Timeline for Complete set of observables in classical mechanics
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2012 at 8:50 | comment | added | Ryan Thorngren | @Yrogirg Now I understand. I will write something up for you. | |
| Sep 11, 2012 at 8:38 | comment | added | Yrogirg | @user404153 I was asking what is the physical meaning of a "complete set of observables" in a classical mechanics. Since classical mechanics knows nothing about QM the meaning "an analogue of a QM concept" is not really a meaning. I assumed "a complete set of observables" in classical mechanics should have to do something with fully describing a classical state. That would mean they are coordinates on a manifold. I was wondering if this is true, if yes, then how to show it? | |
| Sep 11, 2012 at 8:27 | comment | added | Ryan Thorngren | @Yrogirg I am confused by your question then. Are you asking why the ps and qs form a complete set of observables? You can show {p,f} = df/dq and {q,f} = df/dp for any function f(p,q), so certainly if {p,f} = {q,f} = 0, f is constant. | |
| Sep 11, 2012 at 5:13 | comment | added | Yrogirg | @user404153 ok, but what has it to do with classical physics alone, without referencing to QM? That's just math interpretation, if that was what I was looking for, I'd asked the question at Math.SE. | |
| Sep 11, 2012 at 4:59 | comment | added | Ryan Thorngren | @Yrogirg That's not quite what I mean. Certainly if one specifies the position and the momentum of a state, one specifies the classical state completely (after all, positions and momenta are supposed to be the coordinates of the space of classical states). What I mean is that if you specify an observable's Poisson brackets with all the ps and qs, then one has specified the observable up to the addition of a constant. Equivalently, an observable which Poisson commutes with all the ps and qs is a constant. | |
| Sep 11, 2012 at 4:23 | comment | added | Yrogirg | @user404153 so specifying the values of all the classical observables from the complete set won't specify a unique classical state? | |
| Sep 10, 2012 at 21:33 | comment | added | Ryan Thorngren | @Yrogirg indeed another way to say 'complete set of observerables' is to say that these ps and qs form an irreducible representation of the Heisenberg algebra (which is equivalent by Schur's lemma to saying that the only observables which Poisson commute with all the ps and qs are constant). | |
| Sep 10, 2012 at 16:44 | history | answered | Tobias Diez | CC BY-SA 3.0 |