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Angular momentum projection operators $\hat J_z$ and $\hat J_y$ don't commute, as don't the other combinations of different projections. But this means that there's no such state in which the whole angular momentum would be defined. On the other hand, there exists the angular momentum operator:

$$\hat{\vec J}=\hat{\vec r}\times\hat{\vec p}.$$

But since no state with definite angular momentum exists, it seems that then $\hat{\vec J}$ doesn't have any eigenstates! Is it true? If yes, then how can it be that an (Hermitian!) operator has no eigenstates?

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    $\begingroup$ $\hat{\vec J}$ are actually 3 operators, or do you mean $|\hat{\vec J}|$? $\endgroup$ – Wolphram jonny Nov 12 '14 at 5:42
  • $\begingroup$ @julianfernandez no, I mean the full vector. $\endgroup$ – Ruslan Nov 12 '14 at 6:14
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First of all $\vec{J}$ is not an Hermitian operator from a Hilbert space to the same Hilbert space, its three components separately are. Therefore nothing requires that there must exist an orthonormal basis of eigenvectors of $\vec{J}$, that is a orthonormal basis of simultaneous eigenvectors of $J_x$, $J_y$, $J_z$. Otherwise these operators would commute pairwise, which is false.

Nevertheless some common eigenvectors may exist, if they do not form a complete basis.

As a matter of fact, up to factors, there is only one vector of such type, the one usually indicated by $|0,0\rangle$ referred to the common basis of eigenvectors of $J^2, J_z$. In this case $J_i|0,0\rangle =0$ for $i=x,y,z$.

(I am referring here to the orbital angular momentum in the space $L^2(\mathbb S^2)$ (disregarding non-angular degrees of freedom). If you instead consider the total angular momentum including the spin, $|0,0\rangle$ does not exist if the spin is half-integer)

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    $\begingroup$ Not sure why you say that $\vec J$ is not an observable. It's not because they don't commute that they are not all observables at the same time. I'm pretty sure you wouldn't say "X and P are not observables in the sense of QM, but they separately are"... $\endgroup$ – Adam Nov 12 '14 at 9:21
  • $\begingroup$ Well this is just mathematics not physics. For "observable" I meant a Hermitian operator. You are using a theorem saying that a Hermitian operator $A : H \to H$ where $H$ is a (finite dimensional) Hilbert space then there is an orthonormal basis of eigenvectors. In the case you are considering $\vec{J}$ is not an operator from $H$ to $H$, because it associates vectors to triples of vectors. $\endgroup$ – Valter Moretti Nov 12 '14 at 9:31
  • $\begingroup$ But your justification for why it is not an observable is based on the non-commutativity of the components, which is completely different from what you just said... $\endgroup$ – Adam Nov 12 '14 at 9:34
  • $\begingroup$ OK, I have omitted the term "observable" in my answer to avoid any misunderstanding. Are you content with the new version? $\endgroup$ – Valter Moretti Nov 12 '14 at 9:34
  • $\begingroup$ Sure, I think its clearer now. $\endgroup$ – Adam Nov 12 '14 at 9:38
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$\hat{\vec J}$ is not actually a Hermitian operator (well, not even an operator on Hilbert space, but a linear operator from $H$ to $H^3$), but a composition of three of them. One way to see it is that eigenvalues are scalars, not vectors. Each component of $\hat{\vec J}$ is Hermitian, as also is $J^2$, this last commutes with each individual component of $\hat{\vec J}$.

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I think the answer is this:

when we say the angular momentum doesn't have an eigenstate, we actually mean it doesn't in "Cartesian coordinate system". The reason is that you know, i.e., the components can't commute. However, $\vec{J}$ is Hermitian and since any Hermitian matrix can be diagonalised, so it must have an eigenstate, but we don't know how we can find it. If we can resolve $\vec{J}$ into components that commute, then we'll be able to find its eigenstates.

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    $\begingroup$ So you assert that the eigenstates exist but we can't find them? That sounds strange. $\endgroup$ – Ruslan Nov 12 '14 at 6:15
  • $\begingroup$ Yes. I said why, unless $\vec{J}$ is not Hermitian! $\endgroup$ – MEDVIS Nov 12 '14 at 6:25
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As the components of angular operator $\hat J$ does not commute, you couldn't find a vector which is an eigenstate of all three components $\hat J_x, \hat J_y, \hat J_z$, talking about the eigenstates of angular operator usually need to define a reference direction such as "eigenstates of $\hat J \cdot \vec x,\,\, \hat J \cdot \vec z$ or $\hat J \cdot \vec n$"

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  • $\begingroup$ "you couldn't find a vector which is an eigenstate" - this is just restatement of what I've said in the question. $\endgroup$ – Ruslan Nov 12 '14 at 8:57

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