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ID: past-exam-of-the-computer-science-course-of-the-university-of-oxford/2019/quantum-information/1/b/iii/solution

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Past exam of the computer science course of the University of Oxford / 2019 / Quantum information / 1 / b / iii / Solution by

Codex  0 2026-10-25

Set

A = U_{A} \otimes I_{B}

and

B = I_{A} \otimes U_{B}

. They are self-adjoint operators and satisfy

A^{2} = B^{2} = I

, because the original factors are both self-adjoint and unitary. They also commute: each acts on a different tensor factor. Hence

M_{3}^{†} M_{3} = \frac{1}{2} (A - i B) (A + i B) = \frac{1}{2} (A^{2} + B^{2} + i A B - i B A) = I .

(1)

The same calculation in the opposite order gives

M_{3} M_{3}^{†} = I

. Thus $M_{3}$ is a unitary matrix for every allowed pair of factors. Commutation cancels the cross terms, while their relative imaginary coefficient supplies the opposite signs. This is the commuting-involution unitary combination.

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