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Amplitude Amplification — Examples

1 · Prerequisites

2 · Summary

The worked two-dimensional example computes the success probability at three iteration counts, including the overshoot after an exact peak. The counterexample shows that a known-amplitude maximizing count can give zero success for a different success weight.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A prepared state with success amplitude one half

Example

Take an orthonormal pair $(∣ B ⟩, ∣ G ⟩)$ and the unit prepared state $∣ ψ ⟩ = \frac{\sqrt{3}}{2} ∣ B ⟩ + \frac{1}{2} ∣ G ⟩$ with $Π_{g o o d} = ∣ G ⟩ ⟨ G ∣$ . Its success amplitude is $1 / 2$ , so $a = 1 / 4$ and $θ = π / 6$ . Compute the good probabilities after $k = 0, 1, 2$ standard iterations.

Facts & Assumptions

Given: The state, projection, and iteration counts in the Example.

[F1]

Exact success probability after amplitude amplification: after $k$ iterations, $p_{k} = \sin^{2} ((2 k + 1) θ)$ .

[F2]

[F3]

Quarter-turn values and shifts by pi/2 and pi gives $\sin (π / 2) = 1$ , and Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions gives $\sin (5 π / 6) = \sin (π / 6)$ .

Verification

1.1givenF2algebra

Put $c = \cos (π / 3) = \sin (π / 6)$ by [F2]. Since $0 < π / 6 < π$ , [F2] gives $c > 0$ . Also [F2] gives $\cos (2 π / 3) = - c$ , while [F2] gives $\cos (2 π / 3) = 2 c^{2} - 1$ . Hence $2 c^{2} + c - 1 = (2 c - 1) (c + 1) = 0$ ; positivity excludes $c = - 1$ , so $\sin (π / 6) = c = 1 / 2$ . The vector is normalized because $3 / 4 + 1 / 4 = 1$ , and $∥ Π_{g o o d} ψ ∥^{2} = 1 / 4 = \sin^{2} (π / 6)$ .

2.1givenF1F3step 1.1algebra∎

By [F1], [F3], and step 1.1, $p_{0} = \sin^{2} (π / 6) = 1 / 4$ , $p_{1} = \sin^{2} (π / 2) = 1$ , and $p_{2} = \sin^{2} (5 π / 6) = 1 / 4$ . Each value lies in $[0, 1]$ ; the second iteration overshoots the exact peak attained after one iteration.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A fixed Grover iteration count can fail when a is unknown

Statement refuted

For every pair of initial success weights $a_{1}, a_{2} \in (0, 1)$ , an iteration count maximizing success for $a_{1}$ over $0 \leq k \leq ⌈ 1 / \sqrt{a_{1}} ⌉$ gives success probability at least $1 / 2$ for $a_{2}$ when reused unchanged.

Facts & Assumptions

Given: Two prepared success weights and a common fixed iteration count.

[F1]

Exact success probability after amplitude amplification: after $k$ iterations, success is $\sin^{2} ((2 k + 1) θ)$ when $\sin^{2} θ = a$ .

[F2]

Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions gives $\sin (π / 6) = \cos (π / 3)$ and $\cos (2 π / 3) = - \cos (π / 3)$ ; Double-angle and quadratic power-reduction identities gives $\cos (2 x) = 2 \cos^{2} x - 1$ ; Pi is the first positive zero of sine gives $\sin x > 0$ for $0 < x < π$ ; Parity and the Pythagorean identity for sine and cosine gives $\sin^{2} x + \cos^{2} x = 1$ ; Quarter-turn values and shifts by pi/2 and pi gives $\sin (π / 2) = 1$ and $\sin π = 0$ .

Counterexample

1.1givenF1F2algebra

Put $c = \cos (π / 3) = \sin (π / 6)$ by [F2]. Since $0 < π / 6 < π$ , [F2] gives $c > 0$ . Also [F2] gives $\cos (2 π / 3) = - c$ , while [F2] gives $\cos (2 π / 3) = 2 c^{2} - 1$ . Hence $(2 c - 1) (c + 1) = 0$ , and positivity gives $c = 1 / 2$ . By [F2], $\sin^{2} (π / 3) = 1 - \cos^{2} (π / 3) = 3 / 4$ . For $a_{1} = 1 / 4$ , take $θ_{1} = π / 6$ ; the allowed range is $0 \leq k \leq ⌈ 1 / \sqrt{a_{1}} ⌉ = 2$ . The probabilities for $k = 0, 1, 2$ are $1 / 4, 1, 1 / 4$ , respectively, so $k = 1$ is the unique maximizer.

2.1givenF1F2step 1.1algebra∎

For $a_{2} = 3 / 4$ , take $θ_{2} = π / 3$ . Reusing $k = 1$ gives $\sin^{2} (3 θ_{2}) = \sin^{2} π = 0 < 1 / 2$ , while direct sampling at $k = 0$ succeeds with probability $3 / 4$ . Thus the first weight's maximizing count violates the claimed lower bound at the second weight.

Amplitude Amplification — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A prepared state with success amplitude one half

Example

Facts & Assumptions

Verification

A fixed Grover iteration count can fail when a is unknown

Statement refuted

Facts & Assumptions

Counterexample

Sources