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Quantum Signal Processing — Examples

1 · Prerequisites

2 · Summary

The example multiplies the degree-one zero-phase sequence explicitly, checks its parity and complement identity, and verifies both endpoint matrices without dividing by the vanishing off-diagonal signal.

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 →

Degree-one QSP phase identity

Example

Take $d = 1$ and phases $ϕ_{0} = ϕ_{1} = 0$ . Then $Z (0) = I$ and the sequence is $U (x) = W_{R} (x)$ .

Facts & Assumptions

Given: $x \in [- 1, 1]$ , $s = \sqrt{1 - x^{2}} \geq 0$ , and the definitions of $W_{R}$ , $Z$ , and the degree-one QSP matrix in QSP signal matrix and ordered phase sequence.

[F1]

QSP signal matrix and ordered phase sequence: $W_{R} (x) = (\begin{array}{cc} x & s \\ s & - x \end{array})$ , $Z (0) = I$ , and the length-one sequence is $Z (0) W_{R} (x) Z (0)$ .

[F2]

Admissible polynomial pair for Hermitian-reflection QSP: an admissible pair has the declared degree/parity bounds and satisfies $P P^{*} + (1 - x^{2}) Q Q^{*} = 1$ .

Verification

1.1F1algebra

Multiplying $Z (0) W_{R} (x) Z (0)$ gives $U (x) = (\begin{array}{cc} x & s \\ s & - x \end{array})$ . Thus the upper-left polynomial is $P (x) = x$ and the upper-right entry is $s (x) Q (x)$ with $Q (x) = 1$ .

2.1F2step 1.1algebra

The polynomial $P$ has degree one and odd parity; $Q$ is constant with degree zero and even parity. Also $P P^{*} + (1 - x^{2}) Q Q^{*} = x^{2} + (1 - x^{2}) = 1$ , so the pair is admissible by [F2].

3.1F1step 1.1algebra∎

At $x = 1$ and $x = - 1$ , $s (x) = 0$ , so $U (1) = diag (1, - 1)$ and $U (- 1) = diag (- 1, 1)$ . These are exactly the endpoint values from the displayed matrix, with no division by $s (x)$ at either endpoint.