TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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Oblivious-amplification block identity

Statement

Let $J : S \to A \otimes S$ be the clean-ancilla isometry $J x = ∣ 0_{a n c} ⟩ \otimes x$ , let $Π = J J^{†}$ be the orthogonal projection onto its image, and set $R_{0} = I - 2 Π$ . Let $W$ be unitary and assume $J^{†} W J = M / 2$ . For the chronological circuit $W R_{0} W^{†} R_{0} W$ , $J^{†} W R_{0} W^{†} R_{0} W J = - \frac{3}{2} M + \frac{1}{2} M M^{†} M .$ If $M$ is unitary, this projected block is $- M$ . The identity alone gives no bound on leakage outside the clean-ancilla subspace for a nonunitary $M$ .

Facts & Assumptions

Given: A finite-dimensional system, a unitary $W$ , the clean-ancilla isometry $J$ , and $J^{†} W J = M / 2$ .

[L1]

One-step oblivious amplitude amplification: the chronological operation is $W$ , $R_{0}$ , $W^{†}$ , $R_{0}$ , $W$ , and $R_{0} = I - 2 Π$ .

[F1]

Oblivious ancilla reflection: $Π$ projects only the full clean coefficient/work ancilla to zero and is identity on the system.

[F2]

Projector Reflections Are Hermitian Unitary Involutions: for an orthogonal projection $Π$ , $I - 2 Π$ is a unitary involution; the opposite sign is distinct under control.

[F3]

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces and Finite-dimensional linear isometries satisfy the adjoint identity: a unitary is a surjective linear isometry, so $W^{†} W = W W^{†} = I$ ; an isometry $J$ satisfies $J^{†} J = I$ .

[F4]

An isometric quantum encoding gives an orthogonal code projector: $J J^{†}$ is the orthogonal projection onto the image of an isometry $J$ .

[F5]

Finite registers and their ordered computational bases, The computational basis is orthonormal, and Ordered tensor registers, their inner product, and wire permutations: the clean basis vector is unit and the product inner product makes $J x = ∣ 0_{a n c} ⟩ \otimes x$ a linear isometry.

Proof

Given: As in Facts & Assumptions.

1.1givenL1F1F2F3F4F5algebra

Let $B = J^{†} W J = M / 2$ . By [F3]–[F5], $J^{†} J = I$ and $Π = J J^{†}$ is the stated orthogonal projection; hence $J^{†} Π = J^{†}$ and $Π J = J$ . By [L1]–[F2], the three-use circuit is the stated unitary circuit with two reflections.

2.1givenF1step 1.1F3F4algebra

Expand $(I - 2 Π)$ twice inside $J^{†} W (I - 2 Π) W^{†} (I - 2 Π) W J$ . Since $W$ is unitary, [F3] gives $W W^{†} = W^{†} W = I$ , and [F4] gives $Π^{2} = Π$ . The four terms compress to $B - 2 B - 2 B + 4 B B^{†} B = - 3 B + 4 B B^{†} B$ : respectively, use $W^{†} W = I$ , $Π J = J$ , $J^{†} Π = J^{†}$ , and $Π = J J^{†}$ . Substituting $B = M / 2$ gives $- \frac{3}{2} M + \frac{1}{2} M M^{†} M$ .

3.1givenstep 2.1cases∎

If $M$ is unitary, then $M M^{†} M = M$ , so the block is $- M$ . For arbitrary $M$ , step 2.1 proves only the compression identity; it contains no estimate for $(I - Π) W R_{0} W^{†} R_{0} W J$ .

Depends on

Used by

Robust one-step OAA including ancilla leakage Lemma
Segmented Taylor error and resources Theorem

Dependency tree · two levels

42 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 28 §28.3 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, §9.3 and Exercise 9 (standard reference, not scraped)