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Integral for the free propagator

  1. Aug 4, 2010 #1
    This may be more of a maths question, but because I may actually just be interpreting the expression wrong, I think I'd better post it here.

    I'm reading Quantum Field Theory in a Nutshell by A. Zee and I'm stuck on a bit of maths he does. He provides an expression for the free propagator for a particles described by the Klein-Gordon equation,

    D(x-y) = \int \frac{d^4 k}{(2 \pi)^4} \frac{e^{i k (x-y)}}{k^2 - m^2 + i \epsilon}.

    Now, if I am not mistaken, the integral over four counts of [itex]k[/itex] means integrating over [itex]k^0, k^1, k^2, k^3[/itex], each with integration limits [itex]-\infty[/itex] and [itex]\infty[/itex].

    He goes on to perform the integral over [itex]k^0[/itex], and he describes this as a contour integral in the complex plain. He takes this contour to be the real axis and an infinite semicircle to get back to [itex]-\infty[/itex]. My question is, why does he add that semicircle? Once you've integrated over the real line, since the integration limits are [itex]-\infty[/itex] and [itex]\infty[/itex], aren't you done? Or have I perhaps misinterpreted what he means to integrate over?
  2. jcsd
  3. Aug 4, 2010 #2
    Your interpretation of the integral is correct, what your books does is just a trick in order to be able to perform the integral. The idea is to make your integration complex and choose a contour such that the complex integral is zero, except the part on the real line (the integral along the infinite semicircle should give zero). This allows one to use for example the Residue theorem, and therefore one is able to perform the integration.

    See this example http://en.wikipedia.org/wiki/Residue_theorem#Example".
    Last edited by a moderator: Apr 25, 2017
  4. Aug 4, 2010 #3
    Ah, now I see. I was trying to figure out what was happening with the help of my book on Complex Variables, but now I see I was looking in quite the wrong chapters. Thanks!
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