Selection Rules for Dipole Transitions

The electric-dipole approximation hands this subject its central quantity already packaged: from aq-optics-electric-dipole-approximation and aq-optics-spontaneous-emission, the probability of a transition between states $|i\rangle$ and $|f\rangle$ is proportional to the squared modulus of a single complex number, the transition dipole moment $\langle f|\,\mathbf{d}\,|i\rangle$ with $\mathbf{d}=-e\mathbf{r}$ . That number is an integral over all space of two wavefunctions times the vector $\mathbf{r}$ , and the central fact of atomic spectroscopy is that symmetry can force such an integral to vanish exactly. When it does, the transition is forbidden at the dipole level and proceeds — if at all — only through far weaker magnetic-dipole or electric-quadrupole channels. The selection rules are the complete catalogue of when the integral survives. They come in two independent statements, both pure symmetry. Parity: under $\mathbf{r}\to-\mathbf{r}$ the operator changes sign, so the integral survives only if the two states have opposite parity — the Laporte rule, which for a single electron sharpens to $\Delta\ell=\pm1$ . Angular momentum: $\mathbf{r}$ is a rank-1 spherical tensor, so by the Wigner-Eckart theorem it can only connect states whose angular momenta differ by one unit, giving $\Delta J=0,\pm1$ (never $0\to0$ ) and $\Delta m_J=0,\pm1$ . Those three values of $\Delta m_J$ are just the three polarizations of the photon in the atom's language, so the rules are angular-momentum conservation between atom and field — the hidden grammar behind the Zeeman effect, optical pumping, laser cooling, and the ghostly forbidden lines of interstellar nebulae.

Key takeaways

An electric-dipole (E1) transition is governed by $\langle f|\,\mathbf{r}\,|i\rangle$ ; when symmetry forces it to zero the transition is E1-forbidden and must find a weaker route.
Parity must flip: $\pi_f=-\pi_i$ (the Laporte rule), because $\mathbf{r}$ is odd under inversion; for one active electron this becomes $\Delta\ell=\pm1$ .
The photon carries one unit of angular momentum, so $\Delta J=0,\pm1$ (no $0\to0$ ), $\Delta L=0,\pm1$ (no $L=0\to0$ ), and $\Delta m_J=0,\pm1$ set by the photon polarization.
Spin does not couple to $\mathbf{r}$ : $\Delta S=0$ in LS coupling — broken only when spin-orbit mixing lets weak intercombination lines through.
All of it follows from one theorem — Wigner-Eckart applied to the rank-1 tensor $\mathbf{r}$ — plus the parity of the integrand; "forbidden" means the E1 amplitude is zero, not that nothing happens.
Key idea: an E1 transition is allowed only if $\langle f|\mathbf{r}|i\rangle\neq0$ ; because $\mathbf{r}$ is an odd, rank-1 tensor this demands a parity flip and changes in $J,m_J,L,\ell$ compatible with adding one unit of angular momentum, while spin is untouched so $\Delta S=0$ .

The core insight

An E1 transition is allowed only if $\langle f|\mathbf{r}|i\rangle\neq0$ ; since $\mathbf{r}$ is an odd, rank-1 tensor, this demands a parity flip and $\Delta J,\Delta m_J,\Delta L,\Delta\ell$ compatible with adding one unit of angular momentum — while spin is untouched, $\Delta S=0$ .

Where this lives today

Selection rules are the design manual for the most precise instruments ever built. Every optical atomic clock runs on a forbidden transition chosen precisely because the E1 channel is shut: the strontium lattice clock ticks on the $^1S_0\to{}^3P_0$ line at 698 nm, spin- and $0\to0$ -forbidden and therefore breathtakingly narrow, giving the stability that now defines the second. Trapped-ion clocks in ytterbium and aluminium use E2 and intercombination lines for the same reason: a transition the rules make weak has a long coherence time. The rules are equally the grammar of quantum control: laser cooling and magneto-optical traps steer atoms by choosing $\sigma^+$ or $\sigma^-$ light to drive specific $\Delta m_F=\pm1$ channels, herding population into stretched states and building the dark and bright superpositions behind coherent population trapping and Sisyphus cooling. And the same forbidden lines a laboratory could never see radiate freely in space: the green [O III] doublet at 495.9 and 500.7 nm, once mistaken for an element "nebulium," is a magnetic-dipole pair that today reads the density and temperature of planetary nebulae across the galaxy. From the clock on the shelf to the light of a dying star, the question is always Laporte's: has the parity flipped, and can the angular momentum close its triangle?

