# Degree of divergence of a Feynman diagram

I am studying the degrees of divergence of Feynman diagrams. I feel that I miss something but I don't really understand what. Please apologize if this question is silly. Anyway.

As an introduction to systematics of renormalization, Peskin and Schroeder (chapter 10) introduce the counting of ultraviolet divergences. They say that an expression for any diagram typically looks like

$$\sim \int \dfrac{d^4 k_1 d^4 k_2 \dots d^4 k_L}{(\bar{k}_i-m) \dots k^2_j \dots }$$ where $\bar{k}$ is $k$ slash (i.e. $\gamma^{\mu} k_{\mu}$). The number of $d^4 k$ will depend on the number of loops, while the number of $(\bar{k}-m)$ and $k^2$ in the denominator will depend on the number of fermion/boson propagators. Ok.

They then say :

For each loop there is a potentially divergent 4-momentum integral, but each propagator aids the convergence of this integral by putting one or two powers of momentum into the denominator.

I don't understand that. To be more clear, I'll develop a little bit more...

They then introduce the superficial degree of divergence $D$ as $$D \equiv (\text{power of } k \text{ in numerator}) - (\text{power of } k \text{ in denominator})$$

and they say that we can expect a diagram to have a divergence proportional to $\Lambda^D$ ($\Lambda$ is a cut-off) when $D>0$, proportional to $\log(\Lambda)$ when $D=0$ and no divergence when $D<0$.

I know that it's a tool and not always exact and so on to evaluate the divergence of a Feynman diagram. That's not the point. What I don't get is why is a diagram with $D<0$ not divergent ? Integrals like

$$\int \dfrac{d^4 k}{k^7}$$ don't converge, so the associated diagram should be divergent too... Or is it because we are only dealing with UV divergences, and that diagrams with $D<0$ diverge because of IR divergences ?

• Your first paragraph is really clear, thank you very much. However, I stumble on the second one. ;-) First of all, are all the Feynman diagrams with a loop divergent, either UV or IR ? Generally speaking, since I know that some divergences can be unexpectedly cancelled. I suppose that the answer is no since you wrote that IR divergences come from massless particles (and so we could consider a $\lambda \phi^4$ diagram with a loop s.t. $D<0$ but $\phi$ is massive --> no UV and no IR divergences). Why ? – AnSy Apr 24 '13 at 13:07
• "are all the Feynman diagrams with a loop divergent, either UV or IR ?" Nope. IR divergences only appear if there is a massless particle in the loop. Massive particles don't give IR divergences because the propagator goes as $1/(k^2+m^2)$... the $m^2$ controls the $k\sim 0$ region of the integral. Also there are certainly UV finite loop diagrams, such as the one loop 6-point function in $\phi^4$, or the one loop 4 photon function in QED (which is superficially log divergent but actually turns out to be finite). – Michael Brown Apr 24 '13 at 13:12
• I think that the propagator goes like $1/(k^2 - m^2)$ ? So even for $k^2 \sim m^2$ they would be divergences ? Thank you very much for the examples of UV finite loop diagrams. – AnSy Apr 24 '13 at 13:17
• Ahh yes, I was thinking the Euclidean space continuation. Of course, in Minkowski space it's $k^2 - m^2$, but you also need the $i\epsilon$ to make the integral well defined in that case. You typically Wick rotate to evaluate the integral and that particular issue goes away. The $i\epsilon$ prescription guarantees the correct analytic continuation, which is the real reason it is used. Physically it is not such an issue because you can't really have a "soft" massive particle: when it is near it's mass shell ($k^2=m^2$) it can travel a long way, blundering into things and leaving tracks. :) – Michael Brown Apr 24 '13 at 13:37