# 2 body Scalar Interaction

1. Dec 10, 2012

### Sekonda

Hey,

I'm rather confused with this interaction. It's depicted by the diagram below (though the directions are meaningless), it represents a scalar interaction of spin zero bosons.

(I know I called it 4 point but I'm sure it's a 3 point function/interaction)

Anyway, to replicate this interaction we have the Klein Gordon equation of form:

$$(\frac{\partial^2 }{\partial t^2}-\nabla^{2}+m^{2})\Phi=V\Phi$$

My professor said we use a potential of the form

$$V=\lambda\Phi^*$$

to replicate this interaction, upon substitution we get:

$$(\frac{\partial^2 }{\partial t^2}-\nabla^{2}+m^{2})\Phi=\lambda\Phi^{*}\Phi$$

He then said it is a matter of putting in external state into the right hand side of the above, such that it becomes (Note the Psi's in the diagram are supposed to be capital Psi's)

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

By inspection we see this solved for

$$\Phi=-\frac{\lambda\Psi_{f'}^{*}\Psi_{i'}}{E^{2}-\mathbf{p}^{2}-m^{2}}$$

Where the energies and momentums in the denominator are the differences between the states f' and i'.

So I'm a little confused with how we assign external states as we do... How can we do this? It doesn't seem to make any sense to me, if it makes any difference he did write a δ in front of the V at the top of this post.

Thanks guys,
SK

Last edited: Dec 10, 2012