# Mermin-Wagner theorem in the presence of hard-core interactions

It seems quite common in the theoretical physics literature to see applications of the "Mermin-Wagner theorem" (see wikipedia or scholarpedia for some limited background) to systems with hard-core interactions; for example to conclude that genuine crystal phases for a system of hard disks (with possible additional interactions) cannot exist in 2 dimensions. If this particular claim has been proved rigorously a few years ago (see this paper), it is known in general that the presence of hard-core interactions can lead to phases with broken continuous symmetry (a specific example is given below).

To keep things simple, let me focus on nearest-neighbor spin models on the square lattice, with spins taking values in the unit circle. So let us consider a formal Hamiltonian of the form $$\sum_{i\sim j} V(S_i,S_j),$$ with $V$ is a continuous function, assumed to be invariant under the action of $SO(2)$: $V(r_\theta S_i, r_\theta S_j) = V(S_i,S_j)$, where $r_\theta$ rotates the spin by an angle $\theta$. In that case, it is known that all pure phases of the model are invariant under the action of $SO(2)$. (Note that we do not even require $V$ to be smooth, so that the usual, both heuristic and rigorous, arguments relying on a Taylor expansion and a spin-wave argument do not apply immediately; that one can do so was proved here). Substantially more general results are actually known, but this will suffice for my question.

What I am interested in is what happens for models of this type in the presence of hard-core interactions. No general results are known, and the situation is proved to be subtle. Indeed, consider for example the Patrascioiu-Seiler model, in which $$V(S_i,S_j) = \begin{cases} -S_i\cdot S_j & \text{if }|\delta(S_i,S_j)|\leq\delta_0,\\ +\infty & \text{otherwise,} \end{cases}$$ where $\delta(S_i,S_j)$ denotes the angle between the spins $S_i$ and $S_j$, and $\delta_0>0$ is some parameter. In other words, this model coincides with the classical XY model apart from the additional constraint that neighboring spins cannot differ too much. For this model, it is proved here that, when $\delta_0<\pi/4$, there exist (non-degenerate) phases in which rotation invariance is broken. Nevertheless, one expects that phases obtained, say, by taking the thermodynamic limit along square boxes with free, periodic or constant boundary conditions should be rotation invariant.

So, now, here is my question: Are there any heuristic physical arguments supporting the validity of a version of the "Mermin-Wagner theorem" in such situations? All the heuristic arguments I know of fail in such a context. Having good heuristic arguments might help a lot in extending the rigorous proofs to cover such situations.

Edit: Let me precise my question, as the (quite long) discussion with Ron Maimon below shows that I haven't stated it in a clear enough way. I am not interested in a discussion of why the counter-example given above leads to a violation of MW theorem and whether it is physically realistic (as far as I am concerned, its main relevance was to show that one has to make some assumptions on the interaction $V$ in order to have rotation invariance of all infinite-volume Gibbs states, and this is exactly what this example does). What I am really interested in is the following: does there exist heuristic (but formulated in a mathematical way, not just some vague remarks) arguments with which physicists can derive the MW theorem in the presence of hard-core interactions? I would even be interested in arguments that apply in the absence of hard-core interactions, but when $V$ is not differentiable (even though this case is treated rigorously in the reference given above).

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The question was 100% clear, and the answer I gave you is the correct one--- there is no difference in the heuristic argument for Mermin Wagner in the presence of hard-core interactions or in their absence, because your counterexample is nonsense. The argument is the usual establishing of averaged spin-wave states, and finding that the free energy is the free-field, which has only log divergent configurations. The only way this fails is if you have frozen out all fluctuations, so that there are no spin-waves at all, and this is your example. This is not essentially to do with hard-walls. –  Ron Maimon Oct 11 '11 at 2:16
The reason your methods do not work for the hard-potential case is not because the argument of Mermin and Wagner is failing, because your methods are not in the spirit of Mermin-Wagner. The reason they fail is because you are not defining an averaged long-distance spin-wave field, but trying to do things with the bare configurations, which is hopeless. The way to define a long-distance theory is renormalization (in the Wilson sense, but simpler because the limit is free). –  Ron Maimon Oct 11 '11 at 2:20
I am looking forward to seeing your argument. Note that in the rigorous statistical mechanics community, the extension beyond smooth interaction potentials (as we did) remained open for some 20 years (read the mathscinet review if you don't believe that). So, if a simple approach can solve this can kind of problems and more, that would be great. –  Yvan Velenik Oct 11 '11 at 6:59
@RonMaimon: I think that you're somewhat unfair (or misinformed) when you say "your methods are not in the spirit of Mermin-Wagner". The methods we used (which were introduced in the late 1970s by Dobrushin&Shlosman and by Pfister) seem to me very much in the spirit of MW: roughly speaking, what they do is compare the (finite-volume) Gibbs measure before and after deforming the configurations by a global spin wave, and show that the effect of this deformation becomes negligible when the system size gets big (and thus the wave-length large). [to be continued] –  Yvan Velenik Oct 11 '11 at 14:22
[continuation] The difficulty when hard-core is present is that most configurations become forbidden (have infinite energy) after deformation by the spin wave. One can then try to resort to a configuration dependent deformation (this is what is done by Richthammer in his treatment of the hard discs model), but this is technically quite difficult, unfortunately. –  Yvan Velenik Oct 11 '11 at 14:26

First, I will translate the relevant passages in your paper from mathematese.

