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Branch-averaged decoded state fidelity

PDF p. 2: Eq. (4); PDF p. 8: Eq. (S1.4)

Statement

Assume the four-level model with real $ω_{0}, Δ$ , rate $γ \geq 0$ , and finite time $t \geq 0$ . Its code is spanned by the orthonormal states $∣ 0 ⟩, ∣ 1 ⟩$ . For a normalized initial code state $∣ ψ ⟩ = a ∣ 0 ⟩ + b ∣ 1 ⟩$ , apply the specified recovery $R_{4}$ after the model evolution. Define the branch-averaged decoded state fidelity $F_{ψ} (t)$ as the sum, over recovery outcomes, of each outcome probability times the overlap of its normalized output with $∣ ψ ⟩$ ; a zero-probability outcome contributes zero.

If $γ t > 0$ , let $ρ_{d} (t)$ be the normalized decoded decay output and let $χ_{d} (t) = γ (e^{i Δ t} - e^{- γ t}) / [(γ + i Δ) (1 - e^{- γ t})]$ . Then

$F_{ψ} (t) = p_{n u l l} + p_{d e c a y} ⟨ ψ ∣ ρ_{d} (t) ∣ ψ ⟩ = 1 - 2 ∣ a ∣^{2} ∣ b ∣^{2} (1 - e^{- γ t}) (1 - Re χ_{d} (t)),$

where $p_{n u l l} = e^{- γ t}$ and $p_{d e c a y} = 1 - e^{- γ t}$ . The normalized null output is the initial state, so only the decay branch can contribute fidelity loss. If $γ t = 0$ , the decay probability is zero, no normalized decay output is assigned, and $F_{ψ} (t) = 1$ .

Statement references

Four-level spontaneous-decay model

Four-level syndrome recovery

Conditional decoded state and decay coherence

Finite operator and ket conventions

Finite instruments and conditional states

Facts & Assumptions

[F1]

In the postulated four-level model, $∣ 0 ⟩, ∣ 1 ⟩$ are orthonormal code states and $ω_{0}, Δ$ are real with $γ \geq 0$ . Four-level spontaneous-decay model

[F2]

The recovery $R_{4}$ has a null operation $X \mapsto P X P$ and a decay operation $X \mapsto U X U^{†}$ , where $P = ∣ 0 ⟩ ⟨ 0 ∣ + ∣ 1 ⟩ ⟨ 1 ∣$ and $U = ∣ 0 ⟩ ⟨ 2 ∣ + ∣ 1 ⟩ ⟨ 3 ∣$ . Four-level syndrome recovery

[F3]

Under the stated model and recovery, the branch probabilities are $e^{- γ t}$ and $1 - e^{- γ t}$ . The normalized null output is $∣ ψ ⟩ ⟨ ψ ∣$ . When $γ t > 0$ , the normalized decay output is $ρ_{d} = ∣ a ∣^{2} ∣ 0 ⟩ ⟨ 0 ∣ + ∣ b ∣^{2} ∣ 1 ⟩ ⟨ 1 ∣ + [a b^{*} χ_{d} ∣ 0 ⟩ ⟨ 1 ∣ + h . c .]$ , with the displayed $χ_{d}$ ; h.c. means the Hermitian conjugate of the preceding off-diagonal term. Conditional decoded state and decay coherence

[F4]

Bras, adjoints and outer products use the finite orthonormal-basis conventions. Finite operator and ket conventions

[F5]

An instrument outcome has probability equal to the trace of its unnormalized output; its conditional state is defined by division by that probability only when it is positive. Finite instruments and conditional states

Proof

Given: The model, finite time, normalized initial state and recovery in the statement.

Goal: Calculate the branch-averaged fidelity and treat the zero-decay boundary.

1.1givenF1F3F4algebra

Suppose $γ t > 0$ . Using the orthonormal basis and the state in [F3], direct evaluation of the overlap gives $⟨ ψ ∣ ρ_{d} ∣ ψ ⟩ = ∣ a ∣^{4} + ∣ b ∣^{4} + 2 ∣ a ∣^{2} ∣ b ∣^{2} Re χ_{d} = 1 - 2 ∣ a ∣^{2} ∣ b ∣^{2} (1 - Re χ_{d})$ , since $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ .

2.1givenF2F3F5step 1.1algebra∎

The null output in [F3] has overlap one with $∣ ψ ⟩$ . Weighting the two conditional overlaps by the probabilities in [F3], as in the definition of $F_{ψ}$ , and substituting step 1.1 gives the displayed formula. Thus the null branch contributes no fidelity loss. If $γ t = 0$ , [F3] gives probabilities one and zero; by [F5] the zero-probability branch has no assigned conditional state and contributes zero, while the null overlap remains one. Hence $F_{ψ} (t) = 1$ at this boundary.

Depends on

Used by · 3

Dependency tree · 2 levels

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Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 2, Eq. (4); PDF p. 8, Eq. (S1.4)