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Cubic transpose-fidelity loss

PDF p. 3: Eq. (7)

Statement

Assume the postulated four-level model with real parameters $ω_{0}, Δ$ and decay rate $γ \geq 0$ , and code $C = span {∣ 0 ⟩, ∣ 1 ⟩}$ . Apply the model channel for time $t \geq 0$ , followed by its support-completed transpose recovery for the code and logical decoding. Let $F_{t r} (t)$ be the entanglement fidelity of the resulting logical channel at the maximally mixed input $I / 2$ . At fixed parameters, as $t \to 0$ through nonnegative values,

$F_{t r} (t) = 1 - \frac{γ Δ^{2} t^{3}}{24} + O (t^{4}) .$

Here $O (t^{4})$ means an absolute bound $C t^{4}$ for sufficiently small nonnegative $t$ , with $C$ allowed to depend on the fixed parameters. The leading loss is half the worst-case pure-state loss coefficient $γ Δ^{2} / 12$ for the syndrome recovery $R_{4}$ . In the decay branch the transpose recovery cancels the deterministic coherence phase but also applies dephasing: the resulting coherence factor is $∣ χ_{d} ∣^{2}$ , whereas a phase-only correction gives $∣ χ_{d} ∣$ . The latter has leading entanglement-fidelity loss $γ Δ^{2} t^{3} / 48$ .

Statement references

Four-level spontaneous-decay model

Transpose (Petz) recovery on a code

Four-level syndrome recovery

Exact four-level transpose fidelity

Cubic short-time loss and recoverable limits

Transpose recovery for a flagged coherence channel

Qubit coherence channels and fidelity functionals

Exact state evolution from a four-level codeword

Conditional decoded state and decay coherence

Branch-averaged decoded state fidelity

Facts & Assumptions

[F1]

The model specifies the orthonormal four-level basis, code and finite channel, with real $ω_{0}, Δ$ , $γ \geq 0$ and $t \geq 0$ . Four-level spontaneous-decay model

[F2]

Transpose recovery uses the inverse square root of the channel output on the code projector on its support; its completion does not change the recovered channel on encoded inputs. Transpose (Petz) recovery on a code

[F3]

For $γ t > 0$ , put $p = 1 - e^{- γ t}$ and $c = γ (e^{i Δ t} - e^{- γ t}) / (γ + i Δ)$ . The decay coherence is $χ_{d} = c / p$ , and the transpose entanglement fidelity is $F_{t r} = 1 - p (1 - ∣ χ_{d} ∣^{2}) / 2$ . At $γ t = 0$ , it is one. Exact four-level transpose fidelity

[F4]

The syndrome recovery $R_{4}$ measures the code and decay spans and maps the decay basis back to the code basis. Its pure-state loss has leading coefficient $∣ a ∣^{2} ∣ b ∣^{2} γ Δ^{2} / 3$ for a normalized code state $a ∣ 0 ⟩ + b ∣ 1 ⟩$ . Four-level syndrome recovery Cubic short-time loss and recoverable limits For $p > 0$ its exact loss is $1 - F_{ψ} = 2 ∣ a ∣^{2} ∣ b ∣^{2} p (1 - Re χ_{d})$ ; for $γ t = 0$ the fidelity is one. Branch-averaged decoded state fidelity

[F5]

For orthogonally flagged identity and coherence-channel branches, transpose recovery produces decay-branch coherence factor $∣ χ ∣^{2}$ ; a phase-correcting unitary instead produces $∣ χ ∣$ . A zero-weight branch contributes zero. Transpose recovery for a flagged coherence channel

[F6]

A qubit coherence channel preserves diagonal entries and multiplies its $01$ entry by its coherence factor. Its entanglement fidelity at $I / 2$ is $(1 + Re λ) / 2$ , and fidelity is affine under weighted channel sums. Qubit coherence channels and fidelity functionals

[F7]

