Composition of angular momentum
When we studied the hyperfine structure of the hydrogen atom in Chapter 12 we had to work out the internal states of a system composed of two particles—the electron and the proton—each with a spin of one-half. We found that the four possible spin states of such a system could be put together into two groups—a group with one energy that looked to the external world like a spin-one particle, and one remaining state that behaved like a particle of zero spin. That is, putting together two spin one-half particles we can form a system whose “total spin” is one, or zero. In this section we want to discuss in more general terms the spin states of a system which is made up of two particles of arbitrary spin. It is another important problem about angular momentum in quantum mechanical systems.  Let’s first rewrite the results of Chapter 12 for the hydrogen atom in a form that will be easier to extend to the more general case. We began with two particles which we will now call particle $a$ (the electron) and particle $b$ (the proton). Particle $a$ had the spin $j_a$ ($=\tfrac{1}{2}$), and its $z$-component of angular momentum $m_a$ could have one of several values (actually $2$, namely $m_a=+\tfrac{1}{2}$ or $m_a=-\tfrac{1}{2}$). Similarly, the spin state of particle $b$ is described by its spin $j_b$ and its $z$-component of angular momentum $m_b$. Various combinations of the spin states of the two particles could be formed. For instance, we could have particle $a$ with $m_a=\tfrac{1}{2}$ and particle $b$ with $m_b=-\tfrac{1}{2}$, to make a state $\ket{a,+\tfrac{1}{2};b,-\tfrac{1}{2}}$. In general, the combined states formed a system whose “system spin,” or “total spin,” or “total angular momentum” $J$ could be $1$, or $0$. And the system could have a $z$-component of angular momentum $M$, which was $+1$, $0$, or $-1$ when $J=1$, or $0$ when $J=0$. In this new language we can rewrite the formulas in (12.41) and (12.42) as shown in Table 18–3.  In the table the left-hand column describes the compound state in terms of its total angular momentum $J$ and the $z$-component $M$. The right-hand column shows how these states are made up in terms of the $m$-values of the two particles $a$ and $b$.  We want now to generalize this result to states made up of two objects $a$ and $b$ of arbitrary spins $j_a$ and $j_b$. We start by considering an example for which $j_a=\tfrac{1}{2}$ and $j_b=1$, namely, the deuterium atom in which particle $a$ is an electron (e) and particle $b$ is the nucleus—a deuteron (d). We have then that $j_a=j_{\text{e}}=\tfrac{1}{2}$. The deuteron is formed of one proton and one neutron in a state whose total spin is one, so $j_b=j_{\text{d}}=1$. We want to discuss the hyperfine states of deuterium—just the way we did for hydrogen. Since the deuteron has three possible states $m_b=m_{\text{d}}=+1$, $0$, $-1$, and the electron has two, $m_a=m_{\text{e}}=+\tfrac{1}{2}$, $-\tfrac{1}{2}$, there are six possible states as follows (using the notation $\ket{\text{e},m_{\text{e}};\text{d},m_{\text{d}}}$): \begin{equation} \begin{aligned} &\ket{\text{e},+\tfrac{1}{2};\text{d},+1},\\[.5ex] &\ket{\text{e},+\tfrac{1}{2};\text{d},0}; \ket{\text{e},-\tfrac{1}{2};\text{d},+1},\\[.5ex] &\ket{\text{e},+\tfrac{1}{2};\text{d},-1}; \ket{\text{e},-\tfrac{1}{2};\text{d},0},\\[.5ex] &\ket{\text{e},-\tfrac{1}{2};\text{d},-1}. \end{aligned} \label{Eq:III:18:42} \end{equation} You will notice that we have grouped the states according to the values of the sum of $m_{\text{e}}$ and $m_{\text{d}}$—arranged in descending order.  Now we ask: What happens to these states if we project into a different coordinate system? If the new system is just rotated about the $z$-axis by the angle $\phi$, then the state $\ket{\text{e},m_{\text{e}};\text{d},m_{\text{d}}}$ gets multiplied by \begin{equation} \label{Eq:III:18:43} e^{im_{\text{e}}\phi}e^{im_{\text{d}}\phi}= e^{i(m_{\text{e}}+m_{\text{d}})\phi}. \end{equation} (The state may be thought of as the product $\ket{\text{e},m_{\text{e}}}\ket{\text{d},m_{\text{d}}}$, and each state vector contributes independently its own exponential factor.) The factor (18.43) is of the form $e^{iM\phi}$, so the state $\ket{\text{e},m_{\text{e}};\text{d},m_{\text{d}}}$ has a $z$-component of angular momentum equal to \begin{equation} \label{Eq:III:18:44} M=m_{\text{e}}+m_{\text{d}}. \end{equation} The $z$-component of the total angular momentum is the sum of the $z$-components of angular momentum of the parts.
