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+ Ionization
+ Theoretical chemistry
+ Dft calculations
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Nor Reza

How do electron affinity and ionization potential values from DFT...

Andrew Kiener  Follow

Just a hint. Indirect results (that is, calculating the energy diference between the neutral molecule and the cation or anion) is used to get better results than using Koopman's theorem. 

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C. Michael Turner  Follow

Can someone let me know please, how to calculate electron affinity from effective mass?

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David Shanen  Follow

Dear Dragos Palace
But I got almost the same answers from both hf and dft methods. As I told a deviation of only about 0.2 eV.

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Billy Chang  Follow

     
Hi Lokesh,
My molecule is not a simple one. It contains 56 atoms.

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Brian Good  Follow

Oh..., sorry, I misread your question. I suppose that the explanation is that the functional for the exchange correlation energy that is used in your code reproduces very closely the non-local exchange from HF calculations.

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Diane Fiorito  Follow

In the DFT formalism there is Janak theorem, which states that the negative value of an orbital energy is the EXACT ionization potential (or electron affinity, if the orbital is unoccupied), provided that one uses the EXACT exchange-correlation functional. In practice the differences between the IP values from DFT and the experimental values can be substantial (especially for pure GGA functionals). On the other hand, Koopmans theorem can give valueas that are also largely in error (+- 1.5 eV), since this theorem is based merely on fortuitous cancellation of errors. All in all, I would say that the agreement you have observed is probably accidental. Also, for DFT I would agree with David, that computing the genuine energy of the ionization process would give results closer to experimental values. Note, however, that for kations or anions there may be severe spin contamination, especially for functionals with large non-local exchange contribution. 

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Christopher Imoe  Follow

Actually if the molecule under investigation is too simple(small molecule) then the result obtained from DFT and HF may be same. So, plz investigate your molecule under investigation weather it is......

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Bryan Katz  Follow

If your question is why are the results different given the Koopmann theorem, the answer is that in general the HF results are different from DFT because they are not the same theory from the start. While HF includes an "exact" exchange operator and no correlations, the DFT tries to include both but it does it in an approximative manner. Therefore, the different results are kind of natural 

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Crow72  Follow

It''s true that Koopmans theory is valid for HF only. Remember, though, that in DFT we also have (Kohn-Sham) orbitals and (Kohn-Sham) energies, so there ways to formally introduce Koopmans theory for DFT. Some rather old papers by Perdew and Parr give all the details, while a more recent paper (http://scitation.aip.org/content/aip/journal/jcp/131/23/10.1063/1.3269030) gives some examples of actual use and discusses obtained results.

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Alice Twain  Follow

This reference (Pritkova, et al.) may help:
http://pubs.acs.org/doi/abs/10.1021/jp0457491

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