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Feb 1, 2012 at 10:17 history edited shu CC BY-SA 3.0
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Jan 31, 2012 at 12:25 history edited shu CC BY-SA 3.0
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Jan 30, 2012 at 20:51 vote accept shu
Jan 27, 2012 at 22:46 answer added user1504 timeline score: 8
Jan 27, 2012 at 10:32 comment added Alexander Chervov Remark. If you consider 2d picture - similar measure on the space of function f(x,y) it will not live on continuos paths but on distributions ! I guess it is related to the "regularization, renormalization" which is present in QFT, but absent in QM
Jan 27, 2012 at 10:30 comment added Alexander Chervov @Shu thanks for thanks :) Shame on me I forget why it is on continuos paths ... As I remember it was simple - the same as why Wiener measure is on them... I'll try to remember. But actually the measure I mentioned is "more or less" standard Wiener measure ...
Jan 27, 2012 at 9:46 history edited shu CC BY-SA 3.0
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Jan 26, 2012 at 19:57 comment added shu @Alexander Chervov, indeed, this measure works formally. I want to make it rigourous. It is not even clear for me that its support is the contious path.
Jan 26, 2012 at 17:39 comment added Alexander Chervov Periodic function can be decomposed to Fourier Series - on each component you have Gaussian measure scaled by "n" - so we get measure on periodic paths... Is there something wrong with it ?
Jan 26, 2012 at 16:56 history asked shu CC BY-SA 3.0