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I am trying to determine the scattering cross section for the potential

$$ V= \frac{a}{r} + \frac{b}{r^2}$$

As I am completely stuck while determining the scattering angle (I don't see a way to simplify the equations I get), I wanted to know if it is actually possible to do so?

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You can obtain the trajectory starting from the conserved quantities, which are the total energy and the angular momentum. By parametrizing the motion using polar coordinates in the plane of the orbit (the orbit is in a plane owing to conservation of angular momentum) you get

$$E=\frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\theta}^2 +\frac{a}{r}+\frac{b}{r^2}$$

for the energy and

$$L_z = mr^2 \dot{\theta}$$

for the component of angular momentum perpendicular to the plane of the orbit (which is also the direction of the angular momentum). Using this last equation you can write the total energy as a function of radial coordinate only

$$E=\frac{1}{2} m \dot{r}^2 + \frac{a}{r} + \left(b + \frac{L_z^2}{2m} \right)\frac{1}{r^2} $$

Now you can parametrize $r$ as a function of $\theta$ and using

$$\dot{r}=\frac{dr}{dt} = \frac{dr}{d\theta} \dot{\theta}=\frac{L_z}{mr^2} \frac{dr}{d\theta}$$

you can rewrite the energy as

$$E=\frac{L_z^2}{2m}\left(\frac{1}{r^2} \frac{dr}{d\theta} \right)^2 + \frac{a}{r} + \left(b + \frac{L_z^2}{2m} \right)\frac{1}{r^2} $$

If you introduce the new coordinate $u=1/r$ this becomes

$$E=\frac{L_z^2}{2m}\left(\frac{du}{d\theta} \right)^2 + a u + \left(b + \frac{L_z^2}{2m} \right)u^2 $$

and from the conservation $dE/d\theta=0$ so you get the equation

$$\frac{d^2u}{d\theta^2} + \frac{am}{L_z^2} + \left(\frac{2 bm}{L_z^2} + 1 \right)u =0$$

which determine the trajectory.

From now on I will suppose $b>0$. This is a sufficient condition to avoid a trajectory which falls in the centre of attraction. In this case the equation for the trajectory is formally identical to the equation for an harmonic oscillator with an external constant force. The general solution is

$$u=A\cos \left(p\theta+\phi\right) + \kappa$$

where

$$p=\sqrt{1+\frac{2 bm}{L_z^2}}$$

$$\kappa=-\frac{\frac{am}{L_z^2}}{1+\frac{2 bm}{L_z^2}}$$ and $A$, $\phi$ are determined by initial conditions: in particular $A$ can be found as a function of $E$ and $L_z$ by substituting the solution for $u$ inside the energy. Restoring the radial coordinate and setting $\phi=0$ (which amounts to a rigid rotation of the trajectory) this means

$$r=\frac{1}{A\cos p\theta + \kappa}$$

The values of $\theta$ which sets to zero the denominator correspond to the directions of the ingoing and outgoing particle, and you can easily obtain the scattering angle as

$$\psi = \pi-\frac{2}{p} \arccos \left(-\frac{\kappa}{A} \right) $$

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