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Repeated four-level recovery and fixed-time error bound

PDF p. 3: Paragraph after Eq. (8), Eq. (9)

Statement

Assume the postulated four-level model with real fixed parameters $ω_{0}, Δ$ and decay rate $γ \geq 0$ , code $C = span {∣ 0 ⟩, ∣ 1 ⟩}$ , and the support-completed transpose recovery for that code. Let $T > 0$ , $N$ be a positive integer, and $τ = T / N$ . Assume that each interval is described by the same model channel for time $τ$ , followed by ideal instantaneous transpose recovery, and that the resulting logical cycles compose as identical channels. Let $H_{L} = C^{2}$ have ordered orthonormal basis $∣ 0 ⟩_{L}, ∣ 1 ⟩_{L}$ , and let $V : H_{L} \to C^{4}$ be the isometric encoding $V ∣ j ⟩_{L} = ∣ j ⟩$ for $j = 0, 1$ . Write $N_{τ}$ for the model channel and $T_{τ}$ for its support-completed transpose recovery. The decoded logical cycle is $E_{τ} (X) = V^{†} T_{τ} (N_{τ} (V X V^{†})) V$ for $X \in L (H_{L})$ , and its $N$ -fold composition is $E_{τ}^{N}$ . Set $F_{t r} (τ) = F_{e} (E_{τ})$ and $F_{t r}^{(N)} = F_{e} (E_{τ}^{N})$ , where $F_{e}$ is entanglement fidelity at the maximally mixed logical input $I_{H_{L}} / 2$ .

Each cycle is a logical dephasing channel: it preserves the diagonal entries in the ordered code basis and multiplies the $01$ entry by a real factor $q \in (0, 1]$ . Consequently,

$1 - F_{t r}^{(N)} \leq N [1 - F_{t r} (τ)] = \frac{γ Δ^{2} T τ^{2}}{24} + O (τ^{3}) .$

The remainder is for $N \to \infty$ at fixed $T$ and fixed model parameters. In fact the left-hand side has the same leading term. If $γ Δ^{2} T > 0$ , the leading-order scale for a small target infidelity $ϵ > 0$ is $τ ≲ \sqrt{24 ϵ / (γ Δ^{2} T)}$ ; this asymptotic rule is not a finite- $τ$ guarantee without a remainder bound. If $γ Δ^{2} = 0$ , the cycle is the identity and its infidelity is zero. Under these assumptions, checking and decoding the decay branch at each cycle recover every encoded pure state with fidelity tending to one as $τ \to 0$ at fixed $T$ .

Statement references

Four-level spontaneous-decay model

Transpose (Petz) recovery on a code

Cubic transpose-fidelity loss

Explicit four-level transpose decoder

Qubit coherence channels and fidelity functionals

Telescoping error budget for repeated finite channels

Code restriction and decoded channel types

Entanglement fidelity on a finite logical space

Facts & Assumptions

[F1]

The postulated model specifies the four-level basis, code and channel for real $ω_{0}, Δ$ , $γ \geq 0$ and nonnegative interval length. Four-level spontaneous-decay model

[F2]

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

[F3]

For one recovered cycle, put $p = 1 - e^{- γ τ}$ . When $p > 0$ , the logical channel has coherence factor $q = 1 - p + p ∣ χ ∣^{2}$ , where $χ = γ (e^{i Δ τ} - e^{- γ τ}) / [(γ + i Δ) p]$ and $0 < ∣ χ ∣ \leq 1$ . When $p = 0$ the logical channel is the identity. Explicit four-level transpose decoder

[F4]

At fixed model parameters, the one-cycle entanglement fidelity obeys $1 - F_{t r} (τ) = γ Δ^{2} τ^{3} / 24 + O (τ^{4})$ as $τ ↓ 0$ . Cubic transpose-fidelity loss

[F5]

A qubit coherence channel with real factor $q \in [0, 1]$ has entanglement fidelity $(1 + q) / 2$ , and composition multiplies its coherence factors. Qubit coherence channels and fidelity functionals

[F6]

