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Detection-code short-time checking law

PDF p. 17: Fig. S5.2 caption

Statement

Assume the finite equal-rate downward-jump benchmark, including its emitter, bath, spectral and Markov hypotheses, with rate $γ > 0$ and zero degeneracy lift. Use the spin-two detection encoding $V ∣ 0 ⟩ = (∣ 2, - 2 ⟩ + ∣ 2, 2 ⟩) / \sqrt{2}$ , $V ∣ 1 ⟩ = ∣ 2, 0 ⟩$ . At each time use the channel-specific transpose recovery and let $F_{t r} (t)$ be the entanglement fidelity of the recovered logical channel at the maximally mixed logical input. Then, with $x = γ t$ ,

$F_{t r} (t) = \frac{1 + (3 + x) e^{- x}}{4} .$

Consequently $1 - F_{t r} (t) = γ t / 2 + O (t^{2})$ as $t ↓ 0$ , for fixed $γ$ . For $F_{0} \in (0, 1]$ , define the checking time as $τ (F_{0}) = \inf {t \geq 0 : F_{t r} (t) < F_{0}}$ , with empty-set infimum $+ \infty$ . As $F_{0} ↑ 1$ through values below one, $τ (F_{0}) \sim 2 (1 - F_{0}) / γ$ ; the ratio of the two sides tends to one.

Statement references

J equals two detection encoding

Equal-rate rigid-rotor benchmark parameters

Operational checking time

Transpose (Petz) recovery on a code

Facts & Assumptions

[F1]

The detection encoding is the displayed isometry. In the benchmark the restricted dynamics is the finite downward cascade with rate $γ$ ; the model retains the same jumps when specifying the lift, and here the lift is zero. J equals two detection encoding Equal-rate rigid-rotor benchmark parameters Detection-code errors retain partial logical information

[F2]

In the cascade, the output sectors with zero, one and two decays are orthogonal. Their weights are $p_{0} = e^{- x}$ , $p_{1} = x e^{- x}$ and $p_{2} = 1 - (1 + x) e^{- x}$ . The one-decay channel $C_{t}$ extends continuously to zero. Exact two-step equal-rate cascade restricted to the detection isometry

[F3]

With $a = V ∣ 0 ⟩$ , $b = V ∣ 1 ⟩$ , the normalized first-jump matrices satisfy $B_{- 1} a = ∣ 1, 1 ⟩ / \sqrt{2}$ , $B_{+ 1} a = ∣ 1, - 1 ⟩ / \sqrt{2}$ , $B_{- 1} b = ∣ 1, - 1 ⟩ / \sqrt{6}$ , $B_{+ 1} b = ∣ 1, 1 ⟩ / \sqrt{6}$ , $B_{0} b = - \sqrt{2 / 3} ∣ 1, 0 ⟩$ , and $B_{0} a = 0$ . Spin-two downward coefficients and detection-code output populations

[F4]

For a one-decay channel $C$ with $C (I_{2}) = 2 I_{3} / 3$ and $C (Z) = diag (1 / 3, - 2 / 3, 1 / 3)$ , its transpose-recovered entanglement fidelity is $\frac{3}{8} \sum_{j = 0}^{3} ∥ C (Q_{j}) ∥_{H S}^{2}$ , where $Q_{0} = I_{2} / \sqrt{2}$ , $Q_{1} = Z / \sqrt{2}$ , $Q_{2} = X / \sqrt{2}$ , $Q_{3} = Y / \sqrt{2}$ and $X, Y, Z$ are the Pauli matrices. The squared Hilbert–Schmidt norm is $Tr (A^{†} A)$ . A uniform transpose-fidelity bound for the one-decay channel

[F5]

For this cascade, transpose fidelity is $p_{0} + p_{1} F_{e} (T_{t} C_{t}) + p_{2} / 4$ , and is continuous with value one at zero. Orthogonal decay sectors yield a strict fidelity margin Transpose (Petz) recovery on a code

[F6]

The finite exponential expansion has a controlled remainder, and a continuous fidelity with $1 - F (t) = a t + O (t^{2})$ , $a > 0$ , has first strict-crossing time asymptotic to $(1 - F_{0}) / a$ as $F_{0} ↑ 1$ . Finite exponential and smooth quotient remainder control Local inversion of a fidelity first crossing

[A1]

Assume the Axiom of Choice, Countable Choice and Dependent Choice, together with the physical and spectral hypotheses in [F1]. The Axiom of Choice The Axiom of Countable Choice (AC_ω) The axiom of dependent choice: a relation in which every element is related to something admits an ℕ-indexed chain

Proof

Given: The benchmark, encoding, recovery and choice assumptions in the statement and [A1].

Goal: Calculate the fidelity and its short-time checking-time asymptotic.

1.1F1F2F3

At zero lift, the Hamiltonian is scalar on each spin block. Thus the phases accumulated before and after a single jump are common to all magnetic components for a fixed jump time. They cancel in the channel, so the normalized one-decay channel is $C (X) = \sum_{δ = - 1}^{1} B_{δ} V X V^{†} B_{δ}^{†}$ , independently of $t > 0$ . In output order $m = - 1, 0, 1$ , [F3] gives $C (I_{2}) = 2 I_{3} / 3$ and $C (Z) = diag (1 / 3, - 2 / 3, 1 / 3)$ . It also gives $C (∣ 0 ⟩ ⟨ 1 ∣) = (∣ 1, 1 ⟩ ⟨ 1, - 1 ∣ + ∣ 1, - 1 ⟩ ⟨ 1, 1 ∣) / \sqrt{12}$ , which is Hermitian.

2.1F4F5step 1.1

The squared Hilbert–Schmidt norms of $C (Q_{0}), C (Q_{1}), C (Q_{2}), C (Q_{3})$ are respectively $2 / 3, 1 / 3, 1 / 3, 0$ : the last vanishes because $C (∣ 0 ⟩ ⟨ 1 ∣) = C (∣ 1 ⟩ ⟨ 0 ∣)$ . Their sum is $4 / 3$ , so [F4] gives $F_{e} (T_{t} C_{t}) = 1 / 2$ for $t > 0$ . Substitution in [F5] yields $F_{t r} = e^{- x} + x e^{- x} / 2 + [1 - (1 + x) e^{- x}] / 4 = [1 + (3 + x) e^{- x}] / 4$ . At $t = 0$ the same formula equals one, as does the fidelity by [F5].

3.1F5F6step 2.1∎

Expanding $e^{- x} = 1 - x + O (x^{2})$ gives $1 - F_{t r} (t) = x / 2 + O (x^{2}) = γ t / 2 + O (t^{2})$ at fixed $γ$ . The fidelity is continuous, starts at one, and has positive linear coefficient $a = γ / 2$ . Applying the strict first-crossing result to the checking-time definition gives $τ (F_{0}) \sim (1 - F_{0}) / a = 2 (1 - F_{0}) / γ$ as $F_{0} ↑ 1$ .

Source notes

The article's caption also reports numerical fits for the Aydin–Barg and Æ codes. Those fits are attributed to the article and are not derived or independently reproduced here.

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Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 17, Fig. S5.2 caption