Can someone please tell me how to calculate the ground state oxidation potential (GSOP)?

I understand that according to Koopmans' theorem, the negative of HOMO can be considered as the GSOP. Is there any other way to calculate GSOP?

Also, are GSOP and ionization potential same?

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    $\begingroup$ Bear in mind that Koopmans' theorem, as you stated, applies specifically to Hartree-Fock theory. $\endgroup$ Commented Jun 17, 2017 at 15:07

1 Answer 1


Short answer - compute the relative energies of the neutral and cation species

Typically, when people discuss the "oxidation potential," they're considering electrochemistry, so there's solvent around. Keep this in mind if you're trying to compare a calculation done in gas phase - the best comparison would be with an appropriate solvent calculation, although calibrating between a calculation and an experimental electrochemical potential requires work.

On the other hand, ionization potential is typically used for a gas phase measurement - for example, with XPS or UPS spectroscopy, where an electron is removed from a species.

Finally, there's a difference between the vertical ionization potential when an electron is ripped out of a species, but the cation hasn't had time to relax (change geometry) and the adiabatic ionization potential, which is the difference of energy between the neutral and the relaxed cation.

vertical and adiabatic ionization potentials

That's a long way of saying that if you want to compute an accurate ionization potential, you need to:

1) Optimize the geometry of the neutral species using an accurate method.

2) Take the optimized geometry, and run a geometry optimization with one fewer electron. (You will get both the vertical ionization energy as the first SCF energy, and the adiabatic ionization energy of the optimized cation.)

Subtract the total energies and convert to your unit of choice.

  • $\begingroup$ Note that I don't say anything about orbital eigenvalues because there's no guarantee that for DFT/DFA methods, that these will correlate with ionization potentials. Koopmans' theorem is not guaranteed to apply. $\endgroup$ Commented Feb 1, 2018 at 20:54

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