ExampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-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.

Small hypercube transfer

Example

For the one-dimensional Boolean hypercube, $A_{1} = X$ . For the two-dimensional Boolean hypercube in the ordered basis $(∣ 00 ⟩, ∣ 01 ⟩, ∣ 10 ⟩, ∣ 11 ⟩)$ , $A_{2} = X \otimes I + I \otimes X = (\begin{array}{cccc} 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \end{array}) .$ Compute the exact evolution from $∣ 0^{n} ⟩$ at time $π / 2$ and verify the measured opposite corner.

Facts & Assumptions

Given: The edge adjacency $X$ , the ordered tensor register, and the finite-matrix evolution $U_{H} (t)$ .

[F1]

Hypercube continuous-time state transfer: $A_{n} = \sum_{j < n} X_{j}$ and the input is $e_{0^{n}}$ .

[F2]

Finite registers and their ordered computational bases: the qubit basis is the ordered standard basis $(e_{0}, e_{1})$ and the tensor-register basis is ordered lexicographically.

[F3]

Ordered tensor registers, their inner product, and wire permutations: specified factor bases give their lexicographically ordered product basis; the product inner product is the product of the factor inner products, and the empty tensor is $C$ .

[F4]

The computational basis is orthonormal: the listed register basis is orthonormal.

[F5]

Finite Hermitian matrix dynamics: $U_{H} (t)$ acts on a $λ$ eigenvector of $H$ by $e^{- i t λ}$ .

[F6]

$\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential: $e^{z + w} = e^{z} e^{w}$ .

[F7]

$\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ : $e^{i t} = \cos t + i \sin t$ for real $t$ .

[F8]

Quarter-turn values and shifts by pi/2 and pi: $\cos (π / 2) = 0$ and $\sin (π / 2) = 1$ .

[F9]

The measurement law for a finite continuous-time walk: basis measurement probability is the squared modulus of the corresponding output coordinate.

[F10]

Exact state transfer on the Boolean hypercube: for each $n \geq 0$ , $U_{A_{n}} (π / 2) e_{0^{n}} = (- i)^{n} e_{1^{n}}$ and the measured output is $1^{n}$ with probability one.

[F11]

Tensor products of finitely many linear maps: tensor maps act factorwise on elementary tensors, including the empty tensor map as the identity.

Verification

technique · direct

1.1givenF2F3F4F5F6F7F8algebra

Put $f_{+} = (e_{0} + e_{1}) / \sqrt{2}$ and $f_{-} = (e_{0} - e_{1}) / \sqrt{2}$ . By [F2]–[F4] they form an orthonormal basis, and direct multiplication gives $X f_{+} = f_{+}$ and $X f_{-} = - f_{-}$ . By [F5], $U_{X} (t)$ acts on these vectors by $e^{- i t}$ and $e^{i t}$ . On $f_{+}$ , $\cos t I - i \sin t X$ acts by $\cos t - i \sin t = e^{- i t}$ ; on $f_{-}$ it acts by $\cos t + i \sin t = e^{i t}$ , using [F6]–[F7]. Thus the two maps agree on a basis. At $t = π / 2$ , [F8] yields $U_{X} (π / 2) = - i X$ , so $U_{X} (π / 2) e_{0} = - i e_{1}$ .

2.1F1F2F3F5F6F11step 1.1algebra

The displayed matrix for $A_{2}$ follows by applying $X \otimes I + I \otimes X$ to each of the four basis vectors in [F2] using the tensor-map action [F11]. The four vectors $f_{+} \otimes f_{+}$ , $f_{+} \otimes f_{-}$ , $f_{-} \otimes f_{+}$ , and $f_{-} \otimes f_{-}$ form a basis by [F2]–[F3]. On $f_{ϵ} \otimes f_{η}$ with $ϵ, η \in {1, - 1}$ , $A_{2}$ has eigenvalue $ϵ + η$ ; [F5]–[F6] give evolution coefficient $e^{- i π (ϵ + η) / 2} = e^{- i π ϵ / 2} e^{- i π η / 2}$ , which is exactly the coefficient of $U_{X} (π / 2) \otimes U_{X} (π / 2)$ by [F11]. The operators therefore agree on a basis, and step 1.1 gives $U_{A_{2}} (π / 2) = - X \otimes X$ . Hence $U_{A_{2}} (π / 2) e_{00} = - e_{11}$ .

3.1step 1.1step 2.1F9F10algebra∎

In the $n = 1$ case, step 1.1 gives outcome $1$ amplitude $- i$ ; in the $n = 2$ case, step 2.1 gives outcome $11$ amplitude $- 1$ . Both have squared modulus $1$ , so [F9] gives the opposite-corner outcome with probability one in each case. These calculations are the $n = 1, 2$ instances of [F10].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms, Chapter 17 §17.2 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, Chapter 9 §9.1 (standard reference, not scraped)