The ammonia molecule
We want now to show you how the dynamical equation of quantum mechanics can be used to describe a particular physical circumstance. We have picked an interesting but simple example in which, by making some reasonable guesses about the Hamiltonian, we can work out some important—and even practical—results. We are going to take a situation describable by two states: the ammonia molecule. The ammonia molecule has one nitrogen atom and three hydrogen atoms located in a plane below the nitrogen so that the molecule has the form of a pyramid, as drawn in Fig. 8–1(a). Now this molecule, like any other, has an infinite number of states. It can spin around any possible axis; it can be moving in any direction; it can be vibrating inside, and so on, and so on. It is, therefore, not a two-state system at all. But we want to make an approximation that all other states remain fixed, because they don’t enter into what we are concerned with at the moment. We will consider only that the molecule is spinning around its axis of symmetry (as shown in the figure), that it has zero translational momentum, and that it is vibrating as little as possible. That specifies all conditions except one: there are still the two possible positions for the nitrogen atom—the nitrogen may be on one side of the plane of hydrogen atoms or on the other, as shown in Fig. 8–1(a) and (b). So we will discuss the molecule as though it were a two-state system. We mean that there are only two states we are going to really worry about, all other things being assumed to stay put. You see, even if we know that it is spinning with a certain angular momentum around the axis and that it is moving with a certain momentum and vibrating in a definite way, there are still two possible states. We will say that the molecule is in the state $\ketsl{\slOne}$ when the nitrogen is “up,” as in Fig. 8–1(a), and is in the state $\ketsl{\slTwo}$ when the nitrogen is “down,” as in (b). The states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$ will be taken as the set of base states for our analysis of the behavior of the ammonia molecule. At any moment, the actual state $\ket{\psi}$ of the molecule can be represented by giving $C_1=\braket{\slOne}{\psi}$, the amplitude to be in state $\ketsl{\slOne}$, and $C_2=\braket{\slTwo}{\psi}$, the amplitude to be in state $\ketsl{\slTwo}$. Then, using Eq. (8.8) we can write the state vector $\ket{\psi}$ as \begin{equation} \ket{\psi} = \ketsl{\slOne}\braket{\slOne}{\psi}+ \ketsl{\slTwo}\braket{\slTwo}{\psi}\notag \end{equation}  or \begin{equation} \label{Eq:III:8:44} \ket{\psi} =\ketsl{\slOne}C_1+\ketsl{\slTwo}C_2. \end{equation}  Now the interesting thing is that if the molecule is known to be in some state at some instant, it will not be in the same state a little while later. The two $C$-coefficients will be changing with time according to the equations (8.43)—which hold for any two-state system. Suppose, for example, that you had made some observation—or had made some selection of the molecules—so that you know that the molecule is initially in the state $\ketsl{\slOne}$. At some later time, there is some chance that it will be found in state $\ketsl{\slTwo}$. To find out what this chance is, we have to solve the differential equation which tells us how the amplitudes change with time. The only trouble is that we don’t know what to use for the coefficients $H_{ij}$ in Eq. (8.43). There are some things we can say, however. Suppose that once the molecule was in the state $\ketsl{\slOne}$ there was no chance that it could ever get into $\ketsl{\slTwo}$, and vice versa. Then $H_{12}$ and $H_{21}$ would both be zero, and Eq. (8.43) would read \begin{equation*} i\hbar\,\ddt{C_1}{t}=H_{11}C_1,\quad i\hbar\,\ddt{C_2}{t}=H_{22}C_2. \end{equation*}  We can easily solve these two equations; we get \begin{equation} \label{Eq:III:8:45} C_1=(\text{const})e^{-(i/\hbar)H_{11}t},\quad C_2=(\text{const})e^{-(i/\hbar)H_{22}t}. \end{equation}  These are just the amplitudes for stationary states with the energies $E_1=H_{11}$ and $E_2=H_{22}$. We note, however, that for the ammonia molecule the two states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$ have a definite symmetry. If nature is at all reasonable, the matrix elements $H_{11}$ and $H_{22}$ must be equal. We’ll call them both $E_0$, because they correspond to the energy the states would have if $H_{12}$ and $H_{21}$ were zero. But Eqs. (8.45) do not tell us what ammonia really does. It turns out that it is possible for the nitrogen to push its way through the three hydrogens and flip to the other side. It is quite difficult; to get half-way through requires a lot of energy. How can it get through if it hasn’t got enough energy?  