# Basis Functions for Hartree Fock, and Configuration Interaction

I'm about halfway through the book Szabo and Ostlund, and while I think I understand the rough idea of Hartree-Fock and configuration interaction, there is something I would like to clarify.

For one, I'd like to note the fact that while the book often refers to the s, p, and such orbitals, the actual basis functions used to find the SCF are completely arbitrary, and picking a basis of any old functions (don't even have to be orthonormal) yield a valid Slater-Determinant where I can minimize coefficients to obtain a Hartree-Energy. This is true even if I picked really stupid functions, such as Gaussian orbitals centered 100's of A.U. away from any of the nuclei. Then the Hartree-Energy is just a really bad approximation to the ground state.

Furthermore, unless I only have basis functions equal to the number of electrons (minimal basis), my actual orbital solutions ($$|\psi_i\rangle$$) that I obtain in the end can be linear combinations of the basis sets ($$|\phi_j\rangle$$) through coefficients $$c_{ij}$$.

$$| \psi_i \rangle = \sum_{j}c_{ij}|\phi_j \rangle$$

By orbitals here I suppose I mean the eigenfunctions to the Fock-operator. These are ultimately found by being put into a slater determinant $$|\Psi\rangle = A|\psi_1\psi_2...\psi_N\rangle$$ (where $$A$$ is the anti-symmetrizer) and after acted on by the Fock-operator solved iteratively until convergence for the coefficients.

Now, of course we'd expect something closer to the actual energy if we use a larger basis set, but I'm a bit confused about the distinction between this and the configuration interaction. I read that the configuration ground state is give by:

$$|\Phi\rangle =c_0|\Psi\rangle+\sum_ac_a|\Psi_a\rangle + \sum_{ab} c_{ab}|\Psi_{ab}\rangle +\cdots$$

Where $$a$$ denotes a replacement of one of the orbitals by some single other (presumably orthogonal to rest) orbital. Never mind the fact that the book starts calling these "excitations" which to me invokes some sort of physical picture.

My question is that didn't this already happen when we expanded $$| \psi_i \rangle$$ in terms of the basis functions? Why do we need further determinants to correct the ground state? If we used a complete basis before it should be exact? It's not immediately clear to me that summing over Slater determinants creates a bigger basis than taking a Slater determinant over arbitrary sums. I read that the energy is lowered due to 'correlation' effects, but I'm afraid I don't understand the nature of these or encountered a good simple example of what's really happening.

Edit: Disregard spin for this question

• Hartree-Fock method with its variational slater determinental ansatz gives an approximate many electron wave function, which is not a true eigen function of many electron Hamiltonian. True eigen function can be expanded (CI) in terms of Hartree Fock determinants. This expansion can be organized in terms of number of excitations (in terms of Roothans orbitals) over hartree fock ground state. Feb 11, 2019 at 20:59
• You construct a different Hamiltonian in CI and the groundstate of this Hamiltonian is closer to the true energy of the groundstate, in principle exact if you are doing full CI plus infinite basis set. The CI matrix does restore the correlation/coupling/interaction of the electrons which is lost in Hartree-Fock by using the effective average potentials. So what matters is not the expansion alone but the fact that this expansion leads to a CI-Matrix(= representation of the Hamilton operator with determinants as basis) which contains new interactions that are completely missing in normal HF. Oct 1, 2020 at 14:53