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Fast polarization toward poor commuting-output channels

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QAlgorithms.Papers.Independent.Intakethird100.P10.fastPolarization_poor

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

prove2quantum

Let W:x↦ρxW:x\mapsto\rho_xW:x↦ρx​ be a binary cq channel on a finite nonempty register, and suppose ρ0ρ1=ρ1ρ0\rho_0\rho_1=\rho_1\rho_0ρ0​ρ1​=ρ1​ρ0​. For N=2nN=2^nN=2n, synthesize channels with the prescribed polar transform and let fi=F(WN(i))f_i=\sqrt{F(W_N^{(i)})}fi​=F(WN(i)​)​. For every real β<1/2\beta<1/2β<1/2,

lim⁡n→∞12n∣{i:fi>1−2−(2n)β}∣=1−I(W).\lim_{n\to\infty}\frac{1}{2^n}\left|\left\{i:f_i>1-2^{-(2^n)^\beta}\right\}\right|=1-I(W).n→∞lim​2n1​​{i:fi​>1−2−(2n)β}​=1−I(W).

The strict inequality uses root fidelity. Uniform counting is the marginal probability of the fair synthesis process. This result supplies poor-channel polarization under commuting outputs and requires no degradation assumption.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P10
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P10

open scoped BigOperators Matrix ComplexConjugate

open Filter

-- Accepted definitions, without paper-local aliases.

open QAlgorithms.Papers.Independent.Information100.P58 (basisDensity)

open QAlgorithms.Papers.Independent.Information100.P67 (squaredFidelity)

open QAlgorithms.Papers.Independent.Claimed50.P48 (indexedKrausDilation)

/-- PDF pp.4–5, Proposition 4: strict poor threshold for ROOT fidelity; degradation
is not a hypothesis. Counting leaves is the independent fair birth-process marginal. -/
theorem fastPolarization_poor {H : Type*} [Fintype H] [DecidableEq H] [Nonempty H]
    (W : BinaryCQ H) (hW : IsBinaryCQ W) (hc : Commute (W false) (W true))
    (beta : ℝ) (hbeta : beta < 1 / 2) :
    Tendsto (fun n => leafProbability (poorSet W n beta)) atTop
      (nhds (1 - symmetricInformation W)) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P10

end
Source
Polar codes for degradable quantum channels, arXiv:1109.5346v3, PDF p. 4, Proposition 4: fast polarization toward poor commuting channels
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let HHH be a finite, nonempty set with decidable equality, and let W0,W1W_0,W_1W0​,W1​ be positive semidefinite complex matrices indexed by HHH, each having trace 111, such that W0W1=W1W0W_0W_1=W_1W_0W0​W1​=W1​W0​. Let β∈R\beta\in\mathbb Rβ∈R satisfy β<12\beta<\tfrac12β<21​; no positivity assumption on β\betaβ is imposed. Write S(X)=−Re⁡tr⁡(f(X))S(X)=-\operatorname{Re}\operatorname{tr}(f(X))S(X)=−Retr(f(X)), where f(t)=tlog⁡2tf(t)=t\log_2tf(t)=tlog2​t is applied by continuous functional calculus, and put I(W)=S((W0+W1)/2)−(S(W0)+S(W1))/2I(W)=S((W_0+W_1)/2)-(S(W_0)+S(W_1))/2I(W)=S((W0​+W1​)/2)−(S(W0​)+S(W1​))/2. The matrices used to define the asserted limit are as follows. For n∈Nn\in\mathbb Nn∈N, put N=2nN=2^nN=2n, and let dt(j)d_t(j)dt​(j) denote bit ttt, counted from the most significant end, of the nnn-bit representation of 0≤j<N0\le j<N0≤j<N. Set cjk=1c_{jk}=1cjk​=1 exactly when, for every 0≤t<n0\le t<n0≤t<n, dt(k)=1d_t(k)=1dt​(k)=1 implies dn−1−t(j)=1d_{n-1-t}(j)=1dn−1−t​(j)=1, and otherwise set cjk=0c_{jk}=0cjk​=0. For a Boolean word uuu of length NNN, define vk(u)=∑j=0N−1ujcjk(mod2)v_k(u)=\sum_{j=0}^{N-1}u_jc_{jk}\pmod2vk​(u)=∑j=0N−1​uj​cjk​(mod2), interpreting Boolean values as 0,10,10,1. For 0≤i<N0\le i<N0≤i<N and x∈{0,1}x\in\{0,1\}x∈{0,1}, define a matrix Wn,i(x)W_{n,i}^{(x)}Wn,i(x)​, indexed by pairs (h,a)(h,a)(h,a) with h∈{0,1}ih\in\{0,1\}^ih∈{0,1}i and a∈HNa\in H^Na∈HN, by

Wn,i(x)((h,a),(h′,b))={2−(N−1)∑u∈{0,1}Nui=x,u<i=h∏k=0N−1Wvk(u)(ak,bk),h=h′,0,h≠h′.W_{n,i}^{(x)}((h,a),(h',b)) = \begin{cases} 2^{-(N-1)} \displaystyle\sum_{\substack{u\in\{0,1\}^N\\u_i=x,\;u_{<i}=h}} \prod_{k=0}^{N-1}W_{v_k(u)}(a_k,b_k),&h=h',\\ 0,&h\ne h'. \end{cases}Wn,i(x)​((h,a),(h′,b))=⎩⎨⎧​2−(N−1)u∈{0,1}Nui​=x,u<i​=h​∑​k=0∏N−1​Wvk​(u)​(ak​,bk​),0,​h=h′,h=h′.​

For matrices of the same size, put Fr(X,Y)=Re⁡tr⁡X YXF_{\mathrm r}(X,Y)=\operatorname{Re}\operatorname{tr}\sqrt{\sqrt X\,Y\sqrt X}Fr​(X,Y)=RetrX​YX​​, with square roots taken by continuous functional calculus. Then

lim⁡n→∞#{ 0≤i<2n:1−2−(2n)β<Fr(Wn,i(0),Wn,i(1)) }2n=1−I(W).\lim_{n\to\infty} \frac{\#\{\,0\le i<2^n: 1-2^{-(2^n)^\beta}<F_{\mathrm r}(W_{n,i}^{(0)},W_{n,i}^{(1)})\,\}}{2^n} =1-I(W).n→∞lim​2n#{0≤i<2n:1−2−(2n)β<Fr​(Wn,i(0)​,Wn,i(1)​)}​=1−I(W).

The defining sequence includes n=0n=0n=0: then N=1N=1N=1, the bit condition is vacuous, the transformed word is unchanged, and there is one synthesized index.

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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