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Corollary 15: vanishing Renyi conditional information characterizes short quantum Markov chains

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QAlgorithms.Papers.Independent.Intakethird100.P52.markov_renyi

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

prove2quantum

All Hilbert spaces below are finite dimensional and nonzero over C\mathbb CC. 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⁡20=−∞\log_2 0=-\inftylog2​0=−∞; the operator logarithms in exponential formulas are natural logarithms.

Let ρ\rhoρ be a density operator on HHH, let σ≥0\sigma\ge0σ≥0 be an operator on HHH without a trace constraint, and let N:B(H)→B(K)\mathcal N:\mathcal B(H)\to\mathcal B(K)N:B(H)→B(K) be CPTP. Its adjoint N†\mathcal N^\daggerN† is the Hilbert–Schmidt adjoint. Write

Ka=σ(1−a)/2N† ⁣(N(σ)(a−1)/2N(ρ)1−aN(σ)(a−1)/2)σ(1−a)/2,La=σ(1−a)/(2a)N† ⁣(N(σ)(a−1)/(2a)N(ρ)(1−a)/aN(σ)(a−1)/(2a))σ(1−a)/(2a),Ra=Ka1/(1−a),Sa=Laa/(1−a),Δa=log⁡2Tr⁡(ρaKa)a−1,Δ~a=log⁡2Tr⁡[(ρ1/2Laρ1/2)a]a−1.\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}Ka​La​Ra​Δa​​=σ(1−a)/2N†(N(σ)(a−1)/2N(ρ)1−aN(σ)(a−1)/2)σ(1−a)/2,=σ(1−a)/(2a)N†(N(σ)(a−1)/(2a)N(ρ)(1−a)/aN(σ)(a−1)/(2a))σ(1−a)/(2a),=Ka1/(1−a)​,Sa​=Laa/(1−a)​,=a−1log2​Tr(ρaKa​)​,Δa​=a−1log2​Tr[(ρ1/2La​ρ1/2)a]​.​

Let ρ=ρABC\rho=\rho_{ABC}ρ=ρABC​ be a tripartite density operator. Its marginals ρAC,ρBC,ρC\rho_{AC},\rho_{BC},\rho_CρAC​,ρBC​,ρC​ are computed by partial trace from this same state. Operators on fewer registers are extended by identities. Use tensor order A⊗(C⊗B)A\otimes(C\otimes B)A⊗(C⊗B), and set σ=ρAC⊗IB\sigma=\rho_{AC}\otimes I_Bσ=ρAC​⊗IB​ and N=Tr⁡A\mathcal N=\operatorname{Tr}_AN=TrA​. Define Ia(A;B∣C)ρ=Δa(ρ,σ,N)I_a(A;B\mid C)_\rho=\Delta_a(\rho,\sigma,\mathcal N)Ia​(A;B∣C)ρ​=Δa​(ρ,σ,N) and I~a(A;B∣C)ρ=Δ~a(ρ,σ,N)\widetilde I_a(A;B\mid C)_\rho=\widetilde\Delta_a(\rho,\sigma,\mathcal N)Ia​(A;B∣C)ρ​=Δa​(ρ,σ,N). A short quantum Markov chain A−C−BA-C-BA−C−B means that the whole space HCH_CHC​ decomposes as ⨁jHCL,j⊗HCR,j\bigoplus_j H_{C_{L,j}}\otimes H_{C_{R,j}}⨁j​HCL,j​​⊗HCR,j​​ and ρ=⨁jqjρACL,j⊗ρCR,jB\rho=\bigoplus_j q_j\rho_{AC_{L,j}}\otimes\rho_{C_{R,j}B}ρ=⨁j​qj​ρACL,j​​⊗ρCR,j​B​, where qj≥0q_j\ge0qj​≥0, ∑jqj=1\sum_jq_j=1∑j​qj​=1, and both factors in each block are density operators.

