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?
These comments may get you thinking in the right direction:
- to talk about measuring spin in different directions we have to (at least in principle) be able to go to the rest frame of the particle to take spin measurements.
- for a massless particle, hence travelling at the speed of light, we know from special relativity there is no such rest frame. So it is not meaningful to talk of measuring spin in different directions. The only property we can talk about is helicity (i.e. which way it is spinning relative to the direction of motion).
You can place all this on a mathematical footing by studying the Poincare group representations of massive and massless particles. Massive particles have little group SO(3) while massless particles have little group SO(2).
Want more: read Weinberg.