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Note added afterwards: This question also has excellent answers on the Chemistry Stack Exchange: Is density functional theory an ab initio method?


I have very little experience with DFT, but coming from more of a coupled-cluster background, where the improvements are "systematic", to me it seems that the choice of functional in DFT is somewhat trial-and-error based and also problem specific, perhaps requiring "chemical intuition". Therefore, can DFT truly be considered an ab initio method?

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    $\begingroup$ I suppose functionals like B3LYP are heavily fitted to empirical data, whereas PBE is sometimes called a "parameter-free GGA functional". $\endgroup$ May 11, 2020 at 22:09
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    $\begingroup$ This great post from Chem.SE might be relevant. $\endgroup$ May 12, 2020 at 0:29
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    $\begingroup$ I’m voting to close this question because this is a cross-site duplicate of Is density functional theory an ab initio method?. $\endgroup$ May 12, 2020 at 9:27
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    $\begingroup$ @Martin-マーチン Is closing questions the right way to handle cross-site duplicates. Quoting Robert Cartaino from Stack Exchange Meta: "...where the question is appropriate on more than one site, leave it on both sites and let the users of each community benefit from the information." from the most upvoted answer to What to do with cross-site duplicates?. $\endgroup$ May 12, 2020 at 11:46
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    $\begingroup$ @rashid Maybe, maybe not. There are more nuances to this than the post you cite. SE sites generally think of themselves as 'last resort' options, they don't aim to duplicate the internet (this community might decide differently). You would generally expect from every asker here to at least do a courtesy search on google, maybe even read a Wikipedia article or two on the subject. So in this special case, the information that is looked for is easily found and freely accessible. Yes I think closing is the only viable option here. $\endgroup$ May 12, 2020 at 12:01

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This is a good question. The term ab initio literally means 'from the beginning,' "implying that the only inputs into an ab initio calculation are physical constants." (Wikipedia)

However, this term is often used to describe methods that involve empirical approximations (like LDA, GGA) or derived quantities (like pseudopotentials). And DFT is commonly referred to as an ab initio method, including on the DFT Wikipedia entry:

In the context of computational materials science, ab initio (from first principles) DFT calculations allow the prediction and calculation of material behaviour on the basis of quantum mechanical considerations, without requiring higher-order parameters such as fundamental material properties. In contemporary DFT techniques the electronic structure is evaluated using a potential acting on the system’s electrons.

There are techniques, like quantum Monte Carlo, or directly solving the Schrodinger equation, that are truly from first principles, but these techniques are so computationally expensive that they are rarely useful for modelling an actual material at any scale, and are also rarely referred to as ab initio. DFT is ab initio relative to other more empirical methods like molecular mechanics.

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  • $\begingroup$ It's a good question, but has already been asked, and answered, on another SE forum. chemistry.stackexchange.com/questions/33764/… $\endgroup$ May 12, 2020 at 17:50
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    $\begingroup$ LDA is not empirical, neither are many GGAs etc. Modern pseudopotentials are also not empirical, they are derived directly from all-electron reference calculations (usually atomic calculations). There are some adjustable parameters, but these are primarily to control the computational efficiency. $\endgroup$ May 14, 2020 at 1:45
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As pointed out by several people already, some information can be found elsewhere, as in here. And also the differentiation between DFT (exact) and density functional approximations (DFAs), as pointed out regularly by Mel Levy, can be found there.

However, I think there is one aspect missing, and here I would like to quote my late PhD supervisor Jaap Snijders. The most important aspect to know if a method is ab initio or not, is related to the integrals. If the integrals can be computed from the beginning, the method is ab initio; if not, then not. In DFT, DFAs and wavefunction methods, the integrals can be computed, and hence, these methods are ab initio. In semi-empirical methods (AM1, PM3, DFTB, xtb), some of the integrals are either estimated or approximated (from e.g. DFA results in case of DFTB/xtb), and therefore, these methods are not ab initio. Likewise for e.g. the Empirical Valence Bond method, which like the name already indicates, is empirical.

Whether or not a method gives the exact energy is a different aspect. In that case only Full CI with infinite basis set and DFT give the exact energy, all other methods are approximations. By choosing a basis set of a certain size, one is approximating; by using "only" CCSD(T), one is approximating; by using a density functional like PBE, B3LYP or r2SCAN, one is approximating; etc.

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  • $\begingroup$ +1 for a very insightful answer, especially the part about ab initio referring to the calculation of integrals. I hadn't seen that perspective before, and guessed that "empirical" referred to the DFAs coming from fits to empirical data as opposed to coupled cluster where the approximation doesn't involve such a fit. $\endgroup$ Dec 29, 2020 at 12:09
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In principle, DFT is exact, so it's an ab initio method. However, because we don't have an universal functional and work only with approximated functionals, we have to know a priori what functional performs well with our system in study and also compare results with experiment or high-level wavefunction methods. Then, we could say that is a "semi-empirical" approach, but some could argue with that name.

More information in related posts in ChemSE and here.

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  • $\begingroup$ This seems to borrow a lot from this answer on Chem SE. Could you add some of your own interpretation or at least cite this post? $\endgroup$
    – Tyberius
    May 12, 2020 at 2:19
  • $\begingroup$ @Tyberius I didn't have saw that answer before rashid commented, but I'll refer to that answer and another here in this site (from Martin too). $\endgroup$
    – Verktaj
    May 12, 2020 at 2:24

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