Applications to Atomic Spectra

Atomic spectra turn selection rules into observable patterns. Energy eigenvalues determine possible line positions, while matrix elements determine which lines are bright, weak, or absent in a specified approximation.

We will apply rank, parity, and spin rules to atomic states, then calculate one hydrogen dipole matrix element to show what the symmetry argument leaves undetermined. Atomic Selection Rules extends these examples to hyperfine levels, metastability, and external fields.

Line Positions Versus Line Strengths

A spectral line frequency is set by an energy difference:

$$\hbar\omega_{fi} = E_f-E_i.$$

For absorption, $E_f>E_i$. For emission, the initial state is higher in energy and the emitted photon energy is $E_i-E_f$.

But the existence of an energy difference is not enough. A transition also needs a nonzero matrix element of the interaction operator. In the electric-dipole approximation, the relevant amplitude has the form

$$\langle f| \mathbf d\cdot\boldsymbol\epsilon |i\rangle,$$

where $\mathbf d$ is the electric dipole operator and $\boldsymbol\epsilon$ is the positive-frequency polarization vector for absorption. For emission into that mode, use its complex conjugate; Dipole Transitions fixes the convention. If this matrix element vanishes by symmetry, the electric-dipole line is forbidden at that order even when the energy difference matches the photon frequency.

For rates and resonance factors, use Selection Rules in Transition Rates and Fermi’s Golden Rule. For experimental context, use Spectroscopy.

Hydrogenic Orbital States

For the leading spinless hydrogenic problem, bound states are labeled

$$|n,\ell,m\rangle,$$

with spatial wavefunctions

$$\psi_{n\ell m}(\mathbf r) = R_{n\ell}(r)Y_\ell^m(\hat{\mathbf r}).$$

The electric dipole operator is proportional to $\mathbf r$. Its spherical components are rank-$1$ tensor components $r_q$ with $q=0,\pm1$. A typical matrix element separates schematically as

$$\langle n'\ell'm'|r_q|n\ell m\rangle = \int_0^\infty r^3 R_{n'\ell'}^*(r)R_{n\ell}(r)\,dr \int Y_{\ell'}^{m'*}(\hat{\mathbf r}) C_q^{(1)}(\hat{\mathbf r}) Y_\ell^m(\hat{\mathbf r})\,d\Omega.$$

The radial integral affects the size. The angular integral controls many exact zeros.

The rank-$1$ angular rule gives

$$\ell'=\ell,\ell\pm1$$

where the values are allowed by nonnegative angular momentum and triangle constraints. The magnetic rule gives

$$m'=m+q.$$

Parity then removes the $\ell'=\ell$ option for electric dipole transitions, because orbital parity is

$$\pi_\ell=(-1)^\ell$$

and the electric dipole operator is odd. Therefore the familiar hydrogenic electric-dipole orbital rule is

$$\Delta\ell=\pm1, \qquad \Delta m=0,\pm1.$$

The detailed symmetry derivation is in Dipole Transitions, with the parity part isolated in Parity Selection Rules.

Polarization

The spherical component $q$ is tied to polarization relative to the chosen quantization axis:

Component	Common polarization language	Magnetic rule
$q=0$	linear polarization along the quantization axis	$\Delta m=0$
$q=+1$	one circular component	$\Delta m=+1$
$q=-1$	the opposite circular component	$\Delta m=-1$

The sign convention for circular polarization depends on whether one labels emitted or absorbed photons, active or passive rotations, and the propagation direction. The invariant statement is the tensor one:

$$m_f=m_i+q$$

for the operator component $T_q^{(1)}$ used in the matrix element.

External magnetic fields split the $m$ levels and make this polarization structure visible as Zeeman components. The historical and experimental bridge begins with Zeeman Effect Revisited.

Fine Structure and Total Angular Momentum

Once spin and spin–orbit effects are included, atomic states are usually labeled by total angular momentum:

$$|\alpha,J,M,\pi\rangle.$$

The symbol $\alpha$ stands for additional labels: principal quantum numbers, orbital terms, spin multiplicity, configuration labels, radial labels, or other degeneracy labels.

