Proposition: Non-Cliffordness is neither necessary nor sufficient

There is a Clifford coherent program with

There is a non-Clifford coherent program with

Proof. For the first example, allocate $B$ in $|0⟩$ and apply ${\mathrm{CNOT}}_{C\to B}$, returning the original carrier as $R$. Then

and both marginals are computational dephasing

The two residual basis outputs are orthogonal, giving ${\delta }_{\mathrm{full}}\ge 1/2$. The replacement $I_2/2$ gives the matching upper bound: for every referenced input the two diagonal reference blocks multiply output differences of trace norm one. Thus the half-diamond distance is at most $1/2$.

The inputs $|+⟩$ and $|-⟩$ have identical radiation outputs. A decoder of error $\epsilon$ therefore satisfies $1\le 2\epsilon$. The identity decoder has error $1/2$, proving the recovery formula.

For the second example, let

The complement is ${\mathcal{T}}_{|0⟩⟨0|}$, and $T^{†}$ recovers exactly. The gate is non-Clifford because $TX{T}^{†}=(X+Y)/\sqrt{2}$, which is not a Pauli. For input $|+⟩$, the output has Bloch components $(1/\sqrt{2},1/\sqrt{2},0)$, outside the convex hull of the six Pauli eigenstates.