Manuscript VIII-A@3 · VIII-A-v3.tex · example · 2398a747

Example: Subsystem recovery does not require a constant full complement

Recovery distinctions used only as interface safeguards › Full-code recovery and the complete complement · lines 2705–2712

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Let $N = Tr_{S}$ on $H_{L} \otimes H_{S}$ . The logical subsystem $H_{L}$ is returned exactly. A complementary channel returns the $S$ -reduction, which is not constant when $dim H_{S} > 1$ . The preceding theorem concerns a left inverse on the whole $H_{L} \otimes H_{S}$ , not this subsystem task.

Nearby in this section

theorem: Theorem: Full-code constant-complement criterion
section: Full-code recovery and the complete complement