The hydrogen molecular ion
In the last chapter we discussed some aspects of the ammonia molecule under the approximation that it can be considered as a two-state system. It is, of course, not really a two-state system—there are many states of rotation, vibration, translation, and so on—but each of these states of motion must be analyzed in terms of two internal states because of the flip-flop of the nitrogen atom. Here we are going to consider other examples of systems which, to some approximation or other, can be considered as two-state systems. Lots of things will be approximate because there are always many other states, and in a more accurate analysis they would have to be taken into account. But in each of our examples we will be able to understand a great deal by just thinking about two states.  Since we will only be dealing with two-state systems, the Hamiltonian we need will look just like the one we used in the last chapter. When the Hamiltonian is independent of time, we know that there are two stationary states with definite—and usually different—energies. Generally, however, we start our analysis with a set of base states which are not these stationary states, but states which may, perhaps, have some other simple physical meaning. Then, the stationary states of the system will be represented by a linear combination of these base states. For convenience, we will summarize the important equations from Chapter 9. Let the original choice of base states be $\ketsl{\slOne}$ and $\ketsl{\slTwo}$. Then any state $\ket{\psi}$ is represented by the linear combination  \begin{equation} \label{Eq:III:10:1} \ket{\psi}=\ketsl{\slOne}\braket{\slOne}{\psi}+ \ketsl{\slTwo}\braket{\slTwo}{\psi}= \ketsl{\slOne}C_1+\ketsl{\slTwo}C_2. \end{equation} 
  \begin{align} \ket{\psi}&=\ketsl{\slOne}\braket{\slOne}{\psi}+\ketsl{\slTwo}\braket{\slTwo}{\psi}\notag\\[1ex] \label{Eq:III:10:1} &=\ketsl{\slOne}C_1+\ketsl{\slTwo}C_2. \end{align}   The amplitudes $C_i$ (by which we mean either $C_1$ or $C_2$) satisfy the two linear differential equations \begin{equation} \label{Eq:III:10:2} i\hbar\,\ddt{C_i}{t}=\sum_jH_{ij}C_j, \end{equation}  where both $i$ and $j$ take on the values $1$ and $2$. When the terms of the Hamiltonian $H_{ij}$ do not depend on $t$, the two states of definite energy (the stationary states), which we call \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*}  have the energies \begin{equation} \begin{aligned} E_{\slI}&=\!\frac{H_{11}\!+H_{22}}{2}\!+\! \sqrt{\biggl(\!\frac{H_{11}\!-H_{22}}{2}\!\biggr)^2\!\!\!+ H_{12}H_{21}},\\[1.5ex] E_{\slII}&=\!\frac{H_{11}\!+H_{22}}{2}\!-\! \sqrt{\biggl(\!\frac{H_{11}\!-H_{22}}{2}\!\biggr)^2\!\!\!+ H_{12}H_{21}}. \end{aligned} \label{Eq:III:10:3} \end{equation}  The two $C$’s for each of these states have the same time dependence. The state vectors $\ketsl{\slI}$ and $\ketsl{\slII}$ which go with the stationary states are related to our original base states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$ by \begin{equation} \begin{aligned} \ketsl{\slI}&=\ketsl{\slOne}a_1+\ketsl{\slTwo}a_2,\\[1ex] \ketsl{\slII}&=\ketsl{\slOne}a_1'+\ketsl{\slTwo}a_2'. \end{aligned} \label{Eq:III:10:4} \end{equation}  The $a$’s are complex constants, which satisfy \begin{gather} \abs{a_1}^2+\abs{a_2}^2=1,\notag\\[2ex] \label{Eq:III:10:5} \frac{a_1}{a_2}= \frac{H_{12}}{E_{\slI}-H_{11}},\\[2ex] \abs{a_1'}^2+\abs{a_2'}^2=1,\notag\\[2ex] \label{Eq:III:10:6} \frac{a_1'}{a_2'}= \frac{H_{12}}{E_{\slII}-H_{11}}. \end{gather}  If $H_{11}$ and $H_{22}$ are equal—say both are equal to $E_0$—and $H_{12}=H_{21}=-A$, then $E_{\slI}=E_0+A$, $E_{\slII}=E_0-A$, and the states $\ketsl{\slI}$ and $\ketsl{\slII}$ are particularly simple:  \begin{equation} \label{Eq:III:10:7} \ketsl{\slI}=\frac{1}{\sqrt{2}}\,\biggl[ \ketsl{\slOne}-\ketsl{\slTwo}\biggr],\quad \ketsl{\slII}=\frac{1}{\sqrt{2}}\,\biggl[ \ketsl{\slOne}+\ketsl{\slTwo}\biggr]. \end{equation} 
 \begin{equation} \begin{aligned} \ketsl{\slI}=\frac{1}{\sqrt{2}}\,\biggl[ \ketsl{\slOne}-\ketsl{\slTwo}\biggr],\\[1.5ex] \ketsl{\slII}=\frac{1}{\sqrt{2}}\,\biggl[ \ketsl{\slOne}+\ketsl{\slTwo}\biggr]. \end{aligned} \label{Eq:III:10:7} \end{equation} 
