Theorem: Reference-preserving finite-time approximation

At fixed $\mu ,\Lambda$, fixed packet family, and $T<\infty$, there is a finite-dimensional proxy with prepared isometry ${\hat{V}}_t$, embedded in the same complete output space, such that uniformly for $0\le t\le T$,

The same allowance holds for both reduced channels.

For $T>0$ and $0<b<1$, sufficient choices are

These give $b_{\mathrm{reg}}\le b$.

Proof. Use the interaction picture for $H_S+H_B$:

On sectors of occupation at most $k$,

An $\ell$-fold product starting from vacuum therefore has norm at most $(2G{)}^{\ell }\sqrt{\ell !}$. The ordered simplex has volume $t^{\ell }/\ell !$, so the $\ell$-th Dyson term is bounded by

These bounds are uniform over logical inputs and references.

Multiplication by any fixed power of the occupation number adds at most a polynomial in $\ell$, so the series converges in every fixed occupation-number graph norm on finite intervals. Since $h_B$ is bounded by $\Lambda$, this also controls the prepared energy domains. Termwise differentiation on these prepared vectors gives the Schrödinger equation; uniqueness for the self-adjoint Hamiltonian identifies the series with $U(t)$.

For the compressed Hamiltonian $P_{m}H{P}_m$, the Dyson terms agree with the exact terms through order $m$. Each additional projection is a contraction, and the same term bound applies to the compressed series. The prepared-isometry difference is therefore at most $2Z_m(2GT)$.

Partition $[\mu ,\Lambda ]$ into finitely many intervals $I_b$ of width at most $\Delta$, and replace $\omega$ on each interval by its midpoint, obtaining ${\tilde{h}}_B$. Then

On the truncated space both Hamiltonians are bounded. Duhamel's formula adds at most $Tm\Delta /2$ to the prepared-isometry comparison.

The active one-particle space

is finite dimensional, contains $\mathfrak{r}$ and all coupling vectors, and is invariant under ${\tilde{h}}_B$. Its orthogonal complement remains vacuum. Consequently the active proxy with total occupation cutoff $m$ is finite dimensional.

Its total-cutoff space embeds in

This is an embedding, not an equality: the ambient tensor product also contains states with total occupation greater than $m$. Embedding the remaining inactive modes as vacuum gives the common complete output space. The source and carrier stay in $B$.

Lemma lem:bounds converts the prepared-isometry comparison to (eq:breg), including references and reduced channels.

Finally, let $x=2GT$. The factorial bound gives

Later term ratios are at most $1/2$, whence $Z_m(x)\le 2^{-m}$. The occupation term $2Z_m(x)$ and the mesh term are each at most $b/2$ under the stated choices.