Basis Functions for Hartree Fock, and Configuration Interaction

Question

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

Answer 1

Consider a single Helium atom. Clearly for each electron, the probability of finding it on the "left" or "right" side of the nucleus is the same; 50%. However, the probability that one electron is on the left side given that the other is on the left side is less than 50%. This is correlation, and accounts for the reduction in energy associated with a multi-particle wavefunction ansatz - i.e. one that can take account of particle distributions that are conditional on the location of other particles. The HF solution optimizes the spatial distribution of each electron in the presence of the mean field of the other, and so excludes this effect.

Answer 2

1) Quantum chemists are not interested in bad solutions. That is why they take atomic like basis functions. 2) the minimal basis is half the number of electrons, plus one if the latter is odd. 3) The hfs equations are an approximation to the Schrödinger equation valid for the case of a single determinant. In general a single determinant does not solve the many electron Schrödinger equation. The approximation can therefore be improved by including more determinants in a linear combination. CI is a procedure to select and combine more determinants.