I am a bit confused about the spin direction for a Weyl spinor. So as far as I understand, a Weyl spinor represents a massless fermion and it is an eigenstate of the helicity operator. Now say we have a right handed Weyl spinor traveling in the positive x$x$ direction. This means that his spin will always point in the positive x$x$ direction and the helicity will have an eigenvalue of 1/2. Now, Weyl spinor represents actual particles (at least theoretically, but I think they were used for neutrino, too) so these particles have 2 spin states. As initially the spin is along positive x$x$ it looks like $(1/\sqrt 2,1/\sqrt 2)^T$. If we want to measure the spin along the z$z$ direction we have 50-50 chances to get up and down. Now, if we measure the x$x$ component (after we measured the z$z$ component) we have 50% chances to find the spin in the state $(1/\sqrt 2,-1/\sqrt 2)^T$, so pointing along the negative x$x$ direction. So just by measuring its spin, we have 25% chances to turn a right handed Weyl spinor into a left handed one (as the momentum doesn't change - p$p$ and S$S$ commute). Of course this is not right so something is wrong with my understanding of spin in the context of Weyl spinors. Can someone clarify this for me? Thank you!
