Manuscript VII-F-ST@1 · VII-F-ST.tex · proposition · 9eee7a44

Proposition: Exact finite table probe

Finite destructive clock and an exact evaluation table › Four exact classical evaluations · lines 1292–1315 · proof included or linked

VII-F-ST@1#prop:table-probe, VII-F-ST@1#eq:table-probe

Back to manuscript

The resulting complete channel is

P (X) = a, y \sum ∣ a ⟩ ⟨ a ∣_{R} \otimes ∣ y ⟩ ⟨ y ∣_{Y} \otimes p_{y} (a) D_{a} X D_{a}^{†}, p_{1} (a) = q_{a}, p_{0} (a) = 1 - q_{a} .

It is trace preserving, leaves the payload unacted upon within each clock block, and satisfies

Pr (Y = 1 ∣ a) = q_{a} .

The formula holds with every reference.

Proof. Use Kraus operators $p_{y} (a) ∣ a ⟩_{R} ∣ y ⟩_{Y} D_{a}$ . Completeness is $\sum_{y} p_{y} (a) = 1$ followed by clock completeness. The probability and reference identities follow.

Displayed formulas in this passage

VII-F-ST@1#eq:table-probe · line 1295
$P (X) = a, y \sum ∣ a ⟩ ⟨ a ∣_{R} \otimes ∣ y ⟩ ⟨ y ∣_{Y} \otimes p_{y} (a) D_{a} X D_{a}^{†}, p_{1} (a) = q_{a}, p_{0} (a) = 1 - q_{a} .$
unlabeled · line 1304
$Pr (Y = 1 ∣ a) = q_{a} .$

Proposition: Exact finite table probe

Displayed formulas in this passage

Nearby in this section