The Zeeman splitting
Although we have finished the problem of finding the energy levels of the hydrogen ground state, we would like to study this interesting system some more. In order to say anything more about it—for instance, in order to calculate the rate at which the hydrogen atom absorbs or emits radio waves at $21$ centimeters—we have to know what happens when the atom is disturbed. We have to do as we did for the ammonia molecule—after we found the energy levels we went on and studied what happened when the molecule was in an electric field. We were then able to figure out the effects from the electric field in a radio wave. For the hydrogen atom, the electric field does nothing to the levels, except to move them all by some constant amount proportional to the square of the field—which is not of any interest because that won’t change the energy differences. It is now the magnetic field which is important. So the next step is to write the Hamiltonian for a more complicated situation in which the atom sits in an external magnetic field.  What, then, is the Hamiltonian? We’ll just tell you the answer, because we can’t give you any “proof” except to say that this is the way the atom works.  The Hamiltonian is \begin{equation} \label{Eq:III:12:27} \Hop=A(\FLPsigmae\cdot\FLPsigmap)- \mu_{\text{e}}\FLPsigmae\cdot\FLPB- \mu_{\text{p}}\FLPsigmap\cdot\FLPB. \end{equation} It now consists of three parts. The first term $A\FLPsigmae\cdot\FLPsigmap$ represents the magnetic interaction between the electron and the proton—it is the same one that would be there if there were no magnetic field. This is the term we have already had; and the influence of the magnetic field on the constant $A$ is negligible. The effect of the external magnetic field shows up in the last two terms. The second term, $-\mu_{\text{e}}\FLPsigmae\cdot\FLPB$, is the energy the electron would have in the magnetic field if it were there alone.5 In the same way, the last term $-\mu_{\text{p}}\FLPsigmap\cdot\FLPB$, would have been the energy of a proton alone. Classically, the energy of the two of them together would be the sum of the two, and that works also quantum mechanically. In a magnetic field, the energy of interaction due to the magnetic field is just the sum of the energy of interaction of the electron with the external field, and of the proton with the field—both expressed in terms of the sigma operators. In quantum mechanics these terms are not really the energies, but thinking of the classical formulas for the energy is a way of remembering the rules for writing down the Hamiltonian. Anyway, the correct Hamiltonian is Eq. (12.27).  Now we have to go back to the beginning and do the problem all over again. Much of the work is, however, done—we need only to add the effects of the new terms. Let’s take a constant magnetic field $\FLPB$ in the $z$-direction. Then we have to add to our Hamiltonian operator $\Hop$ the two new pieces—which we can call $\Hop'$: \begin{equation*} \Hop'=-(\mu_{\text{e}}\sigmae_z+\mu_{\text{p}}\sigmap_z)B. \end{equation*} Using Table 12-1, we get right away that \begin{equation} \begin{aligned} \Hop'\,\ket{+\,+}&=-(\mu_{\text{e}}+\mu_{\text{p}})B\,\ket{+\,+},\\[.5ex] \Hop'\,\ket{+\,-}&=-(\mu_{\text{e}}-\mu_{\text{p}})B\,\ket{+\,-},\\[.5ex] \Hop'\,\ket{-\,+}&=-(-\mu_{\text{e}}+\mu_{\text{p}})B\,\ket{-\,+},\\[.5ex] \Hop'\,\ket{-\,-}&=(\mu_{\text{e}}+\mu_{\text{p}})B\,\ket{-\,-}. \end{aligned} \label{Eq:III:12:28} \end{equation} How very convenient! The $\Hop'$ operating on each state just gives a number times that state. The matrix $\bracket{i}{H'}{j}$ has, therefore, only diagonal elements—we can just add the coefficients in (12.28) to the corresponding diagonal terms of (12.13), and the Hamiltonian equations of (12.14) become \begin{equation} \begin{aligned} i\hbar\,dC_1/dt&= \{A-(\mu_{\text{e}}+\mu_{\text{p}})B\}C_1,\\[1ex] i\hbar\,dC_2/dt&= -\{A+(\mu_{\text{e}}-\mu_{\text{p}})B\}C_2+2AC_3,\\[1ex] i\hbar\,dC_3/dt&= 2AC_2-\{A-(\mu_{\text{e}}-\mu_{\text{p}})B\}C_3,\\[1ex] i\hbar\,dC_4/dt&= \{A+(\mu_{\text{e}}+\mu_{\text{p}})B\}C_4. \end{aligned} \label{Eq:III:12:29} \end{equation}  The form of the equations is not different—only the coefficients. So long as $B$ doesn’t vary with time, we can continue as we did before. Substituting $C_i=a_ie^{-(i/\hbar)Et}$, we get—as a modification of (12.18)— \begin{equation} \begin{aligned} Ea_1&=\{A-(\mu_{\text{e}}+\mu_{\text{p}})B\}a_1,\\[1ex] Ea_2&=-\{A+(\mu_{\text{e}}-\mu_{\text{p}})B\}a_2+2Aa_3,\\[1ex] Ea_3&=2Aa_2-\{A-(\mu_{\text{e}}-\mu_{\text{p}})B\}a_3,\\[1ex] Ea_4&=\{A+(\mu_{\text{e}}+\mu_{\text{p}})B\}a_4. \end{aligned} \label{Eq:III:12:30} \end{equation} Fortunately, the first and fourth equations are still independent of the rest, so the same technique works again.  One solution is the state $\ketsl{\slI}$ for which $a_1=1$, $a_2=$ $a_3=$ $a_4=$ $0$, or \begin{align} \notag \ketsl{\slI}&=\ketsl{\slOne}=\ket{+\,+},\\ \label{Eq:III:12:31} \kern{-6em}\text{with}\\ % ebook remove % ebook insert: \makebox[0em]{\kern{-10em}\text{with}}\\ \notag E_{\slI}&=A-(\mu_{\text{e}}+\mu_{\text{p}})B. \end{align}  Another is \begin{align} \notag \ketsl{\slII}&=\ketsl{\slFour}=\ket{-\,-},\\ \label{Eq:III:12:32} \kern{-6em}\text{with}\\ % ebook remove % ebook insert: \makebox[0em]{\kern{-10em}\text{with}}\\ \notag E_{\slII}&=A+(\mu_{\text{e}}+\mu_{\text{p}})B. \end{align}   A little more work is involved for the remaining two equations, because the coefficients of $a_2$ and $a_3$ are no longer equal. But they are just like the pair we had for the ammonia molecule. Looking back at Eq. (9.20), we can make the following analogy (remembering that the labels $1$ and $2$ there correspond to $2$ and $3$ here): \begin{equation} \begin{aligned} H_{11}&\to-A-(\mu_{\text{e}}-\mu_{\text{p}})B,\\[1ex] H_{12}&\to2A,\\[1ex] H_{21}&\to2A,\\[1ex] H_{22}&\to-A+(\mu_{\text{e}}-\mu_{\text{p}})B. \end{aligned} \label{Eq:III:12:33} \end{equation} The energies are then given by (9.25), which was \begin{equation} \label{Eq:III:12:34} E=\!\frac{H_{11}\!+H_{22}}{2}\!\pm\! \sqrt{\frac{(H_{11}\!-H_{22})^2}{4}\!+\!H_{12}H_{21}}. \end{equation} Making the substitutions from (12.33), the energy formula becomes \begin{equation*} E=-A\pm\sqrt{(\mu_{\text{e}}-\mu_{\text{p}})^2B^2+4A^2}. \end{equation*}  Although in Chapter 9 we used to call these energies $E_{\slI}$ and $E_{\slII}$, and we are in this problem calling them $E_{\slIII}$ and $E_{\slIV}$, \begin{equation} \begin{aligned} E_{\slIII}&= A\{-1+2\sqrt{1+(\mu_{\text{e}}-\mu_{\text{p}})^2B^2/4A^2}\},\\[1ex] E_{\slIV}&= -A\{1+2\sqrt{1+(\mu_{\text{e}}-\mu_{\text{p}})^2B^2/4A^2}\}. \end{aligned} \label{Eq:III:12:35} \end{equation}  So we have found the energies of the four stationary states of a hydrogen atom in a constant magnetic field. Let’s check our results by letting $B$ go to zero and seeing whether we get the same energies we had in the preceding section. You see that we do. For $B=0$, the energies $E_{\slI}$, $E_{\slII}$, and $E_{\slIII}$ go to $+A$, and $E_{\slIV}$ goes to $-3A$. Even our labeling of the states agrees with what we called them before. When we turn on the magnetic field though, all of the energies change in a different way. Let’s see how they go.  