The ideia is to show that, because of Goldstone modes, 2d systems are quite different from 3d ones. So, considering the Heisenberg model, I'll post here what I'm asked to and my current thoughts on the subject in hope you can give me further assistance.

1. fluctuations on Heisenberg model should destroy long-range spin order (solved but more info is wellcome)

I can calculate the mean square value of the fluctuations $<(\delta\vec{S}(\vec{r}))^2>$ and get the integral $$\int_0^\Lambda dq \;q^{d-1}\frac{1}{q^2}$$ which tells me the lower critical dimension (next question) and how bidimensional lattices can´t have a phase transition, but how can I relate this with long-range order destruction?

2. a bidimensional lattice can't be stable

It seems I gave the wrong idea here. This here differs from the above in that we want to show that the fluctuations of the atom's position is so that the lattice can't be stable in 2d! I think this is associated with a divergence in phonons momentum, but how can I prove this?

3. lower critical dimension of antiferromagnetic heisenberg model

In this one I assume it is $d=2$ because Heisenberg model has continuous symmetry and it follows directly from Mermin-Wagner theorem. But it seems it is not enough! Antiferromagnetic behaviour is quite different of ferromagnetic one, specially concerning sin waves dependence on momentum (linear and quadratic, respectively, if I'm not wrong). So, following the same calculations I did on topic 1, this linear dependence will give me a lower critical dimension $\neq 2$ and I think I'm missing something, because this can't be right according to the theorem stated above. Any ideas?

Sorry for any typos. Hope someone can enlight me! Thanks in advance.

  • $\begingroup$ When you calculated the integral for the mean-squared fluctuations, you would have found that they diverge in the limit of small wave-vectors if you are below the critical dimension. Do you see this? If yes, then ask yourself what does it mean to say that the fluctuations are divergent at large scales. If there large scale fluctuations, can you have long-ranged order? I think stability here only refers to the facr that an ordered state is unstable to small fluctuations of you below the crtical dimension. $\endgroup$ – Vijay Murthy May 29 '12 at 14:44
  • $\begingroup$ Just added the integral I got for discussion. It is clear that it diverges for low momenta $q \sim 0$ for dimensions $d \leq 2$. If I understood you correctly, then the connection I was missing was that lower momenta divergence corresponds to large scale divergences. And this is what explains large scale fluctuations, though no spin order. $\endgroup$ – Hugo May 29 '12 at 14:56
  • $\begingroup$ Yes. Thats the integral. If you evaluate it for a finite-sized system (taking the lower limit for the integral as $1/L$), you will see that the fluctuations diverge logarithmically with the system size in $d=2$. If the fluctuations are large on the scale of the entire system, how can there be long-ranged order? EDIT; You just said the same above. $\endgroup$ – Vijay Murthy May 29 '12 at 15:05
  • $\begingroup$ About the second topic: we already saw that for $d=2$ there's no long range order, how can one evaluate stability in other ways? For instance, from Bogoliubov inequality one sees that there is no spontaneous magnetization for lower temperatures. Is this analysis correct? What more can be said? $\endgroup$ – Hugo May 30 '12 at 10:45
  • $\begingroup$ You should be more precise. What do you have in mind when you say "... evaluate stability in other ways?" the stability of what...? In this context, I think the point in question is the stability of the ordered state if one can exist below a certain temperature. Suppose you prepare your system in the ordered state and maintain the temperature below the transition temperature. You showed above that this state is unstable to thermal fluctuations if you are below the critical dimension. That is all the stability I had in mind. I would like to know if there is anything more than this. $\endgroup$ – Vijay Murthy May 31 '12 at 8:48

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