# Why helicity for massless particles is Lorentz invariant?

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?

• WP. – Cosmas Zachos Aug 2 '19 at 15:26
• – AccidentalFourierTransform Aug 2 '19 at 20:23
• The answer in the link you gave is good but the definition of helicity is different – amilton moreira Aug 2 '19 at 20:33

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.

• @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...) – AccidentalFourierTransform Aug 6 '19 at 21:38
• sorry computer problem – amilton moreira Aug 7 '19 at 3:48
• 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 $\sigma|\Lambda p,\sigma\rangle$? – amilton moreira Aug 7 '19 at 4:46
• $\frac{{\bf J} \cdot {\mathbf{P }}}{|\mathrm{P}|}$ is evidently a scalar under rotations...Elicity cannot change under rotations, this is also true if the particle is massive. – Valter Moretti Aug 9 '19 at 7:36
• 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.... – Valter Moretti Aug 11 '19 at 8:29