Question about Quantum Effective Action

[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 ]## :
[tex] i \Gamma [ \phi _0 ] = ∫_{1PI, CONNECTED} ∏_{r,x} d\phi^r (x) exp(iI[\phi+\phi_0])[/tex]
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.

 

[andrien]

Weinberg has also written there that at any place where [itex]\phi_0[/itex] appears in vertices or propagators within 1PI graph is also a place where an external [itex]\phi[/itex] line could be attached i.e. you can have reducible graphs made of irreducible ones.You can use exp(il^-[itex]\phi_0[/itex]) 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.

 

[wphysics]

Thank you for your answer, but I don't see why your answer is relevant my question and don't understand either.

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?

 

[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

 

[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.

Thank you for your answer