Base states for a system with two spin one-half particles
In this chapter we take up the “hyperfine splitting” of hydrogen, because it is a physically interesting example of what we can already do with quantum mechanics. It’s an example with more than two states, and it will be illustrative of the methods of quantum mechanics as applied to slightly more complicated problems. It is enough more complicated that once you see how this one is handled you can get immediately the generalization to all kinds of problems.  As you know, the hydrogen atom consists of an electron sitting in the neighborhood of the proton, where it can exist in any one of a number of discrete energy states in each one of which the pattern of motion of the electron is different. The first excited state, for example, lies $3/4$ of a Rydberg, or about $10$ electron volts, above the ground state. But even the so-called ground state of hydrogen is not really a single, definite-energy state, because of the spins of the electron and the proton. These spins are responsible for the “hyperfine structure” in the energy levels, which splits all the energy levels into several nearly equal levels.  The electron can have its spin either “up” or “down,” and the proton can also have its spin either “up” or “down.” There are, therefore, four possible spin states for every dynamical condition of the atom. That is, when people say “the ground state” of hydrogen, they really mean the “four ground states,” and not just the very lowest state. The four spin states do not all have exactly the same energy; there are slight shifts from the energies we would expect with no spins. The shifts are, however, much, much smaller than the $10$ electron volts or so from the ground state to the next state above. As a consequence, each dynamical state has its energy split into a set of very close energy levels—the so-called hyperfine splitting.  The energy differences among the four spin states is what we want to calculate in this chapter. The hyperfine splitting is due to the interaction of the magnetic moments of the electron and proton, which gives a slightly different magnetic energy for each spin state. These energy shifts are only about ten-millionths of an electron volt—really very small compared with $10$ electron volts! It is because of this large gap that we can think about the ground state of hydrogen as a “four-state” system, without worrying about the fact that there are really many more states at higher energies. We are going to limit ourselves here to a study of the hyperfine structure of the ground state of the hydrogen atom.  For our purposes we are not interested in any of the details about the positions of the electron and proton because that has all been worked out by the atom so to speak—it has worked itself out by getting into the ground state. We need know only that we have an electron and proton in the neighborhood of each other with some definite spatial relationship. In addition, they can have various different relative orientations of their spins. It is only the effect of the spins that we want to look into.  The first question we have to answer is: What are the base states for the system? Now the question has been put incorrectly. There is no such thing as “the” base states, because, of course, the set of base states you may choose is not unique. New sets can always be made out of linear combinations of the old. There are always many choices for the base states, and among them, any choice is equally legitimate. So the question is not what is the base set, but what could a base set be? We can choose any one we wish for our own convenience. It is usually best to start with a base set which is physically the clearest. It may not be the solution to any problem, or may not have any direct importance, but it will generally make it easier to understand what is going on.  We choose the following four base states: \begin{align*} &\textit{State 1:}\;\small \text{The electron and proton are both spin “up.”}\\[3pt] &\textit{State 2:}\;\small \text{The electron is “up” and the proton is “down.”}\\[3pt] &\textit{State 3:}\;\small \text{The electron is “down” and the proton is “up.”}\\[3pt] &\textit{State 4:}\;\small \text{The electron and proton are both “down.”} \end{align*} We need a handy notation for these four states, so we’ll represent them this way: \begin{equation} \begin{aligned} &\textit{State 1:}\;\small \ket{++};\;\text{electron}\;\textit{up},\;\text{proton}\;\textit{up}\\ &\textit{State 2:}\;\small \ket{+-};\;\text{electron}\;\textit{up},\;\text{proton}\;\textit{down}\\ &\textit{State 3:}\;\small \ket{-+};\;\text{electron}\;\textit{down},\;\text{proton}\;\textit{up}\\ &\textit{State 4:}\;\small \ket{--};\;\text{electron}\;\textit{down},\;\text{proton}\;\textit{down} \end{aligned} \label{Eq:III:12:1} \end{equation} You will have to remember that the first plus or minus sign refers to the electron and the second, to the proton. For handy reference, we’ve also summarized the notation in Fig. 12-1. Sometimes it will also be convenient to call these states $\ketsl{\slOne}$, $\ketsl{\slTwo}$, $\ketsl{\slThree}$, and $\ketsl{\slFour}$.  You may say, “But the particles interact, and maybe these aren’t the right base states. It sounds as though you are considering the two particles independently.” Yes, indeed! The interaction raises the problem: what is the Hamiltonian for the system, but the interaction is not involved in the question of how to describe the system. What we choose for the base states has nothing to do with what happens next. It may be that the atom cannot ever stay in one of these base states, even if it is started that way. That’s another question. That’s the question: How do the amplitudes change with time in a particular (fixed) base? In choosing the base states, we are just choosing the “unit vectors” for our description.  While we’re on the subject, let’s look at the general problem of finding a set of base states when there is more than one particle. You know the base states for a single particle. An electron, for example, is completely described in real life—not in our simplified cases, but in real life—by giving the amplitudes to be in each of the following states: \begin{equation*} \ket{\text{electron “up” with momentum $\FLPp$}} \end{equation*} or \begin{equation*} \ket{\text{electron “down” with momentum $\FLPp$}}. \end{equation*} There are really two infinite sets of states, one state for each value of $\FLPp$. That is to say that an electron state $\ket{\psi}$ is completely described if you know all the amplitudes \begin{equation*} \braket{+,\FLPp}{\psi}\quad \text{and}\quad \braket{-,\FLPp}{\psi}, \end{equation*} where the $+$ and $-$ represent the components of angular momentum along some axis—usually the $z$-axis—and $\FLPp$ is the vector momentum. There must, therefore, be two amplitudes for every possible momentum (a multi-infinite set of base states). That is all there is to describing a single particle.  When there is more than one particle, the base states can be written in a similar way. For instance, if there were an electron and a proton in a more complicated situation than we are considering, the base states could be of the following kind: \begin{align*} |\,&\text{electron spin “up” with momentum $\FLPp_1$},\\[.5ex] &\text{proton spin “down” with momentum $\FLPp_2$}\rangle. \end{align*} And so on for other spin combinations. If there are more than two particles—same idea. So you see that to write down the possible base states is really very easy. The only problem is, what is the Hamiltonian?  For our study of the ground state of hydrogen we don’t need to use the full sets of base states for the various momenta. We are specifying particular momentum states for the proton and electron when we say “the ground state.” The details of the configuration—the amplitudes for all the momentum base states—can be calculated, but that is another problem. Now we are concerned only with the effects of the spin, so we can take only the four base states of (12.1). Our next problem is: What is the Hamiltonian for this set of states? 