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Selection Rules

How do the normal and anomalous Zeeman formulas account for the details of the Zeeman splitting?

I omitted many transition lines from figures 1 and 2 to avoid cluttering the diagram. Nature seems to share this taste for simplicity. Of the multitude of transition lines one might draw, most are forbidden by selection rules.

For figures 1 and 2, the appropriate selection rules assert that $l$ changes by one unit, and $m$ changes by at most one unit: $\Delta l = \pm 1$, $\Delta m = 0, \pm 1$. Conservation of angular momentum accounts for these rules, as hinted earlier. But before getting into that, let us explore the role of selection rules in the Zeeman effect.

As it happens, the normal and anomalous Zeeman effects call for different selection rules. Four selection rules apply to the normal Zeeman effect:

Plugging this into the formula $E^{\rm mag} = {\rm constant}\,B(M_L +
2M_S)$, we get:
\Delta E^{\rm mag} = {\rm constant}\, B\Delta M_L
\end{displaymath} (3)

Spin has dropped out completely! $\Delta M_L$ has only three possible values, thus the magnetic perturbation $\Delta E^{\rm mag}$ splits each spectral line into three.

For example, consider transitions from $L=2$ states to $L=1$ states. Without the magnetic field, each $L=2$ state belongs to a degenerate quintuplet of states (distinguished by the five possible values of $M_L$) - where ``degenerate'', you may recall, simply means that all states in the quintuplet have the same energy. Likewise, each $L=1$ state belongs to a degenerate triplet. This degeneracy stems from the fact that the energy does not depend on $M_L$, when there is no field.

Had we no selection rules at all, we would have 15 possible transitions from the $L=2$ quintuplet to the $L=1$ triplet. The selection rule cuts this down to 9 (we can draw three lines to each state in the triplet). Since the quintuplet and the triplet are each degenerate, $\Delta E$ is the same for all 9 quintuplet-triplet transitions. We see a single spectral line from 9 transitions.

Now turn on the magnetic field. The degeneracy is lifted - the energy levels in the quintuplet separate slightly, as do those in the triplet (see figures 2). Group the 9 quintuplet-triplet transitions into three groups of three, according to the value of $\Delta M_L$. Each group has the same $\Delta E^{\rm mag}$, and so 9 transitions give rise to three spectral lines.

For the anomalous Zeeman effect, different selection rules apply:

$M_L$ and $M_S$ fade from the picture. We will search for them again when we come to the Paschen-Back effect. ($L$ and $S$ are still with us, though.)

We want to plug this into formula 2 for the anomalous Zeeman $E^{\rm mag}$. We don't need the full complexity of formula 2; the following is enough:

E^{\rm mag}_{JLSM_J} = f(J,L,S)M_J
\end{displaymath} (4)

The function $f(J,L,S)$ contains the complicated fraction we had before, plus constants (like the magnetic field strength $B$). Consider a transition from a $(J,L,S,M_J)$ state to a $(J',L',S',M_J')$ state. Plugging in the selection rules, we get:
\Delta E^{\rm mag} = f(J,L,S)M_J - f(J',L',S')M_J'
\end{displaymath} (5)

As before, we first turn the field off, then on. Without the field, the energy of a state does not depend on $M_J$, and so each state belongs to a degenerate multiplet. We label the multiplet with the quantum numbers $(J,L,S)$. Without the field, all transitions from one multiplet to another have the same $\Delta E$-- shining out one single spectral line.

Turn the field on, and this overloaded spectral line splits according to the different values of $\Delta E^{\rm mag}$. In contrast to normal Zeeman splitting, transitions with the same $\Delta M_J$ generally don't have the same $\Delta E^{\rm mag}$. Count the spectral lines, and you've counted the transitions (apart from accidental coincidences, which are rare).

Figure 3: Sodium D-lines
...{\raisebox{0pt}[0pt][0pt]{\rm 2}}}

Figure 4: Zeeman Effect for Sodium
...ox{0pt}[0pt][0pt]{\rm L=1 J=3/2}}}

For example, consider sodium, a historically important case. Sodium has two bright yellow spectral lines, known as D-lines (D$_1$ and D$_2$, close together).11 In a weak magnetic field, the D$_1$ line splits in four, and the D$_2$ line splits in six.

