Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2A / Section A / 17

Writing $A(\mathbf{r})={\varphi }_{1sA}(\mathbf{r})$ and $B(\mathbf{r})={\varphi }_{1sB}(\mathbf{r})$, the normalized determinant isIts two spin orbitals are orthogonal because $⟨\alpha |\beta ⟩=0$, even if the spatial atomic functions have nonzero overlap. Exchanging both the spatial and spin coordinates of the two electrons reverses the sign, as required for fermionic antisymmetry.

At this separation the orbital overlap and interatomic two-electron exchange integrals are negligible. Each localized electron then contributes essentially the isolated hydrogen energy, while the interatomic electrostatic terms cancel to leading order for neutral fragments. Including nuclear repulsion in the molecular energy givesThe closed-shell RHF determinant instead doubly occupies $g\simeq (A+B)/\sqrt{2}$. It retains ionic configurations with both electrons on the same atom, even at dissociation. Using atomic units and the Hartree energy in the stated unrelaxed hydrogenic 1s basis, the same-atom two-electron Coulomb integral is $J_{AA}=J_{BB}=5E_h/8$. Consequently at infinite separationwhere energies and integrals inside the first expression are in atomic units. It is much too high compared with two neutral atoms. Retaining the leading intersite Coulomb term at $R=10\text{A˚}=18.90a_0$ gives $E_{\mathrm{RHF}}\simeq -11/16-1/(2R)\simeq -0.714E_h$, still about $0.286E_h$ above UHF. The numerical RHF estimates specify the supplied fixed 1s basis; a larger optimized orbital space changes them. If nuclear repulsion is excluded by an electronic-energy convention, subtract the same $1/R$ from both quoted total molecular energies. The restricted/unrestricted comparison and their dissociation behaviour are unchanged.^[ref]

The fixed localized determinant is unsuitable near equilibrium. It does not develop the correct shared bonding orbital and is not a pure singlet: for one alpha and one beta orbital, $⟨S^2⟩/{\hslash }^2=1-|⟨A|B⟩{|}^2$, which is nonzero unless the occupied spatial orbitals coincide. At dissociation it is an equal singlet/triplet mixture, though their energies become degenerate there. Near equilibrium the physical ground state is a bonding singlet. An optimized UHF calculation can recover the restricted, spin-pure solution by making its two occupied spatial orbitals equal; that does not rescue the particular fixed atomic-orbital determinant written here. A spin-adapted multideterminant description is needed to combine good bonding near equilibrium with correct singlet dissociation.