 In the list of (18.42), therefore, the state in the top line has $M=+\tfrac{3}{2}$, the two in the second line have $M=+\tfrac{1}{2}$, the next two have $M=-\tfrac{1}{2}$, and the last state has $M=-\tfrac{3}{2}$. We see immediately one possibility for the spin $J$ of the combined state (the total angular momentum) must be $\tfrac{3}{2}$, and this will require four states with $M=+\tfrac{3}{2}$, $+\tfrac{1}{2}$, $-\tfrac{1}{2}$, and $-\tfrac{3}{2}$.  There is only one candidate for $M=+\tfrac{3}{2}$, so we know already that \begin{equation} \label{Eq:III:18:45} \ket{J=\tfrac{3}{2},M=+\tfrac{3}{2}}= \ket{\text{e},+\tfrac{1}{2};\text{d},+1}. \end{equation} But what is the state $\ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}$? We have two candidates in the second line of (18.42), and, in fact, any linear combination of them would also have $M=\tfrac{1}{2}$. So, in general, we must expect to find that  \begin{equation} \label{Eq:III:18:46} \ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}= \alpha\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}+ \beta\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1}, \end{equation} 
 \begin{equation} \begin{gathered} \ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}=\\[.75ex] \alpha\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}+ \beta\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1}, \end{gathered} \label{Eq:III:18:46} \end{equation}  where $\alpha$ and $\beta$ are two numbers. They are called the Clebsch-Gordan coefficients. Our next problem is to find out what they are.  We can find out easily if we just remember that the deuteron is made up of a neutron and a proton, and write the deuteron states out more explicitly using the rules of Table 18–3. If we do that, the states listed in (18.42) then look as shown in Table 18–4.  We want to form the four states of $J=\tfrac{3}{2}$, using the states in the table. But we already know the answer, because in Table 18–1 we have states of spin $\tfrac{3}{2}$ formed from three spin one-half particles. The first state in Table 18–1 has $\ket{J=\tfrac{3}{2},M=+\tfrac{3}{2}}$ and it is $\ket{+\,+\,+}$, which—in our present notation—is the same as $\ket{\text{e},+\tfrac{1}{2};\text{n},+\tfrac{1}{2};\text{p},+\tfrac{1}{ 2}}$, or the first state in Table 18–4. But this state is also the same as the first in the list of (18.42), confirming our statement in (18.45). The second line of Table 18–1 says—changing to our present notation—that  \begin{align} \label{Eq:III:18:47} \ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}=&\frac{1}{\sqrt{3}}\,\{ \ket{\text{e},+\tfrac{1}{2};\text{n},+\tfrac{1}{2};\text{p},-\tfrac{1}{2}}\\[1ex] &+\;\ket{\text{e},+\tfrac{1}{2};\text{n},-\tfrac{1}{2};\text{p},+\tfrac{1}{2}} +\ket{\text{e},-\tfrac{1}{2};\text{n},+\tfrac{1}{2};\text{p},+\tfrac{1}{2}}\}.\notag \end{align} 
 \begin{align} &\ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}=\notag\\[.5ex] &\quad\frac{1}{\sqrt{3}}\,\{ \ket{\text{e},+\tfrac{1}{2};\text{n},+\tfrac{1}{2};\text{p},-\tfrac{1}{2}}\notag\\[-.5ex] &\qquad+\,\ket{\text{e},+\tfrac{1}{2};\text{n},-\tfrac{1}{2};\text{p},+\tfrac{1}{2}}\notag\\[.75ex] \label{Eq:III:18:47} &\qquad+\,\ket{\text{e},-\tfrac{1}{2};\text{n},+\tfrac{1}{2};\text{p},+\tfrac{1}{2}}\}. \end{align}  The right side can evidently be put together from the two entries in the second line of Table 18–4 by taking $\sqrt{2/3}$ of the first term with $\sqrt{1/3}$ of the second. That is, Eq. (18.47) is equivalent to  \begin{equation} \label{Eq:III:18:48} \ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}= \sqrt{2/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}+ \sqrt{1/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1}. \end{equation} 
