Comprehensive Study Notes

Total Electronic Orbital and Spin Angular Momenta

The total electronic orbital angular momentum of an $n$-electron atom is defined as the vector sum of the orbital angular momenta of the individual electrons:

\(
\begin{equation}
\mathbf{L}=\sum{i=1}^{n} \mathbf{L}{i} \tag{11.41}
\end{equation}
\)

Although the individual orbital-angular-momentum operators $\hat{\mathbf{L}}_{i}$ do not commute with the atomic Hamiltonian (11.1), one can show (Bethe and Jackiw, pp. 102-103) that $\hat{\mathbf{L}}$ does commute with the atomic Hamiltonian [provided spin-orbit interaction (Section 11.6) is neglected]. We can therefore characterize an atomic state by a quantum number $L$, where $L(L+1) \hbar^{2}$ is the square of the magnitude of the total electronic orbital angular momentum. The electronic wave function $\psi$ of an atom satisfies $\hat{L}^{2} \psi=L(L+1) \hbar^{2} \psi$. The total-electronic-orbital-angular-momentum quantum number $L$ of an atom is specified by a code letter, as follows:

$L$	0	1	2	3	4	5	6	7	8
letter	$S$	$P$	$D$	$F$	$G$	$H$	$I$	$K$	$L$

The total orbital angular momentum is designated by a capital letter, while lowercase letters are used for orbital angular momenta of individual electrons.

EXAMPLE

Find the possible values of the quantum number $L$ for states of the carbon atom that arise from the electron configuration $1 s^{2} 2 s^{2} 2 p 3 d$.

The $s$ electrons have zero orbital angular momentum and contribute nothing to the total orbital angular momentum. The $2 p$ electron has $l=1$ and the $3 d$ electron has $l=2$. From the angular-momentum addition rule (11.39), the total-orbital-angularmomentum quantum number ranges from $1+2=3$ to $|1-2|=1$; the possible values of $L$ are $L=3,2,1$. The configuration $1 s^{2} 2 s^{2} 2 p 3 d$ gives rise to $P, D$, and $F$ states. [The Hartree-Fock central-field approximation has each electron moving in a central-field potential, $V=V(r)$. Hence, within this approximation, the individual electronic orbital angular momenta are constant, giving rise to a wave function composed of a single
configuration that specifies the individual orbital angular momenta. When we go beyond the SCF central-field approximation, we mix in other configurations so we no longer specify precisely the individual orbital angular momenta. Even so, we can still use the rule (11.39) for finding the possible values of the total orbital angular momentum.]

EXERCISE Find the possible values of $L$ for states that arise from the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p 4 p$. (Answer: 2, 1, 0.)

The total electronic spin angular momentum $\mathbf{S}$ of an atom is defined as the vector sum of the spins of the individual electrons:

\(
\begin{equation}
\mathbf{S}=\sum{i=1}^{n} \mathbf{S}{i} \tag{11.43}
\end{equation}
\)

The atomic Hamiltonian $\hat{H}$ of (11.1) (which omits spin-orbit interaction) does not involve spin and therefore commutes with the total-spin operators $\hat{S}^{2}$ and $\hat{S}{z}$. The fact that $\hat{S}^{2}$ commutes with $\hat{H}$ is not enough to show that the atomic wave functions $\psi$ are eigenfunctions of $\hat{S}^{2}$. The antisymmetry requirement means that each $\psi$ must be an eigenfunction of the exchange operator $\hat{P}{i k}$ with eigenvalue -1 (Section 10.3). Hence $\hat{S}^{2}$ must also commute with $\hat{P}{i k}$ if we are to have simultaneous eigenfunctions of $\hat{H}, \hat{S}^{2}$, and $\hat{P}{i k}$. Problem 11.19 shows that $\left[\hat{S}^{2}, \hat{P}_{i k}\right]=0$, so the atomic wave functions are eigenfunctions of $\hat{S}^{2}$. We have $\hat{S}^{2} \psi=S(S+1) \hbar^{2} \psi$, and each atomic state can be characterized by a total-electronic-spin quantum number $S$.

EXAMPLE

Find the possible values of the quantum number $S$ for states that arise from the electron configuration $1 s^{2} 2 s^{2} 2 p 3 d$.

Consider first the two $1 s$ electrons. To satisfy the exclusion principle, one of these electrons must have $m{s}=+\frac{1}{2}$ while the other has $m{s}=-\frac{1}{2}$. If $M{S}$ is the quantum number that specifies the $z$ component of the total spin of the $1 s$ electrons, then the only possible value of $M{S}$ is $\frac{1}{2}-\frac{1}{2}=0$ [Eq. (11.34)]. This single value of $M{S}$ clearly means that the total spin of the two $1 s$ electrons is zero. Thus, although in general when we add the spins $s{1}=\frac{1}{2}$ and $s{2}=\frac{1}{2}$ of two electrons according to the rule (11.39), we get the two possibilities $S=0$ and $S=1$, the restriction imposed by the Pauli principle leaves $S=0$ as the only possibility in this case. Likewise, the spins of the $2 s$ electrons add up to zero. The exclusion principle does not restrict the $m{s}$ values of the $2 p$ and $3 d$ electrons. Application of the rule (11.39) to the spins $s{1}=\frac{1}{2}$ and $s{2}=\frac{1}{2}$ of the $2 p$ and $3 d$ electrons gives $S=0$ and $S=1$. These are the possible values of the total spin quantum number, since the $1 s$ and $2 s$ electrons do not contribute to $S$.

EXERCISE Find the possible values of $S$ for states that arise from the electron configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p 4 p$. (Answer: 0 and 1.)

