Question 11.P.7: Consider the elastic scattering of an electron from a hydrog...

(a) By analogy with (11.63)

\frac{d \sigma}{d \Omega}=|f(\theta, \varphi)|^{2}=\frac{\mu^{2}}{4 \pi^{2} \hbar^{4}}\left|\int e^{-i \vec{k} \cdot \vec{r}^{\prime}} \hat{V}\left(\vec{r}^{\prime}\right) \psi\left(\vec{r}^{\prime}\right) d^{3} r^{\prime}\right|^{2}=\frac{\mu^{2}}{4 \pi^{2} \hbar^{4}}|\langle\phi|\hat{V}| \psi\rangle|^{2} .         (11.63)

we may write the differential cross section for this process as

\frac{d \sigma}{d \Omega}=|f(\theta)|^{2}=\left|-\frac{\mu}{2 \pi \hbar^{2}}\left\langle\Psi_{f}|\hat{V}| \Psi_{i}\right\rangle\right|^{2},               (11.182)

where \mu \simeq m_{e} / 2, since this problem can be viewed as the scattering of a particle whose reduced mass is half that of the electron. Assuming the atom to be very massive and that it remains in its ground state after scattering, the initial and final states of the system (incident electron plus the atom) are given by \Psi_{i}\left(\vec{r}, \vec{k}_{0}, \vec{r}^{\prime}\right)=e^{i \vec{k}_{0} \cdot \vec{r}} \psi_{0}\left(\vec{r}^{\prime}\right) and \Psi_{f}\left(\vec{r}, \vec{k}, \vec{r}^{\prime}\right)=e^{i \vec{k} \cdot \vec{r}} \psi_{0}\left(\vec{r}^{\prime}\right), where e^{i \vec{k}_{0} \cdot \vec{r}^{\prime}} and e^{i \vec{k} \cdot \vec{r}^{\prime}} are the states of the incident electron before and after scattering, and \psi_{0}\left(\vec{r}^{\prime}\right)=\left(\pi a_{0}^{3}\right)^{-1 / 2} e^{-r^{\prime} / a_{0}} is the atom’s wave function. We have assumed here that the nucleus is located at the origin and that the position vectors of the incident electron and the atom’s electron are given by \vec{r} and \vec{r}^{\prime}, respectively. Since the incident electron experiences an attractive Coulomb interaction -e^{2} / r with the nucleus and a repulsive interaction e^{2} /\left|\vec{r}-\vec{r}^{\prime}\right| with the hydrogen’s electron, we have

f(\theta)=-\frac{\mu}{2 \pi \hbar^{2}} \int d^{3} r e^{i \vec{q} \cdot \vec{r}} \int d^{3} r^{\prime} \psi_{0}^{*}\left(\vec{r}^{\prime}\right)\left[-\frac{e^{2}}{r}+\frac{e^{2}}{\left|\vec{r}-\vec{r}^{\prime}\right|}\right] \psi_{0}\left(\vec{r}^{\prime}\right),                 (11.183)

with q=\left|\vec{k}_{0}-\vec{k}\right|=2 k \sin (\theta / 2), since k=k_{0} (elastic scattering). Using \int_{0}^{\infty} \sin (q r) d r=1 / q (see (11.76)),

=\frac{1}{2 i} \lim _{\lambda \rightarrow 0}\left[\frac{1}{\lambda-i q}-\frac{1}{\lambda+i q}\right]=\frac{1}{q} .              (11.76)

and since \int_{0}^{\pi} e^{i q r \cos \theta} \sin \theta d \theta=\int_{-1}^{1} e^{i q r x} d x=(2 / q r) \sin (q r), we obtain the following relation:

\int d^{3} r \frac{e^{i \vec{q} \cdot \vec{r}}}{r}=\int_{0}^{\infty} r d r \int_{0}^{\pi} e^{i q r \cos \theta} \sin \theta d \theta \int_{0}^{2 \pi} d \varphi=\frac{4 \pi}{q} \int_{0}^{\infty} d r \sin (q r)=\frac{4 \pi}{q^{2}},               (11.184)

which, when inserted into (11.183) and since \int d^{3} r^{\prime} \psi_{0}^{*}\left(\vec{r}^{\prime}\right) \psi_{0}\left(\vec{r}^{\prime}\right)=1, leads to

f(\theta)=\frac{\mu}{2 \pi \hbar^{2}}\left[\frac{4 \pi e^{2}}{q^{2}}-\int d^{3} r e^{i \vec{q} \cdot \vec{r}} \int d^{3} r^{\prime} \psi_{0}^{*}\left(\vec{r}^{\prime}\right) \frac{e^{2}}{\left|\vec{r}-\vec{r}^{\prime}\right|} \psi_{0}\left(\vec{r}^{\prime}\right)\right].           (11.185)

