In page 78 of David Tong's notes on CFT https://www.damtp.cam.ac.uk/user/tong/string/four.pdf, he finds that the propagator for a theory of free massless scalars is

$$\langle X(\sigma)X(\sigma')\rangle=-\frac{\alpha'}{2}\text{log}(\sigma-\sigma')^2$$

Then he goes on saying that in equation (4.22) the OPE of $X(\sigma)X(\sigma')$ is

$$X(\sigma)X(\sigma')=-\frac{\alpha'}{2}\text{log}(\sigma-\sigma')^2 + ... \label{eq}$$

My question is: what are the $...$ in the above equation?

Certainly they are there because he is saying that we could consider the path integral with other isertions away from $\sigma$ and $\sigma'$, but then equation (4.20) would turn to

$$\langle \partial^2 X(\sigma)X(\sigma')...\rangle=-2 \pi \alpha' \langle \delta^2(\sigma,\sigma') ... \rangle$$

Implying that

$$\langle X(\sigma)X(\sigma')...\rangle =-\frac{\alpha'}{2}\langle \text{log}(\sigma-\sigma')^2... \rangle \\ \hspace{2.8cm}=-\frac{\alpha'}{2}\text{log}(\sigma-\sigma')^2\langle ... \rangle$$

Because all insertions are away from $\sigma$ and $\sigma'$.


1 Answer 1


In the free theory the answer is really simple, $$ X(\sigma)X(\sigma') = - \frac{\alpha'}{2} \log(\sigma - \sigma') ~ + : X(\sigma)X(\sigma'): $$ where $:~:$ denotes normal ordering.

  • $\begingroup$ And the insertions would be multiplying both sides then? $\endgroup$
    – Slayer147
    Apr 8, 2020 at 0:39
  • $\begingroup$ That would be correct. $\endgroup$
    – Prahar
    Apr 8, 2020 at 0:55
  • $\begingroup$ One more question, this only holds if conformal normal ordering, which is the one you are using, is the same as creation annihilation normal ordering, right? (And in the case of the free massless boson we know they are at least) $\endgroup$
    – Slayer147
    Apr 8, 2020 at 2:25
  • $\begingroup$ What I have said holds only in free field theory where both types of normal ordering are equivalent. The formula I have written does not hold in generic CFTs anyway. $\endgroup$
    – Prahar
    Apr 8, 2020 at 2:27

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.