Alphabeta Math
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Two-dimensional diagonal HHL inversion

Example

Let H=diag⁡(1,1/κ) for κ≥1 and ∣b⟩=(∣0⟩+∣1⟩)/2. The support promise holds with gap 1/κ.

Verification

[F1] Ideal inverse amplitude and HHL postselection probability: the ideal well vector is (2κ)−1H+∣b⟩ and its squared norm is the ideal success probability.

Given: The displayed diagonal matrix and normalized input.

1.1givenalgebra

Direct inversion gives H+∣b⟩=(∣0⟩+κ∣1⟩)/2. Its norm is 1+κ2/2, so the target is (∣0⟩+κ∣1⟩)/1+κ2.

2.1F1step 1.1algebra∎

By [F1], the well vector is (∣0⟩+κ∣1⟩)/(22κ) and its probability is (1+κ2)/(8κ2), which satisfies 1/8<p≤1/4. For OP5, the eigenvalue grid spacing is 2π/t0=πϵ/(1024κ). The sine window has nonzero tails, so this spacing is a grid resolution and not a hard error guarantee; the kernel moment and filter proof supply the error bound.

Depends on

Used by

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Dependency tree · two levels

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Sources