Exercises

1. Consider the two-state system examined in Section 8.3. Suppose that

   $$\langle 1 \vert\,H_1\,\vert 1\rangle = e_{11},$$

   $$\langle 2 \vert\,H_1\,\vert 2\rangle = e_{22},$$

   $$\langle 1\vert\,H_1\,\vert 2\rangle = \langle 2\vert\,H_1\,\vert 1\rangle^\ast = \frac{1}{2}\,\gamma\,\hbar\,\exp(\,{\rm i}\,\omega\,t),$$

   
   
   where $e_{11}$ , $e_{22}$ , $\gamma$ , and $\omega$ are real. Show that

   $${\rm i}\,\frac{d\hat{c}_1}{dt} = \frac{\gamma}{2}\,\exp\left[+{\rm i}\,(\omega-\hat{\omega}_{21})\,t\right]\hat{c}_2,$$

   $${\rm i}\,\frac{d\hat{c}_2}{dt} = \frac{\gamma}{2}\,\exp\left[-{\rm i}\,(\omega-\hat{\omega}_{21})\,t\right]\hat{c}_1,$$

   
   
   where $\hat{c}_1 = c_1\,\exp(\,{\rm i}\,e_{11}\,t/\hbar)$ , $\hat{c}_2 = c_2\,\exp(\,{\rm i}\,e_{22}\,t/\hbar)$ , and
   $$\hat{\omega}_{21} = \frac{E_2+e_{22}-E_1-e_{11}}{\hbar}.$$
   Hence, deduce that if the system is definitely in state 1 at time $t=0$ then the probability of finding it in state 2 at some subsequent time, $t$ , is
   $$P_2(t) = \frac{\gamma^{\,2}}{ \gamma^{\,2} + (\omega-\hat{\omega... ...gamma^{\,2}+ (\omega-\hat{\omega}_{21})^{\,2}\right]^{1/2} \frac{t}{2}\right).$$

2. Consider an atomic nucleus of spin-$s$ and $g$ -factor $g$ placed in the magnetic field
   $${\bf B} = B_0\, {\bf e}_z + B_1\left[\cos(\omega\, t) \,{\bf e}_x- \sin(\omega \,t) \,{\bf e}_y\right],$$
   where $B_1\ll B_0$ . Let $\vert s, m\rangle$ be a properly normalized simultaneous eigenket of $S^{\,2}$ and $S_z$ , where ${\bf S}$ is the nuclear spin. Thus, $S^{\,2}\,\vert s,m\rangle = s\,(s+1)\,\hbar^{\,2}\,\vert s,m\rangle$ and $S_z \,\vert s,m\rangle = m\,\hbar\,\vert s,m\rangle$ , where $-s\leq m\leq s$ . Furthermore, the instantaneous nuclear spin state is written
   $$\vert A \rangle = \sum_{m=-s,s} c_m(r)\,\vert s,m\rangle,$$
   where $\sum_{m=-s,s}\vert c_m\vert^{\,2}=1$ .
   1. Demonstrate that
      $$\begin{multline} \frac{dc_m}{dt} = \frac{{\rm i}\,\gamma}{2}\left([s\,(s+1)-m\,(... ...{\rm e}^{-{\rm i}\,(\omega-\omega_0)\,t}\,c_{m+1}\right)\nonumber \end{multline}$$
      
      
      for $-s\leq m\leq s$ , where $\omega_0=g\,\mu_N\,B_0/\hbar$ and $\gamma= g\,\mu_N\,B_1/\hbar$ .

   2. Consider the case $s=1/2$ . Demonstrate that if $\omega=\omega_0$ and $c_{1/2}(0)= 1$ then

      $$c_{1/2}(t)= \cos(\gamma\,t/2), c_{-1/2}(t)= {\rm i}\,\sin(\gamma\,t/2).$$

      
      

   3. Consider the case $s=1$ . Demonstrate that if $\omega=\omega_0$ and $c_{1}(0)= 1$ then

      $$c_{1}(t) = \cos^2(\gamma\,t/2),$$

      $$c_0(t) = {\rm i}\sqrt{2}\,\cos(\gamma\,t/2)\,\sin(\gamma\,t/2),$$

      $$c_{-1}(t) = -\sin^2(\gamma\,t/2).$$

      
      

