I have read at some places that a dilaton field is associated with the spontaneous breaking of scale symmetry in a theory. (While others would be difficult to trace right now, the most easily accessible example is the first paragraph in this brilliant answer, which even won a bounty!) This is of particular interest in QCD, which is the context in which I pose this question, and not string theory which was the context of the linked answer (FYI - I don't know String Theory).

How does the incorporation of a dilaton field represent scale symmetry breaking in a field theory?

(1. Wikipedia doesn't have an answer for this aspect of scale invariance.

  1. I would appreciate a detailed answer, but even meaningful, pedagogic links are welcome.

  2. True context of this question was discussed in chat and can be seen in messages starting from this one. It was pointless to write everything all over again, hence just referring.)


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    $\begingroup$ I'm not quite sure what you are asking for - the answer you cite says that the dilaton is the Goldstone boson of a spontaneously broken scale symmetry. Goldstone bosons do not break the symmetry, they are a consequence of scale symmetry being spontaneously broken by the (ground) states of the theory. $\endgroup$
    – ACuriousMind
    Oct 3, 2014 at 16:29
  • $\begingroup$ @ACuriousMind - You are right. I've edited the question accordingly. I hope it is clearer now, otherwise, please let me know. $\endgroup$ Oct 3, 2014 at 17:35
  • $\begingroup$ I'm still not sure what you want to know - are you simply asking for how spontaneous symmetry breaking makes Goldstone bosons appear generally? I'm confused because I see no reason why one should expect that the dilaton is any different from any other boson origination from any other broken symmetry. $\endgroup$
    – ACuriousMind
    Oct 3, 2014 at 17:41
  • $\begingroup$ @ACuriousMind - My point is, if in a theory/model like QCD has scale symmetry broken (which we know, is the case), it must be possible to write it equivalently as a scale invariant theory + scale symmetry breaking term, accomplished via incorporating a dilaton field. The question is, how does this dilaton's incorporation ensure scale symmetry breaking? $\endgroup$ Oct 3, 2014 at 17:42
  • $\begingroup$ i.e. how does it pick out a particular energy scale in preference over others? $\endgroup$ Oct 3, 2014 at 17:44

1 Answer 1


The dilaton $\sigma$ is the Goldstone boson of scale invariance. Scale transformations $x\rightarrow x/\lambda$ are generated non linearly, e.g. $$ \sigma(x)\rightarrow \sigma(\lambda x)+f \log\lambda\,,\qquad \lambda>0 $$ where $f$ is the dilaton decay constant (see below). An effective field theory for this Goldstone boson can be easily written with the following trick described e.g. in this very nice paper. In particular, define $\chi(x)=f e^{\sigma(x)/f}$ so that under scale transformation $\chi$ behaves like a field of scaling dimension $1$ $$ \chi(x)\rightarrow \lambda\, \chi(\lambda x)\,. $$ With this field at hand now you can build your scale invariant lagrangian that describes the effective theory of the dilaton: it is an expansion in derivatives $\mathcal{L}(\chi)= \sum_{n} \mathcal{L}^{(2n)}$ where the lagrangian $\mathcal{L}^{(n)}$ is \begin{equation} \mathcal{L}^{(2n)}=\sum_{m\geq 0} \frac{a_{n,m}}{(4\pi)^{2n-2}f^{2n-4}}\frac{\partial^{2n}\chi^m}{\chi^{2n+m-4}} \end{equation} where the derivatives are inserted in all possible ways among the $\chi^m$. The logic is to write all operators with scaling dimension 4 so that the action $S=\int d^4 x \mathcal{L}$ will be scale invariant (given that $d^4 x\rightarrow \lambda^{-4}d^4 x$). You can basically think of $\chi$ as conformal compensator: build a Lorentz invariant and divide it by enough powers of $\chi$ to make of dimension 4. Notice that a term with no derivatives, contrary the the usual Goldstone bosons of internal symmetries, is instead allowed: $\mathcal{L}^{(4)}=(4\pi)^2a_{0,0}\chi^4$.

