Theorem: Tensor-free snapshot–channel–adjoint equivalence

For the interface of Section sec:interfaces, the following are equivalent:

1. There exists a UCP map

   $$K:{\mathfrak{A}}_{\mathrm{out}}(Y)⟶{\mathfrak{A}}_{\mathrm{in}}(I)$$

   such that

   $$T\circ {\iota }_Y={\iota }_I\circ K.$$

2. For every pair of global input states,

   $$\omega \circ {\iota }_I=\nu \circ {\iota }_I⟹\omega \circ T\circ {\iota }_Y=\nu \circ T\circ {\iota }_Y.$$

3. The Heisenberg pullback satisfies

   $$T({\iota }_Y({\mathfrak{A}}_{\mathrm{out}}(Y)))\subseteq {\iota }_I({\mathfrak{A}}_{\mathrm{in}}(I)).$$

When these conditions hold, $K$ is unique and

on bounded functionals, and hence on states. No spatial complement or state-extension channel is used.

Proof. Equation (eq:algebrafactor) gives (eq:snapshotdet) by evaluation against states, and also gives (eq:containment).

Assume (ii). For a self-adjoint $b\in {\mathfrak{A}}_{\mathrm{out}}(Y)$, complete positivity makes $T({\iota }_Y(b))$ self-adjoint. Its expectation is equal on every pair of states with equal input snapshots. Lemma lem:separation places it in ${\iota }_I({\mathfrak{A}}_{\mathrm{in}}(I))$. Every output element is a complex linear combination of two self-adjoint elements, proving (iii).

Assume (iii). Define

It is linear and unital. For every matrix size $r$, the inclusion ${{\iota }_I}^{(r)}$ is a complete order embedding. If $[b_{ij}]\ge 0$ in $M_r({\mathfrak{A}}_{\mathrm{out}}(Y))$, then

in $M_r({\mathfrak{M}}_{\mathrm{in}})$. Its entries lie in the input local algebra, where the inherited positive cone is the same. Therefore $K$ is completely positive.

Injectivity of ${\iota }_I$ proves uniqueness. Taking dual maps in (eq:algebrafactor) gives (eq:dual-factor).