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Sep 25, 2012 at 16:16 vote accept Sadiq Ahmed
Sep 21, 2012 at 10:55 answer added Sadiq Ahmed timeline score: 0
Sep 20, 2012 at 19:49 comment added Alexander Chervov mathoverflow.net/questions/19495/…
Sep 20, 2012 at 17:12 answer added Palle Jorgensen timeline score: 2
Sep 19, 2012 at 16:12 comment added Alexander Chervov No go theorem is called van Hove theorem - I remember it in Hurt's Geometric quantization, i do not have this book now. google gives some references - I am not sure what is the best....
Sep 19, 2012 at 13:56 comment added Igor Khavkine This is essentially the same question as the one Joel linked to, made slightly more narrow. Why not just edit the original question and ask for it to be reopened?
Sep 19, 2012 at 13:27 comment added JRN See also mathoverflow.net/questions/107323/…
Sep 19, 2012 at 12:29 comment added Alexander Chervov Under this relaxed condition the case of R^2n is more or less covered by standard correspondence p-> d/dx q->x. Moreover this problem can be generalized to symplectic manifolds - see section 5. Berezin and Berezin-Toeplitz quantization on K¨ahler manifolds in the reference. Actually one hopes something similar for Poisson manifolds - to each symplectic leave should correspond an operator representation. But this is subtle project even in the case of Lie algebras - where this reduces to the "orbit method" there are many problems.
Sep 19, 2012 at 12:22 comment added Alexander Chervov Requirement 1.4 is not enlighting \hat {f,g} = [\hat f ,\hat g ] , this should not hold on any large class of observables... The correct definition is notion of deformation quantization - [\hat f ,\hat g ] = h \hat {f,g} + O(h^2) , where by \hat f I mean correspondence from classical to quantum.
Sep 19, 2012 at 12:00 history edited Sadiq Ahmed CC BY-SA 3.0
deleted 4 characters in body; edited title
Sep 19, 2012 at 11:48 history asked Sadiq Ahmed CC BY-SA 3.0