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

1. Jul 22, 2011

### 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:
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?

2. Jul 23, 2011

### meopemuk

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.

Last edited: Jul 23, 2011
3. Jul 23, 2011

### 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?

4. Jul 24, 2011

### meopemuk

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.

5. Jul 24, 2011

### kof9595995

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.

6. Aug 12, 2011

### 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.

#### Attached Files:

• ###### Polarization vectors.pdf
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7. Aug 16, 2011

### 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.