``Scarcely credible!'' wrote Lorentz of this discovery. Each D-line splits into an even number of lines. It would seem that some multiplet must contain an even number of states. But there are $2J+1$ different $M_J$'s for each value of $J$! Heisenberg and Landé unflinchingly ascribed half-integer values to $J$.

Here are the details. The D-lines come from three multiplets: a quartet and two doublets (see figure 3). The pair of quantum numbers $(L,J)$ specify a multiplet, as can be seen from the figure- $S$ is $\frac{1}{2}$ for all eight states. The D-lines are indicated by the slanted lines. The D$_1$-line actually encompasses all transitions from the higher to the lower doublet; the D$_2$-line encompasses all transitions from the quartet to the lower doublet.

Applying the selection rule for $\Delta M_J$, we find that all four conceivable D$_1$ transitions are permitted; six out of the eight conceivable D$_2$ transitions are permitted. Turn on the field, and these all separate. (See figure 4. To avoid cluttering the figure, the D$_2$-transitions are only partly drawn.) Hence the observed splitting.

I note finally that sodium is a ``hydrogenic'' atom. It has 11 electrons, one more than the noble gas neon. Ten electrons form a relatively inert ``core''-- they make up two complete ``shells'', which contribute collectively zero spin and zero orbital angular momentum (all electron spins are paired in the core.) The remaining electron is called a valence electron, and is far more loosely bound. A good approximation treats sodium as a system with a positively charged, immobile core, and a single orbiting electron-- rather like hydrogen. The eight states we just discussed are (to a high degree of approximation) eight different states of the valence electron (which is why they all have $S=\frac{1}{2}$).

Where do the selection rules come from? Here is one way to view a transition:

atom $_{\rm initial}$ $\rightarrow$ atom$_{\rm final}$ + photon
where ``atom $_{\rm initial}$'' and ``atom$_{\rm final}$'' refer of course to the initial and final states of the atom. $J_{\rm photon}=1$, so our rules for adding angular momenta say that

\vert J_{\rm final}-1\vert \leq J_{\rm initial} \leq J_{\rm final}+1

in steps of 1, or in other words, $\Delta J = 0,\pm 1$ for the atom.

We must supplement the catalog of spin facts to get the rule for $M_J$. Our first ``spin fact'' told us that if we have a composite system:

C = A\oplus B

then $\vert J(A)-J(B)\vert\leq J(C) \leq J(A)+J(B)$, in steps of 1. We now add a new requirment:

\vert M_J(A)-M_J(B)\vert\leq M_J(C) \leq M_J(A)+M_J(B)

in steps of 1.

You may be wondering how to interpret these equations. What does it mean to say $C = A\oplus B$? Modern quantum mechanics gives a clear answer: tensor product. (So $\otimes$ would be a better notation than $\oplus$.) But a muddled view hews closer to history. The old quantum theory did offer derivations of the selection rules, via a strange brew of Maxwell's equations and other ingredients. On the other hand, the Paschen-Back effect presented a severe conceptual stumbling block. How does one set of selection rules fade away, and the other take over? Heisenberg discarded the derivations when it suited his purpose, a matter of some controversy at the time (see Cassidy's biography). So I leave a coherent discussion to the textbooks, and continue with cruder arguments.

To get the selection rules for $\Delta S$ and $\Delta M_S$, we shift gears, and treat light as an electromagnetic wave, not photons. Also, we consider the reverse process, where an atomic state absorbs energy from incident light. An electron is so small (point-like, so far as we know today) that it feels at any moment a uniform electric field, and a uniform magnetic field. Classical physics says that magnetic effects are much smaller than electric effects, so we will ignore the former. (In modern parlance, we are ``deriving'' the selection rule for electric dipole transitions, and ignoring magnetic dipole transitions.)

The uniform electric field can shove the electron this way or that, but cannot flip it over. So S is unaffected, and hence $\Delta S=\Delta
M_S = 0$. On the other hand, if the wavelength of light is short enough, the electric field will be shoving the electron one way in one part of its orbit, and another way for another part, and so may change its orbital angular momentum L.

${\bf J} = {\bf L} + {\bf S}$. We've just seen that S doesn't change, so L and J must change the same way. So the selection rules for $L$ and $M_L$ must be the same as those for $J$ and $M_J$.

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© 2001 Michael Weiss