# First-order EM Feynman diagram?

Is there any 1st order electromagnetic Feynman diagram? I.e. a process whose probability is just $\propto \alpha_{EM}$?

If not, is there any physical reason why? We always need at least two particles in and two out to conserve energy and momentum?

A diagram which is first order in $\alpha_\text{EM}$ would have to have one vertex, because $\alpha_\text{EM}\propto g^2$ where $g$ is the factor associated with each vertex (and the amplitude corresponding to the diagram gets squared). There's only one possible vertex in QED, namely the photon-electron-positron vertex, and it's impossible to arrange this in any way that conserves energy and momentum. There are only two really distinct possibilities:
• initial state photon, final state electron and positron: the final state has a rest frame and the initial state doesn't. Or vice-versa (initial state $e^-e^+$, final state $\gamma$)
• initial state photon and electron, final state electron, or vice versa: in the rest frame of the final state, the total energy is $m_e c^2$, whereas the initial state has energy at least $m_e c^2 + E_\gamma$
So no, there is no first-order diagram. The same argument goes to show that the corresponding process ($\gamma\to e^+e^-$, $e^-\gamma\to e^-$, or any variant) is kinematically forbidden. If you replace the photon with a sufficiently massive particle, like a Z boson, then it's totally fine. Of course, Z bosons aren't stable, so whatever diagram you draw that includes the $Z\to e^+e^-$ vertex should probably also include whatever interaction produced the Z boson in the first place, but in theory if you had a free Z boson propagating through space, it could undergo this decay.
• Exactly, conservation of charge, or lepton number, or angular momentum, etc. requires that you have a fermion going in for each one going out. Or, more fundamentally, there is no $\gamma\gamma e$ vertex in QED. – David Z Apr 14 '15 at 19:15