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Finite mixing, product entropy, and bipartite subadditivity

Statement

For $m \geq 1$ density matrices $ρ_{1}, \dots, ρ_{m}$ on one finite-dimensional complex Hilbert space and weights $t_{a} \geq 0$ with $\sum_{a} t_{a} = 1$ , define their mixture $ρ : = \sum_{a} t_{a} ρ_{a}$ . The exact content of entropy concavity under finite mixing is $S (ρ) \geq \sum_{a = 1}^{m} t_{a} S (ρ_{a}) .$ For density matrices $ρ, σ$ on finite-dimensional complex Hilbert spaces, the exact content of product additivity is $S (ρ \otimes σ) = S (ρ) + S (σ)$ .

For the positive entropy scale $k_{B} > 0$ fixed in Quantum spectral entropy and relative-entropy domains (in SI units J/K), entropy is concave under finite mixing and additive on product states for finite-dimensional density matrices. If $R$ is a density matrix on $H_{A} \otimes H_{B}$ , with marginals $ρ_{A} = {Tr}_{B} R$ , $ρ_{B} = {Tr}_{A} R$ , then $S (R) \leq S (ρ_{A}) + S (ρ_{B}),$ with equality exactly when $R = ρ_{A} \otimes ρ_{B}$ . These finite statements include rank-deficient densities and assert neither strong subadditivity nor general infinite-dimensional data processing.

Facts & Assumptions

Given: Finite-dimensional complex Hilbert spaces $H_{A}, H_{B}$ , finite density matrices, $k_{B} > 0$ , and the entropy and relative-entropy definitions of Quantum spectral entropy and relative-entropy domains.

[A1]

The Axiom of Choice is inherited through the cross-log supplier closure; the displayed finite-dimensional calculations themselves use no infinite-dimensional selection (The Axiom of Choice).

[F1]

Trace-class cross-log entropy inequality: on support inclusion, finite cross-log moment gives nonnegative relative entropy with equality exactly for equal densities. In finite dimension the least positive eigenvalue of the second density bounds $- \log σ$ on its support, so this moment is finite even for rank-deficient densities.

[L1]

Finite partial trace, Schmidt decomposition and finite no-signalling, parts (a)–(b): on nonzero finite complex factor spaces, finite partial trace has the coordinate formula, positivity, trace preservation and intrinsic trace-pairing identities used below. No Schmidt or infinite-dimensional clause is invoked.

Proof

1.1givenalgebra

Let $ρ = \sum_{a} t_{a} ρ_{a}$ , $t_{a} \geq 0$ , $\sum_{a} t_{a} = 1$ . If $v \in \ker ρ$ , then $0 = ⟨ ρ v, v ⟩ = \sum_{a} t_{a} ⟨ ρ_{a} v, v ⟩$ ; positivity forces $ρ_{a} v = 0$ whenever $t_{a} > 0$ . Hence $supp ρ_{a} \subseteq supp ρ$ for every positive-weight term.

1.2givenalgebra

If $ρ, σ$ have eigenvalues $p_{i}, q_{j}$ , then $ρ \otimes σ$ has eigenbasis $u_{i} \otimes v_{j}$ and eigenvalues $p_{i} q_{j}$ . Summing $- k_{B} p_{i} q_{j} \log (p_{i} q_{j})$ and using $\sum_{i} p_{i} = \sum_{j} q_{j} = 1$ yields $S (ρ \otimes σ) = S (ρ) + S (σ)$ .

1.3givenL1algebra

Put $P_{A} = supp ρ_{A}$ . For $v \in \ker ρ_{A}$ , the partial-trace formula in [L1] gives $0 = ⟨ ρ_{A} v, v ⟩ = \sum_{j} ⟨ R (v \otimes f_{j}), v \otimes f_{j} ⟩$ . Each summand is nonnegative, so each is zero; positivity gives $R^{1 / 2} (v \otimes f_{j}) = 0$ , hence $R (v \otimes f_{j}) = 0$ . By linearity, $R$ vanishes on $\ker ρ_{A} \otimes H_{B}$ , so its range is orthogonal to that subspace and lies in $P_{A} \otimes H_{B}$ . Applying the same argument to the other marginal gives range $R \subseteq H_{A} \otimes P_{B}$ . The intersection of these two subspaces is $P_{A} \otimes P_{B}$ , as follows by using orthonormal bases adapted to the two support/kernel decompositions. Therefore $supp R \subseteq P_{A} \otimes P_{B}$ .

2.1givenA1step 1.1F1algebra

Omit zero-weight terms. On the supports from step 1.1, expanding the finite traces gives $S (ρ) - \sum_{a} t_{a} S (ρ_{a}) = k_{B} \sum_{a : t_{a} > 0} t_{a} D (ρ_{a} ∥ ρ) \geq 0$ by [F1]. Thus entropy is concave under finite mixing.

3.1step 1.3F1L1algebra∎

Diagonalize $ρ_{A}, ρ_{B}$ on their supports. Their positive eigenvalues show $\log (ρ_{A} \otimes ρ_{B}) = \log ρ_{A} \otimes I + I \otimes \log ρ_{B}$ on $P_{A} \otimes P_{B}$ . The trace-pairing identities in [L1] then give $D (R ∥ ρ_{A} \otimes ρ_{B}) = [S (ρ_{A}) + S (ρ_{B}) - S (R)] / k_{B} \geq 0$ by [F1]. Its equality condition in [F1] is exactly $R = ρ_{A} \otimes ρ_{B}$ .

Finite mixing, product entropy, and bipartite subadditivity

Statement

Facts & Assumptions

Proof

Depends on

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