ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Finite product environment verifies conditional balance

Physical scope

A finite-dimensional product-state example inside the adopted nested horizon-balance model, including a two-dimensional worked instance. It makes no claim about a black-hole solution, continuum quantum field theory, empirical observation, or actual physical evolution.

Example

Let $H_{2}, H_{E}$ be nonzero finite-dimensional complex inner-product spaces. Let $ρ_{2}, σ_{2}$ be density operators on $H_{2}$ and $ρ_{E}, σ_{E}$ density operators on $H_{E}$ , with $σ_{2}$ and $σ_{E}$ positive definite. Set $H_{1} : = H_{2} \otimes H_{E}$ , $ρ_{1} : = ρ_{2} \otimes ρ_{E}$ , and $σ_{1} : = σ_{2} \otimes σ_{E}$ . Put $K_{E} : = - \log σ_{E}$ and $Δ_{E} : = tr ((ρ_{E} - σ_{E}) K_{E})$ . Then $D (ρ_{1} ∥ σ_{1}) - D (ρ_{2} ∥ σ_{2}) = D (ρ_{E} ∥ σ_{E})$ , and the adopted geometric balance gives $S_{g e n, 2} - S_{g e n, 1} = k_{B} D (ρ_{E} ∥ σ_{E})$ . In the particular case $H_{2} = H_{E} = C^{2}$ with $⟨ z, w ⟩ = z_{0} \overline{w_{0}} + z_{1} \overline{w_{1}}$ , let $I_{2}$ be the identity map, take $ρ_{2} = σ_{2} = I_{2} / 2$ and $σ_{E} = I_{2} / 2$ , and define $ρ_{E} : C^{2} \to C^{2}$ by $ρ_{E} (z_{0}, z_{1}) : = (z_{0}, 0)$ . Then one has $Δ_{E} = 0$ and $D (ρ_{E} ∥ σ_{E}) = \log 2$ , so $a_{2} = a_{1}$ satisfies the balance premise and $S_{g e n, 2} - S_{g e n, 1} = k_{B} \log 2$ .

Example references

Regulated generalized entropy monotonicity under explicit balance

Nested finite regulator and gravitational balance premises

Finite relative entropy equals modular energy minus entropy

Partial trace over a finite tensor factor

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm

Density operator

A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S

The basis-independent trace of an endomorphism of a finite-dimensional vector space

Facts & Assumptions

Given: The finite-dimensional spaces and density operators in the Example, the product assignments $ρ_{1} = ρ_{2} \otimes ρ_{E}$ and $σ_{1} = σ_{2} \otimes σ_{E}$ , and the specified two-qubit instance.

[F1]

Finite relative entropy equals modular energy minus entropy: for finite-dimensional density operators $ρ, σ$ with $σ$ positive definite, $D (ρ ∥ σ) = tr ((ρ - σ) (- \log σ)) - (s (ρ) - s (σ))$ , with $s (τ) = - tr (τ \log τ)$ in nats and $0 \log 0 : = 0$ .

[A1]

Nested finite regulator and gravitational balance premises: the geometric balance is $a_{2} - a_{1} = Δ_{1} - Δ_{2}$ and the model entropy is $S_{g e n, i} = k_{B} [a_{i} + s (ρ_{i}) - s (σ_{i})]$ , where $k_{B} > 0$ .

[F2]

Regulated generalized entropy monotonicity under explicit balance: under [A1] and finite partial-trace consistency, $S_{g e n, 2} - S_{g e n, 1} = k_{B} [D (ρ_{1} ∥ σ_{1}) - D (ρ_{2} ∥ σ_{2})]$ .

[F3]

Partial trace over a finite tensor factor: in ordered product bases, $({Tr}_{E} R)_{a a^{'}} = \sum_{b} R_{(a, b), (a^{'}, b)}$ .

[F4]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: for positive reals $x, y$ , $\log (x y) = \log x + \log y$ .

[F5]

Density operator: a density operator on a nonzero finite-dimensional complex inner-product space is non-negative and has trace one.

[F6]

A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S: each finite-dimensional non-negative operator has an orthonormal eigenbasis with nonnegative eigenvalues.

[F7]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: the trace of a finite-dimensional endomorphism is the basis-independent finite sum of its diagonal entries.

Verification

1.1givenF3F5F6F7algebra

By [F5]–[F6], each factor state has an orthonormal eigenbasis with nonnegative eigenvalues summing to one. The product matrices are diagonal in the product bases with pairwise-product eigenvalues, so they are non-negative and have trace one; $σ_{1}$ is positive definite because both reference factors are. In the ordered product basis, [F3] gives $({Tr}_{E} (ρ_{2} \otimes ρ_{E}))_{a a^{'}} = \sum_{b} (ρ_{2})_{a a^{'}} (ρ_{E})_{b b} = (ρ_{2})_{a a^{'}} tr (ρ_{E}) = (ρ_{2})_{a a^{'}}$ , and the same calculation gives ${Tr}_{E} σ_{1} = σ_{2}$ .

