Born series

Remarks

If the operator norm satisfies $\norm{T} < 1$, the series converges in $X$ and its sum is the unique solution of $\psi = \phi + T \psi$, by the Neumann series. Replacing $V$ by $g V$ with a coupling constant $g$, this holds for all small enough $\abs{g}$.

Example: in one dimension, take the point interaction $V = g \delta_0$ with $g \in \R$; then the outgoing Green's function is $\frac{e^{ik\abs{x}}}{2ik}$, and $G_0 V \psi(x) = \frac{g}{2ik} e^{ik\abs{x}} \psi(0)$. For $\phi(x) = e^{ikx}$ each term is a multiple of $e^{ik\abs{x}}$, and the series sums to $\psi(x) = e^{ikx} + \frac{g/(2ik)}{1 - g/(2ik)} e^{ik\abs{x}}$ when $\abs{g} < 2k$. At $x = 0$ this gives $\psi(0) = \frac{1}{1 - g/(2ik)}$, which solves $\psi(0) = 1 + \frac{g}{2ik} \psi(0)$, as it must.

Non-example: for $\abs{g} > 2k$ in the same model the geometric series diverges, although the equation still has the solution given by the closed formula. So the Born series may fail to converge even when the scattering problem is well posed.