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Exact entanglement-assisted isotropic-qubit curve

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QAlgorithms.Papers.Independent.Intakethird100.P27.isotropic_entanglement_assisted

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

prove2quantum

Consider the memoryless maximally mixed qubit source I/2I/2I/2, with the Bell purification ΦRA\Phi_{RA}ΦRA​ and fidelity-distortion observable I−ΦRAI-\Phi_{RA}I−ΦRA​. Let Reaq(D)R^q_{\mathrm{ea}}(D)Reaq​(D) be its operational quantum rate-distortion function with unlimited prior entanglement and no free classical side channel. For 0≤D≤10\le D\le10≤D≤1,

Reaq(D)={1−12H(1−D,D/3,D/3,D/3),D≤3/4,0,D>3/4.R^q_{\mathrm{ea}}(D)=\begin{cases}1-\tfrac12 H(1-D,D/3,D/3,D/3),&D\le3/4,\\0,&D>3/4.\end{cases}Reaq​(D)={1−21​H(1−D,D/3,D/3,D/3),0,​D≤3/4,D>3/4.​

Here H(p)=−∑ipilog⁡2piH(p)=-\sum_i p_i\log_2p_iH(p)=−∑i​pi​log2​pi​, with 0log⁡20=00\log_2 0=00log2​0=0. The branches agree at D=3/4D=3/4D=3/4.

This gives an exact assisted curve, with rate one at zero distortion and zero from distortion 3/43/43/4 onward.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P27
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P27

open Filter

open scoped Matrix ComplexConjugate ComplexOrder ENNReal

attribute [local instance] Classical.propDecidable Classical.decEq

open QAlgorithms.Papers.Independent.Dependency50.P19
  (first marginalRB mutualInformation bipartiteBlock epr)

open QAlgorithms.Papers.Independent.Information100.P08 (formation)

open QAlgorithms.Papers.Independent.Gap48.P20 (redistributionMarginal)

open QAlgorithms.Papers.Independent.Bridge50.P37 (isotropicState normalizedEntangledProjection)

/-- PDF pp.8--9, Theorem 11: exact isotropic-qubit assisted curve. The Bell
purification and fidelity observable are pinned by equalities, not postulated
entropy values. The two branches agree at D=3/4. The p.9 proof's 'fidelity p'
is a typo; for its Pauli channel fidelity is 1-p and distortion is p. -/
theorem isotropic_entanglement_assisted
    (S : DistortionSource (Fin 2) (Fin 2))
    (hsource : pureDensity S.psi = epr 2)
    (hdelta : S.delta = 1 - epr 2)
    (D : ℝ) (hD0 : 0 ≤ D) (hD1 : D ≤ 1) :
    rateDistortion S .entanglement D =
      ENNReal.ofReal (if D ≤ 3 / 4 then 1 - isotropicEntropy D / 2 else 0) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P27

end
Source
Quantum rate distortion coding with auxiliary resources, arXiv:1212.5316v3, PDF p. 8, Theorem 11
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let SSS be a source on A=B=Fin⁡(2)A=B=\operatorname{Fin}(2)A=B=Fin(2), consisting of a unit vector ψ∈CA×A\psi\in\mathbb C^{A\times A}ψ∈CA×A and a positive semidefinite distortion matrix Δ\DeltaΔ. Assume

ψψ†=P,Δ=I−P,P((i,j),(i′,j′))={12,i=j and i′=j′,0,otherwise.\psi\psi^\dagger=P,\qquad \Delta=I-P, \qquad P((i,j),(i',j'))= \begin{cases} \frac12,&i=j\ \text{and }i'=j',\\ 0,&\text{otherwise}. \end{cases}ψψ†=P,Δ=I−P,P((i,j),(i′,j′))={21​,0,​i=j and i′=j′,otherwise.​

For every real DDD with 0≤D≤10\le D\le10≤D≤1, the rate defined below equals

max⁡ ⁣({1−12H4(D),D≤34,0,D>34,0),H4(D)=−(1−D)log⁡2(1−D)−3D3log⁡2D3,\max\!\left( \begin{cases} 1-\frac12H_4(D),&D\le\frac34,\\ 0,&D>\frac34, \end{cases} 0\right), \qquad H_4(D)=-(1-D)\log_2(1-D)-3\frac D3\log_2\frac D3,max({1−21​H4​(D),0,​D≤43​,D>43​,​0),H4​(D)=−(1−D)log2​(1−D)−33D​log2​3D​,

viewed in [0,∞][0,\infty][0,∞], with zero logarithmic terms taken as zero. The rate is the infimum in [0,∞][0,\infty][0,∞] of all nonnegative real rrr such that, for every ε,s>0\varepsilon,s>0ε,s>0, every sufficiently large positive nnn admits positive integers q,hq,hq,h, a positive semidefinite trace-one shared matrix τ\tauτ on Fin⁡(h)×Fin⁡(h)\operatorname{Fin}(h)\times\operatorname{Fin}(h)Fin(h)×Fin(h), an encoder from An×Fin⁡(h)A^n\times\operatorname{Fin}(h)An×Fin(h) to Fin⁡(q)\operatorname{Fin}(q)Fin(q), and a decoder from Fin⁡(q)×Fin⁡(h)\operatorname{Fin}(q)\times\operatorname{Fin}(h)Fin(q)×Fin(h) to BnB^nBn. Both maps must have finite Kraus representations with Kraus adjoint-products summing to the identity; the classical register has dimension 111. Apply these maps to the second halves of nnn copies of ψ\psiψ and the shared matrix, leaving the reference halves unchanged, to obtain XXX on An×BnA^n\times B^nAn×Bn. The bounds required are

log⁡2q≤n(r+s),1n∑i<nRe⁡Tr⁡((I−P)Xi)≤D+ε,\log_2q\le n(r+s), \qquad \frac1n\sum_{i<n}\operatorname{Re}\operatorname{Tr}((I-P)X_i)\le D+\varepsilon,log2​q≤n(r+s),n1​i<n∑​ReTr((I−P)Xi​)≤D+ε,

where XiX_iXi​ sums over equal row and column indices outside position iii. The shared dimension is unrestricted. The stated formula includes D=0D=0D=0, D=3/4D=3/4D=3/4, and D=1D=1D=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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