Transitions in a time-dependent field
In the ammonia maser, the beam with molecules in the state $\ketsl{\slI}$ and with the energy $E_{\slI}$ is sent through a resonant cavity, as shown in Fig. 9–4. The other beam is discarded. Inside the cavity, there will be a time-varying electric field, so the next problem we must discuss is the behavior of a molecule in an electric field that varies with time. We have a completely different kind of a problem—one with a time-varying Hamiltonian. Since $H_{ij}$ depends upon $\Efield$, the $H_{ij}$ vary with time, and we must determine the behavior of the system in this circumstance. To begin with, we write down the equations to be solved: \begin{equation} \begin{aligned} i\hbar\,\ddt{C_1}{t}&=(E_0+\mu\Efield)C_1-AC_2,\\[1ex] i\hbar\,\ddt{C_2}{t}&=-AC_1+(E_0-\mu\Efield)C_2. \end{aligned} \label{Eq:III:9:36} \end{equation}  To be definite, let’s suppose that the electric field varies sinusoidally; then we can write \begin{equation} \label{Eq:III:9:37} \Efield=2\Efield_0\cos\omega t= \Efield_0(e^{i\omega t}+e^{-i\omega t}). \end{equation}  In actual operation the frequency $\omega$ will be very nearly equal to the resonant frequency of the molecular transition $\omega_0=2A/\hbar$, but for the time being we want to keep things general, so we’ll let it have any value at all. The best way to solve our equations is to form linear combinations of $C_1$ and $C_2$ as we did before. So we add the two equations, divide by the square root of $2$, and use the definitions of $C_{\slI}$ and $C_{\slII}$ that we had in Eq. (9.13). We get \begin{equation} \label{Eq:III:9:38} i\hbar\,\ddt{C_{\slII}}{t}= (E_0-A)C_{\slII}+\mu\Efield C_{\slI}. \end{equation}  You’ll note that this is the same as Eq. (9.9) with an extra term due to the electric field. Similarly, if we subtract the two equations (9.36), we get \begin{equation} \label{Eq:III:9:39} i\hbar\,\ddt{C_{\slI}}{t}= (E_0+A)C_{\slI}+\mu\Efield C_{\slII}. \end{equation}  Now the question is, how to solve these equations? They are more difficult than our earlier set, because $\Efield$ depends on $t$; and, in fact, for a general $\Efield(t)$ the solution is not expressible in elementary functions. However, we can get a good approximation so long as the electric field is small. First we will write \begin{equation} \begin{gathered} C_{\slI}=\gamma_{\slI}e^{-i(E_0+A)t/\hbar}= \gamma_{\slI}e^{-i(E_{\slI})t/\hbar},\\ \\ C_{\slII}=\gamma_{\slII}e^{-i(E_0-A)t/\hbar}= \gamma_{\slII}e^{-i(E_{\slII})t/\hbar}. \end{gathered} \label{Eq:III:9:40} \end{equation}  If there were no electric field, these solutions would be correct with $\gamma_{\slI}$ and $\gamma_{\slII}$ just chosen as two complex constants. In fact, since the probability of being in state $\ketsl{\slI}$ is the absolute square of $C_{\slI}$ and the probability of being in state $\ketsl{\slII}$ is the absolute square of $C_{\slII}$, the probability of being in state $\ketsl{\slI}$ or in state $\ketsl{\slII}$ is just $\abs{\gamma_{\slI}}^2$ or $\abs{\gamma_{\slII}}^2$. For instance, if the system were to start originally in state $\ketsl{\slII}$ so that $\gamma_{\slI}$ was zero and $\abs{\gamma_{\slII}}^2$ was one, this condition would go on forever. There would be no chance, if the molecule were originally in state $\ketsl{\slII}$, ever to get into state $\ketsl{\slI}$.  Now the idea of writing our equations in the form of Eq. (9.40) is that if $\mu\Efield$ is small in comparison with $A$, the solutions can still be written in this way, but then $\gamma_{\slI}$ and $\gamma_{\slII}$ become slowly varying functions of time—where by “slowly varying” we mean slowly in comparison with the exponential functions. That is the trick. We use the fact that $\gamma_{\slI}$ and $\gamma_{\slII}$ vary slowly to get an approximate solution. We want now to substitute $C_{\slI}$ from Eq. (9.40) in the differential equation (9.39), but we must remember that $\gamma_{\slI}$ is also a function of $t$. We have \begin{equation*} i\hbar\,\ddt{C_{\slI}}{t}= E_{\slI}\gamma_{\slI}e^{-iE_{\slI}t/\hbar}+ i\hbar\,\ddt{\gamma_{\slI}}{t}\,e^{-iE_{\slI}t/\hbar}. \end{equation*}  The differential equation becomes  \begin{equation} \label{Eq:III:9:41} \biggl(E_{\slI}\gamma_{\slI}+i\hbar\,\ddt{\gamma_{\slI}}{t}\biggr) e^{-(i/\hbar)E_{\slI}t}= E_{\slI}\gamma_{\slI}e^{-(i/\hbar)E_{\slI}t}+ \mu\Efield\gamma_{\slII}e^{-(i/\hbar)E_{\slII}t}. \end{equation} 
