CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.

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A fixed scalar label and zero energy do not determine finite spectral entropy

Statement refuted

On the complex Hilbert space $H = C^{2}$ , a density matrix means a positive semidefinite $2 \times 2$ matrix $ρ$ with $Tr ρ = 1$ . Let the Hamiltonian be the zero self-adjoint matrix $H = 0$ , define its energy by $E_{H} (ρ) = Tr (ρ H)$ , fix one positive real scalar label $a > 0$ , and fix a positive scalar $k_{B} > 0$ . Define the finite spectral entropy by $S (ρ) = - k_{B} \sum_{λ \in spec (ρ)} λ \log λ,$ where eigenvalues are counted with multiplicity, $\log$ is the natural logarithm, and $0 \log 0 : = 0$ . The statement that the fixed data $(a, E_{H} (ρ)) = (a, 0)$ determine one value of $S (ρ)$ for every density matrix $ρ$ on $H$ is false.

The scalar $a$ is held fixed as a mathematical label; $H$ is only the displayed finite matrix. No physical units, dynamics, horizon state, or coupling to geometry is part of this construction.

Facts & Assumptions

Given: The definitions in the Statement, with fixed $a > 0$ and $k_{B} > 0$ .

[A1]

The space is $H = C^{2}$ , $H = 0$ , $a > 0$ is fixed, and $k_{B} > 0$ is fixed.

[A2]

For a density matrix $ρ$ , $E_{H} (ρ) = Tr (ρ H)$ and $S (ρ) = - k_{B} \sum_{λ \in spec (ρ)} λ \log λ$ , with eigenvalues counted with multiplicity and $0 \log 0 = 0$ .

Counterexample

Given: Use the two explicit diagonal matrices below with the same $a$ , $H$ , and $k_{B}$ .

1.1givenA1constructalgebra

Let $ρ_{1} = diag (1, 0)$ and $ρ_{2} = diag (1 / 2, 1 / 2)$ . Both are positive semidefinite and have trace one, so they are density matrices; since $H = 0$ , both have energy $E_{H} (ρ_{i}) = Tr (ρ_{i} H) = 0$ , and both share the same fixed datum $a$ .

2.1givenA2step 1.1algebra∎

The eigenvalues of $ρ_{1}$ are $1, 0$ , so [A2] gives $S (ρ_{1}) = - k_{B} (1 \log 1 + 0 \log 0) = 0$ . The eigenvalues of $ρ_{2}$ are $1 / 2, 1 / 2$ , so $S (ρ_{2}) = - k_{B} \log (1 / 2) = k_{B} \log 2 > 0$ . Thus the same scalar-label and energy data occur with different finite entropies.

Remarks

This finite-dimensional matrix counterexample does not assert a coupled gravitational solution, backreaction, a horizon state, or a continuum entropy construction. It only shows that the displayed scalar label and energy value do not select a unique finite spectral entropy.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.