Timeline for Lippmann-Schwinger Equation the need for $i\varepsilon$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 15, 2017 at 17:31 | comment | added | Andrey Feldman | @Quantumspaghettification There is no jump in $D>1$, because $r \geq 0$. There is just a singularity analogous to the point charge solution of Poisson equation. | |
| Jan 15, 2017 at 17:03 | comment | added | Quantum spaghettification | I was talking about the 1D case, also the LS equation is of (nearly) the same form. But even for $D>1$, you would just get a jump in the derivative $G'(r-r')$ at $r=r'$? | |
| Jan 15, 2017 at 10:59 | comment | added | Andrey Feldman | @Quantumspaghettification You must be talking about 1D case. Are you sure that in this case the LS equation has the same form? What I was talking about is D>1 case, where $\delta$-function leads to a singular behavior near the origin $r=0$, not to the jump of the first derivative on the $x$ axis. | |
| Jan 15, 2017 at 10:23 | comment | added | Quantum spaghettification | I can see how it may come in when using the Fourier method, but I cannot see how it comes in if we first solve $(H_0-E)G=0$ then apply boundary conditions (i.e. continuity of $G$ and discontinuity of $G'$ defined by $\delta$) at $x=x'$. Any ideas? | |
| Jan 15, 2017 at 9:06 | history | edited | Andrey Feldman | CC BY-SA 3.0 |
added 1 character in body
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| Jan 15, 2017 at 8:52 | history | answered | Andrey Feldman | CC BY-SA 3.0 |