LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generated

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.

Orthogonal marked and unmarked components of the uniform state

Statement

Let $M \subseteq D_{n}$ have size $t$ with $0 < t < N$ . Define $∣ G ⟩ = t^{- 1 / 2} \sum_{x \in M} ∣ x ⟩, ∣ B ⟩ = (N - t)^{- 1 / 2} \sum_{x \notin M} ∣ x ⟩ .$ Then $∣ G ⟩$ and $∣ B ⟩$ are orthonormal. For $θ = \arcsin \sqrt{t / N} \in (0, π / 2)$ , the uniform state is $∣ s ⟩ = \sin θ ∣ G ⟩ + \cos θ ∣ B ⟩,$ with $\sin^{2} θ = t / N$ .

Facts & Assumptions

Given: $n \geq 0$ , $N = 2^{n}$ , a marked set $M \subseteq D_{n}$ , and $0 < t = ∣ M ∣ < N$ . Here the uniform state denotes the equal-positive-amplitude vector $∣ s ⟩ = N^{- 1 / 2} \sum_{x \in D_{n}} ∣ x ⟩$ .

[F1]

Finite marked-item search and its bit oracle: $D_{n}$ is a finite address set with marked subset $M$ and marked count $t$ .

[F2]

The computational basis is orthonormal: distinct address basis vectors are orthogonal and every address basis vector has norm $1$ .

[F3]

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

[F4]

Dirac Notation Conversion: kets denote vectors and the Dirac pairing is the reversed-argument root inner product.

[F5]

Principal inverse sine and inverse cosine: $\arcsin$ takes values in $[- π / 2, π / 2]$ and $\sin (\arcsin y) = y$ for $y \in [- 1, 1]$ .

[F6]

Signs, monotonicity intervals, and ranges of sine and cosine: cosine is strictly decreasing on $[0, π]$ and sine is strictly increasing on $[- π / 2, π / 2]$ .

[F7]

Parity and the Pythagorean identity for sine and cosine: $\sin^{2} u + \cos^{2} u = 1$ for every real $u$ .

[F8]

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : each nonnegative real $a$ has a unique nonnegative square root $\sqrt{a}$ .

[F9]

The derivatives of sine and cosine are cosine and minus sine: $\sin 0 = 0$ .

[F10]

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

[F11]

Squaring is monotone on the nonnegatives: on nonnegative reals, strict and non-strict order are equivalent to the corresponding order of squares; applied to nonnegative roots, this compares their order.

Proof

1.1givenF1F2F3F4algebra

Orthonormality of the address basis [F2] gives $⟨ G, G ⟩ = t^{- 1} \sum_{x, y \in M} ⟨ x, y ⟩ = 1$ and $⟨ B, B ⟩ = (N - t)^{- 1} \sum_{x, y \notin M} ⟨ x, y ⟩ = 1$ . The sets $M$ and $D_{n} ∖ M$ are disjoint, so every cross term in $⟨ G, B ⟩$ is zero. Thus the two vectors are orthonormal.

2.1givenF1F5F6F7F8F9F10F11step 1.1algebra∎

Put $a = t / N \in (0, 1)$ and $θ = \arcsin \sqrt{a}$ . By [F5], $\sin θ = \sqrt{a}$ ; since $\sqrt{a} > 0$ and $\sin 0 = 0$ by [F9], strict increase [F6] implies $θ > 0$ . Also $\sqrt{a} < 1$ by [F11] because $a < 1$ , while $1 = \sin (π / 2)$ by [F10]; strict increase gives $θ < π / 2$ . Since cosine is strictly decreasing on $[0, π]$ by [F6], we have $\cos θ > \cos (π / 2) = 0$ by [F10]. The Pythagorean identity [F7] and square-root uniqueness [F8] give $\cos θ = \sqrt{1 - a}$ . Splitting the uniform sum into $M$ and its complement now gives $∣ s ⟩ = \sqrt{t / N} ∣ G ⟩ + \sqrt{(N - t) / N} ∣ B ⟩ = \sin θ ∣ G ⟩ + \cos θ ∣ B ⟩$ and $\sin^{2} θ = t / N$ .

Depends on

Used by

Dependency tree · two levels

51 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

Ronald de Wolf, Quantum Computing: Lecture Notes, §7.2 (standard reference, not scraped)
Andrew M. Childs, Lecture Notes on Quantum Algorithms, §19.1 (standard reference, not scraped)
Gilles Brassard, Peter Høyer, Michele Mosca, and Alain Tapp, Quantum Amplitude Amplification and Estimation, §2 (standard reference, not scraped)