Manuscript XXII-D@1 · XXII-D.tex · proposition · 17889d18

Proposition: An entanglement-breaking transport floor

Geometry and signalling do not certify quantum transport · lines 540–574 · proof included or linked

XXII-D@1#prop:eb-floor, XXII-D@1#eq:eb-floor

Let $M ≅ C^{d}$ , $d \geq 2$ . If a receiver channel $Λ : L (M) \to L (M)$ is measure-and-prepare, then

d_{⋄} (Λ, id_{M}) \geq 1 - \frac{1}{d} .

Proof. Write

Λ (X) = z \sum Tr (F_{z} X) σ_{z}, F_{z} \geq 0, z \sum F_{z} = I .

For a maximally entangled input

∣ Ω_{d} ⟩ = d^{- 1/2} j = 1 \sum^{d} ∣ j ⟩_{M} ∣ j ⟩_{R},

the receiver–reference output is separable. This is the measure-and-prepare characterization of entanglement-breaking channels HSR.

A pure product vector has squared overlap with $∣ Ω_{d} ⟩$ at most $1/ d$ , by Cauchy–Schwarz. Convexity gives the same bound for a separable state. Testing the projector $∣ Ω_{d} ⟩ ⟨ Ω_{d} ∣$ therefore gives trace distance at least $1 - 1/ d$ from the ideal Bell output. This input is allowed in the diamond norm, proving (eq:eb-floor).

Displayed formulas in this passage

XXII-D@1#eq:eb-floor · line 545
$d_{⋄} (Λ, id_{M}) \geq 1 - \frac{1}{d} .$
unlabeled · line 553
$Λ (X) = z \sum Tr (F_{z} X) σ_{z}, F_{z} \geq 0, z \sum F_{z} = I .$
unlabeled · line 558
$∣ Ω_{d} ⟩ = d^{- 1/2} j = 1 \sum^{d} ∣ j ⟩_{M} ∣ j ⟩_{R},$

Proposition: An entanglement-breaking transport floor

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