Moreover, if the dilaton isn't the only light state in the effective theory left over by the spontaneous symmetry breaking, other terms that seem breaking scale invariance need to be compensated by the insertions of the Goldstone so that dynamics will be invariant (whereas the states of the theory will not). For example, since you have mentioned QCD, one could wonder that the RG evolution comes from spontaneous breaking of scale invariance. Some particles that were contributing to the beta functions (which was zero in an unbroken scale invariant theory) becomes massive after spontaneous symmetry breaking, then decouple and don't contribute to $\beta$ anymore, leaving in fact an imbalance for which $\beta$ start running from the scale of symmetry breaking down to smaller energy scales. In practice one knows that the coupling runs logaritmically from the scale $f$ of symmetry breaking down to an arbitrary IR scale $\mu_{IR}$ $$ \mathcal{L}_{gauge}=-\frac{F^2_{\mu\nu}}{4g^2(\mu_{IR})}=-\frac{F_{\mu\nu}^2}{4}\left[\frac{1}{g^2(\mu_{UV})}+\frac{b_{IR}}{(4\pi)^2}\log\frac{f}{\mu_{IR}}\right] $$ But the scale $f$ appeared spontaneously and needs then to be compensated by dilaton insertion to restore scale invariance, $f\rightarrow f e^{\sigma(x)/f}=\chi(x)$, so that one can immediately read off the coupling among the gauge bosons and the dilaton \begin{align} \mathcal{L}_{gauge} & \rightarrow -\frac{F_{\mu\nu}^2}{4}\left[\frac{1}{g^2(\mu_{UV})}+\frac{b_{IR}}{(4\pi)^2}\log\frac{f e^{\sigma(x)/f}}{\mu_{IR}}\right]\\ =& \mathcal{L}_{gauge}-\frac{b_{IR}}{(4\pi)^2 f}\sigma(x)F_{\mu\nu}^2\,. \end{align} As you can see the dilaton couples to the the $\beta$ function. Analogously, if a mass term is assumed to arise because of the spontaneous symmetry breaking of scale invariance, it must be compensated with the dilaton insersion. For example \begin{equation} m\bar{\psi}\psi\rightarrow \frac{m}{f}\chi(x)\bar{\psi}\psi=\frac{m}{f}\bar{\psi}\psi\left[f+\sigma(x)/f+\ldots\right]. \end{equation} In general, any operator $\mathcal{O}(x)$ in the lagrangian with scaling dimension $\Delta\neq 4$ will get compensated by dilaton insertions to restore scale invariance $$ \mathcal{O}(x)\rightarrow \frac{\chi}{f}^{4-\Delta}(x)\mathcal{O}(x)=\mathcal{O}(x)\left[1+(4-\Delta) \sigma(x)/f+\ldots\right] $$ In other words, the dilaton coupling to any field is proportional to the scaling dimension $4-\Delta$. In the case of the gauge field indeed the beta function is proportional to the anomalous dimension. One can show all there result also in another way: Goldstone bosons couples to the derivative of the current that generates the spontaneously broken symmetry. In this case, the dilaton couples to the dilation current $D^\mu$ as $$ \int d^4x \frac{1}{f}\partial_\mu\sigma D^\mu=-\int d^4x \frac{\sigma(x)}{f} \partial_\mu D^\mu =-\int d^4x \frac{\sigma(x)}{f} T_\mu^\mu $$ where $T_\mu^\mu$ is the trace of the energy-momentum tensor that contains indeed the operators with dimension $\Delta\neq 4$ and coupling given by the `anomalous' dimensions $\propto 4-\Delta$. Many more details can be found in the paper mentioned above and in the reference therein.

  • $\begingroup$ Wow. This general formulation is absolutely fantastic. Thanks for this great answer. But if I'm not getting too greedy, could I draw your attention to this context of the question, actually messages starting from that one. If it is not asking for too much, can I request you to make some general comment regarding how $L_{\rm scalebreaking}$ introduces scale invariance breaking in the model? $\endgroup$ Oct 4, 2014 at 3:34
  • $\begingroup$ The reason why I say that is because this general formulation seems to be inverted by what Greiner and others are trying to do in that reference. At least that's what I got out of a fabulous chat discussion with ACuriousMind, who shared some brilliant general insights there. Sorry, if I'm asking for too much favor. $\endgroup$ Oct 4, 2014 at 3:38
  • $\begingroup$ @UserAnonymous I am sorry but to reconstruct the discussion from a chat is too painful for me. Try to post a well definite question and let's see if I can help. In any case, again, I recommend you to look at the reference I was giving in my answer since it covers all aspects such as the relation between explicit breaking vs spontaneous breaking, dilaton mass, etc $\endgroup$
    – TwoBs
    Oct 4, 2014 at 7:01
  • $\begingroup$ OK. The issue is, in this paper guys have proposed a model where they add a scale-breaking interaction term to an otherwise scale-invariant Lagrangian (density). This term depends on a dilaton field, and the authors claim that this incorporates scale symmetry breaking in the model. So, I was wondering, how can that be possible. That's why I posted this question. If it is possible for you to read anything into it, that will be too good (in addition to this already great answer). Thanks $\endgroup$ Oct 4, 2014 at 18:34
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    $\begingroup$ I took a brief look at that paper, and I think they are proposing nothing but anomaly matching for scale invariance. The anomaly in the UV is reproduced by the degrees of freedom in the IR, the dilaton in this case. But, my humble opinion is that it doesn't make much sense because there is should be no dilaton in the first place in such a theory in the IR exactly because scale invariance is not a good symmetry at all being badly broken at around $\Lambda_{QCD}$ where the $\beta$ function becomes big and the non perturbative effects take place. $\endgroup$
    – TwoBs
    Oct 4, 2014 at 21:19

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