Manuscript En02-B@1 · En02-B-v1.tex · example · ca28dee4

Example: Why an arbitrary extension is not a lower bound

Intrinsic recovery channel and one-shot accounting · lines 1004–1031

En02-B@1#ex:offcode

Consider the restricted encoding $ρ \mapsto ρ \otimes ∣ 0 ⟩ ⟨ 0 ∣$ . Partial trace over the second qubit is a valid decoder of Choi rank two.

Another valid decoder acts as identity when the second qubit is zero, and depolarizes the first when it is one:

R^{'} (σ) = ⟨ 0 ∣ σ ∣ 0 ⟩ + tr (⟨ 1 ∣ σ ∣ 1 ⟩) \frac{I _{2}}{2} .

It has the five independent Kraus operators

I \otimes ⟨ 0 ∣, \frac{1}{2} P \otimes ⟨ 1 ∣, P \in {I, X, Y, Z} .

Its Choi rank is five, despite identical recovery on the restricted encoding. Its larger dilation cost is not an intrinsic recovery lower bound.

Theorem thm:B3 avoids this error because the syndrome in (eq:normalform) is faithful and forces uniqueness.

Displayed formulas in this passage

unlabeled · line 1013
$R^{'} (σ) = ⟨ 0 ∣ σ ∣ 0 ⟩ + tr (⟨ 1 ∣ σ ∣ 1 ⟩) \frac{I _{2}}{2} .$
unlabeled · line 1021
$I \otimes ⟨ 0 ∣, \frac{1}{2} P \otimes ⟨ 1 ∣, P \in {I, X, Y, Z} .$