13
$\begingroup$

The Mermin-Wagner theorem states that

continuous symmetries cannot be spontaneously broken at finite temperature in systems with short-range interactions in dimensions d ≤ 2.

And Goldstone bosons are

new excitations that appear when there are spontaneously-broken continuous symmetries.

Therefore, I would assume no Goldstone bosons would appear in 1D or 2D systems.

However, my professor told me these bosons can arise in a 1D superfluid, and that what cannot exist is the broken symmetry and therefore any condensate.

How is this possible?

$\endgroup$
  • 1
    $\begingroup$ You may find section 4.2.3 to be interesting damtp.cam.ac.uk/user/tong/sft/four.pdf $\endgroup$ – SRS May 6 at 20:24
  • $\begingroup$ The only remark from Tong regarding d=1 I can find is In d = 1, the physics is straightforward: there are no gapless modes. As before, this can also be understood in the language of quantum mechanics, where the spectrum of a particle moving on S^{N-1} is discrete and gapped. Not sure that this answers my question. $\endgroup$ – João Bravo May 8 at 14:29
4
+50
$\begingroup$

$\newcommand{\d}{\mathrm{d}} \newcommand{\U}{\mathrm{U}} \newcommand{\p}{\partial}$ According to the Coleman-Mermin-Wagner theorem indeed a continuous symmetry cannot break in 1D and 2D. However, using the superfluid low energy effective action [1] $$S[\phi] = \int \d^dx \; P\Big(\sqrt{-\p_\mu\phi\p^\mu\phi\;}\Big),$$ where $P(\cdot)$ is an arbitrary smooth function away from the origin (representing the pressure), and $\phi$ is the phase of the charged operator that condenses, it was shown in [2] that this action has two symmetries. One is easy to see, it is a $\U(1)$ shift symmetry $\phi\to\phi+\alpha$ yielding a conserved current $$J = P'\frac{D_\mu \phi}{\sqrt{-D_\mu\phi\,D^\mu\phi}},$$ where $D_\mu\phi = \p_\mu\phi-qA_\mu$, with $A_\mu$ the $\U(1)$ background gauge field for the shift symmetry and $q$ the charge of $\phi$. The other one is a higher-form symmetry, namely a $(d-2)$-form $\U(1)^{[d-2]}$ symmetry conserved simply because of the identity $\d^2=0$; it has a $(d-1)$-form current \begin{align*} K_{\mu_1\mu_2\cdots\mu_{d-1}} &= \epsilon_{\mu_1\mu_2\cdots\mu_d} \p^{\mu_d}\phi \qquad\Longleftrightarrow \\ (\star K)_\mu &= \p_\mu \phi \end{align*} However, when the $\U(1)$ background field, $A_\mu$, is turned on, $K$ is anomalous, because it becomes $(\star K)_\mu=D_\mu\phi$, but then $$\p_{[\mu}(\star K)_{\nu]} = a F_{\mu\nu}\neq 0.$$ This $(d-2)$ form symmetry, is physically a symmetry under which winding planes are charged. The upshot is that if one computes the mixed correlator $$\Pi_{\mu\nu}(p) := \int \d^d x\; e^{ip\cdot x} \Big\langle \mathcal{T}\big[(\star K)_\mu(x) J_\nu(0) \big] \Big\rangle,$$ in the Källén-Lehman representation $$ \Pi_{\mu\nu}(p) = \int_0^\infty \d\mu^2\; \rho\big(\mu^2\big)\, \frac{p_\mu p_\nu - p^2 g_{\mu\nu}}{p^2-\mu^2+i\epsilon}, $$ one finds, due to the mixed anomaly, that there exists a massless mode in the spectrum $$ \rho\big(\mu^2\big) = a \delta\big(\mu^2\big) + \text{finite terms at $\mu\to 0$}.$$ Hence, albeit there is no symmetry breaking in $d\leqslant 2$, these symmetry arguments do go through and there is a Goldstone mode!


