Decay Widths of Neutral Vector Mesons

[BOAS]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

[/B]
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!

Thanks

 

[mfb]

You are nearly done.

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}##.

What about the charges of the involved particles?

 

[BOAS]

You are nearly done.
What about the charges of the involved particles?

You are nearly done.
What about the charges of the involved particles?

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?

 

[mfb]

You shouldn't add the charges.

 

[BOAS]

You shouldn't add the charges.

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?

 

[BOAS]

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.

 

[mfb]

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).