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Question about the Klein Gordon Propagator

  1. Jun 26, 2009 #1

    I'm teaching myself QFT. I don't understand some of the calculations described on pages 29 and 30 of Peskin and Schroeder's book on QFT.

    The next step is where I have a problem

    I think I'm making a mathematical mistake somewhere, but I haven't been able to figure it out yet.

    Making a digression, the propagator can be written in momentum space as

    [tex]\Delta_{F}(p) = \frac{1}{2E_{p}}\left[\frac{1}{p^0 - (E_p - i\epsilon)}-\frac{1}{p^0 + (E_p - i\epsilon)}\right][/tex]

    (Feynman prescription)

    Plugging this back into the expression for the Green's function [itex]G(x-x') = \int \frac{d^4 p}{(2\pi)^4}e^{-ip\cdot(x-x')}\frac{1}{(p^0)^2 - E_{p}^2}[/itex] we get

    [tex]\Delta_{F}(x-x') = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E_{p}}\int \frac{dp^{0}}{2\pi} e^{ip\cdot(x-x')}\left[\frac{1}{p^0 - (E_p - i\epsilon)}-\frac{1}{p^0 + (E_p - i\epsilon)}\right][/tex]

    This doesn't give me what I want, but I think I am close to it as there is a -i and two theta functions that enter into the final expression if this is integrated.


    1. I don't understand how the extra factor of {-i} in the expression in the quote box above (from Peskin & Schroeder) comes in.

    2. What is the motivation for the choice of the contour shown on page 30?

    3. What is the significance of [itex]x^0 > y^0[/itex].

    Last edited: Jun 26, 2009
  2. jcsd
  3. Jun 26, 2009 #2
    didn't really follow your calcuation, but it seems to me that you are trying to abtain formula 2.60.. and this one contains theta functions...

    1. which extra facor?

    2. the motivation is having the last line of 2.54 equal to the precedent one... only with that contour you get the two poles that you need (you need two residues!),,,

    3. the significance is that the last line of 2.54 and the line before are equal only if x_0>y_0 (otherwise, as explained the last line is zero)
    Last edited: Jun 26, 2009
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