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Consider the process $$e^+(p_1)+e^−(p_2) \to S(p_3)S^∗(p_4)\tag{1}$$

$S/S^*$ is scalar particle/antiparticle described by the complex scalar field $\phi$ coupled to QED through the Lagrangian:

$$\mathcal{L}= \mathcal{L}_{QED}+ (D_\mu \phi)^* (D^\mu \phi)-m^2 | \phi|^2$$

and $D$ is the covariant derivative $D=\partial +ieA$.

How do I draw the tree-level Feynman diagram for such a Lagrangian/process if the interaction term only has the scalar particles in it? How do I know what interacts with the electron/positron?

My take on it:

I thought that the last term of the Lagrangian tells us that the interaction term must be two scalar particles (dashed lines) meeting at a vertex:

$\phi(p_3)$------*------$\phi(p_4)$ (don't know how to draw here)

So should the diagram be: Sorry it's a very big image

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The most important rule to always remember about feynman diagrams is, that fermion lines can not start nor end anywhere. This leaves either a closed fermion loop. Or any fermion line that enters from the outside must leave as well and must not be disconnected (because any lorentz invariant structure is a bilinear form in the spinors).

If there is no interaction term between the electron/positrons and the scalar particles, then, there is no such interaction possible in your model.

You could add a simple interaction such as

$$\mathcal{L}_\mathrm{int}=g\bar{\psi}\gamma^\mu\psi\partial_\mu\phi $$

to give rise to an interaction between electrons and scalars.

Then, one possible diagram would look like this (time going from left to right). This would be an annihilation process.

enter image description here

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  • $\begingroup$ This is an old exercise, I don't think I am allowed to simply add an extra term to the Lagrangian. I have just done some further looking into it, and I think the $\mathcal{L}_{QED}$ has an interaction term for two fermions and a photon in it and that is what permits a diagram such as the one you have drawn. $\endgroup$
    – Geop
    Jun 1 '20 at 20:36
  • $\begingroup$ A photon is a not a scalar particle. A photon is a vector particle. $\endgroup$ Jun 1 '20 at 20:38
  • $\begingroup$ However, judging from the interaction, the photon can couple to the scalar particle, so you just need to put a photon in before the dotted lines $\endgroup$ Jun 1 '20 at 20:50
  • $\begingroup$ Could it also be a photon being a propagator, two fermions on the LHS and two scalars on the RHS? This idea comes from an interaction term that arises from the $(D_\mu \phi)^* (D^\mu \phi)$ which possesses $A\phi^2$ and another term from $\mathcal{L}_{QED}$ which possesses a term with two fermions and a photon $-eQ\bar{\psi} A \psi$ $\endgroup$
    – Geop
    Jun 1 '20 at 21:08
  • $\begingroup$ Yes, that is probably the solution $\endgroup$ Jun 1 '20 at 22:04
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You action contains
$$ |D_\mu \phi|^2= g^{\mu\nu}[(\partial_\mu-ie A_\mu)\phi][(\partial_\mu+ie A_\mu)\phi^*] $$ so it has a quartic interaction term $A_\mu A^\mu \phi^*\phi$ as well cubic derivative interactions.

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  • $\begingroup$ My last comment on the answer above mentions that, but I am still uncertain about one thing: The term you have mentioned above would involve two photons and two fermions, I cannot have those many photons in my diagram, I think (Can I?)So does that mean I should use the term that has $A \phi \phi^*$ instead (this term also arises from $|D_\mu \phi|^2$ ,doesn't it?) $\endgroup$
    – Geop
    Jun 1 '20 at 21:16
  • $\begingroup$ Yes, the term with $A^\mu \phi^*\partial_\mu \phi$ and conjugate are in the interactions. $\endgroup$
    – mike stone
    Jun 2 '20 at 12:26

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