Past exam of the quantum information course of the University of Cambridge / 2015 / Paper 1 / 1 / e

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- Solution e

Solution

A = π /4

, define

∣ E_{\leftarrow} ⟩ = \frac{∣ E _{↑} ⟩ - i ∣ E _{↓} ⟩}{2}, ∣ E_{\to} ⟩ = \frac{∣ E _{↑} ⟩ + i ∣ E _{↓} ⟩}{2} .

(1)

They are normalized and orthogonal. The full state is the pure superposition

(∣ ↑ d_{↑} E_{\leftarrow} ⟩ + ∣ ↓ d_{↓} E_{\to} ⟩) / 2

. A nonselective von Neumann process 1 selecting the original detector records gives

ρ_{S D E}^{(1)} = \frac{1}{2} (∣ ↑ d_{↑} E_{\leftarrow} ⟩ ⟨ ↑ d_{↑} E_{\leftarrow} ∣ + ∣ ↓ d_{↓} E_{\to} ⟩ ⟨ ↓ d_{↓} E_{\to} ∣) .

(2)

Without imposing any projection on the full state, tracing out the orthogonal environment states already gives

ρ_{S D} = ρ_{up/down}

from part (a). These two global descriptions share that reduced state but differ in their global coherence. Quantum decoherence does not make a pure global state literally equal to the corresponding statistical mixture.

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Past exam of the quantum information course of the University of Cambridge / 2015 / Paper 1 / 1 / e

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