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Orthogonal decay sectors yield a strict fidelity margin

Statement

Assume the Axiom of Choice (AC) and the full hypotheses of the two-step cascade channel lemma, including $γ > 0$ , $ω_{0} > 0$ and $Δ \in R$ . For its finite cascade and restricted isometry, let $F_{D} (t)$ be the entanglement fidelity after its channel-specific transpose recovery. Then $F_{D}$ is continuous on $[0, \infty)$ , $F_{D} (0) = 1$ , and for $x = γ t$ , $F_{D} (t) = p_{0} + p_{1} F_{e} (T_{t} C_{t}) + p_{2} / 4 \geq \frac{1 + 3 e^{- x}}{4} + \frac{x e^{- x}}{8},$ where $p_{0} = e^{- x}$ , $p_{1} = x e^{- x}$ , $p_{2} = 1 - (1 + x) e^{- x}$ and $T_{t}$ is the transpose recovery for $C_{t}$ . Thus the lower bound strictly exceeds $(1 + 3 e^{- x}) / 4$ for every finite $t > 0$ . Moreover, $F_{D} (t) < 1$ for every $t > 0$ and $F_{D} (t) \to 1 / 4$ as $t \to \infty$ .

Facts & Assumptions

[F1]

The code-restricted noise has three orthogonal output sectors, of weights $p_{0}, p_{1}, p_{2}$ , with a no-decay isometry, one-decay channel $C_{t}$ and ground replacement channel. Exact two-step equal-rate cascade restricted to the detection isometry

[F2]

The one-decay channel has recovered fidelity at least 3/8, and its recovered fidelity is a finite sum of squared Hilbert–Schmidt output norms. A uniform transpose-fidelity bound for the one-decay channel

[F3]

Transpose recovery uses the support inverse square root of the noise image of the code projector. Transpose (Petz) recovery on a code

[F4]

Entanglement fidelity is linear in a sum of recovered channels and lies in [0,1] for a channel. Entanglement fidelity on a finite logical space

[F5]

Use finite adjoints, traces and orthogonal direct-sum output blocks. Finite operator and ket conventions

[F6]

Every finite channel has a finite Kraus family, permitting a family whose members are separately supported in the orthogonal output blocks. Finite channel, Kraus, Choi, and Stinespring equivalence

[F7]

Assume full AC as required by the two-step cascade lemma. Its use here is inherited through that lemma's time-integration construction of the sector decomposition and the continuous family $C_{t}$ ; the blockwise recovery calculation requires no additional choice. The Axiom of Choice Exact two-step equal-rate cascade restricted to the detection isometry

Proof

Given: Full AC, the cascade under its stated hypotheses, and the channel-specific transpose recovery.

Goal: Compute the recovered fidelity by output sectors and derive the margin and endpoint behavior.

1.1givenF3F4F5F6algebra

More generally, for channels $C_{j}$ with mutually orthogonal output spaces and weights $p_{j} > 0$ , the channel $N = ⨁_{j} p_{j} C_{j}$ has Kraus maps $\sqrt{p_{j}} K_{j, a}$ in the separate blocks. Its image of $I_{2}$ is $⨁_{j} p_{j} C_{j} (I_{2})$ . The support inverse square root has block $p_{j}^{- 1 / 2} C_{j} (I_{2})^{- 1 / 2}$ . Therefore its transpose Kraus maps restrict on block $j$ to $K_{j, a}^{†} C_{j} (I_{2})^{- 1 / 2}$ and vanish on all other blocks. Composition with $N$ gives $T N = \sum_{j} p_{j} T_{j} C_{j}$ . Zero-weight sectors contribute nothing and use the support convention. Fidelity is consequently the corresponding weighted sum.

2.1F1F2F3F4F7step 1.1algebra

Under [F7], apply step 1.1 to the cascade decomposition in [F1]. Transpose recovery exactly inverts the no-decay isometry on its image: its support image of $I_{2}$ is the range projector and the recovery is the adjoint isometry. Its recovered fidelity is one. For the ground replacement channel, Kraus maps $∣ g ⟩ ⟨ 0 ∣$ , $∣ g ⟩ ⟨ 1 ∣$ give the transpose output $I_{2} / 2$ on input $∣ g ⟩ ⟨ g ∣$ ; its recovered channel is $X \mapsto Tr (X) I_{2} / 2$ and has entanglement fidelity $1 / 4$ , by inserting the maximally entangled projector. Hence $F_{D} = p_{0} + p_{1} F_{e} (T_{t} C_{t}) + p_{2} / 4$ . By [F2] this is at least $p_{0} + 3 p_{1} / 8 + p_{2} / 4 = (1 + 3 e^{- x}) / 4 + x e^{- x} / 8$ .

3.1F1F2F4F7step 2.1algebra∎

Under the inherited hypothesis [F7], the one-decay maps extend continuously to zero by [F1]; their transpose fidelities are continuous and bounded in $[0, 1]$ by the squared-norm formula in [F2]. The sector formula therefore proves continuity including at zero and gives $F_{D} (0) = 1$ . For $t > 0$ , $p_{2} > 0$ : its derivative in $x$ is $x e^{- x} > 0$ and its value at zero is zero. Since the other recovered-sector fidelities are at most one, $F_{D} \leq 1 - 3 p_{2} / 4 < 1$ . Finally $p_{0}, p_{1} \to 0$ and $p_{2} \to 1$ , while the one-decay fidelity stays bounded, so $F_{D} \to 1 / 4$ . Strict positivity of $x e^{- x}$ at finite $t > 0$ gives the claimed margin.

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