Manuscript En10-A@1 · En10-A.tex · proposition · 0a1d7743

Proposition: Record-only quantum-recovery obstruction

Record-only recovery and the correct residual distinction › No unrestricted quantum return from the record alone · lines 1050–1114 · proof included or linked

En10-A@1#prop:no-return, En10-A@1#eq:no-return-bound

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Let $R : L (H_{C}) \to L (H_{B})$ be any CPTP map whose only input is the classical record. Then neither $R M_{B}$ nor $R I$ is the identity channel on an unrestricted four-dimensional input. In fact,

d_{⋄} (R M_{B}, id_{B}) \geq \frac{3}{4}, d_{⋄} (R I, id_{B}) \geq \frac{3}{4} .

Proof. On diagonal record inputs, any such map has the form

R (r \sum p_{r} ∣ r ⟩ ⟨ r ∣_{C}) = r \sum p_{r} σ_{r}, σ_{r} = R (∣ r ⟩ ⟨ r ∣_{C}) .

Consequently,

(R I) (X) = r \sum Tr (E_{r} X) σ_{r}, E_{r} = ℓ \sum H_{ℓ}^{†} P_{r} H_{ℓ} .

The effects $E_{r}$ are positive and sum to $I_{B}$ . The same expression with $E_{r} = P_{r}$ describes $R M_{B}$ .

Such a measure-and-prepare channel has separable output across an untouched-reference/output cut. This is the entanglement-breaking structure studied in HorodeckiShorRuskai2003; here it follows immediately from the displayed form.

Take a four-dimensional reference and

∣ Ω_{4} ⟩ = \frac{1}{2} j = 0 \sum^{3} ∣ j ⟩_{Q} ∣ j ⟩_{B} .

For any product state $τ_{Q} \otimes σ_{B}$ ,

⟨ Ω_{4} ∣ (τ_{Q} \otimes σ_{B}) ∣ Ω_{4} ⟩ = \frac{1}{4} Tr (τ_{Q}^{T} σ_{B}) \leq \frac{1}{4} .

The same bound holds for every separable state by convexity. The identity channel preserves $∣ Ω_{4} ⟩ ⟨ Ω_{4} ∣$ , whose overlap with itself is one. Testing its projector therefore gives trace distance at least $1 - 1/4 = 3/4$ . This particular referenced input lower-bounds the half-diamond distance, proving both inequalities.

Displayed formulas in this passage

En10-A@1#eq:no-return-bound · line 1058
$d_{⋄} (R M_{B}, id_{B}) \geq \frac{3}{4}, d_{⋄} (R I, id_{B}) \geq \frac{3}{4} .$
unlabeled · line 1068
$R (r \sum p_{r} ∣ r ⟩ ⟨ r ∣_{C}) = r \sum p_{r} σ_{r}, σ_{r} = R (∣ r ⟩ ⟨ r ∣_{C}) .$
unlabeled · line 1076
$(R I) (X) = r \sum Tr (E_{r} X) σ_{r}, E_{r} = ℓ \sum H_{ℓ}^{†} P_{r} H_{ℓ} .$
unlabeled · line 1096
$∣ Ω_{4} ⟩ = \frac{1}{2} j = 0 \sum^{3} ∣ j ⟩_{Q} ∣ j ⟩_{B} .$
unlabeled · line 1102
$⟨ Ω_{4} ∣ (τ_{Q} \otimes σ_{B}) ∣ Ω_{4} ⟩ = \frac{1}{4} Tr (τ_{Q}^{T} σ_{B}) \leq \frac{1}{4} .$

Proposition: Record-only quantum-recovery obstruction

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