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Exact fidelity–Kraus-weight identities and short-time bound saturation

PDF p. 9: Recovery comparison paragraph following Eq. (S1.9)

Statement

Assume the Axiom of Choice and the postulated four-level model with real $ω_{0}, Δ$ , rate $γ \geq 0$ , and finite time $t \geq 0$ . The code is $span {∣ 0 ⟩, ∣ 1 ⟩}$ and the orthogonal decay space is $span {∣ 2 ⟩, ∣ 3 ⟩}$ . Put $p = 1 - e^{- γ t}$ . For $p > 0$ define $χ = γ (e^{i Δ t} - e^{- γ t}) / [(γ + i Δ) p]$ and $r = ∣ χ ∣$ . Let $K_{2}, K_{3}$ be the specifically named operators in Exact four-level Kraus operators, using its convention $K_{2} = K_{3} = 0$ when $p = 0$ , with weights $w_{j} = Tr (K_{j}^{†} K_{j})$ . Let $F_{o p t}$ be the maximum, over quantum-channel decoders, of entanglement fidelity at the maximally mixed logical input; let $F_{t r}$ be that fidelity for support-completed transpose recovery. Entanglement fidelity is the overlap with a normalized maximally entangled logical-reference state after the channel acts on the logical factor.

For $p > 0$ ,

$1 - F_{t r} = (1 + r) (1 - F_{o p t}), 1 - F_{o p t} = \frac{w_{3}}{2}, 1 - F_{t r} = \frac{w_{2} w_{3}}{2 p} .$

For fixed $γ > 0$ and $Δ \neq 0$ , as $t ↓ 0$ the ratio $(1 - F_{t r}) / (1 - F_{o p t})$ tends to $2$ , saturating the factor-two bound in that limit. For fixed $γ > 0$ and real $Δ$ , sufficiently small positive $t$ obeys $Re χ \leq r^{2} \leq r$ . The unrotated recovery $R_{4}$ measures the code and decay spaces, leaves the code branch unchanged, and maps $∣ 2 ⟩, ∣ 3 ⟩$ to $∣ 0 ⟩, ∣ 1 ⟩$ respectively. Its entanglement infidelity has leading coefficient $γ Δ^{2} / 12$ at order $t^{3}$ , while transpose recovery has coefficient $γ Δ^{2} / 24$ and the optimal decoder has coefficient $γ Δ^{2} / 48$ . Thus, for $Δ \neq 0$ , transpose recovery halves the leading infidelity of $R_{4}$ and is within a factor two of the optimum.

If $p = 0$ , the decay branch has zero weight, $w_{2} = w_{3} = 0$ , and all three fidelities are one; conditional $χ$ and expressions dividing by $p$ are not defined. If $Δ = 0$ , all three infidelities vanish and their ratios are undefined.

Statement references

Four-level spontaneous-decay model

Four-level syndrome recovery

Optimal phase-corrected four-level recovery

Exact four-level Kraus operators

Decay Kraus weights and photon overlap

Four-level fidelity measures for the unrotated decoder

Qubit coherence channels and fidelity functionals

Transpose recovery for a flagged coherence channel

Factor-two guarantee for transpose recovery

The Axiom of Choice

Facts & Assumptions

[A1]

Assume the Axiom of Choice, as required by the unrotated-recovery comparison. The Axiom of Choice

[F1]

In the postulated model, the code and decay spaces are orthogonal; $R_{4}$ is the stated syndrome recovery. Four-level spontaneous-decay model Four-level syndrome recovery

[F2]

Write $V_{0} ∣ j_{L} ⟩ = ∣ j ⟩$ and $V_{1} ∣ j_{L} ⟩ = ∣ j + 2 ⟩$ for $j = 0, 1$ , identifying the logical qubit with the code and decay spaces respectively. For $p > 0$ , the restricted noise channel has the orthogonal-block form $N_{t} (V_{0} X V_{0}^{†}) = (1 - p) V_{0} X V_{0}^{†} + p V_{1} D_{χ} (X) V_{1}^{†}$ , where $D_{χ}$ preserves diagonal entries and multiplies the $01$ entry by $χ$ . The Kraus representation supplies the separate output blocks and null action; the unrotated-recovery comparison supplies the decoded decay map, with its $χ_{d}$ equal to $χ$ . Exact four-level Kraus operators Four-level fidelity measures for the unrotated decoder The bound is $0 < r \leq 1$ . The optimal entanglement fidelity is $1 - p (1 - r) / 2$ ; at $p = 0$ it is one without defining $χ$ . Optimal phase-corrected four-level recovery

[F3]

The named $K_{2}, K_{3}$ and their zero-decay convention are those of Exact four-level Kraus operators. For positive rate and time, their weights are $w_{2} = p (1 + r)$ and $w_{3} = p (1 - r)$ . At zero decay probability both operators and both weights vanish. Exact four-level Kraus operators Decay Kraus weights and photon overlap

[F4]

For the flagged coherence channel, support-completed transpose recovery produces the mixture $(1 - p) id + p D_{r^{2}}$ , where $D_{λ}$ preserves diagonal entries and multiplies the $01$ entry by $λ$ . Its maximally mixed-input entanglement fidelity is obtained by branch averaging, with $F_{e} (D_{λ}) = (1 + Re λ) / 2$ . Transpose recovery for a flagged coherence channel Qubit coherence channels and fidelity functionals

[F5]

For fixed $γ, Δ$ , the entanglement infidelity of $R_{4}$ is $γ Δ^{2} t^{3} / 12 + O (t^{4})$ as $t ↓ 0$ . Four-level fidelity measures for the unrotated decoder

[F6]

Fixed-parameter exponential and nonvanishing-denominator quotient expansions have controlled remainders. Finite exponential and smooth quotient remainder control

[F7]

For a finite-dimensional code and channel, transpose infidelity is at most twice the optimal entanglement infidelity at the maximally mixed input. Factor-two guarantee for transpose recovery

[F8]

At $p = 0$ , the null branch is the identity on code inputs and the decay branch has zero weight; $R_{4}$ leaves the null branch unchanged. No normalized decay state is needed. Four-level fidelity measures for the unrotated decoder Four-level syndrome recovery

Proof

Given: The model, parameters, fidelity convention and Axiom of Choice in the Statement.

