Separate nuclei; antisymmetrize electrons
Prerequisites: Variational principle; products and determinants
Prerequisite lessons: Variational principle and finite bases
Conventions
Use fixed nuclear coordinates \(R\), electronic coordinates \(x=(r,\sigma)\) with spin summation, and atomic units: energy in hartree, length in bohr. Molecular orbitals are one-electron functions used to build a many-electron state, not labeled electrons’ trajectories.
1. Separate electronic and nuclear terms
Start with the nonrelativistic Coulomb Hamiltonian. Solve the electronic problem at each chosen nuclear geometry; electronic energy plus nuclear repulsion defines a potential-energy surface. Expand the full state in geometry-dependent electronic eigenstates.
2. See exactly what Born–Oppenheimer drops
Nuclear kinetic energy differentiates both factors. The product rule produces derivative couplings; dropping them at leading order gives a single-surface nuclear equation. Heavy nuclei motivate the approximation, but a small electronic gap can make those couplings important. A crossing is not by itself a universal failure criterion.
3. Antisymmetrize complete electron coordinates
Electrons are identical fermions. Exchanging coordinates including spin must change the many-electron sign. For orthonormal spin orbitals, a normalized determinant enforces this: swapping two electron columns changes sign; duplicate occupied orbital rows give zero.
Teaching schematic drawn from the equations or algorithm steps in this lesson; not measured data.
Worked example: shared spatial orbital
Two electrons can occupy one spatial orbital with opposite spin because the spin orbitals differ. The spatial part is symmetric and the singlet spin part antisymmetric, so their product has the required total antisymmetry.
Check yourself and limits
Is nuclear repulsion omitted from molecular total energy?
Answer and reasoning
No; \(V_{NN}\) is necessary even if constant at one geometry.
Does a determinant capture arbitrary electron correlation?
Answer and reasoning
No; it restricts the many-electron ansatz.
Does heavier nuclear mass guarantee BO validity?
Answer and reasoning
No; the electronic gap and derivative couplings also matter.
References and further reading
Derivations and numerical examples here are original teaching constructions, not copied passages or reported research data.