Fixed-parameter exponential series, products and quotients with nonzero limiting denominator have controlled finite-order remainders; a denominator vanishing at zero requires cancellation first. Finite exponential and smooth quotient remainder control

[F8]

For every normalized pure code input, the null and decay probabilities are $1 - p = e^{- γ t}$ and $p = 1 - e^{- γ t}$ . When $p > 0$ , the recovered null output is the input projector, and the recovered decay output has diagonal entries $∣ a ∣^{2}, ∣ b ∣^{2}$ and $01$ entry $a b^{*} χ_{d}$ . At $p = 0$ no conditional decay output is assigned. Conditional decoded state and decay coherence The evolved density operator itself is block diagonal between the code and decay spans, with precisely these branch matrices before the decay basis is mapped back to the code. Exact state evolution from a four-level codeword

Proof

Given: The postulated model, code, fixed parameters and recovery procedures in the statement.

Goal: Prove the cubic transpose expansion and the stated comparison of recoveries.

1.1givenF3F7algebra

Suppose $γ > 0$ and $t > 0$ . Expanding the exponentials in [F3] using [F7] gives $p = γ t - γ^{2} t^{2} / 2 + γ^{3} t^{3} / 6 + O (t^{4})$ and $c = γ t + γ (i Δ - γ) t^{2} / 2 + γ (γ^{2} - i γ Δ - Δ^{2}) t^{3} / 6 + O (t^{4})$ . Hence $p - Re c = γ Δ^{2} t^{3} / 6 + O (t^{4})$ and $p - c = - i γ Δ t^{2} / 2 + O (t^{3})$ .

2.1givenF3F7step 1.1algebra

Since $p > 0$ , the exact identity $p (1 - ∣ c / p ∣^{2}) = 2 (p - Re c) - ∣ p - c ∣^{2} / p$ applies. Canceling one factor of $t$ in $p$ before using [F7], the second term is $γ Δ^{2} t^{3} / 4 + O (t^{4})$ . Thus $p (1 - ∣ χ_{d} ∣^{2}) = γ Δ^{2} t^{3} / 12 + O (t^{4})$ , and [F3] gives the displayed transpose expansion. If $γ = 0$ , [F3] gives fidelity one, so the same expansion holds; the value at $t = 0$ is also one.

3.1givenF4F8step 1.1step 2.1algebra

For $p > 0$ , use the equal-population input in [F8]: its normalized decay output is $\frac{1}{2} (\begin{array}{cc} 1 & χ_{d} \\ {\overline{χ}}_{d} & 1 \end{array})$ , with eigenvalues $(1 \pm ∣ χ_{d} ∣) / 2$ . Positivity therefore gives $∣ χ_{d} ∣ \leq 1$ , so $p - Re c = p (1 - Re χ_{d}) \geq 0$ . By the exact formula in [F4], $1 - F_{ψ} = 2 ∣ a ∣^{2} ∣ b ∣^{2} (p - Re c)$ . Since $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ , the factor $∣ a ∣^{2} ∣ b ∣^{2}$ is at most $1 / 4$ , attained at equal populations. Hence $\max_{ψ} (1 - F_{ψ}) = (p - Re c) / 2 = γ Δ^{2} t^{3} / 12 + O (t^{4})$ by step 1.1. The scalar remainder is independent of the state, and multiplying it by $2 ∣ a ∣^{2} ∣ b ∣^{2} \leq 1 / 2$ bounds every state's remainder uniformly. This agrees with the leading coefficient in [F4] and is twice the transpose coefficient in step 2.1. At $γ t = 0$ , [F4] gives zero loss for every state, including $γ = 0$ for all times.

