Why is the helicity of a neutrino unchanged by the weak interaction?

[Final]

Hi...

Consider a neutrino with a Dirac mass m_\nu and the weak interaction

{\cal{L}}=\frac{g}{2 \sqrt{2}} \sum_l[{W_{\mu}^+ \cdot \bar{\psi}_{\nu_l} \gamma^{\mu}(1-\gamma_5)\psi_l + W_{\mu}^- \cdot \bar{\psi}_{l} \gamma^{\mu}(1-\gamma_5)\psi_{\nu_l} }\right{]} + \frac{g}{4 \cos(\theta_w)} <br /> \sum_l Z_{\mu}[ \bar{\psi}_{\nu_l} \gamma^{\mu}(1-\gamma_5)\psi_{\nu_l} +\bar{\psi}_{l} \gamma^{\mu}(a+b\gamma_5)\psi_{l} ]
Why this interaction doesn't change the helicity of the neutrino? It is true?

 

[Vanadium 50]

Do you have any elements in that interaction that have a left-handed neutrino on one side of the operator and a right-handed one on the other?

 

[Final]

For massless particle ok, because the elicity is the chirality projector \frac{1 \pm \gamma^5}{2}... But for a massive neutrino? It's the same?

 

[Vanadium 50]

I don't see any 1+\gamma^5 there, do you? Which leads me back to my original point: do you see anything in the Lagrangian which has a left-handed neutrino on one side of the operator and a right-handed one on the other?

 

[Final]

Yes... For example Z_{\mu} \bar{\psi}_{\nu_l} \gamma^{\mu}(1-\gamma_5)\psi_{\nu_l}. You may take \nu \rightarrow \nu +Z with the first neutrino left-handed and the second right-handed.
The amplitude is {\cal{M}}_{fi} \approx \bar{u}&#039; \gamma^{\mu}(1-\gamma_5)u \epsilon_{\mu}
with u^t=\sqrt{\epsilon+m}(\omega_+,\frac{\vec{p}\cdot\vec{\sigma}}{\epsilon+m}\omega_+ ) and u&#039;^t=\sqrt{\epsilon&#039;+m}(\omega_-,\frac{\vec{p&#039;}\cdot\vec{\sigma}}{\epsilon+m}\omega_- ) where \omega_{\pm} are the eigenstates of the elicity...