## The Klein-Gordon equation with a potential

 Quote by andrien Second one makes sense,first one does not.
The KGE is the Euler-Lagrange eqn. for the action defined as $S = \int{d^d x \, \mathcal{L}}$, where the Lagrangian (density) is:
$$\mathcal{L} = \mathcal{L_0} + \mathcal{L}_{\mathrm{int}}$$
Here $\mathcal{L_0} = \partial^{\mu} \Psi^{\ast} \, \partial_{\mu} \Psi - m^2 \Psi^{\ast} \, \Psi$ is the Lagrangian for the free complex scalar field. Varying w.r.t. $\Psi^{\ast}$, one obtains:
$$\frac{\delta S}{\delta \Psi^{\ast}(x)} = -\left\lbrace \partial^2 + m^2 \right\rbrace \Psi(x) - \frac{\lambda}{2! 2!} 2 (\Psi^{\ast} \Psi) \, \Psi$$
Equate this variation to zero and you get:
$$\left\lbrace \partial^2 + m^2 \right\rbrace \Psi = - \frac{\lambda}{2} (\Psi^{\ast} \Psi) \, \Psi$$
Apart from a wrong sign, the r.h.s. has the form I gave in my previous post.
 I thought your first term also some sort of a lagrangian which is not possible.

 Quote by andrien I thought your first term also some sort of a lagrangian which is not possible.
How can a Lagrangian term be present in a Klein-Gordon equation?!
 Yes you are right with the correct potential form, though I assumed it was in relation to the Higgs boson as this is what the topic is on though I'm guessing a potential of this form is just an example to familiarise ourselves.

 Quote by Dickfore How can a Lagrangian term be present in a Klein-Gordon equation?!
I did not read it rather just see.
 Hey I think the form of the potential was supposed to be this: $$\lambda\Psi_{f'}^{*} \Psi_{i'}$$ such that: $$(\frac{\partial^2 }{\partial t^2}-\bigtriangledown^2+m^2)\Psi=\lambda\Psi_{f'}^{*} \Psi_{i'}$$ Is this right?
 Recognitions: Gold Member For a complex scalar field theory you want a partition function of the form $$Z = \int D\phi \exp \left[ i \int d^4 x \frac{1}{2} \{ \partial \phi \partial \phi^{\dagger} + m^2 \phi \phi^{\dagger} \} + V(\lambda , \phi, \phi^{\dagger}) + J^{\dagger}\phi + J\phi^{\dagger} \right]$$ you can then assume weak coupling and expand the potential order by order in lambda, and replace the moment integrals with variations with respect to J, $\delta / \delta J$, like $$\int dx \, x^2 e^{-\alpha x} = \frac{\partial^2}{\partial \alpha^2} \int dx \, e^{-\alpha x}$$ doing this, along with a regularization procedure, you can make all the connected green's functions and feynman diagrams.
 Hmm I don't understand most of that - though it does sound like the route we are taking. Does the equation I just posted make sense at all though? Are you implying that it is and it's a complex scalar field? I haven't actually had any substantial teaching of quantum field theory, we're supposed to be taking a simplified route so forgive me for not understanding most of the terms you speak of (i.e. the partition function - i've heard of that but not applied to this!) Thanks!
 Recognitions: Gold Member 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
 Recognitions: Gold Member 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.
 Recognitions: Gold Member 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*?