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

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.

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 $V$.

The first scattered correction to the incident wave isInserting it into the amplitude gives a second, positive contribution because the two minus signs multiply:Thus the amplitude through second Born order is $f^{(1)}+f^{(2)}+O(V^3)$. The intervening outgoing Green function for the three-dimensional Helmholtz equation describes propagation between the two interactions; replacing it by a real standing-wave kernel would discard the imaginary part needed for unitarity.

In the forward direction ${\mathbf{k}}_f={\mathbf{k}}_i$, the first Born amplitude $-C\int V$ is real. Symmetrize the second-order double integral under $\mathbf{r}\leftrightarrow {\mathbf{r}}^{\prime }$ before taking its imaginary part. This replaces $e^{-i{\mathbf{k}}_i\cdot \mathbf{d}}$ by $\cos ({\mathbf{k}}_i\cdot \mathbf{d})$, and givesThe preceding total-cross-section integral equals the same real integral multiplied by $4\pi /k$. ThereforeThis is the optical theorem at second Born order. Comparing only the real first-order forward amplitude with a nonzero second-order scattering cross-section would mix perturbative orders. Terms of order $V^3$ require both the next amplitude order and interference in the scattering cross-section.