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.

Hermitian block-encoding signal subspaces

Statement

Let $(U, J, α)$ be an exact Hermitian block encoding from Qubitization access model, and choose an orthonormal eigenbasis $K ψ_{j} = λ_{j} ψ_{j}$ . Put $x_{j} = λ_{j} / α$ and $s_{j} = \sqrt{1 - x_{j}^{2}}$ . For $∣ x_{j} ∣ < 1$ , define $η_{j} = (I - Π) U J ψ_{j} / s_{j}, {\tilde{η}}_{j} = (I - Π) U^{*} J ψ_{j} / s_{j} .$ The $η_{j}$ are orthonormal in $\ker Π$ , as are the ${\tilde{η}}_{j}$ . In the paired input basis $(J ψ_{j}, {\tilde{η}}_{j})$ and output basis $(J ψ_{j}, η_{j})$ , $U$ has matrix $(\begin{array}{cc} x_{j} & s_{j} \\ s_{j} & - x_{j} \end{array}) .$ The unitary $U^{*}$ maps in the reverse direction by the same matrix. If $∣ x_{j} ∣ = 1$ , the complement is absent and $U J ψ_{j} = x_{j} J ψ_{j} = U^{*} J ψ_{j}$ ; no division by zero is used. The value $x_{j} = 0$ is an ordinary two-dimensional signal plane.

Facts & Assumptions

Given: A finite-dimensional self-adjoint $K$ , an exact block encoding, and the associated signal projector $Π$ .

[F1]

Qubitization access model: $J^{*} U J = K / α$ and $Π = J J^{*}$ ; taking adjoints and using $K = K^{*}$ gives $J^{*} U^{*} J = K / α$ .

[F2]

PUE singular-space invariant decomposition: the positive singular pairs of a projected unitary give orthogonal paired two-dimensional subspaces; their endpoint complements are absent.

[F3]

Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely: a finite-dimensional self-adjoint endomorphism has an orthonormal eigenbasis with real eigenvalues.

[F4]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ : $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ , so a unitary and its adjoint preserve inner products.

[F5]

Finite block encoding: the normalized compression has operator norm at most one, hence $∣ λ_{j} / α ∣ \leq 1$ .

Proof

Given: As in Facts & Assumptions.

1.1givenF1F4F5algebra

Fix an eigenvector $ψ_{j}$ with unit norm and write $v_{j} = J ψ_{j}$ . From [F1], $Π U v_{j} = x_{j} v_{j}$ and $Π U^{*} v_{j} = x_{j} v_{j}$ , because $K = K^{*}$ . Since $U, U^{*}$ preserve norm, orthogonal decomposition by $Π$ gives $∥ (I - Π) U v_{j} ∥^{2} = ∥ (I - Π) U^{*} v_{j} ∥^{2} = 1 - x_{j}^{2} = s_{j}^{2}$ . For $∣ x_{j} ∣ < 1$ , the displayed $η_{j}, {\tilde{η}}_{j}$ are therefore unit vectors in $\ker Π$ .

2.1F1F3F4step 1.1algebra

If $i \neq j$ , orthogonality of eigenvectors and [F4] give $⟨ U v_{i}, U v_{j} ⟩ = ⟨ v_{i}, v_{j} ⟩ = 0$ , while $⟨ Π U v_{i}, Π U v_{j} ⟩ = x_{i} x_{j} ⟨ v_{i}, v_{j} ⟩ = 0$ . Subtracting the projected components shows $⟨ η_{i}, η_{j} ⟩ = 0$ ; the same calculation for $U^{*}$ proves orthogonality of the ${\tilde{η}}_{j}$ .

2.2F1F4step 1.1algebra

Decomposing $U v_{j}$ into its $Π$ and $I - Π$ parts gives $U v_{j} = x_{j} v_{j} + s_{j} η_{j}$ . Also $s_{j} U {\tilde{η}}_{j} = U (I - Π) U^{*} v_{j} = v_{j} - U Π U^{*} v_{j} = v_{j} - x_{j} U v_{j} = s_{j}^{2} v_{j} - x_{j} s_{j} η_{j} .$ Division by $s_{j} > 0$ yields $U {\tilde{η}}_{j} = s_{j} v_{j} - x_{j} η_{j}$ , which is the second column of the displayed matrix. Taking adjoints gives the reverse-basis action of $U^{*}$ .

3.1F1F2F3step 1.1algebra∎

If $∣ x_{j} ∣ = 1$ , the orthogonal complement of $Π U v_{j} = x_{j} v_{j}$ has squared norm $1 - ∣ x_{j} ∣^{2} = 0$ , so $U v_{j} = x_{j} v_{j}$ ; the same argument for $U^{*}$ gives $U^{*} v_{j} = x_{j} v_{j}$ . Thus the endpoint plane collapses. At $x_{j} = 0$ , step 1.1 has $s_{j} = 1$ , so both normalized vectors are well-defined. The mutually orthogonal paired spaces and endpoint lines exhaust the clean input and output signal spaces by the SVD decomposition in [F2]; their orthogonal residual complements stay in $\ker Π$ and cannot contribute to the clean compression.

Depends on

Used by

Two-dimensional Hermitian block-encoding signal Example
Qubitization lifts scalar QSP Theorem

Dependency tree · two levels

34 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

Andrew M. Childs, Lecture Notes on Quantum Algorithms, Chapter 29 §29.3 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, Chapter 9 §9.4 (standard reference, not scraped)
András Gilyén, Yuan Su, Guang Hao Low, and Nathan Wiebe, Quantum singular value transformation and beyond (standard reference, not scraped)