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Spin - magnitude & direction

  1. Jul 10, 2013 #1
    The below might need lots of corrections/modifications, since spin is a hard concept to grasp and I don't understand it much.

    We know that:

    The angular momentum (spin) of a photon does not have an analogy in classical mechanics.

    however lets use a loose/inaccurate analogy of something that is rotating very fast along the axis in the direction of motion....and that generates angular momentum (?)

    1. Is the spin generated from the EM waves (in a sense "rotating" very fast) across the direction of travel of the photon?

    2. The angular momentum remains same for same type of elementary particle. Why? Can we not have faster or slower rotation? (if we can have different frequencies for a same type of elementary particle, why not "rotation"?

    3. Why is the vector different from classical mechanics?

    4. If the angular momentum is quanitzed, can we assume the "fields/forces" producing it must have some sort of quantization as well?
     
    Last edited: Jul 10, 2013
  2. jcsd
  3. Jul 10, 2013 #2
    Spin is angular momentum a particle just "has." Just like a particle "has" a certain mass or a certain charge, it has some inherent angular momentum that can be measured and is quantized.

    Photons aren't the only particles with spin. Electrons, quarks, mesons, etc also have spin. The spin of particles can change, but it can only take on certain values depending on the system. Spin isn't litteral rotation in the classical sense. We just call it angular momentum because when you do experiments with particles that have no classical angular momentum, they still produce results you would expect to see as if it were "spinning" about an axis.

    The vector is different because it doesn't point in any specific direction.

    Yeah, fields are quantized. For example, the electromagnetic filed quanta is something familiar to you: the photon.
     
  4. Jul 10, 2013 #3
    I would say spin is not angular momentum.

    The photon has l = 1, s = 0, this can be pictured from the resultant vector of the E x B field.

    Experimentally you can measure spin, see stern gerlach experiment.
     
  5. Jul 10, 2013 #4

    Bill_K

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    Spin IS angular momentum.

    The photon has s = 1, and any value ℓ ≥ 0. E x B is the momentum density, whereas the angular momentum density is x x (E x B).

    Good luck doing the Stern-Gerlach experiment on a photon. (It only works for charged particles.)
     
  6. Jul 10, 2013 #5

    kith

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    The spin of the photon corresponds to the polarization of the EM wave. Since the photon has spin 1, one would expect 3 different possibilities to align the spin (+1,0,-1) which would correspond to 3 different polarizations (x,y,z). However, the polarization in the direction of propagation is ruled out due to the transversal nature of EM waves. So we are left with only two polarizations (x,y) which is the reason why photons behave like spin-1/2 particles (possible alignments +1/2, -1/2) in experiments like Bell tests.
     
  7. Jul 10, 2013 #6

    morrobay

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    In the Bell tests the inequality, n[y+z-] + n[x-y+] ≥ n [x-z-] with spin 1/2 photons is derived
    from perfect anti correlations when the detector settings are parallel. Where P(α|λ| is
    considered to be deterministic based on conservation laws - a total spin of zero.
    Since the inequality is compared to experiments where detector settings are not parallel.
    Can P(α|λ| be considered to be non deterministic with respect to magnetic field in detector when
    the detector settings are not parallel, and how this may affect the spin of particle at time of measurement ? Also what happens to the conservation law that is in effect when detector
    settings are parallel but then do not seem to be in effect when inequality is disproven ?
     
    Last edited: Jul 11, 2013
  8. Jul 11, 2013 #7
    Bill, I think you are confusing quantum numbers.

    When I say spin I do not mean total angular momentum j.

    Therefore the stern gerlach is for testing spin, presuming I do not mean fermions without em charge.
     
  9. Jul 11, 2013 #8
    From pdg;

    I(J PC) = 0,1 ( 1 - - )
     
  10. Jul 11, 2013 #9

    Bill_K

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    Particle physicists often use J to denote a particle's spin, as you see in the PDG tables. But in ordinary parlance, J = L + S, where S is the spin and L the orbital angular momentum. The photon has spin S = 1, and may have any value of orbital angular momentum. Consequently given the value of L there are only three possibilities for J: either J = L or J = L ± 1. The various cases are known as multipole radiation, and are important whenever a photon is emitted, such as X-rays from an atom or gamma rays from a nucleus.
     
  11. Jul 11, 2013 #10
    Thanks wotanub, bill k and others. Great answers.

    What is "spinning"? The excitation of the em field? What is the speed of the (hypothetical) spin? ...say relative to c.
     
    Last edited: Jul 11, 2013
  12. Jul 11, 2013 #11
    That's the thing... Nothing is actually spinning, but it still has this angular momentum. Ferimions spin in integer multiples of hbar/2 and Bosons have integer multiples of hbar.
     
  13. Jul 11, 2013 #12

    morrobay

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    Let me sort this out for above : The question about the conservation law and spin applies to entangled photons.The question on detector magnetic field affecting spin applies to electrons.
    Maybe this should be a new topic, but Im looking for why Bells' inequalities are violated from the
    physics of how spin changes.
     
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