I think of a relativistic conformal field theory as basically any theory which does not have a length scale. Say all particles are massless and all couplings are dimensionless. (I could really talk about scale invariance but I'm more used to talk about conformal).

Now I want to break conformal symmetry by introducing a length scale. Say one of the particles gets a mass. It seems to me that such breaking can not ever be considered to be small or big, just because there is no other scale to compare it to. However it seems that both at very low and very high energies the theory should be approximately conformal again. In the example above at very low energies the massive field effectively decouples while at very high energies it becomes nearly massless. So the two types of theories that we get (at low and high energies) are different.

I guess I do not have a very specific question but I would like just to understand this type of situation better. Some things that come to mind

  1. Does one need to use the full power of renormalization group to connect the low- and high-energy theories in this case? Or there may be some workaround if I know a CFT I have started with and maybe for a particular kind of symmetry breaking operator?
  2. What are some solvable (and possibly simple) examples where one can trace this behavior?
  3. I am particularly interested in deforming 2d Liouville CFT so any specific references that might be relevant are welcome.

1 Answer 1


I'm not certain what exactly you're looking for with your first bullet, but I'll address the second bullet by looking a an exactly-solvable model with a nontrivial RG flow which can be determined just by looking directly at correlation functions rather than deriving beta functions. It's the large-$N$ limit of $\phi^4$ theory: $$ \mathcal{S} = \int d^dx \,\left[ \frac{1}{2} \left( \partial_{\mu} \phi_{\alpha} \right)^2 + \frac{\lambda}{2N} \left( \phi^2_{\alpha} - N m^2 \right)^2 \right]. $$ Here I'm using slightly different notation to facilitate a large-$N$ expansion, but by multiplying out the last term you get the usual $\phi^4$ theory with some factors of $N$ and $m^2$ placed differently than you're used to (and an unimportant constant). For $2 < d < 4$, the RG flow of this theory is known to look like the following (picture credit https://arxiv.org/abs/1811.03182):

enter image description here

The flows to $m^2 = \pm \infty$ describe a flow to a gapped theory where all correlation functions decay exponentially at long distances, so they're not quite as interesting as the line connecting the free massless theory $G$ to the massless Wilson-Fisher fixed point $WF$. We will see that although all correlation functions will be algebraic at both large- and small- distances, there is a crossover of critical exponents between the two asymptotic cases. The constant $\lambda$, which has units of $\mathrm{(energy)}^{4-d}$, will play an essential role here.

I won't go into every detail of the large-$N$ solution, but I'll outline it (I'm essentially following Section II of this paper, which itself using a similar method as Polyakov's textbook). The first step is to use a Hubbard-Stratanovich transformation to write $$ \mathcal{Z} = \int \mathcal{D}\phi \, e^{-\mathcal{S}} = \int \mathcal{D}\phi \mathcal{D}\tilde{\sigma} \, e^{-\mathcal{S}'} $$ where $$ \mathcal{S}' = \int d^dx \,\left[ \frac{1}{2} \left( \partial_{\mu} \phi_{\alpha} \right)^2 + \frac{i \tilde{\sigma}}{2\sqrt{N}} \left( \phi^2_{\alpha} - N m^2 \right) - \frac{\tilde{\sigma}^2}{8 \lambda} \right]. $$ At this point, we can integrate out the Gaussian fields $\phi_{\alpha}$, obtaining a theory of the form $\mathcal{Z} = \int \mathcal{D} \tilde{\sigma} \, e^{- N \mathcal{S}[\tilde{\sigma}]}$. This can be solved using a saddle-point method, as I discuss in a previous answer of mine. One expands the field as $i \tilde{\sigma} = \Delta^2 + i \sigma$. Then we will obtain a partition function of the form $$ \mathcal{Z} = e^{- N \mathcal{S}'[\Delta^2]} \int \mathcal{D}\sigma \, \exp \left[ \frac{1}{2} \int \frac{d^d p}{(2 \pi)^d} \left( \frac{\Pi(p)}{2} + \frac{1}{4 \lambda} \right) |\sigma(p)|^2 + O(1/\sqrt{N}) \right]. $$ We can drop corrections to the $N=\infty$ solution, and we have solved the theory in principle. One can show that the value of $\Delta^2$ which minimizes the action is given by $$ m^2 + \frac{\Delta^2}{2\lambda} = \int \frac{d^dp}{(2 \pi)^d} \frac{1}{p^2 + \Delta^2}, $$ and I've also introduced the function $$ \Pi(p) = \int \frac{d^d k}{(2\pi)^d} \frac{1}{(k^2 + \Delta^2)((k+p)^2 +\Delta^2)}. $$

We now consider correlation functions of the fields. These can be comuputed, for example, by coupling a source to the field we are interested in in our original theory, carrying out the saddle-point expansion, and then taking variational derivatives with respect to the sources. For the fields $\phi_{\alpha}$ we find $$ \langle \phi_{\alpha}(x) \phi_{\beta}(0) \rangle = \int \frac{d^dp}{(2 \pi)^d} \frac{\delta_{\alpha \beta} \, e^{i p \cdot x}}{p^2 + \Delta^2}. $$ This implies that correlations of the $\phi$ fields decay exponentially unless $\Delta = 0$, in which case $$ \langle \phi_{\alpha}(x) \phi_{\beta}(0) \rangle \sim \frac{\delta_{\alpha \beta}}{|x|^{d - 2}}. $$ We can tune to $\Delta = 0$ by fine-tuning the mass term, $m_c^2 = \int \frac{d^dp}{(2 \pi)^d} \frac{1}{p^2}$. (We have a positive rather than negative $m^2$ because I defined $m^2$ with a different sign than is usual.) Of course, we should regulate this integral in the UV to get a finite value for $m_c$. The integral is IR divergent for $d \leq 2$; there is no gapless solution to the saddle-point equation in this case. In tuning $m_c$ to this value, we have tuned to the line between $G$ and $WF$ in the above picture.

