Theorem: Intrinsic hybrid observation criterion

Let $\mathfrak{F}:{\mathfrak{X}}_{\mathrm{in}}\to {\mathfrak{X}}_{\mathrm{out}}$ be a channel and let ${\mathcal{O}}_{\mathrm{in}},{\mathcal{O}}_{\mathrm{out}}$ be observation spaces. The following are equivalent:

1. for all hybrid states,

   $$\rho {|}_{{\mathcal{O}}_{\mathrm{in}}}=\sigma {|}_{{\mathcal{O}}_{\mathrm{in}}}⟹\mathfrak{F}(\rho ){|}_{{\mathcal{O}}_{\mathrm{out}}}=\mathfrak{F}(\sigma ){|}_{{\mathcal{O}}_{\mathrm{out}}};$$
2. $${\mathfrak{F}}^{\ast }({\mathcal{O}}_{\mathrm{out}})\subseteq {\mathcal{O}}_{\mathrm{in}}.$$

For the direct-sum spaces ${\mathcal{O}}_{\mathrm{in}}={⨁}_g{\mathcal{A}}_g$ and ${\mathcal{O}}_{\mathrm{out}}={⨁}_h{\mathcal{B}}_h$, this is equivalent to

Proof. The inclusion implies agreement by pulling each output observable back to the input space.

Conversely, let $H=H^{†}\in {\mathfrak{X}}_{\mathrm{in}}$ annihilate the Hermitian part of ${\mathcal{O}}_{\mathrm{in}}$ under the trace pairing. Because $1\in {\mathcal{O}}_{\mathrm{in}}$, $\mathrm{Tr}H=0$. Choose a faithful hybrid state $\omega$; a multiple of the block identity suffices. For sufficiently small $a>0$, $\omega \pm aH$ are states agreeing on ${\mathcal{O}}_{\mathrm{in}}$. The assumed output agreement implies

Finite-dimensional double orthogonal complementation places ${\mathfrak{F}}^{\ast }B$ in the Hermitian part of ${\mathcal{O}}_{\mathrm{in}}$. Complex linearity gives (eq:observation-inclusion).

If the spaces are direct sums, an output observable may have one nonzero block $B_h$. Formula (eq:hybrid-adjoint) then makes (eq:observation-inclusion) imply every (eq:branch-inclusion). The reverse implication follows by summing those branch images.