DefinitionDefinition: Literature-sourcedProof: AI-adaptedPipeline-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.

Two-Level Unitary

Definition

Let a finite ordered register have an orthonormal computational basis $(∣ z ⟩)_{z \in X}$ , using the root convention that the inner product is linear in its first argument. For distinct labels $x, y \in X$ , choose $V \in U (2)$ , where $U (2) = {M \in M_{2} (C) : M^{*} M = M M^{*} = I_{2}}$ . The two-level unitary $T_{x, y} (V)$ acts by the matrix $V$ in the ordered basis $(∣ x ⟩, ∣ y ⟩)$ and acts as the identity on $span {∣ x ⟩, ∣ y ⟩}^{⊥}$ . Its action is exact matrix equality: any scalar phase contained in $V$ is retained.

Facts & Assumptions

Given: The finite-dimensional complex inner-product space, its ordered orthonormal computational basis, distinct labels $x, y$ , and $V \in U (2)$ .

[F1]

For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ : a subspace $S$ of a finite-dimensional inner-product space has the orthogonal direct sum decomposition $H = S \oplus S^{⊥}$ .

[F2]

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

[F3]

The computational basis is orthonormal: the computational basis vectors of the finite ordered register form an orthonormal basis.

Verification

1.1givenF1F3algebra

Put $S = span {∣ x ⟩, ∣ y ⟩}$ . By [F3], the selected pair is orthonormal. By [F1], $H = S \oplus S^{⊥}$ , and relative to this sum the defined operator is $V \oplus I_{S^{⊥}}$ .

2.1step 1.1F2algebra∎

The direct sum preserves norms because $V$ is unitary on $S$ and the identity preserves norms on $S^{⊥}$ . Its inverse is $V^{*} \oplus I_{S^{⊥}}$ , so it is an invertible isometry and hence unitary by [F2]. The scalar phase in $V$ is part of this exact direct-sum matrix.

Depends on

Used by

Exact Two-Level Elimination of a Unitary Algorithm
Gray-Path Synthesis of a Two-Level Gate Algorithm
A Two-Level Rotation on Two Qubits Example
Effective synthesis of fixed-dimensional unitaries Lemma
Correctness of Gray-Path Two-Level Synthesis Theorem

Dependency tree · two levels

26 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

Adriano Barenco et al., Elementary gates for quantum computation, arXiv:quant-ph/9503016v1 (standard reference, not scraped)