This is a question that must be asked a lot but is it possible? And if in theory possible, what are the challenges to do it, and what are the drawbacks?

Adiabatic Quantum Computation is simply the time-evolution of a Hamiltonian where the system is prepared in a particular initial state (the ground state) and the Hamiltonian varies slowly in time.

Simulating a Hamiltonian on a quantum computer is a standard problem. Making that Hamiltonian time varying doesn’t really make it any worse since you break it down into lots of little steps to simulate it anyway, and the slow variation means that it’s not a bad approximation to assume that theHamiltonian is fixed in each of those little steps. Preparing the initial state is also not a problem. By design, it’s something that is easily prepared, such as the all zero state.

The challenge, as with any simulation, is that there’s a polynomial overhead compared to the original scheme. While that doesn’t change computational complexity, it can have a massive impact on the practicality of implementation. What you gain, on the other hand, is the ability to use fault-tolerant techniques.

Adiabatic Quantum Computation is Equivalent to Standard Quantum Computation.

This paper proved the other direction, that adiabatic can simulate circuit model. The direction you ask, which is whether or not circuit model can simulate adiabatic, was proven much earlier. In the paper I linked at the top there is the sentence "It is known that adiabatic computation can be efficiently simulated by standard quantum computers [9, 13]."

One of those papers [13], is basically where Adiabatic Quantum Computation was first introduced:

E. Farhi, J. Goldstone, S. Gutmann, J. Lapan, A. Lundgre n, and D. Preda, A quantum adiabatic evolution algorithm applied to random instances of a n NP-complete problem, Science , 292(5516):472–476, 2001.

So what that says is, that right from the beginning of Adiabatic model, people knew that circuit model could simulate it.

  • I'd be interested to see where the cited Ref. [13] shows that. I don't think they do. (Also, that's not the point of the paper.) – Norbert Schuch Nov 10 at 16:56
  • Aharanov et al. referenced [9,13]. I said that, then said that [13] is one of the first papers on Adiabatic Quantum Computation. – user5062 Nov 10 at 17:08
  • I believe the latter. But I don't think they talk about the circuit model. – Norbert Schuch Nov 10 at 17:25
  • On a different note: Why did you reject my edit inserting a reference to the paper? Do you think having a link to the paper does not improve the answer? (After all, it allows to more easily check what that paper actually does!) Flagged for mod attention. – Norbert Schuch Nov 10 at 17:26
  • I'm not sure. I thought it was fine how it was. Can you show it to me again? – user5062 Nov 10 at 17:32

The first difficulty is to answer whether we can convert all conventional circuit gates into adiabatic ones (or simply geometric), this has been done theoretically and recently experimentally as well (ref: https://arxiv.org/abs/1304.5186). To answer the first question, yes, it is possible. Since this correspondence is now true, we need to tackle the problems arising in the evolution of such gates. The challenges and drawbacks could be:

1) Experimentally constructing such systems which can be easily controlled classically (because the driving parameter space in an adiabatic scenario is mostly a classical parameter such as an external magnetic field) and adiabatically. These are now quite possible and easily exploitable and most such systems are optical cavities which interact with spin systems inside.

2) The time complexity and speed. This is the biggest issue with adiabatic evolution, it is very slowly evolving. This is the biggest price to pay w.r.t other models.

Advantage: Very fault tolerant, the speed of the evolution and path doesn't really matter once a particular operation is targeted. Slower or faster don't matter in the adiabatic domain. For operations, even paths in the driving parameter space don't matter up-to a phase.

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.