Question 11.P.2: Find the differential and total cross sections for the scatt...

In the case where the incident particles have small velocities, only the s-waves, l = 0, contribute to the scattering. The differential and total cross sections are given for l = 0 by (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)

\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).          (11.144)

We need now to find the phase shift \delta_{0}. For this, we need to consider the Schrödinger equation for the radial function:

-\frac{\hbar^{2}}{2 m} \frac{d^{2} u(r)}{d r^{2}}+\left[V_{0} \delta(r-a)+\frac{l(l+1) \hbar^{2}}{2 m r^{2}}\right] u(r)=E u(r),                 (11.145)

where u(r) = r (r). In the case of s states and r ≠ a, this equation yields

\frac{d^{2} u(r)}{d r^{2}}=-k^{2} u(r),                  (11.146)

where k^{2}=2 m E / \hbar^{2}. The acceptable solutions of this equation must vanish at r = 0 and be finite at r → ∞:

u(r)= \begin{cases}u_{1}(r)=A \sin (k r), & 0<r<a, \\ u_{2}(r)=B \sin \left(k r+\delta_{0}\right), & r>a.\end{cases}                 (11.147)

The continuity of u(r) at r=a, u_{2}(a)=u_{1}(a), leads to

B \sin \left(k a+\delta_{0}\right)=A \sin (k a).                       (11.148)

On the other hand, integrating (11.145) (with l = 0) from r = a – ε to r = a + ε, we obtain

-\frac{\hbar^{2}}{2 m} \int_{a-\varepsilon}^{a+\varepsilon} \frac{d^{2} u(r)}{d r^{2}} d r+V_{0} \int_{a-\varepsilon}^{a+\varepsilon} \delta(r-a) u(r) d r=E \int_{a-\varepsilon}^{a+\varepsilon} u(r) d r,                      (11.149)

and taking the limit ε → 0, we end up with

\left.\frac{d u_{2}(r)}{d r}\right|_{r=a}-\left.\frac{d u_{1}(r)}{d r}\right|_{r=a}-\frac{2 m V_{0}}{\hbar^{2}} u_{2}(a)=0 .                  (11.150)

An insertion of u_{1}(r) and u_{2}(r) as given by (11.147) into (11.150) leads to

B\left[k \cos \left(k a+\delta_{0}\right)-\frac{2 m V_{0}}{\hbar^{2}} \sin \left(k a+\delta_{0}\right)\right]=A k \cos (k a).                (11.151)

Dividing (11.151) by (11.148), we obtain

k \cot \left(k a+\delta_{0}\right)-\frac{2 m V_{0}}{\hbar^{2}}=k \cot (k a) \quad \Rightarrow \quad \tan \left(k a+\delta_{0}\right)=\left[\frac{1}{\tan (k a)}+\frac{2 m V_{0}}{k \hbar^{2}}\right]^{-1}.       (11.152)

This equation shows that, when there is no scattering potential, V_{0}=0, the phase shift is zero, since \tan \left(k a+\delta_{0}\right)=\tan (k a). In this case, equations (11.103)

f_{0}=\frac{1}{k} e^{i \delta_{0}} \sin \delta_{0} \quad(l=0),                   (11.103)

and (11.104) imply that the scattering amplitude and the cross sections all vanish.

If the incident particles have small velocities, ka « 1, we have \tan (k a) \simeq k a and \tan (k a+\left.\delta_{0}\right) \simeq \tan \left(\delta_{0}\right). In this case, equation (11.152) yields

\tan \delta_{0} \simeq \frac{k a}{1+2 m V_{0} a / \hbar^{2}} \Rightarrow \sin ^{2} \delta_{0} \simeq \frac{k^{2} a^{2}}{k^{2} a^{2}+\left(1+2 m V_{0} a / \hbar^{2}\right)^{2}}.                (11.153)

Inserting this relation into (11.144), we obtain

\frac{d \sigma}{d \Omega}_{0} \simeq \frac{a^{2}}{k^{2} a^{2}+\left(1+2 m V_{0} a / \hbar^{2}\right)^{2}}, \quad \sigma_{0} \simeq \frac{4 \pi a^{2}}{k^{2} a^{2}+\left(1+2 m V_{0} a / \hbar^{2}\right)^{2}}.            (11.154)