2.1F4F7step 1.1algebra

Let $λ_{a} > 0$ and $μ_{b} > 0$ be eigenvalues of $σ_{2}$ and $σ_{E}$ . In the product eigenbasis, [F4] gives $- \log (λ_{a} μ_{b}) = - \log λ_{a} - \log μ_{b}$ , hence $K_{1} = K_{2} \otimes I_{H_{E}} + I_{H_{2}} \otimes K_{E}$ . For simple tensors, $(A \otimes B) (C \otimes D) = A C \otimes B D$ by evaluation on $x \otimes y$ , and [F7]'s product-basis diagonal sum gives $tr ((A \otimes B) (C \otimes D)) = \sum_{a, b} (A C)_{a a} (B D)_{b b} = tr (A C) tr (B D)$ . It follows that $tr (ρ_{1} K_{1}) = tr (ρ_{2} K_{2}) tr (ρ_{E}) + tr (ρ_{2}) tr (ρ_{E} K_{E}) = tr (ρ_{2} K_{2}) + tr (ρ_{E} K_{E})$ ; the same expansion for $σ_{1}$ gives $tr (σ_{1} K_{1}) = tr (σ_{2} K_{2}) + tr (σ_{E} K_{E})$ . Subtracting proves $Δ_{1} = Δ_{2} + Δ_{E}$ .

3.1F1F4F6step 2.1algebra

If $(λ_{a})_{a}$ and $(μ_{b})_{b}$ are the eigenvalues of $ρ_{2}$ and $ρ_{E}$ , then the product state has eigenvalues $λ_{a} μ_{b}$ . Using [F4], trace-one of each factor, and $0 \log 0 : = 0$ from [F1], the finite sum is $s (ρ_{1}) = - \sum_{a, b : λ_{a} μ_{b} > 0} λ_{a} μ_{b} (\log λ_{a} + \log μ_{b}) = s (ρ_{2}) \sum_{b} μ_{b} + s (ρ_{E}) \sum_{a} λ_{a} = s (ρ_{2}) + s (ρ_{E})$ . The same calculation for $σ_{2}, σ_{E}$ gives $s (σ_{1}) = s (σ_{2}) + s (σ_{E})$ . Apply [F1] to the pairs $(ρ_{1}, σ_{1})$ , $(ρ_{2}, σ_{2})$ , and $(ρ_{E}, σ_{E})$ ; using $Δ_{1} = Δ_{2} + Δ_{E}$ from step 2.1 gives $D (ρ_{1} ∥ σ_{1}) = D (ρ_{2} ∥ σ_{2}) + D (ρ_{E} ∥ σ_{E})$ .

4.1F2step 1.1step 3.1algebra

The consistency in step 1.1 makes [F2] applicable, so $S_{g e n, 2} - S_{g e n, 1} = k_{B} [D (ρ_{1} ∥ σ_{1}) - D (ρ_{2} ∥ σ_{2})]$ . Substitute step 3.1 to obtain $S_{g e n, 2} - S_{g e n, 1} = k_{B} D (ρ_{E} ∥ σ_{E})$ .

5.1givenA1F1step 2.1step 4.1algebra∎

In the stated instance, $K_{E} = (\log 2) I_{2}$ , $tr (ρ_{E} - σ_{E}) = 0$ , and therefore $Δ_{E} = 0$ . The map $ρ_{E}$ fixes $(1, 0)$ and sends $(0, 1)$ to zero, so its eigenvalues are $1, 0$ , while $σ_{E} = I_{2} / 2$ , so [F1] gives $D (ρ_{E} ∥ σ_{E}) = 0 - (- \log 2) = \log 2$ . Also $ρ_{2} = σ_{2}$ , hence $Δ_{2} = 0$ ; step 2.1 then gives $Δ_{1} - Δ_{2} = 0$ , so $a_{2} = a_{1}$ satisfies [A1]. Step 4.1 yields the stated growth $k_{B} \log 2$ .

Remarks

The diagram records the operator-space map that traces out the environment. Its coordinate meaning is the supplier definition [F3]; the product-state conclusion is proved by the finite sum in step 1.1, not by the diagram.

diagram failed to render:
\operatorname{End}(H_2\otimes H_E) \arrow[r, "\operatorname{Tr}_E"] & \operatorname{End}(H_2)