### The argument in your reference

You are studying an X-Y model with the constraint that neighboring spins have to always be within a certain angle of each other. You define the collection of statistical-mechanics Gibbs distributions using a given boundary condition at infinity, as the boundaries get further and further away. Then you note that if the field at the boundary makes the spin turn around from top to bottom the maximum possible amount, then the spins are locked in place--- they can't move, because they need to make a certain winding, and they unless they are at the maximum possible angle, they can't make the winding.

Using these boundary conditions, there is no free energy, there is no thermodynamics, there is no spin-wave limit, and the Mermin Wagner theorem fails.

You also claim that the theorem fails with a translation invariant measure, which is just given by averaging the same thing over different centers. You attempt to make the thing more physical by allowing the boundary condition to fluctuate around the mean by a little bit $\delta$. But in order to keep the boundary winding condition tight, as the size of the box $N$ goes to infinity, $\delta$ must shrink as $1\over N$, and the resulting Free energy of your configuration will always be subextensive in the infinite system limit. If $\delta$ does not shrink, the configurations will always randomize their angles, as the Mermin-Wagner theorem says.

The failures of the Mermin-Wagner theorem are all coming from this physically impossible boundary situation, not really from the singular potentials. By forcing the number of allowed configurations to be exactly 1 for all intents and purposes, you are creating a situation where each different average value of the angle has a completely disjoint representative in the thermodynamic limit. This makes the energy as a function of the average angle discontinuous (actually, the energy is infinite except for near one configuration), and makes it impossible to set up spin waves.

This type of argument has a 1d analog, where the analog to Mermin-Wagner is much easier to prove.

### 1-dimensional mechanical analogy

To see that this result isn't Mermin-Wagner's fault, consider the much easier one-dimensional theorem--- there can be no 1d solid (long range translational order). If you make a potential between points which is infinite at a certain distance D, you can break this theorem too.

What you do is you impose the condition that there are N particles, and the N-th particle is at a distance ND from the first. Then the particles are forced to be right on the edge of the infinite well, and you get the same violation: you form a 1d crystal only by imposing boundary conditions on a translation invariant potential.

The argument in 1d that there can be no crystal order comes from noting that a local defect will shift the average position arbitrarily far out, so as you add more defects, you will wash out the positional order.

### Mermin-Wagner is not affected

The standard arguments for the Mermin-Wagner theorem do not need modification. They are assuming that there is an actual thermoodynamic system, with a nonzero extensive free energy, an entropy proportional to the volume, and this is violated by your example. The case of exactly zero temperature is also somewhat analogous--- it has no extensive entropy, and at exactly zero temperature, you do break the symmetry.

If you have an extensive entropy, there is a marvelous overlap property which is central to how physicists demonstrate the smoothness of the macroscopic free-energy. The Gibbs distribution at two angles infinitesimally separated sum over almost the same exact configurations (in the sense that for a small enough angle, you can't tell locally that it changed, because the local fluctuations swamp the average, so the local configurations don't notice)

The enormous, nearly complete, overlap between the configurations at neighboring angles demonstrates that the thermodynamic average potentials are much much smoother than the possibly singular potentials that enter into the microscopic description. You always get a quadratic spin-wave density, including in the case of the model you mention, whenever you have an extensive free energy.

Once you have a quadratic spin-wave energy, the Mermin Wagner theorem follows.

the Gibbs distributions for orientation $\theta$ and the Gibbs distributions for orientation $\theta'$ always include locally overlapping configurations as $\theta$ approaches $\theta'$. This assumption fails in your example, because even an infinitesimal change in angle for the boundary condition changes the configurations completely, because they do not have extensive entropy, and are locked to within a $\delta$, shrinking with system size, of an unphysically constrained configuration.
Also, how would you argue (heuristically but, please, with some math details) that hard discs in the plane cannot form a crystalline phase (i.e., translation invariance cannot be broken)? To keep things as simple as possible, assume that there is no other interactions beside hard-core. A rigorous proof is known (see the ref in my question) but it is quite intricate. Of course, here it's density waves rather than spin waves, but that's rather irrelevant. Note that here the enery of a configuration is either $0$ or $+\infty$, which makes smooth deformation of configurations tricky. –  Yvan Velenik Oct 7 '11 at 15:37
Although that's not the main point, let me come back to your comment about $\delta$ in your answer. Sure, $\delta$ has to go to $0$ with $N$. But that does not imply that the resulting Gibbs measure has not an extensive entropy, because the spins far away from the boundary have room to wiggle. That's only spins at a finite distance from the origin that you see in the thermodynamic limit. So I am pretty sure (although it would some time to come up with a formal proof) that under the measure we "construct", the angle between two given vertices (say $(0,0)$ and $(1,0)$) has a non-zero variance. –  Yvan Velenik Oct 7 '11 at 15:46