Question 11.P.3: Consider the scattering of a particle of mass m from a hard ...

(a) As the scattering is dominated at low energies by s-waves, l = 0, the radial Schrödinger equation is

-\frac{\hbar^{2}}{2 m} \frac{d^{2} u(r)}{d r^{2}}=E u(r) \quad(r>a),             (11.155)

where u(r) = r R(r). The solutions of this equation are

u(r)= \begin{cases}u_{1}(r)=0, & r<a, \\ u_{2}(r)=A \sin \left(k r+\delta_{0}\right), & r>a,\end{cases}                  (11.156)

where k^{2}=2 m E / \hbar^{2}. The continuity of u(r) at r = a leads to

\sin \left(k a+\delta_{0}\right)=0 \quad \Longrightarrow \quad \tan \delta_{0}=-\tan (k a) \quad \Longrightarrow \quad \sin ^{2} \delta_{0}=\sin ^{2}(k a),             (11.157)

since \sin ^{2} \alpha=1 /\left(1+\cot ^{2} \alpha\right). The lowest value of the phase shift is \delta_{0}=-k a; it is negative, as it should be for a repulsive potential. An insertion of \sin ^{2} \delta_{0}=\sin ^{2}(k a) into (11.104)

\frac{d \sigma}{d \Omega}=\left|f_{0}\right|^{2}=\frac{1}{k^{2}} \sin ^{2} \delta_{0}, \quad \sigma=4 \pi\left|f_{0}\right|^{2}=\frac{4 \pi}{k^{2}} \sin ^{2} \delta_{0} \quad(l=0).              (11.104)

yields

\sigma_{0}=\frac{4 \pi}{k^{2}} \sin ^{2} \delta_{0}=\frac{4 \pi}{k^{2}} \sin ^{2}(k a).              (11.158)

For low energies, ka « 1, we have \sin (k a) \simeq k a and hence \sigma_{0} \simeq 4 \pi a^{2}, which is four times the classical value \pi a^{2}.

To obtain a numerical estimate of (11.158), we need first to calculate k^{2}. For this,we need simply to use the relation E=\hbar^{2} k^{2} /\left(2 m_{p}\right)=5 keV, since the proton moves as a free particle before scattering. Using m_{p} c^{2}=938.27 MeV and \hbar c=197.33 MeV fm, we have

k^{2}=\frac{2 m_{p} E}{\hbar^{2}}=\frac{2\left(m_{p} c^{2}\right) E}{(\hbar c)^{2}}=\frac{2(939.57 MeV )\left(5 \times 10^{-3} MeV \right)}{(197.33 MeV fm )^{2}}=0.24 \times 10^{-3} fm ^{-2}.       (11.159)

Thus k=0.0155 fm ^{-1}; the wave shift is given by \delta_{0}=-k a=-0.093 rad =-5.33^{\circ}.

Inserting these values into (11.158), we obtain

\sigma=\frac{4 \pi}{0.24 \times 10^{-3} fm ^{-2}} \sin ^{2}(5.33)=449.89 fm ^{2}=4.5 \text { barn }.              (11.160)

(b) In the high-energy limit, ka » 1, the number of partial waves contributing to the scattering is large. Assuming that l_{\max } \simeq k a, we may rewrite (11.102)

\sigma=\sum_{l=0}^{\infty} \sigma_{l}=\frac{4 \pi}{k^{2}} \sum_{l=0}^{\infty}(2 l+1) \sin ^{2} \delta_{l},            (11.102)

as

\sigma=\frac{4 \pi}{k^{2}} \sum_{l=0}^{l_{\max }}(2 l+1) \sin ^{2} \delta_{l}.               (11.161)

Since somany values of l contribute in this relation, we may replace \sin ^{2} \delta_{l} by its average value, \frac{1}{2}; hence

\sigma \simeq \frac{4 \pi}{k^{2}} \frac{1}{2} \sum_{l=0}^{l_{\max }}(2 l+1)=\frac{2 \pi}{k^{2}}\left(l_{\max }+1\right)^{2}.              (11.162)

where we have used \sum_{l=0}^{n}(2 l+1)=(n+1)^{2}. Since l_{\max } \gg 1 we have

\sigma \simeq \frac{2 \pi}{k^{2}} l_{\max }^{2}=\frac{2 \pi}{k^{2}}(k a)^{2}=2 \pi a^{2}.                (11.163)

Since a = 6 fm, we have \sigma \simeq 2 \pi(6 fm )^{2}=226.1 fm ^{2}=2.26 \text { barn }. This is almost half the value obtained in (11.160).

In conclusion, the cross section from a hard sphere potential is (a) four times the classical value, \pi a^{2}, for low-energy scattering and (b) twice the classical value for high-energy scattering.