   4. Consider the case $s=3/2$ . Demonstrate that if $\omega=\omega_0$ and $c_{3/2}(0)= 1$ then

      $$c_{3/2}(t) = \cos^3(\gamma\,t/2),$$

      $$c_{1/2}(t) = {\rm i}\sqrt{3}\,\cos(\gamma\,t/2)\,\sin^2(\gamma\,t/2),$$

      $$c_{-1/2}(t) = -\sqrt{3}\,\cos^2(\gamma\,t/2)\,\sin(\gamma\,t/2),$$

      $$c_{-3/2}(t) =-{\rm i}\,\sin^3(\gamma\,t/2).$$

      
      

3. Derive Equation (8.45).

4. Derive Equation (8.83).

5. If
   $$P_{i\rightarrow i}(t)=\exp\left(-\frac{t}{\tau_i}\right)$$
   is the probability that a system initially in state $\vert i\rangle$ at time $t=0$ is still in that state at some subsequent time, deduce that the mean lifetime of the state is $\tau_i$ .

6. Demonstrate that ${\bf p}\cdot{\bf A}={\bf A}\cdot{\bf p}$ when $\nabla\cdot{\bf A} = 0$ , where ${\bf p}$ is the momentum operator, and ${\bf A}({\bf x})$ is a real function of the position operator, ${\bf x}$ . Hence, show that the Hamiltonian (8.128) is Hermitian.

7. Demonstrate that
   $$\left\langle \vert\mbox{\boldmath $\epsilon$}\cdot{\bf d}_{if}\vert^{\,2}\right\rangle = \frac{{\bf d}_{if}^{\,\ast}\cdot {\bf d}_{if}}{3},$$
   where the average is taken over all possible directions of the unit vector $\epsilon$ .

8. Demonstrate that
   $$\oint d{\mit\Omega}\,\frac{dw_{i\rightarrow f}^{\,\rm spn}}{d{\mi... ...a}}= \frac{4\,\alpha\,\omega_{if}^{\,3}\,\vert d_{if}\vert^{\,2}}{3\,c^{\,2}},$$
   where $dw_{i\rightarrow f}^{\,\rm spn}/d{\mit\Omega}$ is specified by Equation (8.165). [Hint: Write

   $${\bf n} = (\sin\theta\,\cos\varphi,\,\sin\theta\,\sin\varphi,\,\cos\theta),$$

   $$\mbox{\boldmath$\epsilon$} _1 =(\cos\theta\,\cos\varphi,\,\cos\theta\,\sin\varphi,\,-\sin\theta),$$

   $$\mbox{\boldmath$\epsilon$} _2 =(-\sin\varphi,\,\cos\varphi,\,0),$$

   
   
   where $d{\mit\Omega} = \sin\theta\,d\theta\,d\varphi$ .]

9. Demonstrate that a spontaneous transition between two atomic states of zero orbital angular momentum is absolutely forbidden. (Actually, a spontaneous transition between two zero orbital angular momentum states is possible via the simultaneous emission of two photons, but takes place at a very slow rate [56,15,102].)

10. Find the selection rules for the matrix elements $\langle n,l,m\vert\,x\,\vert n',l',m'\rangle$ and $\langle n,l,m\vert\,y\,\vert n',l',m'\rangle$ to be non-zero. Here, $\vert n,l,m\rangle$ denotes an energy eigenket of a hydrogen-like atom corresponding to the conventional quantum numbers, $n$ , $l$ , and $m$ .

11. The Hamiltonian of an electron in a hydrogen-like atom is written
    $$H = \frac{p^{\,2}}{2\,m_e} + V(r),$$
    where $r=\vert{\bf x}\vert$ . Let $\vert n\rangle$ denote a properly normalized energy eigenket belonging to the eigenvalue $E_n$ . By calculating $[[H, x_i], \,x_i]$ , derive the Thomas-Reiche-Kuhn sum rule,
    $$\sum_m F_{nm}= 1,$$
    where
    $$F_{nm}= \frac{2\,m_e\,\omega_{mn}}{\hbar}\,\vert\langle n\vert\,x_i\,\vert m\rangle\vert^{\,2},$$
    and $\omega_{mn} = (E_m-E_n)/\hbar$ , and the sum is over all energy eigenstates. Here, $x_i$ stands for either $x$ , $y$ , or $z$ . (See Exercise 6.)

