Perturbations to all orders (Srednicki ch19)

  • Thread starter LAHLH
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Hi,

I have a few questions on the content of this chapter.

A theory is renormalizable in general if all it's Lagrangian coefficents have +ve or 0 mass dimension. So far so good. Now Srednicki says he will use [tex] \phi^3[/tex] theory in d=6 as his example, to see how to construct a finite expression for scattering amplitude to arb high order if coupling 'g'.

Am I correct in thinking by scattering amplitude he just means the [tex] \langle f\mid i\rangle [/tex] as given in the LSZ? which then contains [tex]\langle 0\mid T\phi(x_1).....\phi(x^{'}_1)\mid 0\rangle[/tex]

So as he shows in CH10 we only care about the fully connected diagrams. So e.g. for 4 particles scattering we sum all diagrams with 4 sources with sources removed etc.

Then he goes onto:

So just as [tex] \Pi (k^2) [/tex] is the sum of all 1PI diagrams with two external sources (with the sources removed) (although this is not the exact propagator), and [tex] \b{V}_3 [/tex] is the sum of all 1PI diagrams with three external legs, so [tex] \b{V}_n [/tex] is the sum of all 1PI diagrams with n external legs.

He then constructs these upto E (the self energy and [tex] \b{V}_3 [/tex] plus all the higher n vertices from [tex] 4 \leq n \leq E [/tex].)

I just don't understand why he's doing this? what does he mean exactly by "process of interest", does he mean E particles scattering or something? If so why only sum vertices upto E, because surely you could still have vertices that have higher valence within your 1PI diagram, even if the diagram only has E legs.

Thanks alot for any explainations, really appreciate it.

PS how is [tex] \phi^3 [/tex] in d=6 even being used in this chapter as an example?
 
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Avodyne
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what does he mean exactly by "process of interest", does he mean E particles scattering or something?
Yes; E is the number of incoming particles plus the number of outgoing particles.
If so why only sum vertices upto E, because surely you could still have vertices that have higher valence within your 1PI diagram, even if the diagram only has E legs.
But you cannot have higher valence in a tree diagram. After you have all the vertices, you construct the scattering amplitude from tree diagrams only.
PS how is [tex] \phi^3 [/tex] in d=6 even being used in this chapter as an example?
In the special role played by V_3 (it has to be assigned a value at some set of external momenta in order to define the renormalization scheme), and in that there is only one species of particle, so that a vertex function is characterized entirely by the number of external lines (and not by what kinds of particles these lines represent).
 
  • #3
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But you cannot have higher valence in a tree diagram. After you have all the vertices, you construct the scattering amplitude from tree diagrams only.

We count tree diagrams only because of this skeleton idea? But then doesn't this just neglect the diagrams that have E external legs and have n>E type vertices inside. I just don't see where these typ of diagrams go.

For example when considering E=2 (i.e. just the exact propagator), the first loop correction is ---O--- (hope you know what I mean, with this sketch). This diagram clearly has 2 vertices that of type [tex] V_3 [/tex]. But if we only counted vertices upto [tex] V_2 [/tex], we would be neglecting this diagram?

In the special role played by V_3 (it has to be assigned a value at some set of external momenta in order to define the renormalization scheme), and in that there is only one species of particle, so that a vertex function is characterized entirely by the number of external lines (and not by what kinds of particles these lines represent).

Only one species of particle...you mean because we have just one field [tex] \phi [/tex], nothing to do with the theory being [tex] phi^3 [/tex] in d=6 though, [tex] \phi ^4 [/tex] would be one species too right?

Is the V_3 renormalization condition what he talks about a few chapters earlier when we set kappa to zero. and thus V_3(0,0,0)=g? If so if we were in phi^4 would it be V_4 playing the special role?

Thanks alot for your help
 

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