Theorem: Exact peak two on the declared triple

Consider streaming programmes with access exactly to $\{1,2,3\}$, arbitrary arrival order, no additional source-correlated quantum resource, and no undeclared quantum memory. The following programme recovers exactly with decoder peak two:

1. hold qubits $1,2$ and measure $\{{\Pi }_{pq}\}$;

2. retain $(p,q)$ classically and discard qubits $1,2$;

3. receive qubit $3$ and apply $R_{pq}$.

No legal exact programme in this streaming model has peak one.

Proof. The adjoint of the printed programme sends

It sends $Y=iXZ$ to the corresponding logical $Y$ representative, since ${R_{pq}}^{†}Y{R}_{pq}=pqX$. Together with the identity these operators span the output algebra. The programme therefore equals $D_B$, proving exact recovery. Its live peak is two.

For the lower bound, suppose an exact peak-one programme exists. Before another input qubit can arrive, the first input qubit must have been discarded, leaving only a finite classical record. Operations on that first qubit define a POVM $\{E_m\}$.

By Theorem thm:unique, the complete decoder is $D_B$. Every Kraus operator of $D_B$ has the form $(I_2\otimes ⟨v|){U_B}^{†}$. For a refinement according to the first record $m$, the corresponding effect is therefore in the commutant of $X_B$ and $Z_B$. But that effect is $E_m\otimes I$ on the three inputs. At every individual site the local factors of $X_B$ and $Z_B$ anticommute. Hence $E_m$ commutes with two independent Pauli axes and must equal $p_m{I}_2$.

After discarding the first qubit, record $m$ and the other two qubits are thus described by $p_m$ times the two-share reduced channel. Lemma lem:privacy makes that channel constant. No continuation can recover the logical qubit, a contradiction.