Physics theoremStatement: AI-adaptedProof: AI-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Transient later-time advantage of the unrotated decoder

PDF p. 9: End of recovery-comparison paragraph

Statement

Assume the Axiom of Choice and the postulated four-level model with real detuning $Δ$ , decay rate $γ \geq 0$ , finite time $t \geq 0$ , and the unrotated syndrome recovery $R_{4}$ . Here $R_{4}$ leaves the code branch unchanged and maps the two decay-basis states back to their corresponding code-basis states. Compare entanglement fidelities at the maximally mixed logical-qubit input; the transpose recovery is the support-normalized transpose decoder of the model channel.

When $γ t > 0$ , put $p = 1 - e^{- γ t}$ and $χ_{d} (t) = γ (e^{i Δ t} - e^{- γ t}) / [(γ + i Δ) p]$ . A positive-phase crossing means a time $t_{0} > 0$ at which $χ_{d} (t_{0})$ is positive real, equivalently its argument is a multiple of $2 π$ . If $γ > 0$ and $Δ \neq 0$ , then at every such crossing $0 < χ_{d} (t_{0}) < 1$ , so $Re χ_{d} (t_{0}) > ∣ χ_{d} (t_{0}) ∣^{2}$ . The entanglement fidelity of $R_{4}$ is consequently strictly greater than that of transpose recovery throughout some open time interval containing $t_{0}$ . There are such crossings at arbitrarily large times.

A decoder that corrects the decay-branch phase has fidelity at least as large as $R_{4}$ at every positive-decay time; the comparison with transpose recovery therefore does not assert that $R_{4}$ outperforms phase-corrected recovery. If $Δ = 0$ and $γ t > 0$ , $χ_{d} = 1$ and the two compared fidelities tie. If $γ t = 0$ , the decay branch has zero weight, no normalized decay coherence is assigned, and both recoveries act as the identity on the code.

Statement references

Four-level spontaneous-decay model

Four-level syndrome recovery

Conditional decoded state and decay coherence

Four-level fidelity measures for the unrotated decoder

Explicit four-level transpose decoder

Transpose recovery for a flagged coherence channel

The Axiom of Choice

Entanglement fidelity on a finite logical space

Facts & Assumptions

[F1]

The postulated model has a two-dimensional code, an orthogonal two-dimensional decay space, real $Δ$ , rate $γ \geq 0$ , and the stated recovery $R_{4}$ . For $γ t > 0$ , the decay probability is $p = 1 - e^{- γ t}$ and the conditional recovered decay branch multiplies logical $01$ coherence by the displayed $χ_{d}$ . Four-level spontaneous-decay model Four-level syndrome recovery Conditional decoded state and decay coherence

[F2]

Under the Axiom of Choice, the unrotated recovered channel has entanglement fidelity $F_{4} = 1 - p (1 - Re χ_{d}) / 2$ when $p > 0$ . Four-level fidelity measures for the unrotated decoder The Axiom of Choice

[F3]

For $p > 0$ , the model's transpose-recovered channel has coherence multiplier $(1 - p) + p ∣ χ_{d} ∣^{2}$ , and $0 < ∣ χ_{d} ∣ \leq 1$ . At $p = 0$ the transpose recovery acts as the identity after noise on the code. Explicit four-level transpose decoder

[F4]

A qubit coherence channel with multiplier $λ$ has entanglement fidelity $(1 + Re λ) / 2$ . The phase-corrected flagged decoder has overall multiplier $(1 - p) + p ∣ χ_{d} ∣$ . Qubit coherence channels and fidelity functionals Transpose recovery for a flagged coherence channel

[F5]

The complex exponential is entire with complex derivative equal to itself, and $e^{i x} = \cos x + i \sin x$ and $∣ e^{i x} ∣ = 1$ for real $x$ . The real exponential is positive and tends to zero at negative infinity. 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 The exponential tends to +∞ at +∞ and to 0 at -∞

[F6]

A continuous real function on a closed interval takes every value between its endpoint values. Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on [a,b] takes every value between f(a) and f(b)

[F7]

The real exponential is strictly increasing on $R$ . In particular, if $γ > 0$ and $t \geq a$ , then $e^{- γ t} \leq e^{- γ a}$ . The exponential function is strictly increasing

[F8]

For real $x$ , $\sin (x + π) = - \sin x$ and $\cos (x + π) = - \cos x$ ; moreover $\sin (π / 2) = 1$ , $\cos (π / 2) = 0$ , $\sin π = 0$ and $\cos π = - 1$ . The shifts at $x = 0$ consequently give $\sin 0 = 0$ and $\cos 0 = 1$ . Both sine and cosine have period $2 π$ . Also $\sin (- x) = - \sin x$ and $\cos (- x) = \cos x$ . Quarter-turn values and shifts by pi/2 and pi The zero sets of sine and cosine and the least positive common period 2 pi Parity and the Pythagorean identity for sine and cosine

