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So regarding Koopman's theorem I found the the explanation that you should not use it to find the ionization potential of localized orbitals because, much unlike canonical orbitals they will not be stationary towards small orbital changes. So any minor perturbation will cause a big difference in energy.

Usually I would simply say that the canonical orbitals are the eigenvalues of the diagonalized fock matrix while localized orbitals are not direct solutions of the fock matrix.

But isn't localization simply another transformation of the matrix that keeps the energy but sort of contracts the density more around bonds or atoms to resemble the bonds or lone pairs we know from organic chemistry? How does this effect the way they behave towards small changes to the orbitals?

Justanotherchemist
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    Does this answer your question? – hBy2Py Jun 29 '17 at 14:42
  • Thank you for the link but I pretty much based my explanation given above from said discussion. The Koopman's theorem here is just an example where it can be used but I was rather referring to the question what this 'stationary towards small orbitals changes' of the energy means for the canonical and localized orbitals. – Justanotherchemist Jun 29 '17 at 15:08
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    Hm. I don't think I understand your question then. It's not that you shouldn't use localized orbitals to calculate the IE via Koopmans', it's that you *can't: localized orbitals have undefined energies*. Separately, the electron density field calculated from localized orbitals should be identical to that calculated from canonical orbitals. Can you clarify further what you mean by "the way they behave" and "small changes to the orbitals", e.g., in your last sentence? – hBy2Py Jun 29 '17 at 15:21
  • I'll just quote the sentence directly from the original source: 'this expression should not be used for non-canonical orbitals since these values are not stationary with respect to small orbital changes'. This was the explanation of why you cannot use loc. orb. in Koopman's theorem. I just don't understand what this stationary here means and how it differs from canonical to localized. (The text is from my university script) – Justanotherchemist Jun 29 '17 at 15:35
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    Stationary means the orbitals won't react to removing one electron. So this basically refers to the assumption in Koopmans' theorem. For localized orbitals you cannot make such assumptions, they are not even uniquely defined. – Feodoran Jun 29 '17 at 16:29
  • Ah. Okay thank you. I had the assumption that stationary refers to the energy. Because I remember there were comparisons to the potential energy surface at some point. So I assumed that perhaps the position on the PES or the PES itself might change with small perturbations for only one of the two. So this refers to the orbital relaxation term in Koopman's theorem ? – Justanotherchemist Jun 29 '17 at 17:28
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    Well I guess it refers to the orbital energy not being changed, thus being stationary. PES is a whole different story, it is about the nuclear problem, where nuclear coordinates vary. The orbitals belong to the electronic problem, where nuclear coordinates are fixed. Both problems are treated separately due to the Born-Oppenheimer-Approximation. – Feodoran Jun 29 '17 at 17:36
  • If I understand you correctly, @Feodoran, effectively the problem arises because any given localized orbital can be thought of as being composed of "pieces" of a number of the canonical orbitals. Thus, if you "remove" an electron from any of them, you're guaranteed to be "removing" a portion of the wavefunction that is of lower energy than the HOMO. Thus, such a removal would result in the wavefunction 'relaxing' appreciably afterward. (Is this close?) – hBy2Py Jun 30 '17 at 15:21
  • @hBy2Py I guess yes. But maybe one could even find (or construct) a set of non-canonical orbitals for which Koopmans' Theorem will work even better, then for the canonical ones. But this would only be by chance, not for any physical reason. – Feodoran Jun 30 '17 at 15:47

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