LemmaStatement: 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.

Dense normalized one-qubit Clifford–T package

Statement

Use the exact one-qubit matrices $H$ and $T = diag (1, e^{i π / 4})$ of Exact one-qubit gates and rotations. The matrices $h = i H$ and $t = e^{- i π / 8} T$ , together with their inverses, are determinant-one representatives in $S U (2)$ . Their generated subgroup is dense in $S U (2)$ . Replacing each normalized letter in a word by its underlying physical $H, T, T^{†}$ letter changes the product only by one scalar phase, so an unconditional one-qubit circuit can approximate every $S U (2)$ target up to an irrelevant global phase. A normalized letter must not be placed under a control without correcting its scalar phase.

Facts & Assumptions

Given: The exact one-qubit $H, T$ matrices and the dense-projective Clifford–T result.

[F1]

Clifford+T Is Dense Projectively on One Qubit: $h = i H$ and $t = e^{- i π / 8} T$ lie in $S U (2)$ and generate a dense subgroup there.

[F2]

Exact one-qubit gates and rotations: $H$ and $T$ have their displayed exact phases; a scalar phase is part of a matrix when it is controlled.

Proof

1.1F2algebra

Direct determinants give $\det (i H) = i^{2} \det (H) = 1$ because $\det (H) = - 1$ , and $\det (e^{- i π / 8} T) = e^{- i π / 4} e^{i π / 4} = 1$ . Unitarity is preserved by unit-modulus scalars, so $h, t \in S U (2)$ .

2.1F1F2algebra∎

Density of the group generated by $h, t$ is [F1]. For any fixed finite word, replacing each normalized generator by its physical gate multiplies the whole word by the product of the corresponding scalar phases, which is a single scalar on the entire one-qubit subcircuit. Such a scalar does not affect any outcome probability when the subcircuit is applied unconditionally. If the word is controlled, this scalar acts only on the active control block and is therefore not irrelevant; the statement explicitly excludes that use without correction.

Depends on

Used by

Effective synthesis of fixed-dimensional unitaries Lemma

Dependency tree · two levels

22 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

Christopher M. Dawson and Michael A. Nielsen, The Solovay–Kitaev Algorithm (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes (standard reference, not scraped)