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Synthesized complementary information uncertainty

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QAlgorithms.Papers.Independent.Intakethird100.P23.synthesized_information_uncertainty

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

prove2quantum

Let U:C2→Cb⊗CeU:\mathbb C^2\to\mathbb C^b\otimes\mathbb C^eU:C2→Cb⊗Ce be an isometry, so U†U=IU^\dagger U=IU†U=I, defining a memoryless channel to Bob and its fixed complementary channel to Eve. All binary inputs are uniform and entropies are in bits. At recursion depth n≥0n\ge0n≥0, let N=2nN=2^nN=2n and apply the coherent Arikan encoder GN=BNF⊗nG_N=B_NF^{\otimes n}GN​=BN​F⊗n, where F=(1011)F=\left(\begin{smallmatrix}1&0\\1&1\end{smallmatrix}\right)F=(11​01​). Amplitude synthesis retains previous bits and averages future bits; phase synthesis retains future bits and averages previous bits, with Bob's coherent copy register CNC^NCN retained. Write r(W)=∥ρ0ρ1∥1=F(W)r(W)=\|\sqrt{\rho_0}\sqrt{\rho_1}\|_1=\sqrt{F(W)}r(W)=∥ρ0​​ρ1​​∥1​=F(W)​ for the root fidelity of a binary channel WWW.

For every recursion depth n≥0n\ge0n≥0 and position i∈{1,…,N}i\in\{1,\ldots,N\}i∈{1,…,N}, let WP,N(i)W_{P,N}^{(i)}WP,N(i)​ be Bob's synthesized phase channel and WE,N(i)W_{E,N}^{(i)}WE,N(i)​ Eve's synthesized amplitude channel, both constructed from this same UUU and encoder. Then

I(WP,N(i))+I(WE,N(i))≤1.I(W_{P,N}^{(i)})+I(W_{E,N}^{(i)})\le1.I(WP,N(i)​)+I(WE,N(i)​)≤1.

Both informations are symmetric Holevo informations. This complementarity bounds simultaneous phase information at Bob and amplitude information at Eve.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P23
Formal statement
namespace QAlgorithms.Papers.Independent.Intakethird100.P23

open scoped BigOperators ComplexConjugate Matrix

open Filter

open QAlgorithms (Qubits)

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

attribute [local instance] Classical.propDecidable

/-- Synthesized complementary information uncertainty, Eq.(11), PDF p.8.
Neither degradability nor an uncertainty relation is a hypothesis. -/
theorem synthesized_information_uncertainty {b e : ℕ}
    (U : Matrix (Fin b × Fin e) Bool ℂ) (hU : Uᴴ * U = 1)
    (n : ℕ) (i : Fin (2 ^ n)) :
    symmetricInformation (phaseSplit U n i) +
      symmetricInformation (environmentSplit U n i) ≤ 1 := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P23
Source
Quantum polar codes for arbitrary channels, arXiv:1201.2906v2, PDF p. 8, Appendix C, Eq. (11)
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What the Lean code literally says, in plain math · gpt-6.1-sol

For all natural numbers b,eb,eb,e, every complex matrix UUU with rows indexed by {0,…,b−1}×{0,…,e−1}\{0,\ldots,b-1\}\times\{0,\ldots,e-1\}{0,…,b−1}×{0,…,e−1} and Boolean columns satisfying U†U=I2U^\dagger U=I_2U†U=I2​, every natural number nnn, including zero, and every index i<N=2ni< N=2^ni<N=2n, define the following matrices, identifying bits with 0,10,10,1. Let

Grj(n)=1 ⁣{∀ 0≤k<n, ⌊j/2k⌋ mod 2≤⌊r/2n−1−k⌋ mod 2},pn(u)j=(∑r<NurGrj(n)) mod 2,G^{(n)}_{rj} = \mathbf1\!\left\{ \forall\,0\le k<n,\ \left\lfloor j/2^k\right\rfloor\bmod2 \le \left\lfloor r/2^{n-1-k}\right\rfloor\bmod2 \right\}, \qquad p_n(u)_j=\left(\sum_{r<N}u_rG^{(n)}_{rj}\right)\bmod2,Grj(n)​=1{∀0≤k<n, ⌊j/2k⌋mod2≤⌊r/2n−1−k⌋mod2},pn​(u)j​=(r<N∑​ur​Grj(n)​)mod2, vu(r,z)=∏j<NU(rj,zj),pn(u)j,wx(r,u,z)=2−N/2(−1)∑jxjujvu(r,z),v_u(\mathbf r,\mathbf z)=\prod_{j<N}U_{(r_j,z_j),p_n(u)_j}, \qquad w_x(\mathbf r,u,\mathbf z) = 2^{-N/2}(-1)^{\sum_jx_ju_j}v_u(\mathbf r,\mathbf z),vu​(r,z)=j<N∏​U(rj​,zj​),pn​(u)j​​,wx​(r,u,z)=2−N/2(−1)∑j​xj​uj​vu​(r,z),

and put En(u)=Tr⁡r∣vu⟩⟨vu∣E_n(u)=\operatorname{Tr}_{\mathbf r}|v_u\rangle\langle v_u|En​(u)=Trr​∣vu​⟩⟨vu​∣ and Qn(x)=Tr⁡z∣wx⟩⟨wx∣Q_n(x)=\operatorname{Tr}_{\mathbf z}|w_x\rangle\langle w_x|Qn​(x)=Trz​∣wx​⟩⟨wx​∣. Let mi−(u)m_i^-(u)mi−​(u) retain the bits with j<ij<ij<i and set the other bits to zero, and let mi+(u)m_i^+(u)mi+​(u) retain the bits with i<ji<ji<j and set the others to zero. For a∈{0,1}a\in\{0,1\}a∈{0,1}, define

Pa=2−(N−1)∑x:xi=a∣mi+(x)⟩⟨mi+(x)∣⊗Qn(x),Ea=2−(N−1)∑u:ui=a∣mi−(u)⟩⟨mi−(u)∣⊗En(u).P_a=2^{-(N-1)}\sum_{x:x_i=a} |m_i^+(x)\rangle\langle m_i^+(x)|\otimes Q_n(x), \qquad E_a=2^{-(N-1)}\sum_{u:u_i=a} |m_i^-(u)\rangle\langle m_i^-(u)|\otimes E_n(u).Pa​=2−(N−1)x:xi​=a∑​∣mi+​(x)⟩⟨mi+​(x)∣⊗Qn​(x),Ea​=2−(N−1)u:ui​=a∑​∣mi−​(u)⟩⟨mi−​(u)∣⊗En​(u).

For a square matrix KKK, write S(K)=−Re⁡tr⁡[(x↦xlog⁡2x)(K)]S(K)=-\operatorname{Re}\operatorname{tr}[(x\mapsto x\log_2x)(K)]S(K)=−Retr[(x↦xlog2​x)(K)], with the scalar function applied through real continuous functional calculus. Then

S ⁣(P0+P12)−S(P0)+S(P1)2+S ⁣(E0+E12)−S(E0)+S(E1)2≤1.S\!\left(\frac{P_0+P_1}{2}\right) -\frac{S(P_0)+S(P_1)}2 + S\!\left(\frac{E_0+E_1}{2}\right) -\frac{S(E_0)+S(E_1)}2 \le 1.S(2P0​+P1​​)−2S(P0​)+S(P1​)​+S(2E0​+E1​​)−2S(E0​)+S(E1​)​≤1.

There is no restriction on β\betaβ or any degradation assumption in this statement. If b=0b=0b=0 or e=0e=0e=0, the isometry hypothesis is impossible.

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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