The hydrogen molecule
As our next two-state system we will look at the neutral hydrogen molecule H$_2$. It is, naturally, more complicated to understand because it has two electrons. Again, we start by thinking of what happens when the two protons are well separated. Only now we have two electrons to add. To keep track of them, we’ll call one of them “electron $a$” and the other “electron $b$.” We can again imagine two possible states. One possibility is that “electron $a$” is around the first proton and “electron $b$” is around the second, as shown in the top half of Fig. 10–4. We have simply two hydrogen atoms. We will call this state $\ketsl{\slOne}$. There is also another possibility: that “electron $b$” is around the first proton and that “electron $a$” is around the second. We call this state $\ketsl{\slTwo}$. From the symmetry of the situation, those two possibilities should be energetically equivalent, but, as we will see, the energy of the system is not just the energy of two hydrogen atoms. We should mention that there are many other possibilities. For instance, “electron $a$” might be near the first proton and “electron $b$” might be in another state around the same proton. We’ll disregard such a case, since it will certainly have higher energy (because of the large Coulomb repulsion between the two electrons). For greater accuracy, we would have to include such states, but we can get the essentials of the molecular binding by considering just the two states of Fig. 10–4. To this approximation we can describe any state by giving the amplitude $\braket{\slOne}{\phi}$ to be in the state $\ketsl{\slOne}$ and an amplitude $\braket{\slTwo}{\phi}$ to be in state $\ketsl{\slTwo}$. In other words, the state vector $\ket{\phi}$ can be written as the linear combination \begin{equation*} \ket{\phi}=\sum_i\ket{i}\braket{i}{\phi}. \end{equation*}  To proceed, we assume—as usual—that there is some amplitude $A$ that the electrons can move through the intervening space and exchange places. This possibility of exchange means that the energy of the system is split, as we have seen for other two-state systems. As for the hydrogen molecular ion, the splitting is very small when the distance between the protons is large. As the protons approach each other, the amplitude for the electrons to go back and forth increases, so the splitting increases. The decrease of the lower energy state means that there is an attractive force which pulls the atoms together. Again the energy levels rise when the protons get very close together because of the Coulomb repulsion. The net final result is that the two stationary states have energies which vary with the separation as shown in Fig. 10–5. At a separation of about $0.74$ Å, the lower energy level reaches a minimum; this is the proton-proton distance of the true hydrogen molecule. Now you have probably been thinking of an objection. What about the fact that the two electrons are identical particles? We have been calling them “electron $a$” and “electron $b$,” but there really is no way to tell which is which. And we have said in Chapter 4 that for electrons—which are Fermi particles—if there are two ways something can happen by exchanging the electrons, the two amplitudes will interfere with a negative sign. This means that if we switch which electron is which, the sign of the amplitude must reverse. We have just concluded, however, that the bound state of the hydrogen molecule would be (at $t=0$) \begin{equation*} \ketsl{\slII}=\frac{1}{\sqrt{2}}\,(\ketsl{\slOne}+\ketsl{\slTwo}). \end{equation*}  However, according to our rules of Chapter 4, this state is not allowed. If we reverse which electron is which, we get the state \begin{equation*} \frac{1}{\sqrt{2}}\,(\ketsl{\slTwo}+\ketsl{\slOne}). \end{equation*}  and we get the same sign instead of the opposite one. These arguments are correct if both electrons have the same spin. It is true that if both electrons have spin up (or both have spin down), the only state that is permitted is \begin{equation*} \ketsl{\slI}=\frac{1}{\sqrt{2}}\,(\ketsl{\slOne}-\ketsl{\slTwo}). \end{equation*}  For this state, an interchange of the two electrons gives \begin{equation*} \frac{1}{\sqrt{2}}\,(\ketsl{\slTwo}-\ketsl{\slOne}). \end{equation*}  which is $-\ketsl{\slI}$, as required. So if we bring two hydrogen atoms near to each other with their electrons spinning in the same direction, they can go into the state $\ketsl{\slI}$ and not state $\ketsl{\slII}$. But notice that state $\ketsl{\slI}$ is the upper energy state. Its curve of energy versus separation has no minimum. The two hydrogens will always repel and will not form a molecule. So we conclude that the hydrogen molecule cannot exist with parallel electron spins. And that is right. On the other hand, our state $\ketsl{\slII}$ is perfectly symmetric for the two electrons. In fact, if we interchange which electron we call $a$ and which we call $b$ we get back exactly the same state. We saw in Section 4–7 that if two Fermi particles are in the same state, they must have opposite spins. So, the bound hydrogen molecule must have one electron with spin up and one with spin down.  The whole story of the hydrogen molecule is really somewhat more complicated if we want to include the proton spins. It is then no longer right to think of the molecule as a two-state system. It should really be looked at as an eight-state system—there are four possible spin arrangements for each of our states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$—so we were cutting things a little short by neglecting the spins. Our final conclusions are, however, correct. We find that the lowest energy state—the only bound state—of the H$_2$ molecule has the two electrons with spins opposite. The total spin angular momentum of the electrons is zero. On the other hand, two nearby hydrogen atoms with spins parallel—and so with a total angular momentum $\hbar$—must be in a higher (unbound) energy state; the atoms repel each other. There is an interesting correlation between the spins and the energies. It gives another illustration of something we mentioned before, which is that there appears to be an “interaction” energy between two spins because the case of parallel spins has a higher energy than the opposite case. In a certain sense you could say that the spins try to reach an antiparallel condition and, in doing so, have the potential to liberate energy—not because there is a large magnetic force, but because of the exclusion principle. We saw in Section 10–1 that the binding of two different ions by a single electron is likely to be quite weak. This is not true for binding by two electrons. Suppose the two protons in Fig. 10–4 were replaced by any two ions (with closed inner electron shells and a single ionic charge), and that the binding energies of an electron at the two ions are different. The energies of states $\ketsl{\slOne}$ and $\ketsl{\slTwo}$ would still be equal because in each of these states we have one electron bound to each ion. Therefore, we always have the splitting proportional to $A$. Two-electron binding is ubiquitous—it is the most common valence bond. Chemical binding usually involves this flip-flop game played by two electrons. Although two atoms can be bound together by only one electron, it is relatively rare—because it requires just the right conditions. Finally, we want to mention that if the energy of attraction for an electron to one nucleus is much greater than to the other, then what we have said earlier about ignoring other possible states is no longer right. Suppose nucleus $a$ (or it may be a positive ion) has a much stronger attraction for an electron than does nucleus $b$. It may then happen that the total energy is still fairly low even when both electrons are at nucleus $a$, and no electron is at nucleus $b$. The strong attraction may more than compensate for the mutual repulsion of the two electrons. If it does, the lowest energy state may have a large amplitude to find both electrons at $a$ (making a negative ion) and a small amplitude to find any electron at $b$. The state looks like a negative ion with a positive ion. This is, in fact, what happens in an “ionic” molecule like NaCl. You can see that all the gradations between covalent binding and ionic binding are possible. You can now begin to see how it is that many of the facts of chemistry can be most clearly understood in terms of a quantum mechanical description.