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A question about photon helicity

  1. Oct 14, 2012 #1
    Dear everyone,

    I have a simple question about the helicity of photon. The helicity operator is defined as
    [tex]\hat{\mathbf{S}} \hat{\mathbf{p}}/|\mathbf{p}|[/tex]. How to show the photon has +1/-1 helicity eigenvalue from this definition?

    Thank you ~~
  2. jcsd
  3. Oct 15, 2012 #2


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    Science Advisor


    [tex]\chi = \vec{\mathbf{S}} \hat{\mathbf{p}} = \vec{\mathbf{S}} \frac{\vec{\mathbf{p}}}{|\vec{\mathbf{p}}|}[/tex]

    The argument is rather simple. First we know that for the photon we have spin s = 1 and therefore sz = 0, ±1. But the longitudinal polarization is absent for massles vector particles, therefore only sz = ±1 remains. (*)

    Now the next step is to fix the arbitary axis ez w.r.t. which we define the spin orientation; we set

    [tex]\hat{{e}}_z = \hat{{p}}[/tex]

    Now the photon is moving into z-direction and we have

    [tex]\vec{\mathbf{S}} \vec{\mathbf{p}} = \mathbf{S}_x\mathbf{p}_x + \mathbf{S}_y\mathbf{p}_y + \mathbf{S}_z\mathbf{p}_z \to s_z p_z = \pm p[/tex]

    where on the r.h.s sz and pz are the eigenvalues w.r.t. the new reference frame with px = py = 0

    (*) But I guess this simple algebra is not really your question I and I gues you want to understand why for massless particles only these two spin orientations survive

    EDIT: There are two different ways to find the two physical helicity states of the photon, but I don't see how they are related; therefore I started a new thread https://www.physicsforums.com/showthread.php?t=644149
    Last edited: Oct 15, 2012
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