The Hartree-Fock energy in terms of the occupied MOs (indexed i,j) of the neutral system is given by

If we remove an electron out of orbital k (for example the HOMO but this holds for any occupied orbital), the new total energy becomes

The detachment energy (or ionization potential) is simply defined as the difference

And considering the permutation symmetry of two-electron integrals this reduces to

which is nothing but the negative orbital energy of orbital k with respect to the original n-electron system. This is just Koopmans' theorem.

What happens if we take away an electron from the occupied orbital k and put it into the virtual orbital a? Then the energy of the resulting Slater determinant is given by

Subtracting the ground state Hartree-Fock energy from this (and considering permutation symmetry)

Part of this is of course just ε

_{k}. But the other half is not exactly ε

_{a}, which also includes the interaction with orbital k

However, we can still write the first order expression for the excitation energy in the following way

First, there is the orbital energy gap, as we expected. But second also Coulomb and exchange integrals between the occupied and virtual orbitals have to be considered, yielding a term, which can be identified with the exciton binding energy from last post. The first part can be identified with an attractive Coulomb interaction between the "electron" and the "hole", the second one with an exchange interaction, which is present for singlet excited states (or more generally if the ground and excited states are of the same multiplicity).

Since the Coulomb integral is larger than the exchange integral (which I believe it is always) the first order correction term is negative and the excitation energy is lower than the gap between the corresponding orbitals. The first order energy (i.e. CIS) is usually still too high because it neglects orbital relaxation in the excited state.

For DFT the story is different because of self-interaction and these rigorous equalities do not hold. On the one hand this is part of the reason why TDDFT is so unexpectedly good in many cases. On the other hand the failure to rigorously include the electron-hole interaction is the reason why TDDFT fails so badly for charge transfer states and does not retain the correct 1/r asymptotics [see e.g. this Ref.].

Continue at: HOMO-LUMO gaps and spin eigenfunctions

## 5 comments:

Hello Sir,

I am interested in computing HOMO and LUMO energy levels from methods like DFT and HF.

As you are an expert in this field, so I need a suggestion. Which of these two method will be best suited for this computation?

I want to start everything from the beginning.

Hi, the problem is that the HOMO and LUMO are not well-defined physical quantities. They can be applied in some qualitative theories, but there is no rigiorous way to compute them, since they are method dependent

anyway, you can download a program like GAMESS

http://www.msg.ameslab.gov/gamess/

and play around a little bit. Use a small basis set like 6-31G*. Then do HF and DFT/B3LYP computations. And compare the results to get an idea about what is happening

Thank You Sir...!!!

As I'm amateur in the field,, so I want to start everything from the beginning. I want to write an in-general MATLAB code to compute HOMO/LUMO for organic molecules and polymers. As HF doesn't include e-e correlation, so does it affect calculated HOMO/LUMO energy levels?

Is it a feasible idea to do everything from very scratch?

If you want to start from scratch, you would usually go for something like Hückel theory, see e.g. these posts:

http://chemical-quantum-images.blogspot.co.at/2007/08/h-package.html

http://www.chemical-quantum-images.blogspot.co.at/2007/09/h-package-2.html

This is something you can do with little effort. and it will give you HOMO-LUMO gaps that make sense when you compare them to absorption energies for example. and you can understand aromaticity

ab-initio Hartree-Fock is much more difficult to implement because you need to compute 6-dimensional integrals as shown in the first equation of this post

http://chemical-quantum-images.blogspot.co.at/2013/09/density-fitting-and-tensor-hyper.html

Dear Sir I am using HOMO and LUMO energy in corrosion research. I want to know that I am getting similar pictures both for HOMO and LUMO. Then how can I explain which is site in the molecule is electron donar and which is electron acceptor from metal. Please suggest me the correct answer

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