Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Psi^4 theory

  1. Jun 10, 2008 #1
    I've been analysing the following toy theory which I've called psi^4 theory for want of a better name.

    [itex]\mathcal{L} = :i\bar{\psi}\gamma^\mu\partial_\mu \psi - m\bar{\psi}\psi + \lambda (\bar{\psi}\psi)^2:[/itex].

    Ie a fermion with quartic self-interaction. This interaction can describe contact processes such as [itex]\psi + \bar{\psi} \to \psi + \bar{\psi}[/itex] whose Feynman rule I derived to be [itex]4i\lambda[/itex].

    Interestingly, the process [itex]\psi + \bar{\psi} \to \psi \bar{\psi}[/itex] has a Feynman rule of zero and consequently zero scattering amplitude. This comes about because if one expands the interaction Lagrangian in positive and negative frequency parts, there are four operator contributions which cancel after normal ordering. Does anyone know why this might be expected physically (ignoring the obvious unphysicality of the Lagrangian).
     
  2. jcsd
  3. Jun 10, 2008 #2
    Phi^4 is a scalar theory. I think this is more like a four-Fermi interaction. I was unaware that the amplitude were zero. I would think that it is equal to [tex]4i\lambda[/tex], after renormalization, but I haven't actually done the calculation. It might have to do with antisymmetry after exchange of external legs, but I'm not sure how.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?