Superrenormalizable phi cubed theory

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The discussion focuses on demonstrating that phi^3 theory in four dimensions 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, leading to the conclusion that D=4-E-V. It is established that diagrams with E+V≤4 will diverge, while those above this threshold will converge. Participants confirm that increasing the order of vertices will eventually eliminate divergences, aligning with the definition of superrenormalizability. The analysis concludes that the theory is indeed superrenormalizable.
LAHLH
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Hi,

I'm trying to show that phi^3 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 D=[g_E]-V[g_3]

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

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?
 
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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.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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