Eigenvalues of commuting observables (angular momentum)

[gitano]

Homework Statement

Is $z|lm\rangle$ an eigenstate of $L^{2}$? If so, find the eigenvalue.

Homework Equations

$$L_{z}|lm\rangle = \hbar m|lm\rangle$$
$$L^{2}|lm\rangle = \hbar^{2} l(l+1)|lm\rangle$$

The Attempt at a Solution

So since $L_{z}$ and $L^{2}$ are commuting observables, they have are simultaneously diagonalizable and hence share the same eigenkets. Now, since $z$ and $L_{z}$ commute $z|lm\rangle$ is an eigenstate of $L_{z}$ and hence of $L^{2}$. Now I am just having some issues calculating the eigenvalue.

I have derived that $$[x_{i},L_{j}] = i\hbar \epsilon_{ijk}x_{k}$$ and that
$$[x_{i},L^{2}_{j}] = i\hbar\epsilon_{ijk}(x_{k}L_{j}+L_{j}x_{k})$$.

Now $$L^{2}z|lm\rangle = ([L^2,z]+zL^{2})|lm\rangle$$.
So it remains to calculate $$[L^2,z] = [L^{2}_{x}+L^{2}_{y},z]$$
I have proceeded using the relations I derived above, but I can't seem to get this commutator to give me some constant times $z$, which is what I need to extract an eigenvalue from the whole thing.

 

[gitano]

By the way, how do you get the equations to be inline with the text?

 

[gitano]

Sorry guys I changed the problem to correctly read: Is z|lm> an eigenstate of L^2, not of L_z.

 

[dextercioby]

Use $and [ / itex].$

 

[gitano]

I was just thinking, since $z|lm\rangle$ is an eigenstate of $L_{z}$ and has the same eigenvalue as $|lm\rangle$, does this degeneracy mean that $z|lm\rangle$ is also an eigenstate of $L^{2}$ with the same eigenvalue as $|lm\rangle$, namely $\hbar^{2}l(l+1)$?

In general, if an observable has degenerate eigenstates and commutes with another observable, does this mean that these degenerate eigenstates have eigenvalues with respect to the commuting operator which are equal as well (i.e. are the degenerate eigenstates of the original observable also degenerate eigenstates of the commuting observable)?

 

[gitano]

After more consideration, I'm beginning to think that $z|lm\rangle$ is not an eigenstate of $L^{2}$. Even though $[L_{z},L^{2}] = 0$ and share $|lm\rangle$ as eigenstates, since they are not a complete set of commuting observables, all the eigenstates of one are not necessarily eigenstates of the other. Therefore, I don't think it is necessary that $z|lm\rangle$, being a degenerate eigenstate of $L_{z}$ is an eigenstate of $L^{2}$. This seems to be corroborated by the fact that when you work out the commutator $[z,L^{2}]$, you can't get back $z$ times some constant.

Also, there might be some argument one can use based on the rotational properties, i.e. that the operator $z$ changes the value of $l$ and thus cannot be an eigenstate of $L^{2}$, but I don't know if this is true or not.

Am I correct in this reasoning?