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
$$
\langle\chi\rangle= f_\sigma\,.
$$

This post imported from StackExchange Physics at 2020-10-28 19:06 (UTC), posted by SE-user TwoBs