The Hamiltonian for the ground state of hydrogen
We’ll tell you in a moment what it is. But first, we should remind you of one thing: any state can always be written as a linear combination of the base states. For any state $\ket{\psi}$ we can write \begin{align} \ket{\psi}=\phantom{+}&\ket{+\,+}\braket{+\,+}{\psi}\,+\,\ket{+\,-}\braket{+\,-}{\psi}\notag\\[.5ex] \label{Eq:III:12:2} +\,&\ket{-\,+}\braket{-\,+}{\psi}\,+\,\ket{-\,-}\braket{-\,-}{\psi}. \end{align}  Remember that the complete brackets are just complex numbers, so we can also write them in the usual fashion as $C_i$, where $i=1$, $2$, $3$, or $4$, and write Eq. (12.2) as  \begin{equation} \label{Eq:III:12:3} \ket{\psi}=\ket{+\,+}C_1+\ket{+\,-}C_2+\ket{-\,+}C_3+\ket{-\,-}C_4. \end{equation} 
  \begin{align} \ket{\psi}=\phantom{+}&\ket{+\,+}C_1+\ket{+\,-}C_2\notag\\[.5ex] \label{Eq:III:12:3} +\,&\ket{-\,+}C_3+\ket{-\,-}C_4. \end{align}   By giving the four amplitudes $C_i$ we completely describe the spin state $\ket{\psi}$. If these four amplitudes change with time, as they will, the rate of change in time is given by the operator $\Hop$. The problem is to find the $\Hop$.  There is no general rule for writing down the Hamiltonian of an atomic system, and finding the right formula is much more of an art than finding a set of base states. We were able to tell you a general rule for writing a set of base states for any problem of a proton and an electron, but to describe the general Hamiltonian of such a combination is too hard at this level. Instead, we will lead you to a Hamiltonian by some heuristic argument—and you will have to accept it as the correct one because the results will agree with the test of experimental observation.  You will remember that in the last chapter we were able to describe the Hamiltonian of a single, spin one-half particle by using the sigma matrices—or the exactly equivalent sigma operators. The properties of the operators are summarized in Table 12-1. These operators—which are just a convenient, shorthand way of keeping track of the matrix elements of the type $\bracket{+}{\sigma_z}{+}$—were useful for describing the behavior of a single particle of spin one-half. The question is: Can we find an analogous device to describe a system with two spins? The answer is yes, very simply, as follows. We invent a thing which we will call “sigma electron,” which we represent by the vector operator $\FLPsigmae$, and which has the $x$-, $y$-, and $z$-components, $\sigmae_x$, $\sigmae_y$, $\sigmae_z$. We now make the convention that when one of these things operates on any one of our four base states of the hydrogen atom, it acts only on the electron spin, and in exactly the same way as if the electron were all by itself. Example: What is $\sigmae_y\,\ket{-\,+}$? Since $\sigma_y$ on an electron “down” is $-i$ times the corresponding state with the electron “up”, \begin{equation*} \sigmae_y\,\ket{-\,+}=-i\,\ket{+\,+}. \end{equation*} (When $\sigmae_y$ acts on the combined state it flips over the electron, but does nothing to the proton and multiplies the result by $-i$.) Operating on the other states, $\sigmae_y$ would give \begin{align*} \sigmae_y\,\ket{+\,+}&=i\,\ket{-\,+},\\[1ex] \sigmae_y\,\ket{+\,-}&=i\,\ket{-\,-},\\[1ex] \sigmae_y\,\ket{-\,-}&=-i\,\ket{+\,-}. \end{align*} Just remember that the operators $\FLPsigmae$ work only on the first spin symbol—that is, on the electron spin.  Next we define the corresponding operator “sigma proton” for the proton spin. Its three components $\sigmap_x$, $\sigmap_y$, $\sigmap_z$ act in the same way as $\FLPsigmae$, only on the proton spin. For example, if we have $\sigmap_x$ acting on each of the four base states, we get—always using Table 12-1— \begin{align*} \sigmap_x\,\ket{+\,+}&=\ket{+\,-},\\[1ex] \sigmap_x\,\ket{+\,-}&=\ket{+\,+},\\[1ex] \sigmap_x\,\ket{-\,+}&=\ket{-\,-},\\[1ex] \sigmap_x\,\ket{-\,-}&=\ket{-\,+}. \end{align*} As you can see, it’s not very hard.  Now in the most general case we could have more complex things. For instance, we could have products of the two operators like $\sigmae_y\sigmap_z$. When we have such a product we do first what the operator on the right says, and then do what the other one says.1 For example, we would have that  \begin{equation*} \sigmae_x\sigmap_z\,\ket{+\,-}= \sigmae_x(\sigmap_z\,\ket{+\,-})= \sigmae_x(-\,\ket{+\,-})= -\sigmae_x\,\ket{+\,-}= -\ket{-\,-}. \end{equation*} 