By analogy with (11.184), we have \int d^{3} r e^{i \vec{q} \cdot\left|\vec{r}-\vec{r}^{\prime}\right|} /\left|\vec{r}-\vec{r}^{\prime}\right|=4 \pi / q^{2}; hence we can reduce the integral in (11.185) to

=\frac{4 \pi e^{2}}{q^{2}} \int d^{3} r^{\prime} \psi_{0}^{*}\left(\vec{r}^{\prime}\right) e^{i \vec{q} \cdot \vec{r}^{\prime}} \psi_{0}\left(\vec{r}^{\prime}\right).             (11.186)

The remaining integral of (11.186) can, in turn, be written as

=\frac{4}{q a_{0}^{3}} \int_{0}^{\infty} r^{\prime} e^{-2 r^{\prime} / a_{0}} \sin \left(q r^{\prime}\right) d r^{\prime}=\left(1+\frac{a_{0}^{2} q^{2}}{4}\right)^{-2},             (11.187)

where we have used the expression for \int_{0}^{\infty} r e^{-a r} \sin (q r) d r calculated in (11.130).

=-\frac{4 V_{0} \mu a}{\hbar^{2}} \frac{1}{\left(a^{2}+q^{2}\right)^{2}}=-\frac{4 V_{0} \mu a}{\hbar^{2}} \frac{1}{\left(a^{2}+4 k^{2} \sin ^{2}(\theta / 2)\right)^{2}},              (11.130)

Inserting (11.187) into (11.186), and the resulting expression into (11.185), we obtain

f(\theta)=\frac{2 \mu e^{2}}{\hbar^{2} q^{2}}\left[1-\left(1+\frac{a_{0}^{2} q^{2}}{4}\right)^{-2}\right]=\frac{\mu e^{2}}{2 k^{2} \hbar^{2} \sin ^{2}(\theta / 2)}\left[1-\left(1+a_{0}^{2} k^{2} \sin ^{2} \frac{\theta}{2}\right)^{-2}\right].               (11.188)

We can thus reduce (11.182) to

\frac{d \sigma}{d \Omega}=\frac{4 \mu^{2} e^{4}}{\hbar^{4} q^{4}}\left[1-\left(1+\frac{a_{0}^{2} q^{2}}{4}\right)^{-2}\right]^{2}=\frac{\mu^{2} e^{4}}{4 k^{4} \hbar^{4} \sin ^{4} \frac{\theta}{2}}\left[1-\left(1+a_{0}^{2} k^{2} \sin ^{2}(\theta / 2)\right)^{-2}\right]^{2},              (11.189)

with q=2 k \sin (\theta / 2).

(b) (i) If the electrons are in their spin singlet state (antisymmetric), the spatial wave function must be symmetric; hence the differential cross section is

\frac{d \sigma_{S}}{d \Omega}=|f(\theta)+f(\pi-\theta)|^{2},                  (11.190)

where f (θ) is given by (11.188) and

f(\pi-\theta)=\frac{2 \mu e^{2}}{\hbar^{2} q^{2}}\left[1-\left(1+\frac{a_{0}^{2} q^{2}}{4}\right)^{-2}\right]=\frac{\mu e^{2}}{2 k^{2} \hbar^{2} \cos ^{2} \frac{\theta}{2}}\left[1-\left(1+a_{0}^{2} k^{2} \cos ^{2}(\theta / 2)\right)^{-2}\right],     (11.191)

since \sin (\pi-\theta / 2)=\cos (\theta / 2).

(ii) If, however, the electrons are in their spin triplet state, the spatial wave function must be antisymmetric; hence

\frac{d \sigma_{A}}{d \Omega}=|f(\theta)-f(\pi-\theta)|^{2}.               (11.192)

(iii) Finally, if the electrons are unpolarized, the differential cross section must be a mixture of (11.191) and (11.192):

\frac{d \sigma}{d \Omega}=\frac{1}{4} \frac{d \sigma_{S}}{d \Omega}+\frac{3}{4} \frac{d \sigma_{A}}{d \Omega}=\frac{1}{4}|f(\theta)+f(\pi-\theta)|^{2}+\frac{3}{4}|f(\theta)-f(\pi-\theta)|^{2}.             (11.193)