# Question about Quantum Effective Action

1. Dec 16, 2013

### wphysics

I am working on Quantum Effective Action in Weinberg QFT vol2 (page 67).

In the last paragraph of page 67, the author said
"Equivalently, $i \Gamma [ \phi _0 ]$ for some fixed field ........ with a shifted action $I [ \phi + \phi_0 ]$ :
$$i \Gamma [ \phi _0 ] = ∫_{1PI, CONNECTED} ∏_{r,x} d\phi^r (x) exp(iI[\phi+\phi_0])$$
In this equation, I don't understand two things;
First one is why we have to use a shifted action $I [ \phi + \phi_0 ]$.
Second one is why we only take into account of one-particle irreducible and connected terms to get Quantum Effective Action for some fixed field $\phi^{r}_0 (x)$.

Thank you.

2. Dec 17, 2013

### andrien

Weinberg has also written there that at any place where $\phi_0$ appears in vertices or propagators within 1PI graph is also a place where an external $\phi$ line could be attached i.e. you can have reducible graphs made of irreducible ones.You can use exp(il-$\phi_0$) so that you can include disconnected graphs because when doing calculation for n point function those disconnected terms will come out as phases and cancel in numerator and denominator.

3. Dec 17, 2013

### wphysics

In previous paragraph in Weinberg book, for general field $\phi^r (x)$, $i \Gamma[\phi(x)]$ must be the sum of all one-particle-irreducible connected graphs with arbitrary numbers of external lines, each external line corresponding to a factor $\phi$
I think this is kind of obvious, because we consider $i \Gamma[\phi(x)]$ as action and it gives full amplitudes, so the coupling constants in $i \Gamma[\phi(x)]$ should be the renormalized one, in other words, it has to take into account of all one-particle-irreducible graphs.

But, for fixed field, I don't understand why we have the sum of one-particle-irreducible graphs for the vacuum-vacuum amplitude, which has no external lines, and why we have to use a shifted action.

I am sorry, but could you explain in more detail?

Last edited: Dec 17, 2013
4. Dec 17, 2013

### Avodyne

This section is not up to Weinberg's usual standards of clarity. That "integral" is very weird; in particular, the measure isn't invariant under shifts of the field. (If it was, the result would have to be independent of $\phi_0$.)

Try chapter 21 of Srednicki, draft version available here: http://web.physics.ucsb.edu/~mark/qft.html

5. Dec 17, 2013

### wphysics

I think the measure must not be invariant. If so, $i \Gamma [ \phi_0 ]$ is independent of $\phi_0$, and Weinberg mentioned about this point.

I have already read Srednicki book, but for me, that is not enough.