The Klein-Gordon equation with a potential

[Sekonda]

Hey,

My professor says an equation of this form:
$$(\frac{\partial^2 }{\partial t^2}-\bigtriangledown^2+m^2)\Psi=\lambda\Psi_{f'}^{*} \Psi_{i'}\Psi$$
Will give a 4 particle interaction where I need to make the 'f'' state on the rhs an external (not sure what this means yet) and this useful for looking at the Higgs.

Whereas an equation of form:

$$(\frac{\partial^2 }{\partial t^2}-\bigtriangledown^2+m^2)\Psi=\lambda\Psi_{f'}^{*} \Psi_{i'}$$

Is a 3 particle interaction where they meet at a junction and some internal scalar particle is propagated.

 

[jfy4]

sure, I don't have a problem with that. But I don't think that is a consequence of a potential of the form $V=\lambda \Psi^{\dagger}\Psi$. I think the potential must have a different form if the r.h.s of the equation looks like that.

 

[Sekonda]

Well he included a delta sign next to the potential i.e. it was δV=λψ*ψ, are you supposing it should be δV=λψ*?