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Furry theorem

  1. May 5, 2015 #1
    Hello, everyone!

    I have a short question. I think everybody knows the Furry theorem for QED, which tells, in particular, that fermion loops with odd number of fermions always give 0.

    Does the same true for QCD of weak theory? Can one have non-zero contribution to 3-gluon vertex (or 3-boson vertex) from 3-quark (3-lepton) loop with strong (weak) vertices.

    Thanks,
    Ivan
     
  2. jcsd
  3. May 5, 2015 #2

    mfb

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    "QCD of weak theory"? I guess that should be an "or".

    Gluon fusion to produce a Higgs is an example of strong plus weak interaction with a 3-vertex fermion loop (top is dominating).
     
  4. May 7, 2015 #3

    fzero

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    Yes but that process has two vector currents and a scalar current, so the total expression is even under charge conjugation. A three gluon loop would contribute a ##C## odd combination of vector currents.

    The same logic for Furry's theorem in QED works for Yang-Mills theory. Diagrams with an odd number of (only) gluon external legs connected directly to internal fermion lines will vanish. Your point is well taken that you restrict to the class of diagrams that are nonzero because of scalar and gluon self-interaction vertices.
     
  5. May 7, 2015 #4
    I need to check on paper. But it's the same with the exception that we get a trace of ( ta tb tc) , and the other diagram with a minus sign is (ta tc tb). So I think the symmetric colour structure d_abc piece vanishes and the f_abc piece is left over.
     
  6. May 7, 2015 #5

    fzero

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    If you setup the calculation for the 3 external gluons coming out of a fermion triangle there are really two diagrams with the loop momenta running in opposite directions. Because of the color labels we need to consider both of them. Since the fermion propagators are linear in momenta, you'll find the amplitudes come in with the opposite sign. So the sum of the two graphs actually vanishes.
     
  7. May 17, 2015 #6
    Hi fzero,

    So are you saying what I wrote is wrong?

    In one diagram I should get Tr{ tA tB tC}, while in the other I get Tr{ tA tC tB}.

    So for the first I get 1/4( i fABC + dABD) and in the second I get 1/4( -i fABC + dABD ). Therefore, when I sum up the two diagrams (which have opposite sign like i said in the first place), I get something proportional to 1/2 i fABC. I guess something like this is expected, since in colour structure (if these were very massive fermions which were integrated out as in an effective interaction) this would lead to something like the three gluon vertex.

    Anyway, in QCD its of course never as straightforward, since for the three point function it is also necessary to include gluons and ghosts in the loops. There are also extra topologies compared to QED where there are 4-gluon vertices. [Edit: I notice fzero already wrote about this above]
     
    Last edited: May 17, 2015
  8. May 17, 2015 #7

    fzero

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    You are correct. After setting the calculation up I find the same factors that you do. My post about the calculation using currents is also incorrect, since the color index on the fermions prevents the current from being an eigenstate of charge conjugation. Thank you for following up on this.
     
  9. May 18, 2015 #8
    No problem. I was just a little confused as to whether I'd done something stupid.

    It's actually a bit more interesting than I initially thought, so I might compute the full thing (ghosts and gluons) for the three point. If I find anything interesting I'll post back!
     
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