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NDS fidelity minimization at a prescribed photon law

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QAlgorithms.Papers.Independent.Intakethird100.P54.nds_fidelity_minimization

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

prove2quantum

Let M≥1M\ge1M≥1 be the finite number of signal modes and J≥0J\ge0J≥0 the finite number of retained idler modes, on the full optical Fock spaces. The two channels have equal priors, independent vacuum environments, reflectances 0≤R0≤R1≤10\le R_0\le R_1\le10≤R0​≤R1​≤1, field magnitudes rb=Rbr_b=\sqrt{R_b}rb​=Rb​​ and tb=1−Rbt_b=\sqrt{1-R_b}tb​=1−Rb​​, and distinct complex reflection coefficients rbeiθbr_b e^{i\theta_b}rb​eiθb​. Every signal mode sees the same channel; idlers are retained losslessly and environment outputs are inaccessible. Write F\mathcal FF for squared Uhlmann fidelity, PeP_ePe​ for minimum average error over all joint binary receivers, Qs=Tr⁡(ρ0sρ11−s)Q_s=\operatorname{Tr}(\rho_0^s\rho_1^{1-s})Qs​=Tr(ρ0s​ρ11−s​), and Q=min⁡0≤s≤1QsQ=\min_{0\le s\le1}Q_sQ=min0≤s≤1​Qs​, with zero powers interpreted as support projections. Here the common phase is removed, so θ0=θ1=0\theta_0=\theta_1=0θ0​=θ1​=0 and R0<R1R_0<R_1R0​<R1​. Put μ=r0r1+t0t1\mu=r_0r_1+t_0t_1μ=r0​r1​+t0​t1​.

Let ∣ψ⟩|\psi\rangle∣ψ⟩ be any normalized pure NDS probe, and let pnp_npn​ be its total signal photon law. Then

F(ψ)=(∑n=0∞pnμn)2≤F(ϕ)\mathcal F(\psi)=\left(\sum_{n=0}^\infty p_n\mu^n\right)^2\le\mathcal F(\phi)F(ψ)=(n=0∑∞​pn​μn)2≤F(ϕ)

for every normalized pure competitor ∣ϕ⟩|\phi\rangle∣ϕ⟩ with the same total law, allowing any finite number of retained idler modes. In addition, every probability law on multimode signal occupations is realized by a normalized NDS purification with one freely chosen optical idler.

Thus the bound is attained at a prescribed total law, in particular when the entire multimode occupation law is fixed; no adequate purification is postulated in an arbitrarily small prescribed idler space.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P54
Formal statement
noncomputable section

open scoped BigOperators ComplexConjugate

open MeasureTheory Filter

namespace QAlgorithms.Papers.Independent.Intakethird100.P54

open QAlgorithms.RealRegisters (Hilbert)

open QAlgorithms.Papers.Independent.Claimed50.P47 (Occupation)

open QAlgorithms.Papers.Independent.Dependency50.P39 (coherentCoefficient)

attribute [local instance] Classical.propDecidable

attribute [local instance] Classical.decEq

/-- NDS probes minimize fidelity at fixed TOTAL signal photon law, PDF p.6, (48).
Comparators have arbitrary finite idler count. Realizability is asserted with a freely
chosen single optical idler, rather than a possibly inadequate prescribed idler. -/
theorem nds_fidelity_minimization {M J : ℕ} (hM : 0 < M) (c : Channels)
    (hc : ValidSplitters c) (hr : Reading c) (psi : Hilbert (Occupation (Fin J) × Occupation (Fin M)))
    (hpsi : ‖psi‖ = 1) (hnds : IsNDS psi) :
    outputFidelity c psi = (∑' n : ℕ, totalLaw psi n * (splitterOverlap c)^n)^2 ∧
    (∀ (Jother : ℕ) (other : Hilbert (Occupation (Fin Jother) × Occupation (Fin M))),
      ‖other‖ = 1 → totalLaw other = totalLaw psi →
        outputFidelity c psi ≤ outputFidelity c other) ∧
    (∀ p : Occupation (Fin M) → ℝ, (∀ n, 0 ≤ p n) → Summable p → (∑' n, p n)=1 →
      ∃ z : Hilbert (Occupation (Fin 1) × Occupation (Fin M)),
        ‖z‖=1 ∧ IsNDS z ∧ signalLaw z=p) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P54