 \begin{gather*} \sigmae_x\sigmap_z\,\ket{+\,-}= \sigmae_x(\sigmap_z\,\ket{+\,-})=\\[1ex] \sigmae_x(-\,\ket{+\,-})= -\sigmae_x\,\ket{+\,-}= -\ket{-\,-}. \end{gather*}  Note that these operators don’t do anything on pure numbers—we have used this fact when we wrote $\sigmae_x(-1)=(-1)\sigmae_x$. We say that the operators “commute” with pure numbers, or that a number “can be moved through” the operator. You can practice by showing that the product $\sigmae_x\sigmap_z$ gives the following results for the four states: \begin{align*} \sigmae_x\sigmap_z\,\ket{+\,+}&=+\,\ket{-\,+},\\[1ex] \sigmae_x\sigmap_z\,\ket{+\,-}&=-\,\ket{-\,-},\\[1ex] \sigmae_x\sigmap_z\,\ket{-\,+}&=+\,\ket{+\,+},\\[1ex] \sigmae_x\sigmap_z\,\ket{-\,-}&=-\,\ket{+\,-}. \end{align*}  If we take all the possible operators, using each kind of operator only once, there are sixteen possibilities. Yes, sixteen—provided we include also the “unit operator” $\hat{1}$. First, there are the three: $\sigmae_x$, $\sigmae_y$, $\sigmae_z$. Then the three $\sigmap_x$, $\sigmap_y$, $\sigmap_z$—that makes six. In addition, there are the nine possible products of the form $\sigmae_x\sigmap_y$ which makes a total of 15. And there’s the unit operator which just leaves any state unchanged. Sixteen in all.  Now note that for a four-state system, the Hamiltonian matrix has to be a four-by-four matrix of coefficients—it will have sixteen entries. It is easily demonstrated that any four-by-four matrix—and, therefore, the Hamiltonian matrix in particular—can be written as a linear combination of the sixteen double-spin matrices corresponding to the set of operators we have just made up. Therefore, for the interaction between a proton and an electron that involves only their spins, we can expect that the Hamiltonian operator can be written as a linear combination of the same $16$ operators. The only question is, how?  Well, first, we know that the interaction doesn’t depend on our choice of axes for a coordinate system. If there is no external disturbance—like a magnetic field—to determine a unique direction in space, the Hamiltonian can’t depend on our choice of the direction of the $x$-, $y$-, and $z$-axes. That means that the Hamiltonian can’t have a term like $\sigmae_x$ all by itself. It would be ridiculous, because then somebody with a different coordinate system would get different results.  The only possibilities are a term with the unit matrix, say a constant $a$ (times $\hat{1}$), and some combination of the sigmas that doesn’t depend on the coordinates—some “invariant” combination. The only scalar invariant combination of two vectors is the dot product, which for our $\sigma$’s is \begin{equation} \label{Eq:III:12:4} \FLPsigmae\cdot\FLPsigmap= \sigmae_x\sigmap_x+ \sigmae_y\sigmap_y+ \sigmae_z\sigmap_z. \end{equation} This operator is invariant with respect to any rotation of the coordinate system. So the only possibility for a Hamiltonian with the proper symmetry in space is a constant times the unit matrix plus a constant times this dot product, say, \begin{equation} \label{Eq:III:12:5} \Hop=E_0+A\,\FLPsigmae\cdot\FLPsigmap. \end{equation} That’s our Hamiltonian. It’s the only thing that it can be, by the symmetry of space, so long as there is no external field. The constant term doesn’t tell us much; it just depends on the level we choose to measure energies from. We may just as well take $E_0=0$. The second term tells us all we need to know to find the level splitting of the hydrogen.  