12. 1. Show that the only non-zero $1s\rightarrow 2p$ electric dipole matrix elements for the hydrogen atom take the values

       $$\langle 1,0,0\vert\,x\,\vert 2,1,\pm 1\rangle = \mp\frac{2^{\,7}}{3^{\,5}}\,a_0,$$

       $$\langle 1,0,0\vert\,y\,\vert 2,1,\pm 1\rangle =- {\rm i}\,\frac{2^{\,7}}{3^{\,5}}\,a_0,$$

       $$\langle 1,0,0\vert\,z\,\vert 2,1,0\rangle = \sqrt{2}\,\frac{2^{\,7}}{3^{\,5}}\,a_0,$$

       
       
       where $a_0$ is the Bohr radius.

    2. Likewise, show that the only non-zero $1s\rightarrow 3p$ electric dipole matrix elements take the values

       $$\langle 1,0,0\vert\,x\,\vert 3,1,\pm 1\rangle = \mp\frac{3^{\,3}}{2^{\,7}}\,a_0,$$

       $$\langle 1,0,0\vert\,y\,\vert 3,1,\pm 1\rangle =- {\rm i}\,\frac{3^{\,3}}{2^{\,7}}\,a_0,$$

       $$\langle 1,0,0\vert\,z\,\vert 3,1,0\rangle = \sqrt{2}\,\frac{3^{\,3}}{2^{\,7}}\,a_0.$$

       
       

    3. Finally, show that the only non-zero $2s\rightarrow 2p$ electric dipole matrix elements take the values

       $$\langle 2,0,0\vert\,x\,\vert 2,1,\pm 1\rangle = \pm\,\frac{3}{\sqrt{2}}\,a_0,$$

       $$\langle 2,0,0\vert\,y\,\vert 2,1,\pm 1\rangle ={\rm i}\,\frac{3}{2}\,a_0,$$

       $$\langle 2,0,0\vert\,z\,\vert 2,1,0\rangle = -3\,a_0.$$

       
       

13. 1. Demonstrate that the spontaneous decay rate (via an electric dipole transition) from any $2p$ state to a $1s$ state of a hydrogen atom is
       $$w_{2p\rightarrow 1s} = \left(\frac{2}{3}\right)^8\alpha^{\,5}\,\frac{m_e\,c^{\,2}}{\hbar}=6.27\times 10^8\,{\rm s}^{-1},$$
       where $\alpha$ is the fine structure constant. Hence, deduce that the relative natural width of the associated spectral line is
       $$\frac{{\mit\Delta}\lambda}{\lambda} = 4.0\times 10^{-8},$$
       where $\lambda$ denotes wavelength.

    2. Likewise, show that the spontaneous decay rate from any $3p$ state to a $1s$ state is
       $$w_{3p\rightarrow 1s} = \frac{\alpha^{\,5}}{3\,2^{\,5}}\,\frac{m_e\,c^{\,2}}{\hbar}=1.67\times 10^8\,{\rm s}^{-1}.$$
       Hence, deduce that the relative natural width of the associated spectral line is
       $$\frac{{\mit\Delta}\lambda}{\lambda} = 9.1\times 10^{-9}.$$

14. 1. Demonstrate that the net oscillator strength for $1s\rightarrow 2p$ transitions in a hydrogen atom is
       $$F_{1s\rightarrow 2p}= \frac{2^{\,13}}{3^{\,9}} = 0.4162,$$
       irrespective of the polarization of the radiation.
    2. Likewise, show that the net oscillator strength for $1s\rightarrow 3p$ transitions is
       $$F_{1s\rightarrow 3p}= \frac{3^{\,4}}{2^{\,11}} = 0.07910,$$
       irrespective of the polarization of the radiation.

