# Counterterms in saddle point expansion

• geoduck
In summary, during the saddle point evaluation of the path integral at tree level, the classical solution of the field is plugged into the integrand. However, counterterms are ignored when determining the classical solution. These counterterms only show up after a saddle point expansion is made, which does not include them. This is possible because the counterterms usually contain Planck's constant, which goes to zero in the classical limit. However, there is a debate about the inclusion of counterterms in the classical solution, as they have one more power of the coupling than bare terms. This means that the value of the pure number k can affect the magnitude of higher order perturbative terms compared to tree level. For example, the one-loop 4-pt

#### geoduck

In the saddle point evaluation of the path integral, at tree level, you plug in the classical solution of the field into the integrand. However, when determining the classical solution, we ignore counterterms. The counterterms only show up to renormalize divergences after a saddle point expansion is made about the classical solution that does not include the counterterms.

Why can we get away with this? It seems mathematically we have to include counterterms in the classical solution for a saddle point expansion to be valid.

The counterterms usually contain Planck's constant, so they should go to zero in the classical limit. But there is an interesting discussion of this issue in http://arxiv.org/abs/hep-th/0405239.

Don't counter-terms have one more power of the coupling in them than bare terms? For example, for λΦ4, I worked out the the coupling has dimensions $[\lambda]=\frac{1}{\hbar c}$.

So if we write $\lambda=k*\frac{1}{\hbar c}$, where k is a pure number, then whether your higher order perturbative terms are small compared to tree level depends on the value of the pure number k, and not on Planck's constant.

So for example, the one-loop 4-pt function is $\lambda+k*\lambda^2\log\left( \frac{E}{\mu}\right)*(\hbar c)$.

I assume that $(\hbar c)$ is in the one-loop correction to make the dimensions of the two terms the same. With n-loops I assume you'll get $(\hbar c)^n$. But you can't say this is small, because you'll also get $$\lambda^n$$, which has units of $(\hbar c)$ in the denominators.