Manuscript X-B-ST@1 · X-B-ST.tex · theorem · d03c150c

Theorem: Complete destructive-clock instrument

Installed evaluations and a complete clock instrument › Clock extraction · lines 2316–2349 · proof included or linked

X-B-ST@1#thm:instrument, X-B-ST@1#eq:clock

The channel

M_{C} (X) = k \sum ∣ k ⟩ ⟨ k ∣_{R} \otimes D_{k} X D_{k}^{†} = k \sum ∣ k ⟩ ⟨ k ∣_{R} \otimes Tr_{C} [(P_{k} \otimes I_{S}) X (P_{k} \otimes I_{S})]

is CPTP and satisfies

Tr_{R} M_{C} (X) = Tr_{C} X .

These equalities hold after adjoining every untouched reference. The quantum clock names are consumed; there is no quantum $C$ -output.

Proof. The recorded Kraus operators are $∣ k ⟩_{R} \otimes D_{k}$ . Their squared moduli sum to

k \sum P_{k} \otimes I_{S} = I_{C S} .

Thus the map is completely positive and trace preserving.

Expand $X = \sum_{k, ℓ} ∣ τ_{k} ⟩ ⟨ τ_{ℓ} ∣ \otimes X_{k ℓ}$ . Each output block is $X_{k k}$ , which proves both formulae and the discard identity. The same calculation with $S$ enlarged by any reference proves the complete statement.

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X-B-ST@1#eq:clock · line 2319
$M_{C} (X) = k \sum ∣ k ⟩ ⟨ k ∣_{R} \otimes D_{k} X D_{k}^{†} = k \sum ∣ k ⟩ ⟨ k ∣_{R} \otimes Tr_{C} [(P_{k} \otimes I_{S}) X (P_{k} \otimes I_{S})]$
unlabeled · line 2327
$Tr_{R} M_{C} (X) = Tr_{C} X .$
unlabeled · line 2338
$k \sum P_{k} \otimes I_{S} = I_{C S} .$

Theorem: Complete destructive-clock instrument

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