= Solution
Each one-particle <energy eigenstate> has definite parity about the centre of the well, so $\langle n|x|n\rangle=0$. Consequently, a doubly occupied orbital has $\langle x_1x_2\rangle=0$. For distinct orbitals, the two off-diagonal exchange terms instead survive:
$$
\langle\Phi_{rs}^\pm|x_1x_2|\Phi_{rs}^\pm\rangle
=\pm\langle r|x|s\rangle\langle s|x|r\rangle
=\pm|x_{rs}|^2.
$$
The <position matrix elements in a symmetric infinite square well>, with the present width $a$, give
$$
x_{12}=\int_{-a/2}^{a/2}u_1(x)x u_2(x)\,dx=-\frac{16a}{9\pi^2},
\qquad D=|x_{12}|^2=\frac{256a^2}{81\pi^4}.
$$
The sign of $x_{12}$ depends on the orbital phase convention, while $D$ does not. The perturbation acts only on the spatial part and cannot mix orthogonal <singlet states> and <triplet states>. Within a triplet it is proportional to the identity on the spin factor. Thus <first-order nondegenerate perturbation theory>, with this trivial diagonal treatment of spin degeneracy, gives the shifts corresponding to the three rows in part (b):
$$
\boxed{\begin{array}{c|c}
\text{state or multiplet}&\delta E^{(1)}\\ \hline
\Phi_{11}^+\chi_{00}&0\\
\Phi_{12}^-\chi_{1M}&-\alpha D\\
\Phi_{12}^+\chi_{00}&+\alpha D
\end{array}\qquad D=\frac{256a^2}{81\pi^4}.}
$$
These formulae also apply for $J<0$, with the reordered levels described above. At $J=0$, they resolve the fourfold $5\varepsilon$ level into a shifted triplet and singlet. The next doubly occupied orbital $\Phi_{22}^+$ still has zero first-order shift. The perturbative requirement is $|\alpha|a^2\ll\varepsilon$ for mixing with remote orbital states; the spin-independent perturbation does not require division by the small singlet–triplet gap.
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