The physical picture

Picture the photon as a spin-1 messenger: absorbing or emitting one, the atom must accept or surrender exactly one quantum of angular momentum. Add angular momentum one to a state of total $J$ and the only final values the vector triangle permits run from $|J-1|$ to $J+1$ — precisely $\Delta J=0,\pm1$ . A $J=0$ state has no direction to reorient, so nowhere to store the unit: $0\to0$ is impossible for a single photon of any multipole order — the whole angular-momentum rule, read off a triangle. Parity is the deeper constraint, best seen as an integral rather than a slogan. The operator $\mathbf{r}$ is mirror-odd — reflect the world through the nucleus and it changes sign. The amplitude $\int\psi_f^*\,\mathbf{r}\,\psi_i\,\mathrm{d}^3r$ is a number that cannot depend on whether we describe the atom or its mirror image, so the integrand must be even, forcing $\psi_f^*\psi_i$ to be odd — the two states must have opposite parity. If they share parity the integrand is odd, the $+\mathbf{r}$ and $-\mathbf{r}$ contributions cancel exactly, and the amplitude is zero — which is why hydrogen $1s\to2s$ is E1-dead while $1s\to2p$ blazes.

Figure

The dipole integral survives only when the integrand is even, which needs an opposite-parity pair: 1s → 2p is allowed, 1s → 2s is forbidden because equal-parity states make the z-weighted integrand odd and it cancels to zero

The three sublevel channels $\Delta m_J=0,\pm1$ are the photon's polarization along the quantization axis: a photon along the axis is circularly polarized, carries $\pm\hbar$ of spin, and drives $\Delta m_J=\pm1$ (the $\sigma^{\pm}$ transitions), while one polarized along the axis is $\pi$ and drives $\Delta m_J=0$ . Switch on a magnetic field to lift the $m_J$ degeneracy and each channel appears at its own frequency, so which polarization emerges where reads out the sublevel structure directly — the whole principle of Zeeman spectroscopy, drawn below for the textbook case.

Figure

Normal Zeeman triplet for a J = 1 to J = 0 line: the three E1 channels Δm = +1, 0, −1 emit σ⁺, π, σ⁻ light and split into three lines spaced by μ_B·B/h = 14.0 GHz per tesla, shown for B = 1 T on the cadmium red line at 643.8 nm

Spin sits out the whole affair because $\mathbf{r}$ contains no spin operator: an E1 photon cannot flip the electron's spin, only its orbital motion radiates, so in the ideal light-atom limit the spin numbers are frozen, $\Delta S=0$ and $\Delta m_S=0$ . That freezing is spoiled only when spin-orbit coupling stirs a little of the wrong spin into a state, and the tiny leftover amplitude lights up the intercombination lines heavy atoms wear so prominently.

The core equations

The whole subject rests on one proportionality, inherited intact from the electric-dipole approximation: a radiative rate is set by the squared modulus of the transition dipole moment, a single complex number built from the two states and the position operator.

W_{i\to f}\;\propto\;\big|\langle f|\,\mathbf{d}\,|i\rangle\big|^2,\qquad \mathbf{d}=-e\,\mathbf{r} \qquad (1)

where: $W_{i\to f}$ — transition rate between initial state $i$ and final $f$ ; $\mathbf{d}=-e\mathbf{r}$ — electric dipole operator, $e$ the elementary charge and $\mathbf{r}$ the electron position; $\langle f|\mathbf{d}|i\rangle$ — the transition dipole moment, whose vanishing defines a forbidden transition. Physically: no dipole moment, no dipole radiation. The first symmetry that can kill the moment is parity. Since $\mathbf{r}$ is odd under inversion, sandwiching it between states of definite parity multiplies the two parities and a sign; the amplitude equals itself rather than minus itself only if the parities are opposite — the Laporte rule.

