The states of an ammonia molecule
In this chapter we are going to discuss the application of quantum mechanics to a practical device, the ammonia maser. You may wonder why we stop our formal development of quantum mechanics to do a special problem, but you will find that many of the features of this special problem are quite common in the general theory of quantum mechanics, and you will learn a great deal by considering this one problem in detail. The ammonia maser is a device for generating electromagnetic waves, whose operation is based on the properties of the ammonia molecule which we discussed briefly in the last chapter. We begin by summarizing what we found there. The ammonia molecule has many states, but we are considering it as a two-state system, thinking now only about what happens when the molecule is in any specific state of rotation or translation. A physical model for the two states can be visualized as follows. If the ammonia molecule is considered to be rotating about an axis passing through the nitrogen atom and perpendicular to the plane of the hydrogen atoms, as shown in Fig. 9–1, there are still two possible conditions—the nitrogen may be on one side of the plane of hydrogen atoms or on the other. We call these two states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$. They are taken as a set of base states for our analysis of the behavior of the ammonia molecule. In a system with two base states, any state $\ket{\psi}$ of the system can always be described as a linear combination of the two base states; that is, there is a certain amplitude $C_1$ to be in one base state and an amplitude $C_2$ to be in the other. We can write its state vector as \begin{equation} \label{Eq:III:9:1} \ket{\psi}=\ketsl{\slOne}C_1+ \ketsl{\slTwo}C_2, \end{equation}  where \begin{equation} C_1=\braket{\slOne}{\psi}\quad \text{and}\quad C_2=\braket{\slTwo}{\psi}.\notag \end{equation}  These two amplitudes change with time according to the Hamiltonian equations, Eq. (8.43). Making use of the symmetry of the two states of the ammonia molecule, we set $H_{11}=H_{22}=E_0$, and $H_{12}=H_{21}=-A$, and get the solution [see Eqs. (8.50) and (8.51)] \begin{align} \label{Eq:III:9:2} C_1&=\frac{a}{2}\,e^{-(i/\hbar)(E_0-A)t}+ \frac{b}{2}\,e^{-(i/\hbar)(E_0+A)t},\\[1ex] \label{Eq:III:9:3} C_2&=\frac{a}{2}\,e^{-(i/\hbar)(E_0-A)t}- \frac{b}{2}\,e^{-(i/\hbar)(E_0+A)t}. \end{align}  We want now to take a closer look at these general solutions. Suppose that the molecule was initially put into a state $\ket{\psi_{\slII}}$ for which the coefficient $b$ was equal to zero. Then at $t=0$ the amplitudes to be in the states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$ are identical, and they stay that way for all time. Their phases both vary with time in the same way—with the frequency $(E_0-A)/\hbar$. Similarly, if we were to put the molecule into a state $\ket{\psi_{\slI}}$ for which $a=0$, the amplitude $C_2$ is the negative of $C_1$, and this relationship would stay that way forever. Both amplitudes would now vary with time with the frequency $(E_0+A)/\hbar$. These are the only two possibilities of states for which the relation between $C_1$ and $C_2$ is independent of time. We have found two special solutions in which the two amplitudes do not vary in magnitude and, furthermore, have phases which vary at the same frequencies. These are stationary states as we defined them in Section 7–1, which means that they are states of definite energy. The state $\ket{\psi_{\slII}}$ has the energy $E_{\slII}=E_0-A$, and the state $\ket{\psi_{\slI}}$ has the energy $E_{\slI}=E_0+A$. They are the only two stationary states that exist, so we find that the molecule has two energy levels, with the energy difference $2A$. (We mean, of course, two energy levels for the assumed state of rotation and vibration which we referred to in our initial assumptions.)1 If we hadn’t allowed for the possibility of the nitrogen flipping back and forth, we would have taken $A$ equal to zero and the two energy levels would be on top of each other at energy $E_0$. The actual levels are not this way; their average energy is $E_0$, but they are split apart by $\pm A$, giving a separation of $2A$ between the energies of the two states. Since $A$ is, in fact, very small, the difference in energy is also very small. In order to excite an electron inside an atom, the energies involved are relatively very high—requiring