Lippmann–Schwinger equation

Example

Consider the stationary Schrödinger equation $-\Delta \psi + V \psi = k^2 \psi$ on $\R^3$, with a real potential $V$ of compact support, in units with $\hbar^2 / 2m = 1$. Let $k > 0$ and let $\mathbf{k} \in \R^3$ with $\abs{\mathbf{k}} = k$ be the direction and wavenumber of an incoming plane wave. The scattering solution $\psi$ solves the Lippmann–Schwinger equation $$\psi(x) = e^{i \mathbf{k} \cdot x} - \frac{1}{4\pi} \int_{\R^3} \frac{e^{ik\abs{x - y}}}{\abs{x - y}} \, V(y) \, \psi(y) \, dy, \quad x \in \R^3.$$ It is an integral equation of the second kind for $\psi$, with the weakly singular kernel $-\frac{1}{4\pi} \frac{e^{ik\abs{x - y}}}{\abs{x - y}} V(y)$.

Check: the function $G(x) = -\frac{e^{ik\abs{x}}}{4\pi \abs{x}}$ is the outgoing Green's function of the Helmholtz operator, $(\Delta + k^2) G = \delta_0$. Since $(\Delta + k^2) e^{i\mathbf{k} \cdot x} = 0$, applying $\Delta + k^2$ to the right side gives $V \psi$, which is the Schrödinger equation $(\Delta + k^2)\psi = V \psi$. The choice of $e^{+ik\abs{x - y}}$ makes the scattered part $\psi - e^{i \mathbf{k} \cdot x}$ an outgoing spherical wave at infinity.