Is Every Eigenstate of L^2 an Eigenstate of Lz?

[Henk]

I was wondering: is every eigenstate of L^2 also an eigenstate of Lz?

I know that commuting operators have the same eigenfunctions but if [A,B] = 0 and a is a degenerate eigenfunction of A the the corresponding eigenfunctions of A are not always eigenfunctions of B.

 

[Galileo]

Not necessarily. What [A,B]=0 means (for hermitian operators) is that there EXISTS a basis of eigenvectors common to A and B.
If A|a>=a|a> and a is a nondegenerate eigenvalue of A, then |a> is also an eigenvector of B (this is easy to prove). If a is degenerate, then you can find an orthonormal basis in the eigenspace of a consisting of eigenvectors common to A and B.
So not every eigenstate of L^2 is an eigenstate of Lz.

 

[dextercioby]

You can make it a nice discussion in the following way.

\hat{J}^{2} ,\hat{J}_{3}

form a complete system of commuting observables.It's not difficult to show that.
Therefore,a spectral pair \left(\hbar^{2} j(j+1), \hbar m_{j}\right) determine,up to a phase factor,a vector from the irreducible space of the irreducible finite dimensional linear representation of the angular momentum algebra* (which is isomorphic to su(2) which is isomorphic to so(3)).This is an eigenvector to both operators.

*One can prove that an eigensubspace of \hat{J}^{2} associated to the nondegenerate spectral value \hbar^{2} j(j+1) is invariant to the action of all the linops of the representation of the ang.mom.algebra and does not admit any nontrivial invariant subspaces,therefore it is a subspace of an irred.representation of the ang.mom.algebra.

So all eigenstates of \hat{J}^{2} are eigenstates of \hat{J}_{3} in the simple case of uniparticle systems.

Daniel.

P.S.For the particular case of orbital ang.mom.ops,the things are a bit easier,as one has the particular realization \mathcal{H}=L^{2}\left(\mathbb{R}^{3}\right) and the basis from spherical harmonics.

 

[CarlB]

Since L_z does not commute with any other L_u for \hat{u} any other direction not parallel or antiparallel to \hat{z}, it should be clear that it's pretty easy to arrange for an eigenstate of L_u and L^2 to not be an eigenstate of L_z.

In addition, the reverse is also possible. That is, there are eigenstates of L_z that are not eigenstates of L^2. To make one, just take two eigenstates of both L^2 and L_z that happen to have the same L_z eigenvalue but different L^2 eigenvalues, and mix em up.

Carl

 

[dextercioby]

I have no idea why i didn't think of linear combinations of eigenstates.

Daniel.