Proposition: Access factorization and its invariant

A finite decoder program satisfying Definition def:accesscontract defines a CPTP map $D_A:\mathcal{B}(H_A)\to \mathcal{B}(C)$. Its extension to a complete boundary input is

Equal accessible restrictions therefore give equal returned states, including when an independent reference is retained.

Conversely, for a CPTP map $\Psi :\mathcal{B}(H_A\otimes H_{\bar{A}})\to \mathcal{B}(C)$, the following are equivalent:

1. $\Psi =D{\mathrm{Tr}}_{\bar{A}}$ for a CPTP map $D$;

2. equal $A$-marginals always give equal returned states;

3. ${\Psi }^{\ast }(\mathcal{B}(C))\subseteq \mathcal{B}(H_A)\otimes I_{\bar{A}}$.

Proof. Fresh preparation, accessible gates, complete instruments, classical control, and complete branch summation compose to a CPTP map on the received interface. Induction over the program cannot introduce dependence on an unreceived input.

Factorization gives the adjoint formula and snapshot determination. Conversely, let $h=h^{\ast }$ be an adjoint output observable under statement two. If a traceless Hermitian $M$ satisfies ${\mathrm{Tr}}_{\bar{A}}M=0$, then for sufficiently small positive $t$, the states $I/\dim H\pm tM$ have equal $A$-marginals. Thus $\mathrm{Tr}(hM)=0$. The Hilbert–Schmidt orthogonal complement of $\ker {\mathrm{Tr}}_{\bar{A}}$ is $\mathcal{B}(H_A)\otimes I_{\bar{A}}$, proving statement three. Complex linearity handles all output observables.

Under containment write ${\Psi }^{\ast }(O)=D^{\ast }(O)\otimes I$. The identity inclusion reflects matrix positivity, so $D^{\ast }$ is unital and completely positive. Its trace dual is CPTP and gives statement one. Reference amplification follows from complete positivity and the channel identity.