Superrenormalizable phi cubed theory

  • Context: Graduate 
  • Thread starter Thread starter LAHLH
  • Start date Start date
  • Tags Tags
    Phi Theory
Click For Summary
SUMMARY

The discussion confirms that the phi cubed theory in four dimensions (d=4) is superrenormalizable, meaning it has only a finite number of divergent terms. The superficial degree of divergence is calculated using the formula D=dL-2I, with specific relationships established between vertices (V), external legs (E), and internal propagators (I). The analysis shows that for diagrams with E+V≤4, divergences exist, while all diagrams above this threshold converge, validating the superrenormalizability of the theory.

PREREQUISITES
  • Understanding of quantum field theory concepts
  • Familiarity with the phi cubed model
  • Knowledge of dimensional analysis in physics
  • Proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study the implications of superrenormalizability in quantum field theories
  • Explore the phi cubed theory in different dimensions
  • Learn about the renormalization group and its applications
  • Investigate the relationship between divergences and physical observables in quantum field theories
USEFUL FOR

The discussion is beneficial for theoretical physicists, quantum field theorists, and graduate students focusing on renormalization techniques and the properties of scalar field theories.

LAHLH
Messages
405
Reaction score
2
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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K