4.1givenF1F5F6F8step 3.1algebra

Let $H_{L} = C^{2}$ with basis $∣ 0 ⟩_{L}, ∣ 1 ⟩_{L}$ , and define isometries $V ∣ j ⟩_{L} = ∣ j ⟩$ and $W ∣ j ⟩_{L} = ∣ j + 2 ⟩$ for $j = 0, 1$ . Their ranges are orthogonal by [F1]. For $p > 0$ , [F8] gives the restricted model channel on every pure input projector as the block diagonal expression $(1 - p) V X V^{†} + p W D_{χ_{d}} (X) W^{†}$ , where $D_{χ_{d}}$ is the coherence map of [F6]. This equality extends to every logical matrix by complex linearity of the model channel in [F1]: the projectors onto $∣ 0 ⟩_{L}, ∣ 1 ⟩_{L}, (∣ 0 ⟩_{L} + ∣ 1 ⟩_{L}) / \sqrt{2}, (∣ 0 ⟩_{L} + i ∣ 1 ⟩_{L}) / \sqrt{2}$ span all $2 \times 2$ matrices. Indeed, twice the third projector minus the first two gives $E_{01} + E_{10}$ , and twice the fourth minus the first two gives $- i E_{01} + i E_{10}$ , where $E_{j k} = ∣ j ⟩_{L} ⟨ k ∣_{L}$ ; these two combinations recover both off-diagonal matrix units. Thus the restricted channel is exactly the orthogonally flagged channel in [F5], with $0 < p < 1$ and $∣ χ_{d} ∣ \leq 1$ by step 3.1.

5.1givenF1F2F4F5F8step 4.1algebra

To identify its transpose with the decoded physical recovery, let $K_{k}$ be Kraus matrices of the model channel and put $P = V V^{†}$ and $A = N_{t} (P)$ . The restricted channel in step 4.1 has Kraus matrices $K_{k} V$ and sends $I$ to $A$ . By [F2], decoding the physical transpose gives Kraus matrices $V^{†} P K_{k}^{†} A^{- 1 / 2} = (K_{k} V)^{†} A^{- 1 / 2}$ , because $V^{†} P = V^{†}$ . These are exactly the transpose matrices for the restricted channel with whole input code $H_{L}$ . The completion does not affect its noise outputs. Consequently [F5] applies: transpose recovery gives $(1 - p) id + p D_{∣ χ_{d} ∣^{2}}$ , whereas reading the two spans as in [F4] and applying a phase-correcting unitary on the decoded decay branch gives $(1 - p) id + p D_{∣ χ_{d} ∣}$ . At $χ_{d} = 0$ its phase can be arbitrary. At $p = 0$ , [F8] and the linearity argument of step 4.1 give only the identity branch, and no conditional decay map is assigned.

6.1givenF3F5F6F7step 1.1step 2.1step 4.1step 5.1algebra∎

By [F3] and step 1.1, $χ_{d} \to 1$ as $t \to 0$ with $γ > 0$ . From step 2.1, after canceling $t$ in $p$ , $1 - ∣ χ_{d} ∣^{2} = Δ^{2} t^{2} / 12 + O (t^{3})$ . Since $∣ χ_{d} ∣ \to 1$ , the identity $1 - ∣ χ_{d} ∣ = (1 - ∣ χ_{d} ∣^{2}) / (1 + ∣ χ_{d} ∣)$ first gives $1 - ∣ χ_{d} ∣ = O (t^{2})$ , hence $1 + ∣ χ_{d} ∣ = 2 + O (t^{2})$ and $1 - ∣ χ_{d} ∣ = Δ^{2} t^{2} / 24 + O (t^{3})$ by [F7]. The flagged-channel rule [F5] removes the coherence phase in both recoveries, but transpose recovery leaves the smaller real factor $∣ χ_{d} ∣^{2}$ , representing additional dephasing. Its applicability is established in steps 4.1 and 5.1. Using [F6], the phase-only decoder has loss $p (1 - ∣ χ_{d} ∣) / 2 = γ Δ^{2} t^{3} / 48 + O (t^{4})$ . For $γ = 0$ the decay branch has zero weight, so this loss is zero. This completes the comparison.

Depends on

Used by · 2

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Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 3, Eq. (7); PDF p. 9, Table S1.1