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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Degree-one QSP phase identity

Example

Take d=1 and phases ϕ0=ϕ1=0. Then Z(0)=I and the sequence is U(x)=WR(x).

Facts & Assumptions

Given: x∈[−1,1], s=1−x2≥0, and the definitions of WR, Z, and the degree-one QSP matrix in QSP signal matrix and ordered phase sequence.

[F1]

QSP signal matrix and ordered phase sequence: WR(x)=(xss−x), Z(0)=I, and the length-one sequence is Z(0)WR(x)Z(0).

[F2]

Admissible polynomial pair for Hermitian-reflection QSP: an admissible pair has the declared degree/parity bounds and satisfies PP∗+(1−x2)QQ∗=1.

Verification

1.1F1algebra

Multiplying Z(0)WR(x)Z(0) gives U(x)=(xss−x). 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 PP∗+(1−x2)QQ∗=x2+(1−x2)=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.

Depends on

Used by

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Dependency tree · two levels

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