Past exam of the theoretical physics II course of the University of Cambridge / 2013 / Paper 1 / 3 / b / Solution

Expanding the modulus squared yields an outgoing-angle factor $e^{-i{\mathbf{k}}_f\cdot \mathbf{d}}$ and an incoming factor $e^{i{\mathbf{k}}_i\cdot \mathbf{d}}$, where $\mathbf{d}=\mathbf{r}-{\mathbf{r}}^{\prime }$. Choosing a polar axis along $\mathbf{d}$ givesConsequentlyThe apparent $d=0$ singularity is removable: $\sin (kd)/(kd)\to 1$. Exchanging $\mathbf{r}$ and ${\mathbf{r}}^{\prime }$ cancels the imaginary part, while nonnegativity follows directly from its origin as an integral of $|f^{(1)}{|}^2$. A complex absorbing potential would require $V({\mathbf{r}}^{\prime }{)}^{\ast }$ instead; the lossless optical theorem comparison below uses real $V$.