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

  • 3
    $\begingroup$ 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. $\endgroup$ Commented Oct 22, 2023 at 11:00
  • $\begingroup$ 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. $\endgroup$ Commented Oct 22, 2023 at 22:32
  • $\begingroup$ I had in mind the following paper 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) $\endgroup$
    – Syrocco
    Commented Oct 23, 2023 at 7:21
  • $\begingroup$ 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...) $\endgroup$
    – Syrocco
    Commented Oct 23, 2023 at 10:20
  • $\begingroup$ @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. $\endgroup$ Commented Oct 23, 2023 at 12:31

1 Answer 1


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.


  1. arXiv:2306.00085, §2.
  • $\begingroup$ 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... $\endgroup$
    – Syrocco
    Commented Oct 25, 2023 at 9:55
  • $\begingroup$ 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. $\endgroup$
    – Syrocco
    Commented Oct 25, 2023 at 9:57
  • $\begingroup$ @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. $\endgroup$ Commented Oct 27, 2023 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.