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?

  • $\begingroup$ What is the problem with that? $\endgroup$
    – my2cts
    Jun 14, 2018 at 9:00
  • 3
    $\begingroup$ 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". $\endgroup$
    – jacob1729
    Jun 14, 2018 at 9:03
  • $\begingroup$ Comment to the post (v3): For starters, the LHS of eq. (1) is a 3-vector but the RHS is not?? $\endgroup$
    – Qmechanic
    Jun 14, 2018 at 9:12
  • $\begingroup$ 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? $\endgroup$ Jun 14, 2018 at 9:16
  • $\begingroup$ @jacob1729 Thank you for your comment! I think I agree with you. $\endgroup$ Jun 14, 2018 at 9:24

1 Answer 1


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.

  • $\begingroup$ Thank you for your answer! We can't measure 3 components of angular momentum imultaneously, so I think angular momentum operator itself can't be constructed. We can construt its components, and its square, but not itself. If you suggest angular momentum is observable, can you write down angular momentum operator (not projection) and solve its eigenvalue? $\endgroup$ Jun 14, 2018 at 12:34
  • $\begingroup$ @UniverseMaintainer we can construct any projection we want, along any direction we want, and find the eigenvalues and eigenvectors along any direction we want. Once you have an eigenstate for the projection angular momentum along one direction, these states will not be simulteaneous eigenstates for the projection of angular momentum along any other direction, and that's where the "you can't simultaneously measure" statements come in. $\endgroup$ Jun 14, 2018 at 13:44

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