# Post-Hartree Fock Methods

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).

Thanks!

• – Mithoron Feb 21 '15 at 21:10

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:

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

And

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

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.