Atomic Terms

A given electron configuration gives rise in general to several different atomic states, some having the same energy and others having different energies, depending on whether the interelectronic repulsions are the same or different for the states. For example, the $1 s 2 s$ configuration of helium gives rise to four states: The three states with zeroth-order wave functions (10.27) to (10.29) all have the same energy; the single state (10.30) has a different energy. The $1 s 2 p$ electron configuration gives rise to twelve states: The nine states obtained by replacing $2 s$ in (10.27) to (10.29) by $2 p{x}, 2 p{y}$, or $2 p{z}$ have the same energy; the three states obtained by replacing $2 s$ in (10.30) by $2 p{x}, 2 p{y}$, or $2 p{z}$ have the same energy, which differs from the energy of the other nine states.

Thus the atomic states that arise from a given electron configuration can be grouped into sets of states that have the same energy. One can show that states that arise from the same electron configuration and that have the same energy (with spin-orbit interaction neglected) will have the same value of $L$ and the same value of $S$ (see Kemble, Section 63a). A set of equal-energy atomic states that arise from the same electron configuration and that have the same $L$ value and the same $S$ value constitutes an atomic term. For a fixed $L$ value, the quantum number $M{L}$ (where $M{L} \hbar$ is the $z$ component of the total electronic orbital angular momentum) takes on $2 L+1$ values ranging from $-L$ to $+L$. For a fixed $S$ value, $M{S}$ takes on $2 S+1$ values. The atomic energy does not depend on $M{L}$ or $M_{S}$, and each term consists of $(2 L+1)(2 S+1)$ atomic states of equal energy. The degeneracy of an atomic term is $(2 L+1)(2 S+1)$ (spin-orbit interaction neglected).

Each term of an atom is designated by a term symbol formed by writing the numerical value of the quantity $2 S+1$ as a left superscript on the code letter (11.42) that gives the $L$ value. For example, a term that has $L=2$ and $S=1$ has the term symbol ${ }^{3} D$, since $2 S+1=3$.

EXAMPLE

Find the terms arising from each of the following electron configurations: (a) $1 s 2 p$;
(b) $1 s^{2} 2 s^{2} 2 p 3 d$. Give the degeneracy of each term.
(a) The $1 s$ electron has quantum number $l=0$ and the $2 p$ electron has $l=1$. The addition rule (11.39) gives $L=1$ as the only possibility. The code letter for $L=1$ is $P$. Each electron has $s=\frac{1}{2}$, and (11.39) gives $S=1,0$ as the possible $S$ values. The possible values of $2 S+1$ are 3 and 1 . The possible terms are thus ${ }^{3} P$ and ${ }^{1} P$. The ${ }^{3} P$ term has quantum numbers $L=1$ and $S=1$, and its degeneracy is $(2 L+1)(2 S+1)=3(3)=9$. The ${ }^{1} P$ term has $L=1$ and $S=0$, and its degeneracy is $(2 L+1)(2 S+1)=3(1)=3$. [The nine states of the ${ }^{3} P$ term are obtained by replacing $2 s$ in (10.27) to (10.29) by $2 p{x}, 2 p{y}$, or $2 p_{z}$. The three states of the ${ }^{1} P$ term are obtained by replacing $2 s$ in (10.30) by $2 p$ functions.]
(b) In the two previous examples in this section, we found that the configuration $1 s^{2} 2 s^{2} 2 p 3 d$ has the possible $L$ values $L=3,2,1$ and has $S=1,0$. The code letters for these $L$ values are $F, D, P$, and the terms are

\(
\begin{equation}
{ }^{1} P, \quad{ }^{3} P, \quad{ }^{1} D, \quad{ }^{3} D, \quad{ }^{1} F, \quad{ }^{3} F \tag{11.44}
\end{equation}
\)

The degeneracies are found as in (a) and are $3,9,5,15,7$, and 21 , respectively.

Derivation of Atomic Terms

We now examine how to systematically derive the terms that arise from a given electron configuration.

First consider configurations that contain only completely filled subshells. In such configurations, for each electron with $m{s}=+\frac{1}{2}$ there is an electron with $m{s}=-\frac{1}{2}$. Let the quantum number specifying the $z$ component of the total electronic spin angular momentum be $M{S}$. The only possible value for $M{S}$ is zero ( $M{S}=\sum{i} m{s i}=0$ ). Hence $S$ must be zero. For each electron in a closed subshell with magnetic quantum number $m$, there is an electron with magnetic quantum number $-m$. For example, for a $2 p^{6}$ configuration we have two electrons with $m=+1$, two with $m=-1$, and two with $m=0$. Denoting the quantum number specifying the $z$ component of the total electronic orbital angular momentum by $M{L}$, we have $M{L}=\sum{i} m_{i}=0$. We conclude that $L$ must be zero. In summary, a configuration of closed subshells gives rise to only one term: ${ }^{1} S$. For configurations consisting of closed subshells and open subshells, the closed subshells make no contribution to $L$ or $S$ and may be ignored in finding the terms.

We now consider two electrons in different subshells; such electrons are called nonequivalent. Nonequivalent electrons have different values of $n$ or $l$ or both, and we need not worry about any restrictions imposed by the exclusion principle when we derive the terms. We simply find the possible values of $L$ from $l{1}$ and $l{2}$ according to (11.39); combining $s{1}$ and $s{2}$ gives $S=0,1$. We previously worked out the $p d$ case, which gives the terms in (11.44). If we have more than two nonequivalent electrons, we combine the individual $l$ 's to find the values of $L$, and we combine the individual $s$ 's to find the values of $S$. For example, consider a $p d f$ configuration. The possible values of $L$ are given by (11.40). Combining three spin angular momenta, each of which is $\frac{1}{2}$, gives $S=\frac{3}{2}, \frac{1}{2}, \frac{1}{2}$. Each of the three possibilities in (11.40) with $L=3$ may be combined with each of the two possibilities for $S=\frac{1}{2}$, giving six ${ }^{2} F$ terms. Continuing in this manner, we find that the following terms arise from a $p d f$ configuration: ${ }^{2} S(2),{ }^{2} P(4),{ }^{2} D(6),{ }^{2} F(6),{ }^{2} G(6),{ }^{2} H(4),{ }^{2} I(2)$, ${ }^{4} S,{ }^{4} P(2),{ }^{4} D(3),{ }^{4} F(3),{ }^{4} G(3),{ }^{4} H(2),{ }^{4} I$, where the number of times each type of term occurs is in parentheses.

