# Feynman diagram for annihilation

What is the difference between these two Feynman diagrams? They should both describe the same physical process, annihilation between an electron and a positron.

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But your first diagram doesn't have two photons in the final state. It doesn't represent annihilation, but rather s-channel scattering. – rob Apr 25 '14 at 13:50
Non-physicist's query: What does the e (or e*) in the second diagram mean/indicate? – RBarryYoung Apr 25 '14 at 15:48
@RBarryYoung: Usually you see $e$ for those internal lines. That virtual particle could be an electron or a positron, depending on which $\gamma$ is emitted first in your reference frame; it doesn't matter which, since no experiment can distinguish between them. In scattering from composite particles you'll sometimes see $A^*$ to indicate that $A$ passes briefly through an excited state. But since the electron doesn't have any internal excitations I'm not sure what the author of that figure meant. – rob Apr 25 '14 at 18:32
I'm familiar with the first diagram being drawn differently. Shouldn't electron arrows always point forwards in time and positron arrows backwards? – Geoff Pointer Aug 30 '15 at 5:04

The first process corresponds to $e^{-}e^{+}\to e^{-}e^{+}$ (Bhabha scattering), where the final and initial states are the same, consisting of an electron and positron. However, the second process is $e^{-}e^{+}\to \gamma \gamma$, where instead the final state is that of two photons. The scattering amplitudes will be different. Notice that the first diagram requires an insertion of the photon propagator,

$$-\frac{i\eta_{\mu\nu}}{q^2 +i\epsilon}$$

whereas the second diagram has a fermionic internal line, requiring a propagator,

$$\require{cancel} \frac{i\left(\cancel{q}+m_f\right)}{q^2-m_f^2 +i\epsilon}$$

In addition, the second diagram will contain polarization vectors, as the photons are not internal lines but rather external. For a comprehensive overview of QED, see Peskin and Schroeder's text.

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@innisfree: Thanks for adding the Wiki link. – JamalS Apr 25 '14 at 13:53
so what's the difference? THe first photon is virtual? – SuperCiocia Apr 25 '14 at 13:57
@Harold What physical process you're talking about is identified by the initial and final states. If in an experiment you have two electrons going in and two electrons going out, you can't look at it and say "the intermediate state is a single photon". That is just one diagram that contributes to the process. At tree-level you also have the diagram where a photon is exchanged (the t-channel), and there are higher order diagrams as well. – Tim Goodman Apr 25 '14 at 19:25
For the first one, shouldn't we have two photons to conserve momentum though? – SuperCiocia Apr 25 '14 at 21:29
@Harold, You would need two photons to conserve 4-momentum while satisfying $E = pc$, but because it's a virtual photon (an internal line) you're allowed to violate $E = pc$. – Tim Goodman Apr 27 '14 at 2:33