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Integration of complex exponentials

  1. Oct 18, 2015 #1
    I am trying to understand how to go from the first line to the next:

    ##\frac{1}{(2\pi)^{3}}\int d^{3}p\ e^{-it\sqrt{{\bf{p}}^{2}+m^{2}}}\ e^{i {\bf{p}}({\bf{x}}-{\bf{x}}_{0})} = \frac{1}{2\pi^{2}|{\bf{x}}-{\bf{x}}_{0}|} \int_{0}^{\infty} dp\ p\ sin(p|{\bf{x}}-{\bf{x}}_{0}|)\ e^{-it\sqrt{{p}^{2}+m^{2}}}##.

    As an initial guess, I'm thinking that the vector ##\bf{p}## is decomposed in spherical polar coordinates, evidenced by the facts that ##\frac{1}{(2\pi)^{3}}## has been multiplied by ##4\pi## and that the volume element is afterwards ##dp\ p## with limits from ##0## from ##\infty##. Also, I'm thinking that firstly the exponential ##e^{i {\bf{p}}({\bf{x}}-{\bf{x}}_{0})}## is broken using Euler's relation into its sine and cosine components.

    I wonder what's happened to the cosine component, the factor of ##i## in the sine component, how the polar and azimuthal angles have been separated out from the vector expression ##-it\sqrt{{\bf{p}}^{2}+m^{2}}## and how the factor of ##|{\bf{x}}-{\bf{x}}_{0}|## in the denominator outside the integral appears.
     
  2. jcsd
  3. Oct 18, 2015 #2

    andrewkirk

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    I can help with this bit:
    The expression ##{\bf{p}}^{2}## is more correctly written ##\|\bf{p}\|^2##, which is equal to ##p^2##. Hence
    ##-it\sqrt{{\bf{p}}^{2}+m^{2}}## can simply be replaced by ##-it\sqrt{{p}^{2}+m^{2}}##

    I think this will be something to do with the following:
    Convert the integral into
    $$\int_0^\infty \left[\int_{N_p} f(p,\phi,\theta) dA + \int_{S_p} f(p,\phi,\theta) dA\right]\,dp$$
    where ##N_p## and ##S_p## are respectively the Northern and Southern hemispheres of the sphere of radius ##p##.
    Then think about what cancels out between the two hemispheric integrals when you match antipodal points.
     
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