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?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Feb 3, 2017 at 20:12
  • $\begingroup$ 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. $\endgroup$ Feb 3, 2017 at 20:21
  • $\begingroup$ Well, to see that everything works :) Anyway, the key relevant observation is contained in your answer. $\endgroup$ Feb 3, 2017 at 20:39
  • $\begingroup$ @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. $\endgroup$ Feb 7, 2017 at 8:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.