There is some amplitude that it will penetrate the energy barrier. It is possible in quantum mechanics to sneak quickly across a region which is illegal energetically. There is, therefore, some small amplitude that a molecule which starts in $\ketsl{\slOne}$ will get to the state $\ketsl{\slTwo}$. The coefficients $H_{12}$ and $H_{21}$ are not really zero. Again, by symmetry, they should both be the same—at least in magnitude. In fact, we already know that, in general, $H_{ij}$ must be equal to the complex conjugate of $H_{ji}$, so they can differ only by a phase. It turns out, as you will see, that there is no loss of generality if we take them equal to each other. For later convenience we set them equal to a negative number; we take $H_{12}=H_{21}=-A$. We then have the following pair of equations: \begin{align} \label{Eq:III:8:46} i\hbar\,\ddt{C_1}{t}&=E_0C_1-AC_2,\\[1ex] \label{Eq:III:8:47} i\hbar\,\ddt{C_2}{t}&=E_0C_2-AC_1. \end{align}  These equations are simple enough and can be solved in any number of ways. One convenient way is the following. Taking the sum of the two, we get \begin{equation} i\hbar\,\ddt{}{t}\,(C_1+C_2)=(E_0-A)(C_1+C_2),\notag \end{equation}  whose solution is \begin{equation} \label{Eq:III:8:48} C_1+C_2=ae^{-(i/\hbar)(E_0-A)t}. \end{equation}   Then, taking the difference of (8.46) and (8.47), we find that \begin{equation} i\hbar\,\ddt{}{t}\,(C_1-C_2)=(E_0+A)(C_1-C_2),\notag \end{equation}  which gives \begin{equation} \label{Eq:III:8:49} C_1-C_2=be^{-(i/\hbar)(E_0+A)t}. \end{equation}  We have called the two integration constants $a$ and $b$; they are, of course, to be chosen to give the appropriate starting condition for any particular physical problem. Now, by adding and subtracting (8.48) and (8.49), we get $C_1$ and $C_2$: \begin{align} \label{Eq:III:8:50} C_1(t)&=\frac{a}{2}\,e^{-(i/\hbar)(E_0-A)t}+ \frac{b}{2}\,e^{-(i/\hbar)(E_0+A)t},\\[1ex] \label{Eq:III:8:51} C_2(t)&=\frac{a}{2}\,e^{-(i/\hbar)(E_0-A)t}- \frac{b}{2}\,e^{-(i/\hbar)(E_0+A)t}. \end{align}  They are the same except for the sign of the second term. We have the solutions; now what do they mean? (The trouble with quantum mechanics is not only in solving the equations but in understanding what the solutions mean!) First, notice that if $b=0$, both terms have the same frequency $\omega=(E_0-A)/\hbar$. If everything changes at one frequency, it means that the system is in a state of definite energy—here, the energy $(E_0-A)$. So there is a stationary state of this energy in which the two amplitudes $C_1$ and $C_2$ are equal. We get the result that the ammonia molecule has a definite energy $(E_0-A)$ if there are equal amplitudes for the nitrogen atom to be “up” and to be “down.” There is another stationary state possible if $a=0$; both amplitudes then have the frequency $(E_0+A)/\hbar$. So there is another state with the definite energy $(E_0+A)$ if the two amplitudes are equal but with the opposite sign; $C_2=-C_1$. These are the only two states of definite energy. We will discuss the states of the ammonia molecule in more detail in the next chapter; we will mention here only a couple of things. We conclude that because there is some chance that the nitrogen atom can flip from one position to the other, the energy of the molecule is not just $E_0$, as we would have expected, but that there are two energy levels $(E_0+A)$ and $(E_0-A)$. Every one of the possible states of the molecule, whatever energy it has, is “split” into two levels. We say every one of the states because, you remember, we picked out one particular state of rotation, and internal