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Four-level fidelity measures for the unrotated decoder

PDF p. 9: Opening paragraph and Table S1.1

Statement

Assume the Axiom of Choice. For a logical qubit, let $D_{λ}$ be the channel that leaves both diagonal matrix entries unchanged and multiplies the $01$ entry by $λ$ and the $10$ entry by $\overline{λ}$ , where $∣ λ ∣ \leq 1$ . If a recovered channel is the mixture $(1 - p) D_{1} + p D_{λ}$ , with $0 \leq p \leq 1$ , then its entanglement fidelity at the maximally mixed input is $F_{e} = 1 - p (1 - Re λ) / 2$ . Its pure-state fidelity is the overlap of the input pure state with the channel output, and its worst-case state fidelity is the minimum of that overlap over unit input vectors.

In the postulated four-level model, take fixed real $ω_{0}, Δ$ , fixed $γ \geq 0$ , finite $t \geq 0$ , and the recovery $R_{4}$ . Put $p = 1 - e^{- γ t}$ . If $p > 0$ , its decoded decay branch is $D_{χ_{d}}$ , where $χ_{d} = γ (e^{i Δ t} - e^{- γ t}) / [(γ + i Δ) (1 - e^{- γ t})]$ , and the null branch is $D_{1}$ . If $p = 0$ , the decay branch has zero weight and no normalized decay state is needed. For this recovery, the entanglement infidelity equals the worst-case pure-state infidelity exactly. As $t \to 0^{+}$ at fixed $γ, Δ$ , both equal $γ Δ^{2} t^{3} / 12 + O (t^{4})$ , while the Haar-average pure-state infidelity equals $γ Δ^{2} t^{3} / 18 + O (t^{4})$ and is exactly two thirds of the entanglement infidelity.

Statement references

Four-level spontaneous-decay model

Four-level syndrome recovery

Conditional decoded state and decay coherence

Entanglement fidelity on a finite logical space

Qubit coherence channels and fidelity functionals

The Axiom of Choice

Facts & Assumptions

[F1]

In the postulated model, the code has orthonormal basis $∣ 0 ⟩, ∣ 1 ⟩$ , with real $ω_{0}, Δ$ and $γ \geq 0$ ; $R_{4}$ is the specified null-and-decay recovery. Four-level spontaneous-decay model Four-level syndrome recovery

[F2]

The null and decay probabilities are $1 - p = e^{- γ t}$ and $p = 1 - e^{- γ t}$ . The normalized null output is the input state. For $p > 0$ , the normalized decay output preserves its diagonal entries and multiplies its $01$ entry by the displayed $χ_{d}$ ; for $p = 0$ no normalized decay output is assigned. Conditional decoded state and decay coherence

[F3]

For $∣ λ ∣ \leq 1$ , $D_{λ}$ is a channel with entanglement fidelity $(1 + Re λ) / 2$ and pure-state infidelity $2 ∣ a ∣^{2} ∣ b ∣^{2} (1 - Re λ)$ for $a ∣ 0 ⟩ + b ∣ 1 ⟩$ . For fixed branch weights, these fidelities are averaged by the weights. Qubit coherence channels and fidelity functionals Entanglement fidelity on a finite logical space

[F4]

Under the Axiom of Choice, a qubit channel has Haar-average pure-state fidelity $(2 F_{e} + 1) / 3$ . Haar-average versus entanglement fidelity The Axiom of Choice

[F5]

The finite exponential expansion has a controlled remainder, and finite products and quotients with nonzero constant denominator have the stated truncated expansions. Here $O (t^{4})$ means a quantity bounded in absolute value by $C t^{4}$ for sufficiently small positive $t$ , with $C$ allowed to depend on the fixed parameters. Finite exponential and smooth quotient remainder control

[F6]

For $R_{4}$ , the branch-averaged pure-state fidelity is $1 - 2 ∣ a ∣^{2} ∣ b ∣^{2} p (1 - Re χ_{d})$ when $p > 0$ , and is one when $p = 0$ . Branch-averaged decoded state fidelity

[A1]

Assume the Axiom of Choice for the Haar-average relation. The Axiom of Choice

Proof

Given: The flagged coherence-channel mixture, and for the specialization the stated model, parameters and recovery.

Goal: Calculate the fidelities, establish their exact relationships, and derive the short-time coefficients.

1.1givenF3algebra

By [F3], the identity branch has entanglement fidelity one and the other branch has fidelity $(1 + Re λ) / 2$ . Averaging with weights $1 - p$ and $p$ gives $F_{e} = 1 - p (1 - Re λ) / 2$ . The pure-state infidelity is $2 p ∣ a ∣^{2} ∣ b ∣^{2} (1 - Re λ)$ . Since $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ , the maximum of $∣ a ∣^{2} ∣ b ∣^{2}$ is $1 / 4$ , attained at equal magnitudes; $1 - Re λ \geq 0$ because $∣ λ ∣ \leq 1$ . Thus the worst-case infidelity is also $p (1 - Re λ) / 2$ .

2.1givenF1F2F6step 1.1

For $p > 0$ , [F2] identifies the two recovered branches. Applying its decay output to the equal superposition gives a positive $2 \times 2$ density matrix with diagonal entries $1 / 2$ and off-diagonal entry $χ_{d} / 2$ ; positivity therefore gives $∣ χ_{d} ∣ \leq 1$ . Linearity and the displayed action on pure inputs identify the decay map with $D_{χ_{d}}$ . Step 1.1 applies, in agreement with [F6]. For $p = 0$ , [F2] and [F6] give the identity output, so both infidelities vanish without defining $χ_{d}$ .

3.1givenF5step 2.1algebra

If $γ > 0$ , set $c (t) = p χ_{d} = γ (e^{i Δ t} - e^{- γ t}) / (γ + i Δ)$ . The denominator is nonzero. Expanding the two exponentials through degree three by [F5] and taking real parts gives $p - Re c (t) = γ Δ^{2} t^{3} / 6 + O (t^{4})$ . Explicitly, $p = γ t - γ^{2} t^{2} / 2 + γ^{3} t^{3} / 6 + O (t^{4})$ and $Re c = γ t - γ^{2} t^{2} / 2 + γ (γ^{2} - Δ^{2}) t^{3} / 6 + O (t^{4})$ . Hence step 2.1 gives both infidelities as $γ Δ^{2} t^{3} / 12 + O (t^{4})$ . If $γ = 0$ , $p = 0$ for every $t$ , so the same expansion holds with zero left side and coefficient.

4.1F4A1step 2.1step 3.1algebra∎

Applying [F4] to the recovered qubit channel under [A1] yields $1 - \overset{ˉ}{F} = 2 (1 - F_{e}) / 3$ . Substitution of step 3.1 gives $1 - \overset{ˉ}{F} = γ Δ^{2} t^{3} / 18 + O (t^{4})$ . This also holds at the zero-decay boundary.

Depends on

Used by · 2

Dependency tree · 2 levels

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Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 9, opening paragraph and Table S1.1