 '21 at 22:37
  • The HOMO of N$_2$ obtained through HF has a $\pi$ symmetry, which is therefore inconsistent with experiments. However, the HOMO of this molecule obtained through DFT (hence the concept of occupied orbital is still well-defined) using good functional has $\sigma$ symmetry. In this respect, tunneling ionization calculation on N$_2$ is better calculated using DFT rather than HF. – nougako Dec 27 '21 at 22:43
  • I haven't done correlated wavefunction methods such as CI for a while now. But as I recall even for these multideterminant methods where the number of occupied orbitals can be more than the number of electrons, you still typically have $N$ orbitals (where $N$ is the no. of electrons) whose occupation number is significantly higher than the rest of occupied orbitals, indicating that in these methods the HF characteristic is usually somewhat prominent. – nougako Dec 27 '21 at 22:45

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