TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Conditional mutual information and finite channel monotonicity

Statement

Let $H_{A}, H_{B}, H_{C}$ be nonzero finite-dimensional complex inner-product spaces and let $ρ_{A B C}$ be a density operator on $H_{A} \otimes H_{B} \otimes H_{C}$ . With dimensionless entropy $S$ and conditional mutual information $I (A : C ∣ B)$ defined by the conventions below,

$I (A : C ∣ B)_{ρ} = S (ρ_{A B}) + S (ρ_{B C}) - S (ρ_{B}) - S (ρ_{A B C}) \geq 0.$

Also let $d, e \geq 1$ and let $Φ : M_{d} (C) \to M_{e} (C)$ be a finite completely positive trace-preserving map. For all positive semidefinite $X, Y \in M_{d} (C)$ ,

$D (Φ (X) ∥ Φ (Y)) \leq D (X ∥ Y),$

where $D$ is the support-sensitive finite-dimensional quantum relative entropy: it equals $Tr (X \log X) - Tr (X \log Y)$ when $supp X \subseteq supp Y$ , and $+ \infty$ otherwise. These are finite-dimensional mathematical statements. The channel inequality is one-sided and asserts no equality condition, recovery map, or infinite-dimensional extension.

Statement references

Dimensionless entropy, mutual and conditional information

Finite completely positive trace-preserving maps

Finite-dimensional quantum relative entropy

Strong subadditivity of finite-dimensional quantum entropy

Partial-trace data processing for finite quantum relative entropy

Complete finite Choi-Kraus equivalence

Stinespring factorization diagram

diagram failed to render:
\begin{tikzcd}
M_d(\mathbb C)_+ \arrow[r, "X\mapsto VXV^*"] \arrow[dr, "\Phi"'] & \operatorname{End}(\mathbb C^e\otimes\mathcal E)_+ \arrow[d, "\operatorname{Tr}_{\mathcal E}"] \\
& M_e(\mathbb C)_+
\end{tikzcd}

Here $M_{d} (C)_{+}$ denotes positive semidefinite matrices, $E$ is the finite environment constructed in step 1.2, and $V$ is its isometry. The triangle commutes because step 1.2 computes ${Tr}_{E} (V X V^{*}) = Φ (X)$ entrywise. The proof uses that computation and the stated partial-trace inequality, not the diagram.

Facts & Assumptions

Given: The spaces and state in the first clause of the Statement; integers $d, e \geq 1$ and the finite channel $Φ$ in the second clause. Use natural logarithms and the ordered finite tensor conventions of the cited definitions.

[A1]

Dimensionless entropy, mutual and conditional information: $I (A : C ∣ B)_{ρ} = S (ρ_{A B}) + S (ρ_{B C}) - S (ρ_{B}) - S (ρ_{A B C})$ , with each marginal formed by partial trace.

[A2]

Finite completely positive trace-preserving maps: a finite channel is a complex-linear completely positive and trace-preserving map on matrix algebras.

[F1]

Strong subadditivity of finite-dimensional quantum entropy: for a density operator on any ordered finite triple $H_{A} \otimes H_{B} \otimes H_{C}$ , $S (ρ_{A B}) + S (ρ_{B C}) \geq S (ρ_{B}) + S (ρ_{A B C})$ .

[F2]

Partial-trace data processing for finite quantum relative entropy: for positive semidefinite $R, S$ on a finite tensor product, $D ({Tr}_{E} R ∥ {Tr}_{E} S) \leq D (R ∥ S)$ , including the extended-real support-failure cases.

[F3]

Complete finite Choi-Kraus equivalence: a finite CPTP map has a finite Kraus representation $Φ (X) = \sum_{a = 1}^{q} K_{a} X K_{a}^{*}$ , and trace preservation entails $\sum_{a} K_{a}^{*} K_{a} = I_{d}$ .

[F4]

Partial trace over a finite tensor factor: in product bases, the partial trace over the second factor sums diagonal blocks in that factor.

[F5]

Finite-dimensional quantum relative entropy: $D (R ∥ S)$ is finite on support inclusion and $+ \infty$ otherwise; the finite branch uses support-restricted spectral logarithms.

[F6]

Finite matrix spectral and state calculus: positive finite matrices have finite spectral decompositions and spectral logarithms on their positive supports; the finite trace is basis-independent and satisfies $Tr (A B) = Tr (B A)$ .

Proof

1.1givenA1F1

By [A1], the left side of the first assertion is the displayed four-entropy difference. Apply [F1] to the given state with the same factor order; the resulting inequality is exactly $I (A : C ∣ B)_{ρ} \geq 0$ .

1.2givenA2F3F4constructalgebra

By [F3], choose a finite Kraus list $K_{1}, \dots, K_{q}$ with $\sum_{a} K_{a}^{*} K_{a} = I_{d}$ ; because $d \geq 1$ , this identity rules out $q = 0$ . Let $E = C^{q}$ with orthonormal basis $(f_{a})$ and define $V : C^{d} \to C^{e} \otimes E$ by $V x = \sum_{a} K_{a} x \otimes f_{a}$ . Orthogonality of the $f_{a}$ gives $⟨ V x, V y ⟩ = \sum_{a} ⟨ K_{a} x, K_{a} y ⟩ = ⟨ x, (\sum_{a} K_{a}^{*} K_{a}) y ⟩ = ⟨ x, y ⟩$ , so $V^{*} V = I_{d}$ . The coordinate formula [F4] then gives ${Tr}_{E} (V X V^{*}) = \sum_{a} K_{a} X K_{a}^{*} = Φ (X)$ for every $X \in M_{d} (C)$ .

2.1givenF5F6step 1.2algebra

The isometry $V$ maps each positive-eigenvalue eigenspace of $X \geq 0$ onto the corresponding eigenspace of $V X V^{*}$ , with the same eigenvalue; the orthogonal complement of $ran V$ is in the kernel. Hence $supp (V X V^{*}) = V (supp X)$ , and likewise for $Y$ . Because $V$ is injective, support inclusion holds for $X, Y$ exactly when it holds for $V X V^{*}, V Y V^{*}$ . If it fails, both relative entropies are $+ \infty$ by [F5]. If it holds, the support-restricted spectral logarithms satisfy $\log (V X V^{*}) = V (\log X) V^{*}$ and $\log (V Y V^{*}) = V (\log Y) V^{*}$ on the corresponding supports. The trace identity in [F6] therefore yields $D (V X V^{*} ∥ V Y V^{*}) = D (X ∥ Y)$ .

3.1givenF2step 1.2step 2.1algebra∎

Apply [F2] to $R = V X V^{*}$ and $S = V Y V^{*}$ , which are positive semidefinite because $V$ is linear and $X, Y \geq 0$ . By step 1.2 their partial traces are $Φ (X), Φ (Y)$ , and by step 2.1 their relative entropy equals $D (X ∥ Y)$ . Thus $D (Φ (X) ∥ Φ (Y)) \leq D (X ∥ Y)$ , including when the right side is $+ \infty$ .

Conditional mutual information and finite channel monotonicity

Statement

Statement references

Stinespring factorization diagram

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources