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 :)


1 Answer 1


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.

  • 1
    $\begingroup$ Strictly in HF theory you don't need to introduce LCAO at all, or more generally a basis - the HF Hamiltonian is a well defined approximation to the exact Hamiltonian, introduction of a basis is a further step. $\endgroup$
    – Ian Bush
    Commented Dec 23, 2022 at 23:06
  • $\begingroup$ Thank you for your answer! So you're essentially saying that LCAO is indenpendent of Hartree-Fock theory (the fundamental idea, it is useful for computing the coefficients you've mentionned) and is for one-electron wavefunction, right ? Now, if I want to know the many-electron wavefunctions for independent electrons, whether it is for an atom or a molecule, I may use Slater determinant to account for both the separablity and the antisymmetric principle ? $\endgroup$ Commented Dec 24, 2022 at 17:50
  • $\begingroup$ and while I am it, LCAO is merely a sort of basis change from molecular orbitals to atomic orbitals, right ? hence the fact that n AOs lead to n MOs, right ? $\endgroup$ Commented Dec 24, 2022 at 17:55
  • $\begingroup$ @QMLSorbonne Yes, that's my understanding of it. And yes, LCAO is precisely that- you can call it a basis change, but I think a choice of basis is more precise, because it is the only basis we're actually using. $\endgroup$ Commented Dec 24, 2022 at 20:20
  • $\begingroup$ There's an infinite number of basis choices, MO and AO are just two options, but because they all span the same n dimensional Hilbert space there must be n entries in whatever basis you choose. And then as you say you can apply matrices to transform from the representation of the wavefunction in one basis to that in another $\endgroup$
    – Ian Bush
    Commented Dec 24, 2022 at 22:37

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