Proposition: Two exact token-acquisition refusals

1. If $q_J$ has infinitely many values on a preparation class, no finite token with a fixed decoder determines all those values exactly.

2. Let $\rho ,\sigma$ be two positive definite states. No finite-copy measurement can deterministically return different exact labels for $\rho$ and $\sigma$. This applies to every fixed copy number and to finite adaptive instruments using at most that many copies.

Proof. For (i), a decoder $d:\mathcal{T}\to \mathbb{R}$ has finite image. It cannot equal $q_J$ on a class with infinitely many expectation values.

For (ii), for a fixed number $n$, both ${\rho }^{\otimes n}$ and ${\sigma }^{\otimes n}$ are positive definite. If a nonzero positive effect $E$ occurs, then

Indeed ${\rho }^{\otimes n}\ge aI$ for some $a>0$, so the first trace is at least $a\mathrm{Tr}E>0$, and similarly for $\sigma$. An outcome occurring with probability one for one state and zero for the other would contradict this property.

A finite adaptive procedure with at most $n$ copies is a POVM on the $n$-copy input: unused copies may be supplied and ignored. Its output-label effects therefore obey the same argument.