For a many-electron atom, the operators for individual angular momenta of the electrons do not commute with the Hamiltonian operator, but their sum does. Hence we want to learn how to add angular momenta.
Suppose we have a system with two angular-momentum vectors $\mathbf{M}{\mathbf{1}}$ and $\mathbf{M}{\mathbf{2}}$. They might be the orbital angular-momentum vectors of two electrons in an atom, or they might be the spin angular-momentum vectors of two electrons, or one might be the spin and the other the orbital angular momentum of a single electron. The eigenvalues of $\hat{M}{1}^{2}, \hat{M}{2}^{2}, \hat{M}{1 z}$, and $\hat{M}{2 z}$ are $j{1}\left(j{1}+1\right) \hbar^{2}, j{2}\left(j{2}+1\right) \hbar^{2}, m{1} \hbar$, and $m{2} \hbar$, where the quantum numbers obey the usual restrictions. The components of $\hat{\mathbf{M}}{1}$ and $\hat{\mathbf{M}}{\mathbf{2}}$ obey the angular-momentum commutation relations [Eqs. (5.46), (5.48), and (5.107)]
\(
\begin{equation}
\left[\hat{M}{1 x}, \hat{M}{1 y}\right]=i \hbar \hat{M}{1 z}, \text { etc. }\left[\hat{M}{2 x}, \hat{M}{2 y}\right]=i \hbar \hat{M}{2 z}, \text { etc. } \tag{11.19}
\end{equation}
\)
We define the total angular momentum $\mathbf{M}$ of the system as the vector sum
\(
\begin{equation}
\mathbf{M}=\mathbf{M}{\mathbf{1}}+\mathbf{M}{\mathbf{2}} \tag{11.20}
\end{equation}
\)
$\mathbf{M}$ is a vector with three components:
\(
\begin{equation}
\mathbf{M}=M{x} \mathbf{i}+M{y} \mathbf{j}+M_{z} \mathbf{k} \tag{11.21}
\end{equation}
\)
The vector equation (11.20) gives the three scalar equations
\(
\begin{equation}
M{x}=M{1 x}+M{2 x}, \quad M{y}=M{1 y}+M{2 y}, \quad M{z}=M{1 z}+M_{2 z} \tag{11.22}
\end{equation}
\)
For the operator $\hat{M}^{2}$, we have
\(
\begin{align}
& \hat{M}^{2}=\hat{\mathbf{M}} \cdot \hat{\mathbf{M}}=\hat{M}{x}^{2}+\hat{M}{y}^{2}+\hat{M}{z}^{2} \tag{11.23}\
& \hat{M}^{2}=\left(\hat{\mathbf{M}}{1}+\hat{\mathbf{M}}{2}\right) \cdot\left(\hat{\mathbf{M}}{1}+\hat{\mathbf{M}}{2}\right) \
& \hat{M}^{2}=\hat{M}{1}^{2}+\hat{M}{2}^{2}+\hat{\mathbf{M}}{1} \cdot \hat{\mathbf{M}}{2}+\hat{\mathbf{M}}{2} \cdot \hat{\mathbf{M}}_{1} \tag{11.24}
\end{align}
\)
If $\hat{\mathbf{M}}{1}$ and $\hat{\mathbf{M}}{2}$ refer to different electrons, they will commute with each other, since each will affect only functions of the coordinates of one electron and not the other. Even if $\hat{\mathbf{M}}{1}$ and $\hat{\mathbf{M}}{2}$ are the orbital and spin angular momenta of the same electron, they will commute, as one will affect only functions of the spatial coordinates while the other will affect functions of the spin coordinates. Thus (11.24) becomes
\(
\begin{gather}
\hat{M}^{2}=\hat{M}{1}^{2}+\hat{M}{2}^{2}+2 \hat{\mathbf{M}}{1} \cdot \hat{\mathbf{M}}{2} \tag{11.25}\
\hat{M}^{2}=\hat{M}{1}^{2}+\hat{M}{2}^{2}+2\left(\hat{M}{1 x} \hat{M}{2 x}+\hat{M}{1 y} \hat{M}{2 y}+\hat{M}{1 z} \hat{M}{2 z}\right) \tag{11.26}
\end{gather}
\)
We now show that the components of the total angular momentum obey the usual angular-momentum commutation relations. We have [Eq. (5.4)]
\(
\begin{aligned}