[F9]

Sine is increasing on $[- π / 2, π / 2]$ , and cosine is decreasing on $[0, π]$ . With the endpoint values in [F8], this gives $\sin θ \geq 0$ and $\cos θ \geq 0$ for $0 \leq θ \leq π / 2$ . Signs, monotonicity intervals, and ranges of sine and cosine

[F10]

For $A \subseteq R$ , continuity of $f : A \to R$ at $s \in A$ means that for every $ε > 0$ some $δ > 0$ makes $∣ f (t) - f (s) ∣ < ε$ whenever $t \in A$ and $∣ t - s ∣ < δ$ . Constants, finite sums, scalar multiples and products of continuous real functions are continuous. Their quotient is continuous on the subset where its denominator is nonzero. Continuity of f : A → ℝ at a point of A and on A: the ε-δ condition, its agreement with lim_x → c f(x) = f(c) at a limit point, and continuity at an isolated point Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function

[F11]

For a logical-qubit channel $E$ , its entanglement fidelity at $I / 2$ is $F_{e} (E) = ⟨ Φ ∣ (i d \otimes E) (∣ Φ ⟩ ⟨ Φ ∣) ∣ Φ ⟩$ , where $∣ Φ ⟩ = (∣ 00 ⟩ + ∣ 11 ⟩) / \sqrt{2}$ in the ordered reference and logical bases. Entanglement fidelity on a finite logical space

[A1]

Assume the Axiom of Choice as required by the fidelity-comparison supplier. The Axiom of Choice

Proof

Given: The model, recoveries and parameters in the Statement, together with [A1].

Goal: Establish the strict comparison at positive-phase crossings, its persistence and occurrence at arbitrarily large times, and the stated boundary and phase-corrected comparisons.

1.1givenA1F2F3F4F5F11algebra

Using the entanglement-fidelity convention [F11], for $p > 0$ , [F2]–[F4] give $F_{4} - F_{t r} = p (Re χ_{d} - ∣ χ_{d} ∣^{2}) / 2$ . At a positive-phase crossing put $r = χ_{d} (t_{0}) > 0$ . By [F3], $r \leq 1$ . If $r = 1$ , the displayed formula for $χ_{d}$ gives $e^{i Δ t_{0}} = 1 + i (Δ / γ) (1 - e^{- γ t_{0}})$ . The left side has modulus one by [F5], while the right side has modulus strictly greater than one because $γ > 0$ , $t_{0} > 0$ and $Δ \neq 0$ . Thus $0 < r < 1$ and $F_{4} - F_{t r} = p r (1 - r) / 2 > 0$ .

1.2givenF5F7F8F9algebra

It remains to show that positive-phase crossings occur arbitrarily late. First suppose $Δ > 0$ and set $a_{n} = 2 π n / Δ$ , $b_{n} = (2 π n + π / 2) / Δ$ for positive integers $n$ , and $q (t) = e^{- γ t}$ . Since $a_{n} \to \infty$ , [F5] gives $q (a_{n}) \to 0$ , so every sufficiently large $n$ satisfies $q (a_{n}) < \min (1, Δ / γ)$ . For $t \in [a_{n}, b_{n}]$ , set $θ = Δ t - 2 π n \in [0, π / 2]$ . By [F8], $\cos (Δ t) = \cos θ$ and $\sin (Δ t) = \sin θ$ . By [F9], both are nonnegative; by [F5], $\cos^{2} θ + \sin^{2} θ = ∣ e^{i θ} ∣^{2} = 1$ . Hence $(\cos θ + \sin θ)^{2} = 1 + 2 \cos θ \sin θ \geq 1$ , so $\cos θ + \sin θ \geq 1$ . It follows that $γ \cos θ + Δ \sin θ \geq \min (γ, Δ) (\cos θ + \sin θ) \geq \min (γ, Δ)$ . Also $q (t) \leq q (a_{n})$ by [F7].

