# Why is a LCAO necessary within Hartree Fock?

As I understand it, the electronic Schrödinger equation cannot be solved for polyelectronic systems. To circumvent this problem in the Hartree-Fock method, it is assumed that the polyelectronic wavefunction can be written as a combination of single-electron, molecular orbitals. To satisfy the anti-symmetry requirement, this combination takes the form of a Slater determinant.

What is the value of then going on to express these molecular orbitals as a linear combination of atomic orbitals? Would a Hamiltonian written to describe a single electron in the molecular environment (a molecular orbital) not lead to an analytically solvable Schrödinger equation?

• (1) The value is that you know what the atomic orbitals are. (2) n-body problems are not analytically solvable (except in trivial cases). – Jon Custer May 6 '19 at 13:02
• LCAO is not necessary for HF! There are alternatives like plane waves or representing the orbitals on a spatial grid. You may even make up your own functions as a basis of the orbitals. Atomic orbitals are simply an established choice because it works well. – Feodoran May 7 '19 at 8:02

$$\hat{1} = \sum_i |i\rangle\langle i| \quad \Longleftrightarrow \quad |f\rangle = \hat{1}|f\rangle = \sum_i |i\rangle\langle i | f \rangle = \sum_i c_i|i\rangle \text{ where }c_i = \langle i | f \rangle$$