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How would the non-interacting electron orbitals from a perfect DFT solution for a given potential shape differ from the 'true' electron wavefunctions? Or can you only really talk about the total wavefunction? Would they be less localised as they do not include interaction effects?

Since Hohenberg-Kohn showed that a functional of the wavefunction could be transformed into a functional of the density, any 'measurements' on the total density would have to yield the same results as measurements on the total wavefunction, I think.

Intuitively, if we imagine a 3-electron system, the ground state consists of spin-up and -down electrons in the lowest state, and a singlet electron in the next state. Since there is correlation and exchange, the total wavefunction would be a superposition of the various possible combinations of these.

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In reality there's just one many-body-wavefunction for all the electrons at once, not "electron wavefunctionS" in the plural. Are you trying to approximate the true many-body-wavefunction as a single Slater determinant and asking about the component wavefunctions of the Slater determinant? Or something else? –  Steve B Jun 5 '11 at 20:29
    
Well, perhaps - DFT gives you a single slater determinant and a set of component orbitals. Consider the lowest orbital from DFT - how would that compare to the delta between an occupied total wavefunction and where 1 electron has been extracted with maximum energy? –  Phil H Jun 6 '11 at 6:10
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1 Answer

Assuming that:

  1. by "a perfect DFT solution" you mean the set of Kohn-Sham orbitals for the exact (presently unknown) denistiy functional that existance of which Hohenberg and Kohn have proven;

  2. you are asking whether the individual energies of these "perfect" Kohn-Sham orbitals can be used as estimates of exact ionization and/or excitation energies,

the answer is yes, but only approximately with the sign of error depending on the system.

The DFT method with the perfect functional will give exactly the total energy (that's the functional) and the single-electron properties of the ground state only. As far as I know, there are no exact DFT theorems for excited states. As for the estimation of the ionisation energy, the answer is not exact because the perfect DFT solution for (n-1) electrons will give different orbitals than the the perfect DFT solution of n electrons. Thus the difference between the two perfect total energies will not be equal to any particular orbital energy.

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For ionization energy, there’s Koopmans’ theorem –  F'x Mar 16 '12 at 10:32
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