LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck pass

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Flagged trace ratio returns a Gibbs-state observable mean

Statement

Let $G = G^{*}$ and $M = M^{*}$ be $d \times d$ matrices with $d \geq 1$ and $∥ M ∥ \leq 1$ . Set $g = e^{- G / 2}$ , $B > 0$ , and $ρ_{G} = e^{- G} / tr (e^{- G})$ . Prepare $∣ Φ_{d} ⟩$ , apply a block whose top-left amplitude is $g / B$ to the first register, then a block whose squared top-left compression is $(I + M / 2) / 4$ to that register. Let $p_{M}$ be the probability that all work flags are zero. Let $p_{0}$ be the corresponding probability when $M = 0$ . Then $p_{0} > 0$ and $\frac{p_{M}}{p_{0}} = 1 + \frac{1}{2} tr (M ρ_{G}) .$

Facts & Assumptions

Given: The maximally entangled vectorization identity, trace cyclicity, and the finite positive square root for the observable block.

[F1]

SDP maximally entangled reference-state access: $∣ Φ_{d} ⟩ = d^{- 1 / 2} \sum_{i} ∣ i, i ⟩$ .

[F2]

For $A \in M_{m \times n} (F)$ and $B \in M_{n \times m} (F)$ , $tr (A B) = tr (B A)$ : trace is cyclic for compatible matrices.

[F3]

Trace is a linear functional on $M_{n} (F)$ : trace is linear.

[F4]

Trace of a product of Hermitian matrices is real: the trace ratio terms involving Hermitian matrices are real.

Proof

technique · direct

1.1givenF1F2algebra

For any $d \times d$ matrix $X$ , direct expansion of [F1] gives $⟨ Φ_{d} ∣ (X \otimes I) ∣ Φ_{d} ⟩_{D} = d^{- 1} \sum_{i} ⟨ i ∣ X ∣ i ⟩_{D} = tr (X) / d$ . If $F$ is the second block's top-left amplitude, then $F^{*} F = (I + M / 2) / 4$ . The composite success operator on the first register is $F g / B$ , so $p_{M} = \frac{1}{d B^{2}} tr (g F^{*} F g) = \frac{1}{4 d B^{2}} tr (e^{- G} (I + M / 2)),$ where the second equality follows from [F2].

2.1givenF3F4step 1.1algebra∎

With $M = 0$ , step 1.1 gives $p_{0} = tr (e^{- G}) / (4 d B^{2}) > 0$ , since every eigenvalue of $e^{- G}$ is strictly positive. Linearity [F3] and the definition of $ρ_{G}$ now give $p_{M} / p_{0} = 1 + tr (e^{- G} M) / (2 tr (e^{- G})) = 1 + \frac{1}{2} tr (M ρ_{G})$ . The traces are real by [F4].

Depends on

Used by

Flagged Gibbs trace estimation: errors, probability floors and charged access Theorem

Dependency tree · two levels

31 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

Joran van Apeldoorn et al., Quantum SDP-solvers: Better upper and lower bounds (standard reference, not scraped)