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Pinching raises finite entropy

Statement

Let $H$ be a complex Hilbert space of finite dimension $d \geq 1$ . Fix an ordered orthonormal basis $(e_{a})_{a = 1}^{d}$ of $H$ and the unitary coordinate map $U : H \to C^{d}$ given by $U (\sum_{a} z_{a} e_{a}) = (z_{1}, \dots, z_{d})$ . Represent each endomorphism $A : H \to H$ by $U A U^{*} \in M_{d} (C)$ , using the same symbol for the operator and its matrix in these fixed coordinates. Let $P_{0}, \dots, P_{r - 1}$ be pairwise orthogonal projections, meaning $P_{j} = P_{j}^{*} = P_{j}^{2}$ and $P_{j} P_{k} = 0$ for $j \neq k$ , with $r \geq 1$ and $\sum_{j = 0}^{r - 1} P_{j} = I$ . Define the pinching map $Δ : M_{d} (C) \to M_{d} (C)$ by $Δ (X) = \sum_{j = 0}^{r - 1} P_{j} X P_{j}$ . For every positive semidefinite trace-one matrix $ρ \in M_{d} (C)$ , $Δ$ is a finite channel and

$S (Δ (ρ)) \geq S (ρ),$

where $S (ρ) = - k_{B} \sum_{λ} λ \log λ$ , the sum is over eigenvalues with multiplicity, $0 \log 0 = 0$ , the logarithm is natural, and $k_{B} > 0$ has units joule per kelvin. The theorem concerns only this finite pinching operation; it does not say that every channel increases entropy.

Facts & Assumptions

Given: A finite-dimensional complex Hilbert space with the fixed orthonormal coordinates in the Statement, the stated orthogonal projections, $k_{B} > 0$ , and a positive semidefinite trace-one matrix $ρ \in M_{d} (C)$ .

[A1]

The finite entropy supplier used below carries the full Axiom of Choice through its current trace-cross-log dependency closure (The Axiom of Choice). The phase-average and spectral calculations in this item use only finite sums; no stronger choice principle is introduced.

[F1]

Finite completely positive trace-preserving maps: a finite channel is completely positive and trace-preserving.

[F2]

Complete finite Choi-Kraus equivalence: a map with Kraus operators $P_{j}$ is completely positive, and it is trace-preserving when $\sum_{j} P_{j}^{*} P_{j} = I$ .

[F3]

Quantum spectral entropy and relative-entropy domains: for a positive trace-one matrix with eigenvalues $λ$ , $S (ρ) = - k_{B} \sum_{λ} λ \log λ$ , with zero terms interpreted as $0$ .

[F4]

Finite mixing, product entropy, and bipartite subadditivity: entropy is concave under finite mixtures of finite-dimensional density matrices, including rank-deficient inputs. The current draft has a hash-current Step 1 ready record after owner resolution, but has not received independent proof acceptance; this deduction is conditional on that exact finite-mixing clause and its stated Axiom of Choice.

[F5]

Finite matrix spectral and state calculus: finite Hermitian matrices have orthonormal spectral decompositions, and unitary conjugation preserves their eigenvalues.

[F6]

The complex exponential by its power series: for $z \in C$ , $\exp z$ is defined by its convergent complex power series and $\exp 0 = 1$ .

[F7]

$\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential: $\exp (z + w) = \exp z \exp w$ for all $z, w \in C$ .

[F8]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ : $\exp z = 1$ exactly when $z \in 2 π i Z$ .

Proof

1.1givenF1F2algebra

The map $Δ$ has Kraus matrices $P_{j}$ , and $\sum_{j} P_{j}^{*} P_{j} = \sum_{j} P_{j} = I$ . Therefore [F2] gives complete positivity and trace preservation, which is the channel property in [F1].

1.2givenF6F7F8algebra

Let $ω = \exp (2 π i / r)$ and $V = \sum_{j = 0}^{r - 1} ω^{j} P_{j}$ . By [F7], $ω^{n} = \exp (2 π i n / r)$ for positive integers $n$ ; for negative $n$ , addition and $\exp 0 = 1$ give the corresponding reciprocal identity. The kernel fact [F8] then gives $ω^{r} = 1$ and $ω^{n} \neq 1$ for every integer $n$ with $0 < ∣ n ∣ < r$ . Orthogonality gives $V V^{*} = V^{*} V = \sum_{j} P_{j} = I$ , so $V$ is unitary. For $X \in M_{d} (C)$ , $\frac{1}{r} \sum_{k = 0}^{r - 1} V^{k} X V^{- k} = \sum_{j, ℓ = 0}^{r - 1} (\frac{1}{r} \sum_{k = 0}^{r - 1} ω^{k (j - ℓ)}) P_{j} X P_{ℓ} = \sum_{j} P_{j} X P_{j} = Δ (X),$ Indeed the inner sum is $1$ when $j = ℓ$ ; otherwise $q = ω^{j - ℓ}$ satisfies $q^{r} = 1$ and $q \neq 1$ , so $\sum_{k = 0}^{r - 1} q^{k} = (q^{r} - 1) / (q - 1) = 0$ . When $r = 1$ , there are no off-diagonal terms and the same identity holds.

2.1givenA1F3F4F5step 1.2algebra∎

Each $V^{k} ρ V^{- k}$ has the same eigenvalues as $ρ$ , by [F5], hence the same entropy by [F3]. Applying the finite-mixture concavity [F4] with equal weights $1 / r$ and using step 1.2 gives $S (Δ (ρ)) \geq r^{- 1} \sum_{k} S (V^{k} ρ V^{- k}) = S (ρ)$ , conditional on [A1] and the stated draft supplier interface.

Pinching raises finite entropy

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources