Commuting operators

1. Jun 27, 2005

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.

2. Jun 27, 2005

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.

3. Jun 27, 2005

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.

Last edited: Jun 27, 2005
4. Jun 28, 2005

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

5. Jun 28, 2005

dextercioby

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

Daniel.