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I have problems understanding constraints on correlation functions of quasi primary fields (QPF) following DiFrancesco's Conformal field theory book. In chapter 4, section 4.2.1, a QFP is defined as a field with the following transformation law under conformal transformations

$$ \phi(x) \rightarrow \phi' (x') = \Biggl\vert{\frac{\partial x'}{\partial x}} \Biggl\vert ^{-\Delta/d} \phi(x) \tag{1} $$

In section 4.3.1 constraints on 2-point correlation functions are found, we have: $$ \langle \phi_1(x_1) \, \phi_2(x_2) \rangle = \Biggl\vert{\frac{\partial x'}{\partial x}} \Biggl\vert ^{-\Delta_1/d}_{x=x_1} \, \, \Biggl\vert{\frac{\partial x'}{\partial x}} \Biggl\vert ^{-\Delta_2/d}_{x=x_2} \langle \phi_1(x'_1) \, \phi_2(x'_2) \rangle \tag{2} $$

Specializing to a scale transformation $x'=\lambda x$ we have

$$ \langle \phi_1(x_1) \, \phi_2(x_2) \rangle \, = \,\lambda^{\Delta_1 + \Delta_2} \, \langle \phi_1(\lambda x_1) \, \phi_2(\lambda x_2) \rangle \tag{3} $$

Rotations and translations invariance require $$ \langle \phi_1(x_1) \, \phi_2(x_2) \rangle \, = \, f(\mid x_1 -x_2 \mid) \tag{4} $$

Where $$ f(x)\, = \, \lambda^{\Delta_1 + \Delta_2} \, f(\lambda x) \tag {5} $$

In other words $$ \langle \phi_1(x_1) \, \phi_2(x_2) \rangle \, = \, \frac{C_{12}}{\mid x_1-x_2 \mid^{\Delta_1+ \Delta_2}} \tag{6} $$

This passage really confuses me, why isn't it just

$$ \langle \phi_1(x_1) \, \phi_2(x_2) \rangle \, = \, \lambda^{\Delta_1+ \Delta_2} \, f(\lambda \mid x_1 -x_2 \mid) \tag{7} $$

As I think it should follow from $(5)$?

I don't get why it has that specific form showed in $(6)$ and not the general one showed in $(7)$.

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  • $\begingroup$ In $(7)$, what is $\lambda$? Is it an arbitrary parameter? If the l.h.s. is to be $\lambda$-independent, then only one possible form for $f$ can work. Hint: it is the one given by Francesco. $\endgroup$ – AccidentalFourierTransform Dec 6 '18 at 20:31
  • $\begingroup$ @AccidentalFourierTransform i didn't tag you earlier, I copy my old comment: I think $\lambda$ in $(7)$ is the scale parameter, isn't it? to be $\lambda$ independent can't $f$ be $(1/\lambda)^{\Delta_1+\Delta_2} \, g(\mid x_1-x_2 \mid)$ with a generic $g$? $\endgroup$ – Run like hell Dec 7 '18 at 0:07
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    $\begingroup$ @AccidentalFourierTransform just before tagging you and writing the comment again I may have found the flaw in it: I have something like $f(\lambda x)$ and not $f(\lambda,x)$, so if I want to have a $(1/\lambda)^{\Delta_1+\Delta_2} \, I am forced to have the term (\mid x_1-x_2 \mid)^{-\Delta_1-\Delta_2}$. Is this right? $\endgroup$ – Run like hell Dec 7 '18 at 0:13
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You're correct that rotations, translations, and scaling force $\langle \phi(x_1)\phi(x_2)\rangle=f(|x_1-x_2|)$, where $f(|x_1-x_2|)=\lambda^{\Delta_1+\Delta_2}f(\lambda|x_1-x_2|)$. But the only such $f(x)$ that obeys that last condition is $(1/x)^{\Delta_1+\Delta_2}$, so the result follows.

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  • $\begingroup$ Oop sorry left off a sign there that's what I meant! Edited - thanks for catching that! $\endgroup$ – user213887 Dec 7 '18 at 15:29
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    $\begingroup$ One other point I'd add in case it helps, is that since for special conformal transformations $|\tfrac{\partial x'}{\partial x}|=\left(1+2b\cdot x +b^2x^2\right)^{-1}$ for some constant $b$, the dimensions must be the same $\Delta_1=\Delta_2$. $\endgroup$ – user213887 Dec 7 '18 at 15:40

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