3 added 10 characters in body edited Apr 1 '18 at 15:32 Andrew O 1,45911 gold badge77 silver badges1717 bronze badges One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $$T \to \infty$$ which is impractical. In addition if $$T$$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $$\tau$$, using quantum annealing on a 5-qubit system, you only likely to find 2 of the 3 possible states. One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $$T \to \infty$$ which is impractical. In addition if $$T$$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $$\tau$$, using quantum annealing on a 5-qubit system, you only find 2 of the 3 possible states. One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $$T \to \infty$$ which is impractical. In addition if $$T$$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $$\tau$$, using quantum annealing on a 5-qubit system, you only likely to find 2 of the 3 possible states. 2 added 109 characters in body edited Apr 1 '18 at 5:25 Sanchayan Dutta 8,86844 gold badges1919 silver badges6666 bronze badges One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $$T \to \infty$$ which is impractical. In addition if $$T$$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $$\tau$$, using quantum annealing on a 5-qubit system, you only find 2 of the 3 possible states. One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $$T \to \infty$$ which is impractical. In addition if $$T$$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $$\tau$$, using quantum annealing on a 5-qubit system, you only find 2 of the 3 possible states. Matsuda et al. https://arxiv.org/abs/0808.0365v3 One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $$T \to \infty$$ which is impractical. In addition if $$T$$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $$\tau$$, using quantum annealing on a 5-qubit system, you only find 2 of the 3 possible states. 1 answered Apr 1 '18 at 5:17 Andrew O 1,45911 gold badge77 silver badges1717 bronze badges One of the advantages, as stated in the paper you linked, is that with QAOA you can increase the precision arbitrarily, whereas QA will only find the solution with probability 1 as $$T \to \infty$$ which is impractical. In addition if $$T$$ is too long you're likely to not find the solution as the probability is not monotonic. I believe an example of this can be found in a fair-sampling paper by Matsuda et al. Figure 4 shows that for large $$\tau$$, using quantum annealing on a 5-qubit system, you only find 2 of the 3 possible states. Matsuda et al. https://arxiv.org/abs/0808.0365v3