Now consider two electrons in the same subshell (equivalent electrons). Equivalent electrons have the same value of $n$ and the same value of $l$, and the situation is complicated by the necessity to avoid giving two electrons the same four quantum numbers. Hence not all the terms derived for nonequivalent electrons are possible. As an example, consider the terms arising from two equivalent $p$ electrons, an $n p^{2}$ configuration. (The carbon groundstate configuration is $1 s^{2} 2 s^{2} 2 p^{2}$.) The possible values of $m$ and $m{s}$ for the two electrons are listed in Table 11.1, which also gives $M{L}$ and $M_{S}$.

Note that certain combinations are missing from this table. For example, $m{1}=1, m{s 1}=\frac{1}{2}, m{2}=1, m{s 2}=\frac{1}{2}$ is missing, since it violates the exclusion principle. Another missing combination is $m{1}=1, m{s 1}=-\frac{1}{2}, m{2}=1, m{s 2}=\frac{1}{2}$. This combination differs from $m{1}=1, m{s 1}=\frac{1}{2}, m{2}=1, m{s 2}=-\frac{1}{2}$ (row 1) solely by interchange of electrons 1 and 2. Each row in Table 11.1 stands for a Slater determinant, which when expanded

TABLE 11.1 Quantum Numbers for Two Equivalent $\boldsymbol{p}$ Electrons

$m_{1}$	$m_{s 1}$	$m_{2}$	$m_{s 2}$	$M{L}=m{1}+m_{2}$	$M{S}=m{s 1}+m_{s 2}$
1	$\frac{1}{2}$	1	$-\frac{1}{2}$	2	0
1	$\frac{1}{2}$	0	$\frac{1}{2}$	1	1
1	$\frac{1}{2}$	0	$-\frac{1}{2}$	1	0
1	$-\frac{1}{2}$	0	$\frac{1}{2}$	1	0
1	$-\frac{1}{2}$	0	$-\frac{1}{2}$	1	-1
1	$\frac{1}{2}$	-1	$\frac{1}{2}$	0	1
1	$\frac{1}{2}$	-1	$-\frac{1}{2}$	0	0
1	$-\frac{1}{2}$	-1	$\frac{1}{2}$	0	0
1	$-\frac{1}{2}$	-1	$-\frac{1}{2}$	0	-1
0	$\frac{1}{2}$	0	$-\frac{1}{2}$	0	0
0	$\frac{1}{2}$	-1	$\frac{1}{2}$	-1	1
0	$\frac{1}{2}$	-1	$-\frac{1}{2}$	-1	0
0	$-\frac{1}{2}$	-1	$\frac{1}{2}$	-1	0
0	$-\frac{1}{2}$	-1	$-\frac{1}{2}$	-1	-1
-1	$\frac{1}{2}$	-1	$-\frac{1}{2}$	-2	0

contains terms for all possible electron interchanges among the spin-orbitals. Two rows that differ from each other solely by interchange of two electrons correspond to the same Slater determinant, and we include only one of them in the table.

The highest value of $M{L}$ in Table 11.1 is 2 , which must correspond to a term with $L=2$, a $D$ term. The $M{L}=2$ value occurs in conjunction with $M_{S}=0$, indicating that $S=0$ for the $D$ term. Thus we have a ${ }^{1} D$ term corresponding to the five states

\(
\begin{array}{lllrr}
M{L}=2 & 1 & 0 & -1 & -2 \tag{11.45}\
M{S}=0 & 0 & 0 & 0 & 0
\end{array}
\)

The highest value of $M{S}$ in Table 11.1 is 1 , indicating a term with $S=1 . M{S}=1$ occurs in conjunction with $M_{L}=1,0,-1$, which indicates a $P$ term. Hence we have a ${ }^{3} P$ term corresponding to the nine states

\(
\begin{array}{rlrrrrrrrr}
M{L} & =1 & 1 & 1 & 0 & 0 & 0 & -1 & -1 & -1 \tag{11.46}\
M{S} & =1 & 0 & -1 & 1 & 0 & -1 & 1 & 0 & -1
\end{array}
\)

Elimination of the states of (11.45) and (11.46) from Table 11.1 leaves only a single state, which has $M{L}=0, M{S}=0$, corresponding to a ${ }^{1} S$ term. Thus a $p^{2}$ configuration gives rise to the terms ${ }^{1} S,{ }^{3} P,{ }^{1} D$. (In contrast, two nonequivalent $p$ electrons give rise to six terms: ${ }^{1} S,{ }^{3} S,{ }^{1} P,{ }^{3} P,{ }^{1} D,{ }^{3} D$.)

Table 11.2a lists the terms arising from various configurations of equivalent electrons. These results may be derived in the same way that we found the $p^{2}$ terms, but this procedure can become quite involved. To derive the terms of the $f^{7}$ configuration would require a table with 3432 rows. More efficient methods exist [R. F. Curl and J. E. Kilpatrick, Am. J. Phys., 28, 357 (1960); K. E. Hyde, J. Chem. Educ., 52, 87 (1975)].

