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Question about Quantum Effective Action

  1. Dec 16, 2013 #1
    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.
     
  2. jcsd
  3. Dec 17, 2013 #2
    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.
     
  4. Dec 17, 2013 #3
    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?
     
    Last edited: Dec 17, 2013
  5. Dec 17, 2013 #4

    Avodyne

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    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
     
  6. Dec 17, 2013 #5
    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
     
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