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The rigid rotator and Angular momentum.

  1. Jun 2, 2014 #1
    If spherical harmonics are simultaneous eigenfunctions of [itex]\hat{L}[/itex] and [itex]\hat{L}_{z}[/itex], then that means for a state at which l=1, and where you have three possible values of m (1, 0 , -1) that the value of L and L[itex]_{z}[/itex] cannot really be determined simultaneously. Because the three fold degeneracy of the state implies that the rigid rotator exists in a three dimensional subspace with the eigenkets given by the three spherical harmonics determined by l=1. Is this true, or am I getting something wrong? My textbook says that they can be determined simultaneously, but I'm pretty sure this is only true if the particle exists in a state given by one of the eigen-kets of the degenerate subspace.
     
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  3. Jun 2, 2014 #2

    DrClaude

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    Staff: Mentor

    I've having a hard time understanding your argument here. But the rotation of a rigid rotor is described by [itex]\hat{L}^2[/itex], and since [itex]\hat{L}^2[/itex] and [itex]\hat{L}_z[/itex] commute, it is always possible to measure both ##l## and ##m##. That said, if the system is in a superposition of different ##m## states, with the same value of ##l##, then of course upon measurement only a single value of ##m## will be obtained and the rotor will "collapse" to that particular ##l,m## state, which does not affect the possibility of measuring ##l## independently.
     
  4. Jun 2, 2014 #3
    yeah. that makes sense. I was confusing myself. Thanks.
     
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