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Four-level high-fidelity checking time

PDF p. 3: Eq. (8)

Statement

Assume the postulated four-level model with decay rate γ≥0, real degeneracy-lifting parameter Δ, code span⁡{∣0⟩,∣1⟩}, and its support-completed transpose recovery at each elapsed time. Let Ftr(t) be the recovered entanglement fidelity at the maximally mixed logical input, and define the checking time τ(F0)=inf⁡{t≥0:Ftr(t)<F0}, with inf⁡∅=+∞.

For fixed parameters with γΔ2>0, as F0↑1 through values below 1,

τ(F0)∼[24(1−F0)γΔ2]1/3.

The symbol ∼ means that the ratio of the two sides tends to 1. For fixed F0<1, the displayed leading approximation decreases as γ or ∣Δ∣ increases while the other is positive. This comparison is of leading approximations at fixed parameters; it does not assert global monotonicity of the exact checking time. If γ=0 or Δ=0, then Ftr(t)=1 for every t≥0, and τ(F0)=+∞ for every F0∈(0,1].

Facts & Assumptions

[F1]

The model specifies a finite four-level channel with γ≥0, real Δ, the stated code, and the transpose recovery is applied to that channel and code. Four-level spontaneous-decay model Transpose (Petz) recovery on a code

[F2]

The checking time is the infimum of the strict fidelity-crossing set, with value +∞ when that set is empty. Operational checking time

[F3]

At fixed model parameters, Ftr(t)=1−γΔ2t3/24+O(t4) as t↓0, where the remainder is bounded in absolute value by a parameter-dependent constant times t4 near zero. Cubic transpose-fidelity loss

[F4]

If γt>0, put p=1−e−γt and χ=γ(eiΔt−e−γt)/[(γ+iΔ)p]. Then Ftr=1−p(1−∣χ∣2)/2. If γt=0, the fidelity is 1, and the formula has limit 1 as γt→0. Exact four-level transpose fidelity

[F5]

If a continuous F:[0,∞)→R satisfies F(0)=1 and 1−F(t)=atp+O(tp+1) for fixed a,p>0, its strict first-crossing time at 1−ϵ is asymptotic to (ϵ/a)1/p as ϵ↓0. lem-qec-prereq-first-crossing-asymptotics

[F6]

For a bounded complex operator A, the exponential etA is continuous in t in operator norm. In particular this applies to scalar complex exponentials. The exponential series of a bounded operator

[F7]

For positive bases and real exponents, the product, quotient and iterated-power laws hold: (ab)r=arbr, (a/b)r=ar/br and (ar)s=ars. The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents

[F8]

For every real α, the function x↦xα is continuous and differentiable on (0,∞), with derivative αxα−1. Continuity and derivatives of positive-base real powers

Proof

Given: The fixed model parameters, code, recovery and target described in the statement.

Goal: Establish the first-crossing asymptotic, its leading parameter dependence, and the zero-parameter cases.

1.1givenF4F6algebra

If γ>0, the exponentials in [F4] are continuous by [F6]. For t>0, p>0 and γ+iΔ≠0, so the displayed algebraic expression is continuous there. Its value at zero is 1 and its right limit is 1 by [F4]. If γ=0, [F4] gives the constant function 1. Thus Ftr is continuous on [0,∞) and Ftr(0)=1.

2.1givenF2F3F5step 1.1algebra

Suppose γΔ2>0 and put a=γΔ2/24 and ϵ=1−F0. By [F3], 1−Ftr(t)=at3+O(t4) with a>0. Applying [F5] to the continuous function from step 1.1 and the crossing convention in [F2] yields τ(1−ϵ)∼(ϵ/a)1/3 as ϵ↓0, which is the displayed formula.

3.1givenF7F8F9step 2.1algebra

For fixed ϵ>0, put d=∣Δ∣>0. By [F7], the leading approximation from step 2.1 is (24ϵ/(γd2))1/3=(24ϵ)1/3γ−1/3d−2/3 for γ,d>0. By [F8], x↦x−1/3 and x↦x−2/3 are continuous and differentiable on (0,∞) with derivatives −x−4/3/3<0 and −2x−5/3/3<0, respectively. By [F9] both functions are strictly decreasing on that interval. Holding either parameter fixed leaves a positive constant times the corresponding strictly decreasing function, so the leading approximation strictly decreases as either positive γ or positive d increases. The asymptotic in step 2.1 is pointwise in the parameters, so this calculation makes no global monotonicity assertion about the exact crossing time.

4.1givenF2F4algebra∎

If γ=0, [F4] gives Ftr=1 at every time. If Δ=0 and γ>0, then for t>0 the numerator of χ in [F4] is γ(1−e−γt)=γp, so χ=1 and the same fidelity equals 1; the value at t=0 is also 1. For any F0∈(0,1] the strict crossing set is therefore empty, and [F2] gives τ(F0)=+∞.

Depends on

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