# Weyl spinor's spin direction

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 $$x$$ direction. This means that his spin will always point in the positive $$x$$ 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 $$x$$ it looks like $$(1/\sqrt 2,1/\sqrt 2)^T$$. If we want to measure the spin along the $$z$$ direction we have 50-50 chances to get up and down. Now, if we measure the $$x$$ component (after we measured the $$z$$ component) we have 50% chances to find the spin in the state $$(1/\sqrt 2,-1/\sqrt 2)^T$$, so pointing along the negative $$x$$ 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 - $$p$$ and $$S$$ 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?