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I Is J=L+S fundamental?

  1. May 18, 2017 #1
    My textbook "The physics of quantum mechanics" by James Binney and David Skinner, describes the pseudo-vector operators [itex]\vec{J}[/itex], [itex]\vec{L}[/itex] & [itex]\vec{S}[/itex] as generators of various transformations of the system. [itex]\vec{J}[/itex] is the generator of rotations of the system as a whole, [itex]\vec{L}[/itex] is the generator of displacement of the system around circles without rotations, and [itex]\vec{S}[/itex] is the generator for changes of orientation that are not accompanied by any motion of the system as a whole.

    In the book, [itex]\vec{J}[/itex] is defined as the generator of rotations, while [itex]\vec{L}[/itex] is defined by [itex]\vec{L} = \vec{x} \times \vec{p}[/itex], and then [itex]\vec{S}[/itex] is defined by [itex]\vec{S} = \vec{J} - \vec{L}[/itex]. My issue with this is that [itex]\vec{L}[/itex] seems to have been defined by classical analogy, which leads me to question how fundamental the division of angular momentum into orbital and spin is. The book later describes spin as intrinsic to the particle, which implies that it is, in fact, a fundamental division. Assuming it is fundamental, can you predict that mathematically, or is it an observational fact?
     
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  3. May 18, 2017 #2

    dextercioby

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    This split makes sense for massive particles whose states are covariant under the space-time isometry group, be it the Galilei group or the restricted Poincare group, so in a sense it is fundamental.
     
  4. May 19, 2017 #3
    So is it divided up differently for less massive particles?
     
  5. May 19, 2017 #4

    DrDu

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    Spin is the angular momentum of a particle in its rest frame. This presupposes that the rest frame exists. For massless particles, like the photon, this is not the case.
    For massive particles, spin, mass and charge are the defining properties of the particles identity. For massless particles, spin gets replaced by helicity.
     
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