\hat{P}\,\mathbf{r}\,\hat{P}^{-1}=-\mathbf{r}\;\Rightarrow\;\pi_i\,\pi_f=-1 \qquad (2)

where: $\hat{P}$ — the parity (spatial-inversion) operator; $\pi_i,\pi_f=\pm1$ — the parities of the initial and final states. An E1 photon always flips parity, so any two states of the same parity are E1-disconnected whatever their other quantum numbers. The second symmetry is rotational. The position operator is a rank-1 spherical tensor, and the Wigner-Eckart theorem factors any of its matrix elements into a Clebsch-Gordan coefficient — holding all the magnetic-quantum-number dependence — times a reduced matrix element blind to orientation. Every angular-momentum rule is a statement about when that coefficient is non-zero.

\langle \alpha' j' m'|T^{(k)}_q|\alpha j m\rangle=\langle j\,m;k\,q|j'\,m'\rangle\,\frac{\langle\alpha' j'\|T^{(k)}\|\alpha j\rangle}{\sqrt{2j'+1}} \qquad (3)

where: $T^{(k)}_q$ — the $q$ component of a rank- $k$ spherical tensor, $k=1$ for $\mathbf{r}$ ; $\langle j\,m;k\,q|j'\,m'\rangle$ — Clebsch-Gordan coefficient carrying all $m$ -dependence; $\langle\alpha' j'\|T^{(k)}\|\alpha j\rangle$ — reduced matrix element, independent of $m,m',q$ . Geometry (the coefficient) decides whether a transition is allowed, dynamics (the reduced element) how strong. To use the theorem the vector is written in the spherical basis, where its three components are eigenstates of $z$ -projection with $q=0,\pm1$ — the $\pi$ and $\sigma^{\pm}$ photon polarizations.

r_0=z,\qquad r_{\pm1}=\mp\tfrac{1}{\sqrt2}(x\pm i y) \qquad (4)

where: $r_q$ — the $q=0,\pm1$ spherical components of the position vector; $x,y,z$ — Cartesian coordinates about the quantization axis. $r_0=z$ transfers no $z$ -projection (the $\pi$ channel), $r_{\pm1}$ transfer $\pm\hbar$ (the $\sigma^{\pm}$ channels). Feeding $k=1$ into the Clebsch-Gordan triangle condition yields the angular-momentum rules that hold for any atom, however coupled: the total angular momentum changes by at most one unit, its projection by the photon's $q$ .

\Delta J=0,\pm1\ \ (J{=}0\nrightarrow J{=}0),\qquad \Delta m_J=0,\pm1 \qquad (5)

where: $\Delta J$ — change in the total electronic angular momentum; $\Delta m_J$ — change in its projection, fixed by the photon polarization. $J$ and $J'$ must form a triangle with one, and $J=0\to J=0$ is the triangle that closes to nothing — universally forbidden. When LS coupling holds, two further rules appear — one from spin's immunity to $\mathbf{r}$ , one from the single-electron parity argument sharpened by the triangle. They are approximate exactly to the degree LS coupling is.

\Delta S=0,\qquad \Delta L=0,\pm1\ \ (L{=}0\nrightarrow L{=}0),\qquad \Delta\ell=\pm1 \qquad (6)

where: $\Delta S$ — change in total spin, zero because $\mathbf{r}$ is spin-independent; $\Delta L$ — change in total orbital angular momentum; $\Delta\ell$ — change in the active electron's orbital number. Reading: the active electron hops to an adjacent- $\ell$ shell ( $s\leftrightarrow p\leftrightarrow d\leftrightarrow f$ ), the term's $L$ obeys the same triangle as $J$ , and spin is a spectator.

