"Stoquastic" Hamiltonians are sign-problem free Hamiltonians.


On attending a meeting last year, I heard the term stoquastic hamiltonian in a talk and noticed the term stoquastic hamiltonian was mentioned in a poster of quantum annealing. I myself work on a different area so I did not pay enough attention to the name of the talk and poster. So on two occasions I noticed the two terms stoquastic hamiltonian and quantum annealing were mentioned together. I also search the two terms together in google search engine and they do indeed appear together. This seems interesting to me at least.

Research effort:

  1. My initial guess is that by the use of stoquastic hamiltonian, the first order quantum phase transition during the minimum gap is eliminated (since the first order phase quantum transition is a difficulty in quantum annealing). However I cannot find evidence supporting this.

  2. The ability to perform quantum monte carlo with it. Reference of this reason: Monte Carlo simulation of stoquastic hamiltonians.

  3. On The Complexity of Stoquastic Local Hamiltonian Problems, Stoquastic Hamiltonian is common. Spin-1/2 models, the well-studied ferromagnetic Heisenberg models and the quantum transverse Ising model have it. So I think since they are sometimes used in quantum annealing, Stoquastic Hamiltonian is mentioned. Stoquastic Hamiltonian is common, while non-stoquastic Hamiltonian is uncommon. Therefore actually doing quantum annealing on non-stoquastic Hamiltonian would be interesting, like Non-Stoquastic Hamiltonians and Quantum Annealing of Ising Spin Glass. This is my another hypothesis why "Stoquastic" is stressed.

  • $\begingroup$ @kyle It is not my downvote, but I think it is deserved. You have asked about 4 pretty similar questions, rather than first asking one, waiting for the answer, learning something, and then continuing to ask based on what you learned. Moreover, the question is badly researched: You ask several things (What is S.H.? Why is it used? Phase transition?), and you just make random claims. Start by focusing on one question, and substantiate it: Give a number of references where S.H. is being used! $\endgroup$ Commented Mar 12, 2017 at 6:41
  • $\begingroup$ ... to iterate: You should really work on improving this question. And it is easy: Just list the places where you observed this! $\endgroup$ Commented Mar 12, 2017 at 6:47
  • $\begingroup$ You shouldn't argue with me. You should improve your question. If you don't know what a stoquastic Hamiltonian is, you should start by asking what this is. Once you receive an answer, you can digest it, and then go on and ask your next question. But you should start by improving the question above. And if you know the answer, you can even answer it yourself. $\endgroup$ Commented Mar 12, 2017 at 7:38
  • $\begingroup$ @Norbert Schuch i have improved it now. $\endgroup$
    – Ka Wa Yip
    Commented Mar 12, 2017 at 7:42
  • $\begingroup$ @Norbert Schuch If you find the question interesting, would you kindly spare an upvote so that I do not receive any ban warning anymore? Thanks. $\endgroup$
    – Ka Wa Yip
    Commented Mar 12, 2017 at 7:45

1 Answer 1


As far as I can see, there are two reasons why stoquastic Hamiltonians appear in the context of quantum annealing:

  • First, the typical scheme for quantum annealing is to have a classical (i.e., diagonal) Hamiltonian $H_0$ which has the solution to the problem as its ground state. Then, one starts with an equal weight superposition of all configurations, which is the ground state of $-\sum \sigma_x^i$, and interpolates $$ H(\theta) = - (1-\theta)\sum\sigma_x^i + \theta H_0\ . $$ This Hamiltonian is stoquastic.

  • Second, the D-Wave device, which is supposed to realize a quantum annealer, is described by a stoquastic Hamiltonian of the above type.

Other than that, I don't see why a quantum annealer should be restricted to stoquastic Hamiltonians (except that any restriction makes the scenario generally easier to analyze).

  • $\begingroup$ Thanks. This is very interesting to me. Can $\sum\sigma_x^i$ be replaced by $\sum\sigma_y^i$, although a bit weird (to me) to start with this? $\endgroup$
    – Ka Wa Yip
    Commented Mar 13, 2017 at 0:11
  • $\begingroup$ Yes. Indeed, the two are equivalent up to a rotation on each qubit (The proof is left as an exercise), so they have all the same properties. Except that the latter is not stoquastic, from which we infer that stoquastic is a basis-dependent concept. $\endgroup$ Commented Mar 13, 2017 at 9:02
  • 1
    $\begingroup$ Thanks. This is again interesting to me that the latter is not stoquastic. Just curious (1 qubit): will $-2(1-\theta)\sum\sigma_x^i$ or $-(1-\theta)(2\left|0\right>\left<1\right|+\left|1\right>\left<0\right|)$ being the first part modify the stoquasticity of $H(\theta)$? $\endgroup$
    – Ka Wa Yip
    Commented Mar 13, 2017 at 12:19
  • $\begingroup$ @kyle If you have understood what stoquastic is, you should be able to figure this out yourself! (Note that what you wrote is not even hermitian, so it cannot be a Hamiltonian. Is that a typo?) $\endgroup$ Commented Mar 13, 2017 at 19:27

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