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Explicit four-level transpose decoder

PDF p. 9: Eq. (S1.9) and following paragraph

Statement

Assume the four-level model and channel $N_{t}$ of Exact four-level Kraus operators, with real $ω_{0}, Δ$ , rate $γ \geq 0$ , and time $t \geq 0$ . The orthonormal physical basis is $∣ 0 ⟩, \dots, ∣ 3 ⟩$ ; write $P = Π_{01} = ∣ 0 ⟩ ⟨ 0 ∣ + ∣ 1 ⟩ ⟨ 1 ∣$ and $Π_{23} = ∣ 2 ⟩ ⟨ 2 ∣ + ∣ 3 ⟩ ⟨ 3 ∣$ . Put $p = 1 - e^{- γ t}$ . Then $N_{t} (P) = (1 - p) P + p Π_{23}$ .

For $p > 0$ define $χ = γ (e^{i Δ t} - e^{- γ t}) / [(γ + i Δ) p]$ , $r = ∣ χ ∣$ , $z = χ / r$ , $Z = ∣ 0 ⟩ ⟨ 0 ∣ - ∣ 1 ⟩ ⟨ 1 ∣$ , and $D_{+} = \overline{z} ∣ 0 ⟩ ⟨ 2 ∣ + ∣ 1 ⟩ ⟨ 3 ∣$ . The number $χ$ is nonzero and $0 < r \leq 1$ . A Kraus representation of the support-normalized transpose recovery is, up to individual operator phases, $R_{1} = P$ , $R_{2} = \sqrt{(1 + r) / 2} D_{+}$ and $R_{3} = \sqrt{(1 - r) / 2} Z D_{+}$ . Thus the decay branch is mapped back with phase cancellation followed by $Z$ dephasing, meaning the mixture of the identity and conjugation by $Z$ with respective probabilities $(1 + r) / 2$ and $(1 - r) / 2$ . Its normalized recovered coherence is multiplied by $r^{2}$ ; the null branch is unchanged. The overall code channel multiplies its $01$ coherence by $(1 - p) + p r^{2}$ .

For fixed $γ > 0$ and real $Δ$ , as $t ↓ 0$ a continuous phase of $χ$ satisfies $\arg χ = Δ t / 2 + O (t^{2})$ . When $p = 0$ there is no conditional decay branch: the support-normalized transpose recovery is $Y \mapsto P Y P$ , and any completion on the orthogonal complement is immaterial after noise on the code.

Facts & Assumptions

Given: The model, channel, code and parameters in the Statement.

Goal: Determine the support-normalized transpose operators and their action on the code.

[F1]

The completed four-level channel has Kraus operators $K_{1}, K_{2}, K_{3}$ as specified in Exact four-level Kraus operators, including its zero-rate and zero-time convention.

[F2]

For $p > 0$ , after applying the syndrome recovery $R_{4}$ , the conditional recovered decay state in the code basis has diagonal entries unchanged and its $01$ entry multiplied by $χ$ as displayed in the Statement. Conditional decoded state and decay coherence

[F3]

For code projector $P$ , transpose-recovery operators are $P K_{j}^{†} N_{t} (P)^{- 1 / 2}$ , where the inverse square root is taken on the support. Transpose (Petz) recovery on a code

[F4]

The inverse square root acts by reciprocal square roots on positive eigenspaces and vanishes on the kernel. Finite support square roots and generalized inverses

[F5]

A transpose recovery can be completed to a channel without changing its action after noise on a code-supported input. Support-normalized transpose recovery and completion

[F6]

For a flagged qubit channel whose decay branch multiplies $01$ coherence by $χ$ , transpose recovery multiplies the conditional recovered coherence by $∣ χ ∣^{2}$ ; a zero-weight branch has no conditional state. Transpose recovery for a flagged coherence channel

[F7]

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

[F8]

The syndrome recovery has decay operation $X \mapsto U X U^{†}$ , where $U = ∣ 0 ⟩ ⟨ 2 ∣ + ∣ 1 ⟩ ⟨ 3 ∣$ , $U^{†} U = Π_{23}$ and $U U^{†} = P$ . Thus it identifies the decay basis with the code basis. Four-level syndrome recovery

[F9]

At fixed parameters, scalar exponential Taylor remainders are controlled, and products and quotients with nonzero constant denominator have controlled expansions. A common vanishing factor must be cancelled before using the quotient rule. Finite exponential and smooth quotient remainder control

[F10]

Principal arctangent is continuous, takes values in $(- π / 2, π / 2)$ , and satisfies $\tan (\arctan v) = v$ . For $∣ v ∣ < 1$ its series is $\arctan v = \sum_{n \geq 0} (- 1)^{n} v^{2 n + 1} / (2 n + 1)$ ; tangent is $\sin θ / \cos θ$ on its natural domain. The principal inverse tangent arctan:ℝ→(-π/2,π/2) Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series Tangent, cotangent, secant, and cosecant on their exact natural domains

[F11]

The complex exponential is entire with derivative itself, so its restriction along $t \mapsto c t$ is continuous for fixed complex $c$ . For real $θ$ , $e^{i θ} = \cos θ + i \sin θ$ and $∣ e^{i θ} ∣ = 1$ . The complex exponential is entire and its complex derivative is itself exp(x+iy)=eˣ(cos y+i sin y), |exp(x+iy)|=eˣ, and e^iπ+1=0

