Corollary 12: zero-dual-witness lower bound

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QAlgorithms.Papers.Independent.Intakethird100.P95.corollary12

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

prove2quantum

Let $\lambda=(\lambda_1,\ldots,\lambda_d)$ be a finite probability vector: every entry is nonnegative and the entries sum to one. Write $\rho_\lambda=\sum_{i,j}\sqrt{\lambda_i\lambda_j}|ii\rangle\langle jj|$ . All logarithms have base two. A PPT operation is a completely positive, trace-preserving map $L$ such that $\Gamma_{\rm out}L\Gamma_{\rm in}$ is completely positive for arbitrary finite ancillary systems; $\Gamma$ transposes Bob in the fixed product basis. Assume in addition that all entries of $\lambda$ are strictly positive, so $d$ is its Schmidt rank. For $K\ge2$ , let $T(K;\lambda)$ be the minimum of the Schmidt-coordinate primal SDP (20), defined independently of operational reachability. Its real variables $s_{ij}$ for $j\le i$ and $a_{ij}$ for $j<i$ obey $s_{ij}\ge\sqrt{\lambda_i\lambda_j}/(K+1)$ and $a_{ij}\ge\sqrt{\lambda_i\lambda_j}/(K-1)$ , and the matrix $\sum_{i\ge j}s_{ij}\sigma_{ij}^{\Gamma}+\sum_{i>j}a_{ij}\alpha_{ij}^{\Gamma}$ is positive semidefinite. The objective is $\sum s_{ij}+\sum a_{ij}$ . Here $\sigma_{ii}=|ii\rangle\langle ii|$ , while the off-diagonal symmetric and antisymmetric projectors are $(|ij\rangle\pm|ji\rangle)(\langle ij|\pm\langle ji|)/2$ . For a finite order $t$ , $S_t$ is the Rényi entropy, with $S_0$ the log support size, $S_1=-\sum_i\lambda_i\log_2\lambda_i$ (using $0\log_2 0=0$ ), and $S_t=(1-t)^{-1}\log_2\sum_i\lambda_i^t$ otherwise. The separate infinity order is $S_\infty=-\log_2\max_i\lambda_i$ . The zero dual witness gives

T(K;\lambda)\ge\frac{K2^{S_{1/2}(\lambda)}-1}{K^2-1}.

This is a universal lower bound on the preparation SDP.

Preamble

import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P95

Formal statement

noncomputable section

open scoped BigOperators Matrix ComplexOrder

open Filter

namespace QAlgorithms.Papers.Independent.Intakethird100.P95

attribute [local instance] Classical.propDecidable Classical.decEq

-- Exact primitives from the pinned accepted library; no local aliases.

open QAlgorithms.Papers.Independent.Claimed50.P28 (maximallyEntangledDensity)

open QAlgorithms.Papers.Independent.Information100.P27 (IsProbability)

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

open QAlgorithms.Papers.Independent.Information100.P19 (outer)

open QAlgorithms.Papers.Independent.Dependency50.P18 (bipartiteTensor)

open QAlgorithms.Papers.Independent.Support50.P11 (Loewner)

/-- PDF p.7, Corollary 12: the numerator is K*2^{S_{1/2}}-1, with K outside the exponent. -/
theorem corollary12 {d : ℕ} (K : ℕ) (hK : 2 ≤ K) (lam : Fin d → ℝ)
    (hprob : IsProbability lam) (hpos : ∀ i, 0 < lam i) :
    ((K : ℝ) * Real.rpow 2 (schmidtRenyi (1 / 2) lam) - 1) / ((K : ℝ)^2 - 1) ≤
      primalValue K lam := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P95

end

Source

Pure-state transformations and catalysis under operations that completely preserve positivity of partial transpose, arXiv:0801.4322v2, PDF p. 7, Corollary 12: zero-dual-witness lower bound

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

For every natural number $d$ , integer $K\ge2$ , and strictly positive probability vector $\lambda=(\lambda_0,\ldots,\lambda_{d-1})$ , meaning $\lambda_i>0$ for every $i$ and $\sum_i\lambda_i=1$ , put $R_\lambda=(\sum_i\sqrt{\lambda_i})^2$ . On $\mathbb C^d\otimes\mathbb C^d$ , write $e_{ij}=|i\rangle\otimes|j\rangle$ , define $S_{ii}=e_{ii}e_{ii}^\dagger$ , $S_{ij}=\frac12(e_{ij}+e_{ji})(e_{ij}+e_{ji})^\dagger$ for $j<i$ , and $A_{ij}=\frac12(e_{ij}-e_{ji})(e_{ij}-e_{ji})^\dagger$ for $j<i$ . Let $\Gamma$ denote partial transpose on the second factor, so $(\Gamma X)_{(i,j),(k,l)}=X_{(i,l),(k,j)}$ , and set $Q(s,a)=\sum_{j\le i}s_{ij}\Gamma S_{ij}+\sum_{j<i}a_{ij}\Gamma A_{ij}$ . Define $V_K(\lambda)$ as the real infimum of $\sum_{j\le i}s_{ij}+\sum_{j<i}a_{ij}$ over real coefficient arrays satisfying $s_{ij}\ge\sqrt{\lambda_i\lambda_j}/(K+1)$ , $a_{ij}\ge\sqrt{\lambda_i\lambda_j}/(K-1)$ , and $Q(s,a)\succeq0$ . Then

\frac{K R_\lambda-1}{K^2-1}\le V_K(\lambda).

The probability hypothesis excludes $d=0$ ; no relation between $K$ and $d$ is assumed.

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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