Question 11.P.5: Find the differential and total cross sections in the first ...

Since $V(\vec{r})$ is not spherically symmetric, the differential cross section can be obtained from (11.66):

$\frac{d \sigma}{d \Omega}=|f(\theta, \varphi)|^{2}=\frac{\mu^{2}}{4 \pi^{2} \hbar^{4}}\left|\int e^{i \vec{q} \cdot \vec{r}^{\prime}} V\left(\vec{r}^{\prime}\right) d^{3} r^{\prime}\right|^{2}$,            (11.66)

$\frac{d \sigma}{d \Omega}=\frac{m^{2}}{4 \pi^{2} \hbar^{4}}\left|\int V_{0}[\delta(\vec{r}-a \vec{k})+\delta(\vec{r}+a \vec{k})] e^{i \vec{q} \cdot \vec{r}} d^{3} r\right|^{2}=\frac{m^{2} V_{0}}{4 \pi^{2} \hbar^{4}}|I|^{2}$.                 (11.173)

Since $\delta(\vec{r} \pm a \vec{k})=\delta(x) \delta(y) \delta(z \pm a)$ we can write the integral I as

$=e^{i a q_{z}}+e^{-i a q_{z}}=2 \cos \left(a q_{z}\right)$.                  (11.174)

The calculation of qz is somewhat different from that shown in (11.67).

$q=\left|\overrightarrow{k_{0}}-\vec{k}\right|=\sqrt{k_{0}^{2}+k^{2}-2 k k_{0} \cos \theta}=k \sqrt{2\left(1-\cos ^{2} \theta\right)}=2 k \sin \left(\frac{\theta}{2}\right)$.                 (11.67)

Since the incident particle is initially traveling along the z-axis, and since it scatters elastically from the potential $V(\vec{r})$, the magnitudes of its momenta before and after collision are equal. So, as shown in Figure 11.8, we have $q_{z}=q \sin (\theta / 2)=2 k \sin ^{2}(\theta / 2)$, since $q=\left|\vec{k}_{0}-\vec{k}\right|=2 k \sin (\theta / 2)$. Thus, inserting $I=2 \cos \left(a q_{z}\right)=2 \cos \left[2 a k \sin ^{2}(\theta / 2)\right]$ into (11.173), we obtain

$\frac{d \sigma}{d \Omega}=\frac{m^{2} V_{0}}{\pi^{2} \hbar^{4}} \cos ^{2}\left(2 a k \sin ^{2} \frac{\theta}{2}\right)$.                (11.175)

The total cross section can be obtained at once from (11.175):

$=2 \pi \frac{m^{2} V_{0}}{\pi^{2} \hbar^{4}} \int_{0}^{\pi} \sin \theta \cos ^{2}\left(2 a k \sin ^{2} \frac{\theta}{2}\right) d \theta$,              (11.176)

which, when using the change of variable $x=2 a k \sin ^{2}(\theta / 2)$ with $d x=2 a k \sin (\theta / 2) \cos (\theta / 2) d \theta$, leads to

$=\frac{m^{2} V_{0}}{\pi a k \hbar^{4}}$.                        (11.177)