Proposition: One finite channel, two different evolution limits

Fix $0<p<1$, a qubit reset channel

and the finite channel

There are two refinement families whose coarse member is ${\Phi }_{\ast }$, each obeying one-site scheduled locality and exact reference-stable channel semantics, but whose uniform compact-time diamond limits are different.

Proof. Choose $0<\alpha <\beta <1$. Set

For $a\in \{\alpha ,\beta \}$, define

Every channel acts on one site, so every member has the same trivial spatial support ceiling. Every channel is complete.

For $n\ge 2$, idempotence of $\mathcal{R}$ gives

Define

It is a channel and ${{\mathcal{T}}_{t+u}}^a={{\mathcal{T}}_t}^a{{\mathcal{T}}_u}^a$; it is continuous at every $t$.

For $0\le t\le T$, the expansion

and $\left|\lfloor nt\rfloor /n-t\right|\le 1/n$ give

uniformly on that interval. Since ${\Vert \mathrm{id}-\mathcal{R}\Vert }_{\diamond }\le 2$, this is uniform diamond convergence with identity encoders and decoders.

The coarse channels agree: ${{\Phi }_1}^{\alpha }={{\Phi }_1}^{\beta }={\Phi }_{\ast }$. But for $t>0$, on input $|1⟩⟨1|$,

The two outputs have trace distance $\left|e^{-\alpha t}-e^{-\beta t}\right|>0$. Thus their limiting evolutions differ despite the common finite member and identical locality ceiling.