Why helicity for massless particles is Lorentz invariant?

Question

By definition helicity is projection of spin onto the 3 momentum.

$$h={\bf J} \cdot {\mathbf{P }} $$ where ${\mathbf{P }}=(P_1,P_2,P_3)$ is the momentum operator and ${\mathbf{J }}=(J_1,J_2,J_3)$ the angular operator.

Now under a Lorentz transformation massless particles transform like this: $$U(\Lambda)|p,\sigma\rangle=e^{i\theta\sigma}| \Lambda p,\sigma\rangle.$$

As we can see the momentum is changing but the spin not.

Suppose that state $|p,\sigma\rangle$ is a state of helicity $\sigma$ such that we have
$$h|p,\sigma\rangle=J_3P_3|p,\sigma\rangle=\sigma p_3|p,\sigma\rangle $$

But for the state $U(\Lambda)|p,\sigma\rangle=e^{i\theta\sigma}| \Lambda p,\sigma\rangle$, we would have
$$h|\Lambda p,\sigma\rangle=\sigma p'_3e^{i\theta\sigma}| \Lambda p,\sigma\rangle| $$ So for conservation of helicity we would require $p_3=p'_3$ which is not always the case.

So why do people say that helicity is Lorentz invariant?

Answer 1

Your formula for the helicity operator is wrong; this should already be clear at the level of dimensional analysis. The correct formula is (cf. Refs 1&2) $$ h=\frac{{\bf J} \cdot {\mathbf{P }}}{\color{red}{|\mathrm{P}|}} $$ where $|\mathrm{P}|$ denotes the norm of ${\mathbf{P }}$. Acting on your state with $h$ yields no factors of $p_3$, and so the "paradox" is resolved.

References.

1. Schwartz - Quantum Field Theory and the Standard Model §11.1.

2. Ticciati - Quantum Field Theory for Mathematicians §7.8.