0
$\begingroup$

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 $(7)$.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ 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, 2018 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 $g$? $\endgroup$
    – RenatoRenatoRenato
    Dec 7, 2018 at 0:07
  • 1
    $\begingroup$ @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? $\endgroup$
    – RenatoRenatoRenato
    Dec 7, 2018 at 0:13

1 Answer 1

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Oop sorry left off a sign there that's what I meant! Edited - thanks for catching that! $\endgroup$
    – user213887
    Dec 7, 2018 at 15:29
  • 2
    $\begingroup$ 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, 2018 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.