# What does vector operator for angular momentum measure?

The fact that $\hat L_x$, $\hat L_y$ and $\hat L_z$ do not commute means that you cannot have a function which is eigenfunction of the three operators. Your function can only be an eigenfunction of one of the component operators, say $\hat L_z$ (it is the typically chosen one). However, the operator $\hat L^2$ does commute with each of the component operators. So if we have an eigenfunction of $\hat L_z$, it is also an eigenfunction of $\hat L^2$. (But not an eigenfunction of $\hat L_x$ and $\hat L_y$).

Therefore, the answer to your question is that the operator $\hat L$ is used to give us the magnitude of the angular momentum, by operating on the eigenfunction with its square: $\hat L^2$. The eigenvalue is then $l(l+1)ħ^2$, which is the magnitude of the angular momentum squared.

You can only measure operators. A vector operator is a set of *three* operators, which we package into one piece of notation because they transform under rotations in a nice way. You can't measure it all in one go, any more than you can try to measure $\hat{x}$ and $\hat{p}$ at the same time. However, you can measure the components of $\hat{L}$, which are, as you would expect, the components of the angular momentum.

I like this question. What you describe is taking:

$${\bf \hat L}|l\rangle = \hat L_x|l\rangle {\bf \hat x} + \hat L_y \hat L_x|l\rangle {\bf \hat y} ... = {\bf \vec L}|l\rangle$$

which would lead to problems. I think an eigenvalue equation for a vector operator should be:

$${\bf \hat L}|l\rangle = \hat L_x|l\rangle {\bf \hat x} + \hat L_y|l\rangle {\bf \hat y} + \hat L_z|l\rangle {\bf \hat z} = L_x|l\rangle {\bf \hat x} + L_y|l\rangle {\bf \hat y} + L_z|l\rangle {\bf \hat z} = {\bf \vec L}|l>$$

where

$${\bf \hat L} = -i\hbar (y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y}){\bf \hat x} + (permutations) $$

is the operator and

$${\bf \vec L} = L_x {\bf \hat x}+ L_y {\bf \hat y} + L_z {\bf \hat z }$$

is an ordinary vector.

Now with:

$$ |l\rangle \equiv F_l(x, y, z)$$

the equation:

$$ {\bf \hat L}F_l(x, y, z) = {\bf \vec L}F_l(x, y, z)$$

doesn't have any non-trivial solutions: there are no eigenstates of completely known (non zero) angular momentum.

Regarding your question on "meaning": note that:

$$ L_z = -i\hbar\frac{\partial}{\partial\phi} $$

and eigenstates of that function look like:

$$ F(r, \theta, \phi) = R(r)\Theta(\theta)e^{\pm i m \phi} $$

which means that the function is invariant under rotations of $2\pi/m$.

For a wave-function to be an eigenstate of ${\bf \hat L}$ it would have to invariant under some non-trivial rotation about not just the $z$-axis, but the $x$ and $y$ axes also--which would mean it would have to be invariant under some rotation about *any* axis.

The only thing I can think of that can do that is a sphere, and this is, of course, the zero angular momentum eigenstate.