Superrenormalizable phi cubed theory

  • Thread starter LAHLH
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Hi,

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?
 

Answers and Replies

  • #2
DarMM
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It is correct. For any set of external legs ##E \leq 4## increasing order in ##V## will eventually remove divergences, thus one only has divergences in a finite number of graphs. That is the definition of super-renormalizability.
 

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