Discussion
A particle is scattered from a potential \(V(\vec{r}) = g\delta^{3}(\vec{r})\), where \(g\) is a positive constant. Using the first Born approximation, the angular \((\theta, \phi)\) dependence of differential scattering cross section \(\frac{d\sigma}{d\Omega}\) is
Answer: D
The cross section is independent of both \(\theta\) and \(\phi\).
In the first Born approximation the scattering amplitude is the Fourier transform of the potential at the momentum transfer \(\vec{q}\):
\[ \begin{aligned} f(\theta) &= -\frac{m}{2\pi\hbar^{2}}\int e^{-i\vec{q}\cdot\vec{r}}\, g\delta^{3}(\vec{r})\, d^{3}r \\ &= -\frac{mg}{2\pi\hbar^{2}} \end{aligned} \]
The delta function picks out \(\vec{r} = 0\), where the exponential is 1, so \(f\) is a constant and \(\frac{d\sigma}{d\Omega} = |f|^{2} = \frac{m^{2}g^{2}}{4\pi^{2}\hbar^{4}}\) is isotropic.
Any \(\phi\) dependence would need a potential that is not spherically symmetric, and \(\theta\) dependence enters only through \(q = 2k\sin(\theta/2)\), which drops out for a point interaction.
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