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Generator of Rotations and commutation relationships

  1. Sep 16, 2012 #1
    Suppose we have

    [tex][J_i,J_j] = \sum_k \epsilon_{ijk} J_k[/tex]


    [tex][L_i,L_j] = \sum_k \epsilon_{ijk} L_k[/tex]

    1st question, I am right in thinking that [tex]J[/tex] represents Eingavalues for spin 1/2 particles... next...

    Computing the commutation relations, I find that

    [tex]\sum_k \epsilon_{ijk} (J_K + L_K - L_k - L_k)[/tex]

    collapses to simply

    [tex]\sum_k \epsilon_{ijk} S_k[/tex]

    because [tex]S_i \equiv J_i - L_i[/tex]

    2nd question: Now, I believe that taking such a difference means the total angular momentum and the orbital angular momentum just means that [tex]S_i[/tex] will become the generator of rotations for a particle around it's own axis which means we won't be moving the object in this expression... is this right?

    3rd question, is [tex]S[/tex] simply the rotational spin say possibly describing a sphere?
  2. jcsd
  3. Sep 17, 2012 #2


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    First of all, in all your commutation relations a crucial factor i is missing. It must read
    [tex][J_i,J_j]=\mathrm{i} \epsilon_{ijk} J_k.[/tex]
    In nonrelativistic quantum theory you have
    and thus your assumption is correct up to the mentioned factor i, i.e., the orbital and the spin angular momentum fulfill the angular-momentum commutation relation separately.
  4. Sep 17, 2012 #3


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    And, yes, spin is defined as angular momentum in the rest frame.
  5. Sep 17, 2012 #4


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    Yep, but this is only important in the relativistic case. Here, usually one has no unique physical decomposition of total angular momentum into an orbital and a spin part. Only the total angular momentum is a "good quantum number".

    For massless particles special care has to be taken. There not spin but helicity is the right concept. The best textbook treatment can be found in

    S. Weinberg, Quantum Theory of Fields (Vol. 1), Cam. Uni. Press.
  6. Sep 17, 2012 #5


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    I meant this in relation to the OP's second and third question. E.g. a tennis ball flying by me will have non-vanishing angular momentum depending on my position even if it does not rotate. However it will only have spin when it has angular momentum in the coordinate system in which his center of mass (or energy in relativistic context) coincides with the origin. Here the classical meaning of spin (even in sports) coincides with its physical definition.
    The classical picture of rotation around an axis becomes nonsensical for a point particle, however the mathematical formulation as angular momentum in rest frame retains its meaning.
  7. Sep 17, 2012 #6


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    Ok. I thought you were referring to the treatment of angular momentum/spin in relativistic quantum mechanics. Your examples are from non-relativistic (classical) mechanics, and there of course the splitting of total angular momentum in orbital and spin angular momentum have a simple frame-independent notion.

    In relativistic physics, for a massive particle spin is defined as its total angular momentum in its rest frame. A covariant definition of spin for a massive particle can be given in terms of the Pauli-Lubanski vector.

    For a massless particle, spin doesn't make sense at all, but only helicity (projection of the total angular momentum on the direction of the momentum of the particle).
  8. Sep 17, 2012 #7
    Ok, this is what I thought, thank you guys. Vanheese, this is ok, I am not even considering massless particles in the work I am doing, and you are right, I did miss the [tex]i[/tex] coefficient.


    Don't worry, I am not even thinking about pointlike particles, I know what kind of difficulties can arise. :) But thank you for your imput as well. So I have a new question:

    Suppose I have an equation of the form

    [tex]-B \cdot \mu = \nabla \times A \frac{g_s e}{2M} S[/tex]

    Now, since the g-factor equals 2, this equation can reduce simply to this expression

    [tex]\nabla \times A \frac{eS}{M}[/tex]

    Or better written I could give it as

    [tex]\mathcal{H} = \nabla \times A \frac{e}{M}S = \frac{eB}{M}(\mathbb{J} + \mathbb{L})[/tex]

    Now, would I be correct in assuming that I can perform the same idea of the commutation relationships on the total angular momentum, which has replaced our spin operator [tex]S[/tex]?
    Last edited: Sep 17, 2012
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