 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-2/35b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

ρ = ρ^{†}, ρ ⪰ 0, tr ρ = 1.

N \times N

Hermitian matrix has

N^{2}

real parameters, and the trace condition removes one, leaving

N^{2} - 1

A_{H} (t) = U (t)^{†} A_{S} U (t)

, equality of expectation values requires

tr (ρ_{H} U^{†} A_{S} U) = tr (U ρ_{H} U^{†} A_{S}) .

Therefore

ρ_{S} (t) = U (t) ρ_{H} U (t)^{†} .

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