{\left[\hat{M}{x}, \hat{M}{y}\right] } & =\left[\hat{M}{1 x}+\hat{M}{2 x}, \hat{M}{1 y}+\hat{M}{2 y}\right] \
& =\left[\hat{M}{1 x}, \hat{M}{1 y}+\hat{M}{2 y}\right]+\left[\hat{M}{2 x}, \hat{M}{1 y}+\hat{M}{2 y}\right] \
& =\left[\hat{M}{1 x}, \hat{M}{1 y}\right]+\left[\hat{M}{1 x}, \hat{M}{2 y}\right]+\left[\hat{M}{2 x}, \hat{M}{1 y}\right]+\left[\hat{M}{2 x}, \hat{M}{2 y}\right]
\end{aligned}
\)
Since all components of $\hat{\mathbf{M}}{1}$ commute with all components of $\hat{\mathbf{M}}{2}$, we have
\(
\begin{align}
& {\left[\hat{M}{x}, \hat{M}{y}\right]=\left[\hat{M}{1 x}, \hat{M}{1 y}\right]+\left[\hat{M}{2 x}, \hat{M}{2 y}\right]=i \hbar \hat{M}{1 z}+i \hbar \hat{M}{2 z}} \tag{11.27}\
& {\left[\hat{M}{x}, \hat{M}{y}\right]=i \hbar \hat{M}_{z}}
\end{align}
\)
Cyclic permutation of $x, y$, and $z$ gives
\(
\begin{equation}
\left[\hat{M}{y}, \hat{M}{z}\right]=i \hbar \hat{M}{x}, \quad\left[\hat{M}{z}, \hat{M}{x}\right]=i \hbar \hat{M}{y} \tag{11.28}
\end{equation}
\)
The same commutator algebra used to derive (5.109) gives
\(
\begin{equation}
\left[\hat{M}^{2}, \hat{M}{x}\right]=\left[\hat{M}^{2}, \hat{M}{y}\right]=\left[\hat{M}^{2}, \hat{M}_{z}\right]=0 \tag{11.29}
\end{equation}
\)
Thus we can simultaneously quantize $M^{2}$ and one of its components, say $M_{z}$. Since the components of the total angular momentum obey the angular-momentum commutation relations, the work of Section 5.4 shows that the eigenvalues of $\hat{M}^{2}$ are
\(
\begin{equation}
J(J+1) \hbar^{2}, \quad J=0, \frac{1}{2}, 1, \frac{3}{2}, 2, \ldots \tag{11.30}
\end{equation}
\)
and the eigenvalues of $\hat{M}_{z}$ are
\(
\begin{equation}
M{J} \hbar, \quad M{J}=-J,-J+1, \ldots, J-1, J \tag{11.31}
\end{equation}
\)
We want to find out how the total-angular-momentum quantum numbers $J$ and $M{J}$ are related to the quantum numbers $j{1}, j{2}, m{1}, m{2}$ of the two angular momenta we are adding in (11.20). We also want the eigenfunctions of $\hat{M}^{2}$ and $\hat{M}{z}$. These eigenfunctions are characterized by the quantum numbers $J$ and $M{J}$, and, using ket notation (Section 7.3), we write them as $\left|J M{J}\right\rangle$. Similarly, let $\left|j{1} m{1}\right\rangle$ denote the eigenfunctions of $\hat{M}{1}^{2}$ and $\hat{M}{1 z}$ and $\left|j{2} m{2}\right\rangle$ denote the eigenfunctions of $\hat{M}{2}^{2}$ and $\hat{M}{2 z}$. Now it is readily shown (Prob. 11.10) that
\(
\begin{equation}
\left[\hat{M}{x}, \hat{M}{1}^{2}\right]=\left[\hat{M}{y}, \hat{M}{1}^{2}\right]=\left[\hat{M}{z}, \hat{M}{1}^{2}\right]=\left[\hat{M}^{2}, \hat{M}_{1}^{2}\right]=0 \tag{11.32}
\end{equation}
\)
with similar equations with $\hat{M}{2}^{2}$ replacing $\hat{M}{1}^{2}$. Hence we can have simultaneous eigenfunctions of all four operators $\hat{M}{1}^{2}, \hat{M}{2}^{2}, \hat{M}^{2}, \hat{M}{z}$, and the eigenfunctions $\left|J M{J}\right\rangle$ can be more fully written as $\left|j{1} j{2} J M{J}\right\rangle$. However, one finds that $\hat{M}^{2}$ does not commute with $\hat{M}{1 z}$ or $\hat{M}{2 z}$ (Prob. 11.12), so the eigenfunctions $\left|j{1} j{2} J M{J}\right\rangle$ are not necessarily eigenfunctions of $\hat{M}{1 z}$ or $\hat{M}{2 z}$.
