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

  1. Jul 22, 2011 #1
    In page 72, equation (2.5.39) gives
    [itex]J_3\Psi_{k,\sigma}=\sigma\Psi_{k,\sigma}[/itex] (k is the standard momentum (0,0,1,1))
    and he says [itex]\sigma[/itex] will be the helicity. As he explains:
    However, [itex]J_3[/itex] is the generator of rotation along the 3-axis(the z-axis), then why isn't [itex]\sigma[/itex] 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. jcsd
  3. Jul 23, 2011 #2
    The definition of the helicity operator is [itex] (\mathbf{J} \cdot \mathbf{P})/P [/itex]. So, for general momentum value [itex] \mathbf{p} [/itex] you need to prove that

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

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

    Eugene.
     
    Last edited: Jul 23, 2011
  4. Jul 23, 2011 #3
    Now my confusion is what if we interpret [itex]\sigma[/itex] as the angular momentum component along 3-axis? Later in the same page Weinberg plays with generators and [itex]\sigma[/itex] index, and he derived that [itex]\sigma[/itex] 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 [itex]\sigma[/itex] 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?
     
  5. Jul 24, 2011 #4
    If [itex] \mathbf{p} = \mathbf{k} \equiv (0,0,1) [/itex] is standard momentum, then the action of the helicity operator coincides with the action of the operator [itex] J_3 [/itex]

    [tex] (\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}[/tex]

    For other directions of [itex] \mathbf{p} [/itex] there is no such equivalence. So, operator [itex] J_3 [/itex] 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 [itex] J_3 [/itex] you cannot derive any general (=observer-independent) result.

    Eugene.
     
  6. Jul 24, 2011 #5
    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 [itex] J_3 [/itex] (and two other generators which are very trivial when acting on 1-particle state), that's why I'm very disturbed when he claimed [itex]\sigma[/itex] is helicity instead of angular momentum component along a fixed axis(the 3-axis), which ought to be true because he showed [itex]\sigma[/itex] is invariant.
     
  7. Aug 12, 2011 #6
    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:

  8. Aug 16, 2011 #7
    Thank you all, and I think I've figured out my problem, it's more of a notational issue. [itex]\sigma[/itex] is partly defined by [itex]U(L(p))[/itex] via (2.5.5), granted he showed [itex]\sigma[/itex] index is invariant, but if I insist [itex]\sigma[/itex] is angular momentum along 3-axis than helicity as I said in my original post, then this index would be vacuous since in general [itex]J_3\Psi_{p, \sigma}\ne\sigma\Psi_{p, \sigma}[/itex] because of (2.5.5), but helicity will work well.
     
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