Comprehensive Study Notes
The Hamiltonian operator of an atom can be divided into three parts:
\(
\begin{equation}
\hat{H}=\hat{H}^{0}+\hat{H}{\mathrm{rep}}+\hat{H}{\mathrm{S} . \mathrm{O}} \tag{11.68}
\end{equation}
\)
where $\hat{H}^{0}$ is the sum of hydrogenlike Hamiltonians,
\(
\begin{equation}
\hat{H}^{0}=\sum{i=1}^{n}\left(-\frac{\hbar^{2}}{2 m{e}} \nabla{i}^{2}-\frac{Z e^{2}}{4 \pi \varepsilon{0} r_{i}}\right) \tag{11.69}
\end{equation}
\)
$\hat{H}_{\text {rep }}$ consists of the interelectronic repulsions,
\(
\begin{equation}
\hat{H}{\mathrm{rep}}=\sum{i} \sum{j>i} \frac{e^{2}}{4 \pi \varepsilon{0} r_{i j}} \tag{11.70}
\end{equation}
\)
and $\hat{H}_{\text {S.O. }}$ is the spin-orbit interaction (11.64):
\(
\begin{equation}
\hat{H}{\text {S.O. }}=\sum{i=1}^{n} \xi{i} \hat{\mathbf{L}}{i} \cdot \hat{\mathbf{S}}_{i} \tag{11.71}
\end{equation}
\)
If we consider just $\hat{H}^{0}$, all atomic states corresponding to the same electronic configuration are degenerate. Adding in $\hat{H}{\text {rep }}$, we lift the degeneracy between states with different $L$ or $S$ or both, thus splitting each configuration into terms. Next, we add in $\hat{H}{\text {S.O }}$, which splits each term into levels. Each level is composed of states with the same value of $J$ and is $(2 J+1)$-fold degenerate, corresponding to the possible values of $M_{J}$.
We can remove the degeneracy of each level by applying an external magnetic field (the Zeeman effect). If $\mathbf{B}$ is the applied field, we have the additional term in the Hamiltonian [Eq. (6.131)]
\(
\begin{equation}
\hat{H}{B}=-\hat{\mathbf{m}} \cdot \mathbf{B}=-\left(\hat{\mathbf{m}}{L}+\hat{\mathbf{m}}_{S}\right) \cdot \mathbf{B} \tag{11.72}
\end{equation}
\)
where both the orbital and spin magnetic moments have been included. Using $\mathbf{m}{L}=-\left(e / 2 m{e}\right) \mathbf{L}, \mathbf{m}{S}=-\left(e / m{e}\right) \mathbf{S}$, and $\mu{\mathrm{B}} \equiv e \hbar / 2 m{e}$ [Eqs. (6.128), (10.55) with $g_{e} \approx 2$, and (6.130)], we have
\(
\begin{equation}
\hat{H}{B}=\mu{\mathrm{B}} \hbar^{-1}(\hat{\mathbf{L}}+2 \hat{\mathbf{S}}) \cdot \mathbf{B}=\mu{\mathrm{B}} \hbar^{-1}(\hat{\mathbf{J}}+\hat{\mathbf{S}}) \cdot \mathbf{B}=\mu{\mathrm{B}} B \hbar^{-1}\left(\hat{J}{z}+\hat{S}{z}\right) \tag{11.73}
\end{equation}
\)
where $\mu{\mathrm{B}}$ is the Bohr magneton and the $z$ axis is taken along the direction of the field. If the external field is reasonably weak, its effect will be less than that of the spin-orbit interaction, and the effect of the field can be calculated by use of first-order perturbation theory. Since $\hat{J}{z} \psi=M_{J} \hbar \psi$, the energy of interaction with the applied field is
\(
E{B}=\langle\psi| \hat{H}{B}|\psi\rangle=\mu{\mathrm{B}} B M{J}+\mu{\mathrm{B}} B \hbar^{-1}\left\langle S{z}\right\rangle
\)
Evaluation of $\left\langle S_{z}\right\rangle$ (Bethe and Jackiw, p. 169) gives as the final weak-field result:
\(
\begin{equation}
E{B}=\mu{\mathrm{B}} g B M_{J} \tag{11.74}
\end{equation}
\)
where $g$ (the Landé $g$ factor) is given by
\(
\begin{equation}
g=1+\frac{[J(J+1)-L(L+1)+S(S+1)]}{2 J(J+1)} \tag{11.75}
\end{equation}
\)
Thus the external field splits each level into $2 J+1$ states, each state having a different value of $M_{J}$.
