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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 a 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 May 25 '13 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 May 25 '13 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 May 26 '13 at 9:16
  • $\begingroup$ "How faster do I get convergence in the case of SCF orbitals?" Start with a better initial guess. $\endgroup$ – LordStryker Oct 23 '13 at 20:57
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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 May 25 '13 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 May 25 '13 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 May 26 '13 at 0:02
  • $\begingroup$ I think that is how Natural Orbitals are defined: orbital set that puts the most "electron" into the smallest number of MOs (in an aufbau sense). Then one could just use the occupied orbitals. I think the problem is that one has to solve the CI problem to determine the natural orbitals -- sort of a "no free lunch" problem. $\endgroup$ – Eric Brown May 26 '13 at 0:12
  • $\begingroup$ @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. $\endgroup$ – Deathbreath May 30 '13 at 16:05
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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.

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    $\begingroup$ Szabo and Ostlund? :-) (nice to see you on stack exchange) $\endgroup$ – Eric Brown Oct 21 '13 at 16:11
  • $\begingroup$ I usually camp in ComputationalScienceBeta in the land of StackExchange. My error has been corrected. $\endgroup$ – Jeff Oct 23 '13 at 14:53

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