Manuscript Hawk04-C@1 · Hawk04-C-v1.tex · example · 94d46352

Example: A record can assist a surviving quantum system

Row 5: the final-burst and no-remnant analogy › Keeping the residual record changes the access task · lines 1189–1221 · proof included or linked

Hawk04-C@1#ex:assistance

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For a qubit, let

V ∣ ψ ⟩ = \frac{1}{2} (∣ ψ ⟩_{R} ∣ 0 ⟩_{B} + Z ∣ ψ ⟩_{R} ∣ 1 ⟩_{B}),

where $Z = diag (1, - 1)$ . Before residual readout,

N (X) = \frac{1}{2} (X + Z X Z) .

Reading $B$ gives outcome $x$ with probability $1/2$ and branch map $X \mapsto Z^{x} X Z^{x} /2$ . A decoder with $R$ and the record applies $Z^{x}$ and recovers exactly. Dropping the record leaves the original dephasing channel.

Proof. The two residual basis states are orthogonal, so $V^{†} V = I$ . Taking their partial trace gives the displayed channel. The measurement bra $⟨ x ∣_{B}$ extracts $Z^{x} ∣ ψ ⟩ / 2$ . Since $(Z^{x})^{2} = I$ , correction followed by summing the two branches returns $X$ . Every calculation is an operator identity and retains references.

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unlabeled · line 1192
$V ∣ ψ ⟩ = \frac{1}{2} (∣ ψ ⟩_{R} ∣ 0 ⟩_{B} + Z ∣ ψ ⟩_{R} ∣ 1 ⟩_{B}),$
unlabeled · line 1204
$N (X) = \frac{1}{2} (X + Z X Z) .$

Example: A record can assist a surviving quantum system

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