 \begin{align} \ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}&= \sqrt{2/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}\notag\\[.5ex] \label{Eq:III:18:48} &+\sqrt{1/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1}. \end{align}  We have found our two Clebsch-Gordan coefficients $\alpha$ and $\beta$ in Eq. (18.46): \begin{equation} \label{Eq:III:18:49} \alpha=\sqrt{2/3},\quad \beta=\sqrt{1/3}. \end{equation}  Following the same procedure we can find that  \begin{equation} \label{Eq:III:18:50} \ket{J=\tfrac{3}{2},M=-\tfrac{1}{2}}= \sqrt{1/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},-1}+ \sqrt{2/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},0}. \end{equation} 
 \begin{align} \ket{J=\tfrac{3}{2},M=-\tfrac{1}{2}}&= \sqrt{1/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},-1}\notag\\[.5ex] \label{Eq:III:18:50} &+\sqrt{2/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},0}. \end{align}   And, also, of course, \begin{equation} \label{Eq:III:18:51} \ket{J=\tfrac{3}{2},M=-\tfrac{3}{2}}= \ket{\text{e},-\tfrac{1}{2};\text{d},-1}. \end{equation} These are the rules for the composition of spin $1$ and spin $\tfrac{1}{2}$ to make a total $J=\tfrac{3}{2}$. We summarize (18.45), (18.48), (18.50), and (18.51) in Table 18–5.  We have, however, only four states here while the system we are considering has six possible states. Of the two states in the second line of (18.42) we have used only one linear combination to form $\ket{J=\tfrac{3}{2},M=+\tfrac{1}{2}}$. There is another linear combination orthogonal to the one we have taken which also has $M=+\tfrac{1}{2}$, namely \begin{equation} \label{Eq:III:18:52} \sqrt{1/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}- \sqrt{2/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1}. \end{equation} Similarly, the two states in the third line of (18.42) can be combined to give two orthogonal states, each with $M=-\tfrac{1}{2}$. The one orthogonal to (18.52) is \begin{equation} \label{Eq:III:18:53} \sqrt{2/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},-1}- \sqrt{1/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},0}. \end{equation} These are the two remaining states. They have $M=m_{\text{e}}+m_{\text{d}}=\pm\tfrac{1}{2}$; and must be the two states corresponding to $J=\tfrac{1}{2}$. So we have  \begin{equation} \begin{aligned} \ket{J=\tfrac{1}{2},M=+\tfrac{1}{2}}&= \sqrt{1/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}- \sqrt{2/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1},\\[2ex] \ket{J=\tfrac{1}{2},M=-\tfrac{1}{2}}&= \sqrt{2/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},-1}- \sqrt{1/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},0}. \end{aligned} \label{Eq:III:18:54} \end{equation} 
 \begin{equation} \begin{aligned} \ket{J=\tfrac{1}{2},M=+\tfrac{1}{2}}&= \sqrt{1/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},0}\\[.5ex] &-\sqrt{2/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},+1},\\[2.25ex] \ket{J=\tfrac{1}{2},M=-\tfrac{1}{2}}&= \sqrt{2/3}\,\ket{\text{e},+\tfrac{1}{2};\text{d},-1}\\[.5ex] &-\sqrt{1/3}\,\ket{\text{e},-\tfrac{1}{2};\text{d},0}. \end{aligned} \label{Eq:III:18:54} \end{equation} 
 We can verify that these two states do indeed behave like the states of a spin one-half object by writing out the deuterium parts in terms of the neutron and proton states—using Table 18–4. The first state in (18.52) is  \begin{align} \sqrt{1/6}\,\{ \ket{\text{e},+\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}+ \ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}&\}\notag\\\notag\\ \label{Eq:III:18:55} {}-\sqrt{2/3}\, \ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}&, \end{align} 