 Now we will use these results to discuss a number of interesting examples taken from the fields of chemistry and physics. The first example is the hydrogen molecular ion. A positively ionized hydrogen molecule consists of two protons with one electron worming its way around them. If the two protons are very far apart, what states would we expect for this system? The answer is pretty clear: The electron will stay close to one proton and form a hydrogen atom in its lowest state, and the other proton will remain alone as a positive ion. So, if the two protons are far apart, we can visualize one physical state in which the electron is “attached” to one of the protons. There is, clearly, another state symmetric to that one in which the electron is near the other proton, and the first proton is the one that is an ion. We will take these two as our base states, and we’ll call them $\ketsl{\slOne}$ and $\ketsl{\slTwo}$. They are sketched in Fig. 10–1. Of course, there are really many states of an electron near a proton, because the combination can exist as any one of the excited states of the hydrogen atom. We are not interested in that variety of states now; we will consider only the situation in which the hydrogen atom is in the lowest state—its ground state—and we will, for the moment, disregard spin of the electron. We can just suppose that for all our states the electron has its spin “up” along the $z$-axis.1 Now to remove an electron from a hydrogen atom requires $13.6$ electron volts of energy. So long as the two protons of the hydrogen molecular ion are far apart, it still requires about this much energy—which is for our present considerations a great deal of energy—to get the electron somewhere near the midpoint between the protons. So it is impossible, classically, for the electron to jump from one proton to the other. However, in quantum mechanics it is possible—though not very likely. There is some small amplitude for the electron to move from one proton to the other. As a first approximation, then, each of our base states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$ will have the energy $E_0$, which is just the energy of one hydrogen atom plus one proton. We can take that the Hamiltonian matrix elements $H_{11}$ and $H_{22}$ are both approximately equal to $E_0$. The other matrix elements $H_{12}$ and $H_{21}$, which are the amplitudes for the electron to go back and forth, we will again write as $-A$. You see that this is the same game we played in the last two chapters. If we disregard the fact that the electron can flip back and forth, we have two states of exactly the same energy. This energy will, however, be split into two energy levels by the possibility of the electron going back and forth—the greater the probability of the transition, the greater the split. So the two energy levels of the system are $E_0+A$ and $E_0-A$; and the states which have these definite energies are given by Eqs. (10.7). From our solution we see that if a proton and a hydrogen atom are put anywhere near together, the electron will not stay on one of the protons but will flip back and forth between the two protons. If it starts on one of the protons, it will oscillate back and forth between the states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$—giving a time-varying solution. In order to have the lowest energy solution (which does not vary with time), it is necessary to start the system with equal amplitudes for the electron to be around each proton. Remember, there are not two electrons—we are not saying that there is an electron around each proton. There is only one electron, and it has the same amplitude—$1/\sqrt{2}$ in magnitude—to be in either position. Now the amplitude $A$ for an electron which is near one proton to get to the other one depends on the separation between the protons. The closer the protons are together, the larger the amplitude. You remember that we talked in Chapter 7 about the amplitude for an electron to “penetrate a barrier,” which it could not do classically. We have the same situation here. The amplitude for an electron to get across decreases roughly exponentially with the distance—for large distances. Since the transition probability, and therefore $A$, gets larger when the protons are closer together, the separation of the energy levels will also get larger. If the system is in the state $\ketsl{\slI}$, the energy $E_0+A$ increases with decreasing distance, so these quantum mechanical effects make a repulsive force tending to keep the protons apart. On the other hand, if the system is in the state $\ketsl{\slII}$, the total energy decreases if the protons are brought closer together; there is an attractive force pulling the protons together. The variation of the two energies with the distance between the two protons should be roughly as shown in Fig. 10–2. We have, then, a quantum-mechanical explanation of the binding force that holds the $\text{H}_2^+$ ion together. We