Derivation AThe Laporte parity rule and why Δℓ must be odd

The goal is to turn "opposite parity" from a memorized rule into an inevitability, and then, for a single electron, to sharpen it into $\Delta\ell=\pm1$ . We use two facts: that $\mathbf{r}$ reverses sign under spatial inversion, and that a hydrogenic orbital of number $\ell$ has definite parity $(-1)^\ell$ . Everything else follows from one observation — a number cannot equal its own negative unless it is zero — which is how symmetry forbids transitions outright rather than merely disfavouring them.

Step 1. Insert $\hat P^{-1}\hat P=1$ on both sides of the operator, $\langle f|\mathbf{r}|i\rangle=\langle f|\hat P^{-1}\,(\hat P\mathbf{r}\hat P^{-1})\,\hat P|i\rangle$ , then use $\hat P|i\rangle=\pi_i|i\rangle$ , $\langle f|\hat P^{-1}=\pi_f\langle f|$ , and $\hat P\mathbf{r}\hat P^{-1}=-\mathbf{r}$ to collapse it to $-\pi_i\pi_f\langle f|\mathbf{r}|i\rangle$ . Why: this pins the amplitude to itself times $-\pi_i\pi_f$ , so its own symmetry either permits it or annihilates it.

Step 2. Demand consistency: $\langle f|\mathbf{r}|i\rangle=-\pi_i\pi_f\langle f|\mathbf{r}|i\rangle$ can hold with a non-zero amplitude only if $\pi_i\pi_f=-1$ ; if $\pi_i\pi_f=+1$ the amplitude equals minus itself and is exactly zero. Why: this is the Laporte rule in one line — opposite parity, or the dipole channel is dead.

Step 3. Specialize to one electron, parity $(-1)^\ell$ : opposite parity means $\ell_i+\ell_f$ odd, so $\Delta\ell$ is odd, while the rank-1 triangle allows only $\ell_f\in\{\ell_i-1,\ell_i,\ell_i+1\}$ . Why: parity throws out $\Delta\ell=0$ and the triangle throws out $|\Delta\ell|\ge2$ , so only $\Delta\ell=\pm1$ survives both filters.

Step 4. Read it on hydrogen: $1s\to2p$ is allowed and drives the bright Lyman- $\alpha$ line, while $1s\to2s$ ( $\ell:0\to0$ ) violates both parity and $\Delta\ell=\pm1$ and is E1-dead. Why: the same abstract argument now predicts, correctly, which hydrogen lines you see.

Go deeper · Derivation B · 1 minAngular momentum from the Wigner-Eckart theorem

Derivation BAngular momentum from the Wigner-Eckart theorem

The aim is to extract the two rules that hold for every atom, coupled however you like — the change in total angular momentum $J$ and its projection $m_J$ — from the geometry of the Clebsch-Gordan coefficient alone, without evaluating a radial integral. The Wigner-Eckart theorem quarantines all orientation dependence in that one coefficient, so "is this transition allowed?" reduces to "is this coefficient non-zero?", settled by two bookkeeping conditions on angular momenta.

Step 1. From equation (3) the matrix element is proportional to $\langle J\,m;1\,q|J'\,m'\rangle$ , the coefficient for adding the photon's angular momentum one to the atom's $J$ ; the amplitude is non-zero only where this coefficient is. Why: the reduced element sets strength but never vanishing, so allowedness is entirely a property of the coefficient.

Step 2. Apply the triangle condition $|J-1|\le J'\le J+1$ — the requirement that $J$ , $J'$ , and one form a closed vector triangle — giving $\Delta J=0,\pm1$ . Why: you cannot build a triangle whose sides violate the triangle inequality, and a single photon supplies exactly one unit to the third side.

Step 3. Apply the projection condition $m'=m+q$ with $q\in\{0,\pm1\}$ from the spherical components of $\mathbf{r}$ , giving $\Delta m_J=0,\pm1$ . Why: the $z$ -projection is strictly additive, so the atom changes it by exactly what the photon carries — which is what the polarization measures.

Step 4. Handle $J=J'=0$ : the only coefficient is $\langle0\,0;1\,q|0\,0\rangle$ , which vanishes, and $\langle0\,0;k\,q|0\,0\rangle=0$ for every rank $k\ge1$ . Why: no single-photon process of any multipole can connect two states that both have zero angular momentum.

