 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1/3/b/solution

Separate the defining matrix exponential into even and odd powers and use

τ_{j}^{2} = I_{2}

e^{i α_{j} τ_{j}} = I_{2} \sum_{m = 0}^{\infty} \frac{( - 1 ) ^{m} α _{j}^{2 m}}{( 2 m )!} + i τ_{j} \sum_{m = 0}^{\infty} \frac{( - 1 ) ^{m} α _{j}^{2 m + 1}}{( 2 m + 1 )!} = I_{2} cos α_{j} + i τ_{j} sin α_{j} .

Therefore

A_{1} = (cos α_{1} i sin α_{1} i sin α_{1} cos α_{1}), A_{2} = (cos α_{2} - sin α_{2} sin α_{2} cos α_{2}),

A_{3} = (e^{i α_{3}} 0 0 e^{- i α_{3}}) .

Since each

τ_{j}

A_{j}^{†} = cos α_{j} I_{2} - i sin α_{j} τ_{j} = A_{j}^{- 1}

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