ExampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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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.

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Capacity regimes of the qubit erasure channel

Example

For $d = 2$ , the erasure formulas give the following capacities. Entries are shown in bits per channel use; multiply by $\ln 2$ to obtain nats.

$ϵ$	$Q = P$	$C_{E}$	Regime
$0$	$1$	$2$	degradable
$1 / 4$	$1 / 2$	$3 / 2$	degradable
$1 / 2$	$0$	$1$	degradable and antidegradable; self-complementary
$3 / 4$	$0$	$1 / 2$	antidegradable
$1$	$0$	$0$	antidegradable

Facts & Assumptions

Given: A qubit erasure channel, so $d = 2$ , at the five probabilities in the table.

[F1]

Exact quantum, private, and entanglement-assisted erasure capacities: $Q = P = \max {1 - 2 ϵ, 0} \ln d$ and $C_{E} = 2 (1 - ϵ) \ln d$ .

[F2]

Complement and degradation maps of the erasure channel: the receiver and complement outputs exchange the data and erasure probabilities; the channel is degradable for $ϵ \leq 1 / 2$ and antidegradable for $ϵ \geq 1 / 2$ .

[F3]

Finite-dimensional quantum erasure channel: the two outputs use the same data space with a one-dimensional orthogonal erasure flag.

[F4]

Information units and logarithm-base convention: capacities in bits are obtained by dividing nats by $\ln 2$ .

Verification

1.1F1F4givenalgebra

Substituting $d = 2$ into [F1] and dividing by $\ln 2$ gives $Q = P = \max {1 - 2 ϵ, 0}$ bits and $C_{E} = 2 (1 - ϵ)$ bits. At $ϵ = 0, 1 / 4, 1 / 2, 3 / 4, 1$ , these pairs are respectively $(1, 2), (1 / 2, 3 / 2), (0, 1), (0, 1 / 2), (0, 0)$ , which are the table entries.

2.1F2F3step 1.1constructalgebra∎

By [F2], the first two probabilities lie in the degradable region, the last two lie in the antidegradable region, and the half-erasure point lies in both. At $ϵ = 1 / 2$ , [F2] gives $E^{c} (X) = \frac{1}{2} J_{E} X J_{E}^{*} + \frac{1}{2} tr (X) ∣ e_{E} ⟩ ⟨ e_{E} ∣$ , while [F3] gives $E (X) = \frac{1}{2} J_{B} X J_{B}^{*} + \frac{1}{2} tr (X) ∣ e_{B} ⟩ ⟨ e_{B} ∣$ . The unitary identification $J_{B} x \mapsto J_{E} x$ and $e_{B} \mapsto e_{E}$ carries one map to the other, proving self-complementarity up to flag relabeling.

Depends on

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Sources

John Watrous, The Theory of Quantum Information, §8.3.2 (standard reference, not scraped)
Mark M. Wilde, From Classical to Quantum Shannon Theory, §24.7.1 (standard reference, not scraped)