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Confusing integral in Zee's QFT

  1. Dec 27, 2009 #1
    This is probably really simple. In chapter I.4 the jump from (4) -> (5) is sort of eluding

    [tex] W(J) = - \iint dx^0 dy^0 \int \frac{dk^0}{2\pi} e^{i k^0(x - y)^0} \int \frac{d^3k}{(2 \pi)^3} \frac{e^{i \vec{k}(\vec{x_1} - \vec{x_2})}} {k^2 - m^2 + i\epsilon} [/tex]

    and

    [tex] \omega^2 = \vec{k}^2 + m^2 [/tex]


    He got

    [tex] W(J) = \int dx^0 \int \frac{d^3k}{(2\pi)^3} \frac{e^{i \vec{k} (\vec{x_1} - \vec{x_2})}}{\vec{k}^2 + m^2} [/tex]


    the way I see it - the middle term is the delta function

    [tex] W(J) = - \iint dx^0 dy^0 \delta(x^0 - y^0) \int \frac{d^3k}{(2 \pi)^3} \frac{e^{i \vec{k}(\vec{x_1} - \vec{x_2})}} {k^2 - m^2 + i\epsilon} [/tex]

    but how does it disappear, and how does

    [tex]k^2 - m^2 + i\epsilon [/tex] turn into

    [tex]\vec{k}^2 + m^2 [/tex]

    [tex] k^0 [/tex] would be the [tex]\omega [/tex]

    but somehow this doesn't add up.

    so just wondering if anyone could give a pointer on how to solve this
     
  2. jcsd
  3. Dec 27, 2009 #2
    [tex]
    W(J) = - \iint dx^0 dy^0 \int \frac{dk^0}{2\pi} e^{i k^0(x - y)^0} \int \frac{d^3k}{(2 \pi)^3} \frac{e^{i \vec{k}(\vec{x_1} - \vec{x_2})}} {k^2 - m^2 + i\epsilon}
    [/tex]

    To do this integral, he integrated over y0 first. That produces a delta function [tex]\delta(k_0) [/tex]. Then he integrated over k0, and because of the delta function, this just sets k0 equal to zero everywhere.

    k^2-m^2 if written out is k0^2-k^2-m^2, so if k0 is zero, then that writes out to -(k^2+m^2), which cancels the negative sign.
     
  4. Dec 28, 2009 #3
    Took a while to convince myself, but yes it makes sense. Thanks.
     
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