The molecule in a static electric field
If the ammonia molecule is in either of the two states of definite energy and we disturb it at a frequency $\omega$ such that $\hbar\omega=$ $E_{\slI}-E_{\slII}=$ $2A$, the system may make a transition from one state to the other. Or, if it is in the upper state, it may change to the lower state and emit a photon. But in order to induce such transitions you must have a physical connection to the states—some way of disturbing the system. There must be some external machinery for affecting the states, such as magnetic or electric fields. In this particular case, these states are sensitive to an electric field. We will, therefore, look next at the problem of the behavior of the ammonia molecule in an external electric field. To discuss the behavior in an electric field, we will go back to the original base system $\ketsl{\slOne}$ and $\ketsl{\slTwo}$, rather than using $\ketsl{\slI}$ and $\ketsl{\slII}$. Suppose that there is an electric field in a direction perpendicular to the plane of the hydrogen atoms. Disregarding for the moment the possibility of flipping back and forth, would it be true that the energy of this molecule is the same for the two positions of the nitrogen atom? Generally, no. The electrons tend to lie closer to the nitrogen than to the hydrogen nuclei, so the hydrogens are slightly positive. The actual amount depends on the details of electron distribution. It is a complicated problem to figure out exactly what this distribution is, but in any case the net result is that the ammonia molecule has an electric dipole moment, as indicated in Fig. 9–1. We can continue our analysis without knowing in detail the direction or amount of displacement of the charge. However, to be consistent with the notation of others, let’s suppose that the electric dipole moment is $\FLPmu$, with its direction pointing from the nitrogen atom and perpendicular to the plane of the hydrogen atoms. Now, when the nitrogen flips from one side to the other, the center of mass will not move, but the electric dipole moment will flip over. As a result of this moment, the energy in an electric field $\Efieldvec$ will depend on the molecular orientation.2 With the assumption made above, the potential energy will be higher if the nitrogen atom points in the direction of the field, and lower if it is in the opposite direction; the separation in the two energies will be $2\mu\Efield$. In the discussion up to this point, we have assumed values of $E_0$ and $A$ without knowing how to calculate them. According to the correct physical theory, it should be possible to calculate these constants in terms of the positions and motions of all the nuclei and electrons. But nobody has ever done it. Such a system involves ten electrons and four nuclei and that’s just too complicated a problem. As a matter of fact, there is no one who knows much more about this molecule than we do. All anyone can say is that when there is an electric field, the energy of the two states is different, the difference being proportional to the electric field. We have called the coefficient of proportionality $2\mu$, but its value must be determined experimentally. We can also say that the molecule has the amplitude $A$ to flip over, but this will have to be measured experimentally. Nobody can give us accurate theoretical values of $\mu$ and $A$, because the calculations are too complicated to do in detail. For the ammonia molecule in an electric field, our description must be changed. If we ignored the amplitude for the molecule to flip from one configuration to the other, we would expect the energies of the two states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$ to be $(E_0\pm\mu\Efield)$. Following the procedure of the last chapter, we take \begin{equation} \label{Eq:III:9:14} H_{11}=E_0+\mu\Efield,\quad H_{22}=E_0-\mu\Efield. \end{equation}  Also we will assume that for the electric fields of interest the field does not affect appreciably the geometry of the molecule and, therefore, does not affect the amplitude that the nitrogen will jump from one position to the other. We can then take that $H_{12}$ and $H_{21}$ are not changed; so \begin{equation} \label{Eq:III:9:15} H_{12}=H_{21}=-A. \end{equation}  We must now solve the Hamiltonian equations, Eq. (8.43), with these new values of $H_{ij}$. We could solve them just as we did before, but since we are going to have several occasions to want the solutions for two-state systems, let’s solve the equations once and for all in the general case of arbitrary $H_{ij}$—assuming only that they do not change with time. We want the general solution of the pair of Hamiltonian