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Exponent of disconnected feynman diagrams

  1. Mar 21, 2009 #1


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    hello, I am trying to follow the arguments in Peskin pages 96-97 where the exponentiation of disconnected diagrams are performed. I think the 'proof' is too 'not detailed' and was wondering if there exists a better explanation elsewhere?

    My biggest question is the sentence on page 97 "The sum of the connected pieces factors out..." and what n_i is/means. I understand better if one starts with the most basic case, then do it for one more case, and then generalizes it - I think the generalization comes immediately here =/

    best regards
  2. jcsd
  3. Mar 21, 2009 #2
    I studied this in the context of field theory in statistical physics quite some time ago. If the set of all conneceted diagrams is c_1, c_2, c_3,...., then any disconnected diagram can be written by specifying how many diagrams c_j it contains. If there are n_j diagrams c_j, then the amplitude is

    Product over j of 1/n_j! c_j^n_j

    Summing over all the disconnected diagrams, amounts to summing over all the possible values for n_j, which gives the result

    exp[sum over j of c_j]
  4. Mar 21, 2009 #3


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    Thanx for your answer, but I still don't get it 100%, guess I need to make up an own example of this and then generalize it ;-)

    I also had this in quantum field theory in statistical physics, but did not bother so much to understand this back then.. we also had a quite advanced book, it was not introductory (Tseliks book on QFT in condensed matter). But now I am studying more in Peskin, which is nicer, but not nice enough, for me.
  5. Mar 21, 2009 #4
    Well, it all boils down to the 1/n! symmetry factor if you have n times the same connected diagram. So, you get a factor of c^n/n! if you have n copies of diagram c as part of your disconnected diagram.... I don't think there is anything more to it than just this.
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