12.E: Time-Dependent Perturbation Theory (Exercises)

1. Consider the two-state system examined in Section 1.3.[ex8.1] Suppose that $$\begin{aligned} \langle 1 |\,H_1\,|1\rangle &= e_{11},\nonumber\\[0.5ex] \langle 2 |\,H_1\,|2\rangle &= e_{22},\nonumber\\[0.5ex] \langle 1|\,H_1\,|2\rangle = \langle 2|\,H_1\,|1\rangle^\ast &= \frac{1}{2}\,\gamma\,\hbar\,\exp(\,{\rm i}\,\omega\,t),\nonumber\end{aligned}$$ where $e_{11}$, $e_{22}$, $\gamma$, and $\omega$ are real. Show that $$\begin{aligned} {\rm i}\,\frac{d\hat{c}_1}{dt} &= \frac{\gamma}{2}\,\exp\left[+{\rm i}\,(\omega-\hat{\omega}_{21})\,t\right]\hat{c}_2,\nonumber\\[0.5ex] {\rm i}\,\frac{d\hat{c}_2}{dt} &= \frac{\gamma}{2}\,\exp\left[-{\rm i}\,(\omega-\hat{\omega}_{21})\,t\right]\hat{c}_1,\nonumber\end{aligned}$$ 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 $$P_2(t) = \frac{\gamma^{\,2}}{ \gamma^{\,2} + (\omega-\hat{\omega}_{21})^{\,2}}\, \sin^2\left(\left[\gamma^{\,2}+ (\omega-\hat{\omega}_{21})^{\,2}\right]^{1/2} \frac{t}{2}\right).$$
2. Consider an atomic nucleus of spin-$s$ and gyromagnetic ratio $g$ 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 $\chi_{s,m}$ be a properly normalized simultaneous eigenstate of $S^{\,2}$ and $S_z$, where ${\bf S}$ is the nuclear spin. Thus, $S^{\,2}\,\chi_{s,m} = s\,(s+1)\,\hbar^{\,2}\,\chi_{s,m}$ and $S_z \,\chi_{s,m} = m\,\hbar\,\chi_{s,m}$, where $-s\leq m\leq s$. Furthermore, the instantaneous nuclear spin state is written $$\chi= \sum_{m=-s,s} c_m(r)\,\chi_{s,m},$$ where $\sum_{m=-s,s}|c_m|^{\,2}=1$.
   1. Demonstrate that $$\begin{gathered} \frac{dc_m}{dt} = \frac{{\rm i}\,\gamma}{2}\left([s\,(s+1)-m\,(m-1)]^{1/2}\,{\rm e}^{\,{\rm i}\,(\omega-\omega_0)\,t}\,c_{m-1}\right.\\[0.5ex]\left. + [s\,(s+1)-m\,(m+1)]^{1/2}\,{\rm e}^{-{\rm i}\,(\omega-\omega_0)\,t}\,c_{m+1}\right)\nonumber\end{gathered}$$ for $-s\leq m\leq s$, where $\omega_0=g\,\mu_N\,B_0/\hbar$, $\gamma= g\,\mu_N\,B_1/\hbar$, and $\mu_N=e\,\hbar/(2\,m_p)$.
   2. Consider the case $s=1/2$. Demonstrate that if $\omega=\omega_0$ and $c_{1/2}(0)= 1$ then $$\begin{aligned} c_{1/2}(t)= \cos(\gamma\,t/2), && c_{-1/2}(t)= {\rm i}\,\sin(\gamma\,t/2).\nonumber\end{aligned}$$
   3. Consider the case $s=1$. Demonstrate that if $\omega=\omega_0$ and $c_{1}(0)= 1$ then $$\begin{aligned} c_{1}(t)&= \cos^2(\gamma\,t/2),\nonumber\\[0.5ex] c_0(t)&= {\rm i}\sqrt{2}\,\cos(\gamma\,t/2)\,\sin(\gamma\,t/2),\nonumber\\[0.5ex] c_{-1}(t)&= -\sin^2(\gamma\,t/2).\nonumber\end{aligned}$$
   4. Consider the case $s=3/2$. Demonstrate that if $\omega=\omega_0$ and $c_{3/2}(0)= 1$ then $$\begin{aligned} c_{3/2}(t)&= \cos^3(\gamma\,t/2),\nonumber\\[0.5ex] c_{1/2}(t)&= {\rm i}\sqrt{3}\,\cos(\gamma\,t/2)\,\sin^2(\gamma\,t/2),\nonumber\\[0.5ex] c_{-1/2}(t)&= -\sqrt{3}\,\cos^2(\gamma\,t/2)\,\sin(\gamma\,t/2),\nonumber\\[0.5ex] c_{-3/2}(t)&=-{\rm i}\,\sin^3(\gamma\,t/2).\nonumber\end{aligned}$$
3. 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 .)

Contributors and Attributions

• Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)

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