Born approximation

Remarks

For a plane wave $u_i(x) = e^{ik \hat{d} \cdot x}$, the far field of $u_B - u_i$ is $f_B(\hat{x}) e^{ikr}/r$ with $$f_B(\hat{x}) = \frac{k^2}{4\pi} \int_{\R^3} e^{-ik(\hat{x} - \hat{d}) \cdot y} \, m(y) \, dy.$$ So $f_B$ is, up to a factor, the Fourier transform of $m$ at the momentum transfer $k(\hat{x} - \hat{d})$.

Example: let $m = \varepsilon \chi_B$, where $\chi_B$ is the indicator of a ball $B$ and $\varepsilon$ is small. Then $u_B$ is given by the formula above, and $u - u_B = O(\varepsilon^2)$ as $\varepsilon \to 0$ for fixed $k$ and $B$. Check: the Lippmann–Schwinger operator $T w = k^2 G * (m w)$ has norm $O(\varepsilon)$ on bounded continuous functions on $B$, so $u = \sum_{j \ge 0} T^j u_i$ and $u - u_B = \sum_{j \ge 2} T^j u_i$.

Non-example: the exact field $u$ is in general not the Born approximation, since $u - u_B = \sum_{j \ge 2} T^j u_i$ need not vanish. Consider a strong scatterer, such as a ball with $k^2 \abs{m}$ times the squared radius of order $1$ or more. Then the terms $T^j u_i$ need not decrease, and $u_B$ can be far from $u$.

The iterates $u_i + T u_i + \cdots + T^n u_i$ are the higher Born approximations, and the full sum is the Born series.