I've been looking at an example model recently of a massive spin 1 vector field coupled to the conserved current of a massive complex scalar field. My previous question was far too long so I deleted it (it's what I get for posting when I haven't slept in over two days). I'm calling the massive scalar particles phi particles/antiparticles and the massive vector field particles X-Boson's.

Here's my problem; when I compute the tree-level total cross section for the equivalent of Compton Scattering in this theory (one X-boson and one phi particle as the initial state, one X-boson and one phi particle as the final state), I find that the total cross section diverges as we go to arbitrarily high COM three-momenta IF we consider the possibility that the incoming X-boson has a longitudinal polarization. This sort of behavior doesn't happen with a massless spin 1 vector field since such a vector field doesn't have a longitudinal polarization mode. How exactly am I supposed to interpret this? The total cross section is supposed to be bounded between 0 and 1, right? Is perturbation theory just breaking down at high energies? If it is breaking down at high energies, how are you supposed to compute cross sections at those energies?


  • $\begingroup$ The total cross section has dimensions -2 so it cannot be bounded by "0 and 1". The growth of the cross section is just an artifact of perturbation theory. Resum the diagrams for the rescattering of the longitudinally polarized bosons to keep that growth under control. $\endgroup$
    – QuantumDot
    May 12, 2017 at 13:26
  • $\begingroup$ The result of the calculation due to the longitudinal polarization at high energy can be understood by looking at the would-be Goldstone bosons in the limit of vanishing gauge coupling (indeed, this leading behavior is idependent of the gauge coupling). The scattering of Goldstone Bosons (GBs) does indeed grows with energy, since the GBs are derivatively coupled. You've just rediscovered the need for a Higgs boson to unitarize the scattering of the Goldstobe bosons, that is the longitudinal polarizations. $\endgroup$
    – TwoBs
    May 13, 2017 at 23:08


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