# The Klein-Gordon equation with a potential

something doesn't seem right... If I understand you correctly you are writing
$$V(\Psi, \Psi^{\dagger}) = \lambda \Psi \Psi^{\dagger}$$
then the lagrangian would be
$$\mathcal{L} = \frac{1}{2}\partial \Psi^{\dagger} \partial \Psi + \frac{1}{2}m^2 \Psi^{\dagger}\Psi - \lambda \Psi^{\dagger} \Psi$$
If you solve Lagrange's equations for $\Psi$ or $\Psi^{\dagger}$ you will not get what you have posted. If you want to get a vertex with three lines the potential needs to be cubic in field quantities. but with a complex field it will be asymmetrical too since you will write down a term like $\Psi\Psi^{\dagger}\Psi$ which is uneven in the two fields. For real fields it would be like $\phi^{3}/3!$.

Well it probably isn't right knowing me, my professor said we model the perturbation as the interaction between some 2 scalar particles which is of the form of some coupling strength lambda and the product of the complex conjugated final state wavefunction of particle 2 and the initial state wavefunction of particle 2...

I shall send him an email tomorrow and ask!
I apologise if I'm not coherent/making sense.

Thanks,
Tom

please report back! :) I want to know too.

haha will do!

I think that interaction term posted by dickfore is the right one,it seems that in post #27 that is what is being said.

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.

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.

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*?