# Decay Widths of Neutral Vector Mesons

• BOAS
In summary, the decays of neutral vector mesons with ##J^{PC} = 1^{−−}## to hadronic final states have non-zero branching ratios to leptonic states via the process ##qq¯ → γ → l+l-##, where ##l## represents a lepton. The strengths of the interactions at each vertex in the Feynman diagrams for these decays should be proportional to the electromagnetic coupling constant, α, but not the charges of the particles involved. The ratios of leptonic decay widths for the ##\phi##, ##J/\psi##, and ##\Upsilon## are approximately 2:8:3, which can be explained by considering the number of leptonic states each meson
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.

## 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

You are nearly done.
BOAS said:
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?

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

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?

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.

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).

## What are neutral vector mesons?

Neutral vector mesons are subatomic particles that have a spin of 1 and carry no electric charge. They are made up of a quark and an antiquark, and are typically denoted by the letter "J" followed by a superscript indicating their spin state.

## What is the decay width of a neutral vector meson?

The decay width of a neutral vector meson is a measure of how quickly it decays into other particles. It is defined as the rate at which the number of particles in a given state decreases over time.

## How is the decay width of a neutral vector meson calculated?

The decay width of a neutral vector meson is calculated using the principles of quantum mechanics. It involves determining the probability that the meson will decay into specific particles, and then integrating over all possible decay channels to get the total decay width.

## What factors affect the decay width of a neutral vector meson?

The decay width of a neutral vector meson is influenced by the properties of the particles it can decay into, such as their masses and spin states. It is also affected by the strong and weak nuclear forces, which determine the probability of a decay occurring.

## Why is studying the decay widths of neutral vector mesons important?

Studying the decay widths of neutral vector mesons can provide valuable insights into the fundamental properties of matter and the interactions between subatomic particles. It can also help verify the predictions of various theoretical models and inform our understanding of the universe at a fundamental level.

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