References

[1] D. T. Son, Low-Energy Quantum Effective Action for Relativistic Superfluids, arXiv:0204199

[2] Luca V. Delacrétaz, Diego M. Hofman, Grégoire Mathys, Superfluids as Higher-form Anomalies, arXiv:1908.06977

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This answer is quite beyond my knowledge. Does your conclusion mean that the wikipedia definition “Goldstone bosons are bosons that appear necessarily in models exhibiting spontaneous breakdown of continuous symmetries” is not correct? It seems that they appear in a 1D superfluid without them... $\endgroup$ – João Bravo May 6 at 14:54
  • 1
    $\begingroup$ Yes, that's what my conclusion means. What protects the masslessness of the Goldstone modes is the mixed anomaly, rather than the non-linear realisation of the symmetry. $\endgroup$ – ɪdɪət strəʊlə May 6 at 15:11
  • 1
    $\begingroup$ I edited the answer to avoid the use of differential forms, so that you don't have two presumably unknown territories but one. $\endgroup$ – ɪdɪət strəʊlə May 6 at 15:20
  • $\begingroup$ Ok, I understood the general approach. Thank you! $\endgroup$ – João Bravo May 8 at 14:33
6
$\begingroup$

I don't know much about superfluidity, but I think I can explain a little bit about the statement of no Goldstone bosons in two dimensions.

Goldstone modes always show up in a semiclassical expansion, because they simply follow from general considerations about the effective potential. If you write down the most general potential consisting of only operators relevant in the infrared invariant under whatever continuous symmetry you like, its minimum set will be some $G/H$ coset manifold that the Goldstone modes live in. Since symmetry bans a mass term for them, they will remain massless to all orders in perturbation theory. Moreover, by integrating out the massive modes of the original field, one obtains an effective action for the Goldstone modes, which is dictated by symmetry to be a nonlinear sigma model on the target manifold $G/H$.

In $d > 2$, the nonlinear sigma model is purely nonrenormalizable and has no relevant perturbations consistent with the symmetry of the problem. Therefore, one can conclude that the semiclassical description of the low-energy dynamics is good, and in fact increasingly good at low energies by RG improvement. Therefore, the Goldstone bosons are real, and we can identify them with the ones that show up in the perturbative expansion.

In $d=2$, the action is classically conformally invariant so there is a question as to what happens when we turn on fluctuations. In the cases relevant to spontaneous symmetry breaking the sigma model is asymptotically free and some strong-coupling dynamics takes over in the infrared, which causes the true description of the theory at long distances to be in terms of some new collective fields with gapped excitations.

This is the general story, but there is an interesting counterexample when the target of the sigma model is $U(1)$. In this case, the low-energy theory is a free field theory and it appears that there is no dynamical way for the Goldstone bosons to die. The point is that the would-be Goldstone field $\theta(x)$ is circle-valued, so it is not a good local operator in the field theory. So there is "no Goldstone boson"--the well-defined operators are things like $e^{i \theta(x)}$ which respect the circle-valuedness. One can show by cluster decomposition that $\langle e^{i\theta(x)} \rangle = 0 $ which means that there is no symmetry breaking, and therefore no associated Goldstone bosons.

Hope this helped!

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. I must say I don't fully understand lot of those concepts, but I get the general picture. What do you mean by nonlinear sigma model? $\endgroup$ – João Bravo Jun 4 '19 at 20:21
  • $\begingroup$ Also, any idea about the 1D case? $\endgroup$ – João Bravo Jun 4 '19 at 20:21
  • 1
    $\begingroup$ Nonlinear sigma models are field theories where the basic variables are scalars taking values in some Riemannian manifold (target space), and the kinetic terms are determined by the metric on it (en.wikipedia.org/wiki/Non-linear_sigma_model). $\endgroup$ – Spencer Tamagni Jun 6 '19 at 16:56
  • 1
    $\begingroup$ For one-dimensional systems, the field theory is isomorphic to some quantum mechanics problem (in imaginary time) and you can usually show what you want by appealing to standard facts. For symmetry breaking in particular, it's related to domain walls in 1D, which in field theory language are the instantons in QM. Polyakov's book has a pretty good discussion of this. $\endgroup$ – Spencer Tamagni Jun 6 '19 at 16:58
  • 2
    $\begingroup$ No, there are no Goldstone bosons or symmetry breaking, but the reason is different. In 1D it has to do with the domain walls (instantons) that dominate the path integral and restore symmetry. Nothing to do with sigma model dynamics in this case. $\endgroup$ – Spencer Tamagni Jun 6 '19 at 23:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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