(a)
H^{\prime}=-q E \cdot r =-q\left( E _{0} \cdot r \right)( k \cdot r ) \sin (\omega t) . \quad \text { Write } E _{0}=E_{0} \hat{n}, k =\frac{\omega}{c} \hat{k} . Then
H^{\prime}=-q \frac{E_{0} \omega}{c}(\hat{n} \cdot r )(\hat{k} \cdot r ) \sin (\omega t) . \quad H_{b a}^{\prime}=-\frac{q E_{0} \omega}{c}\langle b|(\hat{n} \cdot r )(\hat{k} \cdot r )| a\rangle \sin (\omega t) .
This is the analog to Eq. 11.40: H_{b a}^{\prime}=-q E_{0}\langle b|\hat{n} \cdot r | a\rangle \cos \omega t . The rest of the analysis is identical to the dipole case (except that it is \sin (\omega t) \text { instead of } \cos (\omega t) , but this amounts to resetting the clock, and clearly has no effect on the transition rate). We can skip therefore to Eq. 11.63, except for the factor of 1/3, which came from the averaging in Eq. 11.53:
H_{b a}^{\prime}=-\wp E_{0} \cos (\omega t), \quad \text { where } \wp \equiv q\left\langle\psi_{b}|z| \psi_{a}\right\rangle (11.40).
A=\frac{\omega_{0}^{3}| s |^{2}}{3 \pi \epsilon_{0} \hbar c^{3}} (11.63).
| \wp \cdot \hat{n}|_{ ave }^{2}=\frac{1}{4 \pi} \int| \wp |^{2} \cos ^{2} \theta \sin \theta d \theta d \phi
=\left.\frac{| \rho |^{2}}{4 \pi}\left(-\frac{\cos ^{3} \theta}{3}\right)\right|_{0} ^{\pi}(2 \pi)=\frac{1}{3}| \wp |^{2} (11.53).
A=\frac{\omega^{3}}{\pi \epsilon_{0} \hbar c^{3}} \frac{q^{2} \omega^{2}}{c^{2}}|\langle b|(\hat{n} \cdot r )(\hat{k} \cdot r )| a\rangle|^{2}=\frac{q^{2} \omega^{5}}{\pi \epsilon_{0} \hbar c^{5}}|\langle b|(\hat{n} \cdot r )(\hat{k} \cdot r )| a\rangle|^{2} .
(b) Let the oscillator lie along the x direction, so (\hat{n} \cdot r )=\hat{n}_{x} x \text { and } \hat{k} \cdot r =\hat{k}_{x} x \text {. For a transition from } n \text { to } n^{\prime} \text {, } we have
A=\frac{q^{2} \omega^{5}}{\pi \epsilon_{0} \hbar c^{5}}\left(\hat{k}_{x} \hat{n}_{x}\right)^{2}\left|\left\langle n^{\prime}\left|x^{2}\right| n\right\rangle\right|^{2} . From Example 2.5, \left\langle n^{\prime}\left|x^{2}\right| n\right\rangle=\frac{\hbar}{2 m \bar{\omega}}\left\langle n^{\prime}\left|\left(a_{+}^{2}+a_{+} a_{-}+a_{-} a_{+}+a_{-}^{2}\right)\right| n\right\rangle .
where \bar{\omega} is the frequency of the oscillator, not to be confused with ω, the frequency of the electromagnetic wave. Now, for spontaneous emission the final state must be lower in energy, so n^{\prime}<n and hence the only surviving term is a_{-}^{2} . Using Eq. 2.67:
\hat{a}_{+} \psi_{n}=\sqrt{n+1} \psi_{n+1}, \quad \hat{a}_{-} \psi_{n}=\sqrt{n} \psi_{n-1} (2.67).
