Timeline for A universe of angular momentum?
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Jun 5, 2016 at 14:13 | answer | added | Stéphane Rollandin | timeline score: 1 | |
| Mar 6, 2013 at 16:57 | comment | added | anna v | @JerrySchirmer did you check Lubos' answer to a similar question linked by user10001 above? also www2.ph.ed.ac.uk/~gja/qp/partialwave.pdf | |
| Mar 6, 2013 at 16:23 | comment | added | Zo the Relativist | @annav: I don't see how that would happen at all. And if you want a more trivial example of the same thing, take the scattering states of a $\alpha\delta(\beta r)$ potential, with $\alpha$, $\beta$ constant. You're definitely going to have a continua of momenta and position, which will give you a continuum of allowed angular momentum states. | |
| Mar 6, 2013 at 13:48 | vote | accept | JoeRocc | ||
| Mar 5, 2013 at 18:47 | comment | added | user10001 | Relevant: is-all-angular-momentum-quantized? | |
| Mar 5, 2013 at 18:18 | comment | added | anna v | I think you would find that the impulse that created the angular momentum would be quantized, just because of the form of the angular momentum QM operator. | |
| Mar 5, 2013 at 18:01 | comment | added | Zo the Relativist | @annav: You could have orbital angular momentum in a scattering state of the hydrogen atom, right? ${\vec r} \times {\vec p}$ is perfectly well defined, since both $\vec r$ and $\vec p$ are observables of the theory, so you have neither straight line motion an upper bound to anything. | |
| Mar 5, 2013 at 17:57 | comment | added | anna v | @user10001 see my answer | |
| Mar 5, 2013 at 17:50 | answer | added | anna v | timeline score: 3 | |
| Mar 5, 2013 at 17:34 | comment | added | anna v | I think that in order to have angular momentum a potential/force is needed since otherwise the particle would go in a straight line. In that sense there cannot be a continuous angular momentum at the micro level since it will be a solution of a potential QM problem that will be an eigenfunction of the angular momentum operator . | |
| Mar 5, 2013 at 17:02 | comment | added | user10001 | @JohnRennie Sorry I didn't know that. Do you mean angular momentum observables can have states with continuous eigenvalues? | |
| Mar 5, 2013 at 16:42 | comment | added | John Rennie | Ah good point, I was only thinking about orbital angular momentum. | |
| Mar 5, 2013 at 16:32 | comment | added | Zo the Relativist | @JohnRennie: do you know of a case where spin is unbound? I could think of some for orbital angular momentum, but not spin. | |
| Mar 5, 2013 at 16:31 | history | tweeted | twitter.com/#!/StackPhysics/status/308977981332791296 | ||
| Mar 5, 2013 at 16:26 | comment | added | John Rennie | @user10001: surely angular momentum is not quantised for an unbound state any more than position and momentum are. | |
| Mar 5, 2013 at 15:39 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 23 characters in body; edited tags
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| Mar 5, 2013 at 15:35 | answer | added | John Rennie | timeline score: 0 | |
| Mar 5, 2013 at 14:20 | comment | added | user10001 | @Cheeku Observables in QM are matrices rather than functions so they do not commute in general(ie AB=!BA). In more simple terms suppose S is a physical system (say a particle) and A and B be two (quantum) observables associated with S. Say A= position and B=momentum. Now to know the value of A and B you have two choices - i) first measure A and then measure B ii) first measure B and then measure A. Classically both the ways are equivalent and will give the same result while for quantum systems these two ways of measurement are in general not equivalent and give different values of A and B. | |
| Mar 5, 2013 at 13:48 | comment | added | Cheeku | @user10001 Can you please explain what do you mean by "observables don't commute"? What is "commute"? Thanks in advance! | |
| Mar 5, 2013 at 12:41 | review | First posts | |||
| Mar 5, 2013 at 14:35 | |||||
| Mar 5, 2013 at 12:39 | comment | added | user10001 | However real essence of QM is not in the fact that angular momentum takes discrete values but in the fact that observables generally don't commute (whereas in classical mechanics they always commute). | |
| Mar 5, 2013 at 12:30 | comment | added | user10001 | Angular momentum is the most basic observable where QM differs from classical mechanics. In QM angular momentum takes only discrete values whereas (unless the particle is trapped in a finite region of space) momentum and position take continuous values in QM just as in classical mechanics. | |
| Mar 5, 2013 at 12:22 | history | asked | JoeRocc | CC BY-SA 3.0 |