Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 35A / c / Solution

Using the states from part (a),The Born rule therefore givesAfter the equal-sign outcomes the state is respectively $|\uparrow \uparrow ⟩$ or $|\downarrow \downarrow ⟩$. These are triplet states, hence eigenstates of ${\mathbf{S}}^2$ with eigenvalueThe opposite-sign product states are each a nontrivial linear combination of the singlet and the $m=0$ triplet, so they are not eigenstates of ${\mathbf{S}}^2$.