# Hartree Fock for an atom (Beryllium)?

I'm working on solution of Hartree-Fock equation for Beryllium in configuration $$1s^2 2s^2$$, expanding the orbital with STO-3G basis $$\phi_k=\sum_{p=0}^3 C_{kp}e^{-\alpha_{kp}r^2}$$ I know the values of $$\alpha_k$$ both for 1s and 2s. All books speak about solving molecules and the only case with atom is the helium case, so I have some doubts. In molecular case they expand the molecular orbital as $$\phi_k=\sum_{p} C_{kp}\chi_p$$ without index $$k$$ on $$\chi_p$$ and so also all matrices I need to solve the eigenvalue problem. For this reason I can't understand the difference between the basis set used in molecular case and atom case and consequently how to solve the Fock equation in the case of an atom, and in particular of Beryllium.

• Your definitions of $\phi_k$ and $\chi_p$ are unclear, but maybe that's part of your confusion? If your first equation is correct, then $\phi_k$ are the AOs. But the second equations looks like $\phi_k$ are MOs and $\chi_p$ are AOs. That doesn't add up. Please double check what those are supposed to be. – Feodoran Apr 14 '19 at 21:51
• Exactly! That's the part of my confusion! In the case of a molecule I expand my MOs as combinations of AOs and for closed shells I find Roothaan equation (that's what books treat and what I understood). But in the case of a simple atom, like beryllium, what happens? – Liuuuuk Apr 14 '19 at 22:19
• Formally, the very same thing. But the result may be that the MO coefficient matrix becomes the identity matrix, so MOs equal AOs. But this depends on your AO basis set of course. – Feodoran Apr 15 '19 at 6:06
• The problem is that I have no idea of the form of the matrices in the case of beryllium. In fact my book expand each MOs with a given basis and then create the Fock and overlap matrix. In my case I have an expansion of each AO with ITS OWN STO-3G basis and I don't know how to use the expression for the Fock and overlap matrix given. – Liuuuuk Apr 15 '19 at 7:59
• The matrices in HF (Overlap, Fock etc.) are in the basis of AOs, not the actual basis functions. Whether you choose STO-3G, STO-6G, or maybe even actual Slater type orbitals does not matter for those matrices. This will affect how the AOs look like, and therefore indirectly the numbers in those matrices. But the matrices itself only depend AOs and do not care about the basis function the AOs are expaned in. If you only have 1s and 2s orbitals, then these matrices will be of dimensions $2\times 2$ (using a minimal basis, no p orbitals and spatial orbitals). – Feodoran Apr 15 '19 at 9:12