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Corrected exact Fock ideal-memory error

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QAlgorithms.Papers.Independent.Intakethird100.P54.fock_ideal_error

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​.

Assume R1=1R_1=1R1​=1. For any multimode Fock occupation with total signal photon number KKK,

Pe=12R0K=F2=Q2.P_e=\frac12R_0^K=\frac{\mathcal F}{2}=\frac Q2.Pe​=21​R0K​=2F​=2Q​.

Photon counting followed by decision one exactly when the total returned count equals KKK is optimal, including the K=0K=0K=0 tie.

Source correction. Printed equation (71) contains an extraneous ln⁡\lnln before 1/21/21/2. The counting rule and the next paragraph's equality Pe=F/2P_e=\mathcal F/2Pe​=F/2 force the displayed correction; the printed formula would be negative. Independent moderation of this correction remains required.

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

/-- Exact Fock ideal-memory error, PDF p.9, (71)–(72). SOURCE CORRECTION:
printed (71) has an extraneous 'ln' before 1/2. It is deleted here because the
printed counting rule and the following equality Pe=F/2 force R0^K/2; a logarithm
would make a probability negative. Preserve the printed formula in items/HARD. -/
theorem fock_ideal_error {M : ℕ} (hM : 0 < M) (c : Channels)
    (hc : ValidSplitters c) (hr : Reading c) (h1 : (c true).reflectance=1)
    (n : Occupation (Fin M)) :
    outputError c (fock n)=(c false).reflectance^photons n/2 ∧
    outputError c (fock n)=outputFidelity c (fock n)/2 ∧
    outputError c (fock n)=outputChernoff c (fock n)/2 ∧
    ∃ E, IsOpticalEffect E ∧ CountsAs E (fun q => decide (photons q=photons n)) ∧
      sectorRisk c (fock n) E (1/2) (1/2)=outputError c (fock n) := 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. 9, fock-ideal-error
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let M∈NM\in\mathbb NM∈N satisfy M>0M>0M>0, and let ccc specify splitters with reflectances Rb∈[0,1]R_b\in[0,1]Rb​∈[0,1] and phases ϕb\phi_bϕb​, satisfying R0≤R1R_0\le R_1R0​≤R1​, unequal fields R0eiϕ0≠R1eiϕ1\sqrt{R_0}e^{i\phi_0}\ne\sqrt{R_1}e^{i\phi_1}R0​​eiϕ0​=R1​​eiϕ1​, ϕ0=ϕ1=0\phi_0=\phi_1=0ϕ0​=ϕ1​=0, and R1=1R_1=1R1​=1. Fix n∈NMn\in\mathbb N^Mn∈NM, put K=∑mnmK=\sum_mn_mK=∑m​nm​, and let ∣n⟩|n\rangle∣n⟩ be the unit coordinate vector at (0,n)(0,n)(0,n) in ℓ2(N0×NM;C)\ell^2(\mathbb N^0\times\mathbb N^M;\mathbb C)ℓ2(N0×NM;C). Set ab(q,k)=∏m(qmkm)(Rb)qm−km(1−Rb)kma_b(q,k)=\prod_m\sqrt{\binom{q_m}{k_m}}(\sqrt{R_b})^{q_m-k_m}(\sqrt{1-R_b})^{k_m}ab​(q,k)=∏m​(km​qm​​)​(Rb​​)qm​−km​(1−Rb​​)km​ for k≤qk\le qk≤q, and 000 otherwise, and define zb,k(i,q)=∣n⟩(i,q+k)ab(q+k,k)z_{b,k}(i,q)=|n\rangle(i,q+k)a_b(q+k,k)zb,k​(i,q)=∣n⟩(i,q+k)ab​(q+k,k), ρb(x,y)=∑kzb,k(x)zb,k(y)‾\rho_b(x,y)=\sum_kz_{b,k}(x)\overline{z_{b,k}(y)}ρb​(x,y)=∑k​zb,k​(x)zb,k​(y)​, and sb,ks_{b,k}sb,k​ as the coefficient function zb,kz_{b,k}zb,k​ if square summable, and as the zero ℓ2\ell^2ℓ2 vector otherwise. A resolution of a kernel HHH is H(x,y)=∑apaua(x)ua(y)‾H(x,y)=\sum_a p_a u_a(x)\overline{u_a(y)}H(x,y)=∑a​pa​ua​(x)ua​(y)​, with nonnegative summable weights totaling 111, unit ℓ2\ell^2ℓ2 vectors at positive weights, and orthogonality for distinct positive-weight vectors; zero-weight vectors are unrestricted. Define e(ρ0,ρ1)e(\rho_0,\rho_1)e(ρ0​,ρ1​) as the real infimum over resolutions (pa,ua),(qa,va)(p_a,u_a),(q_a,v_a)(pa​,ua​),(qa​,va​) and bounded complex-linear effects EEE, satisfying ⟨u,Ev⟩=⟨v,Eu⟩‾\langle u,Ev\rangle=\overline{\langle v,Eu\rangle}⟨u,Ev⟩=⟨v,Eu⟩​ and 0≤Re⁡⟨v,Ev⟩≤∥v∥20\le\operatorname{Re}\langle v,Ev\rangle\le\|v\|^20≤Re⟨v,Ev⟩≤∥v∥2, of 12∑apa(1−Re⁡⟨ua,Eua⟩)+12∑aqaRe⁡⟨va,Eva⟩\frac12\sum_a p_a(1-\operatorname{Re}\langle u_a,Eu_a\rangle)+\frac12\sum_a q_a\operatorname{Re}\langle v_a,Ev_a\rangle21​∑a​pa​(1−Re⟨ua​,Eua​⟩)+21​∑a​qa​Re⟨va​,Eva​⟩. Define F(ρ0,ρ1)F(\rho_0,\rho_1)F(ρ0​,ρ1​) as the real supremum of ∣⟨z,w⟩∣2|\langle z,w\rangle|^2∣⟨z,w⟩∣2 over norm-one z,w∈ℓ2(X×N;C)z,w\in\ell^2(X\times\mathbb N;\mathbb C)z,w∈ℓ2(X×N;C), where X=N0×NMX=\mathbb N^0\times\mathbb N^MX=N0×NM, whose marginals ∑ez(x,e)z(y,e)‾\sum_ez(x,e)\overline{z(y,e)}∑e​z(x,e)z(y,e)​ and ∑ew(x,e)w(y,e)‾\sum_ew(x,e)\overline{w(y,e)}∑e​w(x,e)w(y,e)​ equal ρ0,ρ1\rho_0,\rho_1ρ0​,ρ1​, respectively. Finally put

