Constraining the 2-point correlation function

Consider the two-point function $$\langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\rangle=f(x_1,x_2)$$ If the operators are in a CFT, we can constrain this function using the symmetries of the theory. Using translational symmetries and the symmetries of the Lorentz group we have $$f(x_1,x_2) = f(X_{12})$$ where $$X_{12} := (x_1-x_2)$$

When we impose dilatation symmetry we get $$\langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\rangle=\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}$$ where $$\Delta_1,\Delta_2$$ the dilatation weights of the operators.

Now if we impose special conformal symmetries we have$$\left(-2x_{1\mu}\Delta_1-2x_{2\mu}\Delta_2+k_{1\mu}+k_{2\mu}\right)\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}}=0$$ If we make use of $$(k_{1\mu}+k_{2\mu})|x_1-x_2|=-(x_{1\mu}+x_{2\mu})|x_1-x_2|$$ we should be able to derive $$\langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\rangle=\frac{C_{12}}{|x_1-x_2|^{2\Delta}}$$ for $$\Delta_1=\Delta_2=\Delta$$ and $$0$$ otherwise.

I can't derive the final equation from the given identity of $$k$$'s, is there something else that I'm missing?

Edit:$$k_\mu=x^2\partial_\mu-2x_\mu x^\nu\partial_\nu$$ The operator associated to special conformal transformations.

• Where exactly is the problem? Did you compute the action of $k_{1\mu}$ and $k_{2\mu}$ on $C_{12}/|x_1 - x_2|^{\Delta_1 + \Delta_2}$? It should be pretty straightforward after that... – M.Jo Feb 3 '20 at 11:29
• Could you show that explicitly in an answer? I want to see explicitly the calculations that lead to the correlation being zero for $\Delta_1\ne\Delta_2$ – redhood Feb 4 '20 at 7:41

As you wrote, from the definition of $$k_\mu$$ you have $$(k_{1\mu}+k_{2\mu})|x_1-x_2|=-(x_{1\mu}+x_{2\mu})|x_1-x_2|,$$ and so $$(k_{1\mu}+k_{2\mu})\frac{C_{12}}{|x_1-x_2|^{\Delta_1 + \Delta_2}} = (\Delta_1 + \Delta_2) (x_{1\mu}+x_{2\mu}) \frac{C_{12}}{|x_1-x_2|^{\Delta_1 + \Delta_2}}$$ This means that the constraint from special conformal symmetry becomes $$\left(-2x_{1\mu}\Delta_1-2x_{2\mu}\Delta_2+k_{1\mu}+k_{2\mu}\right)\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}} = (\Delta_2 - \Delta_1) (x_{1\mu} - x_{2\mu})\frac{C_{12}}{|x_1-x_2|^{\Delta_1+\Delta_2}} =0.$$ The only way the last equality can be valid for all points $$x_1$$ and $$x_2$$ is that $$\Delta_1 = \Delta_2$$. Now you call this $$\Delta$$ and you have your result...
• I don't see how your last step follows. I think I'm not sure how $(k_{1\mu}+k_{2\mu})$ acts on $|x_1-x_2|^{-(\Delta_1+\Delta_2)}$ – redhood Feb 4 '20 at 11:58
• $k_\mu$ is a simple differential operator, so the chain rule applies, $\partial_\mu |x_1 - x_2|^\alpha = \alpha |x_1 - x_2|^{\alpha - 1} \partial_\mu |x_1 - x_2|$ – M.Jo Feb 4 '20 at 12:12