Is there a difference between Hartree-Fock method and LCAO?

I have to dive in some quantum chemistry for a quantum machine learning project and I came across the so-called Hartree-Fock method.
In one of the reference I used, they considered electrons as independent so that the many-electron wave function (**is it of the full molecule or only for one atomic nucleus ?**) can be described as a Slater determinant.


So far no problem but then I came across the linear combination of atomic orbitals method and I wanted to know how it relates to Hartree-Fock. Because looking at some quantum chemistry textbook, it appears that for $n$ AO we can produce $n$ MO, say in the case of $\ce{H2O}$ it would *a priori* be 3 (for the oxygen) and 2 (for the $\ce{H2}$ fragment). But the Slater determinant accounts for all doesn't it ?


Thank you for your answer :)




Obviously, this is a topic which much can be written on. I'll try to keep it short and simple. The LCAO concept merely says that MOs are formed from AOs through linear combination, i.e.


$$\psi\_n = \sum\_i c\_{ni}\phi\_i$$


where $\phi\_i$'s are AOs, $\psi\_n$'s are MOs, and $c\_{ni}$'s are the coefficients of the $i$-th AO in the $n$-th MO.


It offers no way of determining the coefficients $c\_{ni}$. Sometimes it can be determined through symmetry (let's say $\ce{H2}$, the bonding MO obviously has equal contributions from both hydrogen 1s orbitals). And sometimes you can get qualitative results (in $\ce{HF}$, the bonding orbital is mostly fluorine 2p).


But in general, simply knowing the *form* of the MOs isn't enough, and you need some kind of *quantitative* method to determine the coefficients — which is where methods such as Hartree–Fock come in. Using the iterative method prescribed in Hartree–Fock theory you can actually obtain values for $c\_{ni}$ (and from there, orbital energies, etc.)


The concept of a Slater determinant is completely separate. It is just a convenient way of representing the *total* electronic wavefunction as a product of one-electron wavefunctions (the $\psi\_n$'s above are one-electron wavefunctions), but it still has nothing to do with finding the coefficients $c\_{ni}$. The total electronic wavefunction refers to all electrons in the molecule, by the way—it would not be correct to say that an electron belongs to one specific nucleus.


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