# Reconcile three-point function with OPE in CFT

Three-point function of primary operators in conformal field theory can be fixed up to a constant by symmetry considerations $$\left\langle V_{h_1}(z_1)V_{h_2}(z_2)V_{h_3}(z_3)\right\rangle=\frac{C_{h_1h_2h_3}}{(z_1-z_2)^{h_1+h_2-h_3}(z_1-z_3)^{h_1+h_3-h_2}(z_2-z_3)^{h_2+h_3-h_1}}$$ On the other hand, it should be possible to derive this result using the operator product expansion $$V_{h_1}(z_1)V_{h_2}(z_2)=\sum_{h,K}C^{h,K}_{h_1h_2}(z_1-z_2)^{h+|K|-h_1-h_2}L_{-K}V_h(z_2)$$ here the sum is performed over all primary dimensions $h$ and their descendants parametrized by multi-index $K=\{k_1,...,k_2\}, |K|=k_1+\dots+k_n$. The OPE expansion is valid inside correlation function given that there are no other fields between points $z_1$ and $z_2$. For the three-point function this implies that we must have $|z_3-z_2|>|z_1-z_2|$. With this assumption we can rewrite the three-point function as $$\left\langle V_{h_1}(z_1)V_{h_2}(z_2)V_{h_3}(z_3)\right\rangle=\sum_{h,K}C^{h,K}_{h_1h_2}(z_1-z_2)^{h+|K|-h_1-h_2}\left\langle L_{-K}V_h(z_2)V_{h_3}(z_3)\right\rangle$$

Now, I believe that all the two-point functions inside the r.h.s. vanish except for $h=h_3$ and $K=\{\}$, i.e. for a primary field of the weight $h_3$ (this might be the wrong assumption leading to the inconsistency). Then, using the two-point correlator $V_{h}(z_1)V_h(z_2)=C(h)(z_1-z_2)^{-2h}$ one arrives at the following expression

$$\left\langle V_{h_1}(z_1)V_{h_2}(z_2)V_{h_3}(z_3)\right\rangle=(z_1-z_2)^{h_3-h_1-h_2}(z_2-z_3)^{-2h_3} C(h_3)C^{h_3,\{\}}_{h_1h_2}$$

which clearly has a different coordinate dependence. So, how does one derive the correct three-point function using the OPE?

You should sum over all descendents $K$. In general their contributions do not vanish, although they are subleading in the limit $z_1\to z_2$. For example the two-point function $$\langle L_{-1}V_{h_3}(z_2)V_{h_3}(z_3) \rangle = \frac{\partial}{\partial z_2}\langle V_{h_3}(z_2)V_{h_3}(z_3) \rangle$$ is clearly nonzero.
• Thanks, now I see that. Given that there is no closed form for descendant coefficients $C^{h,K}_{h_1h_2}$ at arbitrary $K$ (correct me if I'm wrong), do you think that it is possible to perform the whole sum analytically? – Weather Report Feb 3 '17 at 20:12
• I do not think you can perform the whole sum over Virasoro descendents, and I also do not see why you would want to do that. What is probably doable is to forget the Virasoro symmetry and sum over $s\ell_2$ descendents i.e. powers of $L_{-1}$. This should give you the same three-point function. – Sylvain Ribault Feb 3 '17 at 20:21
• @WeatherReport, indeed only the $sl_2$ descendants will have a non-zero two-point function (since the two-point function is non-vanishing only between $sl_2$-descendants of quasi-primaries of the same dimensions, and non-$sl_2$ Virasoro descendants come from quasi-primaries of a higher dimension). In 2d the coefficients $C$ are easy to derive, while in higher-d you can view this exercise as a way of computing them. – Peter Kravchuk Feb 7 '17 at 8:06