Figure 11.6 shows what happens when we consider successive interactions in an atom, using the $1 s 2 p$ configuration of helium as the example.
FIGURE 11.6 Effect of inclusion of successive terms in the atomic Hamiltonian for the $1 s 2 p$ helium configuration. $\hat{H}_{B}$ is not part of the atomic Hamiltonian but is due to an applied magnetic field.
We have based the discussion on a scheme in which we first added the individual electronic orbital angular momenta to form a total-orbital-angular-momentum vector and did the same for the spins: $\mathbf{L}=\sum{i} \mathbf{L}{i}$ and $\mathbf{S}=\sum{i} \mathbf{S}{i}$. We then combined $\mathbf{L}$ and $\mathbf{S}$ to get $\mathbf{J}$. This scheme is called Russell-Saunders coupling (or $L-S$ coupling) and is appropriate where the spin-orbit interaction energy is small compared with the interelectronic repulsion energy. The operators $\hat{\mathbf{L}}$ and $\hat{\mathbf{S}}$ commute with $\hat{H}^{0}+\hat{H}{\text {rep }}$, but when $\hat{H}{\text {s.o }}$ is included in the Hamiltonian, $\hat{\mathbf{L}}$ and $\hat{\mathbf{S}}$ no longer commute with $\hat{H}$. ( $\hat{\mathbf{J}}$ does commute with $\hat{H}^{0}+\hat{H}{\text {rep }}+\hat{H}{\text {S.o. }}$.) If the spin-orbit interaction is small, then $\hat{\mathbf{L}}$ and $\hat{\mathbf{S}}$ "almost" commute with $\hat{H}$, and $L-S$ coupling is valid.
As the atomic number increases, the average speed $v$ of the electrons increases. As $v / c$ increases, relativistic effects such as the spin-orbit interaction increase. For atoms with very high atomic number, the spin-orbit interaction exceeds the interelectronic repulsion energy, and we can no longer consider $\hat{\mathbf{L}}$ and $\hat{\mathbf{S}}$ to commute with $\hat{H}$; the operator $\hat{\mathbf{J}}$, however, still commutes with $\hat{H}$. In this case we first add in $\hat{H}{\text {s.o. }}$ to $\hat{H}^{0}$ and then consider $\hat{H}{\text {rep }}$. This corresponds to first combining the spin and orbital angular momenta of each electron to give a total angular momentum $\mathbf{j}{i}$ for each electron: $\mathbf{j}{i}=\mathbf{L}{i}+\mathbf{S}{i}$. We then add the $\mathbf{j}{i}$ 's to get the total electronic angular momentum: $\mathbf{J}=\sum{i} \mathbf{j}_{i}$. This scheme is called $j-j$ coupling. For most heavy atoms, the situation is intermediate between $j-j$ and $L-S$ coupling, and computations are difficult.
Several other effects should be included in the atomic Hamiltonian. The finite size of the nucleus and the effect of nuclear motion slightly change the energy (Bethe and Salpeter, pp. 102, 166, 351). There is a small relativistic term due to the interaction between the spin magnetic moments of the electrons (spin-spin interaction). We should also take into account the relativistic change of electronic mass with velocity. This effect is significant for inner-shell electrons of heavy atoms, where average electronic speeds are not negligible in comparison with the speed of light.
If the nucleus has a nonzero spin, the nuclear spin magnetic moment interacts with the electronic spin and orbital magnetic moments, giving rise to atomic hyperfine structure. The nuclear spin angular momentum $\mathbf{I}$ adds vectorially to the total electronic angular momentum $\mathbf{J}$ to give the total angular momentum $\mathbf{F}$ of the atom: $\mathbf{F}=\mathbf{I}+\mathbf{J}$. For example, consider the ground state of the hydrogen atom. The spin of a proton is $\frac{1}{2}$, so $I=\frac{1}{2}$; also, $J=\frac{1}{2}$. Hence the quantum number $F$ can be 0 or 1 , corresponding to the proton and electron spins being antiparallel or parallel. The transition $F=1 \rightarrow 0$ gives a line at 1420 MHz , the $21-\mathrm{cm}$ line emitted by hydrogen atoms in outer space. In 1951, Ewen and Purcell stuck a horn-shaped antenna out the window of a Harvard physics laboratory and detected this line. The frequency of the hyperfine splitting in the ground state of hydrogen is one of the most accurately measured physical constants: $1420.405751768 \pm 0.000000002$ MHz [S. G. Karshenboim, Can. J. Phys., 78, 639 (2000); arxiv.org/abs/physics/0008051].