Pion decay via the weak interaction

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Discussion Overview

The discussion centers on pion decay via the weak interaction, specifically the decay process of a negatively charged pion into a muon and an anti-muon neutrino. Participants explore the underlying mechanisms, including Feynman diagrams and the role of quarks and W bosons in the decay process. The conversation also touches on helicity and parity violation in weak interactions compared to electromagnetic interactions.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Ray expresses confusion about the pion decay process and the role of quarks in producing the decay products via W boson exchange.
  • Some participants clarify that in the Feynman diagram, the u-bar quark is represented as a u-quark moving backwards in time, indicating a reaction that produces a W boson rather than annihilation.
  • Ray questions the implications of helicity and parity violation in weak interactions, particularly how negative helicity states relate to parity transformations in electromagnetic interactions.
  • Another participant notes that the parity of the photon is negative, suggesting that overall parity is conserved in electromagnetic interactions despite changes in helicity states.

Areas of Agreement / Disagreement

Participants generally agree on the mechanics of pion decay and the role of W bosons, but there is some uncertainty regarding the implications of helicity and parity violation, particularly in comparing weak and electromagnetic interactions. The discussion remains unresolved on these points.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about particle interactions and the definitions of helicity and parity, which are not fully explored or clarified.

rayveldkamp
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Hi, i have a particle physics exam in 2 days and am confused on pion decay via the weak interaction, namely:

pi[-] ->muon + anti-muno neutrino

Thankyou
Ray Veldkamp
 
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What's the question? :confused:
 
indeed Tom, you are right, i don't get the point either...
marlon
 
Sorry guys i should have been more specific. In lectures we were shown to draw Feynnman diagrams to represent these weak interaction questions. My question is, how do the two quarks in the pion produce the muon and anti-muon-neutrino via virtual W+/W- exchange?
In lectures we were told that emission or absorption of the W bosons will only change a particle to its doublet partner, so I am not seeing how there can be no quarks after the interaction, unless some kind of annhilation takes place.

Thanks
Ray
 
In the Feynman diagram, note that the u-bar quark is drawn as a u-quark going backwards in time. So the incoming d-quark and the outgoing u-quark do in fact react to produce a W boson, which decays into the products you mentioned. I wouldn't call this an "annihilation" of the quarks, since that term is normally reserved for the mutual cancellation of a particle with its own antiparticle to produce photons. In this instance, the quarks are destroyed to produce not a photon, but a W.
 
Thanks that makes sense.
I have one more question, concerning helicity and parity violation of the weak interaction. I understand that fermions participating in weak interactions are emitted and absorbed predominantly in negative helicity states, and we have seen that parity transformations change a negative helicity state to a positive helicity.
What i don't understand, is that if we consider say an electromagnetic interaction, if a particle is in a negative helicity state, wouldn't a parity transformation change it to a positive, and hence the electromagnetic interaction would violate parity?

Sorry if that's vague
Thanks
Ray
 
rayveldkamp said:
Thanks that makes sense.
What i don't understand, is that if we consider say an electromagnetic interaction, if a particle is in a negative helicity state, wouldn't a parity transformation change it to a positive, and hence the electromagnetic interaction would violate parity?

Don't forget that the parity of the photon is negative. So if you emit a γ in an EM interaction, and there is a parity change in the material system, the overall parity is conserved. Remember that parity is multiplicative, so πparticleπphoton=(-1)(-1)=1.
 

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