Manuscript Oai280-B@1 · Oai280-B.tex · proposition · 6d92ab75

Proposition: A fixed-spectator counterexample

Finite-factor subsystem certification › Why fixed-spectator validation is weaker · lines 2287–2306 · proof included or linked

Oai280-B@1#prop:spectatortrap

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There is a two-input channel $Φ$ such that

Φ (ρ_{C} \otimes ∣ 0 ⟩ ⟨ 0 ∣_{S}) = ρ_{C}

for every referenced logical input, but

d_{⋄} (Φ, Tr_{S}) = 1.

Proof. Apply a CNOT controlled by $S$ and targeted on $C$ , then trace out $S$ . With $S = ∣ 0 ⟩$ , the induced logical channel is identity. With $S = ∣ 1 ⟩$ and $C = ∣ 0 ⟩$ , the actual output is $∣ 1 ⟩$ whereas the target is $∣ 0 ⟩$ . Their trace distance is one. Channel half-diamond distance cannot exceed one, so equality follows.

Displayed formulas in this passage

unlabeled · line 2290
$Φ (ρ_{C} \otimes ∣ 0 ⟩ ⟨ 0 ∣_{S}) = ρ_{C}$
unlabeled · line 2294
$d_{⋄} (Φ, Tr_{S}) = 1.$

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section: Why fixed-spectator validation is weaker