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Since the major problem with DFT is that the exact functional for exchange and correlation is not known, except for the free-electron gas (LDA from Thomas-Fermi model). The discovery of "the" exact exchange-correlation functional would be (in my opinion):

  • A Nobel prize winning discovery.
  • Would transform the matter modelling world.

I wanted to ask what are the advancements on this front? How do physicists/mathematicians/SSP's and even chemists even go about solving this issue?

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  • $\begingroup$ Indeed ! Will delete it $\endgroup$
    – Elie H
    Commented Jan 6, 2022 at 19:42
  • $\begingroup$ I'll just mark it as a duplicate so someone can find that question even if they use search terms closer to what are in your version of the question. $\endgroup$
    – Tyberius ♦
    Commented Jan 6, 2022 at 19:44
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    $\begingroup$ Note that a key factor is the cost of evaluating the functional. The exact functional will only be a Nobel prize-level discovery if it is significantly cheaper than basis set extrapolated approximate full CI methods, or more or less equivalently, if it is not significantly more expensive than contemporary approximate functionals. But this is very unlikely. In particular, if the exact functional can be evaluated in polynomial time, the discoverer can get another $1,000,000 from the Clay Institute, due to the result implying P=NP. $\endgroup$
    – wzkchem5
    Commented Jan 6, 2022 at 20:35
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    $\begingroup$ @wzkchem5 I think they would also have proven not only that P = NP, but also that NP = QMA. Considering how many quantum computing algorithms we know to be in QMA but don't know to be in NP, I'd be quite surprised and astonished if someone were to prove that NP = QMA, whereas P=NP wouldn't be too surprising for some people (for example Donald Knuth still seemed to lean in this favor as recently as 2018 when he most recently visited my city). $\endgroup$
    – Nike Dattani - No Free Time
    Commented Jan 7, 2022 at 5:05
  • $\begingroup$ @NikeDattani Yes of course. I was mentioning P=NP simply because this is much more well-known, and the amount of prize money for proving P=NP is also very well-known :) $\endgroup$
    – wzkchem5
    Commented Jan 7, 2022 at 8:48

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