photons in the optical or ultraviolet range. To excite the vibrations of the molecules involves photons in the infrared. If you talk about exciting rotations, the energy differences of the states correspond to photons in the far infrared. But the energy difference $2A$ is lower than any of those and is, in fact, below the infrared and well into the microwave region. Experimentally, it has been found that there is a pair of energy levels with a separation of $10^{-4}$ electron volt—corresponding to a frequency $24{,}000$ megacycles. Evidently this means that $2A=hf$, with $f=24{,}000$ megacycles (corresponding to a wavelength of $1\tfrac{1}{4}$ cm). So here we have a molecule that has a transition which does not emit light in the ordinary sense, but emits microwaves. For the work that follows we need to describe these two states of definite energy a little bit better. Suppose we were to construct an amplitude $C_{\slII}$ by taking the sum of the two numbers $C_1$ and $C_2$: \begin{equation} \label{Eq:III:9:4} C_{\slII}=C_1+C_2=\braket{\slOne}{\phi}+\braket{\slTwo}{\phi}. \end{equation}  What would that mean? Well, this is just the amplitude to find the state $\ket{\phi}$ in a new state $\ketsl{\slII}$ in which the amplitudes of the original base states are equal. That is, writing $C_{\slII}=\braket{\slII}{\phi}$, we can abstract the $\ket{\phi}$ away from Eq. (9.4)—because it is true for any $\phi$—and get \begin{equation} \bra{\slII}=\bra{\slOne}+\bra{\slTwo},\notag \end{equation}  which means the same as \begin{equation} \label{Eq:III:9:5} \ketsl{\slII}=\ketsl{\slOne}+\ketsl{\slTwo}. \end{equation}  The amplitude for the state $\ketsl{\slII}$ to be in the state $\ketsl{\slOne}$ is \begin{equation*} \braketsl{\slOne}{\slII}=\braketsl{\slOne}{\slOne}+ \braketsl{\slOne}{\slTwo}, \end{equation*}  which is, of course, just $1$, since $\ketsl{\slOne}$ and $\ketsl{\slTwo}$ are base states. The amplitude for the state $\ketsl{\slII}$ to be in the state $\ketsl{\slTwo}$ is also $1$, so the state $\ketsl{\slII}$ is one which has equal amplitudes to be in the two base states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$. We are, however, in a bit of trouble. The state $\ketsl{\slII}$ has a total probability greater than one of being in some base state or other. That simply means, however, that the state vector is not properly “normalized.” We can take care of that by remembering that we should have $\braketsl{\slII}{\slII}=1$, which must be so for any state. Using the general relation that \begin{equation*} \braket{\chi}{\phi}=\sum_i\braket{\chi}{i}\braket{i}{\phi}, \end{equation*}  letting both $\phi$ and $\chi$ be the state $\slII$, and taking the sum over the base states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$, we get that \begin{equation*} \braketsl{\slII}{\slII}=\braketsl{\slII}{\slOne}\braketsl{\slOne}{\slII}+ \braketsl{\slII}{\slTwo}\braketsl{\slTwo}{\slII}. \end{equation*}  This will be equal to one as it should if we change our definition of $C_{\slII}$—in Eq. (9.4)—to read \begin{equation*} C_{\slII}=\frac{1}{\sqrt{2}}\,[C_1+C_2]. \end{equation*}  In the same way we can construct an amplitude \begin{equation} C_{\slI}=\frac{1}{\sqrt{2}}\,[C_1-C_2],\notag \end{equation}  or \begin{equation} \label{Eq:III:9:6} C_{\slI}=\frac{1}{\sqrt{2}}\,[\braket{\slOne}{\phi}-\braket{\slTwo}{\phi}]. \end{equation}  This amplitude is the projection of the state $\ket{\phi}$ into a new state $\ketsl{\slI}$ which has opposite amplitudes to be in the states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$. Namely, Eq. (9.6) means the same as \begin{equation} \bra{\slI}=\frac{1}{\sqrt{2}}\,[\bra{\slOne}-\bra{\slTwo}],\notag \end{equation}  or \begin{equation} \label{Eq:III:9:7} \ketsl{\slI}=\frac{1}{\sqrt{2}}\,[\ketsl{\slOne}-\ketsl{\slTwo}], \end{equation}  from which it follows that \begin{equation} \braketsl{\slOne}{\slI}=\frac{1}{\sqrt{2}}=-\braketsl{\slTwo}{\slI}.