 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-302/4/vi/solution

For

J_{i}^{\pm} = (J_{i} \pm i K_{i}) /2

[J_{i}^{+}, J_{j}^{+}] = ϵ_{ijk} J_{k}^{+}, [J_{i}^{-}, J_{j}^{-}] = ϵ_{ijk} J_{k}^{-}, [J_{i}^{+}, J_{j}^{-}] = 0 .

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