Manuscript X-L-ST@1 · X-L-ST.tex · theorem · 20a32a22

Theorem: Complete residual-report readout

Recovery of complete residual-report instruments › The destructive instrument · lines 1534–1569 · proof included or linked

X-L-ST@1#thm:instrument, X-L-ST@1#eq:instrument

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The channel

M_{C} (X) = k \sum ∣ k ⟩ ⟨ k ∣_{R} \otimes D_{k} X D_{k}^{†}

is CPTP and equals

k \sum ∣ k ⟩ ⟨ k ∣_{R} \otimes Tr_{C} [(P_{k} \otimes I_{S}) X (P_{k} \otimes I_{S})] .

It satisfies

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

for every untouched reference $E$ .

The quantum clock names are consumed. A positive-probability block is normalized only by dividing by its own trace; no normalized state is assigned to a zero-probability branch.

Proof. The recorded Kraus maps are $∣ k ⟩_{R} \otimes D_{k}$ , and their squared moduli sum to $\sum_{k} P_{k} \otimes I_{S} = I_{C S}$ . Expansion of $X$ in the calibration basis shows that both formulas return its $k k$ -block. Summing gives the partial trace. Tensoring the Kraus maps with $I_{E}$ proves the reference statement. The codomain has no quantum $C$ -factor.

Displayed formulas in this passage

X-L-ST@1#eq:instrument · line 1537
$M_{C} (X) = k \sum ∣ k ⟩ ⟨ k ∣_{R} \otimes D_{k} X D_{k}^{†}$
unlabeled · line 1543
$k \sum ∣ k ⟩ ⟨ k ∣_{R} \otimes Tr_{C} [(P_{k} \otimes I_{S}) X (P_{k} \otimes I_{S})] .$
unlabeled · line 1548
$Tr_{R} [(M_{C} \otimes id_{E}) (X)] = Tr_{C} X$

Theorem: Complete residual-report readout

Displayed formulas in this passage

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