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Proving that a propagator decays exponentially for spacelike separations

  1. Aug 8, 2010 #1
    I'm trying to show that the propagator for spacelike separation decays like [itex]e^{-m r}[/itex] and I'm stuck. At some point I hit the integral

    [tex]
    \int_{-\infty}^{\infty} \frac{dk}{\sqrt{k^2 + m^2}} e^{i k r}.
    [/tex]

    Integration of complex functions not being my forte, I only managed to get to this point using provided answers, but I don't understand the next step. Apparently the above integral is equal to

    [tex]
    2 \int_0^{\infty} dy e^{-(y+m) r} \frac{1}{\sqrt{(y+m)^2 - m^2}},
    [/tex]

    where the substitution [itex]k = i(m+y)[/itex] has been made. Now there are several things I don't understand. Firstly, I see how the integrand has changed, but since the integrand is not an even function of [itex]k[/itex] or [itex]y[/itex], how is it possible to change the integration limits to [itex]0[/itex] and [itex]\infty[/itex]? Secondly, shouldn't the integration now be along the complex axis, i.e. shouldn't the upper limit be [itex]i \infty[/itex], due to the [itex]i[/itex] in the substitution? Finally, there is also a bit of text where the author says the integrand has a branch cut going from [itex]i m[/itex] to [itex]i \infty[/itex], and one has to fold the contour of the integral around this cut. I don't see what that cut has to do with anything whatsoever, or exactly what contour he is using.

    Could someone help me along with this? I think I understand the rest of the derivation, this is the last link I need.
     
  2. jcsd
  3. Aug 9, 2010 #2
    Consider Fa(k) = 1/sqrt(k^2 +m^2) for |k|<=a & 0 otherwise.
    Applying the Riemann-Lebesgue lemma , the integral of exp(irk)Fa(k) =o(r) as r-> inf.
    If you argue carefully & let a->inf. , you'll get the desired result.
     
  4. Aug 9, 2010 #3
    I don't quite follow. I'm not familiar with this lemma, but it seems to say that the integral goes to 0 as [itex]r[/itex] goes to infinity, not to [itex]O(r)[/itex] (I'm assuming that's what you meant). Could you elaborate a little? This is not homework, I'm just trying to understand the author's derivation.
     
  5. Aug 10, 2010 #4
    I figured it out. Thanks, though.
     
  6. Aug 10, 2010 #5
    Could you elaborate on this a little bit more?
    Since the integral of NanakiXIII's question is the same as
    https://www.physicsforums.com/showthread.php?t=420759

    But, there is a problem to use Jordan's lemma.
    To perform the integral along the keyhole contour, we need some regularization as weejee did in that series of post.
     
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