 \begin{align} \sqrt{1/6}\,\{ &\ket{\text{e},+\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\,+\notag\\ &\ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\notag\\[1ex] \label{Eq:III:18:55} -\,\sqrt{2/3}\, &\ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}, \end{align}  which can also be written  \begin{align} &\sqrt{1/3}\,\bigl[\sqrt{1/2}\,\{ \ket{\text{e},+\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}- \ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\notag\\\notag\\ \label{Eq:III:18:56} &{}+\sqrt{1/2}\{ \ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}- \ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\bigr]. \end{align} 
 \begin{align} \sqrt{1/3}\biggl[\sqrt{1/2}\,\{ &\ket{\text{e},+\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\,-\notag\\ &\ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\notag\\[1ex] +\,\sqrt{1/2}\{ &\ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\,-\notag\\[-.5ex] &\ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\biggr]. \label{Eq:III:18:56} \end{align}  Now look at the terms in the first curly brackets, and think of the e and p taken together. Together they form a spin-zero state (see the bottom line of Table 18–3), and contribute no angular momentum. Only the neutron is left, so the whole of the first curly bracket of (18.56) behaves under rotations like a neutron, namely as a state with $J=\tfrac{1}{2}$, $M=+\tfrac{1}{2}$. Following the same reasoning, we see that in the second curly bracket of (18.56) the electron and neutron team up to produce zero angular momentum, and only the proton contribution—with $m_{\text{p}}=\tfrac{1}{2}$—is left. The terms behave like an object with $J=\tfrac{1}{2}$, $M=+\tfrac{1}{2}$. So the whole expression of (18.56) transforms like $\ket{J=\tfrac{1}{2},M=+\tfrac{1}{2}}$ as it should. The $M=-\tfrac{1}{2}$ state which corresponds to (18.53) can be written down (by changing the proper $+\tfrac{1}{2}$'s to $-\tfrac{1}{2}$'s) to get  \begin{align} &\sqrt{1/3}\,\bigl[\sqrt{1/2}\,\{ \ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}- \ket{\text{e},-\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\notag\\\notag\\ \label{Eq:III:18:57} &{}+\sqrt{1/2}\{ \ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}- \ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\}\bigr]. \end{align} 
 \begin{align} \sqrt{1/3}\biggl[\sqrt{1/2}\,\{ &\ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\,-\notag\\ &\ket{\text{e},-\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},+\tfrac{1}{2}}\}\notag\\[1ex] \label{Eq:III:18:57} +\sqrt{1/2}\{ &\ket{\text{e},+\tfrac{1}{2}; \text{n},-\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\,-\notag\\[-.5ex] &\ket{\text{e},-\tfrac{1}{2}; \text{n},+\tfrac{1}{2}; \text{p},-\tfrac{1}{2}}\}\biggr]. \end{align}  You can easily check that this is equal to the second line of (18.54), as it should be if the two terms of that pair are to be the two states of a spin one-half system. So our results are confirmed. A deuteron and an electron can exist in six spin states, four of which act like the states of a spin $\tfrac{3}{2}$ object (Table 18–5) and two of which act like an object of spin one-half (18.54).  The results of Table 18–5 and of Eq. (18.54) were obtained by making use of the fact that the deuteron is made up of a neutron and a proton. The truth of the equations does not depend on that special circumstance. For any spin-one object put together with any spin one-half object the composition laws (and the coefficients) are the same. The set of equations in Table 18–5 means that if the coordinates are rotated about, say, the $y$-axis—so that the states of the spin one-half particle and of the spin-one particle change according to Table 17–1 and Table 17–2—the linear combinations on the right-hand side will change in the proper way for a spin $\tfrac{3}{2}$ object. Under the same rotation the states of (18.54) will change as the states of a spin one-half object. The results depend only on the rotation properties (that is, the spin states) of the two original particles but not in any way on the origins of their angular momenta. We have only made use of this fact to work out the formulas by choosing a special case in which one of the component parts is itself made up of two spin one-half particles in a symmetric state. We have put all our results together in Table 18–6, changing the notation “e” and “d” to “$a$” and “$b$” to emphasize the generality of the conclusions.  