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Let $X$ be a unitary operator with $X^2=I$, and let $P$ be the orthogonal projection operator onto an unperturbed energy eigenspace. Suppose $X$ acts as the same sign $\epsilon =\pm 1$ throughout that eigenspace and $XVX=-V$. For any $u,v$ in the eigenspace, preservation of the inner product gives $⟨u,Vv⟩=⟨Xu,XVv⟩=-⟨\epsilon u,V\epsilon v⟩=-⟨u,Vv⟩$. Hence every projected matrix element vanishes, so $PVP=0$ and degenerate perturbation theory gives zero first-order energy corrections. A zero diagonal alone would be insufficient. For equal spins, the exchange parity of two coupled identical spins is constant on each total-spin sector, while their spin-projection difference changes sign under exchange.

Put $\mathbf{J}={\mathbf{S}}_1+{\mathbf{S}}_2$. Angular momentum addition gives $j=0,1,2,3$, and the spin-dot-product eigenvalue isThus the spin-three-halves pair spectrum isFor identical fermions, exchange parity of two coupled identical spins makes $j=0,2$ spin-antisymmetric and $j=1,3$ spin-symmetric. Corresponding spatial exchange symmetries make the full state antisymmetric; without a specified spatial state, none of these total-spin sectors is excluded.

Let $X$ exchange the two equal-spin factors, so $X^{†}=X=X^{-1}$. For the unperturbed Hamiltonian operator, addition of angular momentum givesFor nonzero $\alpha$, different $J$ have distinct energy eigenvalues, while each eigenspace has the $2J+1$ values of $M$. The stated symmetry of the Clebsch-Gordan coefficients gives the exchange parity of two coupled identical spins:However, $D=(S_3^{(1)}-S_3^{(2)})/\hslash$ changes sign: $XDX=-D$. For any two basis vectors in the same $J$ eigenspace, insertion of $X^2=I$ therefore givesThus the entire projected perturbation $P_J\beta D{P}_J$ is zero, including its off-diagonal matrix elements. Degenerate perturbation theory now proves that every first-order-in-$\beta$ energy correction vanishes. This is first-order cancellation for an exchange-odd perturbation; checking only the individual diagonal expectations would not by itself settle a degenerate problem.