When doing a population analysis using density functional theory (specifically as implemented in Gaussian), what are the orbital energies (i.e. alpha and beta eigenvalues) with respect to? If I have two different but closely related compounds, can I directly compare their HOMO (or LUMO) energies? It should be noted that I am not referring to comparing their HOMO-LUMO gaps, but rather the energy of a single molecular orbital.
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3 Answers
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As mentioned by others, the zero of the system is a free electron.
As for comparing energies, people do this, but they shouldn't. People love to compare HOMO-LUMO gaps and LUMO energies but they are almost entirely arbitrary. In DFT (and Hartree-Fock), the energy of the system is entirely a function of your occupied orbitals. You could delete the bottom 10 unoccupied orbitals from your basis set and you would get the exact same energy but your LUMO energies (and hence HOMO-LUMO gaps) will be completely different. So making your basis smaller can increase your LUMO energies. What about the flip side? If you use a bigger basis, your LUMO will get lower and lower in energy. In the basis set limit, you will have a bunch of orbitals at the zero (you will basically be replicating the "orbital" that corresponds to a free, unbound electron). There's a reason that the electron affinities people try to calculate from LUMO energies are terrible. You are basically banking on the chance that your basis set is just the right size for your molecule in just the right geometry.
To summarize, you can basically tune your basis set to produce whatever LUMO energy you want. I wouldn't compare LUMO energies between the same molecule in different basis sets or even geometries, let alone different molecules.
All hope is not lost though, if you really want some sort of LUMO-like thing. You can try working in some sort of localized basis: NBO produces "anti-bonding" orbital energies that are no where near as sensitive to basis set as canonical orbitals. However, that might not suit your purpose.
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$\begingroup$ this is what I also thought, but (unless I'm missing something) it seems to somewhat disagree with the answer provided by @user213305 . Could we try to reconcile the answers? $\endgroup$– jezzoMar 14, 2019 at 13:34
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2$\begingroup$ Yes, they are wrong. Some quotes from the wikipedia article cited in their answer: "The LUMO energy shows little correlation with the electron affinity with typical approximations.[7]" Zhang, Gang; Musgrave, Charles B. (2007). The Journal of Physical Chemistry A. 111 (8): 1554–1561. "Calculations of electron affinities using this statement of Koopmans' theorem have been criticized[11]" Jensen, Frank (1990). Introduction to Computational Chemistry. Wiley. pp. 64–65. It is an extremely common misconception, but LUMOs are meaningless for DFT/HF. $\endgroup$– levinedsMar 14, 2019 at 20:22
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1$\begingroup$ Unoccupied orbitals are important for correlated methods, MP2 (and therefore double hybrid DFT) because they are actively used in the determination of the energy. For normal DFT/HF, they are the part of the Hilbert space granted to the molecule by our limited basis set that was determined to be useless for improving the energy. All methods are basis set dependent to some degree, so it is not surprising that changing the basis set can change your answer. However, all decent methods converge in to some answer in the complete basis set limit; the LUMO of DFT/HF does not. $\endgroup$– levinedsMar 14, 2019 at 20:27
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$\begingroup$ Unfortunately, for my systems, the NBO code often has a lot of trouble finding a sufficiently "optimal" Lewis-like structure such that NBO-computed properties are often unreasonable. Nonetheless, this answer is the one that makes the most sense to me. $\endgroup$– ArgonMar 16, 2019 at 0:08
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In molecular electronic structure, electronic/orbital energies are measured with respect to an energy zero of an isolated electron in free space.
Hence a lower LUMO energy does correspond to an orbital that is more apt to accept an electron/a more negative electron affinity compared to another molecule.
Correspondingly a higher HOMO energy corresponds to an orbital that is more liable to give up an electron/a less positive ionisation energy compared to another molecule.
This is the basis of Koopmans' theorem in Hartree–Fock theory and it's equivalent in DFT.
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$\begingroup$ From your link: "Ionization energies calculated from DFT orbital energies are usually poorer than those of Koopmans' theorem, with errors much larger than two electron volts possible depending on the exchange-correlation approximation employed." $\endgroup$– jezzoMar 8, 2019 at 23:39
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2$\begingroup$ Yes, that is true. Koopman's theorem for Hartree-Fock is a rough approximation. As the orbitals of DFT are only used to form a density (unless you are using some kind of hybrid functional) you would expect orbitals/HOMO/LUMO from non-hybrid DFT to be less accurate than those from hybrid DFT which are less accurate than those from HF. $\endgroup$ Mar 8, 2019 at 23:43
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EDIT: After some thought, I've summarized the following: we can use Koopmans' theorem (approximately) within DFT to compare single point MO energies, as a reference value is indeed used (that of a free, or fully ionized electron). Therefore, we can (somewhat justifiably) compare the LUMOs of two different systems, as they share the same reference energy.
However, single point energy calculations of other kinds cannot be directly compared in a valid or consistent manner between systems of differing number of atoms and/or electrons. Instead, reference values (from literature or another experiment) must be used to compare between systems. This makes MO energies a special case.
That being said, DFT works with densities that happen to reflect MOs quite well, and so Koopmans' theorem is not exact for DFT and should be used carefully.
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$\begingroup$ It should be noted that I'm not talking about the electronic energy of the system. $\endgroup$– ArgonMar 8, 2019 at 23:25