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Complete attainable Pauli-damping region, including endpoints

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QAlgorithms.Papers.Independent.Intakethird100.P13.attainable_complete

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

prove2quantum

Let AγA_\gammaAγ​ denote amplitude damping, χγ=(id⊗Aγ)(∣Φ+⟩⟨Φ+∣)\chi_\gamma=(\mathrm{id}\otimes A_\gamma)(|\Phi^+\rangle\langle\Phi^+|)χγ​=(id⊗Aγ​)(∣Φ+⟩⟨Φ+∣) its normalized Choi state, and PqP_qPq​ the Pauli map with Bloch action (x,y,z)↦(q1x,−q2y,q3z)(x,y,z)\mapsto(q_1x,-q_2y,q_3z)(x,y,z)↦(q1​x,−q2​y,q3​z). For 0<γ<10<\gamma<10<γ<1, a qubit quantum channel EEE is simulable by noisy Bell/Pauli teleportation over χγ\chi_\gammaχγ​ if and only if

E=Xu∘Aη∘Pq,u∈{0,1},0≤η≤γ,q∈conv⁡(Vγ,η),E=X^u\circ A_\eta\circ P_q,\qquad u\in\{0,1\},\quad0\le\eta\le\gamma,\quad q\in\operatorname{conv}(V_{\gamma,\eta}),E=Xu∘Aη​∘Pq​,u∈{0,1},0≤η≤γ,q∈conv(Vγ,η​),

where XXX denotes conjugation by the Pauli XXX, a=(1−γ)/(1−η)a=\sqrt{(1-\gamma)/(1-\eta)}a=(1−γ)/(1−η)​, b=1−η/γb=1-\eta/\gammab=1−η/γ, and the eight vertices are

Vγ,η={(a,±ab,∓a2b),(±ab,a,∓a2b),(−a,±ab,±a2b),(±ab,−a,±a2b)}.V_{\gamma,\eta}=\{(a,\pm ab,\mp a^2b),(\pm ab,a,\mp a^2b),(-a,\pm ab,\pm a^2b),(\pm ab,-a,\pm a^2b)\}.Vγ,η​={(a,±ab,∓a2b),(±ab,a,∓a2b),(−a,±ab,±a2b),(±ab,−a,±a2b)}.

Signs within each triple are paired. At γ=0\gamma=0γ=0 the simulable channels are exactly all Pauli channels. At γ=1\gamma=1γ=1 they are exactly the input-independent diagonal states with Bloch vector (0,0,z)(0,0,z)(0,0,z), −1≤z≤1-1\le z\le1−1≤z≤1. This characterizes the whole attainable set without asserting uniqueness of parameters.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P13
Formal statement
noncomputable section

open scoped BigOperators Matrix ComplexConjugate ComplexOrder

namespace QAlgorithms.Papers.Independent.Intakethird100.P13

attribute [local instance] Classical.propDecidable Classical.decEq

/-- arXiv:1706.05384v2 PDF pp.7--8, Theorem 16, converse paragraph and vertex proof.
Corrected endpoint clauses use the physical simulator rather than undefined 0/0 vertices.
No decomposition uniqueness is asserted. -/
theorem attainable_complete :
    (∀ gamma : ℝ, 0 < gamma → gamma < 1 → ∀ E : QubitMap, IsChannel E →
      (Simulable (normalizedQubitChoi (blochAmplitudeDamping gamma)) E ↔
        ∃ u : Bool, ∃ eta : ℝ, ∃ q : BlochCoordinates,
          0 ≤ eta ∧ eta ≤ gamma ∧ q ∈ attainableRegion gamma eta ∧
          E = pauliDamping u eta q)) ∧
    (∀ E : QubitMap, IsChannel E →
      (Simulable (normalizedQubitChoi (blochAmplitudeDamping 0)) E ↔ IsPauli E)) ∧
    (∀ E : QubitMap, IsChannel E →
      (Simulable (normalizedQubitChoi (blochAmplitudeDamping 1)) E ↔
        ∃ z : ℝ, -1 ≤ z ∧ z ≤ 1 ∧ E = axialReplacement z)) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P13

end
Source
Simulation of non-Pauli Channels, arXiv:1706.05384v2, PDF p. 7, Theorem 16, Eqs.(59)–(67)
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What the Lean code literally says, in plain math · gpt-6.1-sol

Write σ0=I2,σ1=X,σ2=Y,σ3=Z\sigma_0=I_2,\sigma_1=X,\sigma_2=Y,\sigma_3=Zσ0​=I2​,σ1​=X,σ2​=Y,σ3​=Z; every complex qubit matrix has a unique expression R=(tI2+r0X+r1Y+r2Z)/2R=(tI_2+r_0X+r_1Y+r_2Z)/2R=(tI2​+r0​X+r1​Y+r2​Z)/2, with complex coefficients. Define

Dγ(R)=12[tI2+1−γ r0X+1−γ r1Y+((1−γ)r2+γt)Z],τγ=(id⊗Dγ)(∣Ω⟩⟨Ω∣),∣Ω⟩=(∣00⟩+∣11⟩)/2.D_\gamma(R)=\tfrac12[tI_2+\sqrt{1-\gamma}\,r_0X+\sqrt{1-\gamma}\,r_1Y+((1-\gamma)r_2+\gamma t)Z], \quad \tau_\gamma=(\mathrm{id}\otimes D_\gamma)(|\Omega\rangle\langle\Omega|), \quad |\Omega\rangle=(|00\rangle+|11\rangle)/\sqrt2.Dγ​(R)=21​[tI2​+1−γ​r0​X+1−γ​r1​Y+((1−γ)r2​+γt)Z],τγ​=(id⊗Dγ​)(∣Ω⟩⟨Ω∣),∣Ω⟩=(∣00⟩+∣11⟩)/2​.

