Proposition: Explicit refinement non-determination

Fix $0<p<1$ and

There are two families of one-site complete channels with the same coarse member ${\Phi }_{\ast }$ and different uniform compact-time diamond-norm limits.

Proof. This is the witness of XIX-C CheungXIXC (prop:two-limits), whose calculation is included to specify the exact extent of the test.

Choose $0<a<b<1$. For $c\in \{a,b\}$, set

and define

Every channel acts on one site and has its complete reference extension. Since ${\mathcal{R}}^2=\mathcal{R}$, for $n\ge 2$,

Set

It is a channel, ${{\mathcal{T}}_0}^c=\mathrm{id}$, and ${{\mathcal{T}}_{t+s}}^c={{\mathcal{T}}_t}^c{{\mathcal{T}}_s}^c$.

For every finite $T$,

show that $(1-c/n{)}^{\lfloor nt\rfloor }\to e^{-ct}$ uniformly on $0\le t\le T$. Because $\Vert \mathrm{id}-\mathcal{R}{\Vert }_{\diamond }\le 2$, this gives

But ${{\Phi }_1}^a={{\Phi }_1}^b={\Phi }_{\ast }$, while for $t>0$,

The trace distance between the $a$- and $b$-outputs is $|e^{-at}-e^{-bt}|>0$.