Timeline for Green's function for a linear PDE initial value problem
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 21 at 23:59 | comment | added | Michael Engelhardt | @ChristianRemling - sure, but what will typically happen when you evaluate $K(x,y;E)$ is that the formal expressions will be very similar, just that the poles corresponding to the bound states disappear because certain denominators cannot become zero anymore. At energies $E$ corresponding to scattering states, one cannot tell from the scattering amplitude whether the potential is attractive or repulsive! | |
| Jan 21 at 23:35 | comment | added | Christian Remling | I don't think this is similar to the potential with the opposite sign. For starters, there are no eigenvalues now. | |
| Jan 21 at 23:18 | comment | added | Michael Engelhardt | @ChristianRemling - I think most of the calculation of $K(x,y;E)$ will still go through with a repulsive potential, just towards the end one will have to be careful with the contours (which one will have to be anyway to connect the imaginary time to the real time case). But it needs to be checked carefully. | |
| Jan 21 at 22:52 | comment | added | Christian Remling | The operator is not the hydrogen operator in the OP, though; the potential has the opposite sign. | |
| Jan 21 at 9:15 | history | edited | Michael Engelhardt | CC BY-SA 4.0 |
added 13 characters in body
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| Jan 21 at 8:22 | history | answered | Michael Engelhardt | CC BY-SA 4.0 |