The rigid rotator and Angular momentum.

[siddharth5129]

If spherical harmonics are simultaneous eigenfunctions of \hat{L} and \hat{L}_{z}, 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_{z} 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.

 

[DrClaude]

If spherical harmonics are simultaneous eigenfunctions of \hat{L} and \hat{L}_{z}, 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_{z} 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.

I've having a hard time understanding your argument here. But the rotation of a rigid rotor is described by \hat{L}^2, and since \hat{L}^2 and \hat{L}_z 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.

 

[siddharth5129]

yeah. that makes sense. I was confusing myself. Thanks.