Manuscript Oai279-A@1 · Oai279-A.tex · lemma · a1abc73f

Lemma: Exact amplification requires the correct probability

Stage 1: audit the five interfaces without importing them › Interface 4: completion, deferred filters, and exact output · lines 606–635 · proof included or linked

Oai279-A@1#lem:amplification, Oai279-A@1#eq:amplification

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Let $ψ$ be a unit vector, let $Π$ be a success projector, and put

g = Π ψ, b = (I - Π) ψ, p = ∥ g ∥^{2} .

Then

(2 ∣ ψ ⟩ ⟨ ψ ∣ - I) (I - 2Π) ψ = (3 - 4 p) g + (1 - 4 p) b .

For $p = 1/4$ , the result is $2 g$ , a unit vector wholly supported in the success subspace.

Proof. The first sign flip gives $- g + b$ , whose inner product with $ψ = g + b$ is $1 - 2 p$ . Apply the second reflection:

2 (1 - 2 p) (g + b) - (- g + b) = (3 - 4 p) g + (1 - 4 p) b .

If $p = 1/4$ , the bad component vanishes and $∥2 g ∥^{2} = 4 p = 1$ .

Displayed formulas in this passage

Oai279-A@1#eq:amplification · line 614
$(2 ∣ ψ ⟩ ⟨ ψ ∣ - I) (I - 2Π) ψ = (3 - 4 p) g + (1 - 4 p) b .$
unlabeled · line 610
$g = Π ψ, b = (I - Π) ψ, p = ∥ g ∥^{2} .$
unlabeled · line 628
$2 (1 - 2 p) (g + b) - (- g + b) = (3 - 4 p) g + (1 - 4 p) b .$

Lemma: Exact amplification requires the correct probability

Displayed formulas in this passage

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