Theorem: QND radiation is entanglement breaking

If

for all $a,b$, the reservoir channel from any inserted finite code is entanglement breaking. Every deterministic radiation decoder has error at least

Proof. Choose common spectral projectors $P_{\lambda }$ of the source operators. The propagator is block diagonal:

Tracing the source kills cross-projector terms. The accessible channel therefore has the form

The $F_{\lambda }$ form a POVM. Thus the channel is measure-and-prepare.

A decoder preserves separability of its Bell-input output. Every separable state on $Q\otimes C$ has Bell overlap at most $1/d$, by Cauchy–Schwarz for product vectors and convexity. The Bell-projector test proves the lower bound.