Go deeper · Derivation C · 1 minSpin is a spectator, and how intercombination lines sneak in

Derivation CSpin is a spectator, and how intercombination lines sneak in

This derivation shows why the spin rule $\Delta S=0$ is exact in ideal LS coupling and — crucially — why it is only approximate in real atoms, the loophole that gives heavy elements their intercombination lines. The engine is simple: the dipole operator acts on spatial coordinates alone, so in a state that factorizes cleanly into spatial and spin parts the spin overlap factors straight out and enforces orthogonality. Spin-orbit coupling is exactly what spoils that factorization.

Step 1. In LS coupling write the state as a product $|i\rangle=|\text{space}\rangle\otimes|S,m_S\rangle$ , spatial and spin factors separate. Why: this separation is the defining assumption of LS coupling and is what lets the next step happen.

Step 2. Since $\mathbf{r}$ touches only the spatial factor, $\langle f|\mathbf{r}|i\rangle=\langle\text{space}_f|\mathbf{r}|\text{space}_i\rangle\,\langle S_f m_{S_f}|S_i m_{S_i}\rangle$ and the spin overlap is $\delta_{S_f S_i}\delta_{m_{S_f}m_{S_i}}$ . Why: the spin bra-ket is an inner product of orthonormal states, non-zero only when spin is unchanged — hence $\Delta S=0$ , $\Delta m_S=0$ .

Step 3. Break the factorization: spin-orbit interaction mixes a small amplitude of a neighbouring spin- $S'$ level into a nominal spin- $S$ level, so the "wrong" spin gains a foothold and an otherwise forbidden E1 transition proceeds weakly, with the admixture growing steeply in $Z$ . Why: this is the origin of intercombination lines — He $2\,^3S_1\to1\,^1S_0$ is extraordinarily weak, but mercury's $6\,^1S_0\to6\,^3P_1$ at 253.7 nm is strong enough to run every fluorescent lamp on Earth.

Limiting cases

The rules tighten or loosen with the coupling regime and multipole order — the same atom obeys a stricter version when light, a looser one when heavy:

Case	Result	Meaning
Single active electron	$\Delta\ell=\pm1,\ \ \Delta m_\ell=0,\pm1$	Hydrogen-like and alkali atoms; $\Delta\ell=0$ is killed by parity, so $s\to s$ never radiates.
LS coupling (light atoms)	$\Delta S=0,\ \Delta L=0,\pm1,\ \Delta J=0,\pm1$	$L=0\to L=0$ and $J=0\to J=0$ excluded; spin conserved to excellent approximation.
Heavy atoms toward jj coupling	$\Delta S=0$ breaks down	spin-orbit mixing turns on intercombination lines, e.g. Hg $6\,^1S_0\to6\,^3P_1$ at 253.7 nm becomes strong.
E1 forbidden, higher multipoles	M1: $\Delta J=0,\pm1$ , no parity change; E2: $\Delta J=0,\pm1,\pm2$ , no parity change	when the dipole vanishes, M1 and E2 take over with rates smaller by $\sim\alpha^2$ or $\sim(a_0/\lambda)^2$ .

Common trapForbidden means the dipole amplitude is zero, not that the transition never happens

"Forbidden" is shorthand for "the E1 matrix element vanishes," and it is routinely misread as "impossible." A forbidden transition simply falls back on weaker channels — M1, E2, or two-photon — typically $10^{-5}$ to $10^{-8}$ times slower, which in a collisionless nebula is plenty fast to light up the sky. Two further traps: $\Delta\ell=\pm1$ is a single-electron rule (the many-electron statements are $\Delta L,\Delta J=0,\pm1$ ), and the $0\to0$ exclusion applies to $J$ , $L$ , and $m$ — for a $\Delta J=0$ transition even $m=0\to m=0$ is forbidden, since $\langle0\,0;1\,0|0\,0\rangle=0$ .

