Manuscript XVII-A@1 · XVII-A.tex · section · 711d7ee4

The allowed reconstruction channel

Exact reconstruction of the named observable › The allowed reconstruction channel · lines 578–602

XVII-A@1#eq:F, XVII-A@1#eq:RB, XVII-A@1#eq:ME

Let $H_{C} = C^{2}$ be a classical output register in the basis ${∣ 0 ⟩, ∣ 1 ⟩}$ . Declare

F_{1} R_{B} (σ) M_{O_{bulk}} (ρ) = O_{B}, F_{0} = 1 - O_{B}, = r = 0 \sum^{1} Tr (F_{r} σ) ∣ r ⟩ ⟨ r ∣, = Tr ((1 - O_{bulk}) ρ) ∣ 0 ⟩ ⟨ 0 ∣ + Tr (O_{bulk} ρ) ∣ 1 ⟩ ⟨ 1 ∣ .

These are complete, deterministic binary record channels: both outcomes are retained, and no postselection is used.

For a state $ω_{B}$ on the boundary algebra, the same definition is

R_{B} (ω_{B}) = ω_{B} (1 - O_{B}) ∣ 0 ⟩ ⟨ 0 ∣ + ω_{B} (O_{B}) ∣ 1 ⟩ ⟨ 1 ∣ .

The map class is fixed: exact observable reconstruction with an explicit completely positive trace-preserving record channel. It is not an unrestricted quantum-state recovery problem.

Displayed formulas in this passage

XVII-A@1#eq:F · line 582
$F_{1} R_{B} (σ) M_{O_{bulk}} (ρ) = O_{B}, F_{0} = 1 - O_{B}, = r = 0 \sum^{1} Tr (F_{r} σ) ∣ r ⟩ ⟨ r ∣, = Tr ((1 - O_{bulk}) ρ) ∣ 0 ⟩ ⟨ 0 ∣ + Tr (O_{bulk} ρ) ∣ 1 ⟩ ⟨ 1 ∣ .$
XVII-A@1#eq:RB · line 582
$F_{1} R_{B} (σ) M_{O_{bulk}} (ρ) = O_{B}, F_{0} = 1 - O_{B}, = r = 0 \sum^{1} Tr (F_{r} σ) ∣ r ⟩ ⟨ r ∣, = Tr ((1 - O_{bulk}) ρ) ∣ 0 ⟩ ⟨ 0 ∣ + Tr (O_{bulk} ρ) ∣ 1 ⟩ ⟨ 1 ∣ .$
XVII-A@1#eq:ME · line 582
$F_{1} R_{B} (σ) M_{O_{bulk}} (ρ) = O_{B}, F_{0} = 1 - O_{B}, = r = 0 \sum^{1} Tr (F_{r} σ) ∣ r ⟩ ⟨ r ∣, = Tr ((1 - O_{bulk}) ρ) ∣ 0 ⟩ ⟨ 0 ∣ + Tr (O_{bulk} ρ) ∣ 1 ⟩ ⟨ 1 ∣ .$
unlabeled · line 595
$R_{B} (ω_{B}) = ω_{B} (1 - O_{B}) ∣ 0 ⟩ ⟨ 0 ∣ + ω_{B} (O_{B}) ∣ 1 ⟩ ⟨ 1 ∣ .$

The allowed reconstruction channel

Displayed formulas in this passage

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