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

$${\cal{L}}=\frac{g}{2 \sqrt{2}} \sum_l[{W_{\mu}^+ \cdot \bar{\psi}_{\nu_l} \gamma^{\mu}(1-\gamma_5)\psi_l + W_{\mu}^- \cdot \bar{\psi}_{l} \gamma^{\mu}(1-\gamma_5)\psi_{\nu_l} }\right{]} + \frac{g}{4 \cos(\theta_w)} \sum_l Z_{\mu}[ \bar{\psi}_{\nu_l} \gamma^{\mu}(1-\gamma_5)\psi_{\nu_l} +\bar{\psi}_{l} \gamma^{\mu}(a+b\gamma_5)\psi_{l} ]$$
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 $${\cal{M}}_{fi} \approx \bar{u}' \gamma^{\mu}(1-\gamma_5)u \epsilon_{\mu}$$
with $$u^t=\sqrt{\epsilon+m}(\omega_+,\frac{\vec{p}\cdot\vec{\sigma}}{\epsilon+m}\omega_+ )$$ and $$u'^t=\sqrt{\epsilon'+m}(\omega_-,\frac{\vec{p'}\cdot\vec{\sigma}}{\epsilon+m}\omega_- )$$ where $$\omega_{\pm}$$ are the eigenstates of the elicity...