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Origin of infrared divergences in perturbation theory

  1. Aug 17, 2014 #1
    If you have a momentum integral over the product of propagators of the form [itex]\frac{1}{k_o^2-E_k^2+i\epsilon} [/itex], why are there divergences associated with setting m=0?

    Factoring you get: [itex]\frac{1}{k_o^2-E_k^2+i\epsilon}=\frac{1}{(k_o-E_k+i\epsilon)
    (k_o+E_k-i\epsilon)} [/itex]. This expression has simple poles at [itex]k_0=\pm E_k [/itex]. These two poles do merge to form a pole of order 2 when m=0 at the special value of [itex]\vec{k}=0 [/itex]. But this special value of [itex]\vec{k}=0 [/itex] is only one point in the integration region, and the value of an integral doesn't depend on the behavior at a single point. Everywhere else besides this single point the integrand only has simple poles of order 1 and hence should be convergent around these points.

    So it seems to me that the only divergences should be UV divergences and not IR divergences.
     
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  3. Aug 18, 2014 #2

    Orodruin

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    So what is
    $$
    \int_0^1 \frac{dx}{x}?
    $$
    It only diverges in the lower limit and that is only a point.
     
  4. Aug 18, 2014 #3
    But there are i epsilons in the denominator. If you put a plus i epsilon in your denominator, and take your lower limit to be below zero, then your integral is finite.
     
  5. Aug 18, 2014 #4

    Avodyne

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    It's finite but epsilon dependent. At the end of the complete calculation, you must be able to take the epsilon->0 limit and get a finite result. The point of IR divergences is that some things that you might have thought would be finite (such as the cross section for Compton scattering with exactly one incoming photon and exactly one outgoing photon) are, in fact, divergent. Removing the divergences requires taking into account the fact that no detector has perfect energy or angular resolution.
     
  6. Aug 18, 2014 #5
    Consider s-channel scattering in a phi^4 theory: $$\int d^4k \frac{1}{(k_0-p_0)^2-E_{k}^2+i \epsilon}\frac{1}{k^2_0-E_{k}^2+i \epsilon}$$

    in the COM frame. There are poles at [itex]k_0=p_0 \pm E_{k}[/itex] and [itex]k_0=\pm E_{k}[/itex]

    Assume [itex]p_0 \neq 0[/itex] because if this is not true then you'd have double poles at all values of the three-vector [itex] \vec{k}[/itex] and you would get divergences from an integration over [itex]k_0 [/itex]. If your particle has mass, [itex]E_k \neq 0[/itex] and integrating over the four separate poles should be no problem. For example, as [itex]k_0 \rightarrow E_k[/itex] then in the expression $$\int d^4k \frac{1}{(k_0-p_0)^2-E_{k}^2+i \epsilon}\frac{1}{k^2_0-E_{k}^2+i \epsilon}$$ you can set [itex]k_0[/itex] equal to [itex]E_k[/itex] except at the simple pole and get an integral of the form:
    $$
    \int d^3k \int^{E_k+a}_{E_k-a} dk_0\frac{1}{(E_k-p_0)^2-E_{k}^2}\frac{1}{2E_{k}}\frac{1}{k_0-E_k+i \epsilon}
    $$

    which ought to be perfectly finite and not dependent on [itex]i \epsilon [/itex] (the integral should be a real number times [itex]i \pi[/itex]). Outside the region [itex](E_k-a,E_k+a)[/itex] you don't need the [itex]i \epsilon [/itex]'s until you reach another pole.

    The only trouble I see is the massless case when [itex] \vec{k}=0[/itex], when your simple pole becomes double poles as [itex]+E_k=-E_k=0[/itex], and then you'd get an infinite result when the integration over [itex]k_0[/itex] is done. I'm just having a hard time believing a single point [itex] \vec{k}=0[/itex] is causing all the trouble. For example, in three dimensions 1/r doesn't cause any problems at the origin in an integral. In 1 dimensions, 1/x does cause problems, but I think that's because you can't go around the singuarlity like you can in multiple dimensions.
     
  7. Aug 18, 2014 #6

    Avodyne

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    Exactly right. IR divergences only arise when there are massless particles. See any QFT text for more details.
     
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