Goal: Establish the identities, short-time comparisons and boundary cases.

1.1givenF1F2F3F4algebra

For $p > 0$ , restrict the Kraus representation in [F3] to inputs $V_{0} X V_{0}^{†}$ . Its $K_{1}$ is a common scalar of squared modulus $e^{- γ t} = 1 - p$ on the code, so this term is $(1 - p) V_{0} X V_{0}^{†}$ . Its $K_{2}, K_{3}$ take the code into the decay space, so the sum has no off-diagonal blocks between code and decay. Their adjoint products on the code sum to $p I$ , as follows directly from the displayed Kraus operators; hence the decay branch has input-independent weight $p$ . The decoder in [F1] acts on this block as $Y \mapsto V_{1}^{†} Y V_{1}$ . By the unrotated branch statement in [F2], its normalized decoded map is $D_{χ}$ . Since $V_{1} V_{1}^{†}$ is the identity on the decay space, the unnormalized physical decay map is therefore $p V_{1} D_{χ} (X) V_{1}^{†}$ . In the ordered output basis $∣ 0 ⟩, ∣ 1 ⟩, ∣ 2 ⟩, ∣ 3 ⟩$ , $V_{0}, V_{1}$ are exactly the two coordinate inclusions of the flagged-channel lemma in [F4]; expressing the input code in its logical basis makes that lemma's whole-input-code convention identical to this restriction. Its hypotheses $0 \leq p \leq 1$ and $∣ χ ∣ \leq 1$ follow from the definition of $p$ and [F2]. Thus its transpose-recovery conclusion applies.

1.2givenF2F6algebra

For fixed $γ > 0$ , cancel the common factor $t$ in the numerator and denominator of $χ$ . The cancelled denominator has nonzero constant term $γ (γ + i Δ)$ , so [F6] applies. The normalized numerator coefficients through degree two are $1$ , $- γ / 2 + i Δ / 2$ , and $(γ^{2} - i γ Δ - Δ^{2}) / 6$ , while the normalized denominator is $1 - γ t / 2 + γ^{2} t^{2} / 6 + O (t^{3})$ . Quotient expansion therefore gives $χ = 1 + i Δ t / 2 + (- Δ^{2} / 6 + i γ Δ / 12) t^{2} + O (t^{3})$ . Hence $r^{2} = 1 - Δ^{2} t^{2} / 12 + O (t^{3})$ and $r = 1 - Δ^{2} t^{2} / 24 + O (t^{3})$ , using $r - 1 = (r^{2} - 1) / (r + 1)$ . Thus $r^{2} - Re χ = Δ^{2} t^{2} / 12 + O (t^{3})$ . If $Δ \neq 0$ this is positive for sufficiently small positive $t$ ; if $Δ = 0$ , the formula for $χ$ gives $χ = r = 1$ . In both cases $r^{2} \leq r$ follows from $0 < r \leq 1$ in [F2].

2.1givenF2F3F4step 1.1algebra

By step 1.1, the restricted noise has the flagged coherence form in [F4]. Applying [F4] gives $1 - F_{t r} = p (1 - r^{2}) / 2$ . Together with [F2] and $1 - r^{2} = (1 + r) (1 - r)$ , this proves the first identity. Substituting the weights in [F3] gives $1 - F_{o p t} = w_{3} / 2$ and $1 - F_{t r} = w_{2} w_{3} / (2 p)$ .

3.1givenF5F6F7step 2.1step 1.2algebra

Since $p = γ t + O (t^{2})$ , steps 2.1 and 1.2 give $1 - F_{t r} = γ Δ^{2} t^{3} / 24 + O (t^{4})$ and $1 - F_{o p t} = γ Δ^{2} t^{3} / 48 + O (t^{4})$ . Together with [F5], these give the three coefficients. For $Δ \neq 0$ , the optimal loss is positive for sufficiently small positive $t$ , so the ratio in step 2.1 is defined and tends to $2$ as $r \to 1$ . This is saturation in the limiting ratio of the bound in [F7].

4.1givenF1F2F3F4F8step 2.1algebra∎

If $p = 0$ , [F8] gives an identity null branch with no decay contribution, and [F3] gives zero decay weights. Thus $R_{4}$ has fidelity one; the optimal fidelity is one by the boundary case of the optimal-decoder supplier in [F2]. For transpose recovery, apply the generic flagged-channel result in [F4] with zero decay weight and any auxiliary coherence parameter, for example zero: its recovered map is the identity, independently of that parameter. This assigns no conditional model coherence at $p = 0$ . If $p > 0$ and $Δ = 0$ , the formula for $χ$ gives $r = χ = 1$ , so step 2.1 gives zero optimal and transpose losses, while the unrotated branch is also the identity by [F2] and the definition of $R_{4}$ in [F1]. Consequently the stated ratios are undefined in these zero-loss cases.

Depends on

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Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 9, paragraph following Eq. (S1.9)