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Why the rank of an irreducible tensor must be an integer?

  1. Nov 8, 2009 #1
    why not half-integer?

    according to the definition, such as [J_z,T^k_q]=q T^k_q

    it is quite possible that k can be a half-integer.
  2. jcsd
  3. Nov 9, 2009 #2


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    Well, in that context, a tensor has usually two indices, i.e. it transforms as the product of two equal irreducible representations of the rotation group these two representations may be either both integer or both half integer. In any case, the Clebsch Gordon decomposition will yield only integer representations.
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