end
Source
Discriminating quantum-optical beam-splitter channels with number-diagonal signal states: Applications to quantum reading and target detection, arXiv:1105.4063v3, PDF p. 6, nds-fidelity-minimization
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let M,J∈NM,J\in\mathbb NM,J∈N with M>0M>0M>0, allowing J=0J=0J=0. Let two splitters have reflectances Rb∈[0,1]R_b\in[0,1]Rb​∈[0,1], both phases equal to zero, rb=Rbr_b=\sqrt{R_b}rb​=Rb​​, and tb=1−Rbt_b=\sqrt{1-R_b}tb​=1−Rb​​, with R0≤R1R_0\le R_1R0​≤R1​ and r0≠r1r_0\ne r_1r0​=r1​. Let ψ∈ℓ2(NJ×NM;C)\psi\in\ell^2(\mathbb N^J\times\mathbb N^M;\mathbb C)ψ∈ℓ2(NJ×NM;C) have norm 111 and satisfy ∑iψ(i,n)ψ(i,q)‾=0\sum_i\psi(i,n)\overline{\psi(i,q)}=0∑i​ψ(i,n)ψ(i,q)​=0 for every distinct n,q∈NMn,q\in\mathbb N^Mn,q∈NM. For any probe ϕ\phiϕ, put pnϕ=∑i∣ϕ(i,n)∣2p_n^\phi=\sum_i|\phi(i,n)|^2pnϕ​=∑i​∣ϕ(i,n)∣2 and PNϕ=∑n:∑mnm=NpnϕP_N^\phi=\sum_{n:\sum_mn_m=N}p_n^\phiPNϕ​=∑n:∑m​nm​=N​pnϕ​. Define Lb(n,k)=∏m(nmkm)rbnm−kmtbkmL_b(n,k)=\prod_m\sqrt{\binom{n_m}{k_m}}r_b^{n_m-k_m}t_b^{k_m}Lb​(n,k)=∏m​(km​nm​​)​rbnm​−km​​tbkm​​ if k≤nk\le nk≤n componentwise and zero otherwise; define gb,kϕ(i,p)=ϕ(i,p+k)Lb(p+k,k)g_{b,k}^\phi(i,p)=\phi(i,p+k)L_b(p+k,k)gb,kϕ​(i,p)=ϕ(i,p+k)Lb​(p+k,k) and ρbϕ(x,y)=∑kgb,kϕ(x)gb,kϕ(y)‾\rho_b^\phi(x,y)=\sum_k g_{b,k}^\phi(x)\overline{g_{b,k}^\phi(y)}ρbϕ​(x,y)=∑k​gb,kϕ​(x)gb,kϕ​(y)​. Let F(ϕ)F(\phi)F(ϕ) be the real supremum of ∣⟨z,w⟩∣2|\langle z,w\rangle|^2∣⟨z,w⟩∣2 over norm-one square-summable vectors on the output labels times N\mathbb NN whose respective marginals ∑ez(x,e)z(y,e)‾\sum_e z(x,e)\overline{z(y,e)}∑e​z(x,e)z(y,e)​ and ∑ew(x,e)w(y,e)‾\sum_e w(x,e)\overline{w(y,e)}∑e​w(x,e)w(y,e)​ equal ρ0ϕ,ρ1ϕ\rho_0^\phi,\rho_1^\phiρ0ϕ​,ρ1ϕ​; the supremum of an empty defining set is 000. With o=r0r1+t0t1o=r_0r_1+t_0t_1o=r0​r1​+t0​t1​, the theorem asserts

F(ψ)=(∑N=0∞PNψoN)2.F(\psi)=\left(\sum_{N=0}^{\infty}P_N^\psi o^N\right)^2.F(ψ)=(N=0∑∞​PNψ​oN)2.

It also asserts that, for every Jother∈NJ_{\mathrm{other}}\in\mathbb NJother​∈N and every norm-one ϕ∈ℓ2(NJother×NM;C)\phi\in\ell^2(\mathbb N^{J_{\mathrm{other}}}\times\mathbb N^M;\mathbb C)ϕ∈ℓ2(NJother​×NM;C) satisfying PNϕ=PNψP_N^\phi=P_N^\psiPNϕ​=PNψ​ for every NNN, one has F(ψ)≤F(ϕ)F(\psi)\le F(\phi)F(ψ)≤F(ϕ), without requiring ϕ\phiϕ to satisfy the orthogonality condition imposed on ψ\psiψ. Finally, for every summable function p:NM→Rp:\mathbb N^M\to\mathbb Rp:NM→R with pn≥0p_n\ge0pn​≥0 and ∑npn=1\sum_n p_n=1∑n​pn​=1, there exists z∈ℓ2(N×NM;C)z\in\ell^2(\mathbb N\times\mathbb N^M;\mathbb C)z∈ℓ2(N×NM;C) of norm 111 such that ∑iz(i,n)z(i,q)‾=0\sum_i z(i,n)\overline{z(i,q)}=0∑i​z(i,n)z(i,q)​=0 for all distinct n,qn,qn,q, and ∑i∣z(i,n)∣2=pn\sum_i|z(i,n)|^2=p_n∑i​∣z(i,n)∣2=pn​ for every nnn. Zero occupations and endpoint reflectances are 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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