Computing the amplitude squared for $e^-\mu^-\rightarrow e^-\mu^-$ at tree level we get \begin{equation} \frac{1}{4}\sum_\mathrm{spins}|\mathcal{M}(s,t)|^2=2e^4\frac{s^2+u^2}{t^2} \end{equation} which is IR divergent as $t\rightarrow 0$. The KLN theorem tells us that all IR divergences cancel when computing cross sections, so what cancels this divergence?


1 Answer 1


The singularity of the differential cross section as $t \rightarrow 0$ has the physical interpretation that an on-shell photon is created and travels an infinite distance before interacting.

For massless electrons t takes the form $t \sim (1-\cos\theta)$, and we see that $t\rightarrow 0$ corresponds to forward scattering . If we then use the amplitude to calculate the total cross section we find $$\sigma \sim \int d(\cos\theta)\frac{1}{\sin(\theta/2)^4}$$ (this is basically just the Rutherford formula), and the cross section is infinite.

To see what this means, recall that classically forward scattering corresponds to an infinite impact parameter. The infinite cross section can then be interpreted as that the charged electron and muon will always scatter, even if they are on the other side of the universe, i.e the EM force is long-ranged

  • $\begingroup$ Thanks! So would using wave packets cutoff this divergence? If not, does this not pose a serious problem for defining the 'in/out' states, as these states are by definition supposed to describe non-interacting particles in the asymptotic past and future? $\endgroup$
    – Luke
    Jan 16, 2018 at 8:29
  • $\begingroup$ Also, would'nt an infinite cross section violate unitarity? $\endgroup$
    – Luke
    Jan 16, 2018 at 8:43
  • $\begingroup$ I'll add a comment on the correct @Lunaron answer. Yes, it would violate unitarity and one way to cure that might be using wavepackets, I've seen that somewhere with neutrino physics. Another usually employed way is to use thermal field theory, which is QFT at finite temperature. In this way, you give the mediating photon a "thermal mass" through thermal loops and the divergence is cured. The physical meaning is that you cannot really have zero transfer momentum in a real situation and you always have interaction with particles at temperatures comparable to the electron mass or larger. $\endgroup$
    – Lele
    Dec 9, 2020 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.