PropositionStatement: 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.

Unitary conjugation is invariant under a global phase

Statement

Let $H, K$ be nonzero finite-dimensional complex inner-product spaces of equal dimension, let $U : H \to K$ be a unitary map, and let $z \in C$ satisfy $∣ z ∣ = 1$ . Then $z U$ is unitary and ${Ad}_{z U} = {Ad}_{U} .$ This equality concerns conjugation by a single unitary. It makes no assertion about controlled uses, where a scalar on one control branch may be a relative phase.

Facts & Assumptions

Given: Nonzero finite-dimensional complex inner-product spaces $H, K$ , a surjective linear isometry $U : H \to K$ , and a scalar $z$ with $∣ z ∣ = 1$ .

[F1]

Unitary conjugation map: a unitary map is a surjective linear isometry, and ${Ad}_{U} (X) = U X U^{†}$ .

[F2]

Inner products separate vectors, and the induced norm is homogeneous: $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ : $∥ λ x ∥ = ∣ λ ∣ ∥ x ∥$ .

[F3]

Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ : $(z U)^{†} = \overline{z} U^{†}$ .

[F4]

Finite-dimensional linear isometries satisfy the adjoint identity: every surjective linear isometry between finite-dimensional spaces satisfies both adjoint identities.

Proof

1.1givenF1F2F4choosealgebra

For each $x \in H$ , [F2] and $∣ z ∣ = 1$ give $∥ z U x ∥ = ∣ z ∣ ∥ U x ∥ = ∥ x ∥$ . The map $z U$ is surjective: for any $y \in K$ , choose $x \in H$ with $U x = \overline{z} y$ , then $z U x = y$ . Hence $z U$ is a surjective linear isometry and is unitary by [F1]; [F4] supplies its adjoint identities.

2.1givenF1F3step 1.1algebra∎

For every $X \in L (H)$ , [F1] and [F3] give ${Ad}_{z U} (X) = (z U) X (z U)^{†} = z \overline{z} U X U^{†} = ∣ z ∣^{2} {Ad}_{U} (X) = {Ad}_{U} (X)$ . This holds on every input operator, so the maps are equal. The proof concerns only a single conjugation map and yields no controlled-operation identity.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

16 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

John Watrous, The Theory of Quantum Information (2018), §2.2.1 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 3, §3.2.1 (standard reference, not scraped)