Assuming a vector-axial vector form of the weak interaction,...

Assuming a vector-axial vector form of the weak interaction, explain why the electron emitted in the $\mu^{-}$-decay process, $\mu^{-} \rightarrow \mathrm{e}^{-} \bar{\nu}_{e} \nu_{\mu}$, must be left-handed. What is the helicity of $\mathrm{e}^{+}$from $\mu^{+}$decay?

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Key Concepts

V-A Structure of the Weak Interaction

The weak force is characterized by a vector minus axial vector (V-A) structure, meaning it only couples to left-handed fermions and right-handed antifermions. This chiral nature of the weak interaction is why, in decay processes, only particles of a specific handedness participate. In the case of ?? decay into an electron, electron antineutrino, and muon neutrino, the interaction selects for a left-handed electron, as only left-handed components are involved in the vertex of weak charged current interactions.

Helicity and Chirality in Weak Decays

Helicity is the projection of a particle’s spin along its momentum direction and, particularly for nearly massless or high-energy particles, relates directly to chirality, a fundamental property dictating the particle's behavior under weak interactions. Since the weak force only acts on left-handed chirality states for fermions, the electron produced in ?? decay must be left-handed. Conversely, due to charge conjugation, positrons (antiparticles of electrons) inherently participate as right-handed states in ?? decay.

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