Timeline for Estimate of a solution of Schroedinger equation for a free particle
Current License: CC BY-SA 4.0
16 events
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Nov 15, 2019 at 17:34 | comment | added | Carlo Beenakker | @WillieWong --- indeed, that was my original answer, which I deleted because the OP explicitly wanted $V=0$ --- I have now reinserted that text. | |
Nov 15, 2019 at 17:33 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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Nov 15, 2019 at 15:22 | comment | added | Willie Wong | I have some not entirely rigorous/complete notes based on what I learned, in a slightly different context, here. | |
Nov 15, 2019 at 15:11 | comment | added | Willie Wong | @MKO: adding potential can actually make things better. The free evolution has a resonance at zero which gives the $1/\sqrt{t}$ decay. The perturbed evolution can break that (in fact, frequently does) and you can get (locally uniformly) $t^{-3/2}$ decay. (I only just learned about this a few months ago and it is still extremely surprising for me.) | |
Nov 15, 2019 at 13:43 | comment | added | Carlo Beenakker | I have worked out this example from the paper, with an initial square wave packet. | |
Nov 15, 2019 at 13:42 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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Nov 15, 2019 at 12:21 | comment | added | asv | I have not understood the details of your answer. This would be interesting but bad news for me. I am trying to understand the scattering theory in the baby version, on the line. If what you wrote is correct, there will be a problem to define the Moller operators even for compactly supported potential. I could not find in the literature discussion of scattering theory on the line. | |
Nov 15, 2019 at 11:36 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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Nov 15, 2019 at 11:13 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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Nov 15, 2019 at 10:55 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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Nov 15, 2019 at 10:50 | comment | added | Carlo Beenakker | ah, OK, I will modify the answer accordingly. | |
Nov 15, 2019 at 10:50 | history | undeleted | Carlo Beenakker | ||
Nov 15, 2019 at 10:50 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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Nov 15, 2019 at 10:34 | history | deleted | Carlo Beenakker | via Vote | |
Nov 15, 2019 at 10:24 | comment | added | asv | I consider a different question: the potential vanishes on the whole line. | |
Nov 15, 2019 at 10:22 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |