I read that the process (where the last is the antineutrino of electron): $$ \pi^{-} \Rightarrow e^{-} + \hat{\nu}$$

is disadvantage because of spins and parity argumentations, but why? In the 99% of the case the pion decay into muon and antineutrino of muon, I don't understand the reasons.
In weak interactions like that only left fermions and right antifermions interacts; this mean that the electron spin is in the opposite direction of the motion, right?

Second example (that I don't understand): in the following interaction: $$ \hat\nu_{\mu} + e^{-} \Rightarrow \mu^+ + \nu_e$$ the electron neutrino, in system of the center of mass, is products in isotropy way or in the opposite directions compare to the electron?


1 Answer 1


The pion is a pseudoscalar, with spin-parity $J^\pi=0^-$. In decay of a pion at rest the two decay products have equal and opposite momentum. Choose a coordinate system so that the electron goes to the left, and the antineutrino goes to the right:

e <--------- π -----------> ν

The weak interaction, as you say, involves particles with left-handed chirality and antiparticles with right-handed chirality. For massless particles, chirality is the same as helicity, while for massive particles the two phenomena are independent. So if the electron and neutrino were both massless, the antineutrino spin would point parallel to its momentum (right-handed), which is to the right, and the electron spin would point antiparallel to its momentum (left-handed), which is also to the right. If the electron and antineutrino spins are parallel you cannot have $J=0$; a $0 \to 1$ transition violates conservation of angular momentum and is forbidden. (Orbital angular momentum doesn't help much, for reasons too complex for a parenthesis.)

Since the electron and antineutrino are more-or-less equally sharing 140 MeV of kinetic energy, the electron has relativistic factor $\gamma = E/m \approx 100$, and the approximation that it's massless is pretty good. So the $\pi\to e\nu$ decay is approximately forbidden. In $\pi\to\mu\nu$, on the other hand, the muon is nonrelativistic and its chirality is less strongly related to its helicity.

I don't understand your second interaction (v1 of question), which violates conservation of charge.

  • $\begingroup$ Thank you so much for the answer! Okay now I understand the electron-neutrino products; I understand that for massless particle chirality is egual to helicity but, be patient, in the dacay of pion the spin of muon is in the direction of the motion? In the muon case I don't talk about helicity but I consider only chirality, correct? The second examples I write wrong, it is the exchange of a W-boson, like a neutrino that absorb a W+ or muon that emit a W- and became a muon-neutrino and W- interact with neutrino of electron and give arise to electron. $\endgroup$ Sep 7, 2016 at 17:55
  • $\begingroup$ The muon comes out polarized (helicity) right-handed, to make the angular momentum work out. Since the muon is approximately in its rest frame its chirality is a mixture of left- and right-handed; the W interacts with the left-handed part. $\endgroup$
    – rob
    Sep 8, 2016 at 3:00
  • $\begingroup$ Okay, now I totally understand! Thank you so much. $\endgroup$ Sep 8, 2016 at 10:40
  • $\begingroup$ May I ask you another thing ? Why the $\rho^0 $ cannot decay into a neutral pions pairs ? One reason is because of the charge conjugation but why that decay violates isospin? Rho and neutral pion had the third component of isospin equal to zero, the difference between them is that rho is a vector meson and pion a pseudoscalar meson. $\endgroup$ Sep 13, 2016 at 21:32

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