Manuscript XVI-O1-ST@5 · XVI-O1-ST-v5.tex · theorem · dd45e0d1

Theorem: Complete destructive report

Complete destructive report instruments · lines 530–558 · proof included or linked

XVI-O1-ST@5#thm:instrument

The map (eq:instrument) is CPTP, and for every finite untouched reference $E$ ,

Tr_{R} [(M_{P} \otimes id_{E}) (X)] = Tr_{C} X .

For a state $ρ$ ,

σ_{j} = (D_{j} \otimes I_{E}) ρ (D_{j}^{†} \otimes I_{E}), p_{j} = Tr σ_{j} .

The positive-probability conditional state is $σ_{j} / p_{j}$ . No normalized state is assigned to a zero branch. The returned quantum interface excludes $C$ .

Proof. The Kraus maps $K_{j} = ∣ j, q_{j} ⟩ \otimes D_{j}$ satisfy

j \sum K_{j}^{†} K_{j} = j \sum ∣ τ_{j} ⟩ ⟨ τ_{j} ∣ \otimes I = I .

Writing $X = \sum_{j, k} ∣ τ_{j} ⟩ ⟨ τ_{k} ∣ \otimes X_{j k}$ , the output blocks are $X_{j j}$ , whose sum is $Tr_{C} X$ . The same calculation with $A_{lab} \otimes E$ proves the reference identity. Positivity supplies the branch rule.

Displayed formulas in this passage

unlabeled · line 534
$Tr_{R} [(M_{P} \otimes id_{E}) (X)] = Tr_{C} X .$
unlabeled · line 538
$σ_{j} = (D_{j} \otimes I_{E}) ρ (D_{j}^{†} \otimes I_{E}), p_{j} = Tr σ_{j} .$
unlabeled · line 549
$j \sum K_{j}^{†} K_{j} = j \sum ∣ τ_{j} ⟩ ⟨ τ_{j} ∣ \otimes I = I .$

Theorem: Complete destructive report

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