15. Taking electron spin into account, the wavefunctions of the $(1s)_{1/2}$ states of a hydrogen atom are written $R_{1\,0}\,{\cal Y}_{0\,m}^{\,1/2}$ where $m=\pm 1/2$ . Here, the $R_{n\,l}$ are standard radial wavefunctions, and the ${\cal Y}_{l\,m_j}^{\,j}$ are spin-angular functions. Likewise, the wavefunctions of the $(2p)_{1/2}$ states are written $R_{2\,1}\,{\cal Y}_{1\,m_j}^{\,1/2}$ , where $m_j=\pm 1/2$ . Finally, the wavefunctions of the $(2p)_{3/2}$ states are written $R_{2\,1}\,{\cal Y}_{1\,m_j}^{\,3/2}$ , where $m_j=-3/2$ , $-1/2$ , $1/2$ , $3/2$ . As a consequence of spin-orbit interaction, the four $(2p)_{3/2}$ states lie at a slightly higher energy than the two $(2p)_{1/2}$ states. Demonstrate that the spontaneous decay rates, due to electric dipole transitions, between the various $2p$ and $1s$ states are

    $$w\left({\cal Y}_{1\,\,+3/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,+1/2}^{\,1/2}\right) = w_0,$$

    $$w\left({\cal Y}_{1\,\,+3/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,-1/2}^{\,1/2}\right) =0,$$

    $$w\left({\cal Y}_{1\,\,+1/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,+1/2}^{\,1/2}\right) = \frac{2}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,+1/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,-1/2}^{\,1/2}\right) = \frac{1}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,-1/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,+1/2}^{\,1/2}\right) = \frac{1}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,-1/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,-1/2}^{\,1/2}\right) = \frac{2}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,-3/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,+1/2}^{\,1/2}\right) =0,$$

    $$w\left({\cal Y}_{1\,\,-3/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,-1/2}^{\,1/2}\right) = w_0,$$

    $$w\left({\cal Y}_{1\,\,+1/2}^{\,1/2}\rightarrow {\cal Y}_{0\,\,+1/2}^{\,1/2}\right) = \frac{1}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,+1/2}^{\,1/2}\rightarrow {\cal Y}_{0\,\,-1/2}^{\,1/2}\right) = \frac{2}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,-1/2}^{\,1/2}\rightarrow {\cal Y}_{0\,\,+1/2}^{\,1/2}\right) = \frac{2}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,-1/2}^{\,1/2}\rightarrow {\cal Y}_{0\,\,-1/2}^{\,1/2}\right) = \frac{1}{3}\,w_0,$$

    
    
    where
    $$w_0= \left(\frac{2}{3}\right)^8\alpha^{\,5}\,\frac{m_e\,c^{\,2}}{\hbar}=6.27\times 10^8\,{\rm s}^{\,-1}.$$
    Here, the states have been labeled by their spin-angular functions. Hence, deduce that for an ensemble of hydrogen atoms in thermal equilibrium, the spectral line associated with $(2p)_{3/2}\rightarrow (1s)_{1/2}$ transitions is twice as bright as that associated with $(2p)_{1/2}\rightarrow (1s)_{1/2}$ transitions.

16. Taking electron spin into account, the wavefunctions of the $(2s)_{1/2}$ states of a hydrogen atom are written $R_{2\,0}\,{\cal Y}_{0\,m}^{\,1/2}$ where $m=\pm 1/2$ . The wavefunctions of the $(2p)_{3/2}$ states are specified in the previous exercise. Demonstrate that the spontaneous decay rates, due to electric dipole transitions, between the various $(2p)_{3/2}$ and $(2s)_{1/2}$ states are

    $$w\left({\cal Y}_{1\,\,+3/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,+1/2}^{\,1/2}\right) = w_0,$$

    $$w\left({\cal Y}_{1\,\,+3/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,-1/2}^{\,1/2}\right) =0,$$

    $$w\left({\cal Y}_{1\,\,+1/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,+1/2}^{\,1/2}\right) = \frac{2}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,+1/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,-1/2}^{\,1/2}\right) = \frac{1}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,-1/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,+1/2}^{\,1/2}\right) = \frac{1}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,-1/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,-1/2}^{\,1/2}\right) = \frac{2}{3}\,w_0,$$

    $$w\left({\cal Y}_{1\,\,-3/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,+1/2}^{\,1/2}\right) =0,$$

    $$w\left({\cal Y}_{1\,\,-3/2}^{\,3/2}\rightarrow {\cal Y}_{0\,\,-1/2}^{\,1/2}\right) = w_0,$$

    
    
    where
    $$w_0= \frac{3\,\alpha^{\,11}}{2^{\,13}}\,\frac{m_e\,c^{\,2}}{\hbar}=3.55\times 10^{-6}\,{\rm s}^{\,-1}.$$
    Here, the states have been labeled by their spin-angular functions.

