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Next: Summary Up: The non-equivalence of Weyl Previous: Proof that L L


Weyl neutrinos with Majorana masses

It could be that Lagrangian (4) for a Weyl neutrino contains a small Majorana mass term. In this section I first review the formal reason why this leads to a lepton-number violating theory and then analyze how this is possible for a particle that is never its own antiparticle.
Lepton conservation is induced, according Noether's theorem, by the invariance of (4) under the following continuous transformation group. The charged lepton field e and neutrino field $\nu$ are simultaneously transformed via[11]:
  $\displaystyle \nu^{\prime}= e^{i \alpha} \nu$    
  $\displaystyle \bar{\nu}^{\prime} =\bar{\nu} e^{- i \alpha}$    
(11) $\displaystyle e^{\prime} = e^{i \alpha} e$    

Considering $\alpha$ infinitesimal for the infinitesimal field transformation $\delta \Psi$, Noether's theorem yields lepton conservation. The standard model Lagrangian with the addition of a non standard-model Majorana mass term $m_{Maj}$:
(12) \begin{displaymath}
L_{\rm Weyl}^{\rm SM} =
\bar{\nu}_L \gamma_{\mu} {\partial \...
...right]
+ \left[ m_{Maj} {\bar{\nu}_L} (\nu_L)^c + H.C. \right]
\end{displaymath}

The treatment of section 3 continues to hold. This means:
1. Lagrangian eq.(12) is phenomenologically equivalent to the Lagrangian
  $\displaystyle L_{\rm Maj}^{\rm Weyl-equivalent}=
\bar{\nu}_M
\gamma_{\mu} {\par...
... i g/\sqrt{2} \left[ W_{\mu}^- \bar{e}_L \gamma_{\mu} \nu_M(h=-1)
+ H.C.\right]$    
(13) $\displaystyle + \left[ m_{Maj} \nu_M {\bar{\nu}_M} + H.C. \right]$    

2. assuming the validity of the standard-model gauge sector the neutrino is definitely not a Majorana field.
m$_{Maj}$ violates the invariance of eq.(12) under the transformation group eq.(11) because the Majorana mass term acquires a phase of e$^{2 i \alpha}$ under transformation (11).
What is the mechnanism with which a Weyl field, which is never its own charge conjugate, violates lepton number? Consider the state of a Weyl field with a Majorana mass in an inertial frame at which it is a rest:
(14) \begin{displaymath}
\gamma_4 i {\partial \over {\partial t}} \nu_L(\rm {rest})
= m_{Maj} (\nu_L)^c(\rm {rest})
\end{displaymath}

A solution to this equation in the Weyl representation is: $\nu_L(\rm {rest})$ = $\left( \begin{array}{c} 0 \\ \phi_0+i \sigma_2
\phi_0^*\end{array} \right)$ with $\phi_0= a e^{imt}$. This can be rewritten as:
(15) \begin{displaymath}
\nu_L(rest) = {1\over \sqrt{2}}
(\nu_D(\rm {rest})+\nu_D(\rm {rest})^c)_L = \nu_ L + (\nu^c)_L
\end{displaymath}

where $\nu_D(\rm {rest})$ = $\left( \begin{array}{c} \phi_0 \\ \phi_0
\end{array} \right)$ describes a Dirac particle at rest. This result means: in the rest frame the Weyl spinor consists of the components ``left-handed neutrino $\nu_L$'' (helicity = $-$1) and ``left-handed antineutrino $(\nu^c)_L$'' (helicity=+1) which are not their respective charge conjugates. A Lorentz boost along the z-axis can be shown to transform $\nu_L(\rm {rest})$ (with helicity=0) into a state which is predominantly $\nu_L$ (helicity=$-$1) or $(\nu^c)_L$ (helicity=+1). I.e. depending on the inertial frame, a massive Weyl particle is predominantly particle or antiparticle. In no frame it is its own antiparticle, however.
With a Majorana mass term a Weyl field can thus violate lepton conservation, without being its own antiparticle.


next up previous
Next: Summary Up: The non-equivalence of Weyl Previous: Proof that L L
Rainer Plaga 2001-08-03