TABLE 11.2 Terms Arising from Various Electron Configurations

Configuration	Terms
(a) Equivalent electrons
$s^{2} ; p^{6} ; d^{10}$	${ }^{1} S$
$p ; p^{5}$	${ }^{2} P$
$p^{2} ; p^{4}$	${ }^{3} P,{ }^{1} D,{ }^{1} S$
$p^{3}$	${ }^{4} S,{ }^{2} D,{ }^{2} P$
$d ; d^{9}$	${ }^{2} D$
$d^{2} ; d^{8}$	${ }^{3} \mathrm{~F},{ }^{3} \mathrm{P},{ }^{1} G,{ }^{1} D,{ }^{1} S$
$d^{3} ; d^{7}$	${ }^{4} F,{ }^{4} P,{ }^{2} H,{ }^{2} G,{ }^{2} F,{ }^{2} D(2),{ }^{2} P$
$d^{4} ; d^{6}$	$\left{\begin{array}{l} { }^{5} D,{ }^{3} H,{ }^{3} G,{ }^{3} F(2),{ }^{3} D,{ }^{3} P(2) \ { }^{1} I,{ }^{1} G(2),{ }^{1} F,{ }^{1} D(2),{ }^{1} S(2) \end{array}\right.$
$d^{5}$	$\left{\begin{array}{l} { }^{6} S,{ }^{4} G,{ }^{4} F,{ }^{4} D,{ }^{4} P,{ }^{2} I,{ }^{2} H,{ }^{2} G(2) \ { }^{2} F(2),{ }^{2} D(3),{ }^{2} P,{ }^{2} S \end{array}\right.$

(b) Nonequivalent electrons

$s s$	${ }^{1} S,{ }^{3} S$
$s p$	${ }^{1} P,{ }^{3} P$
$s d$	${ }^{1} D,{ }^{3} D$
$p p$	${ }^{3} D,{ }^{1} D,{ }^{3} P,{ }^{1} P,{ }^{3} S,{ }^{1} S$

Note from Table 11.2a that the terms arising from a subshell containing $N$ electrons are the same as the terms for a subshell that is $N$ electrons short of being full. For example, the terms for $p^{2}$ and $p^{4}$ are the same. We can divide the electrons of a closed subshell into two groups and find the terms for each group. Because a closed subshell gives only a ${ }^{1} S$ term, the terms for each of these two groups must be the same. Table 11.2 b gives the terms arising from some nonequivalent electron configurations.

To deal with a configuration containing both equivalent and nonequivalent electrons, we first find separately the terms from the nonequivalent electrons and the terms from the equivalent electrons. We then take all possible combinations of the $L$ and $S$ values of these two sets of terms. For example, consider an $s p^{3}$ configuration. From the $s$ electron, we get a ${ }^{2} S$ term. From the three equivalent $p$ electrons, we get the terms ${ }^{2} P,{ }^{2} D$, and ${ }^{4} S$ (Table 11.2a). Combining the $L$ and $S$ values of these terms, we have as the terms of an $s p^{3}$ configuration

\(
\begin{equation}
{ }^{3} P,{ }^{1} P,{ }^{3} D,{ }^{1} D,{ }^{5} S,{ }^{3} S \tag{11.47}
\end{equation}
\)

Hund's Rule

To decide which one of the terms arising from a given electron configuration is lowest in energy, we use the empirical Hund's rule: For terms arising from the same electron configuration, the term with the largest value of S lies lowest. If there is more than one term with the largest $S$, then the term with the largest $S$ and the largest $L$ lies lowest.

EXAMPLE

Use Table 11.2 to predict the lowest term of (a) the carbon ground-state configuration $1 s^{2} 2 s^{2} 2 p^{2}$; (b) the configuration $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}$.
(a) Table 11.2a gives the terms arising from a $p^{2}$ configuration as ${ }^{3} P,{ }^{1} D$, and ${ }^{1} S$. The term with the largest $S$ will have the largest value of the left superscript $2 S+1$. Hund's rule predicts ${ }^{3} P$ as the lowest term. (b) Table 11.2a gives the $d^{2}$ terms as ${ }^{3} F,{ }^{3} P,{ }^{1} G,{ }^{1} D,{ }^{1} S$. Of these terms, ${ }^{3} F$ and ${ }^{3} P$ have the highest $S .{ }^{3} F$ has $L=3 ;{ }^{3} P$ has $L=1$. Therefore, ${ }^{3} F$ is predicted to be lowest.
EXERCISE Predict the lowest term of the $1 s^{2} 2 s 2 p^{3}$ configuration using (11.47). (Answer: ${ }^{5}$ S.)

Hund's rule works very well for the ground-state configuration, but occasionally fails for an excited configuration (Prob. 11.30).

Hund's rule gives only the lowest term of a configuration and should not be used to decide the order of the remaining terms. For example, for the $1 s^{2} 2 s 2 p^{3}$ configuration of carbon, the observed order of the terms is

\(
{ }^{5} S<{ }^{3} D<{ }^{3} P<{ }^{1} D<{ }^{3} S<{ }^{1} P
\)

The ${ }^{3} S$ term lies above the ${ }^{1} D$ term, even though ${ }^{3} S$ has the higher spin $S$.
It is not necessary to consult Table 11.2a to find the lowest term of a partly filled subshell configuration. We simply put the electrons in the orbitals so as to give the greatest number of parallel spins. Thus, for a $d^{3}$ configuration, we have

\(
\begin{equation}
m: \frac{\uparrow}{+2} \frac{\uparrow}{+1} \frac{\uparrow}{0} \overline{-1} \overline{-2} \tag{11.48}
\end{equation}
\)

The lowest term thus has three parallel spins, so $S=\frac{3}{2}$, giving $2 S+1=4$. The maximum value of $M_{L}$ is 3 , corresponding to $L=3$, an $F$ term. Hund's rule thus predicts ${ }^{4} F$ as the lowest term of a $d^{3}$ configuration.