If we take the complete set of functions $\left|j{1} m{1}\right\rangle$ for particle 1 and the complete set $\left|j{2} m{2}\right\rangle$ for particle 2 and form all possible products of the form $\left|j{1} m{1}\right\rangle\left|j{2} m{2}\right\rangle$, we will have a complete set of functions for the two particles. Each unknown eigenfunction $\left|j{1} j{2} J M_{J}\right\rangle$ can then be expanded using this complete set:
\(
\begin{equation}
\left|j{1} j{2} J M{J}\right\rangle=\sum C\left(j{1} j{2} J M{J} ; m{1} m{2}\right)\left|j{1} m{1}\right\rangle\left|j{2} m{2}\right\rangle \tag{11.33}
\end{equation}
\)
where the expansion coefficients are the $C\left(j{1} \cdots m{2}\right)$ 's. The functions $\left|j{1} j{2} J M{J}\right\rangle$ are eigenfunctions of the commuting operators $\hat{M}{1}^{2}, \hat{M}{2}^{2}, \hat{M}^{2}$, and $\hat{M}{z}$ with the following eigenvalues:
| $\hat{M}_{1}^{2}$ | $\hat{M}_{2}^{2}$ | $\hat{M}^{2}$ | $\hat{M}_{z}$ |
|---|---|---|---|
| $j{1}\left(j{1}+1\right) \hbar^{2}$ | $j{2}\left(j{2}+1\right) \hbar^{2}$ | $J(J+1) \hbar^{2}$ | $\hat{M}_{J} \hbar$ |
The functions $\left|j{1} m{1}\right\rangle\left|j{2} m{2}\right\rangle$ are eigenfunctions of the commuting operators $\hat{M}{1}^{2}, \hat{M}{1 z}$, $\hat{M}{2}^{2}, \hat{M}{2 z}$ with the following eigenvalues:
| $\hat{M}_{1}^{2}$ | $\hat{M}_{1 z}$ | $\hat{M}_{2}^{2}$ | $\hat{M}_{2 z}$ |
|---|---|---|---|
| $j{1}\left(j{1}+1\right) \hbar^{2}$ | $m_{1} \hbar$ | $j{2}\left(j{2}+1\right) \hbar^{2}$ | $m_{2} \hbar$ |
Since the function $\left|j{1} j{2} J M{J}\right\rangle$ being expanded in (11.33) is an eigenfunction of $\hat{M}{1}^{2}$ with eigenvalue $j{1}\left(j{1}+1\right) \hbar^{2}$, we include in the sum only terms that have the same $j{1}$ value as in the function $\left|j{1} j{2} J M{J}\right\rangle$. (See Theorem 3 at the end of Section 7.3.) Likewise, only terms with the same $j{2}$ value as in $\left|j{1} j{2} J M{J}\right\rangle$ are included in the sum. Hence the sum goes over only the $m{1}$ and $m{2}$ values. Also, using $\hat{M}{z}=\hat{M}{1 z}+\hat{M}_{2 z}$, we can prove (Prob. 11.11) that the coefficient $C$ vanishes unless
\(
\begin{equation}
m{1}+m{2}=M_{J} \tag{11.34}
\end{equation}
\)
To find the total-angular-momentum eigenfunctions, one must evaluate the coefficients in (11.33). These are called Clebsch-Gordan or Wigner or vector addition coefficients. For their evaluation, see Merzbacher, Section 16.6.
Thus each total-angular-momentum eigenfunction $\left|j{1} j{2} J M{J}\right\rangle$ is a linear combination of those product functions $\left|j{1} m{1}\right\rangle\left|j{2} m{2}\right\rangle$ whose $m$ values satisfy $m{1}+m{2}=M{J}$.
We now find the possible values of the total-angular-momentum quantum number $J$ that arise from the addition of angular momenta with individual quantum numbers $j{1}$ and $j{2}$.
