In conformal field theory the operator dimension $\Delta$ determines how fields and thus correlation functions behave under rescaling. I am having trouble seeing how this number arises from a scale transformation of a field, namely, if we rescale

$$x^\mu \rightarrow \lambda x^\mu$$

how do we arrive at the fact that the field should transform as

$$\phi(x)\rightarrow\lambda^\Delta \phi(\lambda x)$$

This is usually stated as a fact in most CFT textbooks I have read (eg. Fradkin) but never shown explicitly. I'm sure it is a straightforward exercise but I cannot seem to do it. I know that a scalar field in 4 dimensions for example should have $\Delta=1$, so I begin by writing

$$\phi(x)=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_p}}(a_pe^{ipx}+a_p^\dagger e^{-ipx})$$

But I cannot see how this would transform into $\lambda\phi(\lambda x)$ if I take $x\rightarrow \lambda x$. Can someone enlighten me?


  • $\begingroup$ Do $x\to\lambda x$ as the scaling trafo and then do $p\to \lambda p$ as a substitution in the integral, which will yield the original integral times an overall factor of $\lambda$. $\endgroup$ – ACuriousMind Mar 24 '15 at 15:16
  • $\begingroup$ This is a well-posed HW question. It gives the problem and solution and asks for an explanation to clarify why the solution is the case. It implies that a conceptual understanding of the strategies and physics is preferred over an analysis of this particular case. I'm not sure why this has a vote to close already, but I think it's a good question $\endgroup$ – Jim Mar 24 '15 at 15:23

You can rescale by $x \to \lambda x$ in that case the integral becomes:

$$ \phi(\lambda x)=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_p}}(a_pe^{i\lambda px}+a_p^\dagger e^{-i\lambda px}) $$

We want this integral now in terms of the original integral. Since we are integrating over p, we are free to redefine the this variable into anything we like. We can use the substitution $\lambda p \to p'$, which in this case yields:

$$ \phi(\lambda x)=\frac{1}{\lambda}\int\dfrac{d^3p'}{(2\pi)^3}\dfrac{1}{\sqrt{2E_{p'}}}(a_{p'}e^{ip'x}+a_{p'}^\dagger e^{-ip'x}) $$

and hence it follows that:

$$ \lambda\phi(\lambda x)=\int\dfrac{d^3p'}{(2\pi)^3}\dfrac{1}{\sqrt{2E_{p'}}}(a_{p'}e^{ip'x}+a_{p'}^\dagger e^{-ip'x}) = \phi(x)$$

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  • $\begingroup$ Shouldn't $a_p = a_{\lambda^{-1} p'}$? $\endgroup$ – Prahar Mar 24 '15 at 16:30
  • $\begingroup$ @Prahar, yeah I think thats correct, thought I don't think it affects the rest of the answer. $\endgroup$ – Constandinos Damalas Mar 24 '15 at 16:34
  • $\begingroup$ How so? You end up getting $\lambda \phi (\lambda x) \sim a_{\lambda^{-1} p'}$ and this $\lambda$ does not come out of the integral. $\endgroup$ – Prahar Mar 24 '15 at 16:35
  • $\begingroup$ Thanks for the help. I have two questions regarding your comment though PhotonicBoom. By $d^3p$ we mean $dp_xdp_ydp_z$ correct? Then are you substituting each dimension like $\lambda^{1/3}dp_i\rightarrow dp'_i$ ? Furthermore, doesn't the answer depend on your choice of substitution? I guess I'm not seeing why you always get a factor $1/\lambda$ out front. $\endgroup$ – pill Mar 24 '15 at 17:01

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