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The KleinGordon equation with a potential 
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#19
Nov2112, 03:05 AM

P: 1,020

I thought your first term also some sort of a lagrangian which is not possible.



#20
Nov2112, 03:08 AM

P: 3,014




#21
Nov2112, 04:50 AM

P: 209

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.



#22
Nov2112, 07:30 AM

P: 1,020




#23
Nov2112, 10:40 AM

P: 209

Hey I think the form of the potential was supposed to be this:
[tex]\lambda\Psi_{f'}^{*} \Psi_{i'}[/tex] such that: [tex](\frac{\partial^2 }{\partial t^2}\bigtriangledown^2+m^2)\Psi=\lambda\Psi_{f'}^{*} \Psi_{i'}[/tex] Is this right? 


#24
Nov2112, 11:41 AM

P: 647

For a complex scalar field theory you want a partition function of the form
[tex] 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] [/tex] 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, [itex]\delta / \delta J [/itex], like [tex] \int dx \, x^2 e^{\alpha x} = \frac{\partial^2}{\partial \alpha^2} \int dx \, e^{\alpha x} [/tex] doing this, along with a regularization procedure, you can make all the connected green's functions and feynman diagrams. 


#25
Nov2112, 03:01 PM

P: 209

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! 


#26
Nov2112, 03:13 PM

P: 647

something doesn't seem right... If I understand you correctly you are writing
[tex] V(\Psi, \Psi^{\dagger}) = \lambda \Psi \Psi^{\dagger} [/tex] then the lagrangian would be [tex] \mathcal{L} = \frac{1}{2}\partial \Psi^{\dagger} \partial \Psi + \frac{1}{2}m^2 \Psi^{\dagger}\Psi  \lambda \Psi^{\dagger} \Psi [/tex] If you solve Lagrange's equations for [itex]\Psi[/itex] or [itex]\Psi^{\dagger}[/itex] 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 [itex]\Psi\Psi^{\dagger}\Psi[/itex] which is uneven in the two fields. For real fields it would be like [itex]\phi^{3}/3![/itex]. 


#27
Nov2112, 03:31 PM

P: 209

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 


#29
Nov2112, 03:46 PM

P: 209

haha will do!



#30
Nov2212, 03:19 AM

P: 1,020

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



#31
Nov2212, 10:41 AM

P: 209

Hey,
My professor says an equation of this form: [tex](\frac{\partial^2 }{\partial t^2}\bigtriangledown^2+m^2)\Psi=\lambda\Psi_{f'}^{*} \Psi_{i'}\Psi[/tex] 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: [tex](\frac{\partial^2 }{\partial t^2}\bigtriangledown^2+m^2)\Psi=\lambda\Psi_{f'}^{*} \Psi_{i'}[/tex] Is a 3 particle interaction where they meet at a junction and some internal scalar particle is propagated. 


#32
Nov2212, 11:22 AM

P: 647

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



#33
Nov2212, 11:47 AM

P: 209

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*?



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