Here an extension id⊗E\mathrm{id}\otimes Eid⊗E applies EEE to each block indexed by the first qubit. For a two-qubit matrix τ\tauτ, put βk(c,a)=2−1/2(σk)ac\beta_k(c,a)=2^{-1/2}(\sigma_k)_{ac}βk​(c,a)=2−1/2(σk​)ac​ and define Tτ,pT_{\tau,p}Tτ,p​ by

[Kτ,k(R)]bb′=∑c,a,d,a′∈{0,1}βk(c,a)‾Rcdτ(a,b),(a′,b′)βk(d,a′),Tτ,p(R)=∑k,ℓ=03pℓkσℓKτ,k(R)σℓ†.[K_{\tau,k}(R)]_{bb'}=\sum_{c,a,d,a'\in\{0,1\}} \overline{\beta_k(c,a)}R_{cd}\tau_{(a,b),(a',b')}\beta_k(d,a'), \qquad T_{\tau,p}(R)=\sum_{k,\ell=0}^3p_{\ell k}\sigma_\ell K_{\tau,k}(R)\sigma_\ell^\dagger.[Kτ,k​(R)]bb′​=c,a,d,a′∈{0,1}∑​βk​(c,a)​Rcd​τ(a,b),(a′,b′)​βk​(d,a′),Tτ,p​(R)=k,ℓ=0∑3​pℓk​σℓ​Kτ,k​(R)σℓ†​.

To be simulable from τ\tauτ means that E=Tτ,pE=T_{\tau,p}E=Tτ,p​ for some real array with pℓk≥0p_{\ell k}\ge0pℓk​≥0 and ∑ℓpℓk=1\sum_\ell p_{\ell k}=1∑ℓ​pℓk​=1 for every kkk. A channel here means a map having a finite representation E(R)=∑hAhRAh†E(R)=\sum_hA_hRA_h^\daggerE(R)=∑h​Ah​RAh†​, with ∑hAh†Ah=I2\sum_hA_h^\dagger A_h=I_2∑h​Ah†​Ah​=I2​, on all complex qubit matrices. The theorem asserts the conjunction of three statements. First, for every 0<γ<10<\gamma<10<γ<1 and every channel EEE, simulability from τγ\tau_\gammaτγ​ is equivalent to the existence of u∈{false,true}u\in\{\mathrm{false},\mathrm{true}\}u∈{false,true}, η∈R\eta\in\mathbb Rη∈R, and q∈R3q\in\mathbb R^3q∈R3 such that 0≤η≤γ0\le\eta\le\gamma0≤η≤γ, qqq belongs to the convex hull of

(a,ab,−a2b), (a,−ab,a2b), (ab,a,−a2b), (−ab,a,a2b), (−a,ab,a2b), (−a,−ab,−a2b), (ab,−a,a2b), (−ab,−a,−a2b),a=1−γ1−η,b=1−ηγ,(a,ab,-a^2b),\ (a,-ab,a^2b),\ (ab,a,-a^2b),\ (-ab,a,a^2b),\ (-a,ab,a^2b),\ (-a,-ab,-a^2b),\ (ab,-a,a^2b),\ (-ab,-a,-a^2b), \quad a=\sqrt{\frac{1-\gamma}{1-\eta}},\quad b=1-\frac{\eta}{\gamma},(a,ab,−a2b), (a,−ab,a2b), (ab,a,−a2b), (−ab,a,a2b), (−a,ab,a2b), (−a,−ab,−a2b), (ab,−a,a2b), (−ab,−a,−a2b),a=1−η1−γ​​,b=1−γη​,

and E(R)=XuDη(Pq(R))XuE(R)=X^uD_\eta(P_q(R))X^uE(R)=XuDη​(Pq​(R))Xu for every RRR, where Xfalse=I2X^{\mathrm{false}}=I_2Xfalse=I2​, Xtrue=XX^{\mathrm{true}}=XXtrue=X, and Pq(R)=(tI2+q0r0X−q1r1Y+q2r2Z)/2P_q(R)=(tI_2+q_0r_0X-q_1r_1Y+q_2r_2Z)/2Pq​(R)=(tI2​+q0​r0​X−q1​r1​Y+q2​r2​Z)/2. Second, for every channel EEE, simulability from τ0\tau_0τ0​ is equivalent to E(R)=∑hwhσhRσh†E(R)=\sum_hw_h\sigma_hR\sigma_h^\daggerE(R)=∑h​wh​σh​Rσh†​ for some real wh≥0w_h\ge0wh​≥0 summing to 111. Third, for every channel EEE, simulability from τ1\tau_1τ1​ is equivalent to the existence of a real z∈[−1,1]z\in[-1,1]z∈[−1,1] such that E(R)=tr⁡(R)(I2+zZ)/2E(R)=\operatorname{tr}(R)(I_2+zZ)/2E(R)=tr(R)(I2​+zZ)/2 for every RRR.

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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