Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 33C / i / Solution

Addition of angular momenta one and $j$ gives $\boxed{J=|j-1|,|j-1|+1,\dots ,j+1}$, with projection condition $\boxed{M=q+m}$. The usual bounds $q\in \{-1,0,1\}$, $-j\le m\le j$, $-J\le M\le J$ must also hold. For $j=0$ only $J=1$ occurs; for $j=1/2$ the values are $3/2,1/2$.

For $j=1/2$, abbreviate the product state $|1,q⟩|1/2,m⟩$ by $|q,\uparrow ⟩$ or $|q,\downarrow ⟩$. The highest-weight state is $|3/2,3/2⟩=|1,\uparrow ⟩$. Applying the total lowering operator, the sum of the two individual lowering operators, and dividing by the required ladder coefficient gives the whole quartet:The normalized vector orthogonal to the $M=1/2$ quartet state is annihilated by total raising, hence is the highest state of the remaining spin-$1/2$ representation. Lowering it givesThese six orthonormal states span the six-dimensional product space, and are therefore all the states. Their relative signs are consistent with the lowering convention; a common phase for either multiplet is immaterial. The coefficients are the corresponding Clebsch-Gordan coefficients.