Theorem: Positive selective TV gap

The complete classical law is

Its distance from the zero-coefficient law is exactly

The complete output states, retaining the matched spectator and reference, also have trace distance $\Delta (d)$.

The gap is positive if and only if $r<1$ and $\nu d\ne 0$.

Proof. Apply the branch maps to the product input. The spectator factor is the same in every branch. The failure probability is unchanged. The two successful probabilities differ from their zero values by opposite quantities of magnitude $(1-r)|\tanh (\nu d/\sigma )|/2$. Summing their absolute differences and dividing by two gives (eq:TV).

For the complete outputs, orthogonality of the classical blocks makes the trace norm the sum of the block norms. The common normalized spectator factor has trace norm one, so the same expression results.