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.

  • $\begingroup$ 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... $\endgroup$ – M.Jo Feb 3 '20 at 11:29
  • $\begingroup$ Could you show that explicitly in an answer? I want to see explicitly the calculations that lead to the correlation being zero for $\Delta_1\ne\Delta_2$ $\endgroup$ – redhood Feb 4 '20 at 7:41

You did all the work but just missed the last step:

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...

  • $\begingroup$ I don't see how your last step follows. I think I'm not sure how $(k_{1\mu}+k_{2\mu})$ acts on $|x_1-x_2|^{-(\Delta_1+\Delta_2)}$ $\endgroup$ – redhood Feb 4 '20 at 11:58
  • $\begingroup$ $k_\mu$ is a simple differential operator, so the chain rule applies, $\partial_\mu |x_1 - x_2|^\alpha = \alpha |x_1 - x_2|^{\alpha - 1} \partial_\mu |x_1 - x_2|$ $\endgroup$ – M.Jo Feb 4 '20 at 12:12
  • $\begingroup$ Of course, thank you very much! $\endgroup$ – redhood Feb 4 '20 at 12:22

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