I Validity of identical eigenfunction of commutating operators

[hokhani]

TL;DR Summary

An example of a function which is eigenstate of $\hat {\mathbf L_z}$ but not $\hat {\mathbf L^2}$

I got confused about an elementary statement in quantum mechanics: As far as I know, the two commuting operators like $\hat {\mathbf L_z}$ and $\hat {\mathbf L^2}$ always have identical eigenfunctions. But, if we consider every function in the form $\psi=f(r,\theta) e^{im\phi}$ where $f(r,\theta)$ is any function including the normalization coefficient, then, this function is always eigenfunction of $\hat {\mathbf L_z}$ ($\hat {\mathbf L_z} \psi=m\hbar \psi$ ) but it is not always the eigenfunction of $\hat {\mathbf L^2}$. What’s wrong with my thinking here?

 

[Nugatory]

What’s wrong with my thinking here?

If two operators commute, then at least some of the eigenfunctions of one will be eigenfunctions of the other - but not necessarily all. It kind of has to be that way; otherwise any eigenfunction of $L^2$ would be an eigenfunction of both $L_x$ and $L_y$ because both commute with $L_2$.

 

[JimWhoKnew]

As far as I know, the two commuting operators like $\hat {\mathbf L_z}$ and $\hat {\mathbf L^2}$ always have identical eigenfunctions.

Consider$$A=\begin{pmatrix}1&0\\0&2\end{pmatrix}$$and the identity $2\times2$ matrix $I$. Surely$$v=\begin{pmatrix}1\\1\end{pmatrix}$$is an eigenvector of $I$.

 

[hokhani]

Consider$$A=\begin{pmatrix}1&0\\0&2\end{pmatrix}$$and the identity $2\times2$ matrix $I$. Surely$$v=\begin{pmatrix}1\\1\end{pmatrix}$$is an eigenvector of $I$.

Thanks, but by operators I meant Hermitian operators.

 

[PeroK]

Thanks, but by operators I meant Hermitian operators.

All Hermitian operators commute with $I$. And every vector is an eigenvector of $I$.

As far as I know, the two commuting operators like $\hat {\mathbf L_z}$ and $\hat {\mathbf L^2}$ always have identical eigenfunctions.

This is wrong. There must be a common eigenbasis. But, one operator might have two (or more) eigenvectors corresponding to the same eigenvalue. This is called degeneracy. In this case, the operator will have a mutli-dimensional eigenspace corresponding to that eigenvalue Whereas, if the other operator does not have this degeneracy, then it has two separate one-dimensional eigenspaces, corresponding to the different eigenvalues.

 

[JimWhoKnew]

Thanks, but by operators I meant Hermitian operators.

So which one is not Hermitian in my example, $I$ or $A$ ?

Edit: for 2 commuting observables there is a basis in which they are both diagonal. Instead of looking at the infinite-dimensional representation, it is easier to consider a $2\times2$ restriction that demonstrates the same basic idea.

 

[hokhani]

If two operators commute, then at least some of the eigenfunctions of one will be eigenfunctions of the other - but not necessarily all. It kind of has to be that way; otherwise any eigenfunction of $L^2$ would be an eigenfunction of both $L_x$ and $L_y$ because both commute with $L_2$.

Thanks, but let's write the commutation in the nondegenerate basis of $\hat {\mathbf L_z}$, i.e., $|m\rangle$. We have: $$\langle m| [\hat {\mathbf L_z}, \hat {\mathbf L^2}]|m' \rangle =0 \rightarrow (m'-m)\hbar \langle m| \hat {\mathbf L^2}|m' \rangle=0.$$ which means that $\hat {\mathbf L^2}$ is diagonal in the basis$|m\rangle$. What is wrong with this statement?

 

[PeroK]

Thanks, but let's write the commutation in the nondegenerate basis of $\hat {\mathbf L_z}$, i.e., $|m\rangle$. We have: $$\langle m| [\hat {\mathbf L_z}, \hat {\mathbf L^2}]|m' \rangle =0 \rightarrow (m'-m)\hbar \langle m| \hat {\mathbf L^2}|m' \rangle=0.$$ which means that $\hat {\mathbf L^2}$ is diagonal in the basis$|m\rangle$. What is wrong with this statement?

Nothing. All eigenvectors of $L_z$ are eigenvectors of $L^2$. But there are eigenvectors of $L^2$ that are not eigenvectors of $L_z$.

 

[hokhani]

So which one is not Hermitian in my example, $I$ or $A$ ?

Edit: for 2 commuting observables there is a basis in which they are both diagonal. Instead of looking at the infinite-dimensional representation, it is easier to consider a $2\times2$ restriction that demonstrates the same basic idea.

Sorry, you are right.

 

[hokhani]

Nothing. All eigenvectors of $L_z$ are eigenvectors of $L^2$. But there are eigenvectors of $L^2$ that are not eigenvectors of $L_z$.

However, in the post #1, I sent the function $\psi=f(r,\theta) e^{im\phi}$ which is always the eigenfunction of $\hat{\mathbf{L_z}}$ but not always the eigenfunction of $\hat{\mathbf{L^2}}$.

 

[JimWhoKnew]

Sorry, you are right.

Hints don't seem to work, so I'll say it explicitly: superpositions of the common diagonalizing basis vectors may still be eigenfunctions of one observable, but not necessarily of the other.

 

[hokhani]

Hints don't seem to work, so I'll say it explicitly: superpositions of the common diagonalizing basis vectors may still be eigenfunctions of one observable, but not necessarily of the other.

Could you please see the post #7? It approves more the discrepancy said in the post #1.

 

[JimWhoKnew]

Could you please see the post #7? It approves more the discrepancy said in the post #1.

As @PeroK replied in #8, nothing is wrong with #7. Only basis states are considered there. But your $\psi$ in OP (#1) is a superposition in general.

Edit: the error in #7 is claiming that $\mathbf{\hat L}_z$ is nondegenerate. There are infinitely many states in the $|l,m\rangle$ basis for which $m=1$ .
$|1,1\rangle$ and $|2,1\rangle$ are eigenstates of both operators. However, $|1,1\rangle+|2,1\rangle$ is an eigenstate of $\mathbf{\hat L}_z$ but not of $\mathbf{\hat L}^2$ (if you'll think about it, it is my example in #3).