have, however, forgotten one thing. In addition to the force we have just described, there is also an electrostatic repulsive force between the two protons. When the two protons are far apart—as in Fig. 10–1—the “bare” proton sees only a neutral atom, so there is a negligible electrostatic force. At very close distances, however, the “bare” proton begins to get “inside” the electron distribution—that is, it is closer to the proton on the average than to the electron. So there begins to be some extra electrostatic energy which is, of course, positive. This energy—which also varies with the separation—should be included in $E_0$. So for $E_0$ we should take something like the broken-line curve in Fig. 10–2 which rises rapidly for distances less than the radius of a hydrogen atom. We should add and subtract the flip-flop energy $A$ from this $E_0$. When we do that, the energies $E_{\slI}$ and $E_{\slII}$ will vary with the interproton distance $D$ as shown in Fig. 10–3. [In this figure, we have plotted the results of a more detailed calculation. The interproton distance is given in units of $1$ Å ($10^{-8}$ cm), and the excess energy over a proton plus a hydrogen atom is given in units of the binding energy of the hydrogen atom—the so-called “Rydberg” energy, $13.6$ eV.] We see that the state $\ketsl{\slII}$ has a minimum-energy point. This will be the equilibrium configuration—the lowest energy condition—for the $\text{H}_2^+$ ion. The energy at this point is lower than the energy of a separated proton and hydrogen ion, so the system is bound. A single electron acts to hold the two protons together. A chemist would call it a “one-electron bond.”  This kind of chemical binding is also often called “quantum mechanical resonance” (by analogy with the two coupled pendulums we have described before). But that really sounds more mysterious than it is, it’s only a “resonance” if you start out by making a poor choice for your base states—as we did also! If you picked the state $\ketsl{\slII}$, you would have the lowest energy state—that’s all. We can see in another way why such a state should have a lower energy than a proton and a hydrogen atom. Let’s think about an electron near two protons with some fixed, but not too large, separation. You remember that with a single proton the electron is “spread out” because of the uncertainty principle. It seeks a balance between having a low Coulomb potential energy and not getting confined into too small a space, which would make a high kinetic energy (because of the uncertainty relation $\Delta p\,\Delta x\approx\hbar$). Now if there are two protons, there is more space where the electron can have a low potential energy. It can spread out—lowering its kinetic energy—without increasing its potential energy. The net result is a lower energy than a proton and a hydrogen atom. Then why does the other state $\ketsl{\slI}$ have a higher energy? Notice that this state is the difference of the states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$. Because of the symmetry of $\ketsl{\slOne}$ and $\ketsl{\slTwo}$, the difference must have zero amplitude to find the electron half-way between the two protons. This means that the electron is somewhat more confined, which leads to a larger energy. We should say that our approximate treatment of the $\text{H}_2^+$ ion as a two-state system breaks down pretty badly once the protons get as close together as they are at the minimum in the curve of Fig. 10–3, and so, will not give a good value for the actual binding energy. For small separations, the energies of the two “states” we imagined in Fig. 10–1 are not really equal to $E_0$; a more refined quantum mechanical treatment is needed. Suppose we ask now what would happen if instead of two protons, we had two different objects—as, for example, one proton and one lithium positive ion (both particles still with a single positive charge). In such a case, the two terms $H_{11}$ and $H_{22}$ of the Hamiltonian would no longer be equal; they would, in fact, be quite different. If it should happen that the difference $(H_{11}-H_{22})$ is, in absolute value, much greater than $A=-H_{12}$, the attractive force gets very weak, as we can see in the following way. If we put $H_{12}H_{21}=A^2$ into Eqs. (10.3) we get \begin{equation*} E=\frac{H_{11}+H_{22}}{2}\pm \frac{H_{11}-H_{22}}{2} \sqrt{1+\frac{4A^2}{(H_{11}-H_{22})^2}}. \end{equation*}  When $H_{11}-H_{22}$ is much greater than $A^2$, the square root is very nearly equal to \begin{equation*} 1+\frac{2A^2}{(H_{11}-H_{22})^2}. \end{equation*}  The two energies are then \begin{equation} \begin{aligned} E_{\slI}&=H_{11}+ \frac{A^2}{(H_{11}-H_{22})},\\[1ex] E_{\slII}&=H_{22}- \frac{A^2}{(H_{11}-H_{22})}. \end{aligned} \label{Eq:III:10:8} \end{equation}  They are now very nearly just the energies $H_{11}$ and $H_{22}$ of the isolated atoms, pushed apart only slightly by the flip-flop amplitude $A$.  