# Feynman Diagrams Pair Annihilation

In Griffith's Introduction to Particle Physics it says that there are two processes mainly contributing to pair annihilation (i.e. $e^++e^-\to\gamma+\gamma$)

1. How does the second process (the one with the twist) differ physically from the first one?

2. Why is the following process not allowed?

1. If the twist in the first set of diagrams is mathematically different. Why is the following process not considered?

• Your last diagram is allowed. Namely, it's the same as the first one. – Chris Jan 23 '18 at 8:48
• @Chris you are correct! That was a mistake on my side. I will fix this Thank you – bodokaiser Jan 23 '18 at 9:02
• To 1: The second differs from the first in the Amplitude function $A(p_1,p_2,p_3,p_4)$. The momenta $p_3$ and $p_4$ are interchanged in this Amplitude function. – kryomaxim Jan 23 '18 at 9:22
• The diagram in 2) is not allowed because you have drawn vertices not permitted in the SM. – CAF Jan 23 '18 at 10:18

1. The physical process is the sum of both diagrams. Remember that Feynmann diagrams are just a pictorial representation to see what is going on. The two diagrams are mathematically different in the ideal world where we could distinguish the two photons (say the red photon and the blue photon). However, we cannot intrinsically distinguish them, and that is why the amplitude of $e^+ \, e^-$ to two photons is given by the sum as if they were distinguishable.
• 3. That process is the same as the second process in 1. (I think that the fermionic arrows are incorrect btw). This is because the initial state is fixed and distinguishable (I have an electron an a positron with some momenta $p_1$ and $p_2$ and some spin z-component $s_1$ and $s_2$) and I want to see the decay into two photons. This is the usual case in particle accelerators, where the momenta of the ingoing particles is known (maybe the spin is not know, so you would have to sum to all the different initial spins and do the average of probabilities). – M. Sasieta Jan 23 '18 at 9:50