Manuscript Oai280-B@1 · Oai280-B.tex · theorem · 6d92ab75

Theorem: Exact complete flagged-erasure optimum

An exactly solved flagged-erasure benchmark · lines 1844–1880 · proof included or linked

Oai280-B@1#thm:erasure, Oai280-B@1#eq:erasureoptimum

Let

p_{m} (q) = s = 3 \sum^{m} (s m) q^{s} (1 - q)^{m - s},

with $p_{m} (q) = 0$ for $m \leq 2$ . The displayed complete decoder induces

Λ_{A, q} = p_{m} (q) id_{2} + (1 - p_{m} (q)) T, T (M) = Tr (M) I_{2} /2.

Moreover,

D in f d_{⋄} (D N_{A, q}^{flag}, id_{2}) = d_{⋄} (Λ_{A, q}, id_{2}) = \frac{3}{4} (1 - p_{m} (q)) .

Proof. The surviving-set branch is the reduction of the same $V$ . Authorized branches return identity and the other branches return depolarization, proving the complete channel formula.

For any decoder, each branch with $∣ S ∣ \leq 2$ has a replacement input and decoded Bell overlap $1/4$ . Any other branch has Bell overlap at most one. Even a decoder acting on the full flagged direct-sum interface therefore has Bell overlap at most

p_{m} (q) + (1 - p_{m} (q)) /4.

The Bell-projector effect gives the lower bound $3 (1 - p_{m} (q)) /4$ . The displayed channel attains it by the depolarizing Pauli decomposition.

Displayed formulas in this passage

Oai280-B@1#eq:erasureoptimum · line 1857
$D in f d_{⋄} (D N_{A, q}^{flag}, id_{2}) = d_{⋄} (Λ_{A, q}, id_{2}) = \frac{3}{4} (1 - p_{m} (q)) .$
unlabeled · line 1847
$p_{m} (q) = s = 3 \sum^{m} (s m) q^{s} (1 - q)^{m - s},$
unlabeled · line 1852
$Λ_{A, q} = p_{m} (q) id_{2} + (1 - p_{m} (q)) T, T (M) = Tr (M) I_{2} /2.$
unlabeled · line 1874
$p_{m} (q) + (1 - p_{m} (q)) /4.$

Theorem: Exact complete flagged-erasure optimum

Displayed formulas in this passage

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