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BSC amplitude root fidelity

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QAlgorithms.Papers.Independent.Intakethird100.P74.bsc_amplitude_fidelity

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

prove2quantum

Let pB∈[0,1]p_B\in[0,1]pB​∈[0,1] and let Bob's computational-basis signal states be those of a BSC: ρx=(1−pB)∣x⟩⟨x∣+pB∣1−x⟩⟨1−x∣\rho_x=(1-p_B)|x\rangle\langle x|+p_B|1-x\rangle\langle1-x|ρx​=(1−pB​)∣x⟩⟨x∣+pB​∣1−x⟩⟨1−x∣. With the source's squared-fidelity convention,

F(ρ0,ρ1)=2pB(1−pB).\sqrt{F(\rho_0,\rho_1)}=2\sqrt{p_B(1-p_B)}.F(ρ0​,ρ1​)​=2pB​(1−pB​)​.

This is the classical Bhattacharyya parameter of Bob's amplitude channel. The expression is root fidelity, rather than its square.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P74
Formal statement
namespace QAlgorithms.Papers.Independent.Intakethird100.P74

open QAlgorithms

open scoped ComplexConjugate Matrix

noncomputable section

local instance (p : Prop) : Decidable p := Classical.propDecidable p

variable {B E S1 S2 : Type}

variable [Fintype B] [Fintype E] [Fintype S1] [Fintype S2]

variable [DecidableEq B] [DecidableEq E] [DecidableEq S1] [DecidableEq S2]

/-- PDF p.7, Appendix C.A: the amplitude ROOT fidelity, with radical over p(1-p). -/
theorem bsc_amplitude_fidelity (pB : ℝ) (hp0 : 0 ≤ pB) (hp1 : pB ≤ 1) :
    rootFidelity (bscOutput pB false) (bscOutput pB true) =
      2 * Real.sqrt (pB * (1 - pB)) := by sorry

end

end QAlgorithms.Papers.Independent.Intakethird100.P74
Source
Polar codes for private classical communication, arXiv:1203.5794v1, PDF p. 7, Appendix C.A, unnumbered: amplitude root fidelity of BSC
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What the Lean code literally says, in plain math · gpt-6.1-sol

For every real number pBp_BpB​ with 0≤pB≤10\le p_B\le10≤pB​≤1, set ρ0=diag⁡(1−pB,pB)\rho_0=\operatorname{diag}(1-p_B,p_B)ρ0​=diag(1−pB​,pB​) and ρ1=diag⁡(pB,1−pB)\rho_1=\operatorname{diag}(p_B,1-p_B)ρ1​=diag(pB​,1−pB​), indexed in the order (false,true)(\mathrm{false},\mathrm{true})(false,true). Their root fidelity, defined by Fr(ρ,σ)=Re⁡tr⁡ρ σρF_{\mathrm r}(\rho,\sigma)=\operatorname{Re}\operatorname{tr}\sqrt{\sqrt{\rho}\,\sigma\sqrt{\rho}}Fr​(ρ,σ)=Retrρ​σρ​​ using spectral functional calculus, satisfies

Fr(ρ0,ρ1)=2pB(1−pB).F_{\mathrm r}(\rho_0,\rho_1)=2\sqrt{p_B(1-p_B)}.Fr​(ρ0​,ρ1​)=2pB​(1−pB​)​.

This includes pB=0p_B=0pB​=0 and pB=1p_B=1pB​=1, where the stated value is zero.

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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