I've seen OPEs commonly used in 2d CFT, it's quite apparent to me that, in that case, it dresses a bridge between the algebraic and the operator formalism especially when combined with radial ordering and the use of contour integral. Even more powerful in the minimal models where it leads to the bootstrap equations and the resolution of the 3 pt functions. I also heard that OPE are sometimes used in other circumstances, for instance in the QCD chapter of Peskin & Schroeder's book but I don't recall the motive. I'd be curious to know what generally it is relevant to decompose the products of operators into an "operator basis" ie. associate an algebra to the space of operators.

  • $\begingroup$ In P&S, the OPE was used to discuss Deep Inelastic scattering. It also discusses $e^+e^-$ scattering using OPEs $\endgroup$ – Prahar Jun 6 '13 at 22:12

My memory of Peskin & Schroder is a little hazy, but they're probably discussing the Shifman-Vainshtein-Zakharov sum rules. The idea is that you can use the OPEs for composite operators representing mesons/hadrons to derive formulae that express meson/hadron n-point functions in terms of the VEVs of various QCD condensates. (Edit: Just discovered that Shifman has some very nice lecture notes on the subject.)

More generally: OPEs are always relevant (if not always easy to use in a given situation) because they carry almost all the information about a field theory. You can actually define a QFT by writing down the set of local observables, the OPEs between them, and the VEVs of the local observables. It's as good a formalism as the Hamiltonian or path integral formalisms -- better in some ways, because it applies when the path integral doesn't.

  • $\begingroup$ Could you expand just a little on your very last point - i.e. what you had in mind for a situation where the PI doesn't apply but OPE does? $\endgroup$ – twistor59 Jun 7 '13 at 6:20
  • $\begingroup$ Hi user1504, a small detail but it seems that even in CFT one likes to think that the path integral exists, even when no Lagrangian description is available. $\endgroup$ – Vibert Jun 7 '13 at 8:14
  • $\begingroup$ @twistor59 I was thinking of the 6d (2,0) theory, chiral 2d CFTs, situations where you can't just guess an action and begin. $\endgroup$ – user1504 Jun 7 '13 at 11:05
  • $\begingroup$ @Vibert Sure, the path integral always exists, at least in the sense that you can insert sums over intermediate states. But without a Lagrangian to start from, you can't do much. $\endgroup$ – user1504 Jun 7 '13 at 11:09
  • $\begingroup$ I was under the impression that the OPEs gave the 3-point functions in a theory. So if the collection of OPEs completely determines the theory, does that mean the theory has no higher (fundamental) vertices i.e. interaction terms in the lagrangian? $\endgroup$ – Siva Jun 8 '13 at 3:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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