# DFT Calculations, Atomic Ionization Potentials — Which Exchange-Correlation Functional to Use, to Preserve Koopmans' Theorem?

I have a program which can perform density-functional calculations for atoms, given a density functional.

Of course the simplest form of exchange potential to use is one relevant for a uniform electron gas (i.e. the original Kohn-Sham exchange, proportional to $n_e^{1/3}$). A good correlation functional is also available. However when I performed calculations for some atoms the energy of the highest occupied orbital differs greatly from the ionization potential (off by a few eVs!).

That the energy of the highest occupied orbital equals the ionisation potential of the neutral atom is known as Koopmans' Theorem (See Wikipedia). I get a feeling from the literature that DFT can be quite accurate. So what's wrong?

Interestingly when I use the Slater exchange functional, coupled with a correction suggested by Skillman (i.e. replacing the potential by $1/r$ when the overall potential drops below that value), the results improve significantly. (See the book "Atomic Structure Calculations" by Herman Skillman). Well, maybe I should follow this recipe, but it seems the procedure is quite ad-hoc.

My question is, are there any functionals which will give reasonable values for the atomic ionisation potentials? I do not want to implement methods like the optimised effective potential method since the latter is not readily generalised to a finite-temperature scenario. Thanks.

You have assumed that Koopman's theorem is correct.

I am afraid that Koopman's theorem is a good first approximation but does not take account of the following issue.

When an atom or molecule is ionized then by Koopman's theorem the energy required is equal to the energy of the orbital the electron is in. What this neglects is that by removing an electron the positively charged ion can 'relax'. That is to say that all the other electrons can slightly lower their energies because of the removal of the ionized electron.

To calculate ionization energies accurately you need to perform a calculation for the neutral atom or molecule and another calculation for the ionized atom or molecule.

I have a question in my mind as to why you are using DFT for an atom. In my experience DFT is used for larger systems with maybe 20 or more atoms where ab-initio type calculations are not possible or difficult. The GAMESS US code might be something to consider as an alternative.

Finally, I am impressed that you calculation is only off by a couple of eV because this type of calcualtion can be quite challenging. The total energy of all the electrons is generally one, two or more orders of magnitude greater than the ionization energy, which makes the calculation difficult as you have to subtract two large numbers from each other to get a small difference (Two large numbers are total energy of electrons in neutral and total energy of electrons in ion).

Frequently researchers who make these calculations will calculate the energy of the ion and the neutral and then adjust the theoretical ionization potential so that it agrees with the experimental value - there are good reasons for doing this. For example, first, the energy differences between the ground state of the ion and excited states of the ion will be more accurate from the caculation than their absolute energies, and secondly, if the calculation of the ion is used to calculate something else, like e.g. an electron impact ionization cross section, the result will be more accurate if the theoretical ion energies are corrected with the experimental ionization energy. An example for a lanthanide atom is here - but I am not sure how easy it is to access full text. Hope it is a useful starting point though.

• You are right. "Finally, I am impressed that you calculation is only off by a couple of eV because this type of calcualtion can be quite challenging. " Haha, indeed I haven't made it clear enough in my original post; I was comparing the energy of the highest occupied orbital (which can be obtained easily) with the experimental ionisation potential. – Jamie Dec 19 '14 at 15:42
• If I'm to calculate the ionization potential theoretically, what should I do? Should I run a calculation on the neutral atom, and calculate the corresponding Hartree-Fock energy (, which differs from the sum of energies of all occupied orbitals), and then run another calculation on the +1 ion, and calculate the corresponding HF energy, and then obtain the difference between the two? Surely this calculation requires high(er) precision for heavy atoms, for which the typical energy of the inner orbitals is ~1000 Hartrees (which is to be compared with the ionisation potential, <~ 1 Hartree). – Jamie Dec 19 '14 at 15:42
• @user54306 Yes I agree with your analysis and it will be more challenging for heavier atoms. Frequently researchers who make these calculations will calculate the energy of the ion and the neutral and then adjust the theoretical ionization potential so that it agrees with the experimental value - the I not a theoretician, but I expect that Hartree Fock is not going to give very accurate energies, though the method you suggest is correct. I think often people might use SSCF or CASSCF type calculations, but I have have the accronyms wrong - I will put small edit in to answer. – tom Dec 20 '14 at 0:15