With normalized states and $J_-^\dagger=J_+$,
$$
J_-|I,m\rangle
=\sqrt{(I+m)(I-m+1)}\,|I,m-1\rangle.
$$
Put $n=I-m$. Iterating from the highest-weight state gives
$$
(d(J_-))^n|I,I\rangle
=\sqrt{\prod_{r=0}^{n-1}(2I-r)(r+1)}\,|I,m\rangle
=\sqrt{\frac{(2I)!(I-m)!}{(I+m)!}}\,|I,m\rangle.
$$
Hence
$$
A(I,m)=\sqrt{\frac{(I+m)!}{(2I)!(I-m)!}},
$$
where the factorial arguments are integers because $I\pm m\in\mathbb Z_{\geq0}$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-302/2/b/solution