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.

Agreement on pure states implies complete agreement

Statement

Let $H, K, R$ be finite-dimensional complex inner-product spaces, with the library convention that the inner product is linear in its first argument. For $x \in H$ , put $P_{x} : H \to H$ , $P_{x} (z) = ⟨ z, x ⟩ x$ (rank one when $x \neq 0$ ). Write $B (H) = {End}_{C} (H)$ . If complex-linear maps $Φ, Ψ : B (H) \to B (K)$ agree on $P_{x}$ for every unit $x \in H$ , then $Φ = Ψ$ on $B (H)$ and $Φ \otimes {id}_{B (R)} = Ψ \otimes {id}_{B (R)}$ on $B (H \otimes R)$ , under the canonical matrix-unit identification.

Facts & Assumptions

Given: Finite-dimensional complex inner-product spaces $H, K, R$ , and complex-linear maps $Φ, Ψ$ satisfying the stated rank-one equalities.

[F1]

Real and complex inner product spaces, with the inner product linear in the first argument: the inner product is linear in the first argument and conjugate-linear in the second.

[F2]

Every finite-dimensional real or complex inner product space has an orthonormal basis: $H$ and $R$ have orthonormal bases, including the empty basis in dimension zero.

[F3]

The elementary tensors of two bases form the product basis of the tensor product: elementary tensors of basis vectors form a basis of the tensor product.

[F4]

Linear map between vector spaces over the same field: a complex-linear map preserves sums and complex scalar multiples.

Proof

1.1givenF1F4algebra

Put $Δ = Φ - Ψ$ . By [F4], $Δ$ is complex-linear. For every nonzero $x \in H$ , $P_{x} = ∥ x ∥^{2} P_{x / ∥ x ∥}$ , and $P_{0} = 0$ , so linearity gives $Δ (P_{x}) = 0$ for every $x \in H$ .

2.1step 1.1F1F4algebra

With $∣ u ⟩ ⟨ v ∣$ denoting the map $z \mapsto ⟨ z, v ⟩ u$ , the linear-first convention gives $P_{u + v} - P_{u - v} = 2 (∣ u ⟩ ⟨ v ∣ + ∣ v ⟩ ⟨ u ∣)$ and $P_{u + i v} - P_{u - i v} = 2 i (∣ v ⟩ ⟨ u ∣ - ∣ u ⟩ ⟨ v ∣) .$ If $S = P_{u + v} - P_{u - v}$ and $T = P_{u + i v} - P_{u - i v}$ , then $∣ u ⟩ ⟨ v ∣ = (S - T / i) / 4$ and $∣ v ⟩ ⟨ u ∣ = (S + T / i) / 4$ . Hence $Δ (∣ u ⟩ ⟨ v ∣) = 0$ for all $u, v$ .

3.1step 2.1F2F4algebrachoose

Choose an orthonormal basis $(e_{j})$ of $H$ using [F2]. Every $T \in B (H)$ has the expansion $T = \sum_{j, k} ⟨ T e_{k}, e_{j} ⟩ ∣ e_{j} ⟩ ⟨ e_{k} ∣$ , so these matrix units form a basis of $B (H)$ . Step 2.1 makes $Δ$ zero on that basis, hence on all of $B (H)$ ; thus $Φ = Ψ$ .

4.1step 2.1step 3.1F3F4algebrachoose∎

Choose an orthonormal basis $(r_{a})$ of $R$ . By [F3], $(e_{j} \otimes r_{a})$ is a basis of $H \otimes R$ and the operators $∣ e_{j} \otimes r_{a} ⟩ ⟨ e_{k} \otimes r_{b} ∣$ form a basis of $B (H \otimes R)$ . Each equals $(∣ e_{j} ⟩ ⟨ e_{k} ∣) \otimes (∣ r_{a} ⟩ ⟨ r_{b} ∣)$ , on which $(Δ \otimes {id}_{B (R)})$ is zero by step 2.1. Therefore the two amplified maps agree on the entire operator space.

Depends on

Used by

Dependency tree · two levels

14 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 Preskill, Quantum Information, Chapter 7: Quantum Error Correction (working draft, March 2026) (standard reference, not scraped)
David W. Kribs, Raymond Laflamme, David Poulin, and Maia Lesosky, Operator Quantum Error Correction (standard reference, not scraped)