$z$ component of angular momentum under Lorentz transformation for massless particle

Question

This question is related to this Helicity states.

Suppose we have $k=[\omega,0,0,\omega]$. In Weinberg's book The Quantum Theory of Fields: Volume I he defines the state $|k,\sigma\rangle$ as an eigenstate of the operator $J_{3}$ that is
\begin{equation} J_{3}|k,\sigma\rangle=\sigma|k,\sigma\rangle \end{equation} where ${\mathbf{J }}=(J_1,J_2,J_3)$ are the rotation generators. Since the 3 momentum and the 3 component of the angular momentum are pointing in the same direction this is a state of helecity $\sigma$.

Then he was able to show that under a Lorentz transformation, a massless particle state should transform like this: $$U(\Lambda)|k,\sigma\rangle=e^{i\theta\sigma}| \Lambda k,\sigma\rangle.$$ Now before the Lorentz transformation we had $$J_3(|k\rangle\otimes |\sigma \rangle)=|k\rangle\otimes J_3|\sigma \rangle=\sigma(|k\rangle\otimes |\sigma \rangle)=\sigma|k,\sigma\rangle$$

Now after Lorentz transformation since parameter $\sigma$ does not change shouldn't we have $$J_3(|\Lambda k\rangle\otimes |\sigma \rangle)=|\Lambda k\rangle\otimes J_3|\sigma \rangle=\sigma(|\Lambda k\rangle\otimes |\sigma \rangle)=\sigma|\Lambda k,\sigma\rangle$$ ?

My main problem is this, if before the Lorentz transformation we had $ | k\rangle\ \otimes J_3|\sigma \rangle= |k\rangle\ \otimes \sigma|\sigma \rangle$ since under Lorentz transformation in the direct product state, only the momentum part change $$\Lambda(| k\rangle\otimes |\sigma \rangle=e^{i\theta\sigma}|\Lambda k\rangle\otimes |\sigma \rangle$$

and since $J_3$ acts only on the spin part, why $$J_3(|\Lambda k\rangle\otimes |\sigma \rangle)=|\Lambda k\rangle\otimes J_3|\sigma \rangle\neq \sigma(|\Lambda k\rangle\otimes |\sigma \rangle)$$ ?

Isn't this like saying that $J_3|\sigma \rangle=\sigma|\sigma \rangle$ and that $J_3|\sigma \rangle \neq \sigma|\sigma \rangle$?

Can anyone give me a mathematical proof why $J_3|\Lambda k,\sigma\rangle \neq \sigma|\Lambda k,\sigma\rangle$?

Answer 1

You have most definitions wrong. First of all, $$ J_3|\vec k,\sigma\rangle=\sigma|\vec k,\sigma\rangle $$ holds only when $\vec k\equiv \vec k_\star:=(0,0,\omega)$ is the standard (reference) momentum for massless particles. For other values $\vec k$ this equation is no longer true. For general $\vec k$, the state is defined (up to a normalisation factor that depends on conventions) as $$ |\vec k,\sigma\rangle:=U(\Lambda)|\vec k_\star,\sigma\rangle $$ where $\Lambda$ is, by definition, the standard Lorentz transformation that takes you from $\vec k_\star$ to $\vec k$: $$ k\equiv\Lambda k_\star $$

In general $|\vec k,\sigma\rangle$ is not an eigenstate of $J_3$, so your calculation is wrong. This is all standard material, and explained very clearly in the textbook.

Another important point that I'd like to stress is that the Poincaré Group is not a direct product (but a semi-direct product instead), and so the reps do not factorise: $$ |\vec k,\sigma\rangle\neq |\vec k\rangle\otimes|\sigma\rangle $$ This is yet another source of errors in your derivation.

With this in mind, what you want to prove is that the helicity of $|\vec k,\sigma\rangle$ is Lorentz invariant. In order to do so, you first have to define the helicity properly. The correct definition was given in a previous question of yours: $$ h=\frac{\vec P\cdot\vec J}{|\vec P|} $$

The proof that $h|\vec k,\sigma\rangle=\sigma|\vec k,\sigma\rangle$ is straightforward (cf. this PSE post). An even simpler proof is to note that $h$ is invariant under rotations, and so you can calculate it in a convenient frame or reference, the standard trick in any relativistic theory (lorentzian or otherwise). In the frame where $\vec k=\vec k_\star$, one has $h=J_3$, and so the claim follows.