Theorem: C3: sharp one-live-qubit recovery error

Under Assumption ass:stream, every deterministic decoder with $Q_{\mathrm{peak}}\le 1$ has

The lower bound permits arbitrary classical storage and arbitrary single-qubit instruments. It is attained in ${\mathfrak{L}}_{\mathrm{Cl}}$. Hence

Proof. Before carrier $2$ can enter a one-slot laboratory, carrier $1$ must cease to be a live quantum system. No intermediate quantum output or external quantum scratch register is permitted. The complete influence of the first carrier is therefore a classical measurement history, described by a POVM effect $E_h$.

Lemma lem:firstmeasurement makes the conditional state of $R23$ separable for every such history. All subsequent processing acts on the observer side, conditioned on classical data. It preserves separability between $R$ and the final recovered output.

A separable two-qubit state has Bell overlap at most $1/2$: for a pure product it is at most $1/2$ by Cauchy–Schwarz, and convexity extends the bound. Testing the Bell projector gives recovery trace distance at least $1/2$, and therefore the same half-diamond lower bound.

For attainment, read $Z_1,X_2,Z_3$ sequentially and report the negative product of their eigenvalues. This is logical $X$-readout by (eq:logicalreps). After consuming all three carriers, allocate and return the corresponding $X$-eigenstate. The unreported quantum channel is

Thus $d_{\diamond }({\Delta }_X,\mathrm{id})\le 1/2$, and the Bell test gives equality. An $X$-readout needs one $H$; preparation needs $H$ and at most one $Z$. The implementation uses peak one and at most three unitary gates.