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Peskin - exponention of disconnected diagrams

  1. Dec 5, 2008 #1

    PJK

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    Hi all,

    I have a question regarding p.97 of Peskin Schroeder and its explantion of disconnected diagram exponentation. I do understand the formula on the buttom of page 96. [tex]\prod{\frac{1}{n_i!}V_{i}^{n_i}} \cdot (value \; of\; connected \; piece)[/tex]

    ButI do not understand the sum over [tex] \{ n_i \}[/tex] in the next step!
    I would think that I have to sum over the values of all diagrams. Each value is given by the formula which I understand. But why are there two sums?
     
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  3. Dec 5, 2008 #2

    Avodyne

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    A complete diagram is specified by (1) the connected piece, and (2) the number n_i of factors of each possible disconnected piece V_i. (The connected piece has the 2 external lines; the disconnected pieces have no external lines.) Summing over all diagrams is implemented by (1) summing over the possible connected pieces, and (2) summing over all possible numbers of factors of the disconnected pieces.
     
  4. Dec 6, 2008 #3

    PJK

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    Ok I got it. Thank you!
    What I find very strange though is that he converts [tex]\sum_{n_i} \frac{1}{n_i!}V_i^{n_i}[/tex] into [tex]exp(V_i)[/tex] - I mean for example n_1 can be an arbitrary number e.g. 999888 and not 1. This doesn't look like an exponential series to me.

    Thank you for your answer.
     
  5. Dec 7, 2008 #4

    Avodyne

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    Yes, he's summing over all possible values of n_i from 0 to infinity. Then he notes that this simply gives the series for the exponential function,

    [tex]\sum_{n=0}^\infty{x^n\over n!}=e^x[/tex]
     
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