When wanting to find the groundstate of this Hamiltonian with QAOA: \begin{equation} H_{C} =\sum_{i }^{n}(1 - Z_{i})/2 + \sum_{\{i,j\}\in \overline{E} } - 2(1 - Z_{i})(1 - Z_{j})/4 \end{equation} whose the Hamiltonian Simulation is: \begin{equation}\label{eq:qubo} e^{ -i\gamma H_{C}} =\prod_{i}^{n} U1(-\gamma)_{i}\prod_{\{i,j\}\in \overline{E} } CU1(2\gamma)_{(i,j)} \end{equation}

is $\gamma \in [0,2 \pi]$ or $\gamma \in [0,\pi]$ because of $2 \gamma$ in $CU1(2\gamma)_{(i,j)} $?

And how are $ U1(2\pi -\gamma)_{i}$ and $U1(-\gamma)_{i}$ distinguished?


1 Answer 1


$\gamma$ should still go from $[0, 2\pi]$, as $U1$ also has domain on $[0, 2\pi]$. See https://qiskit.org/documentation/stubs/qiskit.circuit.library.U1Gate.html. $U1$ is cyclic mod $2\pi$ so in general one cannot distinguish $U1(x \pm 2\pi)$


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