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

Ignoring SSB in massless phi^4 theory

  1. Jul 10, 2014 #1
    If a theory is spontaneously broken, you usually make calculations by expanding the field about the ground-state expectation value [itex]\nu [/itex], so that you define a new field [itex]\rho [/itex] such that [itex]\phi=\nu+\rho [/itex] such that [itex]\langle \rho \rangle =0 [/itex].

    What if you just calculate perturbation theory with [itex]\phi [/itex] instead of the new field [itex]\rho[/itex]?

    It seems that massless [itex]\phi^4[/itex] is spontaneously broken at the 1-loop level. So do you have to do perturbation theory with the field [itex]\rho[/itex] instead of [itex]\phi [/itex]?
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted