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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-51/4/e/solution

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Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 51 / 4 / e / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

Take a normalized environment quantum state

∣ ψ_{E} ⟩ = 1 - ϵ ∣0 ⟩ + ϵ ∣1 ⟩ .

The Controlled-Z gate is symmetric between the two qubits, so it can be written with the environment as the controlling system:

C_{Z} = I \otimes ∣0 ⟩ ⟨ 0∣ + Z \otimes ∣1 ⟩ ⟨ 1∣.

Expanding the joint density operator after the interaction gives

C_{Z} (ρ \otimes ∣ ψ_{E} ⟩ ⟨ ψ_{E} ∣) C_{Z} = (1 - ϵ) ρ \otimes ∣0 ⟩ ⟨ 0∣ + ϵ ZρZ \otimes ∣1 ⟩ ⟨ 1∣ + ϵ (1 - ϵ) (ρZ \otimes ∣0 ⟩ ⟨ 1∣ + Zρ \otimes ∣1 ⟩ ⟨ 0∣) .

In the partial trace, the two off-diagonal environment operators have trace zero, while each diagonal projector has trace one. Hence

tr_{E} [C_{Z} (ρ \otimes ∣ ψ_{E} ⟩ ⟨ ψ_{E} ∣) C_{Z}] = (1 - ϵ) ρ + ϵ ZρZ = D_{ϵ} (ρ) .

Equivalently, the Kraus operators are

K_{0} = 1 - ϵ I

and

K_{1} = ϵ Z

, and

K_{0}^{†} K_{0} + K_{1}^{†} K_{1} = I

. This explicit Stinespring dilation describes the system's dephasing by discarding information in the environment. A fresh environment in this state at each storage interval also realizes the memoryless iteration used in part (a).

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