L6.3 Weak Interactions: Pion Decay | Summary and Q&A

TL;DR

This content explains the decay of a pion and how helicity states are related, highlighting the significance of left-handed chiral states in weak interactions.

Key Insights

• 🦂 Pion decay involves the production of different leptons and antineutrinos.
• 🦂 The helicity states of the leptons in pion decay are determined by the spin and masslessness of the neutrino and the antineutrino.
• 🗨️ The importance of left-handed chiral states in weak interactions is emphasized.
• ☠️ The decay rates and partial decay widths can be calculated using Fermi's golden rule.
• 🦂 Experimental information on pion decay can be obtained using the mass values of the electron, muon, and pion.

Transcript

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Questions & Answers

Q: What are the leading decay modes of a negatively charged pion?

The leading decay modes of a negatively charged pion are the production of an anti-electron neutrino and an electron or a muon and an antimuon neutrino.

Q: Why do the opposite-direction leptons in pion decay have the same helicity state?

The pion having a spin of 0 requires opposite-direction leptons to have the same helicity state. This constraint arises because the pion is a boson.

Q: Why is the antineutrino always right-handed in pion decay?

The antineutrino is always right-handed because both the neutrino and the antineutrino are massless particles. In the case of massless particles, the chiral state and the helicity state are essentially the same.

Q: What would happen if the charged lepton in pion decay were massless?

If the charged lepton were massless, the right-handed helicity state of the charged lepton would not exist. Consequently, the decay would not be allowed.

Summary & Key Takeaways

• The content discusses the decay of a pion and its leading decay modes, which involve the production of an anti-electron neutrino and an electron or a muon and an antimuon neutrino.

• The helicity states of the outgoing leptons are determined by the fact that the pion has a spin of 0, requiring opposite-direction leptons to have the same helicity state.

• While the antineutrino is always right-handed due to being massless, the decay would not be allowed if the charged lepton were also massless.