5

The Mermin-Wagner theorem states that there cannot be any spontaneous symmetry breaking happening in systems with short range interactions below dimension 3. Moreover, we know that Goldstone boson, such as sound in solids, occurs due to the spontaneous breakdown of rotational and translational symmetries. Since in 2D we can't break the translational invariance, do phonons exist in 2D solids?

My thought:

  1. The rotational symmetry is broken in 2D solids due to an effective gapping of the orientational modes (see this link for example). This might allows us to obtain phonons.

  2. We also see phonons in liquid where no symmetry is broken. One might not called these sound waves phonons as in this PSE post. However, it is shown in the book Boulevard of broken symmetries that sound in liquids can be associated to the Goldstone boson linked to the breakdown of Galilean symmetry when one linearize the Navier-Stokes equations (which are Galilean invariant) to obtain the wave equation (not Galilean invariant).

  3. We can very well write down an Hamiltonian in 1D or 2D: $H = \sum |k|u^2$ and find gapless excitations. Obviously, this is wrong but I can't figure out why?

There are some related posts. For example these two: Are there Goldstone bosons in 1D or 2D? Spontaneous symmetry breaking in fluids

Syrocco
  • 973
  • 3
    Minimal comment: in 2D both thermal and quantum fluctuations destroy long-range order, but finite size crystalls can exist. In 2D the decay of long-range order is algebraic, not exponential. – Artem Alexandrov Oct 22 '23 at 11:00
  • Can you substantiate the claim that MW rules out the breaking of any continuous symmetry in 2D? AFAIK the original, fully rigorous proof only applies to the Heisenberg model, and I think that also the more "physics-y" proofs with the divergence due to the density of states rely on some conditions such as the spinon dispersion which should depend on the type of symmetry broken. So while this is possible, I don't think it is evident. – Norbert Schuch Oct 22 '23 at 22:32
  • I had in mind the following paper https://ui.adsabs.harvard.edu/abs/1968PhRv..176..250M/abstract where Mermin derives the impossibility of having translational long range order in soft solids (not orientational order) – Syrocco Oct 23 '23 at 07:21
  • But you are right that is doesn't rule out orientational order for example (as stated in my thought, this might allows us to get phonons...) – Syrocco Oct 23 '23 at 10:20
  • @NorbertSchuch For classical systems, the rigorous proofs are much, much more general than that. They apply basically to any compact Lie group and any model. See, for instance, this answer for an example of the type of statements that can be proved. For translation invariance, see this paper, for instance. I am less knowledgeable about the quantum version, but I don't expect the level of generality to be that different. – Yvan Velenik Oct 23 '23 at 12:31
  • @YvanVelenik Thanks, that's interesting to know. Certainly the version on http://www.scholarpedia.org/article/Mermin-Wagner_Theorem seems to be quite specific about the model. I'd certainly be interested in learning more about the different rigorous ways to prove that, so if you know some good overviews ... (E.g., is there a way to prove this for classical models in the transfer matrix formulation? -- One thing I'd be very interested is whether one can prove an analogous theorem for tensor network wavefunctions, which naturally lead to (complex & completely positive) transfer operators.) – Norbert Schuch Oct 23 '23 at 16:11
  • @NorbertSchuch , have you tried to find connections between tensor networks and simple models in terms of universality classes? It seems that one does not need to prove the SSB-related statement for a particular model, but it is enough to show that this model belongs to the same universality class – Artem Alexandrov Oct 24 '23 at 22:03
  • @ArtemAlexandrov This is indeed one motivation for looking at that -- 2D tensor network wavefunctions give rise to partition functions of 2D classical models with complex weights and some special properties. On the other hand, it is unclear how a connection as you suggest could give rise to a no-go-*theorem*: Connecting individual PEPS models to 2D classical universality classes will not lead to a theorem about general (tinv, say) PEPS. – Norbert Schuch Oct 25 '23 at 09:04

1 Answers1

2

You do have sound in $1+1d$. Mermin-Wagner do not claim that there are no massless particles in low dimensions -- indeed, some of the best understood QFTs are two-dimensional gapless theories, namely the wonderful world of $2d$ CFTs.

What Mermin-Wagner does, is proving that there are no order parameters for symmetry breaking in $2d$. This is not the same as saying there are no Goldstones. There are such particles, just no interpolating field for them.

See ref.1 for a nice recent review.

References:

  1. arXiv:2306.00085, §2.
AccidentalFourierTransform
  • 53,248
  • 20
  • 131
  • 253
  • 1/2. Thanks for you answer! I'm not sure I quite understand the arguments. In your link (and your answer) they argue that: "the absence of symmetry breaking is due to the fact that massless excitations in 2d propagate to arbitrary large distances" and "Coleman’s theorem does not necessarily imply the absence of massless particles in two dimensions. It is crucial that these particles have a non-zero matrix element with a scalar operator to reach a contradiction. In other words, Coleman’s theorem only forbids order parameters." Does this mean that there are gapless particles in these systems... – Syrocco Oct 25 '23 at 09:55
  • 2/2 ...that destroy long range order, but that can't be associated to a SSB? So in 3D, sound would be the goldstone boson associated to SSB and in 2D sound would be a gapless excitation not related to the Goldstone theorem? What troubles me is that in every case they talk about 'goldstone boson, goldstone theorem, ..." as if it was working in 2d. – Syrocco Oct 25 '23 at 09:57
  • @Syrocco Yes, there are massless particles in 2d. And indeed these lead to correlations that do not decay (in higher $d$, the propagator is some power of $1/x$, but in $2d$ it becomes $\log x$ instead). In 2d you can still call these massless particles "Goldstones" if you want, as long as you remember that their mathematics is different from the usual characterization; in particular, you cannot think of them as being created by an order parameter. – AccidentalFourierTransform Oct 27 '23 at 15:10