# Coleman-Weinberg potential: resum at 2 loops?

Say we want to compute the Coleman-Weinberg potential at 2 loops.

The general strategy as we know is to expand the field $\phi$ around some background classical field $\phi \rightarrow \phi_b + \phi$, and do a path integral over the quantum part of the field, $\phi$.

We can retrieve the effective action by doing a path integral, something like eq.42 in this reference.

There are 2 ways to do this at 1 loop, we can either evaluate a functional determinant or do the classic Coleman-Weinberg thing where we sum up all diagrams we get by inserting any number of background fields $\phi_b^2$ into the loop integral. This is eq. (56) of that same reference again.

My question is, why do we not need to do this resummation over background field insertions at 2 loops? For example, in this (quite standard) reference, as well as in chapter 11 in Peskin and Schroeder, the authors seem to claim that the 2 loop contribution to the path integral are simply the "rising sun" and "figure 8" vacuum diagrams, and no summing over classical field insertions is even mentioned.

What am I missing?

EDIT:

To give some more details, in perturbation theory, each diagram contributing to the path integral is spacial integral of some functional derivative acting on the free field path integral with a source: the loop diagram with n insertions of external field $\phi_b$ is the term: $$\left( \phi_b^2 \int dx \left( \frac {\delta}{\delta J(x)}\right)^2 \right)^n Z_0[J]$$

The 2 loop figure 8 is

$$\int dx \left( \frac {\delta}{\delta J(x)}\right)^4 Z_0[J]$$

The 2 loop diagrams that it seems like the papers cited above are excluding are contributions like

$$\left( \phi_b^2 \int dx \left( \frac {\delta}{\delta J(x)}\right)^2 \right)^n\int dx \left( \frac {\delta}{\delta J(x)}\right)^4 Z_0[J]$$

It seems to me that these terms will indeed arise in the exponential expansion of the interacting lagrangian, so it seems that a resummation over $n$, as in the 1 loop case, is still necessary. Where is my error?

• Bechira, in all the cases, one simply writes down all the Feynman diagrams with the appropriate external lines containing the allowed vertices. If the insertions are zero, they're either guaranteed to vanish, or assumed to be subtracted in a different way before you write down the action for the quantum part. Jul 20, 2014 at 5:12
• @LubošMotl yes of course, the question is that it is not obvious to me that the diagrams you get from say, inserting a background field on a line in the setting sun diagram, vanishes.
– zzz
Jul 20, 2014 at 13:55
• Which page in Peskin and Schroeder? Do you mean Fig. 11.8(b)? Feb 15 at 19:10

When computing the effective potential $V(\phi)$ in the ordered phase ($\phi_b>0$), one has to use the classical propagator $G_c[\phi_b]$ given by the inverse of $$\frac{\delta^2 S[\phi]}{\delta\phi^2},$$ which is a functional of $\phi$. The vacuum energy is given by $V(\phi_b)$, where one should remember that $\phi_b$ is also computed consistently in perturbation by $$V'(\phi_b)=0.$$ Using the classical propagator is equivalent to a consistent resummation of the $\phi_b^2$ to all order. In particular, the effective action at two-loops is given by $$\Gamma[\phi]=S[\phi]+\frac{1}{2}Tr \log G_c[\phi]+{\rm 2-loops\; diagrams},$$ where the 2-loops diagrams are the 8 and the rising sun, that have to be computed using the classical propagator.

• why is "Using the classical propagator is equivalent to a consistent resummation of the $\phi_b^2$ to all order"?
– zzz
Jul 22, 2014 at 15:01
• @bechira: Look at the effective potential at one loop. The classical propagator includes the $\lambda\phi_b^2$ term. If you expand the propagator in the log, that will give rise to the $\phi_b^2$ terms that need to be resummed if you use the free propagator.