The energy levels
Now we are ready to work out the energy levels of the ground state of hydrogen by solving the Hamiltonian equations (12.14). We want to find the energies of the stationary states. This means that we want to find those special states $\ket{\psi}$ for which each amplitude $C_i=\braket{i}{\psi}$ in the set belonging to $\ket{\psi}$ has the same time dependence—namely, $e^{-i\omega t}$. Then the state will have the energy $E=\hbar\omega$. So we want a set for which \begin{equation} \label{Eq:III:12:17} C_i=a_ie^{-(i/\hbar)Et}, \end{equation} where the four coefficients $a_i$ are independent of time. To see whether we can get such amplitudes, we substitute (12.17) into Eq. (12.14) and see what happens. Each $i\hbar\,dC/dt$ in Eq. (12.14) turns into $EC$, and—after cancelling out the common exponential factor—each $C$ becomes an $a$; we get \begin{equation} \begin{aligned} Ea_1&=Aa_1,\\[1ex] Ea_2&=-Aa_2+2Aa_3,\\[1ex] Ea_3&=2Aa_2-Aa_3,\\[1ex] Ea_4&=Aa_4, \end{aligned} \label{Eq:III:12:18} \end{equation} which we have to solve for $a_1$, $a_2$, $a_3$, and $a_4$. Isn’t it nice that the first equation is independent of the rest—that means we can see one solution right away. If we choose $E=A$, \begin{equation*} a_1=1,\quad a_2=a_3=a_4=0, \end{equation*} gives a solution. (Of course, taking all the $a$’s equal to zero also gives a solution, but that’s no state at all!) Let’s call our first solution the state $\ketsl{\slI}$:3 \begin{equation} \label{Eq:III:12:19} \ketsl{\slI}=\ketsl{\slOne}=\ket{+\,+}. \end{equation} Its energy is \begin{equation*} E_{\slI}=A. \end{equation*}  With that clue you can immediately see another solution from the last equation in (12.18): \begin{align*} a_1&=a_2=a_3=0,\quad a_4=1,\\[5pt] E&=A. \end{align*} We’ll call that solution state $\ketsl{\slII}$: \begin{align} \label{Eq:III:12:20} \ketsl{\slII}&=\ketsl{\slFour}=\ket{-\,-},\\[5pt] E_{\slII}&=A.\notag \end{align}  Now it gets a little harder; the two equations left in (12.18) are mixed up. But we’ve done it all before. Adding the two, we get \begin{equation} \label{Eq:III:12:21} E(a_2+a_3)=A(a_2+a_3). \end{equation} Subtracting, we have \begin{equation} \label{Eq:III:12:22} E(a_2-a_3)=-3A(a_2-a_3). \end{equation} By inspection—and remembering ammonia—we see that there are two solutions: \begin{alignat}{2} \notag a_2&=a_3,&\quad E& =A\\[-1ex] % ebook remove % ebook insert: =A\\ \label{Eq:III:12:23} \kern{-6em}\text{and}\\ % ebook remove % ebook insert: \makebox[0em]{\kern{-10em}\text{and}}\\ \notag a_2&=-a_3,&\quad E&=-3A. \end{alignat}  They are mixtures of $\ketsl{\slTwo}$ and $\ketsl{\slThree}$. Calling these states $\ketsl{\slIII}$ and $\ketsl{\slIV}$, and putting in a factor $1/\sqrt{2}$ to make the states properly normalized, we have  \begin{equation} \begin{aligned} \ketsl{\slIII}&=\frac{1}{\sqrt{2}}\,(\ketsl{\slTwo}+\ketsl{\slThree})= \frac{1}{\sqrt{2}}\,(\ket{+\,-}+\ket{-\,+}),\\[1ex] E_{\slIII}&=A \end{aligned} \label{Eq:III:12:24} \end{equation} 
 \begin{equation} \begin{aligned} \ketsl{\slIII}&=\frac{1}{\sqrt{2}}\,(\ketsl{\slTwo}+\ketsl{\slThree})\\[1ex] &=\frac{1}{\sqrt{2}}\,(\ket{+\,-}+\ket{-\,+}),\\[1.5ex] E_{\slIII}&=A \end{aligned} \label{Eq:III:12:24} \end{equation}  and  \begin{equation} \begin{aligned} \ketsl{\slIV}&=\frac{1}{\sqrt{2}}\,(\ketsl{\slTwo}-\ketsl{\slThree})= \frac{1}{\sqrt{2}}\,(\ket{+\,-}-\ket{-\,+}),\\[1ex] E_{\slIV}&=-3A. \end{aligned} \label{Eq:III:12:25} \end{equation} 
 \begin{equation} \begin{aligned} \ketsl{\slIV}&=\frac{1}{\sqrt{2}}\,(\ketsl{\slTwo}-\ketsl{\slThree})\notag\\[1ex] &=\frac{1}{\sqrt{2}}\,(\ket{+\,-}-\ket{-\,+}),\\[1.5ex] E_{\slIV}&=-3A. \end{aligned} \label{Eq:III:12:25} \end{equation}  We have found four stationary states and their energies. Notice, incidentally, that our four states are orthogonal, so they also can be used for base states if desired. Our problem is completely solved.  Three of the states have the energy $A$, and the last has the energy $-3A$. The average is zero—which means that when we took $E_0=0$ in Eq. (12.5), we were choosing to measure all the energies from the average energy. We can draw the energy-level diagram for the ground state of hydrogen as shown in Fig. 12-2.  Now the difference in energy between state $\ketsl{\slIV}$ and any one of the others is $4A$. An atom which happens to have gotten into state $\ketsl{\slI}$ could fall from there to state $\ketsl{\slIV}$ and emit light. Not optical light, because the energy is so tiny—it would emit a microwave quantum. Or, if we shine microwaves on hydrogen gas, we will find an absorption of energy as the atoms in state $\ketsl{\slIV}$ pick up energy and go into one of the upper states—but only at the frequency $\omega=4A/\hbar$. This frequency has been measured experimentally; the best result, obtained very recently,4 is  \begin{equation} \label{Eq:III:12:26} f=\omega/2\pi=(1{,}420{,}405{,}751.800\pm0.028)\text{ cycles per second}. \end{equation} 
 \begin{equation} \begin{gathered} f=\omega/2\pi=\\[.5ex] (1{,}420{,}405{,}751.800\pm0.028)\text{ cycles per second}. \end{gathered} \label{Eq:III:12:26} \end{equation}  The error is only two parts in $100$ billion! Probably no basic physical quantity is measured better than that—it’s one of the most remarkably accurate measurements in physics. The theorists were very happy that they could compute the energy to an accuracy of $3$ parts in $10^5$, but in the meantime it has been measured to $2$ parts in $10^{11}$—a million times more accurate than the theory. So the experimenters are way ahead of the theorists. In the theory of the ground state of the hydrogen atom you are as good as anybody. You, too, can just take your value of $A$ from experiment—that’s what everybody has to do in the end.  You have probably heard before about the “$21$-centimeter line” of hydrogen. That’s the wavelength of the $1420$ megacycle spectral line between the hyperfine states. Radiation of this wavelength is emitted or absorbed by the atomic hydrogen gas in the galaxies. So with radio telescopes tuned in to $21$-cm waves (or $1420$ megacycles approximately) we can observe the velocities and the location of concentrations of atomic hydrogen gas. By measuring the intensity, we can estimate the amount of hydrogen. By measuring the frequency shift due to the Doppler effect, we can find out about the motion of the gas in the galaxy. That is one of the big programs of radio astronomy. So now we are talking about something that’s very real—it is not an artificial problem. 