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Szegedy walk invariant subspaces from discriminant eigenvectors

Statement

For any eigenpair $D v = λ v$ with $v \neq 0$ and $λ \in [- 1, 1]$ , put $a : = T v$ and $b : = S T v$ . Then $span {a, b}$ is invariant under $U$ , and

$U a = b, U b = - a + 2 λ b .$

All eigenvalues of $D$ satisfy the displayed interval hypothesis by The Szegedy discriminant is self-adjoint and contractive. If $λ = 1$ or $λ = - 1$ , then $b = λ a$ and the span is one-dimensional. If $- 1 < λ < 1$ , the two vectors are linearly independent and the invariant subspace is two-dimensional.

Facts & Assumptions

Given: A finite row-stochastic kernel, a nonzero vector $v$ with $D v = λ v$ and real $λ \in [- 1, 1]$ , and $T, S, D, U, Π$ as defined in Szegedy row-preparation map and discriminant and Szegedy reflection walk and its two-reflection square.

[F1]

The Szegedy discriminant is self-adjoint and contractive: $D = T^{*} S T$ is self-adjoint with $∥ D ∥ \leq 1$ , and $T^{*} T = I$ .

[F2]

Szegedy reflection walk and its two-reflection square: $U = S (2 Π - I)$ and $Π = T T^{*}$ .

[F3]

Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism: $D v = λ v$ is an eigenpair when $v \neq 0$ .

[F4]

Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ : $span {a, b}$ consists of all linear combinations of $a, b$ .

[F5]

Szegedy row-preparation map and discriminant: the ordered-product basis is orthonormal and $S ∣ x, y ⟩ = ∣ y, x ⟩$ .

Proof

1.1F1F2given

Since $Π T v = T v$ , [F2] gives $U a = S R_{A} T v = S T v = b$ .

2.1F1F2F5step 1.1algebra

The adjoint identity gives $Π b = T T^{*} S T v = T (T^{*} S T) v = T D v = λ T v = λ a$ . Hence $U b = S (2 Π - I) S T v = S (2 λ a - b) = 2 λ b - a$ , because [F5] gives $S^{2} = I$ .

3.1F1F2F3F4F5step 1.1step 2.1algebra∎

Steps 1.1–2.1 and [F4] show the span is invariant. Since $T$ is an isometry by [F1], $∥ a ∥ = ∥ v ∥$ . Write $T v = \sum_{x, y} c_{x y} ∣ x, y ⟩$ in the orthonormal product basis. By [F5], $S$ bijectively permutes that basis, so $∥ b ∥^{2} = \sum_{x, y} ∣ c_{x y} ∣^{2} = ∥ T v ∥^{2}$ and $∥ b ∥ = ∥ v ∥$ . The adjoint identity gives $⟨ a, b ⟩ = ⟨ v, D v ⟩ = λ ∥ v ∥^{2}$ , which is real. If $λ = \pm 1$ , expansion gives $∥ b - λ a ∥^{2} = 0$ , hence $b = λ a$ . If $∣ λ ∣ < 1$ and $b = c a$ , equality of norms gives $∣ c ∣ = 1$ , while $⟨ a, b ⟩ = \overline{c} ∥ v ∥^{2} = λ ∥ v ∥^{2}$ gives $∣ c ∣ = ∣ λ ∣ < 1$ , a contradiction; thus $a, b$ are independent.

Depends on

Used by

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Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms, §18.3 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, §8.2 (standard reference, not scraped)