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Adjoint block encoding from inverse access

Statement

Suppose $U$ is an $(α, a, ε)$ block encoding of a square matrix $A$ with $α > 0$ . If an implementation of $U^{†}$ is separately supplied, then its zero-ancilla compression is $B_{U}^{†}$ and

${∥ B_{U}^{†} - \frac{A^{†}}{α} ∥}_{o p} = {∥ B_{U} - \frac{A}{α} ∥}_{o p} \leq ε .$

For the declared rectangular padding $(E_{i n}, E_{o u t})$ of Finite block encoding, the same unitary $U^{†}$ encodes $A^{†}$ with the input and output embeddings swapped to $(E_{o u t}, E_{i n})$ . The inverse circuit is an access requirement, not a free consequence of a matrix description.

Facts & Assumptions

Given: A finite-dimensional block encoding with positive real normalization and the separately supplied inverse access specified in Block-encoding oracle and implementation cost model.

[F1]

Finite block encoding: the compressed block is $B_{U} = J_{a}^{†} U J_{a}$ and the rectangular padded matrix is $\hat{A} = E_{o u t} A E_{i n}^{†}$ .

[F2]

Block-encoding oracle and implementation cost model: $U^{†}$ is a separate supplied operation with its own access cost.

[F3]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ : $⟨ T x, y ⟩ = ⟨ x, T^{†} y ⟩$ .

[F4]

Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ : $(S T)^{†} = T^{†} S^{†}$ and $(T^{†})^{†} = T$ .

[F5]

The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||: on a nonzero domain the operator norm is the maximum over unit vectors, and the zero-domain norm is zero.

[F6]

Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ .

Proof

1.1F1F2F4algebra

By [F1] and [F4], the supplied operation $U^{†}$ from [F2] has compression $J_{a}^{†} U^{†} J_{a} = (J_{a}^{†} U J_{a})^{†} = B_{U}^{†}$ . In the rectangular case, ${\hat{A}}^{†} = E_{i n} A^{†} E_{o u t}^{†}$ , which is exactly the declared padding of $A^{†}$ with the two embeddings exchanged.

1.2F3F5F6givenalgebra

For any vector $v$ , [F5] and [F6] imply $∥ v ∥ = \max_{∥ y ∥ = 1} ∣ ⟨ v, y ⟩ ∣$ : Cauchy–Schwarz gives the upper bound, and if $v \neq 0$ the unit vector $y = v / ∥ v ∥$ attains it; for $v = 0$ both sides are zero. Therefore, for any linear map $T : V \to W$ between nonzero finite-dimensional inner-product spaces, $∥ T ∥ = \max_{∥ x ∥ = ∥ y ∥ = 1} ∣ ⟨ T x, y ⟩ ∣ = \max_{∥ x ∥ = ∥ y ∥ = 1} ∣ ⟨ x, T^{†} y ⟩ ∣ = ∥ T^{†} ∥$ by [F3] and maximizing over the unit vectors of $V$ and $W$ . If either space is zero, both maps are zero and both norms are zero by [F5].

2.1step 1.1step 1.2F1F4algebra∎

In the rectangular case apply step 1.2 to $T = B_{U} - \hat{A} / α$ ; in the square case $\hat{A} = A$ . Since $α$ is real, [F4] gives $T^{†} = B_{U}^{†} - {\hat{A}}^{†} / α$ , whose norm equals $∥ T ∥ \leq ε$ . Also step 1.2 applied to $A$ gives $∥ A^{†} ∥ = ∥ A ∥ \leq α$ . Together with the swapped padding of step 1.1, these are exactly the normalization and error requirements of [F1].

Depends on

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Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms, Chapter 29 (17 April 2025) (standard reference, not scraped)
András Gilyén, Yuan Su, Guang Hao Low, and Nathan Wiebe, Quantum singular value transformation and beyond, Definition 43 and closure discussion (standard reference, not scraped)