Theorem: Reference-stable finite-volume influence tail

For the unitary evolution channel ${\Psi }_{t,s}$, and $d=d_G(A,B)\ge 1$,

For every $\mu >0$, the tail is also bounded by

The bounds are independent of the dimension of an untouched reference. They are not claimed uniform in volume.

Proof. Pass to the interaction picture for the onsite terms. Their propagator is a product of onsite unitaries and preserves every support. It conjugates each two-site interaction without changing its norm or support.

For an output observable $O_{BR}={O_{BR}}^{†}$ with $\Vert O_{BR}\Vert \le 1$, expand the interaction-picture Heisenberg evolution in its norm-convergent time-ordered nested-commutator series. An order-$n$ term has norm at most

Summing the ordered integrals gives $(2\Lambda {)}^n/n!$.

A nonzero commutator with an edge term requires that edge to meet the current support. Each commutator can extend the graph distance from $B$ by at most one. Thus every term of order $n<d$ acts as the identity on $A$. The same statement holds when the coefficients are arbitrary operators on $R$.

Let $O_{<d}$ be the partial sum of those terms, including the onsite conjugation. Its support excludes $A$, and the remainder has norm at most the tail in (eq:tail). The two intervention adjoints agree on $O_{<d}$ and are contractions in operator norm. Their difference on the remainder has norm at most twice the tail. Duality with the trace norm and the factor $1/2$ in (eq:signal) give (eq:tail).

For $n\ge d$, insert $1\le e^{\mu (n-d)}$ and sum the exponential series.