Assume ρABC>0\rho_{ABC}>0ρABC​>0. If ρ\rhoρ is a short quantum Markov chain A−C−BA-C-BA−C−B, then

Ia(A;B∣C)ρ=I~a(A;B∣C)ρ=0for every a∈(0,1)∪(1,∞).I_a(A;B\mid C)_\rho=\widetilde I_a(A;B\mid C)_\rho=0 \quad\text{for every }a\in(0,1)\cup(1,\infty).Ia​(A;B∣C)ρ​=Ia​(A;B∣C)ρ​=0for every a∈(0,1)∪(1,∞).

Conversely, either Ia(A;B∣C)ρ=0I_a(A;B\mid C)_\rho=0Ia​(A;B∣C)ρ​=0 for some a∈(0,1)∪(1,2)a\in(0,1)\cup(1,2)a∈(0,1)∪(1,2) or I~a(A;B∣C)ρ=0\widetilde I_a(A;B\mid C)_\rho=0Ia​(A;B∣C)ρ​=0 for some a∈(1/2,1)∪(1,∞)a\in(1/2,1)\cup(1,\infty)a∈(1/2,1)∪(1,∞) implies the Markov property.

This gives the Rényi zero-information characterization of short quantum Markov chains.

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)

/-- Corollary 15, PDF p.17. Positive definite tripartite densities;
HasMarkovBlocks describes the entire conditioning register C. -/
theorem markov_renyi {A C B : Type} [Fintype A] [Fintype C] [Fintype B]
    [DecidableEq A] [DecidableEq C] [DecidableEq B]
    [Nonempty A] [Nonempty C] [Nonempty B]
    (rho : Mat (A × (C × B))) (hrho : IsDensityMatrix rho)
    (hrhoPD : rho.PosDef) :
    (HasMarkovBlocks rho → ∀ a : ℝ, (0 < a ∧ a < 1) ∨ 1 < a →
      conditionalRenyiInformation a rho = 0 ∧ conditionalRenyiTilde a rho = 0) ∧
    (((∃ a : ℝ, ((0 < a ∧ a < 1) ∨ (1 < a ∧ a < 2)) ∧ conditionalRenyiInformation a rho = 0) ∨
      (∃ a : ℝ, ((1 / 2 < a ∧ a < 1) ∨ 1 < a) ∧ conditionalRenyiTilde a rho = 0)) →
      HasMarkovBlocks rho) := 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. 17, Corollary 15: vanishing Renyi conditional information characterizes short quantum Markov chains
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let A,C,BA,C,BA,C,B be finite, nonempty index types with decidable equality, and let ρ\rhoρ be a positive definite complex matrix on A×(C×B)A\times(C\times B)A×(C×B) that is positive semidefinite and has trace 111. Write ρAC=Tr⁡Bρ\rho_{AC}=\operatorname{Tr}_B\rhoρAC​=TrB​ρ, σ=ρAC⊗IB\sigma=\rho_{AC}\otimes I_Bσ=ρAC​⊗IB​, N=Tr⁡AN=\operatorname{Tr}_AN=TrA​, R=N(ρ)R=N(\rho)R=N(ρ), and S=N(σ)S=N(\sigma)S=N(σ), where partial traces sum diagonal entries over the indicated register. For a positive semidefinite matrix XXX, write X[t]X^{[t]}X[t] for spectral functional calculus with scalar function x↦xtx\mapsto x^tx↦xt for x≠0x\ne0x=0 and value 000 at x=0x=0x=0, including when t=0t=0t=0; write tr⁡X=Re⁡Tr⁡X\operatorname{tr}X=\operatorname{Re}\operatorname{Tr}XtrX=ReTrX, and let ℓ(x)=log⁡2x\ell(x)=\log_2xℓ(x)=log2​x when x>0x>0x>0 and ℓ(x)=−∞\ell(x)=-\inftyℓ(x)=−∞ otherwise. Define