The traditional explanation of Hund's rule is as follows: Electrons with the same spin tend to keep out of each other's way (recall the idea of Fermi holes), thereby minimizing the Coulombic repulsion between them. The term that has the greatest number of parallel spins (that is, the greatest value of $S$ ) will therefore be lowest in energy. For example, the ${ }^{3} S$ term of the helium $1 s 2 s$ configuration has an antisymmetric spatial function that vanishes when the spatial coordinates of the two electrons are equal. Hence the ${ }^{3} S$ term is lower than the ${ }^{1} S$ term.

This traditional explanation turns out to be wrong in most cases. It is true that the probability that the two electrons are very close together is smaller for the helium ${ }^{3} S 1 s 2 s$ term than for the ${ }^{1} S 1 s 2 s$ term. However, calculations with accurate wave functions show that the probability that the two electrons are very far apart is also less for the ${ }^{3} S$ term. The net result is that the average distance between the two electrons is slightly less for the ${ }^{3} S$ term than for the ${ }^{1} S$ term, and the interelectronic repulsion is slightly greater for the ${ }^{3} S$ term. The calculations show that the ${ }^{3} S$ term lies below the ${ }^{1} S$ term because of a substantially greater electron-nucleus attraction in the ${ }^{3} S$ term as compared with the ${ }^{1} S$ term. Similar results are found for terms of the atoms beryllium and carbon. [See J. Katriel and R. Pauncz, Adv. Quantum Chem., 10, 143 (1977).] The following explanation of these results has been proposed [I. Shim and J. P. Dahl, Theor. Chim. Acta, 48, 165 (1978)]. The Pauli "repulsion" between electrons of like spin makes the average angle between the radius vectors of the two electrons larger for the ${ }^{3} S$ term than for the ${ }^{1} S$ term. This reduces the electronic screening of the nucleus and allows the electrons to get closer to the nucleus in the ${ }^{3} S$ term, making the electron-nucleus attraction greater for the ${ }^{3} S$ term. [See also R. E. Boyd, Nature, 310, 480 (1984); T. Oyamada et al., J. Chem. Phys., 133, 164113 (2010).]

Eigenvalues of Two-Electron Spin Functions

The helium atom $1 s 2 s$ configuration gives rise to the term ${ }^{3} S$ with degeneracy $(2 L+1)(2 S+1)=1(3)=3$ and to the term ${ }^{1} S$ with degeneracy $1(1)=1$. The three helium zeroth-order wave functions (10.27) to (10.29) must correspond to the triply degenerate ${ }^{3} S$ term, and the single function (10.30) must correspond to the ${ }^{1} S$ term. Since $S=1$ and $M{S}=1,0,-1$ for the ${ }^{3} S$ term, the three spin functions in (10.27) to (10.29) should be eigenfunctions of $\hat{S}^{2}$ with eigenvalue $S(S+1) \hbar^{2}=2 \hbar^{2}$ and eigenfunctions of $\hat{S}{z}$ with eigenvalues $M{S} \hbar=\hbar, 0$, and $-\hbar$. The spin function in (10.30) should be an eigenfunction of $\hat{S}^{2}$ and $\hat{S}{z}$ with eigenvalue zero in each case, since $S=0$ and $M_{S}=0$ here. We now verify these assertions.

From Eq. (11.43), the total-electron-spin operator is the sum of the spin operators for each electron:

\(
\begin{equation}
\hat{\mathbf{S}}=\hat{\mathbf{S}}{1}+\hat{\mathbf{S}}{2} \tag{11.49}
\end{equation}
\)

Taking the $z$ components of (11.49), we have

\(
\begin{align}
& \hat{S}{z}=\hat{S}{1 z}+\hat{S}{2 z} \tag{11.50}\
\hat{S}{z} \alpha(1) \alpha(2)= & \hat{S}{1 z} \alpha(1) \alpha(2)+\hat{S}{2 z} \alpha(1) \alpha(2) \
= & \alpha(2) \hat{S}{1 z} \alpha(1)+\alpha(1) \hat{S}{2 z} \alpha(2) \
= & \frac{1}{2} \hbar \alpha(1) \alpha(2)+\frac{1}{2} \hbar \alpha(1) \alpha(2) \
\hat{S}_{z} \alpha(1) \alpha(2)= & \hbar \alpha(1) \alpha(2) \tag{11.51}
\end{align}
\)

where Eq. (10.7) has been used. Similarly, we find

\(
\begin{equation}
\hat{S}_{z} \beta(1) \beta(2)=-\hbar \beta(1) \beta(2) \tag{11.52}
\end{equation}
\)

\(
\begin{align}
& \hat{S}{z}[\alpha(1) \beta(2)+\beta(1) \alpha(2)]=0 \tag{11.53}\
& \hat{S}{z}[\alpha(1) \beta(2)-\beta(1) \alpha(2)]=0 \tag{11.54}
\end{align}
\)

Consider now $\hat{S}^{2}$. We have [Eq. (11.26)]

