DefinitionDefinition: AI-adaptedProof: 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.

Multi-Control Activation Pattern

Definition

Let $m \geq 1$ , let the ordered control register be $Q_{2}^{\otimes m}$ with computational basis $∣ x ⟩$ indexed by $x = (x_{1}, \dots, x_{m}) \in {0, 1}^{m}$ , and let $T$ be a finite-dimensional target register disjoint from the controls. Let $U : T \to T$ be unitary, and let $s = (s_{1}, \dots, s_{m}) \in {0, 1}^{m}$ be the activation string. Define the multi-controlled operation by

$C_{s} (U) (∣ x ⟩ \otimes v) = {\begin{cases} ∣ s ⟩ \otimes U v, & x = s, \\ ∣ x ⟩ \otimes v, & x \neq s . \end{cases}$

Extend this rule linearly. The all-one control is $C_{1^{m}} (U)$ . For negative controls in $s$ , let $N = {j \in {1, \dots, m} : s_{j} = 0}$ and let $X_{N}$ be the tensor product of Pauli $X$ on controls in $N$ and identity on the other controls. Then

$C_{s} (U) = (X_{N} \otimes I_{T}) C_{1^{m}} (U) (X_{N} \otimes I_{T}),$

with the rightmost $X_{N}$ applied first and the leftmost $X_{N}$ restoring the original control string. The matrix definition does not itself grant an implementation or assign a cost to the operation.

Facts & Assumptions

Given: The ordered control basis, $m \geq 1$ , a finite-dimensional target register $T$ , an activation string $s \in {0, 1}^{m}$ , and a unitary $U : T \to T$ .

[L1]

Controlled Unitary Access fixes the positive-control block convention $C (U) = diag (I, U)$ and treats controlled access as an explicit model capability. Its carrier is present in this batch.

[L2]

The Pauli matrices defines the exact Pauli- $X$ matrix. Its current draft definition body was read; the QC07 page carrier exists but remains draft.

[L3]

Tensor products of finitely many linear maps defines the tensor map on elementary tensors and its linear extension. Its current draft proof was read and supplies the simultaneous control-flip/target-identity map.

[L4]

Ordered tensor registers, their inner product, and wire permutations defines the ordered tensor register, computational product basis, and product pairing. Its current draft proof was read.

[F1]

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces: a unitary endomorphism is an invertible linear norm-preserving map.

Verification

1.1givenL4construct

By [L4], the vectors $∣ x ⟩ \otimes v$ over the computational control basis span the joint register. The cases in the Definition therefore determine a unique linear operator and give the activation behavior on every superposition.

2.1givenL1L2L3L4step 1.1algebra

Let $n = (n_{1}, \dots, n_{m}) \in {0, 1}^{m}$ have $n_{j} = 1$ exactly when $s_{j} = 0$ . By [L2] and [L4], $X_{N} \otimes I_{T}$ sends $∣ x ⟩ \otimes v$ to $∣ x \oplus n ⟩ \otimes v$ . The middle positive-control block activates exactly when $x \oplus n = 1^{m}$ , which coordinatewise is equivalent to $x = s$ . The leftmost $X_{N}$ restores the original control string because $n \oplus n = 0$ . Thus the conjugated operation has exactly the defining action of $C_{s} (U)$ , including $N = \emptyset$ and $N = {1, \dots, m}$ .

3.1givenL4F1algebra∎

Write any joint vector uniquely as $v = \sum_{x \in {0, 1}^{m}} ∣ x ⟩ \otimes v_{x}$ using the ordered product basis in [L4]. The Definition maps each block to $W_{x} v_{x}$ , where $W_{s} = U$ and $W_{x} = I_{T}$ otherwise. Each $W_{x}$ is an invertible isometry by [F1], so orthogonality of the control basis and the product pairing give $∥ C_{s} (U) v ∥^{2} = \sum_{x} ∥ W_{x} v_{x} ∥^{2} = \sum_{x} ∥ v_{x} ∥^{2} = ∥ v ∥^{2}$ . The block map with $W_{s} = U^{- 1}$ and other blocks identity is a two-sided inverse. Hence $C_{s} (U)$ is an invertible isometry, therefore unitary by [F1].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 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)