Worked examples

1 · The two faces of hydrogen n = 2 (confidence builder)

Hydrogen has two ways to sit in $n=2$ : $2p$ ( $\ell=1$ ) and $2s$ ( $\ell=0$ ). Decide which decays to the ground state $1s$ by an electric-dipole photon, compute the wavelength of the allowed line, and contrast the two lifetimes.

Step 1. Apply the parity/orbital rule. For $2p\to1s$ , $\Delta\ell=-1$ : allowed, opposite parity, $\Delta J=0$ or $1$ (both $2p_{1/2,3/2}\to1s_{1/2}$ fine). For $2s\to1s$ , $\Delta\ell=0$ : forbidden, same even parity, Laporte-dead. Only $2p\to1s$ radiates by E1.

Step 2. Compute the allowed wavelength from the Rydberg formula, $E=13.6\,\text{eV}\,(1-\tfrac14)=10.2\,\text{eV}$ , so $\lambda=hc/E=1240/10.2=\mathbf{121.6\ nm}$ — the Lyman- $\alpha$ line, one of the brightest features in the Universe.

Step 3. Contrast the lifetimes. The allowed $2p$ decays at $A\approx6.3\times10^8\,\text{s}^{-1}$ , so $\tau_{2p}\approx\mathbf{1.6\ ns}$ . The forbidden $2s$ has no E1 route and limps down by two-photon emission, $\tau_{2s}\approx\mathbf{0.12\ s}$ . The ratio is about $10^8$ : one selection rule, obeyed or broken, changes the decay time by eight orders of magnitude. Only $2p\to1s$ is E1-allowed (121.6 nm, τ ≈ 1.6 ns); $2s$ is metastable at τ ≈ 0.12 s.

2 · The normal Zeeman triplet (a real number on the sky)

Cadmium's red line at 643.8 nm is the singlet transition $5\,^1D_2\to5\,^1P_1$ . Confirm it is E1-allowed, and find how far its Zeeman components split at $B=1.0\ \text{T}$ , in frequency and wavelength.

Step 1. Check the rules: both terms are singlets so $\Delta S=0$ ; $D\to P$ gives $\Delta L=-1$ ; $\Delta J=-1$ ; parity flips — fully allowed. And being singlets they share $g=1$ , exactly the condition for a normal Zeeman pattern.

Step 2. Each sublevel shifts by $\Delta E=g\,m\,\mu_B B=m\,\mu_B B$ , so the photon frequency changes by $\Delta\nu=\Delta m\,\mu_B B/h$ . With $\mu_B/h=13.996\times10^9\ \text{Hz/T}$ , the channels $\Delta m=-1,0,+1$ land at $\nu_0$ and $\nu_0\pm\mathbf{14.0\ GHz}$ for $B=1.0\ \text{T}$ — a clean triplet $\sigma^-$ , $\pi$ , $\sigma^+$ .

Step 3. As a wavelength offset, $\Delta\lambda=\lambda^2\,\Delta\nu/c=(643.8\times10^{-9})^2\,(1.40\times10^{10})/(3.00\times10^8)=\mathbf{0.019\ nm}$ — a whisker of a shift, but its polarization signature is unmistakable and it now measures magnetic fields on the Sun. Allowed; triplet at ν₀ and ν₀ ± 14.0 GHz, a σ split of 0.019 nm at 1 T.

3 · Screening a list of candidate lines

Which of these are E1-allowed? (a) Na $3\,^2S_{1/2}\to3\,^2P_{3/2}$ ; (b) Ca $4s^2\,^1S_0\to4s4p\,^3P_1$ ; (c) $^2D_{5/2}\to{}^2S_{1/2}$ ; (d) $^3P_0\to{}^3P_0$ .

Step 1. (a) The sodium D2 line: $\Delta S=0$ , $\Delta L=+1$ , $\Delta J=+1$ , parity flips — allowed, one of the brightest lines in nature at 589.0 nm ( $2.10\,\text{eV}$ ). (b) Calcium's intercombination line: $\Delta S=1$ violates the spin rule, so it is E1-forbidden in pure LS coupling — yet it shows weakly at 657 nm through spin-orbit mixing and now serves as an optical-clock transition.