First, we have to remember that for the electron, $\mu_{\text{e}}$ is negative, and about $1000$ times larger than $\mu_{\text{p}}$—which is positive. So $\mu_{\text{e}}+\mu_{\text{p}}$ and $\mu_{\text{e}}-\mu_{\text{p}}$ are both negative numbers, and nearly equal. Let’s call them $-\mu$ and $-\mu'$: \begin{equation} \label{Eq:III:12:36} \mu=-(\mu_{\text{e}}+\mu_{\text{p}}),\quad \mu'=-(\mu_{\text{e}}-\mu_{\text{p}}). \end{equation} (Both $\mu$ and $\mu'$ are positive numbers, nearly equal to magnitude of $\mu_{\text{e}}$—which is about one Bohr magneton.) Then our four energies are \begin{equation} \begin{aligned} E_{\slI}&=A+\mu B,\\[1.3ex] E_{\slII}&=A-\mu B,\\[1ex] E_{\slIII}&=A\{-1+2\sqrt{1+\mu'^2B^2/4A^2}\},\\[1ex] E_{\slIV}&=-A\{1+2\sqrt{1+\mu'^2B^2/4A^2}\}. \end{aligned} \label{Eq:III:12:37} \end{equation} The energy $E_{\slI}$ starts at $A$ and increases linearly with $B$—with the slope $\mu$. The energy $E_{\slII}$ also starts at $A$ but decreases linearly with increasing $B$—its slope is $-\mu$. These two levels vary with $B$ as shown in Fig. 12-3. We show also in the figure the energies $E_{\slIII}$ and $E_{\slIV}$. They have a different $B$-dependence. For small $B$, they depend quadratically on $B$, so they start out with horizontal slopes. Then they begin to curve, and for large $B$ they approach straight lines with slopes $\pm\mu'$, which are nearly the same as the slopes of $E_{\slI}$ and $E_{\slII}$.  The shift of the energy levels of an atom due to a magnetic field is called the Zeeman effect. We say that the curves in Fig. 12-3 show the Zeeman splitting of the ground state of hydrogen. When there is no magnetic field, we get just one spectral line from the hyperfine structure of hydrogen. The transitions between state $\ketsl{\slIV}$ and any one of the others occurs with the absorption or emission of a photon whose frequency $1420$ megacycles is $1/h$ times the energy difference $4A$. When the atom is in a magnetic field $\FLPB$, however, there are many more lines. There can be transitions between any two of the four states. So if we have atoms in all four states, energy can be absorbed—or emitted—in any one of the six transitions shown by the vertical arrows in Fig. 12-4. Many of these transitions can be observed by the Rabi molecular beam technique we described in Volume II, Section 35-3.  What makes the transitions go? The transitions will occur if you apply a small disturbing magnetic field that varies with time (in addition to the steady strong field $\FLPB$). It’s just as we saw for a varying electric field on the ammonia molecule. Only here, it is the magnetic field which couples with the magnetic moments and does the trick. But the theory follows through in the same way that we worked it out for the ammonia. The theory is the simplest if you take a perturbing magnetic field that rotates in the $xy$-plane—although any horizontal oscillating field will do. When you put in this perturbing field as an additional term in the Hamiltonian, you get solutions in which the amplitudes vary with time—as we found for the ammonia molecule. So you can calculate easily and accurately the probability of a transition from one state to another. And you find that it all agrees with experiment. 