# Dependence on UV cut off of some $\phi^4$ diagrams

Consider the one loop corrections to the propagator and the vertex in $\phi^4$-theory:

The former gives an integral representation proportional to $\int d^4 k/k^2$ in $4$ dimensions while the latter gives a representation $\int d^4 k/k^2 (k+p)^2$ where $p$ is the momenta input into the vertex from the external legs. Power counting tells us that both diagrams are UV divergent. Can we predict a priori what the dependence on some UV cut off scale would like for both diagrams? I know that it would be a finite function of the cut off such that in the limit that this cut off is extended to infinity, the function diverges. But can we say anything about the functional dependence just by this power counting?

In the former case, the integral is quadratically divergent. So maybe something like $\Lambda_{UV}^2$ or $\log (\Lambda_{UV}^2)$? The latter diagram is maybe just $\propto \Lambda_{UV}$?

• Simply do dimensional analysis. – Javier Mar 14 '16 at 13:22
• @Javier: Ok, so in the first case, it would go simply as $\Lambda_{UV}^2$ as I wrote? But for the second case, would the cut off have to be incorporated into a log for example alongside some mass scale? – CAF Mar 14 '16 at 13:27
• Or could it not simply be $\Lambda_{UV}^2/m^2$ in the latter case? So based on dimensional analysis I get these two options so I am not sure how to constrain it further. Any ideas? – CAF Mar 14 '16 at 17:40

You can calculate the superficial degree of divergence, i.e. the dependence on the UV cutoff $\Lambda$ easily by introducing polar coordinates in your integrals. Consider your first diagram

$$\int^\Lambda d^4k \frac{1}{k^2} \sim \int^\Lambda dk \frac{k^3}{k^2} \sim k^2 \bigg|^\Lambda \sim \Lambda^2\ .$$

$$\int^\Lambda d^4k \frac{1}{k^2 (k+p)^2} \sim \int^\Lambda dk \frac{k^3}{k^4} \sim \log k \bigg|^\Lambda \sim \log\Lambda\ .$$
Since you are only interested in the UV behavior, $(k+p)^2 \approx k^2$. In a more complicated quantum field theory (even QED) you can still compute this superficial degree of divergence, however, since multiple diagrams are summed, the actual degree of divergence might be different.
• Thanks! In the second example, don't you have a dimensionful argument for the logarithm? I suppose this is compensated by the lorentz invariant $p^2$ for example so you get something like $\log(\Lambda^2/p^2)$ for example? – CAF Mar 16 '16 at 9:38
• And therefore in the far UV limit, $\Lambda^2 >> p^2$ and so we are left with what you write? – CAF Mar 16 '16 at 9:55
• @CAF: This should just show the degree of divergence (here logarithmically). Sure you would have some dimensionless ratio $\Lambda / p$ or $\Lambda / m$ if you have a mass as a IR cutoff. – Clever Mar 17 '16 at 8:43