Lippmann-Schwinger Equation the need for $i\varepsilon$


• the $\epsilon$ is what fixes our in and out scattering states. See Chapter 6.1 in Sakurai for a full derivation. – I.E.P. Jan 15 '17 at 8:31

Actually, one has to introduce $\epsilon$ in the latter method as well, while solving the equation on the Green's function. There are may be many methods to solve it, but in the simplest one, which includes Fourier transformation, one has to integrate over the real axis on which there are poles of the integrand, so one is forced to deform a contour. There is a few ways to do it, and the choice influences the boundary condition for the equation. The simplest cases are ingoing and outgoing wave conditions which require introducing $\epsilon$ in precisely the way you cited.
• 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? – Quantum spaghettification Jan 15 '17 at 10:23
• @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. – Andrey Feldman Jan 15 '17 at 10:59
• 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 $r=r'$? – Quantum spaghettification Jan 15 '17 at 17:03
• @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. – Andrey Feldman Jan 15 '17 at 17:31