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Support-normalized transpose recovery and completion

Statement

Let $H_{P}, H_{O}$ be finite-dimensional complex inner-product spaces, let $P$ be the orthogonal projector onto a nonzero code subspace of $H_{P}$ , and let $N : L (H_{P}) \to L (H_{O})$ be a quantum channel with finite Kraus representation $N (X) = \sum_{a} E_{a} X E_{a}^{†}$ . Put $A = N (P)$ , let $S$ be the orthogonal projector onto $supp A = (\ker A)^{⊥}$ , and let $A^{- 1 / 2}$ be the inverse square root on this support and zero on $\ker A$ . Define $R_{a} = P E_{a}^{†} A^{- 1 / 2}$ and $T_{0} (Y) = \sum_{a} R_{a} Y R_{a}^{†}$ .

The map $T_{0}$ is completely positive, has output supported on the code, and preserves trace for every $Y$ supported on $supp A$ , meaning $Y = S Y S$ . For any code state $σ$ —a positive semidefinite operator of trace one with $P σ P = σ$ —the map $T (Y) = T_{0} (Y) + tr ((I - S) Y) σ$ is a quantum channel with code-supported output. For every $X = P X P$ , the output $N (X)$ is supported on $supp A$ ; consequently $T (N (X)) = T_{0} (N (X))$ . In particular, completion does not change any recovered code-state output or the value of any function of that output with all other arguments fixed. Explicitly, let $ρ$ be any code state and let $τ$ be any density operator on $H_{P}$ , meaning a positive semidefinite operator of trace one. With the root-fidelity convention $F (B, C) = ∥ \sqrt{B} \sqrt{C} ∥_{1}$ for density operators $B, C$ on the same space, where $∥ \cdot ∥_{1}$ denotes the trace norm, one has $F (τ, T (N (ρ))) = F (τ, T_{0} (N (ρ)))$ . The same equality holds for squared fidelity $F^{2}$ .

Facts & Assumptions

Given: The finite-dimensional spaces, nonzero code projector, channel, Kraus family and code state in the Statement.

Goal: Establish the support normalization, the completed channel, and equality on outputs of code-supported inputs.

[F1]

A quantum channel is linear, completely positive and trace-preserving; a code-supported physical recovery has outputs $Z$ satisfying $P Z P = Z$ . Code restriction and decoded channel types

[F2]

For positive semidefinite $A$ , the support inverse square root is zero on the kernel and satisfies $A^{- 1 / 2} A A^{- 1 / 2} = S$ . Finite support square roots and generalized inverses

[F3]

The given Kraus representation means $N (X) = \sum_{a} E_{a} X E_{a}^{†}$ for every input operator. Finite Kraus representation

[F4]

For compatible finite matrices, $tr (B C) = tr (C B)$ , including rectangular products. For A∈ M_m× n(F) and B∈ M_n× m(F), tr(AB)=tr(BA)

[F5]

For density operators $B, C$ on the same nonzero finite-dimensional complex inner-product space, root fidelity is $F (B, C) = ∥ \sqrt{B} \sqrt{C} ∥_{1}$ , where $∥ \cdot ∥_{1}$ is the trace norm; squared fidelity is $F (B, C)^{2}$ . Fidelity

Proof

technique · direct

Given: The finite-dimensional spaces, nonzero code projector, channel, Kraus family and code state in the Statement.

Goal: Establish the support normalization, the completed channel, and equality on outputs of code-supported inputs.

Proof technique: direct.

1.1givenF2F3F4

Each $E_{a} P E_{a}^{†}$ is positive semidefinite, since its quadratic form at $y$ is the squared norm of $P E_{a}^{†} y$ . Thus $A = \sum_{a} E_{a} P E_{a}^{†}$ is positive semidefinite. By [F2], $\sum_{a} R_{a}^{†} R_{a} = A^{- 1 / 2} (\sum_{a} E_{a} P E_{a}^{†}) A^{- 1 / 2} = S$ . Each $R_{a}$ has range in the code, so $P T_{0} (Y) P = T_{0} (Y)$ . Conjugation by any operator remains positive after tensoring with a finite ancilla, and finite sums preserve positivity; hence $T_{0}$ is completely positive. By [F4], $tr T_{0} (Y) = tr (S Y)$ , which equals $tr Y$ when $Y = S Y S$ .

2.1givenF1F2F4step 1.1

Write $Q = I - S$ . Choose an orthonormal basis $(w_{j})$ of the range of $Q$ and a spectral decomposition $σ = \sum_{i} p_{i} ∣ u_{i} ⟩ ⟨ u_{i} ∣$ , where $p_{i} \geq 0$ , $\sum_{i} p_{i} = 1$ , and the vectors with nonzero weights lie in the code. The operators $C_{i j} = \sqrt{p_{i}} ∣ u_{i} ⟩ ⟨ w_{j} ∣$ satisfy $\sum_{i, j} C_{i j} Y C_{i j}^{†} = tr (Q Y) σ$ and $\sum_{i, j} C_{i j}^{†} C_{i j} = Q$ . Therefore the completion term is completely positive and code-supported. Together with step 1.1, it makes $T$ completely positive and code-supported; its trace is $tr (S Y) + tr (Q Y) = tr Y$ . Thus $T$ is a channel by [F1].

3.1givenF2F3step 1.1step 2.1

If $y \in \ker A$ , then $0 = ⟨ A y, y ⟩ = \sum_{a} ∥ P E_{a}^{†} y ∥^{2}$ . Every summand is nonnegative, so $P E_{a}^{†} y = 0$ for every $a$ . Taking adjoints gives $Q E_{a} P = 0$ and $P E_{a}^{†} Q = 0$ . Hence, for $X = P X P$ , both $Q N (X) = 0$ and $N (X) Q = 0$ ; equivalently $N (X) = S N (X) S$ . Its completion term therefore vanishes.

4.1givenF1F5step 2.1step 3.1∎

For a code state $ρ$ , the operator $Z = T (N (ρ))$ is positive semidefinite and has trace one because $N$ and $T$ are channels. By step 3.1 it equals $T_{0} (N (ρ))$ , which is therefore also a density operator. For any density operator $τ$ on $H_{P}$ , both sides of the asserted root-fidelity identity are consequently defined by [F5] and equal $F (τ, Z)$ . Squaring gives the squared-fidelity identity. More generally, substitution of the equal recovered operators into any function defined at their common value, with all other arguments fixed, gives equal values.

Depends on

Used by · 7

Dependency tree · 2 levels

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