For an electric dipole operator, the rotational rank is $1$, so

$$|J_i-1| \le J_f \le J_i+1,$$

and

$$M_f=M_i+q.$$

Equivalently, one often writes

$$\Delta J=0,\pm1,$$

subject to the triangle condition. In particular, $J_i=0$ to $J_f=0$ is excluded for a rank-$1$ operator. The parity rule requires

$$\pi_f=-\pi_i.$$

These are angular and parity rules. They do not by themselves compute the line strength; the reduced matrix element contains the atomic dynamics.

Term Symbols and Spin Rules

In an $LS$-coupling description, atomic levels are often labeled by terms of the form

$${}^{2S+1}L_J.$$

The multiplicity $2S+1$ labels total spin, the letter $L$ labels total orbital angular momentum, and the subscript $J$ labels their resultant. See Atomic Term Symbols for parity and configuration notation.

The electric dipole operator acts on spatial charge coordinates and does not act directly on spin in the simplest nonrelativistic approximation. This gives the approximate spin rule

$$\Delta S=0.$$

Spin–orbit and other spin-dependent interactions can mix different nominal $S$ labels while preserving total $J$ and parity. They can therefore lend E1 strength to an opposite-parity intercombination line. Spin-independent electrostatic configuration mixing preserves $S$ and parity; it can alter radial strengths or configuration-based rules, but cannot alone remove a spin or parity zero. This distinction matches the separate levels of approximation in the NIST selection-rule table.

Multipole Competition

When an $E1$ matrix element vanishes, another multipole may become relevant. The leading alternatives are often:

• magnetic dipole $M1$, rank $1$ and parity even;
• electric quadrupole $E2$, rank $2$ and parity even;
• higher electric or magnetic multipoles;
• two-photon processes;
• symmetry-breaking or mixing mechanisms.

The rank and parity table is collected in Multipole Operators. For example:

Mechanism	Parity rule	Typical angular rank
$E1$	opposite parity	$1$
$M1$	same parity	$1$
$E2$	same parity	$2$
$M2$	opposite parity	$2$

This explains why a line called forbidden may still appear weakly. It is often forbidden only for the leading $E1$ operator.

Many-Electron Atoms

In many-electron atoms, the same symmetry logic survives but the labels become richer. A configuration has a parity determined by the occupied orbital angular momenta:

$$\pi_{\rm config} = (-1)^{\sum_a \ell_a}.$$

Term labels organize angular momentum and spin, but electron-electron interactions, antisymmetrization, and configuration mixing affect the reduced matrix elements. Selection rules provide exact or approximate zeros before a detailed many-electron calculation, but they do not replace that calculation.

For an $E1$ transition in a parity-conserving atom, the total parity must change. In a pure $LS$ description one also expects $\Delta S=0$, with angular constraints on $L,J,M$ depending on the coupling scheme and operator component. In intermediate coupling, the strongest statement is made in terms of total $J,M$, parity, and the actual mixed eigenstates.

Worked Hydrogen Dipole Matrix Element

Ignore spin and take hydrogen with an infinitely heavy nucleus, so $a_0$ is the Bohr radius. For $2p,m=0\to1s$, the nonzero $q=0$ matrix element is $\langle1s|z|2p,0\rangle$. Using

$$R_{10}=2a_0^{-3/2}e^{-r/a_0},\qquad R_{21}=\frac{1}{2\sqrt6}a_0^{-3/2}\frac r{a_0}e^{-r/(2a_0)},$$

the angular factor is

$$\int Y_{00}^*\cos\theta\,Y_{10}\,d\Omega=\frac1{\sqrt3}.$$

The radial integral is

$$\int_0^\infty r^3R_{10}R_{21}\,dr =\frac{1}{\sqrt6a_0^4}\int_0^\infty r^4e^{-3r/(2a_0)}dr =\frac{256}{81\sqrt6}a_0,$$

where $\int_0^\infty r^4e^{-\beta r}dr=4!/\beta^5$. Therefore

$$\langle1s|z|2p,0\rangle=\frac{128\sqrt2}{243}a_0.$$

For an electron, multiply by $-e$ to obtain the dipole matrix element. Symmetry predicted that this component could be nonzero; the radial calculation supplies its scale. The result has units of length before multiplication by charge, as it should.

