# Constraints on correlation functions of Quasi Primary Fields

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 $$(7)$$.

• 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. – AccidentalFourierTransform Dec 6 '18 at 20:31
• @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 $g$? – Run like hell Dec 7 '18 at 0:07
• @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? – Run like hell Dec 7 '18 at 0:13

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.
• 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 $\Delta_1=\Delta_2$. – user213887 Dec 7 '18 at 15:40