The operators $X_i$ act on different <qubit>[qubits] and therefore <commuting operators>[commute], so
$$
e^{-itJ_X}=\prod_{i=1}^ne^{-itX_i}.
$$
Since $X=HZH$,
$$
e^{-itX}=H e^{-itZ}H
=e^{-it}H\operatorname{diag}(1,e^{2it})H.
$$
The scalar $e^{-it}$ is a <global phase>. Thus each factor uses two <Hadamard gates> and one <phase gate>, and applying the factors in parallel or sequentially gives an exact <quantum circuit> of $\boxed{3n}$ elementary gates.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 324 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-324/3/b/solution