Feynman diagram for ##\mu^+\mu^-## production in ##p\bar{p}## reaction

In summary, in a proton-antiproton collision, only two quarks combine to create a virtual photon, which then produces a muon and anti-muon. The other quarks likely combine to form mesons, although it is uncertain which specific meson would be produced. It is possible that both the production of two ##\pi^0## mesons and the production of a ##\pi^+ \pi^-## meson pair are equally probable.
  • #1
Nirmal Padwal
41
2
Homework Statement
Remembering that helicity is conserved at high energies,
a) Draw a typical diagram for ##\mu^+ \mu^-##-pair production, with an invariant mass around 30 GeV, in unpolarised ##p\bar{p}## collisions.
b) Derive an expression for the angular distribution (with respect to the ##\bar{p}## direction) of ##\mu^+## in the ##\mu^+\mu^-##, centre-of-mass system.
(Hint: You will need to look up the appropriate ##d^j_{m'm}## rotation matrix elements)
Relevant Equations
1) ##d^1_{11} = \frac{1}{2}(1+\cos\theta)##
2) ##d^1_{-11} = \frac{1}{2}(1-\cos\theta)##
I was able to solve b) but I am confused for a). I understand that in the proton-antiproton collision, only two quarks (one from proton and other from anti-proton) can be combined to get a virtual photon that in turn creates muon and anti-muon. I don't understand what would happen to the other quarks? Since single quarks cannot exist independently, I think maybe they combine to form mesons. Is that correct? But which meson? If I take ##u## and ##\bar{u}## from ##p## and ##\bar{p}## respectively (please check the feynman diagram below), I am still left with ##u,d,\bar{u},\bar{d}##. Do they combine to give two ##\pi^0##s or ##\pi^+\pi^-##?
pp collision feynman diag.jpeg
 
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  • #2
Nirmal Padwal said:
Homework Statement:: Remembering that helicity is conserved at high energies,
a) Draw a typical diagram for ##\mu^+ \mu^-##-pair production, with an invariant mass around 30 GeV, in unpolarised ##p\bar{p}## collisions.
b) Derive an expression for the angular distribution (with respect to the ##\bar{p}## direction) of ##\mu^+## in the ##\mu^+\mu^-##, centre-of-mass system.
(Hint: You will need to look up the appropriate ##d^j_{m'm}## rotation matrix elements)
Relevant Equations:: 1) ##d^1_{11} = \frac{1}{2}(1+\cos\theta)##
2) ##d^1_{-11} = \frac{1}{2}(1-\cos\theta)##

I was able to solve b) but I am confused for a). I understand that in the proton-antiproton collision, only two quarks (one from proton and other from anti-proton) can be combined to get a virtual photon that in turn creates muon and anti-muon. I don't understand what would happen to the other quarks? Since single quarks cannot exist independently, I think maybe they combine to form mesons. Is that correct? But which meson? If I take ##u## and ##\bar{u}## from ##p## and ##\bar{p}## respectively (please check the feynman diagram below), I am still left with ##u,d,\bar{u},\bar{d}##. Do they combine to give two ##\pi^0##s or ##\pi^+\pi^-##?
View attachment 323288
It's not so much a matter of which happens, it's a matter of which is more likely. I haven't checked the tables but I would suspect that both versions are about equally probable

-Dan
 

FAQ: Feynman diagram for ##\mu^+\mu^-## production in ##p\bar{p}## reaction

What is a Feynman diagram?

A Feynman diagram is a graphical representation of the interactions between particles in quantum field theory. It uses lines and vertices to depict particle trajectories and interactions, providing a visual way to calculate and understand particle processes.

What particles are involved in the ##\mu^+\mu^-## production in a ##p\bar{p}## reaction?

In the ##\mu^+\mu^-## production in a ##p\bar{p}## reaction, the initial particles are a proton (##p##) and an antiproton (##\bar{p}##). The final particles are a positively charged muon (##\mu^+##) and a negatively charged muon (##\mu^-##).

What are the key features of the Feynman diagram for ##\mu^+\mu^-## production in a ##p\bar{p}## reaction?

The key features of this Feynman diagram include the initial state with a proton and antiproton, the intermediate state often involving a virtual photon (##\gamma##) or Z boson (##Z##), and the final state with a muon and antimuon pair. The diagram typically shows the proton and antiproton annihilating, producing the intermediate boson, which then decays into the muon pair.

Why is the intermediate particle often a virtual photon or Z boson in the Feynman diagram for this reaction?

The intermediate particle is often a virtual photon (##\gamma##) or Z boson (##Z##) because these particles mediate the electromagnetic and weak interactions, respectively. In high-energy proton-antiproton collisions, the annihilation can produce these bosons, which then decay into a muon-antimuon pair.

How do you calculate the probability of ##\mu^+\mu^-## production in a ##p\bar{p}## reaction using the Feynman diagram?

To calculate the probability (or cross-section) of ##\mu^+\mu^-## production, you use the Feynman diagram to write down the corresponding mathematical expression (amplitude) for the process. This involves applying Feynman rules to each part of the diagram, including propagators for intermediate particles and vertices for interactions. The amplitude is then squared and integrated over the relevant phase space to obtain the probability or cross-section.

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