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Wick rotation and contour integral

  1. Oct 5, 2008 #1
    We have an integral over q from -[tex]\infty[/tex] to +[tex]\infty[/tex] as a contour integral in the complex q plane. If the integrand vanishes fast enough as the absolute value of q goes to infinity, we can rotate this contour counterclockwise by 90 degrees, so that it runs from -i[tex]\infty[/tex] to +i[tex]\infty[/tex].
    In making this, the contour does not pass over any poles. Why?

    When I make a contour integral, the big semi-circle that I use in the upper half has the -x+iy pole in it. How does the fast vanishing of the integral and the rotation make the integral not pass over the pole? I don't see it.

    thank you
    Last edited: Oct 5, 2008
  2. jcsd
  3. Oct 5, 2008 #2
    I think the fast vanishing of the integral is a required condition for performing Wick rotation. Only when the fast vanishing boundary condition is satisfied, one can perform the Wick rotation.
    The condition that the rotation of the contour does not sweep through any poles is not a consequence of fast vanishing condition. Fast vanishing does not imply that the rotation sweeps no poles. Correctly speaking, the rotation passing through no poles is another condition one needs in order to be able to perform Wick rotation.
  4. Oct 5, 2008 #3
    thanks ismaili!

    I misread something in my book. It says the rotation does not pass over no pole ( in the specific integral given). I misunderstood that as after the rotation there are no poles in the contour.

    But why is the fast vanishing of the integral necessary for Wick rotation?
  5. Oct 5, 2008 #4
    If the integral vanishes fast toward the infinity, the big semicircles contribute nothing so that the contour integral from [tex]-\infty[/tex] to [tex]\infty[/tex] in the real axis can be analytic continuated to the imaginary axis.
    (This is my understanding, everyone is welcome to correct me or supplement.)
  6. Oct 5, 2008 #5
    Makes sense.

    thanks again
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