Before discussing the general case, we consider the case with $j{1}=1, j{2}=2$. The possible values of $m{1}$ are $-1,0,1$, and the possible values of $m{2}$ are $-2,-1,0,1,2$. If we
describe the system by the quantum numbers $j{1}, j{2}, m{1}, m{2}$, then the total number of possible states is fifteen, corresponding to three possibilities for $m{1}$ and five for $m{2}$. Instead, we can describe the system using the quantum numbers $j{1}, j{2}, J, M{J}$, and we must have the same number of states in this description. Let us tabulate the fifteen possible values of $M{J}$ using (11.34):
| $m_{1}=-1$ | 0 | 1 | ||
|---|---|---|---|---|
| -3 | -2 | -1 | ||
| -2 | -1 | 0 | ||
| -1 | 0 | 1 | ||
| 0 | 1 | -2 | ||
| 1 | 2 | 3 | ||
| :---: | ||||
| -1 | ||||
| 0 | ||||
| 2 |
where each $M{J}$ value in the table is the sum of the $m{1}$ and $m{2}$ values at the top and side. The number of times each value of $M{J}$ occurs is
| value of $M_{J}$ | 3 | 2 | 1 | 0 | -1 | -2 | -3 |
|---|---|---|---|---|---|---|---|
| number of occurrences | 1 | 2 | 3 | 3 | 3 | 2 | 1 |
The highest value of $M{J}$ is +3 . Since $M{J}$ ranges from $-J$ to $+J$, the highest value of $J$ must be 3 . Corresponding to $J=3$, there are seven values of $M_{J}$ ranging from -3 to +3 . Eliminating these seven values, we are left with
| value of $M_{J}$ | 2 | 1 | 0 | -1 | -2 |
|---|---|---|---|---|---|
| number of occurrences | 1 | 2 | 2 | 2 | 1 |
The highest remaining value, $M{J}=2$, must correspond to $J=2$. For $J=2$, we have five values of $M{J}$, which when eliminated leave
| value of $M_{J}$ | 1 | 0 | -1 |
|---|---|---|---|
| number of occurrences | 1 | 1 | 1 |
These remaining values of $M{J}$ clearly correspond to $J=1$. Thus, for the individual angular-momentum quantum numbers $j{1}=1, j_{2}=2$, the possible values of the total-angular-momentum quantum number $J$ are 3,2 , and 1 .
Now consider the general case. There are $2 j{1}+1$ values of $m{1}$ (ranging from $-j{1}$ to $\left.+j{1}\right)$ and $2 j{2}+1$ values of $m{2}$. Hence there are $\left(2 j{1}+1\right)\left(2 j{2}+1\right)$ possible states $\left|j{1} m{1}\right\rangle\left|j{2} m{2}\right\rangle$ with fixed $j{1}$ and $j{2}$ values. The highest possible values of $m{1}$ and $m{2}$ are $j{1}$ and $j{2}$, respectively. Therefore, the maximum possible value of $M{J}=m{1}+m{2}$ is $j{1}+j{2}$ [Eq. (11.34)]. Since $M{J}$ ranges from $-J$ to $+J$, the maximum possible value of $J$ must also be $j{1}+j{2}$ :
\(
\begin{equation}
J{\max }=j{1}+j_{2} \tag{11.35}
\end{equation}
\)
The second-highest value of $M{J}$ is $j{1}+j{2}-1$, which arises in two ways: $m{1}=j{1}-1$, $m{2}=j{2}$ and $m{1}=j{1}, m{2}=j{2}-1$. Linear combinations of these two states must give one state with $J=j{1}+j{2}, M{J}=j{1}+j{2}-1$ and one state with $J=j{1}+j{2}-1, M{J}=$ $j{1}+j_{2}-1$. Continuing in this manner, we find that the possible values of $J$ are
\(
j{1}+j{2}, \quad j{1}+j{2}-1, \quad j{1}+j{2}-2, \quad \ldots, \quad J_{\min }
\)
where $J_{\text {min }}$ is the lowest possible value of $J$.
