Transitions at resonance
Let’s take the case of perfect resonance first. If we take $\omega=\omega_0$, the exponentials are equal to one in both equations of (9.45), and we have just \begin{equation} \label{Eq:III:9:46} \ddt{\gamma_{\slI}}{t}=-\frac{i\mu\Efield_0}{\hbar}\,\gamma_{\slII},\quad \ddt{\gamma_{\slII}}{t}=-\frac{i\mu\Efield_0}{\hbar}\,\gamma_{\slI}. \end{equation}  If we eliminate first $\gamma_{\slI}$ and then $\gamma_{\slII}$ from these equations, we find that each satisfies the differential equation of simple harmonic motion: \begin{equation} \label{Eq:III:9:47} \frac{d^2\gamma}{dt^2}=-\biggl( \frac{\mu\Efield_0}{\hbar} \biggr)^2\gamma. \end{equation}  The general solutions for these equations can be made up of sines and cosines. As you can easily verify, the following equations are a solution: \begin{equation} \begin{aligned} \gamma_{\slI}&=a\cos\biggl(\frac{\mu\Efield_0}{\hbar}\biggr)t+ b\sin\biggl(\frac{\mu\Efield_0}{\hbar}\biggr)t,\\[1.5ex] \gamma_{\slII}&=ib\cos\biggl(\frac{\mu\Efield_0}{\hbar}\biggr)t- ia\sin\biggl(\frac{\mu\Efield_0}{\hbar}\biggr)t, \end{aligned} \label{Eq:III:9:48} \end{equation}  where $a$ and $b$ are constants to be determined to fit any particular physical situation. For instance, suppose that at $t=0$ our molecular system was in the upper energy state $\ketsl{\slI}$, which would require—from Eq. (9.40)—that $\gamma_{\slI}=1$ and $\gamma_{\slII}=0$ at $t=0$. For this situation we would need $a=1$ and $b=0$. The probability that the molecule is in the state $\ketsl{\slI}$ at some later $t$ is the absolute square of $\gamma_{\slI}$, or \begin{equation} \label{Eq:III:9:49} P_{\slI}=\abs{\gamma_{\slI}}^2= \cos^2\biggl(\frac{\mu\Efield_0}{\hbar}\biggr)t. \end{equation}  Similarly, the probability that the molecule will be in the state $\ketsl{\slII}$ is given by the absolute square of $\gamma_{\slII}$, \begin{equation} \label{Eq:III:9:50} P_{\slII}=\abs{\gamma_{\slII}}^2= \sin^2\biggl(\frac{\mu\Efield_0}{\hbar}\biggr)t. \end{equation}  So long as $\Efield$ is small and we are on resonance, the probabilities are given by simple oscillating functions. The probability to be in state $\ketsl{\slI}$ falls from one to zero and back again, while the probability to be in the state $\ketsl{\slII}$ rises from zero to one and back. The time variation of the two probabilities is shown in Fig. 9–5. Needless to say, the sum of the two probabilities is always equal to one; the molecule is always in some state!  Let’s suppose that it takes the molecule the time $T$ to go through the cavity. If we make the cavity just long enough so that $\mu\Efield_0T/\hbar=\pi/2$, then a molecule which enters in state $\ketsl{\slI}$ will certainly leave it in state $\ketsl{\slII}$. If it enters the cavity in the upper state, it will leave the cavity in the lower state. In other words, its energy is decreased, and the loss of energy can’t go anywhere else but into the machinery which generates the field. The details by which you can see how the energy of the molecule is fed into the oscillations of the cavity are not simple; however, we don’t need to study these details, because we can use the principle of conservation of energy. (We could study them if we had to, but then we would have to deal with the quantum mechanics of the field in the cavity in addition to the quantum mechanics of the atom.) In summary: the molecule enters the cavity, the cavity field—oscillating at exactly the right frequency—induces transitions from the upper to the lower state, and the energy released is fed into the oscillating field. In an operating maser the molecules deliver enough energy to maintain the cavity oscillations—not only providing enough power to make up for the cavity losses but even providing small amounts of excess power that can be drawn from the cavity. Thus, the molecular energy is converted into the energy of an external electromagnetic field. Remember that before the beam enters the cavity, we have to use a filter which separates the beam so that only the upper state enters. It is easy to demonstrate that if you were to start with molecules in the lower state, the process will go the other way and take energy out of the cavity. If you put the unfiltered beam in, as many molecules are taking energy out as are putting energy in, so nothing much would happen. In actual operation it isn’t necessary, of course, to make $(\mu\Efield_0T/\hbar)$ exactly $\pi/2$. For any other value (except an exact integral multiple of $\pi$), there is some probability for transitions from state $\ketsl{\slI}$ to state $\ketsl{\slII}$. For other values, however, the device isn’t $100$ percent efficient; many of the molecules which leave the cavity could have delivered some energy to the cavity but didn’t. In actual use, the velocity of all the molecules is not the same; they have some kind of Maxwell distribution. This means that the ideal periods of time for different molecules will be different, and it is impossible to get $100$ percent efficiency for all the molecules at once. In addition, there is another complication which is easy to take into account, but we don’t want to bother with it at this stage. You remember that the electric field in a cavity usually varies from place to place across the cavity. Thus, as the molecules drift across the cavity, the electric field at the molecule varies in a way that is more complicated than the simple sinusoidal oscillation in time that we have assumed. Clearly, one would have to use a more complicated integration to do the problem exactly, but the general idea is still the same. There are other ways of making masers. Instead of separating the atoms in state $\ketsl{\slI}$ from those in state $\ketsl{\slII}$ by a Stern-Gerlach apparatus, one can have the atoms already in the cavity (as a gas or a solid) and shift atoms from state $\ketsl{\slII}$ to state $\ketsl{\slI}$ by some means. One way is one used in the so-called three-state maser. For it, atomic systems are used which have three energy levels, as shown in Fig. 9–6, with the following special properties. The system will absorb radiation (say, light) of frequency $\hbar\omega_1$ and go from the lowest energy level $E_{\slII}$, to some high-energy level $E'$, and then will quickly emit photons of frequency $\hbar\omega_2$ and go to the state $\ketsl{\slI}$ with energy $E_{\slI}$. The state $\ketsl{\slI}$ has a long lifetime so its population can be raised, and the conditions are then appropriate for maser operation between states $\ketsl{\slI}$ and $\ketsl{\slII}$. Although such a device is called a “three-state” maser, the maser operation really works just as a two-state system such as we are describing. A laser (Light Amplification by Stimulated Emission of Radiation) is just a maser working at optical frequencies. The “cavity” for a laser usually consists of just two plane mirrors between which standing waves are generated.