feynman-diagramsparticle-physicsquantum-electrodynamics
In Griffith's Introduction to Particle Physics it says that there are two processes mainly contributing to pair annihilation (i.e. $e^++e^-\to\gamma+\gamma$)
How does the second process (the one with the twist) differ physically from the first one?
Why is the following process not allowed?
Experimentally speaking, it is found we detect only $\mathrm{SU}(3)$ colour singlets in nature so the gluons we draw in our feynman diagrams will ultimately hadronise to form such states at time scales longer than the hard scattering with which they were involved. So at the end of the day the resultant states will carry $\mathrm{SU}(3)$ colour quantum numbers with a net red/green/blue configuration. Colour combinations other than this, as carried by the gluons in your picture, are not discernible to detectors so from this viewpoint we expect to consider the two topologically distinct diagrams as drawn.
BTW those two diagrams are also equivalent to
Time in this diagram (as the arrow shows) goes from left to right instead of bottom to top. Thus the particles on the left (an electron and photon) are the initial particles and those on the right (an electron and photon) are the final particles. Hence this is Compton scattering instead of pair production.
Best Answer
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.
This s-channel process you are drawing is not allowed in the Standard Model because the internal particle must have zero electric charge due to the left vertex, and therefore it cannot be coupled to the photon (so the right vertex is not possible).