 Solution

For elastic scattering by a real short-range scalar potential, write $C=m/(2\pi {\hslash }^2)$, ${\mathbf{k}}_i=\mathbf{k}$ and $|{\mathbf{k}}_f|=k$. In the far field, $|\mathbf{r}-{\mathbf{r}}^{\prime }|=r-\hat{\mathbf{r}}\cdot {\mathbf{r}}^{\prime }+O(r^{-1})$ and ${\mathbf{k}}_f=k\hat{\mathbf{r}}$. The outgoing spherical-wave coefficient isSubstitution of the incident plane wave inside this integral givesThe incoming and outgoing flux factors cancel because their speeds agree. Here $\mathbf{q}={\mathbf{k}}_f-{\mathbf{k}}_i$ is a wavevector transfer; the actual momentum transfer is $\hslash \mathbf{q}$. This distinction preserves dimensions. The approximation requires the scattered correction to the interior wave to be small, and does not in general apply to an unscreened long-range potential.