Why is the matrix element of Compton scattering in leading order of perturbation theory equal to zero? Why can this process only be described in second order of perturbation theory, i.e. with exchange of a virtual photon? Is there a physical reason?

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    $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Commented Apr 9, 2015 at 16:20

1 Answer 1



If Compton scattering occurred in first order in $e$, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the $z$ direction. Then the electron 4-momentum is $$p^\mu_{\text{in}} = (m,0,0,0)$$ while the photon 4-momentum is $$k^\mu_{\text{in}} = (\omega,0,0,\omega)$$

Post scattering, there is only an electron moving upwards in the $z$ direction. Its momentum must be the same as the initial momentum, so if this process makes sense it should be

$$p^\mu_{\text{f}} = (\sqrt{m^2 + \omega^2},0,0,\omega)$$

However, $\sqrt{m^2 + \omega^2} \neq m + \omega$ so the process is forbidden by conservation of energy.

Another way to see that this amplitude must vanish is to realize that this diagram is related by "crossing" symmetry to a diagram that represents the decay of a single photon into an electron-positron pair. If this process were allowed at all we'd be in big trouble because we can always make the photon energy arbitrarily small/large by boosting. So we'd be able to freely move between a reference frame in which $\omega \geq 2m$, where the decay is "allowed", and one in which $\omega < 2m$, where it is forbidden. Lorentz invariance just died a painful death.

For the same reason, it's impossible for an electron positron pair to annihilate and produce a single photon. You must make at least two.

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    $\begingroup$ Thanks! Why do you assume that we have only an electron after the scattering, and not an electron plus a photon, both with different momenta? And the decay of a single photon into an electron-positron pair is actually possible, but only in a material which recoils, right? $\endgroup$
    – LCF2
    Commented Apr 9, 2015 at 9:34
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    $\begingroup$ You can't produce more than one photon in lowest order because the only diagram you have available is the one with two electron lines and one photon line. If you want to make two photons you must go to at least second order in $e$. As far as your second question, if you have a material medium you no longer have Lorentz invariance to worry about and certain things become possible. If you were analyzing the experiment "microscopically" however you'd see that the "recoil" is really a higher order process in disguise. $\endgroup$
    – Leandro M.
    Commented Apr 9, 2015 at 9:37
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    $\begingroup$ Ah, okay, I get it. Thank you very much, your answer was very helpful! $\endgroup$
    – LCF2
    Commented Apr 9, 2015 at 9:39

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