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.

Adaptive J angle and Pauli-frame update

Statement

Let a one-qubit input carry the Pauli frame $X^{x} Z^{z}$ , where $x, z \in {0, 1}$ , and let $β \in R$ . Choose the next equatorial measurement angle $(- 1)^{x} β$ . For outcome $b \in {0, 1}$ , the one-bit teleportation branch is

$2^{- 1 / 2} γ (x, z, β) X^{b \oplus z} Z^{x} J (β), γ (x, z, β) = {\begin{cases} 1, & x = 0, \\ (- 1)^{z} e^{- i β}, & x = 1. \end{cases}$

Thus the updated Pauli frame is $(x^{'}, z^{'}) = (b \oplus z, x)$ , and the exact branch scalar has unit modulus.

Facts & Assumptions

Given: A real angle $β$ , frame bits $x, z \in {0, 1}$ , and an outcome $b \in {0, 1}$ .

[F1]

The one-bit branch operator is $2^{- 1 / 2} X^{b} J (α)$ for measurement angle $α$ (One-bit J-gate teleportation branch identity).

[F2]

The exact phase identity is $J (- β) X = e^{- i β} Z J (β)$ (J-gate basis and exact phase identities).

[F3]

The current matrix definitions and Hadamard action give $J (β) Z = X J (β)$ and $Z X = - X Z$ (J gate and ordered equatorial measurement vectors, Exact one-qubit gates and rotations, The Pauli matrices, Hadamard pauli action).

Proof

1.1F1F3algebra

When $x = 0$ , the chosen angle is $β$ . By [F3], $J (β) Z^{z} = X^{z} J (β)$ , so the branch in [F1] is $2^{- 1 / 2} X^{b} X^{z} J (β) = 2^{- 1 / 2} X^{b \oplus z} Z^{0} J (β)$ . This has the asserted phase and frame.

1.2F1F2F3algebra

When $x = 1$ , the chosen angle is $- β$ . By [F2] and [F3], $J (- β) X Z^{z} = e^{- i β} Z J (β) Z^{z} = (- 1)^{z} e^{- i β} X^{z} Z J (β)$ . Multiplying by the outcome factor $X^{b}$ gives $2^{- 1 / 2} (- 1)^{z} e^{- i β} X^{b \oplus z} Z J (β)$ , as claimed.

2.1step 1.1step 1.2algebra∎

The two cases give $x^{'} = b \oplus z$ and $z^{'} = x$ , while $∣ γ ∣ = 1$ because $β$ is real. Thus the angle adaptation changes only the known Pauli frame and a branch-global phase.

Depends on

Used by

Dependency tree · two levels

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Sources

Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415, §20.6, p. 179 (standard reference, not scraped)
Robert Raussendorf, Daniel E. Browne, and Hans J. Briegel, Measurement-based quantum computation on cluster states, arXiv:quant-ph/0301052, §§II B and II E, pp. 5–13 (standard reference, not scraped)