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Rank-One Reflection Is Ray-Invariant and Its Signs Differ Under Control

Statement

Let $H$ be a finite-dimensional complex-inner-product space and let $ψ \in H$ be a unit vector. Then $Π_{ψ} = ∣ ψ ⟩ ⟨ ψ ∣$ is a rank-one orthogonal projection, and $Π_{λ ψ} = Π_{ψ}$ for every $∣ λ ∣ = 1$ . Therefore $R_{ψ} = 2 Π_{ψ} - I$ and $S_{ψ} = I - 2 Π_{ψ}$ are well-defined on the ray. Standalone they differ by the global scalar $- 1$ ; when controlled in control-first order,

$C (S_{ψ}) = (Z_{C}) C (R_{ψ}), Z_{C} = diag (I, - I) .$

Let $C = Q_{2}$ be the control qubit with ordered basis $∣ 0 ⟩, ∣ 1 ⟩$ , and set $∣ \pm ⟩ = (∣ 0 ⟩ \pm ∣ 1 ⟩) / \sqrt{2}$ . For input $∣ + ⟩ \otimes ψ$ , $C (S_{ψ})$ produces $∣ - ⟩ \otimes ψ$ and $C (R_{ψ})$ produces $∣ + ⟩ \otimes ψ$ , so the control sign is relative rather than a target-wide global phase. If $a \in H$ is a specified unit input denoted $∣ 0 ⟩$ in an access model, $U : H \to H$ is unitary, and $ψ = U a$ , define $R_{a} = 2 ∣ a ⟩ ⟨ a ∣ - I$ and $S_{a} = I - 2 ∣ a ⟩ ⟨ a ∣$ as in Orthogonal-Projector Reflection and Access Contract. Then $U R_{a} U^{†} = R_{ψ}$ and $U S_{a} U^{†} = S_{ψ}$ . This conjugation uses the separately supplied inverse call $U^{†}$ ; the equality itself grants no such access.

Facts & Assumptions

Given: A unit vector $ψ$ in a finite-dimensional complex-inner-product space, with the root pairing linear in its first argument.

[L1]

Dirac Notation Conversion defines $∣ ψ ⟩ ⟨ ψ ∣ x = ⟨ x, ψ ⟩_{r o o t} ψ$ . Its current draft definition body was read.

[L2]

Pure state and ray defines rays under nonzero scalar multiplication. Its current draft definition body was read; this theorem proves the projector phase identity directly, without relying on that item's separate projector-well-definedness result.

[L3]

Orthogonal-Projector Reflection and Access Contract fixes $R_{Π} = 2 Π - I$ , $S_{Π} = I - 2 Π$ , and the explicit direct, controlled, and $U R_{0} U^{†}$ access conventions. Its carrier is present in this batch.

[L4]

Projector Reflections Are Hermitian Unitary Involutions proves the projector identities and controlled-sign relation. Its carrier is present in this batch; QC07's page and items are now present as draft manifest entries, but not independently accepted.

[L5]

Inverse Unitary Access makes availability and charge of the inverse call a separate access-model declaration. Its current draft definition and proof were read.

[L6]

The computational basis is orthonormal proves the control basis $∣ 0 ⟩, ∣ 1 ⟩$ is orthonormal.

[L7]

Ordered tensor registers, their inner product, and wire permutations defines the product pairing after ordered orthonormal bases are specified for the factors and proves the pairing is independent of those choices.

[F1]

Real and complex inner-product spaces and their induced length: $⟨ \cdot, \cdot ⟩$ is linear in the first argument and conjugate-linear in the second.

[F2]

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

[F3]

Every linear map between finite-dimensional inner product spaces has a unique adjoint: every linear map between finite-dimensional inner-product spaces has a unique adjoint.

[F4]

An endomorphism is an orthogonal projection exactly when it is idempotent and self-adjoint: a self-adjoint idempotent is the orthogonal projection onto its range.

[F5]

The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product: $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ .

[F6]

For an endomorphism in finite dimension, preserving lengths, preserving inner products, carrying orthonormal bases to orthonormal bases, and $T^{*} T = I$ are equivalent: a finite-dimensional isometry is unitary and its adjoint is its inverse.

[F7]

Every finite-dimensional real or complex inner product space has an orthonormal basis: every finite-dimensional inner-product space has an orthonormal basis; this supplier occurs in the dependency closure of [F6].