If you want to, you can think of the Hamiltonian in a different way. If there are two magnets near each other with magnetic moments $\FLPmu_{\text{e}}$ and $\FLPmu_{\text{p}}$, the mutual energy will depend on $\FLPmu_{\text{e}}\cdot\FLPmu_{\text{p}}$—among other things. And, you remember, we found that the classical thing we call $\FLPmu_{\text{e}}$ appears in quantum mechanics as $\mu_{\text{e}}\FLPsigmae$. Similarly, what appears classically as $\FLPmu_{\text{p}}$ will usually turn out in quantum mechanics to be $\mu_{\text{p}}\FLPsigmap$ (where $\mu_{\text{p}}$ is the magnetic moment of the proton, which is about $1000$ times smaller than $\mu_{\text{e}}$, and has the opposite sign). So Eq. (12.5) says that the interaction energy is like the interaction between two magnets—only not quite, because the interaction of the two magnets depends on the radial distance between them. But Eq. (12.5) could be—and, in fact, is—some kind of an average interaction. The electron is moving all around inside the atom, and our Hamiltonian gives only the average interaction energy. All it says is that for a prescribed arrangement in space for the electron and proton there is an energy proportional to the cosine of the angle between the two magnetic moments, speaking classically. Such a classical qualitative picture may help you to understand where it comes from, but the important thing is that Eq. (12.5) is the correct quantum mechanical formula.  The order of magnitude of the classical interaction between two magnets would be the product of the two magnetic moments divided by the cube of the distance between them. The distance between the electron and the proton in the hydrogen atom is, speaking roughly, one half an atomic radius, or $0.5$ angstrom. It is, therefore, possible to make a crude estimate that the constant $A$ should be about equal to the product of the two magnetic moments $\mu_{\text{e}}$ and $\mu_{\text{p}}$ divided by the cube of $1/2$ angstrom. Such an estimate gives a number in the right ballpark. It turns out that $A$ can be calculated accurately once you understand the complete quantum theory of the hydrogen atom—which we so far do not. It has, in fact, been calculated to an accuracy of about $30$ parts in one million. So, unlike the flip-flop constant $A$ of the ammonia molecule, which couldn’t be calculated at all well by a theory, our constant $A$ for the hydrogen can be calculated from a more detailed theory. But never mind, we will for our present purposes think of the $A$ as a number which could be determined by experiment, and analyze the physics of the situation.  Taking the Hamiltonian of Eq. (12.5), we can use it with the equation \begin{equation} \label{Eq:III:12:6} i\hbar\dot{C}_i=\sum_jH_{ij}C_j \end{equation} to find out what the spin interactions do to the energy levels. To do that, we need to work out the sixteen matrix elements $H_{ij}=\bracket{i}{H}{j}$ corresponding to each pair of the four base states in (12.1).  We begin by working out what $\Hop\,\ket{j}$ is for each of the four base states.  For example,  \begin{equation} \label{Eq:III:12:7} \Hop\,\ket{+\,+}=A\FLPsigmae\cdot\FLPsigmap\,\ket{+\,+}= A\{\sigmae_x\sigmap_x+ \sigmae_y\sigmap_y+ \sigmae_z\sigmap_z\}\,\ket{+\,+}. \end{equation} 
  \begin{align} \Hop\,\ket{+\,+}&=A\FLPsigmae\cdot\FLPsigmap\,\ket{+\,+}\notag\\[1ex] \label{Eq:III:12:7} &=A\{\sigmae_x\sigmap_x+\sigmae_y\sigmap_y+\sigmae_z\sigmap_z\}\,\ket{+\,+}. \end{align}   Using the method we described a little earlier—it’s easy if you have memorized Table 12-1—we find what each pair of $\sigma$’s does on $\ket{+\,+}$. The answer is \begin{equation} \begin{aligned} \sigmae_x\sigmap_x\,\ket{+\,+}&=+\,\ket{-\,-},\\[1ex] \sigmae_y\sigmap_y\,\ket{+\,+}&=-\,\ket{-\,-},\\[1ex] \sigmae_z\sigmap_z\,\ket{+\,+}&=+\,\ket{+\,+}. \end{aligned} \label{Eq:III:12:8} \end{equation} So (12.7) becomes  \begin{equation} \label{Eq:III:12:9} \Hop\,\ket{+\,+}=A\{\ket{-\,-}-\ket{-\,-}+\ket{+\,+}\}=A\,\ket{+\,+}. \end{equation} 