For $q \in [0, 1]$ , $1 - q^{N} \leq N (1 - q)$ ; its $N$ -cycle pure-state infidelity at $a ∣ 0 ⟩ + b ∣ 1 ⟩$ is $2 ∣ a ∣^{2} ∣ b ∣^{2} (1 - q^{N})$ . Telescoping error budget for repeated finite channels

[F7]

For an isometric encoding $V$ and a physical recovery $T$ with outputs supported on $V V^{†}$ , the map $Y \mapsto V^{†} T (Y) V$ is a decoder, and composing it with noise and encoding gives a channel on the logical operator space. Code restriction and decoded channel types

[F8]

For a channel $E$ on $H_{L} = C^{2}$ , its entanglement fidelity at $I_{H_{L}} / 2$ is $F_{e} (E) = ⟨ Φ_{2} ∣ (i d \otimes E) (∣ Φ_{2} ⟩ ⟨ Φ_{2} ∣) ∣ Φ_{2} ⟩$ , where $∣ Φ_{2} ⟩ = (∣ 0 ⟩_{L} \otimes ∣ 0 ⟩_{L} + ∣ 1 ⟩_{L} \otimes ∣ 1 ⟩_{L}) / \sqrt{2}$ . Entanglement fidelity on a finite logical space

Proof

Given: The model, recovery, fixed parameters, total time and identical-cycle composition in the statement.

Goal: Establish the repeated-cycle fidelity bound, its asymptotic scale and the pure-state limit.

1.1givenF2F3F5F7F8algebra

The ordered-basis encoding $V$ is isometric. By [F2] and [F7], $E_{τ}$ is a channel on $L (H_{L})$ . Decoding and re-encoding any recovered output $Y$ gives $V V^{†} Y V V^{†} = Y$ because the output is code-supported; hence the decoded $N$ -cycle map is $E_{τ}^{N}$ . Thus [F8] supplies the two entanglement-fidelity functionals used here. By [F3], the recovered logical map preserves code-basis populations and has the displayed real coherence factor. If $p > 0$ , then $p < 1$ for finite $γ τ$ and $0 < ∣ χ ∣ \leq 1$ , so $0 < q \leq 1$ ; if $p = 0$ , $q = 1$ . Thus it is the claimed dephasing channel. Repeating the same cycle gives coherence factor $q^{N}$ , and [F5] gives $1 - F_{t r}^{(N)} = (1 - q^{N}) / 2$ and $1 - F_{t r} (τ) = (1 - q) / 2$ .

2.1givenF4F6step 1.1algebra

Apply [F6] and then [F4]. Since $N = T / τ$ , multiplying the $O (τ^{4})$ one-cycle remainder by $N$ gives an $O (τ^{3})$ fixed- $T$ remainder. This proves the displayed bound.

3.1givenF4step 1.1step 2.1algebra

Put $x = 1 - q$ . The geometric sum gives $0 \leq N x - (1 - q^{N}) = x \sum_{j = 0}^{N - 1} (1 - q^{j}) \leq N (N - 1) x^{2} / 2$ . By [F4] and step 1.1, $x = O (τ^{3})$ , so the difference is $O (τ^{4})$ at fixed $T$ . Hence the actual repeated infidelity has the same leading $γ Δ^{2} T τ^{2} / 24$ term as the bound. When this coefficient is positive, solving that leading term for a small target $ϵ$ gives the displayed checking-interval scale; the uncontrolled sign of the remainder precludes interpreting it alone as an exact finite-interval threshold.

4.1givenF3F6step 1.1step 2.1algebra∎

If $γ = 0$ , [F3] gives the identity. If $Δ = 0$ and $p > 0$ , its formula gives $χ = 1$ , so $q = 1$ ; the $p = 0$ case is already covered. For any normalized encoded pure state, [F6] bounds its $N$ -cycle infidelity by $N (1 - q) / 2 = N [1 - F_{t r} (τ)]$ , which tends to zero by step 2.1. Thus its fidelity tends to one under the stated ideal cycle and decay-branch decoding assumptions.

Depends on

Used by · 1

Checking speed, lift control, and code must be chosen jointly

Dependency tree · 2 levels

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Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 3, paragraph after Eq. (8) and Eq. (9)