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Symmetries and Spin orbit interaction

  1. Oct 8, 2012 #1
    I know that the generators of the Poncaire group that are associated with *orbital* angular momentum belong to an infinite dimensional representation, i.e.

    L = \frac{\partial}{\partial \theta}

    Also the spin generators are associated with some finite dimensional representation of a lie algebra (such as the fundamental rep of SU(2) for spin 1/2 particles). Both of the groups commute with one another and should have their own symmetry charges.

    Now my question is, in spin orbit coupling why is it L+S which is the conserved quantity and not L and S separately?
  2. jcsd
  3. Oct 8, 2012 #2


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    Orbital angular momentum is unrelated to either the Poincare Group or the Lorentz Group. Angular momentum is the conserved quantity resulting from (3-dimensional) rotational invariance. Orbital angular momentum is L = x x p, the angular momentum associated with a function of position ψ(x). When you have a function with internal spin, both L and S contribute to the angular momentum as a direct product, J = LS. This is all a standard part of a quantum course. :frown:

    All of the commonly used representations of the Lorentz Group are finite dimensional.
  4. Oct 8, 2012 #3
    The poncaire group contains, as a subgroup, the group of 3-rotations. Thus any poncaire invariant system is also rotationally invariant.

    Right, the bound states belong to representations of the full symmetry group and are labeled by the groups corresponding Casimir operators. Oh!! yeah that's the answer, the casimir operator for the full symmetry group is not the same as the sum of Casimir operators for the composite groups. Thanks.

    All non-unitary representations are finite-dimensional. If one considers the space of functions (field space) then one has to consider the infinite dimensional representation of the lie algebra. This group will act to transform the coordinates of the field while a finite group will rotate elements of the field.
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