Solving Scattering Problem: Find Diff Cross-Sec in Limit of a→∞

[ehrenfest]

Homework Statement

A particle with mass m scatters off of the potential

$V=A\delta(z)$ for $-a \leq x \leq a$ and $-a \leq y \leq a$

V= 0 otherwise

Find the differential scattering cross-section using the Born approximation and show that in the limit where a goes to infinity the resulting differential cross-section corresponds to only two possible results: either the particle will pass through without any scattering or else it will scatter with angle of incidence equal to angle of reflection.

Homework Equations

The Attempt at a Solution

So, I need to prove that K_x=K_y=0 where K is the momentum transfer.
I got a differential cross-section of:
$$4\left(\frac{Am}{K_x K_y \pi \hbar^2}\right)^2\sin^2(K_y a)\sin^2(K_x a)$$
and I don't even see how that is well-defined when a goes to infinity.

 

[Jhero]

Thank you for your question. First, I will explain the Born approximation and how it can be used to calculate the differential scattering cross-section for the given potential.

The Born approximation is a method used to calculate the scattering amplitude of a particle interacting with a potential. It is based on the assumption that the potential is weak and the interaction between the particle and the potential can be treated as a perturbation. This allows us to use the first-order approximation in the scattering amplitude, which simplifies the calculations.

To use the Born approximation, we need to know the potential and the initial and final states of the particle. In this case, the potential is a delta function, and the initial and final states can be described by plane waves with momenta K and K'. The scattering amplitude is then given by:

f(\theta) = -\frac{2m}{\hbar^2}\int V(\vec{r})e^{i\vec{K}\cdot\vec{r}}d\vec{r}

Here, \theta is the scattering angle, m is the mass of the particle, and \hbar is the reduced Planck's constant. The differential scattering cross-section can be calculated from the scattering amplitude as:

\frac{d\sigma}{d\Omega} = |f(\theta)|^2

Now, let's apply the Born approximation to the given potential. Since the potential is zero outside the region -a \leq x \leq a and -a \leq y \leq a, the integral can be simplified to:

f(\theta) = -\frac{2m}{\hbar^2}A\int_{-a}^{a}\int_{-a}^{a}e^{i(K_xx+K_yy)}dxdy

Using the limits of integration and solving the integral, we get:

f(\theta) = -\frac{4am}{\hbar^2}\frac{A}{K_xK_y}\sin(K_xa)\sin(K_ya)

Substituting this into the expression for the differential cross-section, we get:

\frac{d\sigma}{d\Omega} = \left(\frac{4am}{\hbar^2}\frac{A}{K_xK_y}\sin(K_xa)\sin(K_ya)\right)^2

Now, we need to consider the limit as a goes to