Timeline for Geometric Quantization
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Jul 23, 2014 at 23:25 | answer | added | Tuna Yildirim | timeline score: 1 | |
| May 13, 2014 at 5:36 | review | Close votes | |||
| May 13, 2014 at 15:35 | |||||
| Mar 30, 2014 at 15:54 | vote | accept | CommunityBot | moved from User.Id=62675 by developer User.Id=36770 | |
| Mar 26, 2014 at 20:27 | comment | added | Eric O. Korman | @SanathDevalapurkar Sorry, what I meant is that for a general configuration space there is no known canonical way to quantize it. The point is that geometric quantization gives one method to quantize a space that satisfies certain conditions. That this is a "correct" approach comes down to it satisfying certain axioms a quantization should have and agreeing in the simple cases with what physicists expect (e.g. on $\mathbb R^n$ or $S^2$). | |
| Mar 26, 2014 at 19:45 | answer | added | Carlo Beenakker | timeline score: 4 | |
| Mar 26, 2014 at 19:22 | comment | added | user62675 | @EricO.Korman I meant for a general configuration space $\mathcal E$. | |
| Mar 26, 2014 at 15:05 | comment | added | Eric O. Korman | @SanathDevalapurkar this is no general method of quantization. I would check out these mathoverflow posts mathoverflow.net/questions/6200/what-is-quantization mathoverflow.net/questions/8606/… | |
| Mar 26, 2014 at 14:38 | comment | added | user62675 | @EricO.Korman Yes, I am able to see that. What about the general case? | |
| Mar 26, 2014 at 11:31 | history | edited | Francois Ziegler | CC BY-SA 3.0 |
last edit "of" produced a nonsensical sentence
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| S Mar 26, 2014 at 11:27 | history | suggested | user38200 | CC BY-SA 3.0 |
fixed grammar
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| Mar 26, 2014 at 11:17 | review | Suggested edits | |||
| S Mar 26, 2014 at 11:27 | |||||
| Mar 26, 2014 at 4:49 | comment | added | Eric O. Korman | @SanathDevalapurkar are you able to see why if you start with $X = \mathbb R^n$ you get the usual quantization of first year quantum courses? | |
| Mar 26, 2014 at 0:31 | comment | added | user62675 | @IgorKhavkine I understand that the answer is "by definition". However, could you elaborate on the statement "comparing GQ with some traditional approach"? | |
| Mar 26, 2014 at 0:29 | comment | added | Igor Khavkine | @SanathDevalapurkar, the answer to your literal question is "by definition". Within geometric quantization, the "quantum Hilbert space" is a primitive notion, defined as you have outlined it. A deeper answer is that this terminology happens to agree with what physicists have done in a number of key examples (like those discussed in basic QM books). This agreement is best seen by explicit calculation, comparing GQ with some traditional approach. Whether GQ is "right" or "works" in all possible situations, is still somewhat up for debate. | |
| Mar 25, 2014 at 23:24 | history | edited | user62675 | CC BY-SA 3.0 |
added 459 characters in body
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| Mar 25, 2014 at 23:23 | comment | added | user62675 | @AndréHenriques I expect an answer that explains why the process works. Let $\mathcal{E}$ be the configuration space of the classical theory on $X$. I.e., if $T^{*}\mathcal{E}$ denotes the phase space, why is the space of square-integrable sections of the complex line bundle over $T^{*}\mathcal{E}$ that gives zero when we their covariant derivative at any point $x∈T^{*}\mathcal{E}$ in the direction of any vector in a Lagrangian subspace modeled by the polarization the quantum Hilbert space? | |
| Mar 25, 2014 at 23:18 | comment | added | André Henriques | Dear Sanath. It is not clear to me what kind of answer you want. Could you be more specific, and mention some properties of the geometric quantization construction that you are interested in, and about which you'd like to ask "why"? | |
| Mar 25, 2014 at 23:09 | review | First posts | |||
| Mar 25, 2014 at 23:11 | |||||
| Mar 25, 2014 at 22:53 | history | asked | user62675 | CC BY-SA 3.0 |