Consider the muon decay process: enter image description here

We assign the chirality according to the $W$ boson current: (i.e. P&S eq.(20.80))

$$J_W^{\mu+}=\frac{1}{\sqrt{2}}\bar{\nu}_{\mu L}\gamma^{\mu}\mu_L \quad J_W^{\mu-}=\frac{1}{\sqrt{2}}\bar{e}_L\gamma^{\mu}\nu_{e L} $$

And apply the tree-level diagram Feynman rules (in four-fermi effective theory):

$$ \frac{1}{2 m_W^2}\langle \nu_{\mu} e^- \bar{\nu}_e|\bar{e}_L \gamma^{\mu} \nu_{e L} \bar{\nu}_{\mu L}\gamma^{\mu}\mu_L | \mu^- \rangle $$ with contractions: (sorry that I don't know how to draw that in this website) enter image description here From the contraction it seems that:

$$\mu_L^- \rightarrow \nu_{\mu L}+ e^-_L + \bar{\nu}_{e L} $$

But this conflict with the fact that there is not left-handed anti-neutrino in SM, so what's wrong?


1 Answer 1


You are misreading the notation, a point of notational conflict discussed on this site several times. The point is that the subscripts of the text you are using, like many of them, has counterintuitive L,R subscripts for the fields, outside the bar, opposite to that of the state they destroy. (There are some honorable exceptions of texts using the under-the-bar notation, $\overline{\psi_L}=(P_L \psi)^\dagger \gamma^0=\overline{\psi} P_R$ , but the mainstream uses P&S conventions.)

The field that destroys a right-handed antineutrino is actually $\overline{ \nu_L}= \bar \nu P_R$, as the left projector flips in going through the $\gamma^0$ of the bar, to flip again going past the $\gamma^\mu$ of the current, and harmonize with the $P_L$ of the charged lepton operator; otherwise the current would vanish. Thus, your operator $$ J_W^{\mu-}=\frac{1}{\sqrt{2}}\overline {e_L}\gamma^{\mu}\nu_{e L} $$ destroys a Left-chiral neutrino and creates a Right-chiral antineutrino; and creates a Left- chiral electron while destroying a Right-chiral positron.

So, the actual product particles in your final reaction expression are a L muon neutrino, a L electron, and R electron antineutrino, denoted by the bar! I would personally use a R subscript for it, to indicate the actual chirality of the particle, and prevent the misreading you evince here, but then I'd confuse others...

  • $\begingroup$ Hello professor, thank you very much, your bar notion really helpful! I don't notice that $\gamma^{\mu}$ also have the function to flip chiral. Also, I am wondering how can we assign $\gamma^{\mu}$ to $\overline{e_L}$ or $\nu_{eL}$. If we assign $\gamma^{\mu}$ to $\overline{e_L}$, it seems that we can destroy Left-chiral position and create Left-chiral anti-neutrino. This situation should also been prohibited. So when $\nu_{eL}$ want to create something, we must assign $\gamma^{\mu}$ to it. Would my understanding right? $\endgroup$
    – Daren
    Nov 13, 2022 at 12:57
  • $\begingroup$ Sorry for my last comment, If we assign $\gamma^{\mu}$ to $\overline{e_L}$, we can destroy Right-chiral position (under bar) and create Right-chiral anti-neutrino ($Under\ bar$)! $\endgroup$
    – Daren
    Nov 13, 2022 at 13:07
  • $\begingroup$ I'm really not sure what you are saying... Your QFT course should have cleared all this up. $\gamma^\mu$ scrambles the components of a Dirac spinor according to its rules. When in doubt, write your expressions explicitly, using projectors, and slide them through. E.g., $\overline{\psi_L}= (P_L\psi)^\dagger \gamma^0= \bar\psi ~P_R$, etc... $\endgroup$ Nov 13, 2022 at 13:08
  • $\begingroup$ Thank you very much! $\endgroup$
    – Daren
    Nov 13, 2022 at 13:14

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