Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 33B / a / iii / Solution

Put $J=j_1+j_2$. The only state with $m=J$ is the maximal highest-weight stateApplying $J_{-}=J_{1-}+J_{2-}$ and normalizing givesThe orthonormal vector in the same two-dimensional $m=J-1$ subspace isIt is annihilated by $J_{+}$, so it is the highest-weight state of the $j=J-1$ multiplet. These are all states with $m\ge J-1$. The last state exists when both subsystem spins are nonzero.