 Solution

ID: past-exam-of-the-quantum-information-course-of-the-university-of-cambridge/2015/paper-1/1/d/solution

Since

(Z_{D} \otimes X_{E})^{2} = I

U (t) = I_{S} \otimes [cos A I_{D E} - i sin A Z_{D} \otimes X_{E}], A (t) = g t /ℏ.

The two detector eigenvalues generate opposite rotations of the environment. Acting on the initial state gives

∣Ψ (t)⟩ = \frac{1}{2} [∣ ↑ d_{↑} ⟩ (cos A ∣ E_{↑} ⟩ - i sin A ∣ E_{↓} ⟩) + ∣ ↓ d_{↓} ⟩ (cos A ∣ E_{↑} ⟩ + i sin A ∣ E_{↓} ⟩)] .

These are the indicated

∣ E_{-} ⟩

and

∣ E_{+} ⟩

⟨ E_{+} ∣ E_{-} ⟩ = cos 2 A

. The partial trace therefore multiplies the coherence between

∣ ↑ d_{↑} ⟩

and

∣ ↓ d_{↓} ⟩

cos 2 A

. This is an exactly solvable single-qubit environment dephasing model, with fully reversible global evolution.

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