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Dimensional boundary of a rank-one KL obstruction

Statement

Let $H_{L}, H_{P}, H_{O}$ be finite-dimensional complex inner-product spaces, with inner products linear in the first argument, and let $V : H_{L} \to H_{P}$ be an isometry. Write $P = V V^{*}$ for the orthogonal projector onto the code $V (H_{L})$ , and suppose $\dim H_{L} \geq 2$ . Let a channel $N : L (H_{P}) \to L (H_{O})$ be exactly correctable on this code and have a finite Kraus representation containing the nonzero operator $E : H_{P} \to H_{O}$ given by $E x = ⟨ x, s ⟩ f$ for vectors $s \in H_{P}$ and $f \in H_{O}$ . Then $P s = 0$ .

The dimension hypothesis is essential: in dimension one, exact correctability and the same rank-one form need not imply $P s = 0$ .

Facts & Assumptions

[F1]

The code projector is $P = V V^{*}$ , and exact correctability means that a decoded recovery channel composed with the noise and encoding is the identity on the logical operator space. Code restriction and decoded channel types

[F2]

A finite Kraus representation has the form $N (X) = \sum_{a} E_{a} X E_{a}^{*}$ for every input operator $X$ . Finite Kraus representation

[F3]

The adjoint is characterized by $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ . The adjoint T^*:W→ V is characterised by ⟨ Tv,w⟩_W=⟨ v,T^*w⟩_V

[F4]

Exact correction of a finite-dimensional isometric code for a channel with finite Kraus operators $E_{a}$ implies $P E_{a}^{*} E_{b} P = C_{a b} P$ for scalars $C_{a b}$ . Exact recovery implies the Knill–Laflamme condition

Proof

Given: The finite-dimensional isometric code, its exact recovery, and a finite Kraus family containing the stated nonzero $E$ .

Goal: Prove $P s = 0$ and exhibit the stated one-dimensional boundary.

1.1givenF3F4algebra

By [F4], applied to the index of $E$ twice, $P E^{*} E P = c P$ for a scalar $c$ . By [F3] and the displayed formula for $E$ , for every $x \in H_{P}$ one has $P E^{*} E P x = ∥ f ∥^{2} ⟨ x, P s ⟩ P s$ . Thus the image of $P E^{*} E P$ lies in the span of $P s$ , which has dimension at most one.

2.1givenstep 1.1algebra

The image of $P$ is the code and has dimension at least two, so $c P$ cannot have image of dimension at most one unless $c = 0$ . Hence $P E^{*} E P = 0$ . Evaluating its quadratic form at $s$ gives $0 = ∥ f ∥^{2} ∥ P s ∥^{4}$ . Since $E$ is nonzero, $f \neq 0$ , and therefore $P s = 0$ .

3.1F1F2constructalgebra∎

For the boundary, take all three spaces to be $C$ , let $V = P = I$ , and let $N$ be the identity channel with its single Kraus operator $E = I$ . With $s = f = 1$ , the stated rank-one formula holds, the identity channel is its own decoder, and $P s = 1 \neq 0$ . This establishes the claimed one-dimensional exception.

Depends on

Used by · 3

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