\(
\begin{array}{r}
\hat{S}^{2}=\left(\hat{\mathbf{S}}{1}+\hat{\mathbf{S}}{2}\right) \cdot\left(\hat{\mathbf{S}}{1}+\hat{\mathbf{S}}{2}\right)=\hat{S}{1}^{2}+\hat{S}{2}^{2}+2\left(\hat{S}{1 x} \hat{S}{2 x}+\hat{S}{1 y} \hat{S}{2 y}+\hat{S}{1 z} \hat{S}{2 z}\right) \tag{11.55}\
\hat{S}^{2} \alpha(1) \alpha(2)=\alpha(2) \hat{S}{1}^{2} \alpha(1)+\alpha(1) \hat{S}{2}^{2} \alpha(2)+2 \hat{S}{1 x} \alpha(1) \hat{S}{2 x} \alpha(2) \
+2 \hat{S}{1 y} \alpha(1) \hat{S}{2 y} \alpha(2)+2 \hat{S}{1 z} \alpha(1) \hat{S}{2 z} \alpha(2)
\end{array}
\)

Using Eqs. (10.7) to (10.9) and (10.72) and (10.73), we find

\(
\begin{equation}
\hat{S}^{2} \alpha(1) \alpha(2)=2 \hbar^{2} \alpha(1) \alpha(2) \tag{11.56}
\end{equation}
\)

Hence $\alpha(1) \alpha(2)$ is an eigenfunction of $\hat{S}^{2}$ corresponding to $S=1$. Similarly, we find

\(
\begin{aligned}
\hat{S}^{2} \beta(1) \beta(2) & =2 \hbar^{2} \beta(1) \beta(2) \
\hat{S}^{2}[\alpha(1) \beta(2)+\beta(1) \alpha(2)] & =2 \hbar^{2}[\alpha(1) \beta(2)+\beta(1) \alpha(2)] \
\hat{S}^{2}[\alpha(1) \beta(2)-\beta(1) \alpha(2)] & =0
\end{aligned}
\)

Thus the spin eigenfunctions in (10.27) to (10.30) correspond to the following values for the total spin quantum numbers:

\(
\begin{align}
& \tag{11.57}\
& \text { triplet }\left{\begin{array}{crr}
& S & M_{S} \
2^{-1 / 2}[\alpha(1) \beta(2)+\beta(1) \alpha(2)] & 1 & 1 \
\beta(1) \beta(2) & 1 & 0 \
\text { singlet }\left{2^{-1 / 2}[\alpha(1) \beta(2)-\beta(1) \alpha(2)]\right. & 1 & -1 \
0 & 0
\end{array}\right.
\end{align}
\)

[In the notation of Section 11.4, we are dealing with the addition of two angular momenta with quantum numbers $j{1}=\frac{1}{2}$ and $j{2}=\frac{1}{2}$ to give eigenfunctions with total angular-momentum quantum numbers $J=1$ and $J=0$. The coefficients in (11.57) to (11.60) correspond to the coefficients $C$ in (11.33) and are examples of Clebsch-Gordan coefficients.]

Figure 11.3 shows the vector addition of $\mathbf{S}{1}$ and $\mathbf{S}{2}$ to form $\mathbf{S}$. It might seem surprising that the spin function (11.58), which has the $z$ components of the spins of the two electrons pointing in opposite directions, could have total spin quantum number $S=1$. Figure 11.3 shows how this is possible.

Atomic Wave Functions

In Section 10.6, we showed that two of the four zeroth-order wave functions of the $1 s 2 s$ helium configuration could be written as single Slater determinants, but the other two functions had to be expressed as linear combinations of two Slater determinants. Since $\hat{L}^{2}$ and $\hat{S}^{2}$ commute with the Hamiltonian (11.1) and with the exchange operator $\hat{P}{i k}$, the zeroth-order functions should be eigenfunctions of $\hat{L}^{2}$ and $\hat{S}^{2}$. The Slater determinants $D{2}$ and $D_{3}$ of Section 10.6 are not eigenfunctions of these operators and so are not suitable zeroth-order functions. We have just shown that the linear combinations (10.44) and (10.45) are eigenfunctions of $\hat{S}^{2}$, and they can also be shown to be eigenfunctions of $\hat{L}^{2}$.

For a configuration of closed subshells (for example, the helium ground state), we can write only a single Slater determinant. This determinant is an eigenfunction of $\hat{L}^{2}$ and $\hat{S}^{2}$ and is the correct zeroth-order function for the nondegenerate ${ }^{1} S$ term. A configuration

$\alpha(1) \alpha(2)$

$\beta(1) \beta(2)$

\(
\frac{1}{\sqrt{2}}[\alpha(1) \beta(2)-\beta(1) \alpha(2)]
\)

with one electron outside closed subshells (for example, the boron ground configuration) gives rise to only one term. The Slater determinants for such a configuration differ from one another only in the $m$ and $m_{s}$ values of this electron and are the correct zeroth-order functions for the states of the term. When all the electrons in singly occupied orbitals have the same spin (either all $\alpha$ or all $\beta$ ), the correct zeroth-order function is a single Slater determinant [for example, see (10.42)]. When this is not true, one has to take a linear combination of a few Slater determinants to obtain the correct zeroth-order functions, which are eigenfunctions of $\hat{L}^{2}$ and $\hat{S}^{2}$. The correct linear combinations can be found by solving the secular equation of degenerate perturbation theory or by operator techniques. Tabulations of the correct combinations for various configurations are available (Slater, Atomic Structure, Vol. II). Hartree-Fock calculations of atomic term energies use these linear combinations and find the best possible orbital functions for the Slater determinants.

Each wave function of an atomic term is an eigenfunction of $\hat{L}^{2}, \hat{S}^{2}, \hat{L}{z}$, and $\hat{S}{z}$. Therefore, when one does a configuration-interaction calculation, only configuration functions that have the same $\hat{L}^{2}, \hat{S}^{2}, \hat{L}{z}$, and $\hat{S}{z}$ eigenvalues as the state under consideration are included in the expansion (11.17). For example, the helium $1 s^{2}$ ground term is ${ }^{1} S$, which has $L=0$ and $S=0$. The electron configuration $1 s 2 p$ produces ${ }^{1} P$ and ${ }^{3} P$ terms only and so gives rise to states with $L=1$ only. No configuration functions arising from the $1 s 2 p$ configuration can occur in the CI wave function (11.17) for the He ground state.

