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.

Computational-basis amplitudes of a graph state

Statement

Let $G = (V, E)$ be an ordered finite simple graph with $n = ∣ V ∣$ . In the ordered computational basis indexed by $x \in F_{2}^{V}$ ,

$∣ G ⟩ = 2^{- n / 2} \sum_{x \in F_{2}^{V}} (- 1)^{\sum_{{u, v} \in E} x_{u} x_{v}} ∣ x ⟩,$

where the exponent is evaluated as an integer and only its parity affects the sign.

Facts & Assumptions

Given: The edge-product definition of $∣ G ⟩$ on the ordered register.

[F1]

The graph-state definition gives the ordered finite register, computational basis, and finite edge-gate product (Graph state of an ordered finite simple graph, Big-endian encoding of ordered bit strings, Ordered tensor registers, their inner product, and wire permutations).

[F2]

The controlled-phase definition gives $C Z ∣ x_{u} x_{v} ⟩ = (- 1)^{x_{u} x_{v}} ∣ x_{u} x_{v} ⟩$ at angle $π$ (Controlled Phase Rotation).

Proof

1.1F1algebra

Tensoring $∣ + ⟩ = 2^{- 1 / 2} (∣ 0 ⟩ + ∣ 1 ⟩)$ over the ordered finite vertex set gives $∣ + ⟩^{\otimes V} = 2^{- n / 2} \sum_{x \in F_{2}^{V}} ∣ x ⟩$ . For $n = 0$ this is the single empty tuple with coefficient one.

1.2F1F2algebra

By [F2], applying $C Z_{u v}$ to a basis vector $∣ x ⟩$ multiplies it by $(- 1)^{x_{u} x_{v}}$ and leaves its basis label unchanged. Thus the lexicographically ordered edge product multiplies $∣ x ⟩$ by the product of these signs over all edges.

2.1step 1.1step 1.2algebra∎

Multiplying the signs gives $(- 1)^{\sum_{{u, v} \in E} x_{u} x_{v}}$ , so substituting the expansion from step 1.1 into the graph-state definition proves the formula, including the empty-edge and empty-vertex cases.

Depends on

Used by

One-edge graph state and two-wire CZ pattern Example

Dependency tree · two levels

23 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

M. Hein et al., Entanglement in graph states and its applications, arXiv:quant-ph/0602096, Definition Graph States I, p. 14 (standard reference, not scraped)
Robert Raussendorf, Daniel E. Browne, and Hans J. Briegel, Measurement-based quantum computation on cluster states, arXiv:quant-ph/0301052, pp. 3–5 (standard reference, not scraped)