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I was looking at this paper hep-th/0011040 and I found the following equation:

$$ \langle C_{\mu_1 \dots \mu_l} \mathcal{O}^{\mu_1 \dots \mu_l}(x_1) D_{\nu_1 \dots \nu_l} \mathcal{O}^{\nu_1 \dots \nu_l}(x_2)\rangle $$$$= \dfrac{1}{|x_1 - x_2|^{2\eta}} C_{\mu_1 \dots \mu_l}I^{\mu_1 \nu_1}(x_{12}) \ldots I^{\mu_l \nu_l}(x_{12}) D_{\nu_1 \dots \nu_l}, \tag{2.6}$$

where the $\mathcal{O}$ are spin $\ell$ symmetric traceless primary operators of dimension $\eta$ and

$$ I^{\mu \nu}(x) = \delta^{\mu \nu} - 2\dfrac{x^\mu x^\nu}{x^2}. \tag{2.7}$$

Could anyone tell me how to arrive at this result, or any reference where this has been given? I'm also interested in understanding the analogous derivation for $3$-point functions.

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    $\begingroup$ "some papers"...which papers? Also, hint: The two-point function is completely fixed by symmetry - the $\lvert x_1 - x_2 \rvert^{-2\eta}$ is by conformal symmetry, and the $I^{\mu\nu}$ comes by just plugging in the ansatz $I^{\mu\nu} = a \delta^{\mu\nu} - b\frac{x^\mu x^\nu}{x^2}$. $\endgroup$ – ACuriousMind Apr 17 '15 at 12:23
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    $\begingroup$ Conformal Four Point Functions and the Operator Product Expansion - arxiv.org/abs/hep-th/0011040 $\endgroup$ – Jaswin Apr 17 '15 at 12:25
  • $\begingroup$ arxiv.org/abs/hep-th/0011040 I am looking for the derivation of 2.3 and 2.6 of the above equation. $\endgroup$ – Jaswin Apr 20 '15 at 13:38
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Note: see comments for generalization from scalar fields to higher spins.

I shall do the $2$-point case and leave the $3$-point one to you as an exercise. The canonical reference which I'm using is Di Francesco, Mathieu and Senechal. This is pretty much obligatory reading if you're interested in conformal field theory.

Let $\phi_1$ and $\phi_2$ be two primary fields. Then by definition their conformal transformations are

$$\phi_1(x_1)=\left|\frac{\partial x'}{\partial x}\right|_{x=x_1}^{\Delta_1/d}\phi_1(x_1')$$

where $d$ is the spacetime dimension and $\Delta_1$ the scaling dimension of the field. There's a similar formula for $\phi_2$. Now since the measure and action in the functional integral are conformally invariant we can promote the transformation above to one of the correlation function, viz.

$$\langle\phi_1(x_1)\phi_2(x_2)\rangle=\left|\frac{\partial x'}{\partial x}\right|_{x=x_1}^{\Delta_1/d}\left|\frac{\partial x'}{\partial x}\right|_{x=x_2}^{\Delta_2/d}\langle\phi_1(x_1')\phi_2(x_2')\rangle$$

Now we start specialising to specific symmetries. Already from rotational and translational symmetry we know

$$\langle\phi_1(x_1)\phi_2(x_2)\rangle=f(|x_1-x_2|)$$

Now using a scale transformation $x\to \lambda x$ in our formula above produces

$$f(|x_1-x_2|)=\lambda^{\Delta_1+\Delta_2}f(\lambda|x_1-x_2|)$$

but this fixes the correlation function up to a constant, that is

$$\langle\phi_1(x_1)\phi_2(x_2)\rangle=\frac{C}{|x_1-x_2|^{\Delta_1+\Delta_2}}$$

The final ingredient is invariance under special conformal transformations. These have

$$\left|\frac{\partial x'}{\partial x}\right|=\frac{1}{(1-2b\cdot x+b^2x^2)^d}$$

Substituting this into our formulae above fixes $\Delta_1=\Delta_2$. Therefore we have exactly proved your first result above

$$\langle\phi_1(x_1)\phi_2(x_2)\rangle=\frac{C}{|x_1-x_2|^{2\Delta_1}}$$

if your operators have the same scaling dimension $\Delta_1$.

Now try the $3$-point case using the same symmetries as above! Also, have a think about why this method fails at $4$-point. Hint: you can make conformally invariant cross-ratios when you have $4$ positions.

Endnote on Scaling Dimensions

Remember that the scaling dimension is not necessarily the naive engineering dimension of your field, even in a CFT. This is because loop corrections can introduce divergences even though there's global conformal symmetry. Various theorems guarantee that these don't renormalize masses or couplings but they do cause field strength renormalization $\phi \to \sqrt{Z}\phi$. Typically $Z$ carries some dimensionality, hence we get anomalous dimensions. This argument only falls down if the conformal symmetry is local (as in string theory) or if the operator is protected by supersymmetry.

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  • $\begingroup$ I am looking for a l-spin symmetric traceless primary fields, not for scalars. I know the two-point and three-point function derivation for scalars, as I have gone through the first 6 chapters of the Yellow Book. $\endgroup$ – Jaswin Apr 25 '15 at 11:27
  • $\begingroup$ Ah okay - I'll try to find a good reference for higher spin operators tomorrow and add it to my answer! Incidentally it's useful to put some background on your knowledge into the question so that this confusion doesn't arise :). $\endgroup$ – Edward Hughes Apr 26 '15 at 1:41
  • $\begingroup$ The generalization to spin-$2\ell$ fields is provided by this paper of Osborn and Petkos. The basic idea is to show that conformal invariance forces the general tensor $C^{\mu_1\dots \mu_\ell\nu_1\dots \nu_\ell}$ into the form $I^{\mu_1\nu_1}\dots I^{\mu_\ell\nu_\ell}$. Effectively $I$ is just $C$ constrained by the impact of conformal invariance on the "direction" of tensor fields living on the manifold. $\endgroup$ – Edward Hughes Apr 26 '15 at 22:07
  • $\begingroup$ Thanks a lot for the reference, I'll go thru it, if I find some difficulty, will bug you again :) By the way the solution is still for $l-$spin primary traceless symmetric operators, not $2l$. $\endgroup$ – Jaswin Apr 27 '15 at 7:00
  • $\begingroup$ Sorry - the $2$ was a typo! Do let me know if you have any questions about the paper and I'll try to help you out. $\endgroup$ – Edward Hughes Apr 27 '15 at 7:39

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