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

Superrenormalizable phi cubed theory

  1. Mar 10, 2010 #1

    I'm trying to show that [tex] phi^3 [/tex] theory in d=4 is superrenormalisable (only finite no of terms are power counting divergent).

    In the following I use, d=#dimensions, I=#internal props, E=external legs, V=#number of vertices (of phi^3 type, i.e. three valent)

    The Superficial degree of diveregence is D=dL-2I. Also it can be shown that 3V=E+2I, and also L=I-V+1. Therefore after some plugging in and algebra, I get to [tex] D=[g_E]-V[g_3] [/tex]

    In 4d, [tex] [g_E]=4-E [/tex] and [tex] [g_3] =1 [/tex]. Thus [tex] D=4-E-V [/tex]

    So all diagrams that have E+V<=4, will have D>=0 and these diagrams of the theory will diverge, but above this all diagrams will converge, therefore only finite number divergent and theory is thus superrenorm.

    Does anyone know if this is correct?
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

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