The atomic Hamiltonian (11.1) does not involve electron spin. In reality, the existence of spin adds an additional term, usually small, to the Hamiltonian. This term, called the spin-orbit interaction, splits an atomic term into levels. Spin-orbit interaction is a relativistic effect and is properly derived using Dirac's relativistic treatment of the electron. This section gives a qualitative discussion of the origin of spin-orbit interaction.
If we imagine ourselves riding on an electron in an atom, from our viewpoint, the nucleus is moving around the electron (as the sun appears to move around the earth). This apparent motion of the nucleus produces a magnetic field that interacts with the intrinsic (spin) magnetic moment of the electron, giving the spin-orbit interaction term in the Hamiltonian. The interaction energy of a magnetic moment $\mathbf{m}$ with a magnetic field $\mathbf{B}$ is given by (6.131) as $-\mathbf{m} \cdot \mathbf{B}$. The electron's spin magnetic moment $\mathbf{m}_{S}$ is proportional to its spin $\mathbf{S}$ [Eq. (10.57)], and the magnetic field arising from the apparent nuclear motion is proportional to the electron's orbital angular momentum $\mathbf{L}$. Therefore, the spin-orbit interaction is proportional to $\mathbf{L} \cdot \mathbf{S}$. The dot product of $\mathbf{L}$ and $\mathbf{S}$ depends on the relative orientation of these two vectors. The total electronic angular momentum $\mathbf{J}=\mathbf{L}+\mathbf{S}$ also depends on the relative orientation of $\mathbf{L}$ and $\mathbf{S}$, and so the spin-orbit interaction energy depends on $J$ [Eq. (11.67)].
When a proper relativistic derivation of the spin-orbit-interaction term $\hat{H}_{\text {S.O. }}$ in the atomic Hamiltonian is carried out, one finds that for a one-electron atom (see Bethe and Jackiw, Chapters 8 and 23)
\(
\begin{equation}
\hat{H}{\text {S.O. }}=\frac{1}{2 m{e}^{2} c^{2}} \frac{1}{r} \frac{d V}{d r} \hat{\mathbf{L}} \cdot \hat{\mathbf{S}} \tag{11.63}
\end{equation}
\)
where $V$ is the potential energy experienced by the electron in the atom and $c$ is the speed of light. One way to calculate $\hat{H}{\text {s.o. }}$ for a many-electron atom is first to neglect $\hat{H}{\text {S.O. }}$ and do an SCF calculation (Section 11.1) using the central-field approximation to get an effective potential energy $V{i}\left(r{i}\right)$ for each electron $i$ in the field of the nucleus and the other electrons viewed as charge clouds [Eqs. (11.7) and (11.8)]. One then sums (11.63) over the electrons to get
\(
\begin{equation}
\hat{H}{\text {S.O. }} \approx \frac{1}{2 m{e}^{2} c^{2}} \sum{i} \frac{1}{r{i}} \frac{d V{i}\left(r{i}\right)}{d r{i}} \hat{\mathbf{L}}{i} \cdot \hat{\mathbf{S}}{i}=\sum{i} \xi{i}\left(r{i}\right) \hat{\mathbf{L}}{i} \cdot \hat{\mathbf{S}}{i} \tag{11.64}
\end{equation}
\)
where the definition of $\xi{i}\left(r{i}\right)$ is obvious and $\hat{\mathbf{L}}{i}$ and $\hat{\mathbf{S}}{i}$ are the operators for orbital and spin angular momenta of electron $i$.
Calculating the spin-orbit interaction energy $E{\text {S.o. }}$ by finding the eigenfunctions and eigenvalues of the operator $\hat{H}{(11.1)}+\hat{H}{\text {S.O }}$, where $\hat{H}{(11.1)}$ is the Hamiltonian of Eq. (11.1), is difficult. One therefore usually estimates $E{\text {S.O. }}$ by using perturbation theory. Except for heavy atoms, the effect of $\hat{H}{\text {S.O }}$ is small compared with the effect of $\hat{H}{(11.1)}$, and first-order perturbation theory can be used to estimate $E{\text {S. } O}$.
