# Scattering from the Yukawa potential

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problems 11.11-11.12.

We’ve looked at the Yukawa potential as an example of the variational principle, so here we’ll look at scattering by a Yukawa potential, using the first Born approximation. The Yukawa potential in its general form is

$\displaystyle V\left(r\right)=\beta\frac{e^{-\mu r}}{r} \ \ \ \ \ (1)$

where ${\beta}$ and ${\mu}$ are constants. Since the potential is spherically symmetric, we can use the Born approximation in the form

$\displaystyle f\left(\theta\right)\approx-\frac{2m}{\hbar^{2}\kappa}\int_{0}^{\infty}V\left(r_{0}\right)r\sin\left(\kappa r\right)dr \ \ \ \ \ (2)$

where

 $\displaystyle \kappa$ $\displaystyle =$ $\displaystyle 2k\sin\frac{\theta}{2}\ \ \ \ \ (3)$ $\displaystyle k$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2mE}}{\hbar} \ \ \ \ \ (4)$

We get

 $\displaystyle f\left(\theta\right)$ $\displaystyle \approx$ $\displaystyle -\frac{2m\beta}{\hbar^{2}\kappa}\int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{2m\beta}{\hbar^{2}\left(\kappa^{2}+\mu^{2}\right)} \ \ \ \ \ (6)$

We did the integral using Maple, but if you want to do it by hand, you can do it with two integrations by parts:

 $\displaystyle \int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr$ $\displaystyle =$ $\displaystyle -\left.\frac{\cos\left(\kappa r\right)e^{-\mu r}}{\kappa}\right|_{0}^{\infty}-\frac{\mu}{\kappa}\int_{0}^{\infty}e^{-\mu r}\cos\left(\kappa r\right)dr\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\kappa}-\left.\frac{\sin\left(\kappa r\right)e^{-\mu r}}{\kappa}\right|_{0}^{\infty}-\frac{\mu^{2}}{\kappa^{2}}\int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\kappa}-\frac{\mu^{2}}{\kappa^{2}}\int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr\ \ \ \ \ (9)$ $\displaystyle \left(1+\frac{\mu^{2}}{\kappa^{2}}\right)\int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr$ $\displaystyle =$ $\displaystyle \frac{1}{\kappa}\ \ \ \ \ (10)$ $\displaystyle \int_{0}^{\infty}e^{-\mu r}\sin\left(\kappa r\right)dr$ $\displaystyle =$ $\displaystyle \frac{\kappa}{\mu^{2}+\kappa^{2}} \ \ \ \ \ (11)$

We can find the total cross section by integrating the differential cross section over solid angle:

 $\displaystyle \frac{d\sigma}{d\Omega}$ $\displaystyle =$ $\displaystyle \left|f\left(\theta\right)\right|^{2}\ \ \ \ \ (12)$ $\displaystyle \sigma$ $\displaystyle =$ $\displaystyle \frac{4m^{2}\beta^{2}}{\hbar^{4}}\int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi\;\frac{\sin\theta}{\left(\kappa^{2}+\mu^{2}\right)^{2}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{8\pi m^{2}\beta^{2}}{\hbar^{4}}\int_{0}^{\pi}d\theta\frac{\sin\theta}{\left(4k^{2}\sin^{2}\frac{\theta}{2}+\mu^{2}\right)^{2}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{16\pi m^{2}\beta^{2}}{\mu^{2}\hbar^{4}\left(\mu^{2}+4k^{2}\right)}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{16\pi m^{2}\beta^{2}}{\mu^{2}\hbar^{2}\left(\mu^{2}\hbar^{2}+8mE\right)} \ \ \ \ \ (16)$

Again, we did the integral using Maple. To do it by hand, we use the trig identity

$\displaystyle \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \ \ \ \ \ (17)$

followed by the substitution

 $\displaystyle u$ $\displaystyle =$ $\displaystyle \sin\frac{\theta}{2}\ \ \ \ \ (18)$ $\displaystyle du$ $\displaystyle =$ $\displaystyle \frac{1}{2}\cos\frac{\theta}{2}\; d\theta \ \ \ \ \ (19)$

This gives

 $\displaystyle \int_{0}^{\pi}d\theta\frac{\sin\theta}{\left(4k^{2}\sin^{2}\frac{\theta}{2}+\mu^{2}\right)^{2}}$ $\displaystyle =$ $\displaystyle 4\int_{0}^{1}\frac{u\; du}{\left(4k^{2}u^{2}+\mu^{2}\right)^{2}}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left.-\frac{4}{8k^{2}\left(4k^{2}u^{2}+\mu^{2}\right)}\right|_{0}^{1}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2k^{2}\left(4k^{2}+\mu^{2}\right)}+\frac{1}{2k^{2}\mu^{2}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{\left(4k^{2}+\mu^{2}\right)\mu^{2}} \ \ \ \ \ (23)$