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If we don't consider theory like massive vector field, then the superficial degree of divergence is $$D= 4 - E_b -\frac{3}{2} E_f -\sum_{i}v_i \delta_i$$ where $E_b$ is the number of external boson line, $E_f$ is the number of external fermi line, $i$-th kind of vertice occurs $v_i$ times and the mass dimension of the coefficient of $i$-th kind of vertice is $\delta_i$.

I know that $D<0$ do not means the diagram is convergent, because there may be subdiagram with $D>0$.

My question is:

1 For a diagram with $D= 0$ does it mean that this diagram must be divergent? Does there exist some counterexample?

2 For a diagram with $D> 0$ does it mean that this diagram must be divergent? Does there exist some counterexample?

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Hint: consider the box diagram

enter image description here

(From Wikipedia)

For more details about this diagram, see chapter 10.1 in Peskin&Schroeder (e.g., fig. 10.2, and the discussion around eq. 10.6), and problem 10.1 (you can probably find the solution online).

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  • $\begingroup$ Does it mean for $D\ge0$ only $D=0$ is possible to have convergent result? $D>0$ must not have convergent result? $\endgroup$ – user153663 May 13 '17 at 14:33

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