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A preserved unknown state cannot program its own reflection

Statement

Let $P$ and $T$ be qubit registers. There is no fixed unitary $U$ on $P \otimes T$ that, for every unit program vector $∣ ψ ⟩ \in P$ and target vector $∣ ϕ ⟩ \in T$ , implements $U (∣ ψ ⟩ \otimes ∣ ϕ ⟩) = ∣ ψ ⟩ \otimes (I - 2 ∣ ψ ⟩ ⟨ ψ ∣) ∣ ϕ ⟩,$ while preserving the program register. This only rules out that exact program-state interface; it does not rule out a declared circuit for a known reflection or a separate oracle interface.

Facts & Assumptions

Given: The two-dimensional register with ordered orthonormal basis $∣ 0 ⟩, ∣ 1 ⟩$ and the reflection convention $S_{ψ} = I - 2 ∣ ψ ⟩ ⟨ ψ ∣$ .

[F1]

Finite registers and their ordered computational bases and The computational basis is orthonormal: $∣ 0 ⟩, ∣ 1 ⟩$ are an orthonormal basis of the qubit register.

[F2]

Dirac Notation Conversion: $∣ ψ ⟩ ⟨ ψ ∣$ sends $z$ to $⟨ z, ψ ⟩_{r o o t} ψ$ ; Rank-One Reflection Is Ray-Invariant and Its Signs Differ Under Control proves that for unit $ψ$ it is an orthogonal rank-one projector.

[F3]

Orthogonal-Projector Reflection and Access Contract and Projector Reflections Are Hermitian Unitary Involutions: $S_{ψ} = I - 2 ∣ ψ ⟩ ⟨ ψ ∣$ is unitary.

[F4]

Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces and Finite complex linear isometries preserve inner products: every unitary preserves inner products.

[F5]

Ordered tensor registers, their inner product, and wire permutations: the product inner product satisfies $⟨ x \otimes y, x^{'} \otimes y^{'} ⟩ = ⟨ x, x^{'} ⟩ ⟨ y, y^{'} ⟩$ .

[F6]

Oblivious ancilla reflection: the reflection and its controlled form are declared access operations; a state or matrix description alone supplies neither call.

Proof

Given: As in Facts & Assumptions.

1.1givenF1F5F6assume-contraconstructalgebra

Assume for contradiction that such a unitary $U$ exists as the one fixed implementation replacing the explicitly declared reflection access of [F6]. Compare the unit inputs $∣ 0 ⟩_{P} \otimes ∣ 0 ⟩_{T}$ and $∣ + ⟩_{P} \otimes ∣ 0 ⟩_{T}$ , where $∣ + ⟩ = (∣ 0 ⟩ + ∣ 1 ⟩) / \sqrt{2}$ . Their inner product is $1 / \sqrt{2}$ by [F1] and [F5].

2.1givenF2F3step 1.1algebra

The required output for the first input is $- ∣ 0 ⟩_{P} \otimes ∣ 0 ⟩_{T}$ , since $S_{0} ∣ 0 ⟩ = - ∣ 0 ⟩$ . For the second, $2 ∣ + ⟩ ⟨ + ∣ 0 ⟩ = ∣ 0 ⟩ + ∣ 1 ⟩$ , so $S_{+} ∣ 0 ⟩ = - ∣ 1 ⟩$ and the output is $- ∣ + ⟩_{P} \otimes ∣ 1 ⟩_{T}$ . These computations use [F2]–[F3].

3.1F1F4F5step 1.1step 2.1discharge-contradiction∎

The two required outputs have inner product zero because $⟨ 0 ∣ 1 ⟩ = 0$ on $T$ , by [F1] and [F5]. But [F4] says $U$ preserves the input inner product $1 / \sqrt{2}$ . This contradiction proves that no such fixed unitary exists.

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Sources

Andrew M. Childs, Lecture Notes on Quantum Algorithms, Chapter 28 §28.3 (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, §9.3 (standard reference, not scraped)