1.3givenF5F10algebra

We supply the continuity inputs explicitly. Fix a complex constant $c$ and real $s$ , and put $z_{*} = c s$ . Differentiability in [F5] gives $η > 0$ such that $∣ e^{z_{*} + h} - e^{z_{*}} ∣ \leq M ∣ h ∣$ for $∣ h ∣ < η$ , where $M = ∣ e^{z_{*}} ∣ + 1$ : for $h \neq 0$ the difference quotient is within $1$ of $e^{z_{*}}$ , and for $h = 0$ the inequality is immediate. Given $ε > 0$ , take $δ = \min {η, ε / M} / (∣ c ∣ + 1)$ . If $∣ t - s ∣ < δ$ , substituting $h = c (t - s)$ gives $∣ e^{c t} - e^{c s} ∣ < ε$ . For $w = a + i b$ , $a^{2}, b^{2} \leq a^{2} + b^{2} = ∣ w ∣^{2}$ , so $∣ Re w ∣, ∣ Im w ∣ \leq ∣ w ∣$ . Apply these inequalities to $w = e^{c t} - e^{c s}$ . Taking $c = i Δ$ and $c = - γ$ , the real functions $C (t) = \cos (Δ t)$ , $S (t) = \sin (Δ t)$ and $q (t) = e^{- γ t}$ are continuous by the Cartesian formula in [F5] and the definition in [F10]. Restricting their domain to any interval preserves the same estimates.

2.1givenF5F6F8F10step 1.2step 1.3algebra

Write $W (t) = (γ - i Δ) (e^{i Δ t} - q (t))$ , so $χ_{d} (t) = γ W (t) / [(γ^{2} + Δ^{2}) p (t)]$ . The prefactor of $W$ is positive on $[a_{n}, b_{n}]$ . By [F5] and [F8], $Im W (a_{n}) = - Δ (1 - q (a_{n})) < 0$ and $Im W (b_{n}) = γ + Δ q (b_{n}) > 0$ . Throughout this interval, step 1.2 gives $Re W (t) = γ (\cos θ - q (t)) + Δ \sin θ \geq \min (γ, Δ) - γ q (a_{n}) > 0$ . The real function $Im W (t) = γ S (t) - Δ (C (t) - q (t))$ is continuous by step 1.3 and [F10], so [F6] gives an interior zero, at which $χ_{d}$ is positive real. The endpoint signs show the phase passes from below to above a multiple of $2 π$ within this right-half-plane interval. Every sufficiently large $n$ gives such a crossing, and $a_{n} \to \infty$ . For $Δ < 0$ , [F5] and [F8] give $e^{- i x} = \overline{e^{i x}}$ for real $x$ , so the displayed coherence is the complex conjugate of the one for $∣ Δ ∣$ ; the same construction applies with the phase direction reversed.

2.2givenF5F7F10step 1.1step 1.3algebra

For $γ > 0$ and $t > 0$ , [F7] gives $q (t) < 1$ , so $p (t) = 1 - q (t) > 0$ and $D (t) = (γ^{2} + Δ^{2}) p (t) > 0$ . Rationalizing the displayed formula for $χ_{d}$ gives its real and imaginary parts as $u (t) = γ [γ (C (t) - q (t)) + Δ S (t)] / D (t)$ and $v (t) = γ [γ S (t) - Δ (C (t) - q (t))] / D (t)$ . Step 1.3 and [F10] show that $p, D, u, v$ are continuous on $(0, \infty)$ : each numerator and denominator is a finite sum of scalar multiples and products of the continuous real functions $C, S, q$ and constants, and each denominator is nonzero on this domain. Since $∣ χ_{d} ∣^{2} = u^{2} + v^{2}$ , the fidelity difference is the continuous real function $G (t) = p (t) (u (t) - u (t)^{2} - v (t)^{2}) / 2$ , again by [F10]. At a crossing $t_{0}$ , step 1.1 gives $G (t_{0}) > 0$ . Apply the definition in [F10] with $ε = G (t_{0}) / 2$ , and choose a positive radius smaller than both its $δ$ and $t_{0}$ . On the resulting open interval about $t_{0}$ one has $G (t) > G (t_{0}) / 2 > 0$ .

3.1givenF1F2F3F4algebra∎

By [F4], the phase-corrected fidelity is $1 - p (1 - ∣ χ_{d} ∣) / 2$ . Since $Re χ_{d} \leq ∣ χ_{d} ∣$ , it is at least $F_{4}$ . If $Δ = 0$ and $p > 0$ , direct substitution gives $χ_{d} = 1$ , so [F2] and [F3] give equal fidelity one. If $γ t = 0$ , [F1] gives zero decay weight and an identity null branch, while [F3] gives identity action for transpose recovery; no normalized decay state is needed. These establish all assertions.

Depends on

Dependency tree · 2 levels

Each result is drawn at its shortest distance from this item. Arrows run from prerequisites to the items that use them; this item has a heavier outline. External source nodes link directly to the cited text. The view shows up to two dependency steps; open a prerequisite to continue.

Sources

Monolithic Quantum Error Correction in the Presence of Distinguishability · PDF p. 9, recovery-comparison paragraph immediately before Table S1.1