Theorem: Compositional observation principle

Let $\mathcal{A}\subseteq L({\mathcal{H}}_{\mathrm{in}})$ and $\mathcal{B}\subseteq L({\mathcal{H}}_{\mathrm{out}})$ be unital ${}^{\ast }$-subalgebras, and let $\Theta$ be a channel. On the unrestricted matrix-state spaces, the following are equivalent:

1. for every pair of states,

   $${\omega }_{\rho }{|}_{\mathcal{A}}={\omega }_{\sigma }{|}_{\mathcal{A}}⟹{\omega }_{\Theta (\rho )}{|}_{\mathcal{B}}={\omega }_{\Theta (\sigma )}{|}_{\mathcal{B}};$$
2. $${\Theta }^{\ast }(\mathcal{B})\subseteq \mathcal{A}.$$

Proof. Condition (ii) implies (i) by pulling each output observable back.

Conversely, assume (i). Let $H=H^{†}$ satisfy

Since $I\in \mathcal{A}$, $\mathrm{Tr}H=0$. With $d=\dim {\mathcal{H}}_{\mathrm{in}}$, choose $\epsilon >0$ small enough that

are positive. They are states agreeing on $\mathcal{A}$. For every Hermitian $B\in \mathcal{B}$, assumption (i) gives

Thus ${\Theta }^{\ast }(B)$ is orthogonal, in the real Hilbert space of Hermitian matrices, to the orthogonal complement of the Hermitian part of $\mathcal{A}$. Finite-dimensional double orthogonal complementation puts it in that Hermitian part. Complex linearity proves ${\Theta }^{\ast }(\mathcal{B})\subseteq \mathcal{A}$.