Timeline for Does the angular momentum vector operator $\hat{\vec{J}}$ have no eigenstates?
Current License: CC BY-SA 3.0
10 events
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| Nov 12, 2014 at 19:04 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Nov 12, 2014 at 11:45 | vote | accept | Ruslan | ||
| Nov 12, 2014 at 9:44 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Nov 12, 2014 at 9:38 | comment | added | Adam | Sure, I think its clearer now. | |
| Nov 12, 2014 at 9:35 | history | edited | Valter Moretti | CC BY-SA 3.0 |
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| Nov 12, 2014 at 9:34 | comment | added | Valter Moretti | OK, I have omitted the term "observable" in my answer to avoid any misunderstanding. Are you content with the new version? | |
| Nov 12, 2014 at 9:34 | comment | added | Adam | 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... | |
| Nov 12, 2014 at 9:31 | comment | added | Valter Moretti | 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. | |
| Nov 12, 2014 at 9:21 | comment | added | Adam | 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"... | |
| Nov 12, 2014 at 9:11 | history | answered | Valter Moretti | CC BY-SA 3.0 |