Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 1 / 33C / Solution

The unique maximum-$M$ product state isApply the total lowering operator $J_{-}=J_{1-}+J_{2-}$ and normalize to obtain that multiplet. At the next highest remaining $M$, choose a normalized vector orthogonal to all previously constructed multiplets and annihilated by $J_{+}$. It is the next highest-weight state; lower it in the same way. Iterating produces every allowed total angular momentum and the Clebsch-Gordan coefficients.

For equal constituent spins, a total-spin singlet has $M=0$, so only the product basis vectors with opposite magnetic quantum numbers can occur. Acting with $J_{+}$ on the expansion and examining the coefficient of $|j,m+1{⟩}_1|j,-m{⟩}_2$ givesThus ${\alpha }_{m+1}=-{\alpha }_m$. Normalization requires $(2j+1)|{\alpha }_m{|}^2=1$. Choosing a conventional overall phase givesThe exponent is an integer even for half-integral $j$. Projecting the specified extreme product state onto this unique singlet gives the measurement probability