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The J gate is $J(\alpha )=HP(\alpha )$, where $P(\alpha )=\mathrm{diag}(1,e^{i\alpha })$: the phase gate acts first and the Hadamard gate acts second. Unitary invariance of the operator norm reduces its angle error to the single nonzero diagonal difference of the phase gate. Its norm is $|e^{i\delta }-1|=2|\sin (\delta /2)|\le |\delta |$. The quantum circuit gate-error telescoping bound then makes an angle tolerance of $ϵ/k$ sufficient for $k$ such imperfect gates and exact other gates.

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 isWrite the exact and implemented quantum circuits as ordered products $C=U_m\cdots U_1$ and $C^{\prime }=U_m^{\prime }\cdots U_1^{\prime }$. The quantum circuit gate-error telescoping bound follows fromEvery surrounding factor is unitary, including gates tensored with identities on other qubits, so the triangle inequality and the submultiplicativity of the operator norm giveThe exact Controlled-Z gates contribute zero to that sum. ThusThe endpoint $\eta =ϵ/k$ 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.