 Solution

Suppress the fixed individual quantum numbers $s_1=s_2=1/2$. In the coupled basis, ordered on both rows and columns as $|1,1⟩,|1,0⟩,|1,-1⟩,|0,0⟩$, the total-spin eigenvalues giveHere the first three entries form the triplet state sector and the last is the singlet state. For the product basis, useThe first term has diagonal entries ${\hslash }^2(1,-1,-1,1)/4$. Each nonzero exchange term flips $|\uparrow \downarrow ⟩$ to $|\downarrow \uparrow ⟩$ or back, with coefficient ${\hslash }^2/2$. With rows and columns ordered as $|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩$,The change of basis is made explicit by $|1,0⟩=(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$ and $|0,0⟩=(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$, with $|1,\pm 1⟩=|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩$. The resulting eigenvalues reproduce the diagonal coupled matrix.