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Unitarity angular momentum operators

  1. May 26, 2013 #1

    I'm confused by a sentence in a set of lecture notes I have on quantum mechanics. In it, it is assumed there is some representation [itex]\pi[/itex] of [itex]SO(3)[/itex] on a Hilbert space. This representation is assumed to be irreducible and unitary.

    It is then said that the operators [itex]J_i[/itex], which are said to be the infinitesimal generators of the rotation group satisfying [itex][J_i,J_j]=i \epsilon_{ijk}[/itex], are Hermitian as a consequence of the unitarity of this representation.

    This confuses me. Shouldn't they say that the operators [itex]\pi (J_i)[/itex] are Hermitian? Are they writing [itex]J_i[/itex] for both the infinitesimal generators of the group and the operators they are mapped to?
  2. jcsd
  3. May 26, 2013 #2


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    Yes, they should do so. But it is quite common to write J for some representation.
  4. May 26, 2013 #3
    Thanks for your quick answer.

    Does that, then, imply that the Casimir operator is actually [itex]\sum_i \pi(J_i)^2[/itex]?
    Do the commutation relations satisfied by the Lie group generators carry over to the operators under the representation map? I can see that

    [itex][\pi(J_i),\pi(J_j)] = \pi(J_i)\pi(J_j) - \pi(J_j)\pi(J_i) = \pi(J_i J_j) -\pi(J_j J_i)[/itex],

    but I don't see why I could conclude that the last line equals [itex]\pi([J_i,J_j])[/itex]

    EDIT: The only way I could see that happen if the representation is linear, but I haven't seen that assumed in its definition. Should that be included?
    Last edited: May 26, 2013
  5. May 26, 2013 #4


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    The representation is linear, if the representation space is linear (and a separable Hilbert space is).
  6. May 26, 2013 #5
    How so? I mean : how do you know the representation is linear whenever the representation space is?
  7. May 26, 2013 #6


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    In QM it's the result of the theorem of Wigner. The representation operators can only be linear or antilinear (i.e. conjugate linear).
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