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Total Angular Momentum of Circularly Polarised Light

  1. Feb 11, 2013 #1
    I am considering the occurence of an incident circularly polarised EM wave on a ground state hydrogen atom. The result is that the final state of the atomic electron transition will be at m= ±1 depending on the orientation of polarisation (LCP or RCP).

    I understand that this is due to the conservation of angular momentum. The EM wave carries total angular momentum J of ± hbar. J also equals L+S (orbital angular momentum & spin angular monetum respectively). However the angular momentum of circularly polarised light comes solely from the spin angular momentum, which would imply that L=0 and S= ± hbar.

    My problem is tho, is that S=m_s hbar and I understood that m_s could only equal ± 1/2. So how in the instance of my light, can S be an integer value of hbar?

    Any assistance would be much appreciated. Thanks :)
  2. jcsd
  3. Feb 11, 2013 #2


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    Unlike electrons, photons have spin 1, so m_s = +-1.
  4. Feb 11, 2013 #3

    Meir Achuz

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    The m=+1 or -1 is for L_z of the orbital angular momentum of the electron.
  5. Feb 11, 2013 #4
    Thank you. That makes sense now :)

    I'm thinking that although it is the S component of the photon's J that has a value, because of spin-orbit interaction, it is only J that needs to be conserved and not the individual components S or L. Which is why it can be the electron's L that carries the conservation.
  6. Feb 12, 2013 #5
    one more thing,photons are massless then don't have a zero value for m.
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