Proposition: One coarse channel with two exactly consistent continuous-time completions

The identity qubit channel has two different continuous-time channel groups with the same time-one channel. Each group has exact subdivision channels whose $n$-fold composition equals the common coarse channel. The corresponding tick approximations converge uniformly on compact time intervals in diamond norm, but their limits differ.

Proof. Let

Both families are continuous channel groups. Since $U_0(1)=U_1(1)=I$,

At refinement $n$, use the tick channel

Then $({{\Phi }_n}^a{)}^n=\mathrm{id}$ exactly.

Define

For real $s,t$,

For an arbitrary reference and a trace-class input $M$, expanding the difference of the two unitary conjugations gives a trace-norm bound $2\Vert U_a(t)-U_a(s)\Vert {\Vert M\Vert }_1$. Hence

uniformly in $t$.

At $t=1/2$, ${{\mathcal{T}}_{1/2}}^0$ leaves $|+⟩⟨+|$ unchanged, whereas ${{\mathcal{T}}_{1/2}}^1$ sends it to $|-⟩⟨-|$. The two outputs have trace distance one. The limiting evolutions therefore differ.