The Klein-Gordon equation with a potential

  • Thread starter Sekonda
  • Start date
649
3
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].
 
207
0
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
 
649
3
please report back! :) I want to know too.
 
207
0
haha will do!
 
1,023
30
I think that interaction term posted by dickfore is the right one,it seems that in post #27 that is what is being said.
 
207
0
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.
 
649
3
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.
 
207
0
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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