 \begin{equation} \begin{gathered} \biggl(E_{\slI}\gamma_{\slI}+i\hbar\,\ddt{\gamma_{\slI}}{t}\biggr) e^{-(i/\hbar)E_{\slI}t}=\\[1ex] E_{\slI}\gamma_{\slI}e^{-(i/\hbar)E_{\slI}t}+ \mu\Efield\gamma_{\slII}e^{-(i/\hbar)E_{\slII}t}. \end{gathered} \label{Eq:III:9:41} \end{equation}  Similarly, the equation in $dC_{\slII}/dt$ becomes  \begin{equation} \label{Eq:III:9:42} \biggl(E_{\slII}\gamma_{\slII}+i\hbar\,\ddt{\gamma_{\slII}}{t}\biggr) e^{-(i/\hbar)E_{\slII}t}= E_{\slII}\gamma_{\slII}e^{-(i/\hbar)E_{\slII}t}+ \mu\Efield\gamma_{\slI}e^{-(i/\hbar)E_{\slI}t}. \end{equation} 
 \begin{equation} \begin{gathered} \biggl(E_{\slII}\gamma_{\slII}+i\hbar\,\ddt{\gamma_{\slII}}{t}\biggr) e^{-(i/\hbar)E_{\slII}t}=\\[1ex] E_{\slII}\gamma_{\slII}e^{-(i/\hbar)E_{\slII}t}+ \mu\Efield\gamma_{\slI}e^{-(i/\hbar)E_{\slI}t}. \end{gathered} \label{Eq:III:9:42} \end{equation}  Now you will notice that we have equal terms on both sides of each equation. We cancel these terms, and we also multiply the first equation by $e^{+iE_{\slI}t/\hbar}$ and the second by $e^{+iE_{\slII}t/\hbar}$. Remembering that $(E_{\slI}-E_{\slII})=$ $2A=$ $\hbar\omega_0$, we have finally, \begin{equation} \label{Eq:III:9:43} \begin{aligned} i\hbar\,\ddt{\gamma_{\slI}}{t}&= \mu\Efield(t)e^{i\omega_0t}\gamma_{\slII},\\[1ex] i\hbar\,\ddt{\gamma_{\slII}}{t}&= \mu\Efield(t)e^{-i\omega_0t}\gamma_{\slI}. \end{aligned} \end{equation}  Now we have an apparently simple pair of equations—and they are still exact, of course. The derivative of one variable is a function of time $\mu\Efield(t)e^{i\omega_0t}$, multiplied by the second variable; the derivative of the second is a similar time function, multiplied by the first. Although these simple equations cannot be solved in general, we will solve them for some special cases. We are, for the moment at least, interested only in the case of an oscillating electric field. Taking $\Efield(t)$ as given in Eq. (9.37), we find that the equations for $\gamma_{\slI}$ and $\gamma_{\slII}$ become \begin{equation} \label{Eq:III:9:44} \begin{aligned} i\hbar\,\ddt{\gamma_{\slI}}{t}&= \mu\Efield_0[e^{i(\omega+\omega_0)t}+ e^{-i(\omega-\omega_0)t}]\gamma_{\slII},\\[1ex] i\hbar\,\ddt{\gamma_{\slII}}{t}&= \mu\Efield_0[e^{i(\omega-\omega_0)t}+ e^{-i(\omega+\omega_0)t}]\gamma_{\slI}. \end{aligned} \end{equation}  Now if $\Efield_0$ is sufficiently small, the rates of change of $\gamma_{\slI}$ and $\gamma_{\slII}$ are also small. The two $\gamma$’s will not vary much with $t$, especially in comparison with the rapid variations due to the exponential terms. These exponential terms have real and imaginary parts that oscillate at the frequency $\omega+\omega_0$ or $\omega-\omega_0$. The terms with $\omega+\omega_0$ oscillate very rapidly about an average value of zero and, therefore, do not contribute very much on the average to the rate of change of $\gamma$. So we can make a reasonably good approximation by replacing these terms by their average value, namely, zero. We will just leave them out, and take as our approximation: \begin{equation} \begin{aligned} i\hbar\,\ddt{\gamma_{\slI}}{t}&= \mu\Efield_0e^{-i(\omega-\omega_0)t}\gamma_{\slII},\\[1ex] i\hbar\,\ddt{\gamma_{\slII}}{t}&= \mu\Efield_0e^{i(\omega-\omega_0)t}\gamma_{\slI}. \end{aligned} \label{Eq:III:9:45} \end{equation}  Even the remaining terms, with exponents proportional to $(\omega-\omega_0)$, will also vary rapidly unless $\omega$ is near $\omega_0$. Only then will the right-hand side vary slowly enough that any appreciable amount will accumulate when we integrate the equations with respect to $t$. In other words, with a weak electric field the only significant frequencies are those near $\omega_0$. With the approximation made in getting Eq. (9.45), the equations can be solved exactly, but the work is a little elaborate, so we won’t do that until later when we take up another problem of the same type. Now we’ll just solve them approximately—or rather, we’ll find an exact solution for the case of perfect resonance, $\omega=\omega_0$, and an approximate solution for frequencies near resonance.