Suppose we have the general problem of finding the states which can be formed when two objects of arbitrary spins are combined. Say one has $j_a$ (so its $z$-component $m_a$ runs over the $2j_a+1$ values from $-j_a$ to $+j_a$) and the other has $j_b$ (with $z$-component $m_b$ running over the values from $-j_b$ to $+j_b$). The combined states are $\ket{a,m_a;b,m_b}$, and there are $(2j_a+1)(2j_b+1)$ different ones. Now what states of total spin $J$ can be found?  The total $z$-component of angular momentum $M$ is equal to $m_a+m_b$, and the states can all be listed according to $M$ [as in (18.42)]. The largest $M$ is unique; it corresponds to $m_a=j_a$ and $m_b=j_b$, and is, therefore, just $j_a+j_b$. That means that the largest total spin $J$ is also equal to the sum $j_a+j_b$: \begin{equation*} J=(M)_{\text{max}}=j_a+j_b. \end{equation*} For the first $M$ value smaller than $(M)_{\text{max}}$, there are two states (either $m_a$ or $m_b$ is one unit less than its maximum). They must contribute one state to the set that goes with $J=j_a+j_b$, and the one left over will belong to a new set with $J=j_a+j_b-1$. The next $M$-value—the third from the top of the list—can be formed in three ways. (From $m_a=j_a-2$, $m_b=j_b$; from $m_a=j_a-1$, $m_b=j_b-1$; and from $m_a=j_a$, $m_b=j_b-2$.) Two of these belong to groups already started above; the third tells us that states of $J=j_a+j_b-2$ must also be included. This argument continues until we reach a stage where in our list we can no longer go one more step down in one of the $m$’s to make new states.  Let $j_b$ be the smaller of $j_a$ and $j_b$ (if they are equal take either one); then only $2j_b$ values of $J$ are required—going in integer steps from $j_a+j_b$ down to $j_a-j_b$. That is, when two objects of spin $j_a$ and $j_b$ are combined, the system can have a total angular momentum $J$ equal to any one of the values \begin{equation} J=\left\{ \begin{aligned} &j_a+j_b\\ &j_a+j_b-1\\ &j_a+j_b-2\\ &\;\vdots\\ |&j_a-j_b|. \end{aligned} \right. \label{Eq:III:18:58} \end{equation} (By writing $\abs{j_a-j_b}$ instead of $j_a-j_b$ we can avoid the extra admonition that $j_a\geq j_b$.)  For each of these $J$ values there are the $2J+1$ states of different $M$-values—with $M$ going from $+J$ to $-J$. Each of these is formed from linear combinations of the original states $\ket{a,m_a;b,m_b}$ with appropriate factors—the Clebsch-Gordan coefficients for each particular term. We can consider that these coefficients give the “amount” of the state $\ket{j_a,m_a;j_b,m_b}$ which appears in the state $\ket{J,M}$. So each of the Clebsch-Gordan coefficients has, if you wish, six indices identifying its position in the formulas like those of Tables 18–3 and 18–6. That is, calling these coefficients $C(J,M;j_a,m_a;j_b,m_b)$, we could express the equality of the second line of Table 18–6 by writing \begin{align*} C(\tfrac{3}{2},+\tfrac{1}{2};\tfrac{1}{2},+\tfrac{1}{2};1,0)= \sqrt{2/3},\\[.5ex] C(\tfrac{3}{2},+\tfrac{1}{2};\tfrac{1}{2},-\tfrac{1}{2};1,+1)= \sqrt{1/3}. \end{align*}  We will not calculate here the coefficients for any other special cases.9 You can, however, find tables in many books. You might wish to try another special case for yourself. The next one to do would be the composition of two spin-one particles. We give just the final result in Table 18–7.  These laws of the composition of angular momenta are very important in particle physics—where they have innumerable applications. Unfortunately, we have no time to look at more examples here. 