C(ρ0,ρ1)=inf⁡s∈[0,1]sup⁡resolutions (pa,ua),(qb,vb)∑a,bfs(pa)f1−s(qb)∣⟨ua,vb⟩∣2,C(\rho_0,\rho_1)= \inf_{s\in[0,1]} \sup_{\text{resolutions }(p_a,u_a),(q_b,v_b)} \sum_{a,b}f_s(p_a)f_{1-s}(q_b)|\langle u_a,v_b\rangle|^2,C(ρ0​,ρ1​)=s∈[0,1]inf​resolutions (pa​,ua​),(qb​,vb​)sup​a,b∑​fs​(pa​)f1−s​(qb​)∣⟨ua​,vb​⟩∣2,

where ft(0)=0f_t(0)=0ft​(0)=0 and ft(p)=ptf_t(p)=p^tft​(p)=pt otherwise. All extrema are real, with value 000 for empty defining sets, and infinite sums have value 000 if not summable. The assertion is

e(ρ0,ρ1)=R0K2=F(ρ0,ρ1)2=C(ρ0,ρ1)2,e(\rho_0,\rho_1)=\frac{R_0^K}{2} =\frac{F(\rho_0,\rho_1)}2 =\frac{C(\rho_0,\rho_1)}2,e(ρ0​,ρ1​)=2R0K​​=2F(ρ0​,ρ1​)​=2C(ρ0​,ρ1​)​,

and there exists an effect EEE such that, for every vector vvv and label (i,q)(i,q)(i,q), (Ev)(i,q)=0(Ev)(i,q)=0(Ev)(i,q)=0 if ∑mqm=K\sum_mq_m=K∑m​qm​=K, and (Ev)(i,q)=v(i,q)(Ev)(i,q)=v(i,q)(Ev)(i,q)=v(i,q) otherwise, and such that

12∑k(∥s0,k∥2−Re⁡⟨s0,k,Es0,k⟩)+12∑kRe⁡⟨s1,k,Es1,k⟩=e(ρ0,ρ1).\frac12\sum_k\left(\|s_{0,k}\|^2-\operatorname{Re}\langle s_{0,k},Es_{0,k}\rangle\right) +\frac12\sum_k\operatorname{Re}\langle s_{1,k},Es_{1,k}\rangle =e(\rho_0,\rho_1).21​k∑​(∥s0,k​∥2−Re⟨s0,k​,Es0,k​⟩)+21​k∑​Re⟨s1,k​,Es1,k​⟩=e(ρ0​,ρ1​).

The hypotheses allow R0=0R_0=0R0​=0 and n=0n=0n=0; every zeroth power, including 000^000, is 111.

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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