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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?

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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.

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  • $\begingroup$ @amiltonmoreira Can you try again please? I have no idea what you're trying to say. (First you have an operator, then a state, then a tensor product of states; none of these can be equal to each other...) $\endgroup$ – AccidentalFourierTransform Aug 6 at 21:38
  • $\begingroup$ sorry computer problem $\endgroup$ – amilton moreira Aug 7 at 3:48
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    $\begingroup$ Suppose that we have $\frac{{\bf J} \cdot {\mathbf{P }}}{\color{red}{|\mathrm{P}|}}|p,\sigma\rangle=\sigma|p,\sigma\rangle$. Now suppose that our transformation is a rotation in the $x$ axis than $\frac{{\bf J} \cdot {\mathbf{P }}}{\color{red}{|\mathrm{P}|}}|\Lambda p,\sigma\rangle=(\frac{p'_2\mathbf{J_2 }}{p_3}+\frac{p'_3 \sigma}{p_3})|\Lambda p,\sigma\rangle.$ Is $(\frac{p'_2\mathbf{J_2 }}{p_3}+\frac{p'_3 \sigma}{p_3})|\Lambda p,\sigma\rangle$ equal to $\sigma|\Lambda p,\sigma\rangle$? $\endgroup$ – amilton moreira Aug 7 at 4:46
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    $\begingroup$ $\frac{{\bf J} \cdot {\mathbf{P }}}{|\mathrm{P}|}$ is evidently a scalar under rotations...Elicity cannot change under rotations, this is also true if the particle is massive. $\endgroup$ – Valter Moretti Aug 9 at 7:36
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    $\begingroup$ Just apply the unitary representation of rotations and use the fact that $J$ and $P$ are 3-vectors under the action of it whereas $|P|$ is a scalar.... $\endgroup$ – Valter Moretti Aug 11 at 8:29

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