Why do we say spin/angular momentum is observable even though its components can't be determined simultaneously? Why do we say spin or angular momentum of a particle is observable even though all of its components can't be determined simultaneously?
For example, we can measure the $\hat{L_x}$ of a particle's angular momentum but not $\hat{L_x}$ and $\hat{L_y}$ at the same time due to uncertainty principle. This means that we cannot measure angular momentum at all. Why then do we insist angular momentum is observable? In fact, how do we solve the eigenvalue problem:
$$(\hat{L_x}\vec{i}+\hat{L_y}\vec{j}+\hat{L_z}\vec{k}) \Psi=l \Psi\tag{1}$$
where $\Psi$ is the wavefunction and $l$ is an eigenvalue of the operator  ${\hat{L}}$?
PS:I don't think my equation is correct because the basis vectors show up on the left hand side but not on the right hand side. Maybe I need to erase basis vectors?
 A: The fact that two (or more) hermitian operators do not commute does not mean they are not observables.  For instance, $x$ and $p$ do not commute but are surely observables.  Likewise, the kinetic energy $p^2/2m$ and the potential energy $V(x)$ usually do not commute, but they are observables and finding eigenvectors for their sum $p^2/2m+V(x)=H$ amounts to solving the time-independent Schrodinger equation.
It's not a problem to diagonalize your Eq.(1) either: simply write each operator in matrix form, sum the matrices (the result is still a hermitian matrix) and find the eigenvectors of this hermitian matrix.  You will then get eigenvectors that represents states with definite projection in the $\vec i+\vec j+\vec k$ direction.  Again, the fact that the various projections do not pairwise commute does not mean that the sum is not an observable.  For example, the Pauli matrices $\sigma_x$ and $\sigma_y$ do not commute, but their sum 
$$
\sigma_x+\sigma_y=\left(
\begin{array}{cc}
 0 & 1-i \\
 1+i & 0 \\
\end{array}
\right)
$$
has eigenvectors 
$$
\left(\begin{array}{c}-\frac{1}{2}+\frac{i}{2} \\ 
\frac{1}{\sqrt{2}}\end{array}\right)\qquad 
\left(\begin{array}{c}\frac{1}{2}-\frac{i}{2} \\ \frac{1}{\sqrt{2}}\end{array}\right)
$$
When the Hilbert space is finite dimensional, every hermitian operator corresponds in theory to an observable, although it might be difficult in practice to design a physical experiment that will measure this observable, and it may be difficult to write something like a classical version of the operator you start with.
When the Hilbert space is infinite dimensional, there are additional subtle points and there are operators which are hermitian (or more properly not self-adjoint) but do not correspond to observables. 
