ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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.

Controlled Global Phase Changes an X-Outcome

Example

Take the target qubit unitary $U = I_{2}$ , control input $∣ + ⟩ = (∣ 0 ⟩ + ∣ 1 ⟩) / \sqrt{2}$ , and target input $∣ 0 ⟩$ . Separately grant controlled access to $C (I_{2})$ and $C (e^{i θ} I_{2})$ . For the finite control effects $E_{\pm} = ∣ \pm ⟩ ⟨ \pm ∣ \otimes I_{2}$ , where $∣ \pm ⟩ = (∣ 0 ⟩ \pm ∣ 1 ⟩) / \sqrt{2}$ , the plus outcome has probability $1$ for $C (I_{2})$ and $\cos^{2} (θ / 2)$ for $C (e^{i θ} I_{2})$ .

Facts & Assumptions

Given: A control-first two-qubit register, $U = I_{2}$ , real $θ$ , the stated input, separate controlled access to the two specified matrices, and the two control effects in the Example.

[L1]

A Controlled Global Phase Becomes a Relative Control Phase proves the general eigenvector witness, verifies the finite two-outcome POVM, and computes its plus probability.

[L2]

Controlled Unitary Access defines the control-first block action and requires each capability to be explicitly granted. Its carrier is present in this batch.

[L3]

Dirac Notation Conversion defines the Dirac pairing by reversing the root linear-first pairing and defines the corresponding outer product.

[L4]

A unit-vector outer product is a density operator proves that every unit-vector outer product is a density operator.

[L5]

Finite POVMs and their outcome law defines the finite POVM outcome weight using the density operator and effect.

[L6]

Ordered tensor registers, their inner product, and wire permutations defines the control-first tensor order and product inner product. Its current draft proof was read.

[L7]

The computational basis is orthonormal proves the one-qubit computational basis is orthonormal. Its current draft proof was read.

[L8]

POVM outcome weights are probabilities proves the finite weights of this POVM form a probability distribution.

[F1]

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ : for real $t$ , $∣ e^{i t} ∣ = 1$ and $e^{i t} = \cos t + i \sin t$ .

Verification

1.1givenL1L2L4L6L7algebra

In the ordered basis $∣ 00 ⟩, ∣ 01 ⟩, ∣ 10 ⟩, ∣ 11 ⟩$ , the first controlled block is $C (I_{2}) = (\begin{array}{cc} I_{2} & 0 \\ 0 & I_{2} \end{array}) = I_{4}$ . It sends the input to $ω_{0} = ∣ + ⟩ \otimes ∣ 0 ⟩$ . Both factors are unit vectors by [L7], so their product is unit by [L6], and its rank-one projector is a density operator by [L4]. In [L1], specialize $U = I_{2}$ , $ϕ = ∣ 0 ⟩$ , and $μ = 1$ ; its verified control POVM then gives plus probability $1$ .

1.2givenL1L2L3L4L5L6L7F1algebra

The separately granted second block is $C (e^{i θ} I_{2}) = (\begin{array}{cc} I_{2} & 0 \\ 0 & e^{i θ} I_{2} \end{array}) = diag (1, 1, e^{i θ}, e^{i θ})$ . It sends the input to $ω_{θ} = χ_{θ} \otimes ∣ 0 ⟩$ , where $χ_{θ} = (∣ 0 ⟩ + e^{i θ} ∣ 1 ⟩) / \sqrt{2}$ . By [L3] and [L7], $⟨ + ∣ χ_{θ} ⟩_{D} = ⟨ χ_{θ}, + ⟩_{r o o t} = (1 + e^{i θ}) / 2$ . The phase identity [F1] and orthonormality [L7] give $∥ χ_{θ} ∥^{2} = (1 + ∣ e^{i θ} ∣^{2}) / 2 = 1$ ; then the product pairing [L6] gives $∥ ω_{θ} ∥ = 1$ . Its projector is a density operator by [L4], so the outcome law [L5] applies and gives $∣ (1 + e^{i θ}) / 2 ∣^{2}$ , which equals $\cos^{2} (θ / 2)$ by the derivation in the proof of [L1].

2.1step 1.1step 1.2L1L2L5L8algebra∎

At $θ = 0$ , the second controlled matrix is $I_{4}$ and its plus probability is $1$ . At $θ = π$ , its diagonal is $(1, 1, - 1, - 1)$ and its plus probability is $0$ . The minus probability is the complement by normalization [L8], so these are distinct finite outcome laws under the explicit separate-access assumption.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

53 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 6: Quantum Algorithms (standard reference, not scraped)