Author Topic: Reference state of DFT-calculated energies  (Read 934 times)


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Reference state of DFT-calculated energies
« on: September 10, 2020, 09:30:00 AM »
Dear all,

Some people in my project require me to specify what the reference state for the calculated energies is. Specifically, in a cosmo file, I find the property "  Total energy corrected [a.u.]  =     -538.2691269347" under the keyword "$cosmo_energy".

Well, the question may be basic, but I don't have a straightforward answer to it. To me, this represents the energy level of the molecule in the COSMO continuum. But this energy level has to be calculated as a function of some sort of reference. The question is: what is, exactly, this reference? In which state are the nuclei and atoms before they form the molecule? A connected question is: is this reference state different with different DFT functionals? i. e. are the differences in value between those found in cosmo files obtained from, e. g., TZVP and TZVPD-FINE basis sets only due to the refined description of interactions of the latter, or are they also due to a different reference state being chosen?

Is the answer "free nuclei and electrons in vacuum"? (which would imply that the reference state is the same for any functional, then).

I don't know if my question is clear, but I long for an accurate and unambiguous answer to give to my partners for this question.

Best regards.
« Last Edit: September 10, 2020, 09:36:52 AM by Glxblt76 »


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Re: Reference state of DFT-calculated energies
« Reply #1 on: September 18, 2020, 11:48:53 AM »

COSMO is not changing the reference for the calculated energies. The answer is thus simply: "Same as for gas phase calculations".

But I think a better view on the reference is to assume the same number of nuclei and electrons as in the calculation, but all having an infinite distance to every other particle. For the COSMO energy the zero reference is therefore for the case of 1) no charges at all or b) no cavity.

Thinking of it a bit further, this would mean that the cavity is infinitely far away from any charge. But the cavity has to be there as it is introduced artificially and the exact shape and position is not derived from first principles (well, technically and empirically it is determined by the position of the nuclei of course).

This is my understanding and it might be incorrect. In that case I'd be very happy to learn from experts.