ExampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Two-term LCU postselection is input dependent

Example

On a qubit let $M = I + 2 Z$ , with LCU terms $(1, I)$ and $(2, Z)$ , so $α = 3$ . Use the complete PREPARE unitary $P = (\begin{array}{cc} 1 / \sqrt{3} & - \sqrt{2 / 3} \\ \sqrt{2 / 3} & 1 / \sqrt{3} \end{array}),$ whose first column prepares $3^{- 1 / 2} ∣ 0 ⟩ + \sqrt{2 / 3} ∣ 1 ⟩$ , and explicitly supply inverse access $P^{†}$ . Let SELECT apply $I$ and $Z$ on the corresponding indices. The projected block is $M / 3 = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 / 3 \end{array}) .$ For $∣ ψ ⟩ = a ∣ 0 ⟩ + b ∣ 1 ⟩$ with $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ , the clean-ancilla branch vector is $a ∣ 0 ⟩ - (b / 3) ∣ 1 ⟩$ and its weight is $p_{0} = ∣ a ∣^{2} + ∣ b ∣^{2} / 9.$ The branch is always nonzero, with $1 / 9 \leq p_{0} \leq 1$ ; its conditional state is the branch vector divided by $\sqrt{p_{0}}$ . The least weight occurs at $a = 0$ , and the greatest at $b = 0$ .

Facts & Assumptions

Given: A qubit system, the LCU list $(1, I), (2, Z)$ , the complete PREPARE unitary and its explicitly supplied inverse access, SELECT, and a unit input $a ∣ 0 ⟩ + b ∣ 1 ⟩$ .

[F1]

The Pauli matrices: $Z ∣ 0 ⟩ = ∣ 0 ⟩$ and $Z ∣ 1 ⟩ = - ∣ 1 ⟩$ .

[F2]

Finite registers and their ordered computational bases and The computational basis is orthonormal: the qubit basis is orthonormal.

[F3]

LCU projected-block and postselection correctness: a supplied LCU block has clean branch vector $M ∣ ψ ⟩ / α$ and weight $∥ M ∣ ψ ⟩ ∥^{2} / α^{2}$ .

[F4]

Clean-ancilla zero-branch postselection and Clean-ancilla postselection has normalized branch weights: a positive branch weight gives the normalized conditional system vector.

Verification

Given: As in Facts & Assumptions.

1.1givenF1F2F3constructalgebra

The absolute coefficient sum is $α = ∣ 1 ∣ + ∣ 2 ∣ = 3$ . The specified real rotation $P$ is unitary because its columns have norm one and inner product zero by [F2]; its first column is the required prepared index state, and the inverse access is explicitly supplied. SELECT is unitary because it is block diagonal with blocks $I$ and $Z$ , and direct multiplication of the matrices in [F1] gives $I^{†} I = Z^{†} Z = I$ . Thus the access satisfies [F3].

2.1givenF1F2F3step 1.1algebra

By [F1], $M = I + 2 Z$ is diagonal with entries $3, - 1$ , so [F3] gives the projected block $M / 3 = diag (1, - 1 / 3)$ and branch vector $a ∣ 0 ⟩ - (b / 3) ∣ 1 ⟩$ . Orthonormality in [F2] makes its squared norm $p_{0} = ∣ a ∣^{2} + ∣ b ∣^{2} / 9$ .

3.1givenF2F4step 2.1algebra∎

Since $∣ a ∣^{2} + ∣ b ∣^{2} = 1$ , $p_{0} = 1 - (8 / 9) ∣ b ∣^{2}$ . The bounds $0 \leq ∣ b ∣^{2} \leq 1$ give $1 / 9 \leq p_{0} \leq 1$ ; equality occurs at $a = 0$ and $b = 0$ , respectively. Thus the branch is nonzero for every unit input and [F4] gives the stated conditional vector.

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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)