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

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Two nearest phase labels carry probability at least eight over pi squared

Statement

Let $M = 2^{m}$ with $m \geq 1$ , and let $θ \in [0, 1)$ be an eigenphase measured by the $M$ -label QPE kernel. The two labels $⌊ M θ ⌋$ and $⌈ M θ ⌉$ reduced modulo $M$ (one label when $M θ$ is an integer) have total outcome probability at least $8 / π^{2}$ .

Facts & Assumptions

Given: The QPE eigenphase outcome law with $M = 2^{m}$ , $m \geq 1$ , and phase $θ \in [0, 1)$ .

[F1]

For label $y$ , $P_{θ} (y) = M^{- 2} ∣ \sum_{j = 0}^{M - 1} e^{2 π i j (θ - y / M)} ∣^{2}$ (Phase-estimation finite outcome law).

[F2]

The finite geometric sum reduces to $(1 - z^{M}) / (1 - z)$ when $z \neq 1$ (Finite geometric sum for phase-estimation amplitudes).

[F3]

For $v \geq 0$ , $\sin v \leq v$ (Sine chord and linear bounds).

[F4]

$π$ satisfies $3 < π < 22 / 7$ (Rational bounds for pi used in finite schedules).

[F5]

Sine/cosine derivatives, tangent/secant derivatives, first-quadrant signs, and the Pythagorean identity are given by The derivatives of sine and cosine are cosine and minus sine, Derivatives and fundamental periods of tangent, cotangent, secant, and cosecant, Signs, monotonicity intervals, and ranges of sine and cosine, Pythagorean and parity identities for all six trigonometric functions on their natural domains, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , and Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ .

[F6]

The mean value theorem converts a derivative sign into monotonicity on a compact interval (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[F7]

Cofunction and supplementary identities give $\sin (π / 2 - x) = \cos x$ , $\tan (π / 2 - x) = \cot x$ , and $\cos (π - x) = - \cos x$ on their domains (Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions).

[F8]

$\cos (2 x) = 2 \cos^{2} x - 1$ and $\sin^{2} x + \cos^{2} x = 1$ (Double-angle and quadratic power-reduction identities, Pythagorean and parity identities for all six trigonometric functions on their natural domains).

Proof

1.1givenF1F2F3algebra

If $r = {M θ} = 0$ , write $q = M θ$ . For label $q$ modulo $M$ , every summand in [F1] is $1$ , so its probability is $1$ ; for any other label, the geometric ratio in [F2] is not $1$ but its $M$ th power is $1$ , so the sum is zero. Otherwise $0 < r < 1$ , the floor and ceiling labels have exact probabilities $\sin^{2} (π r) / (M^{2} \sin^{2} (π r / M))$ and $\sin^{2} (π r) / (M^{2} \sin^{2} (π (1 - r) / M))$ by [F1]–[F2]. Applying [F3] to each denominator shows their sum is at least $g (r) / π^{2}$ , where $g (r) = \sin^{2} (π r) (r^{- 2} + (1 - r)^{- 2})$ .

1.2F5F7algebra

The function $g$ is symmetric under $r \mapsto 1 - r$ . For $0 < r < 1 / 2$ , set $t = 1 - 2 r$ , $A = r^{- 2} + (1 - r)^{- 2} > 0$ and $C = r^{- 3} - (1 - r)^{- 3} > 0$ . Differentiation gives $g^{'} (r) = 2 \sin (π r) [π \cos (π r) A - \sin (π r) C]$ , and direct substitution gives $C / A = 2 t (3 + t^{2}) / (1 - t^{4})$ . Since $\sin (π r) > 0$ , $\cos (π r) > 0$ , and $A > 0$ , $g^{'} (r) \leq 0$ exactly when $π \cot (π r) \leq C / A$ . By [F7], this is $\tan (π t / 2) \leq 2 t (3 + t^{2}) / [π (1 - t^{4})]$ .