17. 1. Consider the $2,1,\pm 1\rightarrow 1,0,0$ electric dipole transition of a hydrogen atom. Show that the angular distribution of spontaneously emitted photons is
       $$\frac{dw_{2,1,\pm1\rightarrow 1,0,0}^{\,\rm spn}}{d{\mit\Omega}} ... ...4}\,\alpha^{\,5}}{3^{\,7}\,\pi}\,\frac{m_e\,c^{\,2}}{\hbar}\,(1+\cos^2\theta),$$
       where $d{\mit\Omega} = \sin\theta\,d\theta\,d\varphi$ , and the photon's direction of motion is parallel to ${\bf n} = (\sin\theta\,\cos\varphi,\,\sin\theta\,\sin\varphi,\,\cos\theta)$ .
    2. Consider the $2,1,0\rightarrow 1,0,0$ electric dipole transition of a hydrogen atom. Show that the angular distribution of spontaneously emitted photons is
       $$\frac{dw_{2,1,0\rightarrow 1,0,0}^{\,\rm spn}}{d{\mit\Omega}} = \... ...^{\,5}\,\alpha^{\,5}}{3^{\,7}\,\pi}\,\frac{m_e\,c^{\,2}}{\hbar}\,\sin^2\theta.$$

    3. Hence, show that for an ensemble of hydrogen atoms in thermal equilibrium, the angular distribution of spontaneously emitted photons associated with the $2p\rightarrow 1s$ electric dipole transition is
       $$\frac{dw_{2p\rightarrow 1s}^{\,\rm spn}}{d{\mit\Omega}} = \frac{2^{\,6}\,\alpha^{\,5}}{3^{\,8}\,\pi}\,\frac{m_e\,c^{\,2}}{\hbar}.$$

18. The properly normalized $1s$ triplet and singlet state kets of a hydrogen atom can be written

    $$\vert 1,1\rangle = \vert+\rangle_p\,\vert+\rangle_e,$$

    $$\vert 1,0\rangle =\frac{1}{\sqrt{2}}\left(\vert+\rangle_p\,\vert-\rangle_e+\vert-\rangle_p\,\vert+\rangle_e\right),$$

    $$\vert 1,-1\rangle = \vert-\rangle_p\,\vert-\rangle_e,$$

    
    
    and
    $$\vert,0\rangle = \frac{1}{\sqrt{2}}\left(\vert+\rangle_p\,\vert-\rangle_e-\vert-\rangle_p\,\vert+\rangle_e\right),$$
    respectively. Here, the kets on the left are $\vert j,m_j\rangle$ kets, where $j$ and $m_j$ are the conventional quantum numbers that determine the overall angular momentum, and the projection of the overall angular momentum along the $z$ -axis, respectively. Finally, $\vert\pm \rangle_p$ and $\vert\pm \rangle_e$ are the properly normalized spin-up and spin-down states for the proton and the electron, respectively. As explained in Section 7.9, the triplet states have slightly higher energies than the singlet state, as a consequence of spin-spin coupling between the proton and the electron. Spontaneous magnetic dipole transitions between the triplet and singlet states occur via the interaction of the magnetic component of the emitted photon and the electron magnetic moment. (The magnetic moment of the proton is much smaller than that of the electron, and consequently does not play a significant role in this transition.) In the following, the initial state, $i$ , corresponds to one of the triplet states, and the final state, $f$ , corresponds to the singlet state.
    1. Demonstrate that
       $$\omega_{if} = \frac{E_i-E_f}{\hbar} = \frac{4}{3}\,g_p\,\alpha^{\,4}\,\frac{m_e}{m_p}\,\frac{m_e\,c^{\,2}}{\hbar},$$
       where $E_i$ and $E_f$ are the energies of the initial and final states, and $g_p$ is the proton $g$ -factor.
    2. Let ${\bf M}_{if} = \langle i\vert\,2\,{\bf S}_e\,\vert f\rangle$ be the magnetic dipole matrix element, where ${\bf S}_e$ is the electron spin. Show that
       $${\bf M}_{if} = \frac{\hbar}{\sqrt{2}}\left({\bf e}_x-{\rm i}\,{\bf e}_y\right)$$
       if the initial state is the $\vert 1,1\rangle$ state,
       $${\bf M}_{if} = -\frac{\hbar}{\sqrt{2}}\left({\bf e}_x+{\rm i}\,{\bf e}_y\right)$$
       if the initial state is the $\vert 1,-1\rangle$ state, and
       $${\bf M}_{if} = - \hbar\,{\bf e}_z$$
       if the initial state is the $\vert 1,0\rangle$ state.
    3. Deduce that the angular distribution of the spontaneously emitted photon is
       $$\frac{dw_{i\rightarrow f}^{\,\rm spn}}{d{\mit\Omega}} = \frac{2^{... ...3}\left(\frac{m_e}{m_p}\right)^3\,\frac{m_e\,c^{\,2}}{\hbar}\,(1+\cos^2\theta)$$
       if the initial state is either the $\vert 1,1\rangle$ state or the $\vert 1,-1\rangle$ state. Show that the angular distribution is
       $$\frac{dw_{i\rightarrow f}^{\,\rm spn}}{d{\mit\Omega}} = \frac{2^{... ...{\,13}\left(\frac{m_e}{m_p}\right)^3\,\frac{m_e\,c^{\,2}}{\hbar}\,\sin^2\theta$$
       if the initial state is the $\vert 1,0\rangle$ state. Here, the photon's direction of motion is parallel to ${\bf n} = (\sin\theta\,\cos\varphi,\,\sin\theta\,\sin\varphi,\,\cos\theta)$ .
    4. Finally, show that the overall spontaneous transition rate is
       $$w_{i\rightarrow f}^{\,\rm spn}= \frac{2^{\,6}}{3^{\,4}}\,g_p^{\,3... ..._p}\right)^3\,\frac{m_e\,c^{\,2}}{\hbar}= 2.88\times 10^{-15}\,{\rm s}^{\,-1}.$$

