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IV B: soundness of a candidate degrading-map extension certificate

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QAlgorithms.Papers.Independent.Intakethird100.P30.extensionCertificate

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

prove2quantum

Let P⪰0P\succeq0P⪰0 be an operator on finite nonempty systems A⊗BA\otimes BA⊗B, in a fixed product basis. Make exact copies E1=AE_1=AE1​=A, E2=BE_2=BE2​=B, and construct ∣ψ⟩=vec⁡P|\psi\rangle=\operatorname{vec}\sqrt P∣ψ⟩=vecP​, with the physical ABABAB registers first. Let N:E1→B′N:E_1\to B'N:E1​→B′ be any specified completely positive trace-preserving channel, where B′B'B′ is a copy of BBB, and put E=N∘Tr⁡E2E=N\circ\operatorname{Tr}_{E_2}E=N∘TrE2​​. Let W=Tr⁡B∣ψ⟩⟨ψ∣W=\operatorname{Tr}_B|\psi\rangle\langle\psi|W=TrB​∣ψ⟩⟨ψ∣. If the tested antidegrading equality holds,

(idA⊗E)(W)=P,(\mathrm{id}_A\otimes E)(W)=P,(idA​⊗E)(W)=P,

then Tr⁡E1E2∣ψ⟩⟨ψ∣=P\operatorname{Tr}_{E_1E_2}|\psi\rangle\langle\psi|=PTrE1​E2​​∣ψ⟩⟨ψ∣=P, and the constructed operator

R=(idAB⊗E)(∣ψ⟩⟨ψ∣)R=(\mathrm{id}_{AB}\otimes E)(|\psi\rangle\langle\psi|)R=(idAB​⊗E)(∣ψ⟩⟨ψ∣)

is positive and has both ABABAB and AB′AB'AB′ marginals equal to PPP. Hence PPP is Bob-extendable. The complementary map is not assumed trace preserving. This is a soundness certificate for a candidate NNN, including P=0P=0P=0, and does not specify the particular experimental map or assert nonextendability when a candidate fails.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P30
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P30

open scoped BigOperators Matrix ComplexConjugate ComplexOrder

attribute [local instance] Classical.propDecidable Classical.decEq

variable {I J K L A B : Type}

variable [Fintype I] [Fintype J] [Fintype K] [Fintype L]

variable [Fintype A] [Fintype B]

/-- PDF pp.8–9, IV B steps 1–3: soundness for any candidate channel N satisfying
the tested antidegrading equality. This does not identify the experimental special N. -/
theorem extensionCertificate [Nonempty A] [Nonempty B]
    (P : Matrix (A × B) (A × B) ℂ) (hP : P.PosSemidef)
    (N : Superop A B) (hN : IsChannel N)
    (htest : idTensorSuperop (candidateDegrader N) (complementaryOperator P) = P) :
    traceRight (pureDensity (vectorPurification P)) = P ∧
    IsBobCopyExtension P (candidateExtension P N) ∧ BobExtendible P := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P30

end
Source
Numerical evidence for bound secrecy from two-way post-processing in quantum key distribution, arXiv:1612.07734v2, PDF p. 9, IV B: soundness of a candidate degrading-map extension certificate
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What the Lean code literally says, in plain math · gpt-6.1-sol

For finite, nonempty index types A,BA,BA,B with decidable equality, let PPP be a positive semidefinite complex matrix indexed by A×BA\times BA×B. Let NNN be a function from complex matrices indexed by AAA to complex matrices indexed by BBB that is a channel in the following explicit sense: there exist a natural number rrr and matrices FjF_jFj​, with rows indexed by BBB and columns indexed by AAA, such that

∑j=0r−1Fj∗Fj=IA,N(X)=∑j=0r−1FjXFj∗for every matrix X.\sum_{j=0}^{r-1}F_j^{*}F_j=I_A, \qquad N(X)=\sum_{j=0}^{r-1}F_jXF_j^{*} \quad\text{for every matrix }X.j=0∑r−1​Fj∗​Fj​=IA​,N(X)=j=0∑r−1​Fj​XFj∗​for every matrix X.

Put S=PS=\sqrt PS=P​, obtained by the real continuous functional calculus, and define

v(x,y)=Sx,y,K(a,x),(a′,y)=∑c∈BS(a,c),xS(a′,c),y‾,v(x,y)=S_{x,y},\qquad K_{(a,x),(a',y)}=\sum_{c\in B}S_{(a,c),x}\overline{S_{(a',c),y}},v(x,y)=Sx,y​,K(a,x),(a′,y)​=c∈B∑​S(a,c),x​S(a′,c),y​​,

for x,y∈A×Bx,y\in A\times Bx,y∈A×B. Assume that, for every a,a′∈Aa,a'\in Aa,a′∈A and b,b′∈Bb,b'\in Bb,b′∈B,

[N(Ha,a′)]b,b′=P(a,b),(a′,b′),Hu,u′a,a′=∑c∈BK(a,(u,c)),(a′,(u′,c)).\bigl[N(H^{a,a'})\bigr]_{b,b'}=P_{(a,b),(a',b')}, \qquad H^{a,a'}_{u,u'}=\sum_{c\in B}K_{(a,(u,c)),(a',(u',c))}.[N(Ha,a′)]b,b′​=P(a,b),(a′,b′)​,Hu,u′a,a′​=c∈B∑​K(a,(u,c)),(a′,(u′,c))​.

Define EEE, indexed by (A×B)×B(A\times B)\times B(A×B)×B, by

E(x,b),(y,b′)=[N(Yx,y)]b,b′,Yu,u′x,y=∑c∈BSx,(u,c)Sy,(u′,c)‾.E_{(x,b),(y,b')}=\bigl[N(Y^{x,y})\bigr]_{b,b'}, \qquad Y^{x,y}_{u,u'}=\sum_{c\in B}S_{x,(u,c)}\overline{S_{y,(u',c)}}.E(x,b),(y,b′)​=[N(Yx,y)]b,b′​,Yu,u′x,y​=c∈B∑​Sx,(u,c)​Sy,(u′,c)​​.

Then the partial trace of vv∗vv^{*}vv∗ over its second A×BA\times BA×B factor equals PPP, explicitly

∑y∈A×Bv(x,y)v(x′,y)‾=Px,x′.\sum_{y\in A\times B}v(x,y)\overline{v(x',y)}=P_{x,x'}.y∈A×B∑​v(x,y)v(x′,y)​=Px,x′​.

Furthermore, EEE is positive semidefinite and satisfies both marginal identities

∑cE((a,b),c),((a′,b′),c)=∑cE((a,c),b),((a′,c),b′)=P(a,b),(a′,b′).\sum_cE_{((a,b),c),((a',b'),c)} = \sum_cE_{((a,c),b),((a',c),b')} =P_{(a,b),(a',b')}.c∑​E((a,b),c),((a′,b′),c)​=c∑​E((a,c),b),((a′,c),b′)​=P(a,b),(a′,b′)​.

Consequently, such a positive semidefinite Bob-copy extension of PPP exists. There is no trace-one hypothesis on PPP; in particular, P=0P=0P=0 is included.

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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