Theorem: Exact retained-source channel

The retained-source channel preserves diagonal matrix units and multiplies each off-diagonal one by

Equivalently,

Its half-diamond distance from the identity is $(1-{\chi }_F)/2$, and its Bell-input fidelity is $(1+{\chi }_F)/2$.

Proof. The coefficient of $|s⟩⟨t|$ after tracing the field is

The branch directions are collinear, so there is no Weyl phase. For $s\ne t$, $(s-t{)}^2=4$, giving (eq:sourcecoherence).

The resulting qubit channel is

On a Bell input its two Pauli outputs are orthogonal; the triangle inequality supplies the matching diamond upper bound. Their identity weight is the Bell fidelity.