 Solution

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

The phases of the three basis vectors can be chosen so that

A

and

B

[J_{+}, J_{-}] = diag (A^{2}, B^{2} - A^{2}, - B^{2}) .

2ℏ J_{3} = 2 ℏ^{2} diag (1, 0, - 1)

. Equating diagonal entries yields

A^{2} = 2 ℏ^{2}, B^{2} - A^{2} = 0, B^{2} = 2 ℏ^{2} .

Hence

A = B = 2 ℏ, J_{+} = 2 ℏ 000100010, J_{-} = 2 ℏ 010001000 .

These are also the

j = 1

ℏ j (j + 1) - μ (μ \pm 1)

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