Proposition: Finite-time prepared-channel regulator

For finitely many declared packets, occupation cutoff $m$, frequency bins of width at most $\Delta$, and $0\le t\le T$, one can construct a finite active proxy with

The embedding preserves the accessible–inaccessible partition.

Proof. In interaction picture, on occupation at most $k$,

An $\ell$-fold product starting from vacuum has norm at most $(2G{)}^{\ell }\sqrt{\ell !}$. Ordered integration gives Dyson-term bound $(2Gt{)}^{\ell }/\sqrt{\ell !}$, uniformly over code–reference inputs. These series converge also after multiplication by any fixed occupation power; since $h_B$ is bounded, termwise differentiation on prepared vectors identifies the series with the self-adjoint evolution.

Compression to occupation at most $m$ leaves all terms through order $m$ unchanged. Higher terms in either series obey the same bound, giving prepared-isometry difference at most $2Z_m(2GT)$.

Replace frequency by the midpoint of each bin. The one-particle operator difference has norm at most $\Delta /2$; on occupation at most $m$, its second quantization has norm at most $m\Delta /2$. Duhamel's formula adds $Tm\Delta /2$.

The span of the bin-restricted form factors and packet functions is finite dimensional and invariant under the midpoint operator. Modes orthogonal to it stay in vacuum. Its total-occupation cutoff is therefore a finite active proxy. The total cutoff embeds in separate collected and uncollected cutoff Fock spaces; it is not equal to their full tensor product. The source and displaced carrier remain in the residual.

Lemma lem:distances converts prepared-isometry error to the complete-channel allowance.