Consider an electron emitting a single photon. There is exactly one gamma matrix (corresponding to the photon vertex) between the outgoing spinor $\bar{u}$ and incoming spinor $u$. This implies that both $\bar{u}$ and $u$ have to be right handed electrons (or both left handed).

In other words the non-zero amplitude corresponds to $u^{\dagger}(\frac{1 + \gamma^5}{2}) \gamma^0 \gamma^{\mu} (\frac{1 + \gamma^5}{2}) u$.

The amplitude of $u^{\dagger}(\frac{1 - \gamma^5}{2}) \gamma^0 \gamma^{\mu} (\frac{1 + \gamma^5}{2}) u$ would be 0.

Lets ignore the mass of the electron giving the helicities to be frame independent.

This means that both outgoing and incoming electrons have to have either spin +1/2 or spin -1/2.

The question is: when the +1/2 electron emits a photon (of spin +1) then, should not the outgoing electron have spin -1/2; so that spin is conserved? What/where is the flaw in my analysis?


I think that although spin conservation is a good rule of the thumb, it is not necessary. What should be conserved at all times is the total angular momentum: J=S+L (S:spin,L=angular momentum). Therefore after the emission of the photon J should be the same as before and not S...

  • $\begingroup$ but here, L = 0; there is no bound state. So, this gives that S is conserved. $\endgroup$ – Angela May 1 '18 at 11:55

An electron running along does not radiate. A photon can be emitted if there is an interaction ( what acceleration means for classical electrodynamics), i.e.energy has to be supplied which will be released as radiation. An example is brehmsstrahlung radiation,


The virtual photon in the diagram can be exchanged with a magnetic or electric field in a more general case of acceleration and radiation.

So it is not a simple conservation of angular momentum or an electron and an emitted photon, as more elements have to enter the interaction introducing angular momenta, and it is angular momentum that has to be conserved , as @hal states.

  • $\begingroup$ A free running electron can absorb a photon. They above math does not change with photon absorption/emission. Lets assume that L somehow comes into picture (though not sure how). Now, if one were to add another photon vertex, two gamma matrixes are introduced 1) for the photon vertex, and 2) for the electron propagator. This means that the incoming electron and outgoing electron continue to be spin +1/2 (or -1/2). So, adding more photon vertexes makes no difference. Why is L not playing a role now? If L plays a role for 1 photon vertex, it should play a role for multiple photon vertexes too. $\endgroup$ – Angela May 1 '18 at 19:50
  • $\begingroup$ your question is about emmission . The math sure changes, if no diagram can be written because of momentum and energy conservation. Absorption means compton scattering, here is a calculation personal.soton.ac.uk/ab1u06/teaching/qft/qft1/… . $\endgroup$ – anna v May 2 '18 at 3:29
  • $\begingroup$ if the incoming photon has +1 projection and the electron -1/2 projection, the virtual electron would carry 1/2 so as to conserve angular momentum in compton scattering. and so on for the other vertex, according to the probability calculated for these possibilities. Virtual means an integration which includes probabilites for spin orientation, imo. $\endgroup$ – anna v May 2 '18 at 3:41
  • $\begingroup$ The amplitude and cross-section would obviously change. The expression that I've mentioned does not change with incoming/outgoing photon except for the fact that the spinor ub(p-k) changes to ub(p+k), where k is the photon 4 momentum. The algebra regarding helicity does not change. The input and output spinor remain right handed (or left handed). And as you have mentioned if input electron is -1/2, and photon has +1, the output electron would have +1/2 spin. This does not seem to match the helicity values. Or is it that we cannot talk about the spin orientation of a virtual electron? $\endgroup$ – Angela May 2 '18 at 14:12
  • $\begingroup$ you cannot use the gamma (velocity)to check the virtual exchange, as its fourvector is variable under the integral $\endgroup$ – anna v May 2 '18 at 14:25

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