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Usually textbooks on CI takes the SCF Hartree-Fock orbitals as a starting point. Can one use the configuration interaction method starting directly from the single-particle Slater determinants avoiding the iterative procedure. This question concerns mainly small molecules like the hydrogen molecule. Do I get a high inaccuracy in this case? If yes, why? Could this inaccuracy be decreased taking into account more configurations?

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  • $\begingroup$ ummmm… why would you want to do that? :-) I guess I'm wondering if you are doing singles and doubles or toward full CI? I think you would get toward the Full CI limit faster (with fewer excitations) if you start with more reasonable orbitals. $\endgroup$
    – Eric Brown
    Commented May 25, 2013 at 22:08
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    $\begingroup$ Thank you for the reply. That is actually what I am asking. How faster do I get convergence in the case of SCF orbitals. How many additional excitations I have to take into account in the case of avoiding preliminary HF computations. $\endgroup$
    – freude
    Commented May 25, 2013 at 22:19
  • $\begingroup$ @freude: I don't think it's really possible to make an accurate but general statement about it: it'll vary a lot by system, and will be less important in simple systems where the difference between cheap approximations and accurate models is smaller. $\endgroup$
    – Aesin
    Commented May 26, 2013 at 9:16
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    $\begingroup$ Szabo and Ostlund describes CI calculations using natural orbitals where the convergence is faster than with Hartree-Fock orbitals. You should read that book for more information. $\endgroup$ Commented Oct 20, 2013 at 21:45
  • $\begingroup$ "How faster do I get convergence in the case of SCF orbitals?" Start with a better initial guess. $\endgroup$ Commented Oct 23, 2013 at 20:57

2 Answers 2

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If you were to perform Full CI (within a finite basis) then I don't think that it matters whether you start with SCF orbitals or some other choice. This solution will have the lowest conceivable energy no matter what.

But if you are doing truncated (less than full) CI, then you are very much at the mercy of your starting orbitals. In general, one would want to start with orbitals that resemble those of the actual state that you are trying to describe via CI.

Otherwise, your precious excitations (singles and doubles) may be consumed correcting the "slop" that you provided, rather than moving toward the full CI limit.

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    $\begingroup$ (As an example, I know some people using CI-related approaches (CASSCF and CASPT2, specifically) on heavier elements who prefer to start with DFT orbitals because they produce a better result than starting with HF.) $\endgroup$
    – Aesin
    Commented May 25, 2013 at 22:46
  • $\begingroup$ Right. I had a little trouble answering this question, because practically speaking, there is always optimization of the inital vectors. It's sort of a "textbook" question. Very interesting about DFT orbitals for metals. I've never done MCSCF on metals--they probably need all the help they can get :-) $\endgroup$
    – Eric Brown
    Commented May 25, 2013 at 23:01
  • $\begingroup$ Isn't there a theory that the exact natural orbitals (and approximations to them) should give you the quickest convergence, though? I believe that's the idea behind Iterative Natural Orbital approaches, for example. $\endgroup$
    – Aesin
    Commented May 26, 2013 at 0:02
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    $\begingroup$ @EricBrown: Natural orbitals are the eigenfunctions of the 1st order reduced density matrix of a wavefunction. If you take n strongly orthogonal wavefunctions and mix them, your leading NOs will not make up the Slater determinant with the largest prefactor. Of course that's contrived, since the wavefunctions also are orthogonal in the energy product. $\endgroup$ Commented May 30, 2013 at 16:05
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    $\begingroup$ If by "some other choice" you mean a basis that can be obtained from the other one by unitary rotation then of course you are right. Otherwise it is not true that the Full CI will give you the same result. Energy minimization by Full CI only depends on your basis. $\endgroup$
    – perplexity
    Commented Oct 21, 2013 at 15:16
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Adding to the answer and all the related comments.

Using the HF reference as starting point provides a "clean" interpretation for the different contribution to the total energy. Indeed, the difference between HF and CI energy is, by definition, the electron correlation energy. One can then furthermore break down the correlation into single, double, etc contributions and consider a step-wise convergence to a full CI. When the reference used for the GS already contains different "effects" (by using NOs or DFT orbitals) the lower energy obtained after performing a CI is not easily interpretable.

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  • $\begingroup$ "lead to a final lower full CI energy". Certainly not - full CI is full CI, no matter the starting orbitals (as long as they have the correct number of electrons and similar basic conditions). I have not run such calculations, but I would wager that if one starts with orbitals other than canonical SCF orbitals, a HF-like energy is obtained, thus not the DFT energy is used. $\endgroup$
    – TAR86
    Commented Sep 22, 2022 at 18:37
  • $\begingroup$ Thank you for your comment. I have difficulty to agree with your statement "full CI is full CI, no matter the starting orbitals". In the present state of my understanding, this will only be valid when considering a infinite basis. As soon as you truncated the basis, you also impose a constrain on the way the space is spanned. The calculation should thus become orbital depended (for basis sets of the same size). Isn't it ? I think that some literature search would be needed. $\endgroup$
    – Mils
    Commented Sep 23, 2022 at 7:29
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    $\begingroup$ I assume that the same AO basis is used. It then does not matter how the reference determinant is constructed because "all" determinants will be used in the FCI. The result is the lowest possible energy for that system in that AO basis. $\endgroup$
    – TAR86
    Commented Sep 23, 2022 at 11:28
  • $\begingroup$ Agreed. It is the underlying AO basis that defines the space that can be spannend, not the way the MO basis is constructed. The same full CI will always lead to the same energy, whatever the kind of orthogonal MO basis (HF or other). Thanks for the clarification. $\endgroup$
    – Mils
    Commented Sep 23, 2022 at 11:42

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