QAOA circuit and expected combinatorial objective

Let $G=(V,E)$ be a finite simple undirected graph, $p\ge 1$, and ${\gamma }_1,\dots ,{\gamma }_p,{\beta }_1,\dots ,{\beta }_p\in \mathbb{R}$. On $\mathrm{\mid }V\mathrm{\mid }$ qubits define $C_G={\sum }_{(u,v)\in E}\frac{I-Z_u{Z}_v}{2},B_G={\sum }_{v\in V}{X}_v\mathrm{.}$ Prepare $\mathrm{\mid }+{⟩}^{\otimes \mathrm{\mid }V\mathrm{\mid }}$ and apply, in the declared right-to-left product convention, $U_{G,p}(\gamma ,\beta )={\prod }_{j=p}^1{e}^{-i{\beta }_j{B}_G}{e}^{-i{\gamma }_j{C}_G}\mathrm{.}$ The output state is $\mathrm{\mid }{\psi }_{G,p}(\gamma ,\beta )⟩=U_{G,p}\mathrm{\mid }+{⟩}^{\otimes \mathrm{\mid }V\mathrm{\mid }}$ and the objective is $⟨\psi \mathrm{\mid }{C}_G\mathrm{\mid }\psi {⟩}_D$. Measuring every qubit in the computational basis returns a random bitstring, hence a cut. Each cost and mixer exponential, graph-input access, parameter encoding, state preparation, measurement, repetition, and classical optimizer cost must be supplied and charged; this definition is not a guarantee of a maximum cut or approximation ratio.