Equation (9.22) gives $E{\mathrm{S} . \mathrm{O} .} \approx\langle\psi| \hat{H}{\text {S.O. }}|\psi\rangle$, where $\psi$ is an eigenfunction of $\hat{H}_{(11.1)}$. For a one-electron atom,
\(
\begin{equation}
E_{\mathrm{S} . \mathrm{O} .} \approx\langle\psi| \xi(r) \hat{\mathbf{L}} \cdot \hat{\mathbf{S}}|\psi\rangle \tag{11.65}
\end{equation}
\)
We have
\(
\begin{aligned}
\mathbf{J} \cdot \mathbf{J} & =(\mathbf{L}+\mathbf{S}) \cdot(\mathbf{L}+\mathbf{S})=L^{2}+S^{2}+2 \mathbf{L} \cdot \mathbf{S} \
\mathbf{L} \cdot \mathbf{S} & =\frac{1}{2}\left(J^{2}-L^{2}-S^{2}\right) \
(\hat{\mathbf{L}} \cdot \hat{\mathbf{S}}) \psi=\frac{1}{2}\left(\hat{J}^{2}-\hat{L}^{2}-\hat{S}^{2}\right) \psi & =\frac{1}{2}[J(J+1)-L(L+1)-S(S+1)] \hbar^{2} \psi
\end{aligned}
\)
since the unperturbed $\psi$ is an eigenfunction of $\hat{L}^{2}, \hat{S}^{2}$, and $\hat{J}^{2}$. Therefore,
\(
\begin{equation}
E_{\mathrm{S} . \mathrm{O} .} \approx \frac{1}{2}\langle\xi\rangle \hbar^{2}[J(J+1)-L(L+1)-S(S+1)] \tag{11.66}
\end{equation}
\)
For a many-electron atom, it can be shown (Bethe and Jackiw, p. 164) that the spin-orbit interaction energy is
\(
\begin{equation}
E_{\text {S.O. }} \approx \frac{1}{2} A \hbar^{2}[J(J+1)-L(L+1)-S(S+1)] \tag{11.67}
\end{equation}
\)
where $A$ is a constant for a given term; that is, $A$ depends on $L$ and $S$ but not on $J$. Equation (11.67) shows that when we include the spin-orbit interaction, the energy of an atomic state depends on its total electronic angular momentum $J$. Thus each atomic term is split into levels, each level having a different value of $J$. For example, the $1 s^{2} 2 s^{2} 2 p^{6} 3 p$ configuration of sodium has the single term ${ }^{2} P$, which is composed of the two levels ${ }^{2} P{3 / 2}$ and ${ }^{2} P{1 / 2}$. The splitting of these levels gives the observed fine structure of the sodium D line (Fig. 11.5). The levels of a given term are said to form its multiplet structure.
What about the order of levels within a given term? Since $L$ and $S$ are the same for such levels, their relative energies are determined, according to Eq. (11.67), by $A J(J+1)$. If $A$ is positive, the level with the lowest value of $J$ lies lowest, and the multiplet is said to be regular. If $A$ is negative, the level with the highest value of $J$ lies lowest, and the multiplet is said to be inverted. The following rule usually applies to a configuration with only one partly filled subshell: If this subshell is less than half filled, the multiplet is regular; if this
FIGURE 11.5 Fine structure of the sodium $D$ line.
subshell is more than halffilled, the multiplet is inverted. (A few exceptions exist.) For the half-filled case, see Prob. 11.28.
EXAMPLE
Find the ground level of the oxygen atom.
The ground electron configuration is $1 s^{2} 2 s^{2} 2 p^{4}$. Table 11.2 gives ${ }^{1} S,{ }^{1} D$, and ${ }^{3} P$ as the terms of this configuration. By Hund's rule, ${ }^{3} P$ is the lowest term. [Alternatively, a diagram like (11.48) could be used to conclude that ${ }^{3} P$ is the lowest term.] The ${ }^{3} P$ term has $L=1$ and $S=1$, so the possible $J$ values are 2, 1, and 0 . The levels of ${ }^{3} P$ are ${ }^{3} P{2},{ }^{3} P{1}$, and ${ }^{3} P{0}$. The $2 p$ subshell is more than half filled, so the rule just given predicts the multiplet is inverted, and the ${ }^{3} P{2}$ level lies lowest. This is the ground level of O .
EXERCISE Find the ground level of the Cl atom. (Answer: ${ }^{2} P_{3 / 2}$.)