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Decay Widths of Neutral Vector Mesons

  1. Mar 22, 2017 #1
    1. The problem statement, all variables and given/known data

    Neutral vector mesons with ##J^{PC} = 1^{−−}## include the ##\phi## (mφ = 1020 MeV), ##J/\psi## (mJ/ψ = 3100 MeV), and ##\Upsilon## (mΥ = 9460 MeV), with quark content ss¯, cc¯, and b ¯b respectively. The decays of these mesons go largely to hadronic final states (jets), but there are non zero branching ratios to leptonic states via the process: qq¯ → γ → l+l-.

    (a) Draw a Feynman diagram for the leptonic decay of a neutral vector meson, qq¯ → γ → l+l−, labelling all particles and coupling strengths.

    (b) Show that the decay widths of the φ, J/ψ, and Υ to (all) leptonic states can be given approximately by the ratios: φ : J/ψ : Υ = 2 : 8 : 3.

    (c) Compare the ratios of leptonic decay widths estimated in part (b) against experimental data as is found in the PDG (http://pdg.lbl.gov) and comment on the validity of your estimate.

    2. Relevant equations

    3. The attempt at a solution

    I have done parts (a) and (c) of this question, but am struggling to do part (b).

    I think that these decays to leptonic states are governed by the EM interaction, and thus the interaction strength at each vertex in the Feynman diagrams should be ##\sqrt{\alpha}##.

    I have also noticed that only the ##\Upsilon## is able to decay into a ##\tau^+ \tau^-## since they are so massive.

    How do I go about calculating a decay width here?

    I have been trying to read around the subject, but most of the material online seems to rely on QFT and other things I have not studied.

    Any help would be greatly appreciated!

  2. jcsd
  3. Mar 22, 2017 #2


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    You are nearly done.
    What about the charges of the involved particles?
  4. Mar 22, 2017 #3
    The strength at a vertex should also be proportional to the charge at the vertex. But would that not lead to me having zero, since I have a particle and it's anti-particle?
  5. Mar 22, 2017 #4


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    You shouldn't add the charges.
  6. Mar 23, 2017 #5
    considering the first vertex in a ##s \bar{s} \rightarrow \gamma \rightarrow l^+l_-##, the interaction strength should be proportional to ##Q_sQ_{\bar{s}} \sqrt{\alpha}##, and the contribution from the second vertex (##\sqrt{\alpha}##) leads to a total of ##Q_sQ_{\bar{s}}\alpha##.

    I think at this point I would multiply by a mass to get the dimensions correct, but multiplying by the vector mesons does not lead to a correct result. I have noticed however that for my three vector mesons, I have

    ##\phi: 2 \times (\frac{1}{9} \alpha)## (since it can decay to two leptonic states)
    ##J/\psi: 2 \times (\frac{4}{9} \alpha)## (since it can decay to two leptonic states)
    ##\Upsilon: 3 \times (\frac{1}{9} \alpha)## (since it can decay to three leptonic states)

    which gives me my ratio of 2:8:3

    So is it the mass of the final leptons that I want to multiply by in order to get the correct units? That would make sure my ratio remains as desired, but what is the rationale behind using that mass?
  7. Mar 23, 2017 #6
    Or more generally, I think it would make sense to use the center of mass energy, since that is what is available for the reaction.
  8. Mar 23, 2017 #7


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    The question only asks about ratios, and the initial quark masses don't play a role. There is no mass you have to consider. The phase space for the final states will be different, but that is a small effect as no decay is close to a threshold (e.g. close to but above 2 times the muon or tau mass).
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