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Theorem 1: exact conversion from maximally entangled currency

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QAlgorithms.Papers.Independent.Intakethird100.P19.maximallyEntangled_conversion

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

prove2quantum

All Hilbert spaces are finite-dimensional complex spaces with fixed product bases and positive local dimensions. States are positive semidefinite operators of trace one; pure kets have norm one. A non-entangling map is completely positive and preserves the full separable positive cone. A channel is also trace preserving. Let k≥2k\ge2k≥2, let ϕk+\phi_k^+ϕk+​ be the normalized maximally entangled state on Ck⊗Ck\mathbb C^k\otimes\mathbb C^kCk⊗Ck, and let ψ\psiψ be any bipartite pure output state, on possibly different local dimensions. Write R(ρ)R(\rho)R(ρ) for the standard robustness, minimized over separable normalized noise. Then

∃ Λ non-entangling channel:Λ(ϕk+)=ψ⟺R(ψ)≤R(ϕk+).\exists\,\Lambda\text{ non-entangling channel}:\Lambda(\phi_k^+)=\psi\quad\Longleftrightarrow\quad R(\psi)\le R(\phi_k^+).∃Λ non-entangling channel:Λ(ϕk+​)=ψ⟺R(ψ)≤R(ϕk+​).

This characterizes exact preparation from maximally entangled currency.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P19
Formal statement
noncomputable section

open QAlgorithms QAlgorithms.Entanglement Filter

open scoped Matrix ComplexOrder ComplexConjugate ENNReal

attribute [local instance] Classical.propDecidable Classical.decEq

namespace QAlgorithms.Papers.Independent.Intakethird100.P19

open QAlgorithms.Papers.Independent.Information100.P21 (Op FiniteLOCC)

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

open QAlgorithms.Papers.Independent.Information100.P69 (CP)

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

open QAlgorithms.Papers.Independent.Dependency50.P19 (epr)

/-- Theorem 1, PDF p.8: exact conversion from maximally entangled currency
is characterized by standard robustness, with arbitrary positive output dimensions. -/
theorem maximallyEntangled_conversion {a b k : ℕ} (ha : 0 < a) (hb : 0 < b)
    (hk : 2 ≤ k) (psi : UnitBipartiteKet a b) :
    (∃ L : Superop (Fin k × Fin k) (Fin a × Fin b),
      IsChannel L ∧ NE L ∧ L (epr k) = psi.rho) ↔
      standardStateRobustness psi.rho ≤ standardStateRobustness (epr k) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P19

end
Source
Entanglement manipulation and distillability beyond LOCC, arXiv:1711.03835v2, PDF p. 8, Theorem 1
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What the Lean code literally says, in plain math · gpt-6.1-sol

For all natural numbers a,b,ka,b,ka,b,k with a>0a>0a>0, b>0b>0b>0, and k≥2k\ge2k≥2, and every complex Euclidean vector ψ\psiψ on {0,…,a−1}×{0,…,b−1}\{0,\ldots,a-1\}\times\{0,\ldots,b-1\}{0,…,a−1}×{0,…,b−1} satisfying ∥ψ∥=1\|\psi\|=1∥ψ∥=1, put ρψ=ψψ†\rho_\psi=\psi\psi^\daggerρψ​=ψψ†. Call a matrix separable positive if it is a finite sum ∑iwiAi⊗Bi\sum_iw_iA_i\otimes B_i∑i​wi​Ai​⊗Bi​, where wi≥0w_i\ge0wi​≥0 and each Ai,BiA_i,B_iAi​,Bi​ is positive semidefinite with trace 111; the weights need not sum to one, and the empty sum is allowed. Call it a separable state if it is also positive semidefinite with trace 111, and define

Rs(ρ)=inf⁡{s∈R:s≥0, ∃ separable state σ, ρ+sσ is separable positive}.R_s(\rho)=\inf\{s\in\mathbb R:s\ge0,\ \exists\text{ separable state }\sigma,\ \rho+s\sigma\text{ is separable positive}\}.Rs​(ρ)=inf{s∈R:s≥0, ∃ separable state σ, ρ+sσ is separable positive}.

This is a real infimum, with the total real-infimum convention giving 000 for an empty set. Let (Ωk)(i,j),(r,s)=1/k(\Omega_k)_{(i,j),(r,s)}=1/k(Ωk​)(i,j),(r,s)​=1/k when i=ji=ji=j and r=sr=sr=s, and 000 otherwise, for indices less than kkk. Then there exists a map LLL from matrices on the k×kk\times kk×k bipartite system to matrices on the a×ba\times ba×b system that is a channel, maps every separable positive matrix to a separable positive matrix, and satisfies L(Ωk)=ρψL(\Omega_k)=\rho_\psiL(Ωk​)=ρψ​, if and only if Rs(ρψ)≤Rs(Ωk)R_s(\rho_\psi)\le R_s(\Omega_k)Rs​(ρψ​)≤Rs​(Ωk​). The channel requirement means a finite representation L(X)=∑iKiXKi†L(X)=\sum_iK_iXK_i^\daggerL(X)=∑i​Ki​XKi†​ with ∑iKi†Ki=I\sum_iK_i^\dagger K_i=I∑i​Ki†​Ki​=I; the additional condition also explicitly requires a finite Kraus representation without a completeness requirement.

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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