We determine $J{\text {min }}$ by the requirement that the total number of states be $\left(2 j{1}+1\right)\left(2 j{2}+1\right)$. For a particular value of $J$, there are $2 J+1 M{J}$ values, and so $2 J+1$ states correspond to each value of $J$. The total number of states $\left|j{1} j{2} J M{J}\right\rangle$ for fixed $j{1}$ and $j{2}$ is found by summing the number of states $2 J+1$ for each $J$ from $J{\min }$ to $J_{\max }$ :
\(
\begin{equation}
\text { number of states }=\sum{J=J{\min }}^{J_{\max }}(2 J+1) \tag{11.36}
\end{equation}
\)
This sum goes from $J{\min }$ to $J{\max }$. Let us now take the lower limit of the sum to be $J=0$ instead of $J{\min }$. This change adds to the sum terms with $J$ values of $0,1,2, \ldots, J{\min }-1$. To compensate, we must subtract the corresponding sum that goes from $J=0$ to $J=J_{\min }-1$. Therefore, (11.36) becomes
\(
\text { number of states }=\sum{J=0}^{J{\max }}(2 J+1)-\sum{J=0}^{J{\min }-1}(2 J+1)
\)
Problem 6.16 gives $\sum{l=0}^{n-1}(2 l+1)=n^{2}$. Replacing $n-1$ with $b$, we get $\sum{j=0}^{b}(2 J+1)=$ $(b+1)^{2}$. Therefore,
\(
\text { number of states }=\left(J{\max }+1\right)^{2}-J{\min }^{2}=J{\max }^{2}+2 J{\max }+1-J_{\min }^{2}
\)
Replacing $J{\text {max }}$ by $j{1}+j{2}$ [Eq. (11.35)] and equating the number of states to $\left(2 j{1}+1\right)\left(2 j{2}+1\right)=4 j{1} j{2}+2 j{1}+2 j_{2}+1$, we have
\(
\begin{gather}
\left(j{1}+j{2}\right)^{2}+2\left(j{1}+j{2}\right)+1-J{\min }^{2}=4 j{1} j{2}+2 j{1}+2 j{2}+1 \
J{\min }^{2}=j{1}^{2}-2 j{1} j{2}+j{2}^{2}=\left(j{1}-j{2}\right)^{2} \
J{\min }= \pm\left(j{1}-j_{2}\right) \tag{11.37}
\end{gather}
\)
If $j{1}=j{2}$, then $J{\min }=0$. If $j{1} \neq j_{2}$, then one of the values in (11.37) is negative and must be rejected [Eq. (11.30)]. Thus
\(
\begin{equation}
J{\min }=\left|j{2}-j_{1}\right| \tag{11.38}
\end{equation}
\)
To summarize, we have shown that the addition of two angular momenta characterized by quantum numbers $j{1}$ and $j{2}$ results in a total angular momentum whose quantum number J has the possible values
\(
\begin{equation}
J=j{1}+j{2}, j{1}+j{2}-1, \ldots,\left|j{1}-j{2}\right| \tag{11.39}
\end{equation}
\)
EXAMPLE
Find the possible values of the total-angular-momentum quantum number resulting from the addition of angular momenta with quantum numbers $j{1}=2$ and $j{2}=3$.
The maximum and minimum $J$ values are given by (11.39) as $j{1}+j{2}=2+3=5$ and $\left|j{1}-j{2}\right|=|2-3|=1$. The possible $J$ values (11.39) are therefore $J=5,4,3,2,1$.
EXERCISE Find the possible values of the total-angular-momentum quantum number resulting from the addition of angular momenta with quantum numbers $j{1}=3$ and $j{2}=\frac{3}{2}$. (Answer: $J=\frac{9}{2}, \frac{7}{2}, \frac{5}{2}, \frac{3}{2}$.)
EXAMPLE
Find the possible $J$ values when angular momenta with quantum numbers $j{1}=1$, $j{2}=2$, and $j_{3}=3$ are added.
To add more than two angular momenta, we apply (11.39) repeatedly. Addition of $j{1}=1$ and $j{2}=2$ gives the possible quantum numbers 3,2 , and 1 . Addition of $j_{3}$ to each of these values gives the following possibilities for the total-angular-momentum quantum number:
\(
\begin{equation}
6,5,4,3,2,1,0 ; \quad 5,4,3,2,1 ; \quad 4,3,2 \tag{11.40}
\end{equation}
\)
We have one set of states with total-angular-momentum quantum number 6, two sets of states with $J=5$, three sets with $J=4$, and so on.
EXERCISE Find the possible $J$ values when angular momenta with quantum numbers $j{1}=1, j{2}=1$, and $j_{3}=1$ are added. (Answer: $J=3,2,2,1,1,1,0$.)