Manuscript Oai280-C@1 · Oai280-C.tex · theorem · 7d44b8c0

Theorem: Connected subsystem and full-bulk recovery

One connected two-cell encoding › Triple patch decoders · lines 783–826 · proof included or linked

Oai280-C@1#thm:patch, Oai280-C@1#eq:subsystemexact

Let

N_{A}^{W} (ρ) = Tr_{A ˉ} (W ρ W^{†}) .

If $A$ contains a left exterior triple $K$ , then

D_{K} (σ) = Tr_{e, h} (U_{K}^{†} σ U_{K})

satisfies

D_{K} N_{K}^{W} = Tr_{S} .

The equality holds for arbitrary $L S$ inputs, including spectator correlations and arbitrary external references.

If $A$ contains a triple on each side, applying both inverse unitaries and discarding their joining and auxiliary registers recovers the entire two-qubit bulk exactly.

Proof. Trace the entire right exterior output in (eq:W). Because $F_{R}$ is isometric, this is equivalent to tracing $S, e^{'}$ . The left joining register is then $I_{e} /2$ , independent of the remaining state on $L$ and its reference. Applying Lemma lem:patchidentity gives

N_{K}^{W} (ρ) = U_{K} (Tr_{S} (ρ) \otimes I_{e} /2 \otimes I_{h} /2) U_{K}^{†} .

This is an operator identity and proves (eq:subsystemexact).

For two triples, the two inverse unitaries return

ρ_{L S} \otimes Φ_{e e^{'}} \otimes I_{h_{L}} /2 \otimes I_{h_{R}} /2.

Tracing the indicated registers returns $ρ_{L S}$ . All steps commute with adjoining an untouched reference.

Displayed formulas in this passage

Oai280-C@1#eq:subsystemexact · line 794
$D_{K} N_{K}^{W} = Tr_{S} .$
unlabeled · line 786
$N_{A}^{W} (ρ) = Tr_{A ˉ} (W ρ W^{†}) .$
unlabeled · line 790
$D_{K} (σ) = Tr_{e, h} (U_{K}^{†} σ U_{K})$
unlabeled · line 812
$N_{K}^{W} (ρ) = U_{K} (Tr_{S} (ρ) \otimes I_{e} /2 \otimes I_{h} /2) U_{K}^{†} .$
unlabeled · line 821
$ρ_{L S} \otimes Φ_{e e^{'}} \otimes I_{h_{L}} /2 \otimes I_{h_{R}} /2.$

Theorem: Connected subsystem and full-bulk recovery

Displayed formulas in this passage

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