In QAOA 1, why do we pick the initial Hamiltonian $B$ to be $\sigma_x$ applied to each qubit? Would it be possible to pick $B$ to be an application of $\sigma_z$'s instead? Then $C$ and $B$ would be both diagonal in the Z-basis. What is preventing us from taking this choice for $B$? Thanks in advance!


1 Answer 1


We don't really need $B = \sum \sigma_j^x$ in our QAOA algorithm. As long as you pick it in such a way that it doesn't commute with $C$. One of the reason is if they are commute, then they share a common eigenvector. Then if you run into this type of situation, you will never get out, and you will be stuck in this state. You can think of $U(\beta, B)$ as a driver, it helps to navigate the Ansatze from getting stuck.

In term of the reason why $B = \sum \sigma_j^x$ in the first place is because QAOA is sort of a discretization of Quantum Annealing so that is why we see that the Ansatze of QAOA takes the form: $U = e^{-i\beta_p B}e^{-i \gamma_p C} \cdots e^{-i\beta_1 B} e^{-i \gamma_1 C} = \prod_{i} e^{-i\beta_i B} e^{-i \gamma_i C} $ which is a trotter approximation of the time evolution in the quantum annealing.

  • $\begingroup$ I always thought that B and C do commute because we simulate $H(t) = (1 - s(t)) B + s(t) C$ as $e^{-i (1 - s (j\Delta t )) B\Delta t - is(j \Delta t ) C\Delta t }=e^{-i (1 - s (j\Delta t )) B\Delta t } e^{-is(j \Delta t ) C\Delta t } $ . Is that then not right? $\endgroup$
    – Hannah
    Nov 28, 2020 at 11:52
  • $\begingroup$ @Hannah at infinite depth it doesn't matter. quantumcomputing.stackexchange.com/a/14039/9858 $\endgroup$
    – KAJ226
    Dec 2, 2020 at 0:42

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