
#1
Nov2812, 02:02 PM

P: 647

Hi,
I am given an interaction lagrangian piece as [tex] \mathcal{L}_1 = \frac{1}{2} g \phi \partial^\mu \phi \partial_\mu \phi [/tex] Now normally when I have an interaction lagrangian piece I turn the field's into variations with respect to the source [itex]\delta_J[/itex], and take variations of the free partition function to get feynman diagrams, however in this case the partials confuse me. Am I allowed to use the EOM at this stage to simplify the interaction term? Or can I preform the following operations [tex] \frac{1}{2}g \phi \partial^\mu \phi \partial_\mu \phi \rightarrow \frac{1}{2}g \Box (\delta_J)^3 [/tex] since typically the variation and the partial `commute'? EDIT: I realized I can integrate by parts, that maybe helps a little, but I am still almost in the same boat as before. After integration I get [tex] \mathcal{L}_1 = \frac{1}{4}g \phi^2 \Box \phi [/tex] since the total derivative vanishes. Thanks, any help is appreciated. 



#2
Nov2912, 01:41 PM

Sci Advisor
Thanks
P: 3,853

No transformation necessary, each φ is represented by a line coming to the vertex, and for each ∂_{μ} include as a factor the corresponding kvector.




#3
Nov2912, 02:34 PM

P: 647

Thanks a lot Bill.
So I would write something like [tex] \rightarrow \frac{i}{2}g(i k^\mu )(i k_{\mu}) = \frac{i}{2}g k^\mu k_\mu [/tex] for the vertex factor in a diagram. Then for a field redefinition to the scalar lagrangian [tex] \mathcal{L}_0 = \frac{1}{2}\partial^\mu \phi \partial_\mu \phi  \frac{1}{2}m^2 \phi^2 [/tex] of the form [tex] \phi \rightarrow \phi' = \phi +\lambda \phi^2 [/tex] I end up with something of the form [tex] \mathcal{L} = \mathcal{L}_0 2\lambda \phi \partial^\mu \phi \partial_\mu \phi  2 \lambda^2 \phi^2 \partial^\mu \phi \partial_\mu \phi \lambda m^2 \phi^3  \frac{1}{2}\lambda^2 m^2 \phi^4 [/tex] Then I have associated vertex factors of the form [tex] 2i\lambda k^2 \quad 2i \lambda^2 k^2 \quad i \lambda m^2 \quad \frac{i}{2}\lambda^2 m^2 [/tex] 



#4
Nov2912, 04:23 PM

Sci Advisor
Thanks
P: 3,853

Field Redefinition and EOM
Well if the interaction has three φ's, there's three k's also, with k_{1} + k_{2} + k_{3} = 0. So the factor in front of φ ∂_{μ}φ ∂_{μ}φ, instead of k^{2} wouldn't it have to look more like k_{1}·k_{2}?




#5
Nov2912, 05:17 PM

P: 647

So the phi without a partial derivative is acting like a mediator; it takes away [itex](k_1 + k_2)[/itex] from the interaction vertex? However, the assignment of momentum labels is arbitrary, how can I let my convention decided which momenta are the ones I have as factors...?




#6
Nov3012, 08:06 AM

Sci Advisor
Thanks
P: 3,853

You'd have to include all three possibilities. And for the φ^{2} ∂_{μ}φ ∂_{μ}φ term there would be six!




#7
Nov3012, 09:31 AM

P: 977





#8
Nov3012, 12:13 PM

P: 647

So then using my expression for [itex]\mathcal{L}[/itex], I would have
[tex] (i \lambda) (2(k_1 k+k_2 k+k_1 k_2)m^2) \quad \text{and}\quad (i \lambda^2)(2(k_1 k_2 + k_1 k_3 + k_1 k_4 + k_2 k_3 + k_2 k_4 + k_3 k_4)\frac{1}{2}m^2). [/tex] as my vertex factors? Thanks for your guys help. 



#9
Dec112, 07:00 AM

P: 977

I am not getting why are you redefining the field by a transformation,Vertex factors are always sandwiched between free particle states.There you can use eqn of motion to elliminate some k's by onshell condition(it will be an approximation)




#10
Dec112, 09:08 AM

Sci Advisor
Thanks
P: 3,853





#11
Dec212, 01:00 AM

P: 977




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