Daily Archives: Tue, 16 June 2015

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)