The states in a magnetic field
We would like now to discuss the shapes of the curves in Fig. 12-3. In the first place, the energies for large fields are easy to understand, and rather interesting. For $B$ large enough (namely for $\mu B/A\gg1$) we can neglect the $1$ in the formulas of (12.37). The four energies become \begin{equation} \begin{alignedat}{2} E_{\slI}\;&=A+\mu B,&\quad E_{\slII}\;&=A-\mu B,\\[1ex] E_{\slIII}\;&=-A+\mu'B,&\quad E_{\slIV}\;&=-A-\mu'B. \end{alignedat} \label{Eq:III:12:38} \end{equation} These are the equations of the four straight lines in Fig. 12-3. We can understand these energies physically in the following way. The nature of the stationary states in a zero field is determined completely by the interaction of the two magnetic moments.  The mixtures of the base states $\ket{+\,-}$ and $\ket{-\,+}$ in the stationary states $\ketsl{\slIII}$ and $\ketsl{\slIV}$ are due to this interaction. In large external fields, however, the proton and electron will be influenced hardly at all by the field of the other; each will act as if it were alone in the external field. Then—as we have seen many times—the electron spin will be either parallel to or opposite to the external magnetic field.  Suppose the electron spin is “up”—that is, along the field; its energy will be $-\mu_{\text{e}}B$. The proton can still be either way. If the proton spin is also “up,” its energy is $-\mu_{\text{p}}B$. The sum of the two is $-(\mu_{\text{e}}+\mu_{\text{p}})B=\mu B$. That is just what we find for $E_{\slI}$—which is fine, because we are describing the state $\ket{+\,+}=\ketsl{\slI}$. There is still the small additional term $A$ (now $\mu B\gg A$) which represents the interaction energy of the proton and electron when their spins are parallel. (We originally took $A$ as positive because the theory we spoke of says it should be, and experimentally it is indeed so.) On the other hand, the proton can have its spin down. Then its energy in the external field goes to $+\mu_{\text{p}}B$, so it and the electron have the energy $-(\mu_{\text{e}}-\mu_{\text{p}})B=\mu'B$. And the interaction energy becomes $-A$. The sum is just the energy $E_{\slIII}$, in (12.38). So the state $\ketsl{\slIII}$ must for large fields become the state $\ket{+\,-}$.  Suppose now the electron spin is “down.” Its energy in the external field is $\mu_{\text{e}}B$. If the proton is also “down,” the two together have the energy $(\mu_{\text{e}}+\mu_{\text{p}})B=-\mu B$, plus the interaction energy $A$—since their spins are parallel. That makes just the energy $E_{\slII}$ in (12.38) and corresponds to the state $\ket{-\,-}=\ketsl{\slII}$—which is nice. Finally if the electron is “down” and the proton is “up,” we get the energy $(\mu_{\text{e}}-\mu_{\text{p}})B-A$ (minus $A$ for the interaction because the spins are opposite) which is just $E_{\slIV}$. And the state corresponds to $\ket{-\,+}$.  “But, wait a moment!”, you are probably saying, “The states $\ketsl{\slIII}$ and $\ketsl{\slIV}$ are not the states $\ket{+\,-}$ and $\ket{-\,+}$; they are mixtures of the two.” Well, only slightly. They are indeed mixtures for $B=0$, but we have not yet figured out what they are for large $B$. When we used the analogies of (12.33) in our formulas of Chapter 9 to get the energies of the stationary states, we could also have taken the amplitudes that go with them. They come from Eq. (9.24), which is \begin{equation*} \frac{a_2}{a_3}=\frac{E-H_{22}}{H_{21}}. \end{equation*} The ratio $a_2/a_3$ is, of course, just $C_2/C_3$. Plugging in the analogous quantities from (12.33), we get \begin{align} \frac{C_2}{C_3}&= \frac{E+A-(\mu_{\text{e}}-\mu_{\text{p}})B}{2A}\notag\\ \kern{-6em}\text{or}\notag\\ % ebook remove % ebook insert: \makebox[0em]{\kern{-10em}\text{or}}\notag\\ \label{Eq:III:12:39} \frac{C_2}{C_3}&=\frac{E+A+\mu'B}{2A}, \end{align} where for $E$ we are to use the appropriate energy—either $E_{\slIII}$ or $E_{\slIV}$. For instance, for state $\ketsl{\slIII}$ we have \begin{equation} \label{Eq:III:12:40} \biggl(\frac{C_2}{C_3}\biggr)_{\slIII}\approx\frac{\mu'B}{A}. \end{equation} So for large $B$ the state $\ketsl{\slIII}$ has $C_2\gg C_3$; the state becomes almost completely the state $\ketsl{\slTwo}=\ket{+\,-}$. Similarly, if