Spin Glasses are known to converge to their ground state under Simulated Annealing.

The word choice is especially interesting since annealing is also the name of a process performed on actual glass. However, in both cases the basic principle is the same: by gradually cooling the system, we can get it to relax into a stable, low-energy configuration.

Borrowing another concept from glass working, is there a spin glass version of a Prince Rupert's Drop? (Here's a video, too.) In other words, is it possible to "freeze" a spin glass into a higher-energy state by cooling it rapidly? If so, would this metastable state suddenly transition to a much lower energy state if a small perturbation (like a tiny $\vec H$-field) were applied?

Do non-frustrated spin systems like the square-lattice Ising model demonstrate similar properties?

  • $\begingroup$ Title discussed in this meta post. $\endgroup$
    – Qmechanic
    Dec 7, 2015 at 13:07

2 Answers 2


Perhaps this isn't quite the answer you're seeking, but you may be interested in the phenomena of thermoremanent and isothermal remanent magnetization.

Basically, if you perform a deep quench on a spin glass (i.e., freeze the spins) in a uniform external magnetic field and then, after some time, switch off the magnetic field, (or alternatively, quench first and temporarily add a field later) it turns out that the spin glass retains some internal magnetization that slowly decays with time, roughly like some power law $\exp{(-t^\alpha)}$ where $0< \alpha <1$ (as noted in on p.818 of Binder & Young's 1986 paper).

In fact, if you adopt the first procedure and keep track of how long the quenched spin glass sits in the field, it turns out that after removing the field, the decay in magnetization actually depends on how long the field was on. This is known as aging - see this paper for a more thorough investigation.

To relate back to your question about simulated annealing, the spin glass is 'mostly stuck' near whatever configuration it happened to be in just before the deep quench (presumably a configuration with lots of spins attempting to align with the field present). After removing the field, the system does undergo relaxation like in classic annealing, but perhaps on timescales of much greater lengths (hence the slow power law decay).


Yes, yes and yes.

If you cool a spin glass rapidly (or, more likely to happen in a computer model: start it in a completely random state, which corresponds to infinite temperature, and immediately set the temperature to a low value) then it will indeed end up frozen into a higher-energy state than the ground state. This is known as "quenching." There would be no need for simulated annealing if this didn't happen - the purpose of the annealing is to try and avoid it.

In fact you usually end up in a "meta-stable" state - it will eventually decay to a lower-energy state, but this is likely to take a long time, because the system has to climb up a large energy hill before it can fall down the other side. In spin glasses the time taken to reach equilibrium can become infinite.

Making a small perturbation to such a state can indeed cause a rapid transition to a lower energy state. When there is frustration, flipping a spin can reduce the energy of one coupling while raising the energy of another. This can lead to a "cascade" or "avalanche" of many spins flipping one after the other. The dynamics of a spin glass at a low but finite temperature consist of multiple such avalanches, the size of which (I believe) obey a power law distribution.

Finally, something similar can indeed happen in the Ising model. Suppose you start the Ising model with a smallish positive external magnetic field and a high temperature. If you now reduce the temperature to a very low value, you will end up in a state where almost all the spins are aligned in the same way, with only a small probability of fluctuating in the other direction.

Suppose you now change the external field so that it has a smallish negative value. Since the model is frozen, nothing much will happen at first. Small domains of the opposite spin will tend to shrink to zero size, because the cells on the boundary of the domain are on average surrounded by more cells in the original state than the new one. However, you can note that the system is no longer in its ground state - it would have lower energy if all the spins were flipped the other way. It turns out that although small perturbations shrink, a large enough one will grow. So if due to a thermal fluctuation or an external perturbation you end up with a large enough domain of spins flipped the other way, it will grow fairly rapidly until it covers the whole system.

(All of this is assuming there is some sensible dynamics defined on the Ising model and the spin glass models, such as Glauber dynamics.)

I think this Ising model version is probably the best analogue to Prince Rupert's drop, since avalanches come in a variety of sizes whereas this phenomenon always affects the whole system.


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