0
$\begingroup$

I'm trying to go off based on what I could find in Peskin & Schroeder's book (specifically in chapter 11.4). Taking the renormalised Lagrangian, it uses a classical background field to calculate this:

enter image description here

(This is the exponent of the generating functional, in terms of an expansion with the first few terms where the functional derivatives are evaluated at the classical field)

But then it only keeps terms up to quadratic order when it performs the integral over the quantum fluctuation η(x), and says it's a Gaussian integral which can be evaluated in terms of a functional determinant:

enter image description here

As far as I can understand, this will be correct up to 1 loop (this agrees with the 1 loop formula from Wikipedia at least? https://en.wikipedia.org/wiki/Effective_action, it's at the bottom)

Now what do I do about the 2-loop term? My thinking was that it would be given by the cubic η term but I have no idea how to evaluate the integral $$ \int D\eta \text{exp}[i(\frac{1}{3!}\int\eta \eta \eta \frac{\delta^3L_{1}}{\delta\phi\delta\phi\delta\phi})]. $$

Is this how it should be done at all or do I have a major misunderstanding here? On page 375 it is talking about two-loop diagrams as well, so presumably keeping only the quadratic $\eta$ term somehow gives 2-loop as well, but I'm confused about this.

Any insight into this is much appreciated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.