Why does the problem Hamiltonian of QAOA always consist of $Z$ and $I$ gates?

Question

I noticed that in QAOA the problem hamiltonian always consists of $Z$ and $I$ gates. But isn't QAOA a form of Adiabatic Programming? Where the idea is just to go from one ground state to another? Does that not mean that every Hamiltonian can be the problem Hamiltonian?

And the Mixer consists always just of $X$ gates. Why not $Y$ gates for example?

Answer 1

QAOA was first introduced as a quantum algorithm to tackle combinatorial optimization problems. Hence, there is a direct translation of these to Hamiltonians with $I$ and $Z$ operators. And you always see the mixer as just $X$ operators because

1. the equal-superposition state (obtained by Hadamard transform on $| 0\rangle^{\otimes N}$) is a ground state of the mixer (which is a requirement in adiabatic computing)
2. the type of problems tackled do not include constraints.

However, other mixers can be used depending what one wants to solve (see this article on modifying the mixers for constrained combinatorial optimization problems).