Proof

technique · direct

1.1givenF1F2F7F8

The operators in [F1] send the code either to itself or to the orthogonal span of $∣ 2 ⟩, ∣ 3 ⟩$ . Their action on $P$ gives $N_{t} (P) = e^{- γ t} P + (1 - e^{- γ t}) Π_{23}$ . For $p > 0$ and a normalized code state $ρ = ∣ ψ ⟩ ⟨ ψ ∣$ , put $B_{ρ} = p^{- 1} Π_{23} N_{t} (ρ) Π_{23}$ . By [F8], the recovered state in [F2] is $U B_{ρ} U^{†}$ . Since $B_{ρ}$ is supported on the decay span, $B_{ρ} = U^{†} (U B_{ρ} U^{†}) U$ by [F7] and [F8]. Consequently the normalized physical decay block sends $∣ 0 ⟩ ⟨ 0 ∣$ to $∣ 2 ⟩ ⟨ 2 ∣$ , $∣ 1 ⟩ ⟨ 1 ∣$ to $∣ 3 ⟩ ⟨ 3 ∣$ , and $∣ 0 ⟩ ⟨ 1 ∣$ to $χ ∣ 2 ⟩ ⟨ 3 ∣$ . For the last assertion, apply [F2] to $(∣ 0 ⟩ \pm ∣ 1 ⟩) / \sqrt{2}$ and $(∣ 0 ⟩ \pm i ∣ 1 ⟩) / \sqrt{2}$ ; subtracting the corresponding projectors isolates the real and imaginary off-diagonal matrix units, and complex linearity gives the stated action and its conjugate. Positivity of the resulting two-by-two output for an equal superposition gives $r \leq 1$ . Also $χ \neq 0$ , since $∣ e^{i Δ t} ∣ = 1 > e^{- γ t}$ when $p > 0$ .

1.2givenF2F9algebra

Fix $γ > 0$ and real $Δ$ . Cancel the common factor $t$ in the numerator and denominator of the displayed formula for $χ$ . The resulting denominator has constant term $γ (γ + i Δ) \neq 0$ , so [F9] gives $χ = 1 + i Δ t / 2 + O (t^{2})$ . Thus $x = Re χ = 1 + O (t^{2}) > 0$ for sufficiently small positive $t$ , $y = Im χ = Δ t / 2 + O (t^{2})$ , and $v = y / x = Δ t / 2 + O (t^{2})$ . All remainders are at these fixed parameters.

2.1givenF3F4F7step 1.1

Set $q_{\pm} = (1 \pm r) / 2$ and let $V = ∣ 2 ⟩ ⟨ 0 ∣ + ∣ 3 ⟩ ⟨ 1 ∣$ . The code-restricted channel has the Kraus family $\sqrt{1 - p} P$ , $\sqrt{p q_{+}} V diag (z, 1)$ and $\sqrt{p q_{-}} V diag (z, - 1)$ , where the diagonal operators act on the code. Indeed, their conjugation sum has the diagonal and off-diagonal actions found in step 1.1. By [F3] and [F4], its transpose operators are $P$ , $\sqrt{q_{+}} D_{+}$ and $\sqrt{q_{-}} Z D_{+}$ , up to individual phases. This gives the same transpose map as the original Kraus family because the defining sum is the adjoint of the same channel, evaluated between the same support and code projectors.

2.2F9F10F11step 1.2algebra

Define $θ (t) = \arctan v (t)$ for those small times and set $θ (0) = 0$ . The nonzero-denominator quotient defining $χ$ , hence $v$ , is continuous at positive times by [F11], and step 1.2 gives its extension at zero. By [F10], $θ$ is therefore continuous and $\tan θ = v$ . Its series gives $∣ \arctan v - v ∣ \leq \sum_{n \geq 1} ∣ v ∣^{2 n + 1} = ∣ v ∣^{3} / (1 - ∣ v ∣^{2})$ for $∣ v ∣ < 1$ , so $θ = Δ t / 2 + O (t^{2})$ . Moreover $θ \to 0$ , and [F9] applied to $e^{i θ}$ gives $Re e^{i θ} > 0$ near zero. By [F10] and [F11], $\sin θ = v \cos θ$ , $\cos θ > 0$ and $\cos^{2} θ + \sin^{2} θ = 1$ . Therefore $e^{i θ} = (1 + i v) / \sqrt{1 + v^{2}} = χ / ∣ χ ∣$ since $x > 0$ . Hence $θ$ is a continuous phase of $χ$ and proves the asserted argument asymptotic.

3.1givenF5F6F7step 2.1

On the decay block the two operators in step 2.1 apply $D_{+}$ and then mix identity with $Z$ conjugation with weights $q_{+}, q_{-}$ . The phase correction changes the coherence multiplier from $χ$ to $r$ , and that mixture multiplies it by $q_{+} - q_{-} = r$ . Hence the conditional multiplier is $r^{2}$ , in agreement with [F6]. Combining the orthogonal branch probabilities gives $(1 - p) + p r^{2}$ ; the null branch is the identity. By [F5], completion does not affect these outputs.

4.1givenF1F3F4F5step 1.1step 2.1step 3.1step 2.2∎

If $p = 0$ , [F1] gives no decay Kraus operators and a scalar unitary on the code. The support of $N_{t} (P)$ is then the code, and [F3]–[F5] give $Y \mapsto P Y P$ on that support and an arbitrary immaterial completion. Together with the exact operators and coherence action in steps 2.1 and 3.1 and the phase calculation in step 2.2, this proves the statement.

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. 9, Eq. (S1.9) and following paragraph