Manuscript Hawk05-A@1 · Hawk05-A.tex · theorem · 449f85bb

Theorem: Readout-and-drop identity

Destructive Enter, hybrid recovery, and readout-and-drop › Terminal readout followed by discard · lines 799–828 · proof included or linked

Hawk05-A@1#thm:drop, Hawk05-A@1#eq:drop

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For every operator $X_{A M}$ ,

(id_{A} \otimes dropBit_{Y}) (id_{A} \otimes Enter_{M}) (X_{A M}) = Tr_{M} X_{A M} .

The identity remains valid with every untouched reference. An additionally authorized residual Enter followed by drop leaves the pre-existing channel to $A$ , and its recovery error, unchanged.

Proof. Write

X_{A M} = y, z \sum X_{y z} \otimes ∣ y ⟩ ⟨ z ∣_{M} .

Readout retains only the diagonal blocks:

(id_{A} \otimes Enter_{M}) (X_{A M}) = y \sum X_{y y} \otimes ∣ y ⟩ ⟨ y ∣_{Y} .

Dropping $Y$ gives $\sum_{y} X_{y y} = Tr_{M} X_{A M}$ . The calculation holds for operator-valued blocks on $Q A$ . Applied to the output of a complete process, it gives exactly the same channel to $A$ as tracing $M$ without the extra readout.

Displayed formulas in this passage

Hawk05-A@1#eq:drop · line 802
$(id_{A} \otimes dropBit_{Y}) (id_{A} \otimes Enter_{M}) (X_{A M}) = Tr_{M} X_{A M} .$
unlabeled · line 816
$X_{A M} = y, z \sum X_{y z} \otimes ∣ y ⟩ ⟨ z ∣_{M} .$
unlabeled · line 820
$(id_{A} \otimes Enter_{M}) (X_{A M}) = y \sum X_{y y} \otimes ∣ y ⟩ ⟨ y ∣_{Y} .$

Theorem: Readout-and-drop identity

Displayed formulas in this passage

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