Due to conservation laws a free electron,as I understand it, cannot absorb a photon. But in computing QED probabilities, diagrams are drawn showing emission and absorption of photons by an electron. Also QED requires electrons to exchange photons, which means one emits one and the other absorbs one.

Is this because conservation laws are relaxed for virtual particles?

  • $\begingroup$ @probably_someone: A free electron can emit and re-absorb a virtual photon in a one-loop diagram, without a need for anything else nearby. $\endgroup$ – Michael Seifert Apr 28 '18 at 1:31
  • $\begingroup$ @MichaelSeifert That is correct. I was only thinking at tree-level in my earlier comment. $\endgroup$ – probably_someone Apr 28 '18 at 1:33

The four-momentum of any real particle satisifies the relationship $p_\mu p^\mu = -m^2$. This defines a 3-D surface in the 4-D space of all possible four-momenta; this surface is called the mass shell for the particle. A virtual particle, on the other hand, can have any four-momentum vector that you want; a virtual particle is usually "off-shell", because its four-momentum doesn't lie on the mass shell.

However, virtual particles still obey energy and momentum conservation: the four-momentum going into any vertex in a Feynman diagram must equation the four-momentum going out. It is entirely possible for a real electron to emit a virtual photon and remain on its own mass shell; this is exactly what happens in the classic Feynman diagram with two "real" electrons exchanging a virtual photon. The only reason that conservation laws prohibit a "real" electron from emitting a "real" photon is that it is impossible for all three four-momenta (electron before, electron after, and photon) to simultaneously lie on their respective mass shells.

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