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Angular momentum operators as infinitesimal generators

  1. Jun 6, 2013 #1
    Hello everyone,

    I'm going through some lecture notes and there are some things I don't understand about the whole derivation of the angular momentum multiplet.

    It's said that the skew-symmetric 3x3 matrices [itex]J_i[/itex] are the infinitesimal generators of the rotation group [itex]SO(3)[/itex]. Later, however, these are operators acting on a Hilbert space spanned by vectors [itex]|jm\rangle[/itex] which can take on all different kinds of dimensions depending on the spin [itex]j[/itex].

    This doesn't make any sense to me! What is the relation between these operators and the matrices and how does spin 'tell' what the dimensions of the operators should be?
    Can anyone help?
     
  2. jcsd
  3. Jun 6, 2013 #2

    dextercioby

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    There's the subtle difference into what a representation is: Since the group is 3 dimensional, the generators will be 3, namely 3 matrices with 3 columns and 3 rows. You wish to represent this (symmetry) group onto some complex separable Hilbert space. But the dimension of the Hilbert space can be any number from 1 to infinity. So a representation of the group takes a 3x3 matrix into a linear (unitary in case of SO(3), as per Wigner's theorem) operator on a Hilbert space. If the Hilbert space in which you wish to represent the rotation matrix is 2-dimensional (as in the case of spin of an electron, for example), the representation is a mapping from SO(3) to C2.
     
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