We can conclude that, as a function of length scale $x$, the scaling dimension of the $\phi_{\alpha}$ fields does not change; it is equal to the free-field value of $D_{\phi} = (d-2)/2$ in both the $G$ and $WF$ CFTs.

But not all operators behave so trivially. Consider the O($N$) singlet operator, $\phi^2 \equiv \sum_{\alpha} \phi_{\alpha} \phi_{\alpha}$. By coupling this to a source field, one can show the identity $$ \langle \sigma(x) \sigma(0) \rangle = 4 \lambda \delta^d(x) - \frac{4 \lambda^2}{N} \langle \phi^2(x) \phi^2(0) \rangle. $$ So by studying the behavior of the $\sigma$ field using the above Gaussian theory, we can determine the scaling dimension of $\phi^2$. For $\Delta = 0$, it is not hard to show $\Pi(p) = F_d p^{d - 4}$ for an uninteresting dimensionless constant $F_d$, and we can read off the $\sigma$ propagator: $$ G_{\sigma}(p) = \frac{2}{\Pi(p) + 1/(2 \lambda)} = \frac{2}{F_d p^{d - 4} + 1/(2\lambda)}. $$ I will rewrite this to make its dependence on $\lambda$ more apparent: $$ G_{\sigma}(p) = \frac{1}{p^{d - 4}} \frac{4 \lambda p^{d - 4}}{2 F_d \lambda p^{d - 4} + 1}. $$ The point of this rewriting is to single out the dimensionless combination $\lambda p^{d - 4}$, which clearly controls the flow between the IR ($p \rightarrow 0$) and the UV ($p \rightarrow \infty$).

First consider the IR. Since we are assuming $d<4$, we find $G_{\sigma}(p) = 2p^{4 - d}/F_d$, so after a Fourier transform, we expect $$ \langle \phi^2(x) \phi^2(0) \rangle \sim \frac{N/\lambda^2}{|x|^{4}}. $$ We find that the IR scaling dimension is $D_{\phi^2} = 2$, which is not equal to twice the scaling dimension of $\phi_{\alpha}$ as in the free theory. Note that this precise power of $\lambda^2$ which appears on the right-hand side is required for the engineering and scaling dimensions of $\phi^2$ to match.

In contrast, in the UV one has $$ G_{\sigma}(p) = 4 \lambda - 8 \lambda^2 F_d p^{d - 4} + \cdots $$ After a Fourier transform, the first term on the right-hand side generates the delta function indicated above (with the correct factor of 4), and we find $$ \langle \phi^2(x) \phi^2(0) \rangle \sim \frac{N}{|x|^{2(d - 2)}}, $$ indicating that the scaling dimension takes its free-field value $D_{\phi^2} = (d - 2) = 2 D_{\phi}$. So the UV of this theory is at the Gaussian fixed point. Note that the $\lambda$ dependence dropped out.

Of course, for intermediate observation scales, one needs to compute the entire function $$ \int \frac{d^d p}{(2 \pi)^d} \frac{e^{i p \cdot x}}{p^{d - 4}} \frac{4 \lambda p^{d - 4}}{2 F_d \lambda p^{d - 4} + 1} $$ to obtain how the correlator $\langle \phi^2(x) \phi^2(0) \rangle$ behaves as a function of $\lambda |x|^{4 - d}$, and it will not behave as a power-law (and therefore not describe a CFT) until you take the UV or IR scaling limits.

  • $\begingroup$ Hi, Seth. I am the writer of this post and you said that my question was exactly the same in the derivation of the coorelation function in this large N approximation. After reading the derivation of correlation in IR limit, may I ask that why the scaling dimension of of $\phi^{2}$ is just $D_{\phi^{2}} = 2$ but not $D_{\phi^{2}} = 0.5 ( -2D_{\lambda} - 4D_{x} )$? Is it true that the scaling dimension only count the power in $|x|$? $\endgroup$
    – Ricky Pang
    Commented Jul 27, 2021 at 15:20
  • $\begingroup$ @RickyPang I do not understand what you are referring to as $D_{\lambda}$ or $D_x$. By definition, the scaling dimension of an operator $\mathcal{O}$ with definite scaling dimension is given by the constant $\Delta$ appearing in the equation $\langle \mathcal{O}(x) \mathcal{O}(0)\rangle = 1/|x|^{2\Delta}$. $\endgroup$ Commented Jul 29, 2021 at 13:53
  • 1
    $\begingroup$ Thank for your answer. The $D_{\lambda}$ and $D_{x}$ is the engineering dimension of the constant $\lambda$ and length $x$. Now I understand what is different of scaling dimension and engineering dimension. Thank for your detailed derivation! Besides clarifying the concept in scaling dimension, I also learnt something new in large N expansion. I appreciate your help very much. $\endgroup$
    – Ricky Pang
    Commented Jul 29, 2021 at 14:03

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