Confusion (3) from Weinberg's QFT.(photon's angular momentum or helicity)

[kof9595995]

In page 72, equation (2.5.39) gives
J_3\Psi_{k,\sigma}=\sigma\Psi_{k,\sigma} (k is the standard momentum (0,0,1,1))
and he says \sigma will be the helicity. As he explains:

Since the momentum \mathbf{k} is in the three-direction, \sigma gives the component of angular momentum in the direction of motion, or helicty

However, J_3 is the generator of rotation along the 3-axis(the z-axis), then why isn't \sigma the angular momentum component along the 3-axis in general? It is also the helicity in his case because the standard momentum happens to be along the 3-axis, but what a about a state with arbitrary momentum?

 

[meopemuk]

In page 72, equation (2.5.39) gives
J_3 \Psi_{k,\sigma}=\sigma\Psi_{k,\sigma} (k is the standard momentum (0,0,1,1))
and he says \sigma will be the helicity. As he explains:

However, J_3 is the generator of rotation along the 3-axis(the z-axis), then why isn't \sigma the angular momentum component along the 3-axis in general? It is also the helicity in his case because the standard momentum happens to be along the 3-axis, but what a about a state with arbitrary momentum?

The definition of the helicity operator is (\mathbf{J} \cdot \mathbf{P})/P. So, for general momentum value \mathbf{p} you need to prove that

(\mathbf{J} \cdot \mathbf{P})/P \Psi_{\mathbf{p}, \sigma}=\sigma\Psi_{\mathbf{p}, \sigma}

I think this can be done if you take into account how vectors \Psi_{\mathbf{p}, \sigma} transform under rotations and use definition of the Wigner angle.

Eugene.

 

[kof9595995]

Now my confusion is what if we interpret \sigma as the angular momentum component along 3-axis? Later in the same page Weinberg plays with generators and \sigma index, and he derived that \sigma is invariant under any Lorentz transform. But obviously angular momentum component along 3-axis can't be a invariant quantity, only helicity can.
To summarise, the math in page 72 looks just as valid if \sigma were angular momentum component along 3-axis, but then we'll arrive at something absurd. There has to be some loophole in the train of thought, so what did I miss?

 

[meopemuk]

Now my confusion is what if we interpret \sigma as the angular momentum component along 3-axis? Later in the same page Weinberg plays with generators and \sigma index, and he derived that \sigma is invariant under any Lorentz transform. But obviously angular momentum component along 3-axis can't be a invariant quantity, only helicity can.
To summarise, the math in page 72 looks just as valid if \sigma were angular momentum component along 3-axis, but then we'll arrive at something absurd. There has to be some loophole in the train of thought, so what did I miss?

If \mathbf{p} = \mathbf{k} \equiv (0,0,1) is standard momentum, then the action of the helicity operator coincides with the action of the operator J_3

(\mathbf{J} \cdot \mathbf{P})/P \Psi_{\mathbf{k}, \sigma}= (\mathbf{J} \cdot \mathbf{k})/k \Psi_{\mathbf{k}, \sigma}= J_3 \Psi_{\mathbf{k}, \sigma}= \sigma\Psi_{\mathbf{k}, \sigma}

For other directions of \mathbf{p} there is no such equivalence. So, operator J_3 should not play any role in arguments concerning the whole Hilbert space of the particle. This should be clear also from the fact that the direction of the 3-axis is observer-dependent, so by using J_3 you cannot derive any general (=observer-independent) result.

Eugene.

 

[kof9595995]

For other directions of \mathbf{p} there is no such equivalence. So, operator J_3 should not play any role in arguments concerning the whole Hilbert space of the particle. This should be clear also from the fact that the direction of the 3-axis is observer-dependent, so by using J_3 you cannot derive any general (=observer-independent) result.

Eugene.

Usually to avoid confusion I think in terms of active transformation,i.e. transform the physical system rather than the coordinate, so we only have one and unique 3-axis. Anyway, Weinberg did not use any helicity operator, he derived the transformation rules only using J_3 (and two other generators which are very trivial when acting on 1-particle state), that's why I'm very disturbed when he claimed \sigma is helicity instead of angular momentum component along a fixed axis(the 3-axis), which ought to be true because he showed \sigma is invariant.

 

[erler]

Weinberg's sigma is in the three direction for particles at rest.
You can then boost the particle in any direction, using the "standard boost".
The longitudinal polarization (sigma = 0) is then automatically
in the direction of motion, even though we started with the 3-direction.
This is for massive particles but the logic is the same for massless particles.
I attach some notes from me that are based on Weinberg's book, but translated
to the Bjorken metric and use a slightly different notation.

 

[kof9595995]

Thank you all, and I think I've figured out my problem, it's more of a notational issue. \sigma is partly defined by U(L(p)) via (2.5.5), granted he showed \sigma index is invariant, but if I insist \sigma is angular momentum along 3-axis than helicity as I said in my original post, then this index would be vacuous since in general J_3\Psi_{p, \sigma}\ne\sigma\Psi_{p, \sigma} because of (2.5.5), but helicity will work well.