= Solution
Use the <operator norm> induced by the usual vector norm, and assume the input <quantum state> is normalized. Since $J(\alpha)=HP(\alpha)$ and the <Hadamard gate> is unitary, the <J-gate phase-error operator norm> is
$$
\|J(\alpha')-J(\alpha)\|
=\|P(\alpha')-P(\alpha)\|
=|e^{i\alpha'}-e^{i\alpha}|
=2\left|\sin\frac{\alpha'-\alpha}{2}\right|
\leq|\alpha'-\alpha|<\eta.
$$
Write the exact and implemented <quantum circuits> as ordered products $C=U_m\cdots U_1$ and $C'=U'_m\cdots U'_1$. The <quantum circuit gate-error telescoping bound> follows from
$$
C'-C=\sum_{j=1}^mU'_m\cdots U'_{j+1}(U'_j-U_j)U_{j-1}\cdots U_1.
$$
Every surrounding factor is unitary, including gates tensored with identities on other <qubits>, so the <triangle inequality> and the <submultiplicativity of the operator norm> give
$$
\|C'-C\|\leq\sum_j\|U'_j-U_j\|<k\eta.
$$
The exact <Controlled-Z gates> contribute zero to that sum. Thus
$$
\||\psi'_{\rm out}\rangle-|\psi_{\rm out}\rangle\|
\leq\|C'-C\|<k\eta,
\qquad
\boxed{0<\eta\leq\frac{\epsilon}{k}\quad(k\geq1)}.
$$
The endpoint $\eta=\epsilon/k$ is sufficient because each implemented angle error is strictly smaller than $\eta$. If $k=0$, the circuits are identical and any positive $\eta$ works. The bound controls the stated vector distance with actual gate phases retained, so no adjustment of the global phase of one output is needed.
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