Scalar QED - pair annihilation into photon cross section I just spent the last three days trying to compute the cross section of a process of pair annihilation of complex scalars to a pair of photon in scalar QED.
For some reason I don't seem to be able to get the right answer.
I am trying to find back the result from Itzykson and Zuber's Quantum Field Theory, eq 6.66 and 6.67.
The only other "ressource" I found that does that computation is a youtube video. Everything is pretty clear until you hit 10:30, and the guy gives the amplitudes of each diagram.
I do not get how he finds those amplitudes, as in the first one, one of the vertex contribution is stated as $-ie(p_1-k_1+p_2)$, and this does not seem right to me :
The incoming antiscalar has momentum $p_2$, the outgoing photon has momentum $k_2$, and the internal scalar has momentum $k = p_1-k_1 = k_2-p_2$. To me, this vertex contribution should be :
$$-ie(k - p_2) = -ie(p_1 -  k_1 - p_2) = -ie(k_2 - 2p_2)$$
For some reason, the guy in the video switched the sign of the incoming antiscalar momentum. But this seem to violate the prescription for scalar qed (incoming antiscalar momentum should have a negative sign in the vertex contribution).
What drives me mad is that if you do the computation like he does, you do end up with the right cross section as given by Itzykson and Zuber. If you do it my way, you get nonsense (cross section going to infinity in the non relativistic limit).
Is there someone who would have a decent explanation as to why this sign is switched ? Thanks in advance !
By switching the sign and using momentum conservation, the guy in the video shows that the contribution of the t and u channel diagrams vanish, using that the polarization vector and momentum vector of the photon are orthogonal. Indeed, from $-ie(p_1-k_1+p_2)$, you get $-iek_2$, which vanishes when taken as a dot product with the associated polarization vector. That fails to work if you use the right prescription for the vertex contribution [$ie(p_1-k_1+p_2)$].
 A: You have trouble with the signs if I understand correctly. The Feynman rule regarding the vertex with the three prongs (1 boson and 2 charged scalars) is proportional to the sum of the four momenta $p$ and $p'$, where $p$ is the momentum of one of the charged scalars, whereas $p'$ is the momentum of the other.
In your particular case, you have a diagram, in which two charged scalars are incoming, exchanging another charged scalar with the outgoing state particles being the two photons. So, if you focus on each vertex, you will have the momentum of the incoming charged scalar going inside the vertex, the momentum of the gauge boson flowing away from the vertex and another momentum (that of tha scalar propagator) leaving one of the vertex and going to another.
You have to be careful in identifying the momenta you are about to sum such that you apply the Feynman rule correctly. If $p_1$ is the momentum of the first incoming scalar and $k_1$ is the momentum of the outgoing boson, then the propagator will carry momentum $k=\pm(p_1-k_1)$ ($\pm$ depends on how you choose the momentum flow of the propagator). Adding the two, as it should, yields the Feynman rule for one of the vertex
$$ie(p_1+k)^{\mu}=ie(p_1+p_1-k_1)^{\mu}=ie(2p_1-k_1)^{\mu}$$
if I choose the momentum flowing from the first vertex to the second (the one whose momentum flowing in/out from I shall denote with $p_2$ and $k_2$). This is what the guy in the youtube video has I think.
In a similar manner you will work in order to get the vertex that corresponds to the second scalar coming in and the second boson going out, with the only difference being that the momentum of the propagator will flow in the vertex if it was chosen to flow away from the previous vertex and vice versa. Also, do not forget to dot with the respective polarization vectors and also include the factor for the propagator. This should suffice to get your Feynman amplitude!
If anything is unclear, please let me know...
Pro tip: Whenever you are unsure about the vertices Feynman rules, you can always calculate the scattering amplitudes $\langle out|i\int d^4x \mathcal{L}_{\text{int}}|in\rangle$, stripp off any potential polarization vectors or spinors you might find in your calculation and the remaining factors will comprise your Feynman rule! The $|in\rangle$ state will be chosen to contain all the particles, whose momentum flows into the vertex and the $|out\rangle$ state will be chosen to contain all the particles, whose momentum flows away from the respective vertex. $\mathcal{L}_{\text{int}}$ is the interacting Lagrangian in the interaction picture.
