The Lippmann-Schwinger equation is often solved by the addition of the factor $i\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} \newcommand{\bra}[1]{\left< #1\right|}\e$ such that we have: $$\ket{\psi}=\ket{p}+\f{V}{E-H_0\pm i\e}\ket{\psi}$$ which produces a Green's function of the form (in the 1d case): $$G_\pm=\bra{x}\f{1}{E-H_o\pm i\e}\ket{x'}$$ This is a pain to solve, requiring contour integration. It is much simpler to just solve the equation: $$(H_0-E)G=\delta(x-x')$$ which defines $G$ with the appropriate boundary conditions on $G$. This latter method does not involve the introduction of any factors of $\e$, but I have never seen it done like this. Why not? is this method in someway wrong?

  • $\begingroup$ the $\epsilon$ is what fixes our in and out scattering states. See Chapter 6.1 in Sakurai for a full derivation. $\endgroup$
    – I.E.P.
    Jan 15, 2017 at 8:31

1 Answer 1


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.

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    $\begingroup$ 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? $\endgroup$ Jan 15, 2017 at 10:23
  • $\begingroup$ @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. $\endgroup$ Jan 15, 2017 at 10:59
  • $\begingroup$ 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 $r=r'$? $\endgroup$ Jan 15, 2017 at 17:03
  • $\begingroup$ @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. $\endgroup$ Jan 15, 2017 at 17:31

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