# 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?

• What is the problem with that? Jun 14, 2018 at 9:00
• Is the question essentially "If I have three observables, is it legitimate to call the triple of them an observable, even if they don't commute?" I'd be tempted to say no, and that saying "angular momentum is an observable" is shorthand for "each component is". Jun 14, 2018 at 9:03
• Comment to the post (v3): For starters, the LHS of eq. (1) is a 3-vector but the RHS is not?? Jun 14, 2018 at 9:12
• As a matter of fact, they say $\hat{\vec{L}}^2=L(L+1)$ and $L_z$ are observable in QM. Do you see the difference? Jun 14, 2018 at 9:16
• @jacob1729 Thank you for your comment! I think I agree with you. Jun 14, 2018 at 9:24

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 $$\left(\begin{array}{c}-\frac{1}{2}+\frac{i}{2} \\ \frac{1}{\sqrt{2}}\end{array}\right)\qquad \left(\begin{array}{c}\frac{1}{2}-\frac{i}{2} \\ \frac{1}{\sqrt{2}}\end{array}\right)$$