19. Consider the $3,2,0\rightarrow 1,0,0$ electric quadrupole transition of a hydrogen atom.
    1. Show that the angular distribution of spontaneously emitted photons can be written
       $$\frac{dw_{3d\rightarrow 1s}^{\,\rm spn}}{d{\mit\Omega}} =\frac{\a... ...1s}\cdot{\bf n}\right\vert^{\,2}}{a_0^{\,4}}\right)\frac{m_e\,c^{\,2}}{\hbar}.$$

    2. Demonstrate that the only non-zero $3d(m=0)\rightarrow 1s$ electric quadrupole matrix elements takes the values $Q_{xx}=-Q_{zz}/2$ , $Q_{yy}= -Q_{zz}/2$ , and
       $$Q_{zz} = \sqrt{\frac{2}{3}}\,\frac{3^{\,4}}{2^{\,7}}\,a_0^{\,2}.$$

    3. Hence, deduce that
       $$\frac{dw_{3d\rightarrow 1s}^{\,\rm spn}}{d{\mit\Omega}} = \frac{\... ...\,7}}{2^{\,8}\,3\pi}\,\frac{m_e\,c^{\,2}}{\hbar} \,\cos^2\theta\,\sin^2\theta,$$
       and
       $$w_{3d\rightarrow 2s}^{\,\rm spn} = \frac{\alpha^{\,7}}{2^{\,5}\,3^{\,2}\,5}\,\frac{m_e\,c^{\,2}}{\hbar}=594.1\,{\rm s}^{\,-1}.$$
       Here, the spontaneously emitted photon's direction of motion is parallel to ${\bf n} = (\sin\theta\,\cos\varphi,\,\sin\theta\,\sin\varphi,\,\cos\theta)$ .

20. Let
    $$J_n = \int_0^\infty dy\,y^{\,n}\,{\rm e}^{-\beta\,y}\,\sin(b\,y),$$
    where $n$ is a non-negative integer. Demonstrate that
    $$J_n = (-1)^n\left(\frac{\partial}{\partial\beta}\right)^n J_0,$$
    and
    $$J_0 = \frac{b}{\beta^{\,2}+b^{\,2}}.$$
    Hence, deduce that
    $$J_1 = \frac{2\,b\,\beta}{(\beta^{\,2}+b^{\,2})^2},$$
    and
    $$J_2 = \frac{8\,b\,\beta^{\,2}}{(\beta^{\,2}+b^{\,2})^{\,3}} -\frac{2\,b}{(\beta^{\,2}+b^{\,2})^{\,2}}.$$

