Question

When deriving the Hartee-Fock method, we minimize the electronic energy with respect to all molecular orbitals with the constraint of orthonormality of the molecular orbitals by using the method of Lagrange multipliers. Is there a fundamental reason why the molecular orbitals need to be orthogonal? Does it ensure a lower energy compared to any non-orthogonal set of molecular orbitals​

Answer 1

◇ANSWER ◇

☆ Hartree-Fock theory is fundamental to much of electronic structure theory. It is the basis of

molecular orbital (MO) theory, which posits that each electron’s motion can be described by a

single-particle function (orbital) which does not depend explicitly on the instantaneous motions

of the other electrons.

》 The idea is to approximate the many-body wave function with a single Slater determinant. Because of the determinant, any non-orthogonal component of the orbitals is irrelevant and only the orthogonal part survives

》In other words, if you take a non-orthogonal set of orbitals and construct a Slater determinant out of those, you will get the same determinant if you would have first orthogonalised them (Gram-Schmidt, Löwdin, canonical, etc.). You still have the same volume.

》Generally one prefers to work with orthogonal orbitals, since this makes it easier to work out the expectation values of the determinants (e.g. energy) via the Slater-Condon rules. Generalisations also exist for non-orthogonal orbitals, but due to the cross-terms, you get overlap matrices, cofactors and adjugates all over the place. So in this sense, orthonormal orbitals are more a convenience than a necessity. Non-orthogonal orbitals would not add anything (the determinant remains the same).

◇ On the other hand, once you have used the Slater-Condon rules to work out the expectation value of the Hamiltonian, this energy expression is only valid for orthonormal orbitals. So you better enforce orthonormality in the optimisation, to get a physically sensible answer out. And All orbitals could become identical, so all the particles would occupy the same state. So you would get bosons out instead of fermions.

HOPE IT HELPS :)