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## Field Redefinition and EOM

Hi,

I am given an interaction lagrangian piece as
$$\mathcal{L}_1 = \frac{1}{2} g \phi \partial^\mu \phi \partial_\mu \phi$$
Now normally when I have an interaction lagrangian piece I turn the field's into variations with respect to the source $\delta_J$, 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
$$\frac{1}{2}g \phi \partial^\mu \phi \partial_\mu \phi \rightarrow \frac{1}{2}g \Box (\delta_J)^3$$
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
$$\mathcal{L}_1 = -\frac{1}{4}g \phi^2 \Box \phi$$
since the total derivative vanishes.

Thanks, any help is appreciated.
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No transformation necessary, each φ is represented by a line coming to the vertex, and for each ∂μ include as a factor the corresponding k-vector.
 Am I allowed to use the EOM at this stage to simplify the interaction term?
Not in general, the equation of motion holds only for particles on the mass shell.
 Recognitions: Gold Member Thanks a lot Bill. So I would write something like $$\rightarrow \frac{i}{2}g(i k^\mu )(i k_{\mu}) = \frac{-i}{2}g k^\mu k_\mu$$ for the vertex factor in a diagram. Then for a field redefinition to the scalar lagrangian $$\mathcal{L}_0 = -\frac{1}{2}\partial^\mu \phi \partial_\mu \phi - \frac{1}{2}m^2 \phi^2$$ of the form $$\phi \rightarrow \phi' = \phi +\lambda \phi^2$$ I end up with something of the form $$\mathcal{L} = \mathcal{L}_0 -2\lambda \phi \partial^\mu \phi \partial_\mu \phi - 2 \lambda^2 \phi^2 \partial^\mu \phi \partial_\mu \phi -\lambda m^2 \phi^3 - \frac{1}{2}\lambda^2 m^2 \phi^4$$ Then I have associated vertex factors of the form $$2i\lambda k^2 \quad 2i \lambda^2 k^2 \quad -i \lambda m^2 \quad -\frac{i}{2}\lambda^2 m^2$$

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## Field Redefinition and EOM

Well if the interaction has three φ's, there's three k's also, with k1 + k2 + k3 = 0. So the factor in front of φ ∂μφ ∂μφ, instead of k2 wouldn't it have to look more like k1·k2?
 Recognitions: Gold Member So the phi without a partial derivative is acting like a mediator; it takes away $-(k_1 + k_2)$ 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...?
 Blog Entries: 2 Recognitions: Science Advisor You'd have to include all three possibilities. And for the φ2 ∂μφ ∂μφ term there would be six!

 Quote by jfy4 Thanks a lot Bill. So I would write something like $$\rightarrow \frac{i}{2}g(i k^\mu )(i k_{\mu}) = \frac{-i}{2}g k^\mu k_\mu$$ for the vertex factor in a diagram. Then for a field redefinition to the scalar lagrangian $$\mathcal{L}_0 = -\frac{1}{2}\partial^\mu \phi \partial_\mu \phi - \frac{1}{2}m^2 \phi^2$$ of the form $$\phi \rightarrow \phi' = \phi +\lambda \phi^2$$ I end up with something of the form $$\mathcal{L} = \mathcal{L}_0 -2\lambda \phi \partial^\mu \phi \partial_\mu \phi - 2 \lambda^2 \phi^2 \partial^\mu \phi \partial_\mu \phi -\lambda m^2 \phi^3 - \frac{1}{2}\lambda^2 m^2 \phi^4$$ Then I have associated vertex factors of the form $$2i\lambda k^2 \quad 2i \lambda^2 k^2 \quad -i \lambda m^2 \quad -\frac{i}{2}\lambda^2 m^2$$
vertex factor can be calculated directly from interaction term by taking iLint substiuting the plane wave form for different ∅ like exp(ikx) and exp(-ikx) and then throwing out the term already taken care by normalization,exponential factor etc.it is recommended to use lagrangian not hamiltonian in case partials getting involved.
 Recognitions: Gold Member So then using my expression for $\mathcal{L}$, I would have $$(i \lambda) (2(k_1 k+k_2 k+k_1 k_2)-m^2) \quad \text{and}\quad (i \lambda^2)(2(k_1 k_2 + k_1 k_3 + k_1 k_4 + k_2 k_3 + k_2 k_4 + k_3 k_4)-\frac{1}{2}m^2).$$ as my vertex factors? Thanks for your guys help.
 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 on-shell condition(it will be an approximation)

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 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 on-shell condition(it will be an approximation)
No they aren't! What if the vertex is just one part of a larger diagram. Then the lines coming to the vertex will be off-shell propagators.

 Quote by Bill_K No they aren't! What if the vertex is just one part of a larger diagram. Then the lines coming to the vertex will be off-shell propagators.
No,even if the vertex is part of a larger diagram it holds.It is because you are not considering the whole of diagram,if you will do this then you will find that those vertex which are sandwiched b/w off-shell propagators, also sandwiched between free particle states.Since K's does not involve any operators,they can be transferred to extreme left or right to operate on free particle state.If free particle states are spinors then also it can be manipulated with γ matrix algebra so as to get something like -iγ.p which can be set up to m.I think you understand it now.

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