Commutator of ##L^2## with ##L_x,L_y,L_z##...

[Apashanka]

TL;DR

$[L^2,L_i]=0$.....where $i=x,y,z$

For a given state say ${l,m_l}$ where $l$ is the orbital angular momentum quantum no. and $m_l$ be it's $z$ component...a given state $|l,m_l>$ is an eigenstate of $L^2$ but not an eigenstate of $L_x$...therefore all eigenstates of $L_x$ are eigenstates of $L^2$ but the reverse is not true... isn't it??

 

[DrClaude]

therefore all eigenstates of $L_x$ are eigenstates of $L^2$ but the reverse is not true... isn't it??

Correct. When you have a degeneracy, an eigenstate corresponding to the degenerate eigenvalue is not necessarily an eigenstate of another operator that commutes. However, it is always possible to find a linear combination of these degenerate eigenstates that will be eigenstates of both operators.

 

[strangerep]

@DrClaude : Sorry -- I must be missing something...

If we are given only that $L_x \Psi = m \Psi$, how does that imply $\Psi$ is necessarily an eigenstate if $L^2$ ?

Maybe I'm overlooking a tacit assumption?

 

[anuttarasammyak]

OP uses $m_l$ to mean as common eigenstate of Lz, usually z is taken, and L^2.

Otherwise, eigenstates of m in general includes states of l=0,1,2,3,... and eigenstates of l in general includes the states of m=−l,−l+1,0,1,2,3,...,l
One can play a role of sub-index of the other.

For an example, s electron and p_z electron are both eigenstates of m=0. We can distinguish them by value of l as sub-index because of the commutation of L^2 and Lz.

 

[Apashanka]

@DrClaude : Sorry -- I must be missing something...

If we are given only that $L_x \Psi = m \Psi$, how does that imply $\Psi$ is necessarily an eigenstate if $L^2$ ?

Maybe I'm overlooking a tacit assumption?

For a given $l$ the eigenstates of $L_x$ are some linear combinations of $|l,m_l>$ which are also the eigenstates of $L^2$...

 

[DrClaude]

@DrClaude : Sorry -- I must be missing something...

If we are given only that $L_x \Psi = m \Psi$, how does that imply $\Psi$ is necessarily an eigenstate if $L^2$ ?

Maybe I'm overlooking a tacit assumption?

OP was looking at states that are eigenstates of $L^2$.

Of course, this works both ways. eigenstates of $L_x$ are not necessarily eigenstates of $L^2$, but there exists a linear combinations of the degenerate $L_x$ states that form a common basis for $L_x$ and $L^2$.

 

[PeroK]

@DrClaude : Sorry -- I must be missing something...

If we are given only that $L_x \Psi = m \Psi$, how does that imply $\Psi$ is necessarily an eigenstate if $L^2$ ?

Maybe I'm overlooking a tacit assumption?

Starting from $L_+ = L_x +iL_y$ and $L_- = L_x - iL_y$, we get:
$$L^2 = L_x^2 + L_y^2 + L_z^2 = L_+L_- -\hbar L_z + L_z^2$$
And any eigenstate of $L_z$ is an eigenstate of $L^2$. And, by symmetry, the same is true for $L_x$ and $L_y$.

 

[vanhees71]

No! There are eigenstates of $L_z$ that are not eigenstates of $L^2$, e.g., $|l_1,m \rangle+|l_2,m \rangle$ with $l_1 \neq l_2$.

 

[PeroK]

No! There are eigenstates of $L_z$ that are not eigenstates of $L^2$, e.g., $|l_1,m \rangle+|l_2,m \rangle$ with $l_1 \neq l_2$.

Yes, of course. I didn't think of that.

 

[anuttarasammyak]

And any eigenstate of LzLzL_z is an eigenstate of L2L2L^2. And, by symmetry, the same is true for LxLxL_x and LyLyL_y.

More precise saying is
we can find common set of eigenvectors for the two operators, i.e.,
L^2|l,m>=l^2|l,m>
L_z|l,m>=m|l,m>

May I remind that
\sum_m c_m|l,m> is an eigenvector of L^2 with eigenvalue l^2, but not an eigenvector of L_z anymore ?

 

[A. Neumaier]

how does that imply \Psi is necessarily an eigenstate of L^2 ?

L^2 is a central element (Casimir) of the universal enveloping algebra, hence in an irreducible unitary representation, it has every vector as eigenvector.

 

[strangerep]

L^2 is a central element (Casimir) of the universal enveloping algebra, hence in an irreducible unitary representation, it has every vector as eigenvector.

In a specific unirep, yes, but not in general, as Hendrik pointed out in his post #8.

 

[A. Neumaier]

In a specific unirep, yes, but not in general, as Hendrik pointed out in his post #8.

In every irreducible unitary representation. The rep specified in post #1 is of this form, while that in #8 is not.

 

[vanhees71]

We were talking about all eigenvectors of $L^2$ and $L_z$, and there an eigenvector of $L_z$ is not necessarily also an eigenvector of $L^2$.

Of course if you restrict yourself to a subspace, which builds an irreducible representation of the rotation group, than it's an eigenspace of $L^2$ to only one eigenvalue $\hbar^2 l(l+1)$ and thus in this subspace all vectors are eigenvectors of $L^2$ with this eigenvalue.

 

[strangerep]

In every irreducible unitary representation. The rep specified in post #1 is of this form, while that in #8 is not.

Yes, of course. But I was trying to clarify Dr Claude's post#2, which omits some of the OP's context. (Anyway, I think this thread has run its course, and then some. <Exit>)