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In QFT when one works out the cross section between two colliding electrons one gets a formula which is proportional to $\theta^{-4}$ where $\theta$ is the scattering angle which is due to a nearly on-shell photon with momentum $\approx 0$. See for example Peskin and Schroeder pg 155. This is also seen in ordinary Rutherford scattering and apparently is not a problem experimentally as one can't detect particles along the collision beam. Also, in Rutherford scattering the singularity gets removed if one takes into account the screening of the nucleus by the electron cloud (see QM by Messiah (1961), pg 422) which modifies the Coulomb nature of the potential.

I was wondering if this singularity is indicating some problem with the calculation? Is there some effect in the calculation which has been neglected which when included would remove the singularity? Could renormalization effects maybe have a similar effect to the electron cloud in the Rutherford case?

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The divergence is real but does not reflect a problem.

It arises because the potential has infinite range. It is similar to saying that in theory every snowball in the Andromeda galaxy is infinitesimally scattered by the sun, and it is just as meaningless because every really distant particle is more strongly perturbed by some other effect.

This divergence is not observed in interactions with a finite range such as the weak and effective (nuclear) strong interactions.

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    $\begingroup$ Thanks, but doesn't an infinite cross-section imply an infinite number of electrons are scattered with $\theta=0$? Is it just that the electrons are scattered a very long way away from the each other? Would this singularity go away if one used wave packets rather than plane waves? $\endgroup$
    – Virgo
    Sep 24, 2012 at 3:23
  • $\begingroup$ Dear @physicsphile, quite on the contrary, the infinite cross section (long range force) means that there are no electrons at all that get through at $\theta=0$: all electrons get scattered, usually with extremely small angles $\theta$. Well, the angle $\theta$ at which an electron scatters is (approximately) a decreasing, power-law function of the impact parameter, much like in classical physics. ... The singularity should't go away and doesn't go away and you should stop asking about it. It's a true formula with a clear physical, long-distance (infrared) interpretation. $\endgroup$
    – Luboš Motl
    Sep 24, 2012 at 6:09
  • $\begingroup$ Hi @ Luboš, sorry I don't quite understand your comment. Say we have $\frac{d\sigma}{d\Omega}d\Omega=\frac{\mbox{number of particles scattered into $d\Omega$/second} }{\mbox{number of incident/sec/area in $\rho$ plane}}$ where the $\rho$ plane is perpendicular to the beam and $d\Omega$ is the area of the detector. Then if we want to work out the total number of particles scattered, don't we integrate over $\Omega$? But if so and $\sigma\propto\frac{1}{\sin(\theta/2)^4}$ wont the integral be non-convergent? $\endgroup$
    – Virgo
    Sep 24, 2012 at 10:11
  • $\begingroup$ @physicsphile The thing is that your beam has finite dimensions---or at least the part of it you care about has finite dimensions. For any finite range of impact parameters the number of particles scattered into any $\mathrm{d}\Omega$ is finite. So in the real world there is no problem. $\endgroup$
    – dmckee --- ex-moderator kitten
    Sep 25, 2012 at 0:12
  • $\begingroup$ @dmckee sorry I don't quite follow what you are saying. Along the beam $\theta=0$ so the singular region will also contribute as far as I can see. If the singular region contributes the integral will not converge. $\endgroup$
    – Virgo
    Sep 25, 2012 at 2:10

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