Proposition: Finite-promise acquisition obstruction

For distinct $m,n$, no deterministic one-copy channel can produce their different exact token labels on both ${\rho }_m,{\rho }_n$. For a binary label-acquisition attempt with error probabilities $e_m,e_n$,

No finite number of copies makes exact deterministic discrimination of these full-rank states possible.

Proof. The input trace distance is $|z_m-z_n|/2<1$, whereas distinct deterministic output labels have distance one. Contractivity under a channel excludes exact acquisition.

Let the output event “label $m$” have probabilities $1-e_m$ and $e_n$. Their difference is bounded by the input trace distance. Therefore $e_m+e_n\ge 1-d_{\mathrm{tr}}({\rho }_m,{\rho }_n)$, proving (eq:token-error). Tensor powers remain full rank and have overlapping support, so their trace distance remains strictly below one.