# Conformal symmetry with Goldston bosons

I am looking for examples of the physical systems with both conformal symmetry and Goldston bosons. The systems can be in any physical context, like QCD phases, Condensed Matter (Graphene, etc), or SUSY gauge theories.

Naively, in 1+1D gapless system, we can more easily have conformal symmetry. But to have a spontaneous symmetry breaking with Goldston bosons, and together with a (perhaps emergent) conformal symmetry seem to be somehow more challenging.

In 2+1D and higher dimensional system, we may find harder to have conformal symmetry. Are there such examples of conformal symmetry with Goldston bosons.?

• You won't have Goldstone modes in 1+1D because of the Mermin-Wagner theorem (unless you're at the large central charge limit).
– user106422
Apr 28, 2017 at 20:56
• You can't have unbroken conformal symmetry and GBs because their decay constant is dimensionfull in d>2. Even more physically, a theory of GB has a cutoff and its therefore not scale invariant. The only way out I know is to break conformal invariance spontaneously too, with the appearance of a dilaton. Apr 29, 2017 at 23:47

## 1 Answer

You can't have unbroken conformal symmetry and Goldstone Bosons (GB)'s at the same time because the GB decay constant $f_\pi$ is dimensionfull in $d>2$. And in $d=2$ there are no (physical) GB's. Even more physically, a theory of GB has a cutoff $\Lambda$ and its therefore not scale invariant.

The only way out to this is by weakening the requests, e.g. allowing for a non-linearly realized conformal invariance. In this case, conformal symmetry is broken spontaneously too, and a light dilaton appears in the spectrum alongside with the GB's from breaking spontaneously a global internal continuous symmetry. The leading terms in the effective action (ind $d=4$) for a dilaton $\sigma$ and ordinary GBs $\pi$ would take the form $$\mathcal{S}[\sigma,\pi]=\int d^4x \frac{f^2_\sigma}{2}(\partial_\mu e^{\sigma})^2+ \frac{f_\pi^2}{2} e^{2\sigma}(\partial_\mu\pi)+\ldots$$ which is scale invariant under $x\rightarrow x e^{\alpha}$ if the dilaton transforms non-linearly $$\sigma(x)\rightarrow \sigma(x e^{\alpha})+ \alpha\,,\qquad \pi(x)\rightarrow \pi(x e^{\alpha})\,.$$ The best way to see how this invariance is realized is by actually making a field redefinition $$\chi\equiv f_\sigma e^{\sigma/f_\sigma}$$ so that the action becomes $$S=\int d^4x \frac{1}{2}(\partial_\mu\chi)^2+\frac{1}{2}\left(\frac{f_\pi}{f_\sigma}\right)^2\chi^2(\partial_\mu\pi)^2$$ which contains only ratio of scales $f_\pi/f_\sigma$ and it's hence scale invariant. The physical scales are recovered because the new $\chi$ variables has non-vanishing vev that breaks spontaneously conformal invariance $$\langle\chi\rangle= f_\sigma\,.$$