Common Mistakes

• Predicting spectral line positions from selection rules. Selection rules constrain amplitudes, not energy differences.
• Treating $E1$ rules as rules for all possible radiation mechanisms.
• Forgetting parity when using only angular momentum triangle rules.
• Applying $\Delta S=0$ outside the coupling approximation where spin is a good label.
• Ignoring polarization and then misassigning $\Delta M$ components.
• Assuming a symmetry-allowed transition must be experimentally strong.
• Treating many-electron term labels as exact when configuration mixing is important.

Exercises

1. Is the hydrogenic transition $2p,m=0\to1s,m=0$ electric-dipole allowed for the $q=0$ component?

Solution

The transition has $\ell_i=1$, $m_i=0$, $\ell_f=0$, and $m_f=0$. For $q=0$, the magnetic rule gives $\Delta m=0$, which is satisfied. The electric dipole operator has rank $1$, so $\Delta\ell=-1$ is allowed by the triangle rule. Parity changes from $(-1)^1=-1$ to $(-1)^0=+1$, as required for $E1$. Thus the transition is allowed by these symmetry rules.

2. Why is the hydrogenic $2s\to1s$ one-photon electric-dipole transition forbidden?

Solution

Both states have $\ell=0$, so they have the same parity. The electric dipole operator is odd and requires opposite parity. Equivalently, the rank-$1$ angular rule would require $\ell_f=1$ when $\ell_i=0$, not $\ell_f=0$. Thus the $E1$ matrix element vanishes. Other mechanisms, such as two-photon decay, are separate higher-order processes.

3. A pure $LS$ model has an even $^1S_0$ ground level and two odd levels $^1P_1$ and $^3P_1$. Which transition is spin forbidden, and what kind of mixing can open it?

Solution

Both transitions satisfy the $J=0\leftrightarrow1$ and opposite-parity E1 rules. The $^1S_0\leftrightarrow{}^1P_1$ line also preserves $S=0$. The $^1S_0\leftrightarrow{}^3P_1$ line changes $S$ from $0$ to $1$ and vanishes in pure $LS$ coupling. Spin–orbit mixing between the two odd $J=1$ levels can lend it singlet character and E1 strength. Electrostatic mixing that preserves $S$ cannot do this by itself.

4. Two atomic levels have the same parity and satisfy the rank-$1$ angular rule. Which is the more natural leading multipole candidate, $E1$ or $M1$?

Solution

$E1$ is parity odd, so it requires opposite parity and is forbidden between same-parity states. $M1$ is parity even and rank $1$, so it is the more natural leading one-photon multipole candidate by these symmetry checks. Whether the actual $M1$ matrix element is large enough to observe depends on the detailed states and dynamics.

References

• B. H. Bransden and C. J. Joachain, Physics of Atoms and Molecules, 2nd ed., Pearson, 2003.
• E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra, Cambridge University Press, 1935.
• R. D. Cowan, The Theory of Atomic Structure and Spectra, University of California Press, 1981.
• A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1957.
• H. Friedrich, Theoretical Atomic Physics, 4th ed., Springer, 2017.
• National Institute of Standards and Technology, “Atomic Spectroscopy: Spectral Lines,” section 17 of Atomic Spectroscopy: A Compendium of Basic Ideas, Notation, Data, and Formulas. Selection rules and transition strengths, accessed 2026-10-13.
• I. I. Sobelman, Atomic Spectra and Radiative Transitions, 2nd ed., Springer, 1992.