Step 2. (c) $D\to S$ gives $\Delta L=-2$ , violating $\Delta L=0,\pm1$ : forbidden (an E2 line). (d) $\Delta J=0$ is allowed in general, but $J=0\to J=0$ is universally excluded: forbidden for E1, M1, and E2 alike. Allowed: (a). Forbidden: (b) by ΔS, (c) by ΔL, (d) by the 0→0 rule.

Physics of the portrait

The parity rule carries a name because one young researcher, staring at the dense iron spectrum, saw its levels sort into two non-combining families — years before anyone knew the principle was the mirror symmetry of space itself.

Origins · 1 minOtto Laporte and how this was discovered

Otto Laporte

1902–1971 · the parity selection rule and iron spectra

In 1924 Otto Laporte, analysing the bewildering arc spectrum of iron, found that its terms fall into two classes such that transitions occur only between the classes, never within one. He had no name for the property — he called the families simply primed and unprimed — but his empirical rule instantly tamed a spectrum that had defied classification. Three years later Eugene Wigner recognized the two classes as the even- and odd-parity states and traced the rule to invariance of the electromagnetic interaction under spatial inversion: an E1 photon carries away odd parity, so it can only join a state of one parity to a state of the other. This was among the earliest triumphs of symmetry reasoning in quantum mechanics, making parity a physical observable — a status it held, seemingly inviolable, until the 1956 discovery that the weak interaction violates it. Laporte later turned to fluid dynamics at Michigan, but the rule bearing his name is still the first thing every spectroscopist checks: has the parity flipped? If not, the line is dark before any number is computed.

History

1913

— Bohr's model fixes quantized stationary states and radiative jumps between them

1924

— Otto Laporte sorts the iron spectrum into two non-combining parity classes

1925

— Uhlenbeck and Goudsmit introduce electron spin, later grounding the ΔS = 0 rule

1927

— Wigner identifies Laporte's empirical rule as conservation of parity under inversion

1930

— The Wigner-Eckart theorem recasts every selection rule as spherical-tensor algebra

Going further

Texts. Foot, Atomic Physics (2005), Ch. 2 and Appendix C — the cleanest modern derivation of the E1 rules · Bransden and Joachain, Physics of Atoms and Molecules, Ch. 4 — explicit Gaunt/Wigner-3j evaluation of the angular integrals · Sakurai and Napolitano, Modern Quantum Mechanics, Ch. 3 — the Wigner-Eckart theorem in full · Loudon, The Quantum Theory of Light (3rd ed., 2000), Ch. 1–2 · Demtröder, Laser Spectroscopy, Vol. 1 · Hecht, Optics (5th ed.), Ch. 3.

Provenance. The rules are recalled from the transition-dipole and dipole-approximation articles and derived here two ways: from the parity of the integrand ( $\hat P\mathbf{r}\hat P^{-1}=-\mathbf{r}$ , following Sakurai Sec. 3.11 and the spherical-harmonic integrals in Bransden and Joachain Ch. 4) and from the Wigner-Eckart triangle condition. A verifier should recompute the parity argument $\pi_i\pi_f=-1$ , the triangle giving $\Delta J=0,\pm1$ and $\Delta\ell=\pm1$ , and the vanishing of $\langle0\,0;1\,q|0\,0\rangle$ . The anchor numbers — Lyman- $\alpha$ at 121.6 nm with $\tau_{2p}\approx1.6$ ns, the $2s$ two-photon lifetime $\approx0.12$ s, $\mu_B/h=14.0$ GHz/T, Hg 253.7 nm, and [O III] 495.9/500.7 nm — are standard values from Foot, Demtröder, and atomic-data compilations.

Builds on: Electric Dipole Approximation · Transition Dipole Moment · Spontaneous Emission. Leads to: Oscillator Strength · Einstein A and B Coefficients · Radiative Lifetime and Natural Linewidth · Thomas-Reiche-Kuhn Sum Rule.

Ask the tutor: Give me any transition — states, term symbols, or just two orbitals — and I will walk you through the parity, angular-momentum, and spin checks to say whether it glows or stays dark, and by what channel.

