Proposition: Implemented phase allowance

If the residual accessible phase error is $\delta$, the noiseless specified decoded channel is

Its error is at most

With additional channel and decoder certificates $\xi ,\zeta$, the bound becomes

Proof. The logical phase is twice the one-particle phase. For $U=\mathrm{diag}(1,e^{2\mathrm{i}\delta })$, a purified input overlap between the identity and unitary outputs is $\mathrm{Tr}(\rho U)$. The numerical range is the segment between the two eigenvalues; its minimal modulus is $|\cos \delta |$. The pure-state distance is therefore maximized at $|\sin \delta |$. The channel triangle inequality gives the first bound, and telescoping gives the second.