7
$\begingroup$

Single particle states in quantum field theory appear as discrete components in the spectrum of the Poincare group's action on the state space (i.e. in the decomposition of the Hilbert space of quantum states into irreducible representations of the Poincare group). Classification of irreducible unitary representations of the Poincare group leads to the notions of mass and spin.

Now, suppose we have a conformal QFT and are doing the same trick with the conformal group. Which irreducible representations do we have?

We still have the massless particles (at least I'm pretty sure we do although I don't immediately see the action of special conformal transformations). However, all representations for a given spin $s$ and any mass $m > 0$ combine into a single irreducible representation.

  • What sort of physical object corresponds to this representation?
  • Is it possible to construct a scattering theory for such objects?
  • Is it possible to define unstable objects of this sort?
Improve this question
$\endgroup$
2
  • $\begingroup$ Very very naive question: you say there will be (irreducible) representations with a fixed spin $s$ and any mass $m>0$. Since any mass $m$ introduce a length scale $L\sim \frac 1m$, conformal transformations would transform states of different masses into each other. So you would need a theory of uncountable number of particles with any mass $m>0$? If this is correct, doesn't it (naively) seem to be quit hopeless to construct any consistent quantum field theory of this kind? Has such a theory ever been constructed? $\endgroup$
    – Heidar
    Dec 3, 2011 at 1:05
  • $\begingroup$ @Heidar, these states would not be particles. This is because the mass spectrum within each such representation is continuous. $\endgroup$
    – Squark
    Dec 3, 2011 at 8:32

1 Answer 1

Reset to default
10
$\begingroup$

Representation theory of the conformal group is discussed in the canonical reference by Mack. As for physical interpretation of the theory, the construction of asymptotic states and scattering theory does not work in CFT for the reasons you write. Rather, the basic observables are Euclidean correlation functions, and the operators of the theory can be arranged into Hilbert space. This is explained in the classic paper of Mack and Luscher.

Improve this answer
$\endgroup$

Your Answer

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