Proposition: Complete instrument and nontrivial quantum output

Equation (eq:Bellchannel) is CPTP. Its selective quantum maps are the declared local Lueders maps after the source gate. It is not a constant classical sampler.

Moreover, for product inputs between $W$ and the other input factors, its action on $W$ is exactly ${\mathrm{id}}_W$. With an external reference entangled with $W$, that entanglement is retained. Consequently the complete channel is not entanglement breaking.

Proof. The Kraus operators of the full channel are the operators which read the setting basis $⟨st|$, apply $K_{ab\mid st}$, and write the output setting and outcome basis vectors. They satisfy

for every $s,t$. Summing over the orthogonal input setting projectors gives the full identity. Thus the channel is CPTP.

The operators in (eq:BellKraus) contain $I_W$, so the channel factors as the displayed Bell instrument tensored with ${\mathrm{id}}_W$, with the conventional tensor ordering understood. This preserves every state of $W$ and its untouched reference when the other factors are fixed independently.

For an explicit nonconstant record example, take $s=1$, $t$ replaced, if desired, by an admitted right $X$-setting payload modification. The source inputs $|00⟩$ and $|10⟩$ produce ${\Phi }^{+}$ and ${\Phi }^{-}$, whose $X\otimes X$ correlations have opposite signs. More generally the projectors in (eq:BellKraus) directly distinguish input coherences in their admitted measurement bases. Thus the selective channel contains more than one fixed output law.

Finally, fix all other inputs and put $W$ in a Bell state with an untouched reference. The $W$–reference marginal remains that Bell state after the channel. An entanglement-breaking channel could not have this property.