Finite programs and quantum return

This is a semantic existence statement, not finite-gate compilation, efficient synthesis, laboratory realization, or a proof about an unspecified larger calculus. An infinite-dimensional decoder supplied by Theorem thm:exact need not have a finite program.

For finite encodings, the interfaces are:

The local decoder can be written as:

w := new; (Cout,G) := DecoderUnitary(R,w); k := Enter(G); Drop(k); returnQ(Cout);

All unused work wires belong to $G$. The consumed qubits are work or syndrome qubits, not the recovered diary output.

Applying it locally to the coherent body gives the joint observable channel

The observer's first-factor marginal is $\mathcal{DN}$. A strict one-register observable implementation is:

(R,B) := CoherentBody(C); k := Enter(B); Drop(k); Cout := Decode(R); returnQ(Cout);

Its channel is again $\mathcal{DN}$, by Lemma lem:erasure-interface. This added readout is not inferred from an autonomous Hamiltonian.

The joint observable return $C_{\mathrm{out}}B$ does not silently become the complete physical dilation of the decoder. Decoder work, measurement carriers, and omitted apparatus outputs must be added to that complete ledger.

For a prescribed terminal observation $\mathcal{M}$, contraction gives

The converse for one chosen observation is false: computational dephasing preserves computational readout but removes off-diagonal matrix units.