You are correct that for a *massive* spinor, helicity is not Lorentz invariant. For a *massless* spinor, helicity is Lorentz invariant and coincides with chirality. Chirality is always Lorentz invariant.

Helicity defined
$$
\hat h = \vec\Sigma \cdot \hat p,
$$
commutes with the Hamiltonian,
$$
[\hat h, H] = 0,
$$
but is clearly not Lorentz invariant, because it contains a dot product of a three-momentum.

Chirality defined
$$
\gamma_5 = i\gamma_0 \ldots \gamma_3,
$$
is Lorentz invariant, but does not commute with the Hamiltonian,
$$
[\gamma_5, H] \propto m
$$
because a mass term mixes chirality, $m\bar\psi_L\psi_R$. If $m=0$, you can show from the massless Dirac equation that $\gamma_5 = \hat h$ when acting on a spinor.

Your second answer is closest to the truth:

*The weak interaction couples only with left chiral spinors and is not frame/observer dependent.*

A left chiral spinor can be written
$$
\psi_L = \frac12 (1+\gamma_5) \psi.
$$
If $m=0$, the left and right chiral parts of a spinor are independent. They obey separate Dirac equations.

If $m\neq0$, the mass states $\psi$,
$$
m(\bar\psi_R \psi_L + \bar\psi_L \psi_R) = m\bar\psi\psi\\
\psi = \psi_L + \psi_R
$$
are not equal to the interaction states, $\psi_L$ and $\psi_R$. There is a single Dirac equation for $\psi$ that is not separable into two equations of motion (one for $\psi_R$ and one for $\psi_L$).

If an electron, say, is propagating freely, it is a mass eigenstate, with both left and right chiral parts propagating.

This post imported from StackExchange Physics at 2014-04-13 14:39 (UCT), posted by SE-user innisfree