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I calculated the $\mu \rightarrow e^- + \nu_{\mu} + \bar{\nu}_{e}$ both in Fermi's theory (V-A) and Intermediate Vector Boson theory and IVB at first order seems to match with Fermi's theory which is great. But the following process in V-A theory;

$$M^{V-A} = -i\frac{4G_F}{\sqrt{2}}(\bar{u}_{\nu_{\mu}}(q)\gamma^{\alpha}_L {u}_{\mu}(p))(\bar{u}_{e^-}(p')\gamma_{\alpha}^L {v}_{\bar{\nu}_{e}}(q'))$$

creates an electron anti-neutrino, and $\gamma^{\alpha}_L$ implies that my anti-neutrino is left-handed, this seems like a problem.

Possible Explanation

  • In Fermi's theory, even if left-handed anti-neutrino appears as it does in the above process, one can interpret it as a left-handed neutrino (by postulating that neutrinos are Majorana particles.) However, I am not sure if this explanation is The explanation or I am missing something more obvious.
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Nononono.... no Majorana nonsense here!

Looks like you miswrote and misinterpreted the relevant spinors. The correct effective 4-Fermi interaction term involving 4-spinors is proportional to $$ \overline { u_{\nu_{\mu}}(q)}P_R\gamma^{\alpha} P_L {u}_{\mu}(p)~~\overline {u_{e}(p')} P_R \gamma_{\alpha} P_L {u}_{{\nu}_{e}}(q') ~.$$

I've inserted projectors everywhere, so you don't get confused by the essential right-handedness of the particles as the left projector has cleared the implicit $\gamma^0$ in the bar and flipped to a right one.

Specifically, the four spinors, from left to right, do the following for your reaction:

  1. Creates a left-handed muon-neutrino
  2. Eliminates a left-handed muon
  3. Creates a left-handed electron
  4. Creates a right-handed electron-antineutrino (instead of eliminating a left-handed e-neutrino).

It might be less confusing for you if you considered the hermitian conjugate of the second bilinear, which you did not use here, to compare it with the first.

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  • $\begingroup$ Okay now it makes almost perfect sense, however in 4. did you mean "eliminating a left-handed e-neutrino" instead of "left-handed anti-neutrino"? Why do we say the last term creates a right-handed e-anti-neutrino rather than saying it eliminates a left-handed e-neutrino? $\endgroup$
    – Monopole
    Jan 14 at 20:46
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    $\begingroup$ Yes, thanks, corrected. Let's skip the es to save space. $P_L u_\nu(q')$ both eliminates a L neutrino and creates a R antineutrino, depending on the states it acts on: a neutrino or else a vacuum/empty state, as your QFT book should specify for you. $\endgroup$ Jan 14 at 21:04
  • $\begingroup$ Okay I think I got it so would it be correct to say "it creates R anti-neutrino" because it can not eliminate "L neutrino" from vacuum? $\endgroup$
    – Monopole
    Jan 14 at 21:16
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    $\begingroup$ I guess so... This is the rule. You may picture eliminating a L neutrino from a "chiral Dirac sea", creating a R "hole" in it, which Dirac would picture as a R anti-neutrino, but this is a rabbit hole you might not choose to go down in.... $\endgroup$ Jan 14 at 21:23
  • $\begingroup$ I'm not sure I'm not giving you a glimpse of the rabbit hole, but a helpful mnemonic might be $L u(-E,-p) = R v(E,p)$... $\endgroup$ Jan 14 at 21:50

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