energy, and so on. For each possible condition of that kind there is a doublet of energy levels because of the flip-flop of the molecule. Let’s now ask the following question about an ammonia molecule. Suppose that at $t=0$, we know that a molecule is in the state $\ketsl{\slOne}$ or, in other words, that $C_1(0)=1$ and $C_2(0)=0$. What is the probability that the molecule will be found in the state $\ketsl{\slTwo}$ at the time $t$, or will still be found in state $\ketsl{\slOne}$ at the time $t$? Our starting condition tells us what $a$ and $b$ are in Eqs. (8.50) and (8.51). Letting $t=0$, we have that \begin{equation*} C_1(0)=\frac{a+b}{2}=1,\quad C_2(0)=\frac{a-b}{2}=0. \end{equation*}  Clearly, $a=b=1$. Putting these values into the formulas for $C_1(t)$ and $C_2(t)$ and rearranging some terms, we have \begin{align*} C_1(t)&=e^{-(i/\hbar)E_0t}\biggl( \frac{e^{(i/\hbar)At}+e^{-(i/\hbar)At}}{2} \biggr),\\[1ex] C_2(t)&=e^{-(i/\hbar)E_0t}\biggl( \frac{e^{(i/\hbar)At}-e^{-(i/\hbar)At}}{2} \biggr). \end{align*}  We can rewrite these as \begin{align} \label{Eq:III:8:52} C_1(t)&=\phantom{i}e^{-(i/\hbar)E_0t}\cos\frac{At}{\hbar},\\[1ex] \label{Eq:III:8:53} C_2(t)&=ie^{-(i/\hbar)E_0t}\sin\frac{At}{\hbar}. \end{align}  The two amplitudes have a magnitude that varies harmonically with time. The probability that the molecule is found in state $\ketsl{\slTwo}$ at the time $t$ is the absolute square of $C_2(t)$: \begin{equation} \label{Eq:III:8:54} \abs{C_2(t)}^2=\sin^2\frac{At}{\hbar}. \end{equation}  The probability starts at zero (as it should), rises to one, and then oscillates back and forth between zero and one, as shown in the curve marked $P_2$ of Fig. 8–2. The probability of being in the $\ketsl{\slOne}$ state does not, of course, stay at one. It “dumps” into the second state until the probability of finding the molecule in the first state is zero, as shown by the curve $P_1$ of Fig. 8–2. The probability sloshes back and forth between the two. A long time ago we saw what happens when we have two equal pendulums with a slight coupling. (See Chapter 49, Vol. I.) When we lift one back and let go, it swings, but then gradually the other one starts to swing. Pretty soon the second pendulum has picked up all the energy. Then, the process reverses, and pendulum number one picks up the energy. It is exactly the same kind of a thing. The speed at which the energy is swapped back and forth depends on the coupling between the two pendulums—the rate at which the “oscillation” is able to leak across. Also, you remember, with the two pendulums there are two special motions—each with a definite frequency—which we call the fundamental modes. If we pull both pendulums out together, they swing together at one frequency. On the other hand, if we pull one out one way and the other out the other way, there is another stationary mode also at a definite frequency. Well, here we have a similar situation—the ammonia molecule is mathematically like the pair of pendulums. These are the two frequencies—$(E_0-A)/\hbar$ and $(E_0+A)/\hbar$—for when they are oscillating together, or oscillating opposite. The pendulum analogy is not much deeper than the principle that the same equations have the same solutions. The linear equations for the amplitudes (8.39) are very much like the linear equations of harmonic oscillators. (In fact, this is the reason behind the success of our classical theory of the index of refraction, in which we replaced the quantum mechanical atom by a harmonic oscillator, even though, classically, this is not a reasonable view of electrons circulating about a nucleus.) If you pull the nitrogen to one side, then you get a superposition of these two frequencies, and you get a kind of beat note, because the system is not in one or the other states of definite frequency. The splitting of the energy levels of the ammonia molecule is, however, strictly a quantum mechanical effect. The splitting of the energy levels of the ammonia molecule has important practical applications which we will describe in the next chapter. At long last we have an example of a practical physical problem that you can understand with the quantum mechanics!