What is the difference between a "forbidden" and an "impossible" transition?

Forbidden means only that the electric-dipole matrix element $\langle f|\mathbf{r}|i\rangle$ is zero, so the strongest channel is closed. The transition still proceeds through weaker routes — M1, E2, or two-photon emission — typically $10^{-5}$ to $10^{-8}$ times slower. In a dense laboratory gas those slow decays are swamped by collisions, so you never see them; in a tenuous nebula or ion trap, where an atom sits undisturbed for seconds or hours, they radiate freely. So "forbidden" is about the dipole amplitude, not about whether the atom can ever change state — truly impossible would need a conservation law with no loophole at all.

Why does spin not change in an electric-dipole transition?

Because the electric-dipole operator $\mathbf{d}=-e\mathbf{r}$ is built entirely from position and contains no spin operator. In LS coupling a state factorizes into spatial and spin parts, the operator touches only the spatial factor, and the spin overlap factors out as a pure inner product — non-zero only when spin is unchanged. Physically, the charge cloud radiates while the spin is carried along. The rule fails only when spin-orbit coupling mixes a little of a different-spin state into each level; that admixture grows as a power of $Z$ , which is why intercombination lines are invisible in helium but conspicuous in mercury.

How can one rule be exact and another only approximate?

Parity and total-angular-momentum rules ( $\pi_i\pi_f=-1$ , $\Delta J=0,\pm1$ , no $0\to0$ ) are exact because they rest on symmetries of space itself — inversion and rotation — which the full Hamiltonian respects; nothing internal can violate them. The rules $\Delta S=0$ , $\Delta L=0,\pm1$ , and $\Delta\ell=\pm1$ , by contrast, assume LS coupling — that spin and orbit are separately good quantum numbers. Spin-orbit interaction spoils that to a degree that grows with $Z$ , so these are excellent for light atoms and leaky for heavy ones. Rules tied to space are rigid; rules tied to a coupling scheme are only as good as the scheme.

Why do the three Δm values correspond to polarizations?

Because a photon carries angular momentum, and its projection along the quantization axis is exactly what its polarization encodes. A photon along the axis is circularly polarized, carries $\pm\hbar$ of spin projection, and changes the atom's $m_J$ by $\pm1$ — the $\sigma^{\pm}$ channels; one polarized along the axis ( $\pi$ ) has zero projection and drives $\Delta m_J=0$ . Formally this comes from writing $\mathbf{r}$ in the spherical basis, where $r_{\pm1}$ and $r_0$ carry definite $q=\pm1,0$ , so the azimuthal integral forces $m_f-m_i=q$ . Which polarization emerges tells you which sublevel transition occurred — the basis of Zeeman spectroscopy and optical pumping.

If Δℓ = ±1 for one electron, what replaces it in a many-electron atom?

The single-electron rule $\Delta\ell=\pm1$ applies to the one "active" electron and is the right tool for hydrogen and the alkalis. In a many-electron atom the good quantum numbers are the total $L$ and $J$ , and the rules become $\Delta L=0,\pm1$ (no $L=0\to0$ ) and $\Delta J=0,\pm1$ (no $J=0\to0$ ), plus exact parity. Note $\Delta L=0$ is now allowed — a term can radiate without changing $L$ , provided parity still flips through a single electron's $\Delta\ell=\pm1$ . Applying the single-electron $\Delta\ell$ rule to a whole term is a common and costly mistake.

Why are forbidden lines so important in astrophysics if they are so weak?

Precisely because they are weak, and because space is empty. In a laboratory an excited atom collides billions of times per second, so a metastable state that would take seconds to radiate is knocked out of its level long before it can emit — the line is "collisionally quenched." In the near-perfect vacuum of a nebula, collisions are minutes or hours apart, so the atom emits its forbidden photon undisturbed. The green [O III] lines at 495.9 and 500.7 nm are the classic example, once attributed to a fictitious "nebulium." Their brightness relative to allowed lines is a direct thermometer and barometer of the gas — one of the most powerful diagnostics in astrophysics.