The Klein-Gordon equation with a potential


by Sekonda
Tags: equation, kleingordon, potential
andrien
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#19
Nov21-12, 03:05 AM
P: 987
I thought your first term also some sort of a lagrangian which is not possible.
Dickfore
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#20
Nov21-12, 03:08 AM
P: 3,015
Quote Quote by andrien View Post
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?!
Sekonda
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#21
Nov21-12, 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.
andrien
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#22
Nov21-12, 07:30 AM
P: 987
Quote Quote by Dickfore View Post
How can a Lagrangian term be present in a Klein-Gordon equation?!
I did not read it rather just see.
Sekonda
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#23
Nov21-12, 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?
jfy4
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#24
Nov21-12, 11:41 AM
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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.
Sekonda
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#25
Nov21-12, 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!
jfy4
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#26
Nov21-12, 03:13 PM
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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].
Sekonda
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#27
Nov21-12, 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
jfy4
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#28
Nov21-12, 03:41 PM
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P: 647
please report back! :) I want to know too.
Sekonda
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#29
Nov21-12, 03:46 PM
P: 209
haha will do!
andrien
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#30
Nov22-12, 03:19 AM
P: 987
I think that interaction term posted by dickfore is the right one,it seems that in post #27 that is what is being said.
Sekonda
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#31
Nov22-12, 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.
jfy4
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#32
Nov22-12, 11:22 AM
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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.
Sekonda
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#33
Nov22-12, 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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