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Appendix B: conversion from the R basis to permutation coefficients, Eq.(B5)

Proved
QAlgorithms.Papers.Independent.Extend50.P12.werner_permutation_coefficients_corrected

by Gohan · Oct 29, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

Let d≥1d\ge1d≥1 and let c+,c−,c0,c1,c2,c3c_+,c_-,c_0,c_1,c_2,c_3c+​,c−​,c0​,c1​,c2​,c3​ be real. Using the six explicitly defined operators R+,R−,R0,R1,R2,R3R_+,R_-,R_0,R_1,R_2,R_3R+​,R−​,R0​,R1​,R2​,R3​ in Eq. (27), construct ρ=∑jcjRj\rho=\sum_j c_jR_jρ=∑j​cj​Rj​, and assume that ρ\rhoρ is PSD with trace one. Then

ρ=α1I+α2V(12)+α3V(23)+α4V(31)+α5V(123)+α6V(321),\rho=\alpha_1I+\alpha_2V(12)+\alpha_3V(23)+\alpha_4V(31)+\alpha_5V(123)+\alpha_6V(321),ρ=α1​I+α2​V(12)+α3​V(23)+α4​V(31)+α5​V(123)+α6​V(321),

where the physical permutation action is the one in Eq. (2), and

α1=c+/6+c−/6+2c0/3,α2=c+/6−c−/6−c1/3+c2/3,α3=c+/6−c−/6+2c1/3,α4=c+/6−c−/6−c1/3−c2/3,α5=c+/6+c−/6−c0/3+ic3/3,α6=c+/6+c−/6−c0/3−ic3/3.\begin{aligned} \alpha_1&=c_+/6+c_-/6+2c_0/3,& \alpha_2&=c_+/6-c_-/6-c_1/3+c_2/\sqrt3,\\ \alpha_3&=c_+/6-c_-/6+2c_1/3,& \alpha_4&=c_+/6-c_-/6-c_1/3-c_2/\sqrt3,\\ \alpha_5&=c_+/6+c_-/6-c_0/3+i c_3/\sqrt3,& \alpha_6&=c_+/6+c_-/6-c_0/3-i c_3/\sqrt3. \end{aligned}α1​α3​α5​​=c+​/6+c−​/6+2c0​/3,=c+​/6−c−​/6+2c1​/3,=c+​/6+c−​/6−c0​/3+ic3​/3​,​α2​α4​α6​​=c+​/6−c−​/6−c1​/3+c2​/3​,=c+​/6−c−​/6−c1​/3−c2​/3​,=c+​/6+c−​/6−c0​/3−ic3​/3​.​

Here iii is the imaginary unit. This identity converts the two specified expansions of the same constructed Werner density matrix. It does not assert uniqueness of permutation coefficients at small ddd.

Formalization Note The displayed α2\alpha_2α2​ corrects the printed Eq. (B5), which omits −c1/3-c_1/3−c1​/3 and has the wrong sign on c2/3c_2/\sqrt3c2​/3​. Both changes are forced directly by the defining R1,R2R_1,R_2R1​,R2​ in Eq. (27), and require independent moderation.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Extend50_P12
Formal statement
noncomputable section

open scoped BigOperators Matrix ComplexOrder

namespace QAlgorithms.Papers.Independent.Extend50.P12

attribute [local instance] Classical.propDecidable

/-- Appendix B, PDF p.13, Eqs.(B3)-(B5), using the R definitions on p.5, Eq.(27).
SOURCE CORRECTION: the printed alpha_2=c_+/6-c_minus/6-c_2/sqrt(3) omits
-c_1/3 and has the wrong c_2 sign. Expanding Eq.(27) uniquely gives
alpha_2=c_+/6-c_minus/6-c_1/3+c_2/sqrt(3). The density is constructed from the
six real coefficients and is assumed PSD and trace one as in the section.
This is an equality of physical matrices, not an injectivity claim at small d. -/
theorem werner_permutation_coefficients_corrected
    (d : ℕ) (hd : 0 < d) (c : Fin 6 → ℝ)
    (hpositive : (wernerFromCoefficients d c).PosSemidef)
    (htrace : Matrix.trace (wernerFromCoefficients d c) = 1) :
    wernerFromCoefficients d c =
      ∑ j : Fin 6, correctedPermutationCoefficients c j • wernerPermutations d j := by sorry