Parity of Atomic States

Consider the atomic Hamiltonian (11.1). We showed in Section 7.5 that the parity operator $\hat{\Pi}$ commutes with the kinetic-energy operator. The quantity $1 / r{i}$ in (11.1) is $r{i}^{-1}=\left(x{i}^{2}+y{i}^{2}+z{i}^{2}\right)^{-1 / 2}$. Replacement of each coordinate by its negative leaves $1 / r{i}$ unchanged. Also

\(
r{i j}^{-1}=\left[\left(x{i}-x{j}\right)^{2}+\left(y{i}-y{j}\right)^{2}+\left(z{i}-z_{j}\right)^{2}\right]^{-1 / 2}
\)

and inversion has no effect on $1 / r_{i j}$. Thus $\hat{\Pi}$ commutes with the atomic Hamiltonian, and we can choose atomic wave functions to have definite parity.

FIGURE 11.3 Vector addition of the spins of two electrons. For $\alpha(1) \alpha(2)$ and $\beta(1) \beta(2)$, the projections of $\mathbf{S}{1}$ and $\mathbf{S}{2}$ in the $x y$ plane make an angle of $90^{\circ}$ with each other (Prob. 11.18c).

For a one-electron atom, the spatial wave function is $\psi=R(r) Y{l}^{m}(\theta, \phi)$. The radial function is unchanged on inversion, and the parity is determined by the angular factor. In Prob. 7.29 we showed that $Y{l}^{m}$ is an even function when $l$ is even and is an odd function when $l$ is odd. Thus the states of one-electron atoms have even or odd parity according to whether $l$ is even or odd.

Now consider an $n$-electron atom. In the Hartree-Fock central-field approximation, we write the wave function as a Slater determinant (or linear combination of Slater determinants) of spin-orbitals. The wave function is the sum of terms, the spatial factor in each term having the form

\(
R{1}\left(r{1}\right) \cdots R{n}\left(r{n}\right) Y{l{1}}^{m{1}}\left(\theta{1}, \phi{1}\right) \cdots Y{l}^{m{n}}\left(\theta{n}, \phi_{n}\right)
\)

The parity of this product is determined by the spherical-harmonic factors. We see that the product is an even or odd function according to whether $l{1}+l{2}+\cdots+l{n}$ is an even or odd number. Therefore, the parity of an atomic state is found by adding the $l$ values of the electrons in the electron configuration that gives rise to the state: If $\sum{i} l{i}$ is an even number, then $\psi$ is an even function; if $\sum{i} l{i}$ is odd, $\psi$ is odd. For example, the configuration $1 s^{2} 2 s 2 p^{3}$ has $\sum{i} l_{i}=0+0+0+1+1+1=3$, and all states arising from this electron configuration have odd parity. (Our argument was based on the SCF approximation to $\psi$, but the conclusions are valid for the true $\psi$.)

Total Electronic Angular Momentum and Atomic Levels

The total electronic angular momentum $\mathbf{J}$ of an atom is the vector sum of the total electronic orbital and spin angular momenta:

\(
\begin{equation}
\mathbf{J}=\mathbf{L}+\mathbf{S} \tag{11.61}
\end{equation}
\)

The operator $\hat{\mathbf{J}}$ for the total electronic angular momentum commutes with the atomic Hamiltonian, and we may characterize an atomic state by a quantum number $J$, which has the possible values [Eq. (11.39)]

\(
\begin{equation}
L+S, L+S-1, \ldots,|L-S| \tag{11.62}
\end{equation}
\)

We have $\hat{J}^{2} \psi=J(J+1) \hbar^{2} \psi$.
For the atomic Hamiltonian (11.1), all states that belong to the same term have the same energy. However, when spin-orbit interaction (Section 11.6) is included in $\hat{H}$, one finds that the value of the quantum number $J$ affects the energy slightly. Hence states that belong to the same term but that have different values of $J$ will have slightly different energies. The set of states that belong to the same term and that have the same value of $J$ constitutes an atomic level. The energies of different levels belonging to the same term are slightly different. (See Fig. 11.6 in Section 11.7.) To denote the level, one adds the $J$ value as a right subscript on the term symbol. Each level is $(2 J+1)$-fold degenerate, corresponding to the $2 J+1$ values of $M{J}$, where $M{J} \hbar$ is the $z$ component of the total electronic angular momentum $\mathbf{J}$. Each level consists of $2 J+1$ states of equal energy.

EXAMPLE

Find the levels of a ${ }^{3} P$ term and give the degeneracy of each level.
For a ${ }^{3} P$ term, $2 S+1=3$ and $S=1$; also, from (11.42), $L=1$. With $L=1$ and $S=1,(11.62)$ gives $J=2,1,0$. The levels are ${ }^{3} P{2},{ }^{3} P{1}$, and ${ }^{3} P{0}$.
The ${ }^{3} P{2}$ level has $J=2$ and has $2 J+1=5$ values of $M{J}$, namely, $-2,-1,0,1$, and 2 . The ${ }^{3} P{2}$ level is 5 -fold degenerate. The ${ }^{3} P{1}$ level is $2(1)+1=3$-fold degenerate. The ${ }^{3} P{0}$ level has $2 J+1=1$ and is nondegenerate.

