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Recently, I came across some lecture notes on electronic structure theory which had something peculiar about it. I noticed that FCI (configuration interaction) was introduced before Hartree-Fock theory, that got me thinking if its possible to think about FCI as an approach independent of HF theory as starting point (not particularly as post-HF method).

Would such an approach be possible?

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In general can there be a generalized formalism for Full configuration interaction? (say in the form of a particle-hole construction, for example see the introduction section in arXiv:1906.11361)

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  • $\begingroup$ If the question is too naive or off topic, I am happy to take this down. But, seemed like a reasonable question to think about it. $\endgroup$
    – user784
    Mar 28, 2022 at 16:48
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    $\begingroup$ I think the question is reasonable. Someone will almost certainly be able to provide a better/quicker answer than I can, but to spoil some of the surprise: yes, CI can be done without an HF reference. While an HF reference (see page 9) is particularly convenient, CI is fundamentally about solving the Schrodinger equation by building an N-electron basis, for which a lot of possible reference states could be used for the excitation. Multiconfigurational methods could also fit, using more than just HF as the reference. $\endgroup$
    – Tyberius ♦
    Mar 28, 2022 at 16:56
  • $\begingroup$ What is your question, actually? Could you elaborate? I am no chemist, but if FCI simply means: expand a state in a basis then yes, there is no reference to the HF method. $\endgroup$
    – Jakob
    Mar 28, 2022 at 17:03
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    $\begingroup$ @Jakob in general the way CI expansion is defined with reference to a particular determinant or a set of determinant. More often than not, the slater determinant of the HF state forms the reference. But, CI isn't just expanding in any basis, you need a reference and that actually determines if many things. $\endgroup$
    – user784
    Mar 29, 2022 at 2:14

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Yes, you can surely do full CI without Hartree-Fock. However, the reason why one typically starts from Hartree-Fock is that this guarantees a good conditioning of the FCI matrix, which becomes diagonally dominant and thereby iterative diagonalization becomes tractable.

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    $\begingroup$ Doesn't the FCI Energy generated from a HF reference be different from the energy obtained, from say, a reference with multiple determinants. So, in that case how would the correlation energy defined as a difference between $E_{FCI}$ and $E_{HF}$, change? $\endgroup$
    – user784
    Mar 29, 2022 at 2:20
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    $\begingroup$ @EverydayFoolish One will have to do a separate HF calculation to obtain the correlation energy. Otherwise, a FCI calculation from a non-HF reference gives the correct total energy but does not give the correlation energy. $\endgroup$
    – wzkchem5
    Mar 29, 2022 at 10:05
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    $\begingroup$ Note that the above definition for the correlation energy may attain several values in strongly correlated systems due to the local minima problem in HF. $\endgroup$
    – Susi Lehtola
    Mar 31, 2022 at 21:43
  • $\begingroup$ Is there any proof or has anyone shown that full CI without Hartree-Fock, say starting with a guess charge density bond-order matrix will give the true eigenvalues? It would be interesting to try this for Helium atom or so. $\endgroup$
    – Raghunathan Ramakrishnan
    Apr 6, 2022 at 3:36
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    $\begingroup$ @RaghunathanRamakrishnan FCI is invariant to orbital rotations, so yes: it doesn't matter what orbitals you put in, the answer will always be the same, provided you do the diagonalization exactly. The point is just that since the matrices are so humongous, you really do want to do HF first since HF costs nothing compared to FCI and as a result the sparse diagonalization with the Davidson method will converge much, much more rapidly. $\endgroup$
    – Susi Lehtola
    Apr 8, 2022 at 15:46
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Full CI often indicates the exact solution of the electronic Schroedinger equation that can also be determined without starting with the Hartree-Fock reference wavefunction. For example, the Hylleraas method for Helium does not require Hartree-Fock or SCF and one can determine the ground state energy to high precision.

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    $\begingroup$ But Hylleraas isn't FCI. In the Hylleraas method, you actually write down the wave function expansion in closed form in terms of interelectronic variables. In the configuration interaction approach, you diagonalize in terms of electron configurations a.k.a. determinants. In the case of two-electron systems (the Hylleraas method doesn't really generalize to many particles) you should get the same energy and wave function; however, the Hylleraas expansion converges way faster than the determinant expansion. $\endgroup$
    – Susi Lehtola
    Apr 5, 2022 at 0:50
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    $\begingroup$ FCI wavefunction, implying a complete expansion in the configuration state function representation (using a complete set of MOs), and a Hylleraas wavefunction that is expanded completely, are the same in the position representation. When either expansion is truncated, it is a different matter. There are some old works where they have done CI on top of Hueckel/INDO MOs that also do not require Hartree-Fock (in connection to the original question). But their FCI limits do not converge to the exact limit reached by HF-based FCI or the Hylleraas expansion. $\endgroup$
    – Raghunathan Ramakrishnan
    Apr 5, 2022 at 2:44
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    $\begingroup$ Of course, since the Huckel (or Pariser-Pople-Parr for that matter) model Hamiltonians are not the same as the electronic Hamiltonian. And yes, like I said, both Hylleraas and FCI are ways of solving the exact wave function, but that does not mean that Hylleraas == FCI. $\endgroup$
    – Susi Lehtola
    Apr 5, 2022 at 17:23
  • $\begingroup$ Methodology-wise, CI and Hylleraas are, of course, not the same. But what I pointed out in the last comment is: the exact wavefunction (say, in position representation) is unique whether it is obtained by full-CI or a complete expansion in Hylleraas. $\endgroup$
    – Raghunathan Ramakrishnan
    Apr 6, 2022 at 3:31
  • $\begingroup$ Paraphrasing from Szabo, a CI wave function can be written using N-electron configurations obtained from any one-electron basis$^{\dagger}$. I guess if it is exact within that particular N-electron basis then its Full CI? $\\$ ${\dagger}$Note: I think this also extends to say 2e- wave functions like AGP too? $\endgroup$
    – user784
    Apr 6, 2022 at 10:07

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