Appendix A (A.28): comparison of the two Renyi differences

Open

QAlgorithms.Papers.Independent.Intakethird100.P52.renyi_difference_comparison

by Gohan · Nov 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

All Hilbert spaces below are finite dimensional and nonzero over $\mathbb C$ . A density operator is positive semidefinite with trace one. Every spectral power is zero on the kernel, including zero and negative powers. Information quantities use base-two logarithms, with $\log_2 0=-\infty$ ; the operator logarithms in exponential formulas are natural logarithms.

Let $\rho$ be a density operator on $H$ , let $\sigma\ge0$ be an operator on $H$ without a trace constraint, and let $\mathcal N:\mathcal B(H)\to\mathcal B(K)$ be CPTP. Its adjoint $\mathcal N^\dagger$ is the Hilbert–Schmidt adjoint. Write

\begin{aligned} K_a&=\sigma^{(1-a)/2}\mathcal N^\dagger\!\left(\mathcal N(\sigma)^{(a-1)/2}\mathcal N(\rho)^{1-a}\mathcal N(\sigma)^{(a-1)/2}\right)\sigma^{(1-a)/2},\\ L_a&=\sigma^{(1-a)/(2a)}\mathcal N^\dagger\!\left(\mathcal N(\sigma)^{(a-1)/(2a)}\mathcal N(\rho)^{(1-a)/a}\mathcal N(\sigma)^{(a-1)/(2a)}\right)\sigma^{(1-a)/(2a)},\\ R_a&=K_a^{1/(1-a)},\qquad S_a=L_a^{a/(1-a)},\\ \Delta_a&=\frac{\log_2\operatorname{Tr}(\rho^aK_a)}{a-1},\qquad \widetilde\Delta_a=\frac{\log_2\operatorname{Tr}[(\rho^{1/2}L_a\rho^{1/2})^a]}{a-1}. \end{aligned}

Assume $\rho>0$ , $\sigma>0$ , and that $\mathcal N$ is strict: $\mathcal N(X)>0$ for every $X>0$ . The channel is sufficient for $(\rho,\sigma)$ when one CPTP map $\mathcal R$ satisfies $\mathcal R(\mathcal N(\rho))=\rho$ and $\mathcal R(\mathcal N(\sigma))=\sigma$ .

For $a>1$ set $a^\prime=(2a-1)/a$ , so $1<a^\prime<2$ . Then

\widetilde\Delta_a(\rho,\sigma,\mathcal N)\ge\Delta_{a^\prime}(\rho,\sigma,\mathcal N).

This compares the sandwiched difference to a Petz-type difference at the transformed order.

Preamble

import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P52

Formal statement

noncomputable section

open scoped Matrix ComplexOrder

namespace QAlgorithms.Papers.Independent.Intakethird100.P52

attribute [local instance] Classical.propDecidable Classical.decEq

open QAlgorithms.Papers.Independent.Information100.P63 (Mat)

open QAlgorithms.Papers.Independent.Dependency50.P21 (channelAdjoint petz marginalAB)

open QAlgorithms.Papers.Independent.Dependency50.P35 (tr elog maxRelative)

open QAlgorithms.Papers.Independent.Information100.P71 (supportPower)

open QAlgorithms.Papers.Independent.Renew46.P35 (HasMarkovBlocks)

/-- Appendix A, PDF p.22, equation (A.28), inheriting Proposition 20
hypotheses. Primary (1.17)/(A.17) fixes the scale despite the auxiliary typo
in (A.19); the comparison has coefficient one. -/
theorem renyi_difference_comparison {H K : Type} [Fintype H] [Fintype K]
    [DecidableEq H] [DecidableEq K] [Nonempty H] [Nonempty K]
    (rho sigma : Mat H) (hrho : IsDensityMatrix rho)
    (hrhoPD : rho.PosDef) (hsigmaPD : sigma.PosDef)
    (N : Superop H K) (hN : IsStrictChannel N)
    (a : ℝ) (ha : 1 < a) :
    renyiDifference ((2 * a - 1) / a) rho sigma N ≤ deltaTilde a rho sigma N := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P52

end

Source

Quantum Markov chains, sufficiency of quantum channels, and Renyi information measures, arXiv:1501.05636v4, PDF p. 22, Appendix A (A.28): comparison of the two Renyi differences

Read-back

What the Lean code literally says, in plain math · gpt-6.1-sol

Let $H,K$ be finite, nonempty index types with decidable equality. Let $\rho,\sigma$ be positive definite complex matrices on $H$ , with $\rho$ positive semidefinite and of trace $1$ , and with no trace normalization required of $\sigma$ . Let $N$ admit a finite Kraus representation $N(X)=\sum_jE_jXE_j^\dagger$ satisfying $\sum_jE_j^\dagger E_j=I_H$ , and assume that $N(X)$ is positive definite for every positive definite $X$ . Define $(N^*Y)_{ij}=\sum_{u,v\in K}\overline{N(E_{ij})_{uv}}Y_{uv}$ , where $E_{ij}$ is a matrix unit. Write $X^{[t]}$ for spectral functional calculus using $x^t$ at nonzero eigenvalues and $0$ at zero eigenvalues, including at exponent $0$ ; write $\operatorname{tr}X=\operatorname{Re}\operatorname{Tr}X$ , and let $\ell(x)=\log_2x$ for $x>0$ and $-\infty$ otherwise. For real $b$ , define

\begin{aligned} K_b&=\sigma^{[(1-b)/2]}N^*\!\left( (N\sigma)^{[(b-1)/2]}(N\rho)^{[1-b]} (N\sigma)^{[(b-1)/2]}\right)\sigma^{[(1-b)/2]},\\ D_b&=\frac{1}{b-1}\ell\!\left(\operatorname{tr}(\rho^{[b]}K_b)\right), \end{aligned}

and, for real $a>1$ , define

\begin{aligned} \widetilde K_a&=\sigma^{[(1-a)/(2a)]}N^*\!\left( (N\sigma)^{[(a-1)/(2a)]}(N\rho)^{[(1-a)/a]} (N\sigma)^{[(a-1)/(2a)]}\right)\sigma^{[(1-a)/(2a)]},\\ \widetilde D_a&=\frac{1}{a-1}\ell\!\left( \operatorname{tr}\bigl((\rho^{[1/2]}\widetilde K_a\rho^{[1/2]})^{[a]}\bigr)\right). \end{aligned}

The declaration asserts, for every such real $a>1$ , the extended-real inequality

D_{(2a-1)/a}\le\widetilde D_a.

Human review

Endorsed by Shuze Chen · Nov 1, 2026
Confirmed by the moderator at approval.
Endorsed by Gohan · Nov 1, 2026
Confirmed by the mission captain (proposal self-audit).

View graph