= Solution
Let $P_{12}$ exchange the complete labels of the two <electrons>. The spatial factors obey $P_{12}\Phi_\pm=\pm\Phi_\pm$. The <singlet state> <spin> factor is antisymmetric, while each <triplet state> <spin> factor is symmetric. Thus in all four cases
$$
\boxed{P_{12}\Psi=-\Psi,}
$$
as required by the <Pauli exclusion principle>. In particular, the $M_S=0$ <triplet state> remains a triplet despite having opposite measured <spin> projections in each product term.
For a spin-independent <Hamiltonian> and the same fixed spatial <atomic orbitals>, direct evaluation gives
$$
E_{\rm singlet}=h_a+h_b+J_{ab}+K_{ab},\qquad
E_{\rm triplet}=h_a+h_b+J_{ab}-K_{ab},
$$
where $J_{ab}$ is the direct <Coulomb integral> and $K_{ab}>0$ is the <exchange integral> for the <helium> $1s,2s$ pair. Therefore
$$
\boxed{E_{\rm singlet}-E_{\rm triplet}=2K_{ab}>0.}
$$
The antisymmetric spatial <wavefunction> vanishes when the two positions coincide and reduces their Coulomb repulsion. The lower <energy> of the maximum-spin member of this fixed <electron configuration> is the first of <Hund's rules>.
Back to article page
