LemmaStatement: AI-generatedProof: AI-generated

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Transpose recovery for a flagged coherence channel

Statement

Let $H = C^{2}$ have ordered orthonormal basis $∣ 0 ⟩, ∣ 1 ⟩$ . Take $K = H \oplus H = {(x_{0}, x_{1}) : x_{0}, x_{1} \in H}$ with coordinatewise addition and complex scalar multiplication, and with inner product $⟨ (x_{0}, x_{1}), (y_{0}, y_{1}) ⟩_{K} = ⟨ x_{0}, y_{0} ⟩_{H} + ⟨ x_{1}, y_{1} ⟩_{H}$ in the finite-operator convention. Define $ι_{0} x = (x, 0)$ and $ι_{1} x = (0, x)$ , and order the basis of $K$ as $ι_{0} ∣ 0 ⟩, ι_{0} ∣ 1 ⟩, ι_{1} ∣ 0 ⟩, ι_{1} ∣ 1 ⟩$ . For $0 \leq p \leq 1$ and $∣ χ ∣ \leq 1$ , define the flagged channel $N (X) = (1 - p) ι_{0} X ι_{0}^{†} + p ι_{1} D_{χ} (X) ι_{1}^{†}$ , where $D_{χ}$ preserves the diagonal entries and multiplies the $01$ and $10$ entries by $χ$ and $\overline{χ}$ , respectively. Use the whole input space as the code, with maximally mixed input $I / 2$ , and the support-normalized transpose recovery for $N$ .

On each positive-weight branch, the conditional recovered map is the identity for branch $0$ and $D_{∣ χ ∣^{2}}$ for branch $1$ . Consequently, after discarding the flag the recovered channel is $(1 - p) id + p D_{∣ χ ∣^{2}}$ . A decoder that reads the flag, uses the identity on branch $0$ , and applies a phase-correcting unitary on branch $1$ instead gives $(1 - p) id + p D_{∣ χ ∣}$ . At $χ = 0$ the unitary phase can be chosen arbitrarily. A zero-weight branch contributes zero and has no conditional state.

Facts & Assumptions

Given: The spaces, basis, parameters and flagged channel in the Statement.

Goal: Compute the transpose-recovered and phase-corrected channels, including boundary cases.

[F1]

The finite operator conventions specify matrices, adjoints and orthonormal bases. Finite operator and ket conventions

[F2]

The maps $D_{λ}$ for $∣ λ ∣ \leq 1$ are channels and satisfy $D_{μ} \circ D_{λ} = D_{μ λ}$ . Qubit coherence channels and fidelity functionals

[F3]

For code projector $P$ and channel Kraus operators $E_{a}$ , the transpose map has Kraus operators $P E_{a}^{†} N (P)^{- 1 / 2}$ , with the inverse taken on the support; its completion does not change outputs of code-supported inputs. Support-normalized transpose recovery and completion

[F4]

The support inverse square root acts by the reciprocal square root on positive eigenspaces and by zero on the kernel. Finite support square roots and generalized inverses

[F5]

A channel is linear, completely positive and trace-preserving; a decoder is a channel from the noise output space to the logical space. Code restriction and decoded channel types

[F6]

For two complex vector spaces, viewed as modules over $C$ , the indexed direct sum consists of all pairs with coordinatewise operations; its coordinate inclusions insert a vector in one coordinate and zero in the other. The finite-support condition is automatic for the index set ${0, 1}$ . The direct sum of an indexed family of modules

Proof

technique · direct

1.1givenF1F2F5F6

In the coordinates of the Statement, [F6] gives the underlying direct sum, and the displayed inner product is the standard coordinate inner product on $C^{4}$ . By [F1], taking conjugate transposes of the inclusion matrices gives $ι_{0}^{†} (x_{0}, x_{1}) = x_{0}$ and $ι_{1}^{†} (x_{0}, x_{1}) = x_{1}$ . Thus $ι_{j}^{†} ι_{j} = I_{H}$ for $j = 0, 1$ , $ι_{0}^{†} ι_{1} = ι_{1}^{†} ι_{0} = 0$ , and $ι_{0} ι_{0}^{†} + ι_{1} ι_{1}^{†} = I_{K}$ . The two ranges are therefore orthogonal and their block projections exhaust $K$ . Put $r = ∣ χ ∣$ and choose $z = χ / r$ when $r > 0$ , and any unit-modulus $z$ when $r = 0$ . Set $U = diag (1, \overline{z})$ , $Z = diag (1, - 1)$ , $q_{+} = (1 + r) / 2$ and $q_{-} = (1 - r) / 2$ . Direct multiplication gives $D_{χ} (X) = q_{+} U X U^{†} + q_{-} U Z X Z U^{†}$ . Thus a Kraus family for $N$ is $E_{0} = \sqrt{1 - p} ι_{0}$ and $E_{\pm} = \sqrt{p q_{\pm}} ι_{1} U Z$ . In the last expression use $Z$ only for $E_{-}$ : explicitly, $E_{+} = \sqrt{p q_{+}} ι_{1} U$ and $E_{-} = \sqrt{p q_{-}} ι_{1} U Z$ . These operators also show that the two branches form a channel: their adjoint products sum to $I$ , and conjugation by each operator is completely positive after every finite amplification.

2.1givenF1F3F4step 1.1

With $P = I$ , the unitality of $D_{χ}$ gives $A = N (I) = (1 - p) ι_{0} ι_{0}^{†} + p ι_{1} ι_{1}^{†}$ . By [F4], $A^{- 1 / 2}$ acts as $(1 - p)^{- 1 / 2}$ on the first block when $1 - p > 0$ , as $p^{- 1 / 2}$ on the second when $p > 0$ , and as zero on any zero-weight block. Substitution into [F3] gives the identity map on the first positive-weight block and $Y \mapsto q_{+} U^{†} Y U + q_{-} Z U^{†} Y U Z$ on the second. The latter map is $D_{\overline{χ}}$ by direct matrix multiplication. The support completion has no effect on $N (X)$ .

3.1givenF2F5step 2.1

Applying the branch maps from step 2.1 to the noise outputs and using [F2] yields $(1 - p) X + p D_{\overline{χ}} D_{χ} (X) = (1 - p) X + p D_{r^{2}} (X)$ . For a positive branch weight, dividing its output by that weight gives the stated conditional map; a zero-weight branch has zero output and no conditional state. The normalization uses the trace preservation in [F5].

4.1givenF1F2F5step 1.1∎

On the second block, conjugation by $U^{†}$ is $D_{\overline{z}}$ , so its composition with $D_{χ}$ is $D_{r}$ by [F2]. Combining it with the identity decoder on the first block gives the map $R_{p h} (Y) = ι_{0}^{†} Y ι_{0} + U^{†} ι_{1}^{†} Y ι_{1} U$ . Each summand is completely positive by the conjugation argument of step 1.1. For every $Y$ , summing its four diagonal entries in the ordered basis gives $tr Y = tr (ι_{0}^{†} Y ι_{0}) + tr (ι_{1}^{†} Y ι_{1})$ ; conjugation by the diagonal unitary $U$ leaves diagonal entries unchanged. Hence $R_{p h}$ preserves trace and is a channel by [F5]. It therefore produces $(1 - p) id + p D_{r}$ . When $r = 0$ , every unit-modulus choice of $z$ gives the same $D_{0}$ after composition.

Depends on

Used by · 6

Dependency tree · 2 levels

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