 Degenerate orbital exchange splitting from a product perturbation

Two distinguishable particles with equal one-particle spectra have degenerate ordered orbital products $|r,s⟩$ and $|s,r⟩$ for $r\ne s$. Suppose both individual position means vanish, as in a centred parity-symmetric well. The perturbation $\alpha x_1{x}_2$ then has zero diagonal matrix elements, but its restriction to this two-dimensional subspace isDegenerate perturbation theory gives normalized symmetric and antisymmetric eigenfunctions $(|r,s⟩\pm |s,r⟩)/\sqrt{2}$ with shifts $\pm \alpha |x_{rs}{|}^2$. Thus vanishing diagonal expectations in a chosen degenerate basis do not imply vanishing first-order energy shifts. Distinguishability permits both exchange symmetries; it does not forbid their energy splitting.

For an infinite square well of width $a$, the first two orbitals have $|x_{12}{|}^2=256a^2/(81{\pi }^4)$ by the position matrix elements in a symmetric infinite square well. If the particles are identical spin one-half fermions, the spatially symmetric and antisymmetric combinations must instead be paired with the singlet state and triplet states, respectively. A spin-independent perturbation leaves each allowed spin multiplicity intact.