equations \begin{align} \label{Eq:III:9:16} i\hbar\,\ddt{C_1}{t}&=H_{11}C_1+H_{12}C_2,\\[1ex] \label{Eq:III:9:17} i\hbar\,\ddt{C_2}{t}&=H_{21}C_1+H_{22}C_2. \end{align}  Since these are linear differential equations with constant coefficients, we can always find solutions which are exponential functions of the dependent variable $t$. We will first look for a solution in which $C_1$ and $C_2$ both have the same time dependence; we can use the trial functions \begin{equation*} C_1=a_1e^{-i\omega t},\quad C_2=a_2e^{-i\omega t}. \end{equation*}  Since such a solution corresponds to a state of energy $E=\hbar\omega$, we may as well write right away \begin{align} \label{Eq:III:9:18} C_1&=a_1e^{-(i/\hbar)Et},\\[1ex] \label{Eq:III:9:19} C_2&=a_2e^{-(i/\hbar)Et}, \end{align}  where $E$ is as yet unknown and to be determined so that the differential equations (9.16) and (9.17) are satisfied. When we substitute $C_1$ and $C_2$ from (9.18) and (9.19) in the differential equations (9.16) and (9.17), the derivatives give us just $-iE/\hbar$ times $C_1$ or $C_2$, so the left sides become just $EC_1$ and $EC_2$. Cancelling the common exponential factors, we get \begin{equation*} Ea_1=H_{11}a_1+H_{12}a_2,\quad Ea_2=H_{21}a_1+H_{22}a_2. \end{equation*}  Or, rearranging the terms, we have \begin{align} \label{Eq:III:9:20} (E-H_{11})a_1-H_{12}a_2&=0,\\[1ex] \label{Eq:III:9:21} -H_{21}a_1+(E-H_{22})a_2&=0. \end{align}  With such a set of homogeneous algebraic equations, there will be nonzero solutions for $a_1$ and $a_2$ only if the determinant of the coefficients of $a_1$ and $a_2$ is zero, that is, if \begin{equation} \label{Eq:III:9:22} \Det\begin{pmatrix} E-H_{11} & \phantom{E}-H_{12}\\[1ex] \phantom{E}-H_{21} & E-H_{22} \end{pmatrix}=0. \end{equation}  However, when there are only two equations and two unknowns, we don’t need such a sophisticated idea. The two equations (9.20) and (9.21) each give a ratio for the two coefficients $a_1$ and $a_2$, and these two ratios must be equal. From (9.20) we have that \begin{equation} \label{Eq:III:9:23} \frac{a_1}{a_2} =\frac{H_{12}}{E-H_{11}}, \end{equation}  and from (9.21) that \begin{equation} \label{Eq:III:9:24} \frac{a_1}{a_2} =\frac{E-H_{22}}{H_{21}}. \end{equation}  Equating these two ratios, we get that $E$ must satisfy \begin{equation*} (E-H_{11})(E-H_{22})-H_{12}H_{21}=0. \end{equation*}  This is the same result we would get by solving Eq. (9.22). Either way, we have a quadratic equation for $E$ which has two solutions: \begin{equation} \label{Eq:III:9:25} E=\!\frac{H_{11}\!+\!H_{22}}{2}\!\pm\! \sqrt{\frac{(H_{11}\!-\!H_{22})^2}{4}\!+\!H_{12}H_{21}}. \end{equation}  There are two possible values for the energy $E$. Note that both solutions give real numbers for the energy, because $H_{11}$ and $H_{22}$ are real, and $H_{12}H_{21}$ is equal to $H_{12}H_{12}\cconj=\abs{H_{12}}^2$, which is both real and positive. Using the same convention we took before, we will call the upper energy $E_{\slI}$ and the lower energy $E_{\slII}$. We have \begin{align} \label{Eq:III:9:26} E_{\slI}&=\!\frac{H_{11}\!+\!H_{22}}{2}\!+\! \sqrt{\frac{(H_{11}\!-\!H_{22})^2}{4}\!+\!H_{12}H_{21}},\\[1.5ex] \label{Eq:III:9:27} E_{\slII}&=\!\frac{H_{11}\!+\!H_{22}}{2}\!-\! \sqrt{\frac{(H_{11}\!-\!H_{22})^2}{4}\!+\!H_{12}H_{21}}. \end{align}  Using each of these two energies separately in Eqs. (9.18) and (9.19), we have the amplitudes for the two stationary states (the states of definite energy). If there are no external disturbances, a system initially in one of these states will stay that way forever—only its phase changes. We can check our results for two special cases. If $H_{12}=H_{21}=0$, we have that $E_{\slI}=H_{11}$ and $E_{\slII}=H_{22}$. This is certainly correct, because then Eqs. (9.16) and (9.17) are uncoupled, and each represents a state of energy $H_{11}$ and $H_{22}$. Next, if we set $H_{11}=H_{22}=E_0$ and $H_{21}=H_{12}=-A$, we get the solution we found before: \begin{equation*} E_{\slI}=E_0+A\quad \text{and}\quad E_{\slII}=E_0-A. \end{equation*}  For the general case, the two solutions $E_{\slI}$ and $E_{\slII}$ refer to two states—which we can again call the states \begin{equation*} \ket{\psi_{\slI}}=\ketsl{\slI}e^{-(i/\hbar)E_{\slI}t}\quad \text{and}\quad \ket{\psi_{\slII}}=\ketsl{\slII}e^{-(i/\hbar)E_{\slII}t}. \end{equation*}  These states will have $C_1$ and $C_2$ as given in Eqs. (9.18) and (9.19), where $a_1$ and $a_2$ are still to be determined. Their ratio is given by either Eq. (9.23) or Eq. (9.24). They must also satisfy one more condition. If the system is known to be in one of the stationary states, the sum of the probabilities that it will be found in $\ketsl{\slOne}$ or $\ketsl{\slTwo}$ must equal one. We must have that \begin{equation} \label{Eq:III:9:28} \abs{C_1}^2+\abs{C_2}^2=1, \end{equation}  or, equivalently, \begin{equation} \label{Eq:III:9:29} \abs{a_1}^2+\abs{a_2}^2=1. \end{equation}  These conditions do not uniquely specify $a_1$ and $a_2$; they are still undetermined by an arbitrary phase—in other words, by a factor like $e^{i\delta}$. Although general solutions for the $a$’s can be written down,3 it is usually more convenient to work them out for each special case. Let’s go back now to our particular example of the ammonia molecule in an electric field. Using the values for $H_{11}$, $H_{22}$, and $H_{12}$ given in (9.14) and (9.15), we get for the energies of the two stationary states  \begin{equation} \label{Eq:III:9:30} E_{\slI}=E_0+\sqrt{A^2+\mu^2\Efield^2},\quad E_{\slII}=E_0-\sqrt{A^2+\mu^2\Efield^2}. \end{equation} 
 \begin{equation} \begin{aligned} E_{\slI}&=E_0+\sqrt{A^2+\mu^2\Efield^2},\\[1ex] E_{\slII}&=E_0-\sqrt{A^2+\mu^2\Efield^2}. \end{aligned} \label{Eq:III:9:30} \end{equation}  These two energies are plotted as a function of the electric field strength $\Efield$ in Fig. 9–2. When the electric field is zero, the two energies are, of course, just $E_0\pm A$. When an electric field is applied, the splitting between the two levels increases. The splitting increases at first slowly with $\Efield$, but eventually becomes proportional to $\Efield$. (The curve is a hyperbola.) For enormously strong fields, the energies are just  \begin{equation} \label{Eq:III:9:31} E_{\slI}=E_0+\mu\Efield=H_{11},\quad E_{\slII}=E_0-\mu\Efield=H_{22}. \end{equation} 
 \begin{equation} \begin{aligned} E_{\slI}&=E_0+\mu\Efield=H_{11},\\[1ex] E_{\slII}&=E_0-\mu\Efield=H_{22}. \end{aligned} \label{Eq:III:9:31} \end{equation}   
The fact that there is an amplitude for the nitrogen to flip back and forth has little effect when the two positions have very different energies. This is an interesting point which we will come back to again later. We are at last ready to understand the operation of the ammonia maser. The idea is the following. First, we find a way of separating molecules in the state $\ketsl{\slI}$ from those in the state $\ketsl{\slII}$.4 Then the molecules in the higher energy state $\ketsl{\slI}$ are passed through a cavity which has a resonant frequency of $24{,}000$ megacycles. The molecules can deliver energy to the cavity—in a way we will discuss later—and leave the cavity in the state $\ketsl{\slII}$. Each molecule that makes such a transition will deliver the energy $E=E_{\slI}-E_{\slII}$ to the cavity. The energy from the molecules will appear as electrical energy in the cavity. How can we separate the two molecular states? One method is as follows. The ammonia gas is let out of a little jet and passed through a pair of slits to give a narrow beam, as shown in Fig. 9–3. The beam is then sent through a region in which there is a large transverse electric field. The electrodes to produce the field are shaped so that the electric field varies rapidly across the beam. Then the square of the electric field $\Efieldvec\cdot\Efieldvec$ will have a large gradient perpendicular to the beam. Now a molecule in state $\ketsl{\slI}$ has an energy which increases with $\Efield^2$, and therefore this part of the beam will be deflected toward the region of lower $\Efield^2$. A molecule in state $\ketsl{\slII}$ will, on the other hand, be deflected toward the region of larger $\Efield^2$, since its energy decreases as $\Efield^2$ increases. Incidentally, with the electric fields which can be generated in the laboratory, the energy $\mu\Efield$ is always much smaller than $A$. In such cases, the square root in Eqs. (9.30) can be approximated by \begin{equation} \label{Eq:III:9:32} A\biggl( 1+\frac{1}{2}\,\frac{\mu^2\Efield^2}{A^2} \biggr). \end{equation}  So the energy levels are, for all practical purposes, \begin{equation} \label{Eq:III:9:33} E_{\slI} =E_0+A+\frac{\mu^2\Efield^2}{2A} \end{equation}  and \begin{equation} \label{Eq:III:9:34} E_{\slII} =E_0-A-\frac{\mu^2\Efield^2}{2A}. \end{equation}  And the energies vary approximately linearly with $\Efield^2$. The force on the molecules is then \begin{equation} \label{Eq:III:9:35} \FLPF=\frac{\mu^2}{2A}\,\FLPgrad{\Efield^2}. \end{equation}  Many molecules have an energy in an electric field which is proportional to $\Efield^2$. The coefficient is the polarizability of the molecule. Ammonia has an unusually high polarizability because of the small value of $A$ in the denominator. Thus, ammonia molecules are unusually sensitive to an electric field. (What would you expect for the dielectric coefficient of NH$_3$ gas?)