we put $E_{\slIV}$ into (12.39) we get $(C_2/C_3)_{\slIV}\ll1$; for high fields state $\ketsl{\slIV}$ becomes just the state $\ketsl{\slThree}=\ket{-\,+}$. You see that the coefficients in the linear combinations of our base states which make up the stationary states depend on $B$. The state we call $\ketsl{\slIII}$ is a $50$-$50$ mixture of $\ket{+\,-}$ and $\ket{-\,+}$ at very low fields, but shifts completely over to $\ket{+\,-}$ at high fields. Similarly, the state $\ketsl{\slIV}$, which at low fields is also a $50$-$50$ mixture (with opposite signs) of $\ket{+\,-}$ and $\ket{-\,+}$, goes over into the state $\ket{-\,+}$ when the spins are uncoupled by a strong external field.  We would also like to call your attention particularly to what happens at very low magnetic fields. There is one energy—at $-3A$—which does not change when you turn on a small magnetic field. And there is another energy—at $+A$—which splits into three different energy levels  when you turn on a small magnetic field. For weak fields the energies vary with $B$ as shown  in Fig. 12–5. Suppose that we have somehow selected a bunch  of hydrogen atoms which all have the energy $-3A$. If we put them through a Stern-Gerlach  experiment—with fields that are not too strong—we would find that they just go straight through.  (Since their energy doesn’t depend on $B$, there is—according to the principle of virtual  work—no force on them in a magnetic field gradient.) Suppose, on the other hand, we were to  select a bunch of atoms with the energy $+A$, and put them through a Stern-Gerlach apparatus,  say an $S$ apparatus. (Again the fields in the apparatus should not be so great that they disrupt  the insides of the atom, by which we mean a field small enough that the energies vary linearly  with $B$.) We would find three beams. The states $\ketsl{\slI}$ and  $\ketsl{\slII}$ get opposite forces—their energies vary linearly with $B$ with the slopes $\pm\mu$  so the forces are like those on a dipole with $\mu_z=\mp\mu$; but the state  $\ketsl{\slIII}$ goes straight through. So we are right back in Chapter 5.  A hydrogen atom with the energy $+A$ is a spin-one particle.  This energy state is a “particle” for which $j=1$, and it can be described—with respect to some set of axes in space—in terms of the base states $\ket{+S}$, $\ket{\OS}$, and $\ket{-S}$ we used in Chapter 5. On the other hand, when a hydrogen atom has the energy $-3A$, it is a spin-zero particle. (Remember, what we are saying is only strictly true for infinitesimal magnetic fields.)  So we can group the states of hydrogen in zero magnetic field this way: \begin{alignat}{2} & \left. \begin{aligned} \ketsl{\slI}&=\ket{+\,+}\\[2pt] \ketsl{\slIII}&=\dfrac{\ket{+\,-}+\ket{-\,+}}{\sqrt{2}}\\[3pt] \ketsl{\slII}&=\ket{-\,-}\\ \end{aligned} \right\} && \;\text{spin $1$}\; \left\{ \begin{array}{@{}l@{}} \ket{+S}\\[2pt] \vphantom{\dfrac{\ket{+\,-}+\ket{-\,+}}{\sqrt{2}}}\ket{\OS}\\[3pt] \ket{-S}\\ \label{Eq:III:12:41} \end{array} \right.\\[1ex] & \left. % ebook insert: \phantom{0} \ketsl{\slIV}=\dfrac{\ket{+\,-}-\ket{-\,+}}{\sqrt{2}} \right. && \;\text{spin $0$.} \label{Eq:III:12:42} \end{alignat}  We have said in Chapter 35 of Volume II that for any particle its component of angular momentum along any axis can have only certain values always $\hbar$ apart. The $z$-component of angular momentum $J_z$ can be $j\hbar$, $(j-1)\hbar$, $(j-2)\hbar$, $\ldots$, $(-j)\hbar$, where $j$ is the spin of the particle (which can be an integer or half-integer). Although we neglected to say so at the time, people usually write \begin{equation} \label{Eq:III:12:43} J_z=m\hbar, \end{equation} where $m$ stands for one of the numbers $j$, $j-1$, $j-2$, $\ldots$, $-j$. You will, therefore, see people in books label the four ground states of hydrogen by the so-called quantum numbers $j$ and $m$ [often called the “total angular momentum quantum number” ($j$) and “magnetic quantum number” ($m$)]. Then, instead of our state symbols $\ketsl{\slI}$, $\ketsl{\slII}$, and so on, they will write a state as $\ket{j,m}$. So they would write our little table of states for zero field in (12.41) and (12.42) as shown in Table 12-3. It’s not new physics, it’s all just a matter of notation. 