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Feynman Rules for Phi to the 6th theory

  1. Nov 8, 2005 #1
    What are the Feynman rules for phi to the sixth theory? Can anyone please help? Peskin and Schroeder does phi ^ 4th... I can't help thinking the derivation is the same for phi to the sixth, and that the rules are the same. Could that be correct?
    Thanks so much for your time,
    Job
    job@mailinator.com
     
  2. jcsd
  3. Nov 8, 2005 #2

    vanesch

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    Well, lambda^6 would give rise to a vertex with 6 scalars connected to it, not 4. So this changes the pictures (and the calculations). But it would indeed be very similar at first sight. However, drama and catastrophy ! lambda^6 theory is not renormalizable, so from the moment you'd consider loop diagrams, you'd be in deep s**t.
    That's explained later in P&S, around chapter 12 or so, when "power counting" is explained.
     
  4. Nov 8, 2005 #3
    Thanks Vanesch.

    I noticed that Peskin doesn't really "derive" the Feynman rules so much as go from one example in phi ^ 4 theory to the rules. At least in chapter 4 that's what they do. To "derive" the rules, is it better to use the path integral formalism? I think Srednicki (in the notes for his forthcoming QFT book) does something like that for phi psi ^2 theory...

    I'm not so worried about whether the theory is renormalizable, I just need to derive the Feynman rules for the theory, which I think are exactly the same as for the phi to the fourth theory. The derivation would be slightly different, as you mentioned, since each internal vertex will have 6 lines going in, and all the 4 factorials go to 6 factorials, but all those factorials cancel anyway, of course, so one could just replace every 4 with a 6 and then the derivation would be the same?? I just expected this to be tougher I suppose...

    Can someone point me to a formal derivation of the Feynman rules for say the phi ^4 theory or phi ^6, unless I missed such a thing in Peskin? Or even better post such a thing here? Thanks again,

    Job.
     
  5. Nov 8, 2005 #4
    any other thoughts about this one?
     
  6. Nov 8, 2005 #5
    bump

    bumped back up
     
  7. Nov 9, 2005 #6

    SpaceTiger

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    Developing the Feynman rules for scalar field theories is relatively straightforward, as you say, but you can find another example here:

    Michael Luke's Notes

    Be careful when interpreting the arrows on the diagrams between the above notes and Peskin & Schroeder. In the former, the arrows on the lines represent the flow of charge, while the latter uses a confusing convention that combines charge flow and the direction of momentum flow.
     
  8. Nov 9, 2005 #7

    dextercioby

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    The Feynman rules for such theory are found in an elegant manner. You need to Fourier transform to momenta representation the connected Green functions in various orders of perturbation theory. For [itex] \lambda \varphi^{4} [/itex] and [itex] \lambda \varphi^{6} [/itex] you'll need 4 respectively 6 functional derivatives to get the first order approximation of the unconnected Green functions generating functional. From [itex] W_{0} [/itex] and [itex] W_{1} [/itex] u can easily get [itex] iX_{0} [/itex] and [itex] iX_{1} [/itex].

    Usually fisrt order pertubation theory in momentum space gives the Feynman rules for the propagators for both internal & external lines.

    Daniel.
     
  9. Nov 9, 2005 #8

    Hans de Vries

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    It seems that the [itex]\phi^6[/itex] rules are equal to the [itex]\phi^4[/itex] rules as given on P&S page 801
    of appendix A1. with the only difference that the [itex]\phi^6[/itex] vertex now has six instead
    of four connections. (see vanesch's post)

    You may want to look at Zee's "baby problem" on page 42 of "QFT in a nutshell"
    to get the idea. The rest of the chapter then basically handles [itex]\phi^4[/itex]


    Regards, Hans
     
    Last edited: Nov 9, 2005
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