Manuscript VII-E@1 · VII-E.tex · theorem · b713cf3c

Theorem: Residual-state obstruction

Record compatibility, residuals, and composition › Record agreement does not determine residuals · lines 1333–1360 · proof included or linked

VII-E@1#thm:residual-obstruction

Record compatibility does not imply agreement of $F$ and $G$ on $O_{L (H_{c})}$ , even when $T = id$ .

Proof. Take one qubit, computational projectors $P_{0}, P_{1}$ , and a single coarse outcome $*$ . Let the coarse instrument be the identity. Then

E_{*} = I = Π_{*},

so the record error is zero. For $ρ = ∣ + ⟩ ⟨ + ∣$ ,

G (ρ) = ∣ * ⟩ ⟨ * ∣_{R} \otimes ∣ + ⟩ ⟨ + ∣, F (ρ) = ∣ * ⟩ ⟨ * ∣_{R} \otimes I /2.

Their difference has residual block $X /2$ , with eigenvalues $\pm 1/2$ . Therefore

∥ G (ρ) - F (ρ) ∥_{1} = 1, d_{tr} (G (ρ), F (ρ)) = \frac{1}{2} .

The full residual algebra distinguishes the outputs.

Displayed formulas in this passage

unlabeled · line 1343
$E_{*} = I = Π_{*},$
unlabeled · line 1347
$G (ρ) = ∣ * ⟩ ⟨ * ∣_{R} \otimes ∣ + ⟩ ⟨ + ∣, F (ρ) = ∣ * ⟩ ⟨ * ∣_{R} \otimes I /2.$
unlabeled · line 1354
$∥ G (ρ) - F (ρ) ∥_{1} = 1, d_{tr} (G (ρ), F (ρ)) = \frac{1}{2} .$