 Solution

ID: past-exam-of-the-quantum-physics-course-of-the-university-of-oxford/2016/7/b/solution

Expanding the total angular momentum and replacing transverse components by ladder operators gives

S_{z} = s_{1 z} + s_{2 z}, S^{2} = s_{1}^{2} + s_{2}^{2} + 2 s_{1 z} s_{2 z} + s_{1 +} s_{2 -} + s_{1 -} s_{2 +} .

P_{12}^{2} = I

and

P_{12} s_{1 α} P_{12} = s_{2 α}

, with the reverse relation for index

2

. These statements also apply to the ladder operators. Exchanging the indices preserves

s_{1}^{2} + s_{2}^{2}

and

s_{1 z} + s_{2 z}

; it interchanges the last two terms in

S^{2}

. Operators on different tensor factors commute. Hence

P_{12} S^{2} P_{12} = S^{2}, P_{12} S_{z} P_{12} = S_{z}, [P_{12}, S^{2}] = [P_{12}, S_{z}] = 0.

The full particle exchange includes the spin coordinates as well as the spatial coordinates.

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