 \begin{align} \Hop\,\ket{+\,+}&=A\{\ket{-\,-}-\ket{-\,-}+\ket{+\,+}\}\notag\\[1ex] \label{Eq:III:12:9} &=A\,\ket{+\,+}. \end{align}  Since our four base states are all orthogonal, that gives us immediately that \begin{equation} \begin{aligned} \bracket{+\,+}{H}{+\,+}&=A\braket{+\,+}{+\,+}=A,\\[1ex] \bracket{+\,-}{H}{+\,+}&=A\braket{+\,-}{+\,+}=0,\\[1ex] \bracket{-\,+}{H}{+\,+}&=A\braket{-\,+}{+\,+}=0,\\[1ex] \bracket{-\,-}{H}{+\,+}&=A\braket{-\,-}{+\,+}=0. \end{aligned} \label{Eq:III:12:10} \end{equation} Remembering that $\bracket{j}{H}{i}=\bracket{i}{H}{j}\cconj$, we can already write down the differential equation for the amplitudes $C_1$: \begin{equation*} i\hbar\dot{C}_1=H_{11}C_1+H_{12}C_2+H_{13}C_3+H_{14}C_4 \end{equation*} or \begin{equation} \label{Eq:III:12:11} i\hbar\dot{C}_1=AC_1. \end{equation} That’s all! We get only the one term.  Now to get the rest of the Hamiltonian equations we have to crank through the same procedure for $\Hop$ operating on the other states. First, we will let you practice by checking out all of the sigma products we have written down in Table 12-2. Then we can use them to get: \begin{equation} \begin{aligned} \Hop\,\ket{+\,-}&=A\{2\,\ket{-\,+}-\ket{+\,-}\},\\[1ex] \Hop\,\ket{-\,+}&=A\{2\,\ket{+\,-}-\ket{-\,+}\},\\[1ex] \Hop\,\ket{-\,-}&=A\,\ket{-\,-}. \end{aligned} \label{Eq:III:12:12} \end{equation} Then, multiplying each one in turn on the left by all the other state vectors, we get the following Hamiltonian matrix, $H_{ij}$: \begin{equation} \label{Eq:III:12:13} H_{ij}= \kern -1.5ex \raise 30 pt {\scriptstyle i\downarrow \,\raise 7 pt \scriptstyle j\rightarrow}\kern -13pt % ebook remove % ebook insert: \raise{40pt}{\scriptstyle i\downarrow \,\raise{12pt}{\scriptstyle j\rightarrow}\kern -15pt} \begin{pmatrix} A &  0 &  0 & 0\\[1ex] 0 & -A & 2A & 0\\[1ex] 0 & 2A & -A & 0\\[1ex] 0 &  0 &  0 & A \end{pmatrix}. \end{equation} It means, of course, nothing more than that our differential equations for the four amplitudes $C_i$ are \begin{equation} \begin{aligned} i\hbar\dot{C}_1&=AC_1,\\[1ex] i\hbar\dot{C}_2&=-AC_2+2AC_3,\\[1ex] i\hbar\dot{C}_3&=2AC_2-AC_3,\\[1ex] i\hbar\dot{C}_4&=AC_4. \end{aligned} \label{Eq:III:12:14} \end{equation}  Before solving these equations we can’t resist telling you about a clever rule due to Dirac—it will make you feel that you are really advanced—although we don’t need it for our work. We have—from the equations (12.9) and (12.12)—that \begin{equation} \begin{aligned} \FLPsigmae\cdot\FLPsigmap\,\ket{+\,+}&=\ket{+\,+},\\[1ex] \FLPsigmae\cdot\FLPsigmap\,\ket{+\,-}&=2\,\ket{-\,+}-\ket{+\,-},\\[1ex] \FLPsigmae\cdot\FLPsigmap\,\ket{-\,+}&=2\,\ket{+\,-}-\ket{-\,+},\\[1ex] \FLPsigmae\cdot\FLPsigmap\,\ket{-\,-}&=\ket{-\,-}. \end{aligned} \label{Eq:III:12:15} \end{equation}  Look, said Dirac, I can also write the first and last equations as \begin{align*} \FLPsigmae\cdot\FLPsigmap\,\ket{+\,+}&=2\,\ket{+\,+}-\ket{+\,+},\\[1ex] \FLPsigmae\cdot\FLPsigmap\,\ket{-\,-}&=2\,\ket{-\,-}-\ket{-\,-}; \end{align*} then they are all quite similar. Now I invent a new operator, which I will call $P_{\text{spin exch}}$ and which I define to have the following properties:2 \begin{align*} P_{\text{spin exch}}\,\ket{+\,+}&=\ket{+\,+},\\[1ex] P_{\text{spin exch}}\,\ket{+\,-}&=\ket{-\,+},\\[1ex] P_{\text{spin exch}}\,\ket{-\,+}&=\ket{+\,-},\\[1ex] P_{\text{spin exch}}\,\ket{-\,-}&=\ket{-\,-}. \end{align*} All the operator does is interchange the spin directions of the two particles. Then I can write the whole set of equations in (12.15) as a simple operator equation: \begin{equation} \label{Eq:III:12:16} \FLPsigmae\cdot\FLPsigmap=2P_{\text{spin exch}}-1. \end{equation}  That’s the formula of Dirac. His “spin exchange operator” gives a handy rule for figuring out $\FLPsigmae\cdot\FLPsigmap$. (You see, you can do everything now. The gates are open.) 