Proof

1.1givenL1F1F2F3algebra

By [L1], $Π_{ψ} x = ⟨ x, ψ ⟩_{r o o t} ψ$ . Then $Π_{ψ}^{2} x = ⟨ x, ψ ⟩ ⟨ ψ, ψ ⟩ ψ = Π_{ψ} x$ . Also $⟨ Π_{ψ} x, y ⟩ = ⟨ x, ψ ⟩ ⟨ ψ, y ⟩ = ⟨ x, Π_{ψ} y ⟩$ by [F1] and [F2]; uniqueness [F3] gives $Π_{ψ}^{*} = Π_{ψ}$ . Its image is contained in $C ψ$ and $Π_{ψ} ψ = ψ$ , hence its range is exactly one-dimensional.

2.1step 1.1L1F1F4algebra

The self-adjoint idempotent result [F4] makes $Π_{ψ}$ an orthogonal projection. For $∣ λ ∣ = 1$ , [L1] and [F1] give $Π_{λ ψ} x = ⟨ x, λ ψ ⟩ λ ψ = \overline{λ} λ ⟨ x, ψ ⟩ ψ = Π_{ψ} x$ for every $x$ .

3.1L2F1F5step 2.1algebra

If $v$ is any other unit representative of the ray of $ψ$ , [L2] gives $v = c ψ$ for some nonzero scalar $c$ . By [F1] and [F5], $1 = ∥ v ∥^{2} = ⟨ c ψ, c ψ ⟩ = ∣ c ∣^{2} ⟨ ψ, ψ ⟩ = ∣ c ∣^{2}$ , so $∣ c ∣ = 1$ . Step 2.1 then gives $Π_{v} = Π_{ψ}$ , and both reflection signs depend only on the ray.

4.1L3L4step 3.1algebra

Applying [L4] to $Π_{ψ}$ gives both standalone reflections as unitary involutions and $C (S_{ψ}) = (Z_{C}) C (R_{ψ})$ . Since $S_{ψ} = - R_{ψ}$ , their standalone outputs differ only by the scalar $- 1$ and represent the same ray.

5.1givenL2L4L6L7F7step 1.1step 4.1choosealgebra

Choose an ordered orthonormal basis of $H$ using [F7]. Together with the control basis from [L6], it supplies the factor bases required by [L7], whose proof shows the pairing is independent of this choice. Because $ψ \in ran Π_{ψ}$ , [L4] gives $R_{ψ} ψ = ψ$ and $S_{ψ} ψ = - ψ$ . The controlled input $∣ + ⟩ \otimes ψ$ therefore becomes $∣ + ⟩ \otimes ψ$ under $C (R_{ψ})$ and $∣ - ⟩ \otimes ψ$ under $C (S_{ψ})$ . Since $∣ \pm ⟩ = (∣ 0 ⟩ \pm ∣ 1 ⟩) / \sqrt{2}$ , [L6] gives $⟨ +, - ⟩ = \frac{1}{2} (⟨ 0, 0 ⟩ - ⟨ 0, 1 ⟩ + ⟨ 1, 0 ⟩ - ⟨ 1, 1 ⟩) = 0$ and $⟨ +, + ⟩ = ⟨ -, - ⟩ = 1$ . By [L7], the two joint outputs are unit and orthogonal. Orthogonal nonzero vectors are not nonzero scalar multiples, so the ray definition [L2] makes the output rays distinct and the control sign relative.

6.1givenL1L3L5F1F2F6algebra∎

Let $U$ be unitary and let $ψ = U a$ for the specified unit input $a \in H$ . For any $z \in H$ , [L1] gives $U ∣ a ⟩ ⟨ a ∣ U^{†} z = U (⟨ U^{†} z, a ⟩_{r o o t} a) = ⟨ U^{†} z, a ⟩_{r o o t} ψ$ . The adjoint identity [F2] with $v = a, w = z$ , followed by conjugate symmetry [F1], gives $⟨ U^{†} z, a ⟩_{r o o t} = ⟨ z, U a ⟩_{r o o t} = ⟨ z, ψ ⟩_{r o o t}$ . Thus $U ∣ a ⟩ ⟨ a ∣ U^{†} = Π_{ψ}$ as operators. By [F6], $U U^{†} = U^{†} U = I$ , so conjugating $R_{a} = 2 ∣ a ⟩ ⟨ a ∣ - I$ and $S_{a} = I - 2 ∣ a ⟩ ⟨ a ∣$ gives $R_{ψ}$ and $S_{ψ}$ , respectively. The algebraic equality does not grant the calls: [L5] requires inverse availability and its cost to be stated separately.

Depends on

Used by

A preserved unknown state cannot program its own reflection Counterexample

Cited to discharge well-definedness by Orthogonal-Projector Reflection and Access Contract.

Dependency tree · two levels

47 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

Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 6: Quantum Algorithms (standard reference, not scraped)