I'm new to computational chemistry, so perhaps this question has an obvious answer. I'm wondering what the intuitive reason for constructing Slater determinants that involve excited states is, when one tries to describe correlations, starting from the Hartree-Fock picture. I understand that the single determinant in the Hartree-Fock model only describes exchange (antisymmetry of fermionic wave functions). I am trying to understand the logic behind why constructing Slater determinants involving excited states can give us a picture of correlation. I guess I don't understand the relation between excited states and correlations (especially thinking about it from the point of view of DFT, which is interested only in the ground state, but correlations are still important).



I think that here it is not accurate to say "starting from the Hartree-Fock picture", instead of it, I find better to say "starting from the Hartree-Fock method". When this method is carried out we get a lot of orbitals and that allow us to construct a lot of Slater determinants, even if, we can be only interested in one of them.

Given a complete set of functions $\{f_i(x)\}$, any function $g(x)$ can be thought as a linear combination

$g(x) = \sum c_i f_i(x)$

For a two variable function $g(x_1,x_2)$, fixing $x_2$ we have

$g(x_1,x_2) = \sum c_i(x_2)f_i(x_1)$

But as $\{f\}$ is a compete set:

$c_i(x_2) = \sum d_{ij} f_j(x_2)$


$g(x_1,x_2) = \sum d_{ij}f_i(x_1)f_j(x_2)$

This reasoning is trivially extended to a function of $n$ variables.

For an $n$-fermions system, the wavefunction depends of $n$ spin-orbitals ($n$ variables). To get a mathematic valid form, we only need to antisymmetrize the generalization of the expressions above.

Take in mind that none of these determinants are the true state (they are not eigenfunctions of the true hamiltonian). The same happens with the orbitals (that are eigenvectors of the Fock operator).

So, the use of excited determinants arises naturally from a math view point. Just because we already have functions for the determinantal expansion, and we know that they are pretty good in some sense, because the are the best we can get at least for one determinant (the Hartree Fock solution).

It is not needed to think configuration interaction as a mixture of stationary quantum states (ground and excited states), as said above they truly aren't.

Direct to your question: We can think the problem just as a decomposition of the wave function, somewhat like Fourier series, where the functions involved are the Slater determinants, just because it is convenient. But, this doesn't imply a physical meaning.

Edit in response to comments:

I agree: By choosing an effective Hamiltonian one is working on a class of approximations that build up a picture for a set of analogous systems. When considering correlation we have in mind a "picture" of the system. Being that H-F is a mean field method, it belong to some picture, but I feel that when dealing with Coupled Cluster the H-F method is just a part of a bigger procedure consisting in finding some functions in order to obtain a solution for another picture. In such sense, while the H-F method itself can be understood as the method for obtaining a solution of the H-F picture, in this context is just part of method for solving a problem in another picture. I added the word here for strictness.

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    $\begingroup$ Also this discussion is quite old, I would like to add that the term "picture change" is frequently used in relativistic quantum chemistry, when (effective) Hamiltonians are compared to each other. E.g., when approximating the four-component (4c) Dirac equation by an effective two-component (2c) formalism, one usually introduces further approximations. In this respect, "starting from the Hartree-Fock picture" is a correct terminology, as it implies certain approximations on the Hamiltonian. However, the better way is usually to end up with a certain picture, compared to starting from it... $\endgroup$ – TheFox Oct 10 '17 at 8:34
  • $\begingroup$ @TheFox ,I edited my answer in response to your comment. Feel free to edit it if you consider it appropriate $\endgroup$ – user1420303 Oct 11 '17 at 19:34

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