Theorem: One named observable from one declared region

For the model of Sections sec:objects–sec:boundary, the independently declared region $B$, the named observable ${\mathcal{O}}_{\mathrm{bulk}}$ in (eq:named-observable), and the map (eq:RB), the following statements hold.

1. The boundary representative belongs to the declared algebra: ${\mathcal{O}}_B\in \mathcal{A}(B)$.

2. Exact operator reconstruction holds:

   $${\mathcal{O}}_{B}U=U{\mathcal{O}}_{\mathrm{bulk}},U^{†}{\mathcal{O}}_{B}U={\mathcal{O}}_{\mathrm{bulk}}.$$

3. ${\mathcal{R}}_B$ is completely positive and trace preserving.

4. Exact operational reconstruction holds:

   $${\mathcal{R}}_B\circ {\mathcal{N}}_B={\mathcal{M}}_{{\mathcal{O}}_{\mathrm{bulk}}}.$$

   Equivalently, in the Heisenberg picture,

   $${\mathcal{N}}_B^{\ast }\left({\mathcal{R}}_B^{\ast }(|1⟩⟨1|)\right)={\mathcal{O}}_{\mathrm{bulk}}.$$

5. For every finite reference system $R$ and every state ${\rho }_{\mathrm{bulk},R}$,

   $$({\mathcal{R}}_B\otimes {\mathrm{id}}_R)((U\otimes 1_R){\rho }_{\mathrm{bulk},R}(U^{†}\otimes 1_R))=({\mathcal{M}}_{{\mathcal{O}}_{\mathrm{bulk}}}\otimes {\mathrm{id}}_R)({\rho }_{\mathrm{bulk},R}).$$

   In particular,

   $$d_{\diamond }({\mathcal{R}}_B\circ {\mathcal{N}}_B,{\mathcal{M}}_{{\mathcal{O}}_{\mathrm{bulk}}})=0.$$

The metric and the bulk–boundary representation correspondence are inputs. Metric reconstruction and derivation of Einstein dynamics are not conclusions of this theorem.

Proof. Membership was proved after (eq:boundary-representative). Lemma lem:smearing and unitary covariance of finite-matrix functional calculus give

This proves the operator identities.

Since $Q_B$ is self-adjoint, ${\mathcal{O}}_B$ is an effect; hence $F_0,F_1$ are positive and sum to $1$. A Stinespring isometry for (eq:RB) is

Indeed, $W^{†}W=F_0+F_1=1$. Tracing out the last two factors yields (eq:RB). Thus ${\mathcal{R}}_B$ is CPTP.

The operator identity gives, for each outcome,

with $U^{†}{F}_{1}U={\mathcal{O}}_{\mathrm{bulk}}$ and $U^{†}{F}_{0}U=1-{\mathcal{O}}_{\mathrm{bulk}}$. This proves the channel identity.

With a reference retained, the unnormalized reference block associated with outcome $r$ is

Unitary transport of the partial trace gives the corresponding bulk block with effect $U^{†}{F}_{r}U$. Equality of both classical blocks proves (eq:reference-exact). Equality of channels gives zero diamond distance.