 Solution

ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2023/3/a/ii/solution

(∣ \frac{1}{2}, \frac{1}{2} ⟩, ∣ \frac{1}{2}, - \frac{1}{2} ⟩)

J_{i} = (ℏ/2) σ_{i}

J_{1} = \frac{ℏ}{2} (0110), J_{2} = \frac{ℏ}{2} (0 i - i 0), J_{3} = \frac{ℏ}{2} (10 0 - 1) .

J_{+} = ℏ (0010), J_{-} = ℏ (0100) .

Explicit multiplication gives

J_{+} J_{-} = ℏ^{2} diag (1, 0)

and

J_{-} J_{+} = ℏ^{2} diag (0, 1)

. Their commutator is therefore

[J_{+}, J_{-}] = ℏ^{2} (10 0 - 1) = 2ℏ J_{3} .

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