Timeline for Counterexamples to weak dispersion for the Schrödinger group
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 2, 2022 at 18:28 | comment | added | Terry Tao | Thanks, this is now corrected. | |
| Nov 2, 2022 at 18:28 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Nov 2, 2022 at 18:28 | comment | added | Giorgio Metafune | Is a $x$ missing in the exponential in the definition of $U(t)$? | |
| Nov 2, 2022 at 18:22 | comment | added | Terry Tao | @PieroD'Ancona The enemy is singular continuous spectrum; if $A$ only has pure point and absolutely continuous spectrum then one can get strong mixing from the Riemann-Lebesgue lemma. There are certainly PDO type operators with singular continuous spectrum (cf. the Hofstadter butterfly) so I doubt there is an easy general way to enforce a strong RAGE theorem in general (one would have to prevent the singular continuous spectrum from having enough arithmetic structure to have non-decaying Fourier coefficients). | |
| Nov 2, 2022 at 18:17 | comment | added | Nike Dattani | Yea I found that condition on $f$ to be a bit weird. I was only thinking about discrete systems. Still isn't it easier to just give an infinite-dimensional $A$ and $f$ such that we don't get decay to 0? | |
| Nov 2, 2022 at 18:16 | comment | added | Terry Tao | Corrected, thanks. | |
| Nov 2, 2022 at 18:16 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Nov 2, 2022 at 18:14 | comment | added | Giorgio Metafune | A factor $1/N$ is missing in the definition of weakly mixing. | |
| Nov 2, 2022 at 18:11 | comment | added | Terry Tao | @NikeDattani By the spectral theorem for unitary matrices, it is not possible for a non-zero vector to be orthogonal to all eigenstates in a finite dimensional system. | |
| Nov 2, 2022 at 18:06 | comment | added | Nike Dattani | I'm surprised that the answer is involves so many words and symbols. With a simple 2x2 matrix for $A$ and $f$ being the vector [1;0], we have $U(t)f$ oscillating forever w.r.t. $t$ (i.e. not decaying to 0). Can we not just give an example of an $A$ and $f$ with $f$ orthogonal to all eigenstates of $A$ in which $U(t)f$ doesn't decay to 0? | |
| Nov 2, 2022 at 16:56 | vote | accept | Piero D'Ancona | ||
| Nov 2, 2022 at 16:56 | comment | added | Piero D'Ancona | Excellent, this settles the question. When $A$ is a PDO, convergence to 0 usually follows from local energy decay; I was hoping for some more abstract result in the spirit of RAGE, but of course the general case is hopeless | |
| Nov 2, 2022 at 16:03 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| Nov 2, 2022 at 15:56 | history | answered | Terry Tao | CC BY-SA 4.0 |