Question

Find the scattering amplitude, in the Born approximation, for soft sphere scattering at arbitrary energy. Show that your formula reduces to Equation 11.82 in the low-energy limit.

Short answer

For the scattering amplitude,

$$\begin{gathered} r\left(\theta \right)=-\frac{2m{V}_0}{{\sout{h}}^2{k}^3}\left\{\sin \left(\mathbb{k}a\right)-ka\cos \left(\mathbb{k}a\right)\right\} \\ f\left(\theta \right)=-\frac{2}{3}\frac{m{V}_0{a}^3}{{\sout{h}}^2} \end{gathered}$$

Step-by-step solution

Given.

A soft sphere is defined by the potential

$V\left(r\right)=\left\{\begin{array}{ll}{V}_0& r\le a\\ 0& r>a\end{array}\right.$ …… (1)

Where$V_0>0$is a constant.

For a spherical symmetry potential, the scattering amplitude:

role="math" localid="1658305775955" $f\left(\theta \right)=-\frac{2m}{{\sout{h}}^2\mathbb{k}}{\int }_0^{\infty }rV\left(r\right)\sin \left(\mathbb{k}r\right)dr$

To find the scattering amplitude.

For the potential above, we get:

$f\left(\theta \right)=-\frac{2m}{\sout{h^{2}k}}{V}_0{\int }_0^{a}r\sin \left(kr\right)dr$

To do the integral, I used theintegral calculator, we get:

role="math" localid="1658305697027" $$\begin{gathered} f\left(\theta \right)=-\frac{2m{V}_0}{{\sout{h}}^2{\mathbb{k}}^3}{\overline{)\left[\frac{1}{k^2}\sin \left(\mathbb{k}r\right)-\frac{r}{k}\cos \left(\mathbb{k}r\right)\right]}}_0^a \\ f\left(\theta \right)=-\frac{2m{V}_0}{{\sout{h}}^2{\mathbb{k}}^3}\left\{\sin \left(\mathbb{k}a\right)-ka\cos \left(\mathbb{k}a\right)\right\} \end{gathered}$$

Where;

$\mathbb{k}=2k\sin \left(\frac{\theta }{2}\right)$

In the low- energy limit $\mathbb{k}a\ll 1$, and hence,

$\sin \left(\mathbb{k}a\right)\approx \mathbb{k}a-\frac{1}{3!}{\left(\mathbb{k}a\right)}^3\cos \left(\mathbb{k}a\right)=1-\frac{1}{2}{\left(\mathbb{k}a\right)}^2$

Therefore, the scattering amplitude.

Thus, the scattering amplitude is:

$$\begin{gathered} f\left(\theta \right)\approx \frac{2m{V}_0}{{\sout{h}}^2{\mathbb{k}}^3}\left[\mathbb{k}a-\frac{1}{6}{\left(\mathbb{k}a\right)}^3-\mathbb{k}a+\frac{1}{2}{\left(\mathbb{k}a\right)}^3\right] \\ f\left(\theta \right)=-\frac{2}{3}\frac{m{V}_0{a}^3}{{\sout{h}}^2} \end{gathered}$$