21. Treating $\beta$ as a small parameter, and neglecting terms of order $\beta^{\,2}$ , show that
    $$\oint d{\mit\Omega}\, \frac{\sin^2\theta\,\cos^2\varphi}{(1-\beta\,\cos\theta)^{\,4}} = \frac{4\pi}{3}.$$

22. Repeat the calculation of Section 8.15 for the case where the electric polarization vector of the incident photon is $\epsilon$ $={\bf e}_y$ . Hence, show that, in this case, the differential photo-ionization cross-section is
    $$\frac{d\sigma_{i\rightarrow f}^{\,{\rm abs}}}{d{\mit\Omega}} \sim... .../2}a_0^{\,2}\,\frac{\sin^2\theta\,\sin^2\varphi}{(1-\beta\,\cos\theta)^{\,4}}.$$
    Deduce that the photo-ionization cross-section for unpolarized electromagnetic radiation propagating in the $z$ -direction is
    $$\frac{d\sigma_{i\rightarrow f}^{\,{\rm abs}}}{d{\mit\Omega}} \sim... ...{I}\right)^{\,3/2}a_0^{\,2}\,\frac{\sin^2\theta}{(1-\beta\,\cos\theta)^{\,4}}.$$

23. Consider the photo-ionization of a hydrogen atom in the $2s$ state. Demonstrate that formula (8.249) is replaced by
    $$\frac{d\sigma_{i\rightarrow f}^{\,{\rm abs}}}{d{\mit\Omega}} =\frac{2^{\,10}\,\alpha\,\hbar}{m_e\,k\,c}\, k_f\,($$$\epsilon$ $$\cdot{\bf k}_f)^{\,2}\,a_0^{\,3} \,\frac{\left[1-(2\,q\,a_0)^2\right]^{\,2}}{\left[1+(2\,q\,a_0)^2\right]^{\,6}}.$$

24. Consider the inverse process to the photo-ionization of a hydrogen atom investigated in Section 8.15. In this process, which is known as radiative association, an unbound electron of energy $E_i>0$ is captured by a (stationary) proton to form a hydrogen atom in its ground state, with the emission of a photon of energy
    $$\hbar\,\omega = E_i - E_f,$$
    where $E_f=E_0=-I$ . Here, $E_0$ is (negative) hydrogen ground-state energy, and $I$ is the (positive) ground-state ionization energy. Radiative association can be thought of as a form of stimulated emission in which the initial state is unbound. Thus, according to the analysis of Section 8.9,
    $$d\sigma_{i\rightarrow f}^{\,\rm stm} =\frac{4\pi^{\,2}\,\alpha}{m... ...ight\rangle\right\vert^{\,2} \rho(\omega)\right\vert _{E_f=E_i-\hbar\,\omega},$$
    where ${\bf p}$ is the electron momentum, ${\bf k}$ is the wavevector of the emitted photon, and $\rho(\omega)\,d\omega$ is the number of photon states whose angular frequencies lie between $\omega$ and $\omega+d\omega$ , and whose direction of motion lie in the range of solid angles $d{\mit\Omega}$ . Here, $\epsilon$ $_{1,2}$ are two independent unit vectors normal to ${\bf k}$ .
    1. Show that
       $$\rho(\omega) = \frac{V\,k^{\,2}}{(2\pi)^{\,2}\,c}\,d{\mit\Omega},$$
       where the initial electron and final photon are both assumed to be contained in a periodic box of volume $V\gg a_0^{\,3}$ .
    2. Hence, demonstrate that
       $$\frac{d\sigma_{i\rightarrow f}^{\,\rm stm}}{d{\mit\Omega}} = \fra... ...\,2}}\,\frac{k\,k_i^{\,2}\,a_0^{\,2}\,\sin^2\theta}{[1+(q\,a_0)^{\,2}]^{\,4}},$$
       where ${\bf q}= {\bf k}-{\bf k}_i$ , and $\cos\theta= {\bf k}\cdot{\bf k}_i/(k\,k_i)$ .

    3. Finally, show that
       $$\frac{d\sigma_{i\rightarrow f}^{\,\rm stm}}{d{\mit\Omega}}=2^{\,4... ...}{(E_i+I)^{\,3}}\,a_0^{\,2}\,\frac{\sin^2\theta}{(1-\beta\,\cos\theta)^{\,4}},$$
       where $\beta=(2\,E_i/m_e\,c^{\,2})^{1/2}$ .