# w do I draw the tree-level Feynman diagram if the interaction term only represents the scalar particles?

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:

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.

• 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.
– Geop
Jun 1 '20 at 20:36
• A photon is a not a scalar particle. A photon is a vector particle. Jun 1 '20 at 20:38
• 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 Jun 1 '20 at 20:50
• 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 $-eQ\bar{\psi} A \psi$
– Geop
Jun 1 '20 at 21:08
• Yes, that is probably the solution Jun 1 '20 at 22:04

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.

• 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?)
– Geop
Jun 1 '20 at 21:16
• Yes, the term with $A^\mu \phi^*\partial_\mu \phi$ and conjugate are in the interactions. Jun 2 '20 at 12:26