Timeline for Quantization of a classical system (e.g. the case of a billiard)
Current License: CC BY-SA 4.0
20 events
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| S Oct 31, 2018 at 17:52 | history | suggested | jeq | CC BY-SA 4.0 |
Some English corrections.
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| Oct 31, 2018 at 17:32 | review | Suggested edits | |||
| S Oct 31, 2018 at 17:52 | |||||
| Sep 16, 2018 at 2:43 | comment | added | youpilat13 | Some discussion on why certain classical theories cannot be ''quantized'' as one would naively expect: motls.blogspot.com/2018/09/… | |
| Jan 29, 2013 at 14:31 | comment | added | Alexander Chervov | ambiguties appear only if you have the terms like "pq", since \hat p \hat q is different from \hat q \hat p. This ambiguity in general seems cannot be resolved in general, the most optimistic hope may consists in the following - for different choices operators will have same specturm, moreover conjugate | |
| Jan 29, 2013 at 14:29 | comment | added | Alexander Chervov | "I'd like to precise it by asking: when people working in Quantum Theory quantizes a classical physical system (like in the article quoted above),..." Here depends on what do you mean by "system". Usually it is Poisson manifold and function "f" on it, BUT let us look more closer: what kind of manifold - compact/not, boundary/ corners. Do we have some choice of the coordinates on it ? Simples situation is R^2n with canonical coordinates p,q. So if functions is f=p^2+q^2 you quantize it to \hat p^2 + \hat q ^2 - NO problems | |
| Jan 27, 2013 at 19:25 | history | edited | Joël | CC BY-SA 3.0 |
added 789 characters in body
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| Jan 27, 2013 at 14:46 | answer | added | user80744 | timeline score: 1 | |
| Jan 26, 2013 at 21:45 | answer | added | Tobias Diez | timeline score: 1 | |
| Jan 26, 2013 at 21:06 | history | edited | Joël | CC BY-SA 3.0 |
added 162 characters in body; edited title
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| Jan 26, 2013 at 17:18 | answer | added | user21349 | timeline score: 1 | |
| Jan 26, 2013 at 16:28 | comment | added | user21349 | You might want to change "quantification" in the title to "quantization." | |
| Jan 26, 2013 at 12:47 | history | edited | Alexander Chervov |
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| Jan 26, 2013 at 10:30 | answer | added | Alexander Chervov | timeline score: 3 | |
| Jan 26, 2013 at 4:13 | answer | added | Theo Johnson-Freyd | timeline score: 3 | |
| Jan 26, 2013 at 4:07 | answer | added | Theo Johnson-Freyd | timeline score: 4 | |
| Jan 26, 2013 at 0:47 | comment | added | Joël | ...What is not clear to me is how this hypothesis get translated in the quantum setting... I was told that the corresponding quantum system is the space $V$ of function on $B$ vanishing on $\delta B$, and the Hamiltonian is the Laplacian. But how does that translate the hypothesis about bouncing? If we assume a weird bouncing where the the inward angle is said twice the outward angle, what would be the attached quantum system? | |
| Jan 26, 2013 at 0:44 | comment | added | Joël | Dear Alexander, thanks for your comment. Okay, I understand that if the wave function is continuous on $\mathbb R^2$ with support on the billard $B$, it must vanish on its boundary. But I am not sure why the wave function should be defined on $\R^2$ instead of just on $B$, and even while it should be continuous. In other words, there is a physical hypothesis in our classical billiard setting, that our ball bounces on the boundary $\delta B$ of the billiard by making an inward angle at a point $b \in \delta B$ with the tangent of $\delta B$ at $b$ equal to the outward angle... | |
| Jan 25, 2013 at 19:41 | answer | added | Uwe Franz | timeline score: 5 | |
| Jan 25, 2013 at 17:02 | comment | added | Alexander Chervov | |Wave function|^2 is probability. It vanishes outside region by your requrement. That is why "vanishing". About Laplacian it seems you know it. +1 but I am not sure I understand your problem. | |
| Jan 25, 2013 at 16:10 | history | asked | Joël | CC BY-SA 3.0 |