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Second moment of the signed sine-window phase kernel

Statement

Let $t_{0} = 2048 κ / ϵ$ , $0 < ϵ \leq 1$ , $κ \geq 1$ , and let $T$ be the least power of two at least $8 (t_{0} + 1)$ . For any $λ \in [- 1, 1]$ , prepare Finite sine-window phase state, apply the controlled phases $e^{i λ τ t_{0} / T}$ and the negative-sign Fourier transform, then decode $y$ to signed $k$ . With probability $p_{k} (λ)$ , put $δ_{k} = λ t_{0} - 2 π k$ . Then $\sum_{k = - T / 2}^{T / 2 - 1} p_{k} (λ) δ_{k}^{2} < 300.$

Facts & Assumptions

Given: The sine-window circuit and OP5 parameter range.

[F1]

Exact binary QFT circuit correctness: the negative-sign Fourier transform has basis amplitudes $T^{- 1 / 2} e^{- 2 π i k τ / T}$ , so the measured amplitude is the finite Fourier sum of the prepared coefficients.

[F2]

Finite geometric sum for phase-estimation amplitudes: finite exponential sums equal their geometric-series quotient when the ratio is not one; the coincident-ratio case is their continuous finite-sum limit.

[F3]

Rational bounds for pi used in finite schedules: $3 < π < 22 / 7 < 16 / 5$ .

[F4]

Sine chord and linear bounds: $\sin v \leq v$ for $v \geq 0$ .

[F5]

Sine and cosine are $1$ -Lipschitz on $R$ : cosine changes by at most the change in its real argument.

[F6]

The derivatives of sine and cosine are cosine and minus sine: $\cos 0 = 1$ .

[F7]

Sine chord and linear bounds: $\sin (π u) \geq 2 u$ for $0 \leq u \leq 1 / 2$ .

Proof

1.1givenF1F2algebra

Put $θ = λ t_{0} / (2 π)$ and $δ = 2 π (θ - k)$ . Expanding sine into its two exponentials and summing the two geometric series from [F2] gives the amplitude $α_{k} = - \frac{\sqrt{2}}{T} e^{i δ (1 - 1 / T) / 2} \frac{\cos (δ / 2) \cos (δ / (2 T)) \sin (π / (2 T))}{\sin ((δ + π) / (2 T)) \sin ((δ - π) / (2 T))} .$ At a removable zero this formula means its continuous limit. By [F1], $p_{k} = ∣ α_{k} ∣^{2}$ and $\sum_{k} p_{k} = 1$ .

2.1F3F4F5F6F7step 1.1algebra

For $∣ δ ∣ \geq 2 π$ , set $x = ∣ δ ∣ / (2 T)$ and $a = π / (2 T)$ . Signed decoding and $∣ λ ∣ \leq 1$ give $x \leq π / 2 + t_{0} / (2 T) < π / 2 + 1 / 16$ ; also $x \geq 2 a$ . Since $t_{0} \geq 2048$ , $a < 1 / 100$ , hence $x + a < π / 2 + 1 / 10$ . On $[0, π / 2 + 1 / 10]$ one has $\sin u \geq u / 2$ : on $[0, π / 2]$ put $s = v / π \in [0, 1 / 2]$ and apply [F7] to get $\sin v \geq 2 v / π > v / 2$ using [F3]; on the remaining interval, $\sin u = \cos (u - π / 2) \geq 1 - 1 / 10 = 9 / 10$ by [F5] and [F6], while $u / 2 < (16 / 5) / 4 + 1 / 20 = 17 / 20 < 9 / 10$ . Therefore $∣ \sin^{2} x - \sin^{2} a ∣ = \sin (x + a) \sin (x - a) \geq \frac{1}{4} (x^{2} - a^{2}) \geq \frac{3 δ^{2}}{64 T^{2}} .$ Using $∣ \cos ∣ \leq 1$ and $\sin a \leq a$ in step 1.1 yields $p_{k} \leq (2048 π^{2} / 9) δ^{- 4}$ .

3.1F3step 2.1algebra∎

The outcomes with $∣ δ ∣ < 2 π$ contribute at most $4 π^{2}$ because their total probability is at most one. For the rest, group integers by $n \leq ∣ θ - k ∣ < n + 1$ , $n \geq 1$ ; each shell has at most two integers. Step 2.1 gives $\sum_{∣ δ_{k} ∣ \geq 2 π} p_{k} δ_{k}^{2} \leq \frac{2048 π^{2}}{9} \frac{2}{4 π^{2}} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \leq \frac{2048}{9},$ where $\sum_{n \geq 1} n^{- 2} \leq 1 + \int_{1}^{\infty} x^{- 2} d x = 2$ . By [F3], $4 π^{2} < 4 (16 / 5)^{2} < 41$ , so the total is less than $41 + 2048 / 9 < 300$ .

Depends on

Used by

HHL finite sine-window phase-estimation algorithm Algorithm
OP5 HHL returns the QLSP state within the requested error Theorem

Dependency tree · two levels

41 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

Aram W. Harrow, Avinatan Hassidim, and Seth Lloyd, Quantum algorithm for linear systems of equations (standard reference, not scraped)
Andrew M. Childs, Robin Kothari, and Rolando D. Somma, Quantum algorithm for systems of linear equations with exponentially improved dependence on precision (standard reference, not scraped)
Andrew M. Childs, Lecture Notes on Quantum Algorithms (standard reference, not scraped)