# Counterterms in saddle point expansion

1. Dec 21, 2014

### 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.

2. Dec 26, 2014

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Dec 26, 2014

### atyy

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.

4. Dec 27, 2014

### geoduck

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.