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

1. Mar 14, 2016

### CAF123

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

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 $\propto \Lambda_{UV}$?

Thanks!

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2. Mar 15, 2016

### vanhees71

This is what Weinberg's theorem tells you. It's not so easy to prove:

Weinberg, Steven: High-Energy Behavior in Quantum Field Theory, Phys. Rev. 118, 838, 1960

There is however a shortcut by dimensional arguments, which makes the theorem quite plausible: Since you look at the massless theory there are only the momentum cutoff (most simply implemented in the Wick rotated Euclidean theory) and the external momenta of the diagrams. Thus your cut-off regularized self-energy $\Pi(p^2,\Lambda)$ , which is of mass dimension 2 can only go like $p^2$ and $\Lambda^2$. Experience shows that there are also (powers of) logarithms, which can only be of the form $\ln(p^2/\Lambda^2)$, because the argument of the logarithm MUST be dimensionless, because a logarithm with a dimensionful argument are indefined useless expressions.

In your case a specialty of $\phi^4$ theory occurs. At one-loop order there's indeed only the diagram you've drawn, and it's an effective one-point function, i.e., independent of the external momentum. In this case thus your diagram can only be of the form
$\Pi(p^2,\Lambda)=A \Lambda^2,$
because there's nothing to compensate the dimensionful $\Lambda$ in a possible logarithm (with $A$ some dimensionless constant).

The same dimensional argument holds for the four-point function, which however is also again an effective two-point function, i.e., the single dinosaur diagram depends only on one external momentum, the sum of the incoming (and then due to energy-momentum conservation also outgoing) momenta. So the dinosaur diagram must be of the form
$$\Gamma(p^2)=A+B [\ln(p^2/\Lambda^2)]^k,$$
with some power $k$ and dimensionless constants $A$ and $B$. The actual calculation leads to $k=1$ in this case.

3. Mar 15, 2016

### CAF123

Hi vanhees71, many thanks, I think I got the just of what you said. A few follow up questions:
1) In the second example, why can't we have something like $\Lambda^2/p^2$ on its own? (i.e without the logarithm)
2) If we were to have massive propagators, can we have something like the integral being proportional to $m^2 + \Lambda^4/m^2$ in the first example or something like $\log(\Lambda^2/m^2)$ in the second?

Thanks!

4. Mar 16, 2016

### vanhees71

That you don't have such terms is the content of Weinberg's theorem!

5. Mar 16, 2016

### CAF123

I see, thanks! I tried accessing the link but it requires a payment to proceed with viewing the document.