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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-2/10d/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 10D / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

Expanding either Bell state and using

⟨ 0∣1 ⟩_{B} = 0

removes the cross terms:

⟨ Φ^{\pm} ∣ Q \otimes I ∣ Φ^{\pm} ⟩ = \frac{1}{2} (⟨ 00∣ \pm ⟨ 11∣) (Q \otimes I) (∣00 ⟩ \pm ∣11 ⟩) = \frac{1}{2} (⟨ 0∣ Q ∣0 ⟩ + ⟨ 1∣ Q ∣1 ⟩) = \frac{1}{2} tr Q .

(1)

Thus the two expectation values agree. Equivalently, both states have the same reduced density matrix $I /2$ on qubit $A$ ; the positivity of $Q$ makes it a possible measurement effect but is not needed for the algebraic equality.

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