\notag \end{equation}  Now the reason we have done all this is that the states $\ketsl{\slI}$ and $\ketsl{\slII}$ can be taken as a new set of base states which are especially convenient for describing the stationary states of the ammonia molecule. You remember that the requirement for a set of base states is that \begin{equation*} \braket{i}{j}=\delta_{ij}. \end{equation*}  We have already fixed things so that \begin{equation*} \braketsl{\slI}{\slI}=\braketsl{\slII}{\slII}=1. \end{equation*}  You can easily show from Eqs. (9.5) and (9.7) that \begin{equation*} \braketsl{\slI}{\slII}=\braketsl{\slII}{\slI}=0. \end{equation*}  The amplitudes $C_{\slI}=\braket{\slI}{\phi}$ and $C_{\slII}=\braket{\slII}{\phi}$ for any state $\phi$ to be in our new base states $\ketsl{\slI}$ and $\ketsl{\slII}$ must also satisfy a Hamiltonian equation with the form of Eq. (8.39). In fact, if we just subtract the two equations (9.2) and (9.3) and differentiate with respect to $t$, we see that \begin{equation} \label{Eq:III:9:8} i\hbar\,\ddt{C_{\slI}}{t}=(E_0+A)C_{\slI}=E_{\slI}C_{\slI}. \end{equation}  And taking the sum of Eqs. (9.2) and (9.3), we see that \begin{equation} \label{Eq:III:9:9} i\hbar\,\ddt{C_{\slII}}{t}=(E_0-A)C_{\slII}=E_{\slII}C_{\slII}. \end{equation}  Using $\ketsl{\slI}$ and $\ketsl{\slII}$ for base states, the Hamiltonian matrix has the simple form \begin{alignat*}{2} H_{\slI,\slI}&=E_{\slI},&\quad H_{\slI,\slII}&=0,\\[1ex] H_{\slII,\slI}&=0,&\quad H_{\slII,\slII}&=E_{\slII}. \end{alignat*}  Note that each of the Eqs. (9.8) and (9.9) look just like what we had in Section 8–6 for the equation of a one-state system. They have a simple exponential time dependence corresponding to a single energy. As time goes on, the amplitudes to be in each state act independently. The two stationary states $\ket{\psi_{\slI}}$ and $\ket{\psi_{\slII}}$ we found above are, of course, solutions of Eqs. (9.8) and (9.9). The state $\ket{\psi_{\slI}}$ (for which $C_1=-C_2$) has \begin{equation} \label{Eq:III:9:10} C_{\slI}=e^{-(i/\hbar)(E_0+A)t},\quad C_{\slII}=0. \end{equation}  And the state $\ket{\psi_{\slII}}$ (for which $C_1=C_2$) has \begin{equation} \label{Eq:III:9:11} C_{\slI}=0,\quad C_{\slII}=e^{-(i/\hbar)(E_0-A)t}. \end{equation}  Remember that the amplitudes in Eq. (9.10) are \begin{equation*} C_{\slI}=\braket{\slI}{\psi_{\slI}},\quad \text{and}\quad C_{\slII}=\braket{\slII}{\psi_{\slI}}; \end{equation*}  so Eq. (9.10) means the same thing as \begin{equation*} \ket{\psi_{\slI}}=\ketsl{\slI}\,e^{-(i/\hbar)(E_0+A)t}. \end{equation*}  That is, the state vector of the stationary state $\ket{\psi_{\slI}}$ is the same as the state vector of the base state $\ketsl{\slI}$ except for the exponential factor appropriate to the energy of the state. In fact at $t=0$ \begin{equation*} \ket{\psi_{\slI}}=\ketsl{\slI}; \end{equation*}  the state $\ketsl{\slI}$ has the same physical configuration as the stationary state of energy $E_0+A$. In the same way, we have for the second stationary state that \begin{equation*} \ket{\psi_{\slII}}=\ketsl{\slII}\,e^{-(i/\hbar)(E_0-A)t}. \end{equation*}  The state $\ketsl{\slII}$ is just the stationary state of energy $E_0-A$ at $t=0$. Thus our two new base states $\ketsl{\slI}$ and $\ketsl{\slII}$ have physically the form of the states of definite energy, with the exponential time factor taken out so that they can be time-independent base states. (In what follows we will find it convenient not to have to distinguish always between the stationary states $\ket{\psi_{\slI}}$ and $\ket{\psi_{\slII}}$ and their base states $\ketsl{\slI}$ and $\ketsl{\slII}$, since they differ only by the obvious time factors.) In summary, the state vectors $\ketsl{\slI}$ and $\ketsl{\slII}$ are a pair of base vectors which are appropriate for describing the definite energy states of the ammonia molecule. They are related to our original base vectors by \begin{equation} \label{Eq:III:9:12} \ketsl{\slI}=\frac{1}{\sqrt{2}}\,[\ketsl{\slOne}-\ketsl{\slTwo}],\quad \ketsl{\slII}=\frac{1}{\sqrt{2}}\,[\ketsl{\slOne}+\ketsl{\slTwo}]. \end{equation}  The amplitudes to be in $\ketsl{\slI}$ and $\ketsl{\slII}$ are related to $C_1$ and $C_2$ by \begin{equation} \label{Eq:III:9:13} C_{\slI}=\frac{1}{\sqrt{2}}\,[C_1-C_2],\quad C_{\slII}=\frac{1}{\sqrt{2}}\,[C_1+C_2]. \end{equation}  Any state at all can be represented by a linear combination of $\ketsl{\slOne}$ and $\ketsl{\slTwo}$—with the coefficients $C_1$ and $C_2$—or by a linear combination of the definite energy base states $\ketsl{\slI}$ and $\ketsl{\slII}$—with the coefficients $C_{\slI}$ and $C_{\slII}$. Thus, \begin{equation*} \ket{\phi} =\ketsl{\slOne}C_1+\ketsl{\slTwo}C_2 \end{equation*}  or \begin{equation*} \;\ket{\phi} =\ketsl{\slI}C_{\slI}\,+\ketsl{\slII}C_{\slII}. \end{equation*}  The second form gives us the amplitudes for finding the state $\ket{\phi}$ in a state with the energy $E_{\slI}=E_0+A$ or in a state with the energy $E_{\slII}=E_0-A$.