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 $J_3|\Lambda k,\sigma\rangle \neq \sigma|\Lambda k,\sigma\rangle$?

  • $\begingroup$ After a general Lorentz transformation, the 3-momentum won't necessarily point along the z-axis as before. In that case, the first definition(as you have stated it) won't hold anymore for the transformed state. $\endgroup$
    – Mani Jha
    Aug 17, 2019 at 13:55
  • $\begingroup$ what first definition? You mean that $J_3(|\Lambda k\rangle\otimes |\sigma \rangle) \neq |\Lambda k\rangle\otimes J_3|\sigma \rangle$ $\endgroup$ Aug 17, 2019 at 13:56
  • $\begingroup$ As you have stated it, the definition of the state $|k,\sigma>$only holds for massless particles with momentum along z-axis right? $\endgroup$
    – Mani Jha
    Aug 17, 2019 at 13:58
  • $\begingroup$ But why $J_3(|\Lambda k\rangle\otimes |\sigma \rangle) \neq |\Lambda k\rangle\otimes J_3|\sigma \rangle$ ? $\endgroup$ Aug 17, 2019 at 14:01
  • 1
    $\begingroup$ Remember that helicity is not equal to $J_{3}$, it is equal to the projection of $J$ along $p$. So $J_{3}|\sigma>=\sigma|\sigma>$ doesn't make sense in general. Replace $J_{3}$ by the helicity operator, and it is alright $\endgroup$
    – Mani Jha
    Aug 17, 2019 at 14:04

1 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.

  • $\begingroup$ My problem is if you start with a state $ | k\rangle\ \otimes J_3|\sigma \rangle= |k\rangle\ \otimes \sigma|\sigma \rangle$ and if the sigma part does not change under Lorentz transformation why $J_3(|\Lambda k\rangle\otimes |\sigma \rangle) \neq |\Lambda k\rangle\otimes J_3|\sigma \rangle$. My point is that $|\sigma \rangle$ part does not change under Lorentz transformation $\endgroup$ Aug 19, 2019 at 17:10
  • $\begingroup$ @amiltonmoreira please see the updated version. $\endgroup$ Aug 19, 2019 at 17:14
  • $\begingroup$ Can you give me any reference that explain me why Poincaré Group is not a direct product (but a semi-direct product instead), and so the reps do not factorize? $\endgroup$ Aug 19, 2019 at 17:23
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    $\begingroup$ @amiltonmoreira For the Poincare group to be a direct product of the translation group and Lorentz group, the representation of its elements as translation and Lorentz transform $(T,\Lambda)$ would have to satisfy $(T_1,\Lambda_1)\cdot(T_2,\Lambda_2) = (T_1T_2,\Lambda_1\Lambda_2)$. However, it does not. $\endgroup$ Aug 19, 2019 at 17:30
  • $\begingroup$ @amiltonmoreira see also e.g. ncatlab.org/nlab/show/Poincar%C3%A9+group, together with the references in math.stackexchange.com/q/41206/289977, if you want to get mathematical. $\endgroup$ Aug 19, 2019 at 17:32

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