The energy difference $E_{\slI}-E_{\slII}$ is \begin{equation*} (H_{11}-H_{22})+\frac{2A^2}{H_{11}-H_{22}}. \end{equation*}  The additional separation from the flip-flop of the electron is no longer equal to $2A$; it is smaller by the factor $A/(H_{11}-H_{22})$, which we are now taking to be much less than one. Also, the dependence of $E_{\slI}-E_{\slII}$ on the separation of the two nuclei is much smaller than for the $\text{H}_2^+$ ion—it is also reduced by the factor $A/(H_{11}-H_{22})$. We can now see why the binding of unsymmetric diatomic molecules is generally very weak. In our theory of the $\text{H}_2^+$ ion we have discovered an explanation for the mechanism by which an electron shared by two protons provides, in effect, an attractive force between the two protons which can be present even when the protons are at large distances. The attractive force comes from the reduced energy of the system due to the possibility of the electron jumping from one proton to the other. In such a jump the system changes from the configuration (hydrogen atom, proton) to the configuration (proton, hydrogen atom), or switches back. We can write the process symbolically as \begin{equation*} (H,p)\rightleftharpoons (p,H). \end{equation*}  The energy shift due to this process is proportional to the amplitude $A$ that an electron whose energy is $-W_H$ (its binding energy in the hydrogen atom) can get from one proton to the other. For large distances $R$ between the two protons, the electrostatic potential energy of the electron is nearly zero over most of the space it must go when it makes its jump. In this space, then, the electron moves nearly like a free particle in empty space—but with a negative energy! We have seen in Chapter 3 [Eq. (3.7)] that the amplitude for a particle of definite energy to get from one place to another a distance $r$ away is proportional to \begin{equation*} \frac{e^{(i/\hbar)pr}}{r}, \end{equation*}  where $p$ is the momentum corresponding to the definite energy. In the present case (using the nonrelativistic formula), $p$ is given by \begin{equation} \label{Eq:III:10:9} \frac{p^2}{2m}=-W_H. \end{equation}  This means that $p$ is an imaginary number, \begin{equation*} p=i\sqrt{2mW_H} \end{equation*}  (the other sign for the radical gives nonsense here). We should expect, then, that the amplitude $A$ for the $\text{H}_2^+$ ion will vary as \begin{equation} \label{Eq:III:10:10} A\propto\frac{e^{-(\sqrt{2mW_H}/\hbar)R}}{R} \end{equation}  for large separations $R$ between the two protons. The energy shift due to the electron binding is proportional to $A$, so there is a force pulling the two protons together which is proportional—for large $R$—to the derivative of (10.10) with respect to $R$. Finally, to be complete, we should remark that in the two-proton, one-electron system there is still one other effect which gives a dependence of the energy on $R$.  We have neglected it until now because it is usually rather unimportant—the exception is just for those very large distances where the energy of the exchange term $A$ has decreased exponentially to very small values. The new effect we are thinking of is the electrostatic attraction of the proton for the hydrogen atom, which comes about in the same way any charged object attracts a neutral object.  The bare proton makes an electric field $\Efield$ (varying as $1/R^2$) at the neutral hydrogen atom. The atom becomes polarized, taking on an induced dipole moment $\mu$ proportional to $\Efield$. The energy of the dipole is $\mu\Efield$, which is proportional to $\Efield^2$—or to $1/R^4$. So there is a term in the energy of the system which decreases with the fourth power of the distance. (It is a correction to $E_0$.) This energy falls off with distance more slowly than the shift $A$ given by (10.10); at some large separation $R$ it becomes the only remaining important term giving a variation of energy with $R$—and, therefore, the only remaining force. Note that the electrostatic term has the same sign for both of the base states (the force is attractive, so the energy is negative) and so also for the two stationary states, whereas the electron exchange term $A$ gives opposite signs for the two stationary states.