Question 11.P.4: Calculate the total cross section for the low-energy scatter...

Since the scattering is dominated at low energies by the s partial waves, l = 0, the Schrödinger equation for the radial function is given by

-\frac{\hbar^{2}}{2 m} \frac{d^{2} u(r)}{d r^{2}}-V_{0} u(r)=E u(r) \quad(r<a),             (11.164)

-\frac{\hbar^{2}}{2 m} \frac{d^{2} u(r)}{d r^{2}}=E u(r) \quad(r>a),                  (11.165)

where u(r) = r R(r). The solutions of these equations for positive energy states are

u(r)= \begin{cases}u_{1}(r)=A \sin \left(k_{1} r\right), & r<a, \\ u_{2}(r)=B \sin \left(k_{2} r+\delta_{0}\right), & r>a,\end{cases}                     (11.166)

where k_{1}^{2}=2 m\left(E+V_{0}\right) / \hbar^{2} and k_{2}^{2}=2 m E / \hbar^{2}. The continuity of u(r) and its first derivative, u^{\prime}(r)=d u(r) / d r, at r = a yield

\left.\frac{u_{2}(r)}{u_{2}^{\prime}(r)}\right|_{r=a}=\left.\frac{u_{1}(r)}{u_{1}^{\prime}(r)}\right|_{r=a} \Longrightarrow \frac{1}{k_{2}} \tan \left(k_{2} a+\delta_{0}\right)=\frac{1}{k_{1}} \tan \left(k_{1} a\right),                 (11.167)

which yields

\delta_{0}=-k_{2} a+\tan ^{-1}\left[\frac{k_{2}}{k_{1}} \tan \left(k_{1} a\right)\right].                    (11.168)

Since

\tan \left(k_{2} a+\delta_{0}\right)=\frac{\sin \left(k_{2} a\right) \cos \delta_{0}+\cos \left(k_{2} a\right) \sin \delta_{0}}{\cos \left(k_{2} a\right) \cos \delta_{0}-\sin \left(k_{2} a\right) \sin \delta_{0}}=\frac{\tan \left(k_{2} a\right)+\tan \delta_{0}}{1-\tan \left(k_{2} a\right) \tan \delta_{0}},            (11.169)

we can reduce Eq. (11.167) to

\tan \delta_{0}=\frac{k_{2} \tan \left(k_{1} a\right)-k_{1} \tan \left(k_{2} a\right)}{k_{1}+k_{2} \tan \left(k_{1} a\right) \tan \left(k_{2} a\right)}.                (11.170)

Using the relation \sin ^{2} \delta_{0}=1 /\left(1+1 / \tan ^{2} \delta_{0}\right), we can write

\sin ^{2} \delta_{0}=\left[1+\left(\frac{k_{1}+k_{2} \tan \left(k_{1} a\right) \tan \left(k_{2} a\right)}{k_{2} \tan \left(k_{1} a\right)-k_{1} \tan \left(k_{2} a\right)}\right)^{2}\right]^{-1},               (11.171)

which, when inserted 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)

leads to

\sigma_{0}=\frac{4 \pi}{k_{1}^{2}} \sin ^{2} \delta_{0}=\frac{4 \pi}{k_{1}^{2}}\left[1+\left(\frac{k_{1}+k_{2} \tan \left(k_{1} a\right) \tan \left(k_{2} a\right)}{k_{2} \tan \left(k_{1} a\right)-k_{1} \tan \left(k_{2} a\right)}\right)^{2}\right]^{-1}.                   (11.172)

If k_{2} a \ll 1 then (11.170) becomes \tan \delta_{0} \simeq \frac{\tan \left(k_{1} a\right)-k_{1} a}{k_{1} / k_{2}+k_{2} a \tan \left(k_{1} a\right)}, since \tan \left(k_{2} a\right) \simeq k_{2} a. Thus, if k_{2} a \ll 1 and if E (the scattering energy) is such that \tan \left(k_{1} a\right) \simeq k_{1} a, we have \tan \delta_{0}=0; hence there will be no s-wave scattering and the cross section vanishes. Note that if the square well potential is extended to a hard sphere potential, i.e., E → 0 and V_{0} \rightarrow \infty, equation (11.168) yields the phase shift of scattering from a hard sphere \delta_{0}=-k a, since \left(k_{2} / k_{1}\right) \tan \left(k_{1} a\right) \rightarrow 0.