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Background: The quantum annealing process performed by the D-Wave machines is described as follows: We have a problem Hamiltonian, $H_{P}$, whose ground state encodes the solution to a problem of interest, and a 'disordering' Hamiltonian, $H'$, which does not commute with $H_{P}$. The energy of the D-Wave system can be described by the Hamiltonian $H = H_{P} + \Gamma H'$, where $\Gamma$ changes from a large value to $0$ throughout the computation, so that $H=H_{P}$ at our computation's completion. http://vanilla47.com/PDFs/Quantum/1/An%20Introduction%20to%20Quantum%20Annealing.pdf

Question: What state is the system set up in (so what is $H(0)$)? Is it an easy to encode Hamiltonian, such as the one used in adiabatic quantum computing (normally labelled $H_{0}$)? Or my other guess would be that $\Gamma$ is initially $0$, so $H=H_{P}$ initially, and then $\Gamma$ is increased to enable 'quantum tunneling' through barriers separating local minima (to help the system get into its ground state), and then $\Gamma$ is slowly decreased to $0$ again.

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  • $\begingroup$ Isn't your question answered in this answer: physics.stackexchange.com/a/302864/4888 $\endgroup$
    – Norbert Schuch
    Commented Jan 5, 2017 at 0:16
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    $\begingroup$ 1. Annealers are subsets of adiabatic quantum computers. 2. I don't think the term "quantum annealer" is cleanly defined, different people might understand different things. 3. Are you talking of a general annealer or the D-Wave machine? -- I think your questions would profit if you would try to keep them more crisp! (Currently, it feels like every question is about everything, somehow.) $\endgroup$
    – Norbert Schuch
    Commented Jan 5, 2017 at 1:47
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    $\begingroup$ I'd say they are adiabatic quantum computers, since they operate by changing a Hamiltonian adiabatically. I don't think you are required to operate an adiabatic qc in the gapped/adiabatic regime. That's more about the algorithm you run on the device. $\endgroup$
    – Norbert Schuch
    Commented Jan 5, 2017 at 2:32
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    $\begingroup$ Cross-posted to cstheory.stackexchange.com/q/37255 $\endgroup$
    – Norbert Schuch
    Commented Jan 6, 2017 at 18:10
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    $\begingroup$ For the very least, you should declare this yourself, so people don't write answers twice. (Wouldn't you find this annoying to write an answer just to see someone else has written the same on the other site?) $\endgroup$
    – Norbert Schuch
    Commented Jan 9, 2017 at 11:33

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In the D-Wave setup (and most other setups), the initial Hamiltonian $H'$ is a local magnetic field on each spin (in the D-Wave case, in a basis dual to the basis of the Ising glass to be solved), and $\Gamma$ is large. Since the ground state of this Hamiltonian is simple, one expects the system to easily thermalize into this state.

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  • $\begingroup$ Thanks for your answer Norbert. But does your description fit the quantum annealing model of $H=H_{P} + \Gamma H'$ with $\Gamma$ large initially? For no matter how large $\Gamma$ is initially, $H \neq H'$ initially in this model. Your description seems to fit the adiabatic model where one evolves the Hamiltonian like $H = (1-t/T)H_{0} + (t/T)H_{P}$ better. $\endgroup$
    – Alex Michael
    Commented Jan 5, 2017 at 1:39
  • $\begingroup$ I don't think it matters so much. $\endgroup$
    – Norbert Schuch
    Commented Jan 5, 2017 at 1:47
  • $\begingroup$ Maybe it doesn't matter so much, but it does seem strange that D-Wave would publish a paper with an incorrect formula in it $\endgroup$
    – Alex Michael
    Commented Jan 5, 2017 at 2:45
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    $\begingroup$ I'm saying it does not matter so much how you do the interpolation. I don't say the formula is wrong. It's just that slightly different things can be termed quantum annealers. $\endgroup$
    – Norbert Schuch
    Commented Jan 5, 2017 at 2:47

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