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Divergent tree level Feynman diagrams?

  1. Jun 8, 2009 #1
    Hi everybody! I'm a new Physics Forums user and hope someone could help me out with my minor dilemma. I'm a PhD student in mathematical/theoretical physics and I' working on the Boltzmann equation in QFT. Up to now, there was no major emphasis on Feynamn diagrams - the approach was rather more mathematical with the fundamentals of QFT (definition of Wick products, Wick expansion, Wick reordering, Bogoliubov's formula for interacting fields and such). But now I have to make sense of some expression that emerges in my computations.

    To make a long story short, I'm working with \phi^4 theory and need to consider a 3 -> 3 scattering process in perturbation theory. Also, I'm using a (-,+,+,+) signature for the metric - opposite of most textbooks so some signs might appear odd. Now, this can only happen to second order in the coupling constant, and at that order, there are no loop diagrams. With the possible permutations, the diagrams I have to consider are the following

    Code (Text):
         q_1   q_2  q_3
             \   |   /
              \  |  /
               \ | /
               / | \
              /  |  \
             /   |   \
           p_1  p_2  p_3

    Code (Text):
    q_1  q_2   q_3
       \   \  /
        \   \/            
         \  /\            
          \/k \            
          /\   \            
         /  \   \
      p_1  p_2  p_3  
    I have no problems with the first one, but the second one is causing me some confusion. If I write down the amplitude associated to it, I get (neglecting symmetry factors)

    (-i g)^2/(k^2 +m^2 -i \epsilon) = (-i g)^2/[(p_1 + p_2 - q_1)^2 +m^2 -i \epsilon],

    and g is my coupling constant.

    The physical interpretation seems clear: two particles scatter and, subsequently one of them scatters off the third one. And this is one way to see the origin my confusion. The momentum k is associated to a real particle, even though it's not an "external" particle. In other words, k is now on-shell, i.e. k^2 = -m^2, giving me a divergent amplitude!

    I've never heard of divergent tree level amplitudes... What am I missing? Most probably the reason why k is not on-shell but why is it not? This is the physical amplitude for the above process and all the momenta are on-shell. What's going on here?

    Thanks for your time!
  2. jcsd
  3. Jun 8, 2009 #2


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    k is not on-shell. Why do you think it is?
  4. Jun 8, 2009 #3


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    Homework Helper

  5. Jun 8, 2009 #4
    I think k can be on-shell as long as the particles aren't massless. I think the answer is finite however when you integrate over final momenta.

    edit - first sentence should read instead: "k can be on-shell and your final result finite, as long as the particles aren't massless." If the particles are massless, then you have infrared divergences you have to treat.
    Last edited: Jun 8, 2009
  6. Jun 9, 2009 #5
    Thanks for the replies!

    Avodyne -> If I take the idea of "two particles scatter off each other and then one of them re-scatters off the third one" seriously, then I can set up a standard 2 body scattering situation (going to the CM frame and performing explicit computations) to analyze the first scattering which would confirm k being on shell. In other words, when I consider a 2 -> 2 scattering, the final 2 particles are on shell. If this does not apply, why doesn't it?

    malawi_glenn -> Read my reply to Avodyne. As for the example, I think I understand it, but it's really not really analogous to my situation. It's more related to, say, an electron-electron scattering in QED with the exchange of a photon. I think I understand 2 -> 2 scattering fairly well, but as I said, my confusion comes from 3 -> 3 scattering, and the above diagram in particular.

    RedX -> I'm not sure I see what you mean. If k is on-shell, then the propagator is singular and it's not related to its mass.
  7. Jun 9, 2009 #6
    I'm not sure what I said either.

    Your problem seems related to infrared divergences that come up when a final particle line splits up into multiple soft photon lines. You have your original diagram and you calculate it, but then you take the final line of the original diagram and split it up into more particles. What happens is that you get infinities that get canceled by loop diagrams and stuff.

    I don't really know why you're calculating 3 -> 3 scattering. Seems weird.
  8. Jun 9, 2009 #7


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    Science Advisor

    The internal line could be on-shell, but in general it does not have to be. For generic choices of p1, p2, and q1, k will not be on-shell.

    To see this, let's work in the center-of-mass frame of particles p1 and p2 only. Then (assuming all particles are the same type and so have the same mass), the energies E1 and E2 are the same; call this energy E. If q1 and k were both on-shell, then (by conservation of energy and momentum) their energies would also each have to equal E. But this might not be true of q1 in the 3-body process. And if the energy of q1 does not equal E exactly, then k is not on-shell.
  9. Jun 10, 2009 #8
    RedX -> It is kinda weird, but I really need to. But it's too long and also technical to explain. Take it as a simple field theoretical question.

    Avodyne -> I see your point, and it makes sense. that was my original line of thought, but then I confused myself with the idea of "2 particles scattering and one subsequent re-scatter" (as 3 billiards might do). Probably this way of thinking is useful but may be a bit misleading if taken too literally.
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