Theorem: Classical records are not a quantum decoder

If the only retained output of a complete process is classical, every subsequent deterministic reconstruction of $C$ has error at least

The same bound holds for every measure-and-prepare channel followed by a decoder. Computational readout followed by basis-state preparation attains it.

Proof. A finite classical-output channel has the form

A decoder prepares one state for each outcome, giving a measure-and-prepare channel. Its Bell-input output is separable.

For a normalized product vector $u\otimes v$,

Convexity gives the same Bell-overlap upper bound for separable states. Testing the Bell projector gives error at least $1-1/d$.

Computational dephasing is

The $d$ Bell outputs are orthogonal, yielding distance $1-1/d$; the triangle inequality gives the matching upper bound.