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

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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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retagged Mar 24, 2014
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.

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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).

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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](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=MR&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=MR1892461&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq) if you don't believe that). So, if a simple approach can solve this can kind of problems and more, that would be great.

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@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]

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[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.

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@RonMaimon: Any news? If you have thought about it and concluded that it is indeed less trivial than you expected, then I'd also be interested to know it (although, of course, I'd prefer a solution ;) ).

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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.

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answered Oct 7, 2011 by (7,720 points)
Thanks for your answer. I must confess that I don't get all your points, though, and shall probably have to translate everything back to mathematese, as you say. Nevertheless, if I agree that our example is quite contrived and unrealistic (it was given to explain why we _do_ need some regularity on the interaction in our paper, namely continuity or the more general condition (26)), I don't think that the resulting Gibbs state is just a mixture of measures with all spins frozen in a given direction (even though this would be a Gibbs measure, albeit a pathological one). [to be continued]

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[continuation] But in any case, my question was about a heuristic argument that would explain how to implement a spin-wave argument in the presence of hard-core interactions. you seem to give one, but I can't follow it, as it is too imprecise. Could you add some more math? Actually, even expliciting your argument when the interaction is continuous (but not differentiable) would be already nice (even though we do know how to prove that rigorously, and a nice physical argument can be extracted from the proof).

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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.

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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.

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BTW, I would be very interested if you had some reference where the kind of arguments you have in mind are detailed (I mean, with the relevant mathematical derivation; of course, I'm not asking for rigour here!). The ones I know always make expansions to second order, which you can't do in all the cases I mention.

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...and the size limitation in comments is _really_ a pain... >:(

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@Yvan: The Mermin-Wagner theorem is unaffected by singular potentials, such as the ones you use. The reason is that a nonzero local entropy averages out the potential using a smooth averaging function--- namely "$\exp(-\beta E)$", and this averaging procedure smooths out the potential to a usual spin-wave form. There is nothing unusual in your example, it is analogous to the totally frozen zero temperature limit, which also violates the theorem.

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To have a thermodynamic free energy, you need not just that the variance between 0,0 and 0,1 is nonzero, which _will_ be true using a $\delta$, but that the number of configurations is exponential in the total number of lattice vertices. If the entropy is not extensive (meaning if there isn't a nonzero entropy per lattice site), then you aren't averaging over overlapping configurations for neighboring average spin directions, and you don't have macroscopic smoothness.

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No, I don't think it's similar to zero-temperature, as I explained above. But that does not matter. What I really want to know is how you make the arguments you suggest work with a potential which behaves, say, like $1-e^{-1/\theta}$ at $0$, where $\theta$ is the angle difference between two neighbouring spins. Note that a small angle results in a huge energy difference. (Again: this is a case I know how to deal with, but your point of view might be interesting). Similarly, how do you treat the hard disc case?

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"...the number of configurations is exponential in the total number of lattice vertices". Of course, I know that. And I'm saying that this is probably the case in our example _in the limiting Gibbs state_: all the spins at a finite distance from the origin will have room to wiggle, which gives rise to a positive entropy density.

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After some thoughts, it might indeed be the case that the entropy density is zero: it would be sufficient that the wiggling room of the spins decays sufficiently fast with the distance to $0$. Looks unlikely to be the case, but I certainly have no good argument at the moment. But, please, let's rather focus on my main questions. For example, the hard discs, or the following example, which is much closer to the models we discussed in these comments [example in the next comment].

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Here is the example. It is _not_ known how to prove the result, so if you have any idea why it _should_ be true, I'd be very happy. But, please, try to give some math details or a reference to such. The problem is as follows: The spins take values in $\mathbb{R}$; I'll call their values "height". The interaction energy between two neighbours (on $\mathbb{Z}^2$) is equal to $0$ if the heights differ by less than $1$, and $+\infty$ if they differ by $1$ or more. Consider the box $\{-N,\ldots,N\}^2$. MW would imply that the variance of the height at $0$ grows like $\log N$. How would you argue?

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@Yvan: It is similar to zero temperature in that the entropy density _is_ zero, which is obvious from the decay of delta with N. You need constant $\delta$ to have an entropy density. This is the central flaw in the model. The Mermin Wagner argument is all I am using, there is _no modification_ for singular potentials. Why don't you ask the height function case as a question, the comments are too restricting--- I understand heuristically why it works (massless 2d field). The only obstacle to proof are certain probability constructions which I use personally, which I have never written down.

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Ok, I'll ask this as a new question. I disagree with you claim about $\delta$, but this might be because we are using the same words for two different things. In any case, I still don't see how one uses positive entropy density to deal with any of the cases I mentionned in the comments above...

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Let me add that the "Mermin-Wagner theorem" has a very precise (and very strong) meaning in mathematical statistical mechanics: in dimensions $1$ and $2$, provided that the interaction decays fast enough with the distance, a continuous symmetry of the Hamiltonian is always also a symmetry of each infinite-volume Gibbs measure. So, the measure we constructed is a bona-fide counter-example. However, I agree (and this is even written down in my question) that _we expect the theorem to hold for "reasonable" Gibbs states_. Our example shows that one must put assumptions on the interaction, though.

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just a last comment/question about this entropy density business (I'd like to move on to the real question instead, not the discussion of this counterexample ;) ). What you consider seems to be the limit as $N\to\infty$ of the finite-volume entropy densities. This does go to zero, when $\delta\to 0$. However, I don't see why the entropy density associated with the limiting Gibbs state should also be zero. As an example to illustrate this point, the limiting entropy density of the dimer model in the Aztec diamond is not equal to the entropy density of the limiting measure[...]

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[...] because there are macroscopic frozen regions in the finite boxes, while the limiting measure (in the thermodynamic limit, i.e., topology of local convergence) is supported on disordered dimers configurations.

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@Yvan: Your interpretation of Mermin-Wagner is wrong--- in the exact zero temperature limit, there is symmetry breaking in 2d, and the reason is exactly the same as in your model, vanishing extensive entropy. The Mermin Wagner theorem says that the symmetry is a symmetry of the Gibbs state only when the Gibbs state has fluctuations. It is most important to understand that your construction works in 1d, you can get a 1d long-range ordered crystal from your type of thing, and Mermin Wagner is even stronger in 1d, and completely trivial. Since you asked a new question,　I'll go there to answer.

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I understand the Mermin-Wagner theorem very well, thank you. You seem to be misunderstanding the very definition of an infinite-volume Gibbs state, that's the single reason we disagree, I think... Note that the Gibbs state we obtain _has_ fluctuations (non zero variance of difference of angles between distinct spins). (Or at least, this should follow from a more careful analysis along the lines of our proof...)

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Still I think that you misunderstand my very question: _I don't care about the precise counter-example we gave in the paper_ (and which was just a remark in the paper). What I care about is a (not necessarily rigorous) mathematical argument that derives absence of continuous symmetry breaking in situations where you have a hard-core interactions. You have not given me that, unfortunately. As I said above, even a nice heuristic (but mathematical!) argument in the case of non-differentiable interactions would be interesting (even though we treated those in the paper).

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The arguments of Mermin and Wagner do not care about the differentiability in the microscale potentials. The spin waves have the usual free energy in the thermodynamic limit away from situations where the spins are nearly completely frozen. I will try to prove the same result rigorously in your new question. It requires some simple real space renormalization techniques. The methods which do not renormalize to an averaged spin-wave field miss the boat completely, and the hard core limitation is an artifact of Microsystems configuration methods that are renormalization blind.

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great, I'm looking forward to seeing that :) .

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