Consider the vector operator for angular momentum $\hat L=\hat L_x \vec i +\hat L_y \vec j + \hat L_z \vec k$.
Does this mean that if we want to measure the angular momentum of a particle in state $\psi$, we take $\hat L$ and let it act on $\psi$ to give us three possible eigenvalues of $\hat L_x$, $\hat L_y$ and $\hat L_z$, which will correspond to the $x,y,z$ components of the angular momentum?
But since $\hat L_x$, $\hat L_y$ and $\hat L_z$ does not commute, this should not be the meaning of $\hat L$ because if we first make a measurement of $\hat L_x$, the state of the particle will be changed, and $\hat L_y$ should no longer act on the original state $\psi$. What then does the vector operator $\hat L$ gives us? More precisely, what is the measurement this operator $\hat L$ is trying to measure?
Best Answer
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.