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Propagator in Feynman rules

  1. Aug 22, 2010 #1
    I'm probably missing something small but I haven't been able to figure this out. In the Feynman rules (for a scalar field that obeys the Klein-Gordon equation), you write a propagator for internal lines as

    [tex]
    \frac{i}{k^2 - m^2 + i \epsilon}.
    [/tex]

    The propagator integrand is originally

    [tex]
    \frac{e^{i k (x-y)}}{k^2 - m^2 + i \epsilon}.
    [/tex]

    Since we're dealing with an internal line, both exponentials, in [itex]x[/itex] and [itex]y[/itex], are integrated out to delta functions, leaving you with

    [tex]
    \frac{1}{k^2 - m^2 + i \epsilon}.
    [/tex]

    That I see, but where does the [itex]i[/itex] in the numerator of the first expression above come from?
     
  2. jcsd
  3. Aug 23, 2010 #2

    vanhees71

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    This factor i comes from the i in the path-integral formula for the generating functional for (connected Green's functions), [tex]W[J]=\ln Z[J][/tex] with

    [tex]Z[J]=\int \mathcal{D} \phi \exp[\mathrm{i} \int_{\mathbb{R}^4} \mathrm{d}^4 x [\mathcal{L}(\phi,\partial \phi)+J \phi]].[/tex]
     
  4. Aug 23, 2010 #3
    Ah, I think I got it. I had ignored the factor [itex]i[/itex] when I wrote things down as Wick contractions. Thanks.
     
  5. Aug 23, 2010 #4
    Chapter 10 of Srednicki is a good way to see the Feynman rules emerge, including this factor of i.
     
  6. Aug 24, 2010 #5
    He seems to adopt quite a different approach than the author of the book I'm using. I may have a look at that later. Thanks for the tip.
     
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