\left\langle n^{\prime}\left|x^{2}\right| n\right\rangle=\frac{\hbar}{2 m \bar{\omega}}\left\langle n^{\prime}|\sqrt{n(n-1)}| n-2\right\rangle=\frac{\hbar}{2 m \bar{\omega}} \sqrt{n(n-1)} \delta_{n^{\prime}, n-2} .
Evidently transitions only go from |n\rangle \text { to }|n-2\rangle , and hence
\omega=\frac{E_{n}-E_{n-2}}{\hbar}=\frac{1}{\hbar}\left[\left(n+\frac{1}{2}\right) \hbar \bar{\omega}-\left(n-2+\frac{1}{2}\right) \hbar \bar{\omega}\right]=2 \bar{\omega} .
\left\langle n^{\prime}\left|x^{2}\right| n\right\rangle=\frac{\hbar}{m \omega} \sqrt{n(n-1)} \delta_{n^{\prime}, n-2} ; \quad R_{n \rightarrow n-2}=\frac{q^{2} \omega^{5}}{\pi \epsilon_{0} \hbar c^{5}}\left(\hat{k}_{x} \hat{n}_{x}\right)^{2} \frac{\hbar^{2}}{m^{2} \omega^{2}} n(n-1) .
It remains to calculate the average of \left(\hat{k}_{x} \hat{n}_{x}\right)^{2} . It’s easiest to reorient the oscillator along a direction \hat{r} , making angle θ with the z axis, and let the radiation be incident from the z direction \text { (so } \left.\hat{k}_{x} \rightarrow \hat{k}_{r}=\cos \theta\right) .
Averaging over the two polarizations (\hat{i} \text { and } \hat{j}):\left\langle\hat{n}_{r}^{2}\right\rangle=\frac{1}{2}\left(\hat{i}_{r}^{2}+\hat{j}_{r}^{2}\right)=\frac{1}{2}\left(\sin ^{2} \theta \cos ^{2} \phi+\sin ^{2} \theta \sin ^{2} \phi\right)= \frac{1}{2} \sin ^{2} \theta . Now average overall directions:
\left\langle\hat{k}_{r}^{2} \hat{n}_{r}^{2}\right\rangle=\frac{1}{4 \pi} \int \frac{1}{2} \sin ^{2} \theta \cos ^{2} \theta \sin \theta d \theta d \phi=\frac{1}{8 \pi} 2 \pi \int_{0}^{\pi}\left(1-\cos ^{2} \theta\right) \cos ^{2} \theta \sin \theta d \theta .
=\left.\frac{1}{4}\left[-\frac{\cos ^{3} \theta}{3}+\frac{\cos ^{5} \theta}{5}\right]\right|_{0} ^{\pi}=\frac{1}{4}\left(\frac{2}{3}-\frac{2}{5}\right)=\frac{1}{15} .
R=\frac{1}{15} \frac{q^{2} \hbar \omega^{3}}{\pi \epsilon_{0} m^{2} c^{5}} n(n-1) . Comparing Eq. 11.70: \frac{R(\text { forbidden })}{R(\text { allowed })}=\frac{2}{5}(n-1) \frac{\hbar \omega}{m c^{2}} .
A=\frac{n q^{2} \omega^{2}}{6 \pi \epsilon_{0} m c^{3}} (11.70).
For a nonrelativistic system, \hbar \omega \ll m c^{2} ; hence the term forbidden.
(c) If both the initial state and the final state have l = 0, the wave function is independent of angle \left(Y_{0}^{0}=\right. 1 / \sqrt{4 \pi}) , and the angular part of the integral is:
\langle a|(\hat{n} \cdot r )(\hat{k} \cdot r )| b\rangle=\cdots \int(\hat{n} \cdot r )(\hat{k} \cdot r ) \sin \theta d \theta d \phi=\cdots \frac{4 \pi}{3}(\hat{n} \cdot \hat{k}) . (Eq. 6.95).
But \hat{n} \cdot \hat{k}=0 , since electromagnetic waves are transverse. So R = 0 in this case, both for allowed and for forbidden transitions.