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Wick rotation is consistent with caculus?

  1. Aug 23, 2009 #1
    knowing the action
    [tex]S[x(t)]=\int_0^\infty dt'\left[\frac{m}{2}\left(\frac{dx}{dt'}\right)^2-V(x)\right][/tex]
    After the so-called wick's rotation[tex]t'=-i\tau[/tex] with [tex]\tau[/tex] being real,
    the action becomes
    [tex]S[x(t)]=i\int_0^\infty d\tau\left[\frac{m}{2}\left(\frac{dx}{d\tau}\right)^2+V(x)\right][/tex]
    my question is why the upper limit of the integral in the secong equation is still [tex]\infty[/tex]?
    I think ist should be [tex]+i\infty[/tex]
     
  2. jcsd
  3. Aug 23, 2009 #2
    Because it is not merely a change of variable : it is really a rotation in the complex plane. A Wick rotation is usually equivalent to a so-called "i[itex]\epsilon[/itex]" prescription : you want to define a contour in the complex plane avoiding the singularities of your integrand on a precise side (up or down, right or left). Usually you do that because your integrand is well behaved at infinity only in certain directions.

    If I understand correctly, "calculus" refers to "real analysis". A Wick rotation involves complex analysis. If that can cheer you up, complex analysis is very different from real analysis, and in many regards more powerful, so I would say more fun to use.
     
  4. Aug 23, 2009 #3
    Do you mean the origin, the positive real axis and positive imaginary axis constitute a
    closed contour?
     
  5. Aug 23, 2009 #4
    So by "usually" do you mean that a Wick rotation in position space is almost the same as doing the Wick rotation in momentum space, but not always?

    Because when you have the [tex]i\epsilon [/tex] prescription in momentum space, you know where your singularities are, so you can perform a Wick rotation. But in position space you just perform a Wick rotation without knowing where the singularities are.
     
  6. Aug 23, 2009 #5
    No. Wick rotation generally proceeds through a quadrant of the complex plane where the integrand is strictly zero at infinity. So, the contour actually involves (in the case you cite) the positive real axis, a 90 degree arc at infinity and the positive complex axis.
     
  7. Aug 23, 2009 #6
    Well I hope that the potential V can be treated pertubatively around a minimum (steepest descent) as usual in momentum space.
     
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