There are several programs that can calculate the "energies" of the natural atomic orbitals(NAO) that come from NBO analysis. However, I have several evidence that, unlike the energies of the "normal" MO's, the energies of NAO's are not trivially determined:

  1. Practically, they're not always equal to the ground-state energies of the same orbital in the neutral atom, even when the oxidation states are the same; for example, the five 3d orbital energies are not the same in a lone palladium atom([Kr]3d¹⁰) and in tetrakis(phosphine)palladium(0).
  2. Theoretically, NAO's are not true eigenfunctions of the relevant Hamiltonial operator, so their "energies", strictly speaking, should not be able to be well-defined at all.

My question now follows- exactly what methods are used to calculate the NAO "energies", assuming that a wavefunction (from HF, DFT, or any other method) is already given?

  • 1
    $\begingroup$ I would assume that the initial NBO paper discusses this. The newer versions of the program are probably a bit different, but the theory should remain largely the same. Have you read it? $\endgroup$ Nov 18, 2021 at 0:56

1 Answer 1


Comparing single energies of basis functions (or subsets) from different basis sets is not meaningful. There is no reason for the NAO energies of single orbitals to be identical with the energies of functions from a different basis set. The only energy that has to be identical is the energy expectation value the complete electronic wavefunction. This energy has to be the same no matter what basis you chose for the calculation.

Note that the NAO's are specific to the particular wavefunction of your system, since they are based on the concrete one-electron reduced density matrix of your system, which itself is based on the full multiparticle-wavefunction of the system.

This means that the NAO's change as soon as your electronic wavefunction changes. The same oxidation state does not imply the same electronic wavefunction, so there is no reason that the NAO's should be exactly the same in the palladium example of yours and neither will their energies be the same.


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