Ka=σ[(1−a)/2](IA⊗[S[(a−1)/2]R[1−a]S[(a−1)/2]])σ[(1−a)/2],K~a=σ[(1−a)/(2a)](IA⊗[S[(a−1)/(2a)]R[(1−a)/a]S[(a−1)/(2a)]])σ[(1−a)/(2a)],Da=1a−1 ℓ ⁣(tr⁡(ρ[a]Ka)),D~a=1a−1 ℓ ⁣(tr⁡((ρ[1/2]K~aρ[1/2])[a])).\begin{aligned} K_a&=\sigma^{[(1-a)/2]} \bigl(I_A\otimes[S^{[(a-1)/2]}R^{[1-a]}S^{[(a-1)/2]}]\bigr) \sigma^{[(1-a)/2]},\\ \widetilde K_a&=\sigma^{[(1-a)/(2a)]} \bigl(I_A\otimes[S^{[(a-1)/(2a)]}R^{[(1-a)/a]}S^{[(a-1)/(2a)]}]\bigr) \sigma^{[(1-a)/(2a)]},\\ D_a&=\frac{1}{a-1}\,\ell\!\left(\operatorname{tr}(\rho^{[a]}K_a)\right),\\ \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}Ka​Ka​Da​Da​​=σ[(1−a)/2](IA​⊗[S[(a−1)/2]R[1−a]S[(a−1)/2]])σ[(1−a)/2],=σ[(1−a)/(2a)](IA​⊗[S[(a−1)/(2a)]R[(1−a)/a]S[(a−1)/(2a)]])σ[(1−a)/(2a)],=a−11​ℓ(tr(ρ[a]Ka​)),=a−11​ℓ(tr((ρ[1/2]Ka​ρ[1/2])[a])).​

Here Da,D~aD_a,\widetilde D_aDa​,Da​ take values in the extended real numbers. Say that ρ\rhoρ has the stated block decomposition if there are a positive integer rrr, positive integers lj,rjl_j,r_jlj​,rj​, a unitary identification C≅⨁j=1r(Clj⊗Crj)C\cong\bigoplus_{j=1}^{r}(\mathbb C^{l_j}\otimes\mathbb C^{r_j})C≅⨁j=1r​(Clj​⊗Crj​) with inclusions VjV_jVj​, nonnegative weights qjq_jqj​ summing to 111, and positive semidefinite trace-one matrices αj\alpha_jαj​ on A×{1,…,lj}A\times\{1,\ldots,l_j\}A×{1,…,lj​} and βj\beta_jβj​ on {1,…,rj}×B\{1,\ldots,r_j\}\times B{1,…,rj​}×B, such that

ρ=∑j=1rqj (IA⊗Vj⊗IB)(αj⊗βj)(IA⊗Vj⊗IB)†.\rho=\sum_{j=1}^{r}q_j\, (I_A\otimes V_j\otimes I_B)(\alpha_j\otimes\beta_j) (I_A\otimes V_j\otimes I_B)^\dagger.ρ=j=1∑r​qj​(IA​⊗Vj​⊗IB​)(αj​⊗βj​)(IA​⊗Vj​⊗IB​)†.

Zero weights are permitted. The declaration asserts both that this decomposition implies Da=D~a=0D_a=\widetilde D_a=0Da​=Da​=0 for every real a∈(0,1)∪(1,∞)a\in(0,1)\cup(1,\infty)a∈(0,1)∪(1,∞), and that the decomposition exists whenever either Da=0D_a=0Da​=0 for some a∈(0,1)∪(1,2)a\in(0,1)\cup(1,2)a∈(0,1)∪(1,2), or D~a=0\widetilde D_a=0Da​=0 for some a∈(1/2,1)∪(1,∞)a\in(1/2,1)\cup(1,\infty)a∈(1/2,1)∪(1,∞).

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).

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