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

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

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?

$$\gamma$$ 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 $$U1(x \pm 2\pi)$$