1.3F4F5F6F7F8algebra

Put $T (t) = \tan (π t / 2)$ on $[0, 1 / 2]$ . Its second derivative is $T^{''} (t) = (π^{2} / 2) \tan (π t / 2) \sec^{2} (π t / 2) \geq 0$ by [F5]. Thus $T^{'}$ is nondecreasing by [F6]; applying the mean value theorem on $[a, t]$ and $[t, b]$ for $a < t < b$ shows $T (t)$ lies below the chord joining $(a, T (a))$ and $(b, T (b))$ . For endpoint values, let $c = \cos (π / 3) > 0$ . By [F7]–[F8], $2 c^{2} - 1 = \cos (2 π / 3) = - c$ , so $(2 c - 1) (c + 1) = 0$ and $c = 1 / 2$ . Hence $\sin (π / 3) > 4 / 5$ because its square is $3 / 4 > 16 / 25$ , so [F7] gives $\tan (π / 6) = \cot (π / 3) < 5 / 8$ . Also $\sin (π / 4) = \cos (π / 4) > 0$ by cofunction symmetry, hence $\tan (π / 4) = 1$ .

2.1F4step 1.2step 1.3algebra

For $0 \leq t \leq 1 / 3$ , the chord from $0$ to $1 / 3$ in step 1.3 gives $T (t) \leq (15 / 8) t$ . For $t > 0$ , the right side of the inequality in step 1.2 is at least $6 t / π > 21 t / 11 > (15 / 8) t$ , using $π < 22 / 7$ and $21 / 11 > 15 / 8$ ; at $t = 0$ both sides are zero.

2.2F4F5F6step 1.2step 1.3algebra

For $1 / 3 \leq t \leq 1 / 2$ , the chord from $0$ to $1 / 2$ in step 1.3 gives $T (t) \leq 2 t$ . Put $R (t) = (3 + t^{2}) / (1 - t^{4})$ . On this interval its denominator is positive, and differentiation gives $R^{'} (t) = 2 t (1 + 6 t^{2} + t^{4}) / (1 - t^{4})^{2} \geq 0$ , so $R (t) \geq R (1 / 3) = 63 / 20 > 22 / 7 > π$ . The right side in step 1.2 is therefore $2 t R (t) / π \geq 2 t$ .

2.3F4F5F6step 1.2algebra

For $1 / 2 \leq t < 1$ , put $α = π (1 - t) / 2$ . The function $u (x) = \tan x - x$ has $u (0) = 0$ and $u^{'} (x) = \tan^{2} x \geq 0$ on $[0, π / 4]$ , so [F5]–[F6] give $\tan α \geq α$ . Hence $\sin α - α \cos α = \cos α (\tan α - α) \geq 0$ . The function $h (α) = \sin α (1 - α^{2} / 3) - α \cos α$ has derivative $(α / 3) (\sin α - α \cos α) \geq 0$ and $h (0) = 0$ , so $h (α) \geq 0$ and $\cot α \leq 1 / α - α / 3$ . After multiplying the desired comparison by $π (1 - t) > 0$ , its right side minus this upper bound is $(1 - t)^{2} [π^{2} / 6 - 2 / ((1 + t) (1 + t^{2}))] > 0$ , since $(1 + t) (1 + t^{2}) \geq 15 / 8$ and $π > 3$ . This proves the inequality in step 1.2 on the last interval.

3.1step 1.1step 2.1step 2.2step 2.3algebra∎

Steps 2.1–2.3 show $g$ is nonincreasing on $(0, 1 / 2]$ ; symmetry then shows its minimum on $(0, 1)$ is $g (1 / 2) = 8$ . Step 1.1 therefore gives probability at least $8 / π^{2}$ , and the exact-grid case in step 1.1 already has probability one.

Depends on

Used by

Amplitude-estimation additive error and success probability Theorem

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73 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

Brassard, Høyer, Mosca, and Tapp, Quantum Amplitude Amplification and Estimation, complete article, §§4–5, Theorems 11–18 (standard reference, not scraped)
Andrew M. Childs, Lecture Notes on Quantum Algorithms, complete lecture notes, §§19.1–19.3 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, complete 239-page notes, §7.3 (standard reference, not scraped)