The total number of states for the three levels ${ }^{3} P{2},{ }^{3} P{1}$, and ${ }^{3} P{0}$ is $5+3+1=9$. When spin-orbit interaction was neglected, we had a ${ }^{3} P$ term that consisted of $(2 L+1)(2 S+1)=3(3)=9$ equal-energy states. With spin-orbit interaction, the 9-fold-degenerate term splits into three closely spaced levels: ${ }^{3} P{2}$ with five states, ${ }^{3} P{1}$ with three states, and ${ }^{3} P{0}$ with one state. The total number of states is the same for the term ${ }^{3} P$ as for the three levels arising from this term.
EXERCISE Find the levels of a ${ }^{2} D$ term and give the level degeneracies. (Answer: ${ }^{2} D{5 / 2},{ }^{2} D{3 / 2} ; 6,4$.)
The quantity $2 S+1$ is called the electron-spin multiplicity (or the multiplicity) of the term. If $L \geq S$, the possible values of $J$ in (11.62) range from $L+S$ to $L-S$ and are $2 S+1$ in number. If $L \geq S$, the spin multiplicity gives the number of levels that arise from a given term. For $L<S$, the values of $J$ range from $S+L$ to $S-L$ and are $2 L+1$ in number. In this case, the spin multiplicity is greater than the number of levels. For example, if $L=0$ and $S=1$ (a ${ }^{3} S$ term), the spin multiplicity is 3 , but there is only one possible value for $J$, namely, $J=1$. For $2 S+1=1,2,3,4,5,6, \ldots$, the words singlet, doublet, triplet, quartet, quintet, sextet, $\ldots$, are used to designate the spin multiplicity. The level symbol ${ }^{3} P_{1}$ is read as "triplet $P$ one."

For light atoms, the spin-orbit interaction is very small and the separation between levels of a term is very small. Note from Fig. 11.6 in Section 11.7 that the separations between the ${ }^{3} P{0},{ }^{3} P{1}$, and ${ }^{3} P_{2}$ levels of the helium $1 s 2 p{ }^{3} P$ term are far, far less than the separation between the ${ }^{1} P$ and ${ }^{3} P$ terms of the $1 s 2 p$ configuration.

Terms and Levels of Hydrogen and Helium

The hydrogen atom has one electron. Hence $L=l$ and $S=s=\frac{1}{2}$. The possible values of $J$ are $L+\frac{1}{2}$ and $L-\frac{1}{2}$, except for $L=0$, where $J=\frac{1}{2}$ is the only possibility. Each electron configuration gives rise to only one term, which is composed of one level if $L=0$ and two levels if $L \neq 0$. The ground-state configuration $1 s$ gives the term ${ }^{2} S$, which is composed of the single level ${ }^{2} S{1 / 2}$; the level is twofold degenerate $\left(M{J}=-\frac{1}{2}, \frac{1}{2}\right)$. The $2 s$ configuration also gives a ${ }^{2} S{1 / 2}$ level. The $2 p$ configuration gives rise to the levels ${ }^{2} P{3 / 2}$ and ${ }^{2} P{1 / 2}$; the ${ }^{2} P{3 / 2}$ level is fourfold degenerate, and the ${ }^{2} P_{1 / 2}$ level is twofold degenerate. There are $2+4+2=8$ states with $n=2$, in agreement with our previous work.

The helium ground-state configuration $1 s^{2}$ is a closed subshell and gives rise to the single level ${ }^{1} S{0}$, which is nondegenerate $\left(M{J}=0\right)$. The $1 s 2 s$ excited configuration gives rise to the two terms ${ }^{1} S$ and ${ }^{3} S$, each of which has one level. The ${ }^{1} S{0}$ level is nondegenerate; the ${ }^{3} S{1}$ level is threefold degenerate. The $1 s 2 p$ configuration gives rise to the terms ${ }^{1} P$ and ${ }^{3} P$. The levels of ${ }^{3} P$ are ${ }^{3} P{2},{ }^{3} P{1}$, and ${ }^{3} P{0} ;{ }^{1} P$ has the single level ${ }^{1} P{1}$.

Tables of Atomic Energy Levels

The spectroscopically determined energy levels for atoms with atomic number less than 90 are given in the tables of C. E. Moore and others: C. E. Moore, Atomic Energy Levels, National Bureau of Standards Circular 467, vols. I, II, and III, 1949, 1952, and 1958, Washington D.C.; these have been reprinted as Natl. Bur. Stand. Publ. NSRDS-NBS 35, 1971; W. C. Martin et al., Atomic Energy Levels-The Rare-Earth Elements, Natl. Bur. Stand. Publ. NSRDS-NBS 60, Washington, D.C., 1978. The atomic-energy-level data in Moore's tables and subsequent revisions are available online at the NIST Atomic Spectra Database at www.nist.gov/pml/data/asd.cfm.

These tables also list the levels of many atomic ions. Spectroscopists use the symbol I to indicate a neutral atom, the symbol II to indicate a singly ionized atom, and so on. To view the energy level data for the $\mathrm{C}^{2+}$ ion in the NIST online database, click on Levels and then enter C III after Spectrum.

FIGURE 11.4 Some term Energy/eV energies of the carbon atom.

The tables take the zero level of energy at the lowest energy level of the atom and list the level energies $E{i}$ as $E{i} / h c$ in $\mathrm{cm}^{-1}$, where $h$ and $c$ are Planck's constant and the speed of light. The difference in $E / h c$ values for two levels gives the wavenumber [Eq. (4.64)] of the spectral transition between the levels (provided the transition is allowed). An energy $E$ of 1 eV corresponds to $E / h c=8065.544 \mathrm{~cm}^{-1}$ (Prob. 11.29).

Figure 11.4 shows some of the term energies of the carbon atom. The separations between levels of each term are too small to be visible in this figure.