Ryder at the beginning of the chapter about renormalization defines the "superficial degree of divergence" of diagrams in $\lambda \phi^4$ theory. I'll recap the derivation.

A diagram in $\lambda\phi^4$ theory has $E$ external lines, $I$ internal lines of which $L$ are loops and $n$ vertices. Each internal line contributes with a factor $\frac{1}{p^2}$ and each undetermined loop momentum with an integral $\int d^d p$ in $d$ dimensions, so we define the superficial degree of divergence as $$D=dL-2I$$

and we say that if $D\geq0$, the diagram must diverge.

But consider the simplest diagram in $\lambda\phi^4$ with 4 external lines and no loops, just crossing in the center like an $X$. This diagram has no internal lines and no loops, so $D=0$ and the diagram should diverge logarithmically. Nevertheless, the value of the diagram is just $$ -i\lambda(2\pi)^4\delta^{(4)}(p_1+p_2-p_3-p_4)$$

which is finite. What happened?


1 Answer 1


As we start study divergence of diagrams only when we introduce loops, I think this formula for the superficial degree of divergence is meant to be applied only to those cases where at least a loop is present;


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