Manuscript Oai276-A@1 · Oai276-A.tex · proposition · 215068af

Proposition: Readout-and-drop and record-only recovery

Stage 2: implementation, readout, and statistical pricing › Destructive readout is not an unknown-state decoder · lines 1344–1376 · proof included or linked

Oai276-A@1#prop:readout

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For an orthonormal basis ${∣ z ⟩}$ of a measured register, the destructive readout channel

M (X) = z \sum ∣ z ⟩ ⟨ z ∣_{R} \otimes (⟨ z ∣ \otimes I) X (∣ z ⟩ \otimes I)

satisfies

Tr_{R} M (X) = Tr_{measured} X,

including after every untouched reference extension.

If the only input-dependent output available to a quantum decoder is a classical record, the decoded channel is measure-and-prepare and cannot equal the identity on a code of dimension at least two.

Proof. Summing the diagonal input blocks is exactly the partial trace. The same block calculation includes a reference.

A classical-record decoder has form

Λ (ρ) = z \sum Tr (E_{z} ρ) σ_{z} .

On half of a maximally entangled input its output is separable. The identity preserves that entanglement, so the two channels cannot coincide.

Displayed formulas in this passage

unlabeled · line 1348
$M (X) = z \sum ∣ z ⟩ ⟨ z ∣_{R} \otimes (⟨ z ∣ \otimes I) X (∣ z ⟩ \otimes I)$
unlabeled · line 1354
$Tr_{R} M (X) = Tr_{measured} X,$
unlabeled · line 1370
$Λ (ρ) = z \sum Tr (E_{z} ρ) σ_{z} .$

Nearby in this section

section: Destructive readout is not an unknown-state decoder