Timeline for Feynman path integral and Wilsonian renormalization
Current License: CC BY-SA 4.0
9 events
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| Nov 20, 2020 at 15:03 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title, added tag
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| Nov 20, 2020 at 15:00 | history | edited | iolo | CC BY-SA 4.0 |
Try to fix issues raised in the comments
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| Nov 20, 2020 at 14:05 | comment | added | iolo | I see... That is hard to say, as I have been unable to find a formal treatment that I also understand... But you are of course right, that the decomposition of $\v_{0,\infty}$ cannot hold true in a meaningful way, thanks! I will update the question to something more meaningful. | |
| Nov 20, 2020 at 12:54 | comment | added | Abdelmalek Abdesselam | This is not how on usually integrates out degrees of freedom. What reference are you following? | |
| Nov 20, 2020 at 12:53 | comment | added | Abdelmalek Abdesselam | It you split the Gaussian measure with only the Laplacian from the rest treated as interaction, and if you think of the degrees of freedom in a "microlocal way", i.e., as being indexed by phase space ($x$-space times frequency, or AdS) you will see the following. The Gaussian couples different positions but keeps different frequencies independent. By contrast, the interaction vertices couple different frequencies but do not couple different locations. So when you put all together, degrees of freedom are all coupled. | |
| Nov 19, 2020 at 18:41 | comment | added | iolo | I should of course say that the cylinder set measures live on a Hilbert space completion of $\mathcal{S}$. | |
| Nov 19, 2020 at 18:12 | comment | added | iolo | I just meant to emphasize that relevant Gaussian cylinder measures on $\mathcal{S}$ push forward to true Gausian measures on $\mathcal{S}'$. The independence you mention is also puzzling me and it is possible that I have misunderstood something. But isn't it the way we usually integrate out degrees of freedom? | |
| Nov 19, 2020 at 15:09 | comment | added | Abdelmalek Abdesselam | Everything makes sense up until "From a Wilsonian point of view...", except the mention of cylinder measures. You should use Borel probability measures on $\mathcal{S}'$, so you have a good notion of convergence for measures. More problematic: when you decompose $\nu_{0,\infty}$ as a convolution what you are doing is writing the final random distribution as a sum of a random high frequency part and a random low frequency part, where the two parts are independent. This independence holds for a Gaussian measure but has no reason to be true for the interacting QFT $\nu_{0,\infty}$. | |
| Nov 19, 2020 at 14:43 | history | asked | iolo | CC BY-SA 4.0 |