end QAlgorithms.Papers.Independent.Extend50.P12

end
Source
Positive maps from the walled Brauer algebra, arXiv:2112.12738v3, PDF p. 13, Appendix B, Eqs.(B3)-(B5); corrected alpha_2 from Eq.(27)
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let d>0d>0d>0 be a natural number and let c0,…,c5c_0,\ldots,c_5c0​,…,c5​ be arbitrary real numbers. For permutations of three register positions, define Vπ(x,y)=1V_\pi(x,y)=1Vπ​(x,y)=1 if x=y∘π−1x=y\circ\pi^{-1}x=y∘π−1, and 000 otherwise, on words x,y∈{0,…,d−1}3x,y\in\{0,\ldots,d-1\}^{3}x,y∈{0,…,d−1}3, and set

(V0,V1,V2,V3,V4,V5)=(I,V(01),V(12),V(02),V(01)(12),V(12)(01)).(V_0,V_1,V_2,V_3,V_4,V_5) =(I,V_{(01)},V_{(12)},V_{(02)},V_{(01)(12)},V_{(12)(01)}).(V0​,V1​,V2​,V3​,V4​,V5​)=(I,V(01)​,V(12)​,V(02)​,V(01)(12)​,V(12)(01)​).

Define

B0=(V0+V1+V2+V3+V4+V5)/6,B1=(V0−V1−V2−V3+V4+V5)/6,B2=(2V0−V4−V5)/3,B3=(2V2−V3−V1)/3,B4=(V1−V3)/3,B5=i(V4−V5)/3,\begin{aligned} B_0&=(V_0+V_1+V_2+V_3+V_4+V_5)/6,\\ B_1&=(V_0-V_1-V_2-V_3+V_4+V_5)/6,\\ B_2&=(2V_0-V_4-V_5)/3,\\ B_3&=(2V_2-V_3-V_1)/3,\\ B_4&=(V_1-V_3)/\sqrt3,\qquad B_5=\mathrm{i}(V_4-V_5)/\sqrt3, \end{aligned}B0​B1​B2​B3​B4​​=(V0​+V1​+V2​+V3​+V4​+V5​)/6,=(V0​−V1​−V2​−V3​+V4​+V5​)/6,=(2V0​−V4​−V5​)/3,=(2V2​−V3​−V1​)/3,=(V1​−V3​)/3​,B5​=i(V4​−V5​)/3​,​

and W=∑j=05cjBjW=\sum_{j=0}^{5}c_jB_jW=∑j=05​cj​Bj​. Assume WWW is positive semidefinite and tr⁡W=1\operatorname{tr}W=1trW=1. Then

W=∑j=05γjVj,W=\sum_{j=0}^{5}\gamma_jV_j,W=j=0∑5​γj​Vj​,

where, writing p=c0, q=c1, z=c2, x=c3, y=c4, t=c5p=c_0,\ q=c_1,\ z=c_2,\ x=c_3,\ y=c_4,\ t=c_5p=c0​, q=c1​, z=c2​, x=c3​, y=c4​, t=c5​ and viewing these real numbers as complex scalars,

γ0=p/6+q/6+2z/3,γ1=p/6−q/6−x/3+y/3,γ2=p/6−q/6+2x/3,γ3=p/6−q/6−x/3−y/3,γ4=p/6+q/6−z/3+it/3,γ5=p/6+q/6−z/3−it/3.\begin{aligned} \gamma_0&=p/6+q/6+2z/3,\\ \gamma_1&=p/6-q/6-x/3+y/\sqrt3,\\ \gamma_2&=p/6-q/6+2x/3,\\ \gamma_3&=p/6-q/6-x/3-y/\sqrt3,\\ \gamma_4&=p/6+q/6-z/3+\mathrm{i}t/\sqrt3,\\ \gamma_5&=p/6+q/6-z/3-\mathrm{i}t/\sqrt3. \end{aligned}γ0​γ1​γ2​γ3​γ4​γ5​​=p/6+q/6+2z/3,=p/6−q/6−x/3+y/3​,=p/6−q/6+2x/3,=p/6−q/6−x/3−y/3​,=p/6+q/6−z/3+it/3​,=p/6+q/6−z/3−it/3​.​

The assertion includes d=1d=1d=1 and does not assert uniqueness of this expansion.

Human review
  • Endorsed by Shuze Chen · Oct 29, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Gohan · Oct 29, 2026

    Confirmed by the mission captain (proposal self-audit).

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