Spin group for the Weyl equation

I never considered this until now. Dirac particles have a spin representation of ##(1/2,0) \oplus (0,1/2)##. This represents both parts of the 4-spinor. But what is the representation of Weyl particles? Is it still ##(1/2,0) \oplus (0,1/2)## or is it just (1/2,0), since we have a definite helicity?

Thanks!

-Dan

Addendum: I'm going to add to this. The question about helicity may be a bit of a red herring: the photon is ##(1,0) \oplus (0,1)##, which covers both A and B subAlgebras. But Proca particles are (1/2, 1/2). I'm not sure what to do about the m = 0 case for Dirac particles.

topsquark said:

TL;DR Summary: What is the spin group classification for Weyl particles?

I never considered this until now. Dirac particles have a spin representation of ##(1/2,0) \oplus (0,1/2)##. This represents both parts of the 4-spinor. But what is the representation of Weyl particles? Is it still ##(1/2,0) \oplus (0,1/2)## or is it just (1/2,0), since we have a definite helicity?

Thanks!

-Dan

Addendum: I'm going to add to this. The question about helicity may be a bit of a red herring: the photon is ##(1,0) \oplus (0,1)##, which covers both A and B subAlgebras. But Proca particles are (1/2, 1/2). I'm not sure what to do about the m = 0 case for Dirac particles.

After pondering some more I think the only conclusion that I can draw is that the Weyl representation of the Poincare Algebra must be ##(1/2,0) \oplus (0.1/2)##, just like the Dirac representation.

Mathematically, the groups A and B are independent, so if the spin representations aren't the same we need to add equivalent (but switched) groups to make it symmetric. The reason the Proca and Maxwell representations are different is due to the fact that the Maxwell representation admits no spin 0 (well, helicity 0) particles, so we need to change it. The spins of the Weyl particles are not actually restricted, they are just fixed to particles/antiparticles. But the m = 0 condition doesn't actually change the wave equation, so it shouldn't change the Lagrangian, so it shouldn't change the representation, either.

I'm not completely satisfied with this answer. My logic chain only works in one direction: I don't know how to derive this result from first principles, but it may simply be that the extra condition on Weyl particles does not affect the representation we are using, which does make some sense to me.

-Dan