The set-up:

Consider the Ising model on an $L \times L$ square lattice, where the coupling of each bond is chosen to be $+J$ (ferromagnetic) with probability $(1-p)$ and $-J$ (antiferromagnetic) with probability $p$.

For $p=0$ we have the standard Ising ferromagnet with two ground states (all up and all down). For $p \lesssim 0.11$ we will still have a ferromagnet, and the ground states still fall into the classification of 'mostly up' and 'mostly down', but there will be regions of opposing spins. These can take various different shapes while not changing the energy, and so introduce a large degeneracy into the ground state space. But the different shapes can be locally moved between by rotations of $O(1)$ spins at an $O(1)$ energy cost, so they are not like the deep wells of a spin glass.

The question:

Suppose we wish to calculate the thermal average of the energy at a temperature below the Curie temperature (and hence within the ferromagnetic phase). How many Monte Carlo steps are required to do this (equivalenlty: what is the autocorrelation time), assuming that we start in an initially random state? How much does it depend on the Monte Carlo used (such as Metropolis or Heat Bath, and with or without parallel tempering)?

Since energy only depends of the length of domain walls, and so doesn't care whether the system is in a 'mostly up' or 'mostly down' state, my intuition makes me think that it is not necessary to wait for the $\exp(L)$ required for the system to flip between these. So just local changes should be enough, and hence the autocorrelation should be polynomial at most. Is this true? Another issue is that the initially random state will be around 50% spin up and 50% spin down, and hence needs to break the symmetry to move towards a ground state. What is the time required to do this?

The background

There's a lot of statements out there in the literature about the feasibility of Monte Carlo studies in certain cases. They tell you that frustration (which this model has) is bad and will lead to inefficient Monte Carlo. They tell the same for spin glasses (which this model is at $T=0$ and high $p$, but not in the regime discussed above). But definitions of inefficiency seem to vary over time and between authors. For me it means higher than polynomial wih $L$. But for some people even polynomial scaling seems enough to claim inefficiency. It seems difficult to find a consistent overview of the issue. But this is a model that I have a particular interest in, and would really like to know at least what the autocorrelation time for the energy is.

  • $\begingroup$ Not really an answer, just a suggestion, but if you're content to just sample one half of the bimodal ground state distribution and assume the other half to be identical by symmetry (which I agree is a good idea), then you might as well start with an initial state with already broken symmetry (e.g. the all down state). That way, you avoid having to spontaneously break the initial symmetry, which can be slow since the symmetry can locally break in different directions, leaving domain walls that can take time to disappear. I'll see if I can give a better answer after taking a look in Liggett. $\endgroup$ – Ilmari Karonen May 3 '13 at 14:38
  • $\begingroup$ In numerics, you usually don't program the number of steps required to converge, but you check up convergence in real time, by tracking the fluctuations of the acceptance rate for Monte-Carlo moves. In the development stage, I recomment that you should be able to stop at one point and then go on, that is you need to be able to load/save a configuration. $\endgroup$ – Tom-Tom Sep 21 '15 at 13:21

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