= Degenerate orbital exchange splitting from a product perturbation
{title2=$\delta E_\pm=\pm\alpha|\langle r|x|s\rangle|^2$}
Two distinguishable particles with equal one-particle spectra have degenerate ordered orbital products $|r,s\rangle$ and $|s,r\rangle$ for $r\ne s$. Suppose both individual position means vanish, as in a centred parity-symmetric well. The perturbation $\alpha x_1x_2$ then has zero diagonal <matrix elements>, but its restriction to this two-dimensional subspace is
$$
W=\alpha\begin{pmatrix}0&|x_{rs}|^2\\|x_{rs}|^2&0\end{pmatrix},\qquad x_{rs}=\langle r|x|s\rangle.
$$
<Degenerate perturbation theory> gives normalized symmetric and antisymmetric <eigenfunctions> $(|r,s\rangle\pm|s,r\rangle)/\sqrt2$ 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.
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