University Physics V · The Hydrogen Atom · 11.9

Dipole Matrix Elements & Selection Rules

Selection rules are not a table to memorise. They are what survives when you write r as a rank-one spherical tensor and let Wigner-Eckart split the geometry from the one radial integral — which is how you know, before computing anything, which of hydrogen's levels can radiate to which.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Light couples to the electron's charge, and across the 0.1 nm scale of an atom an optical field is essentially uniform. Keep only the leading term of e^(ik⋅r) and the interaction collapses to −d⋅E with d = −e r, so Fermi's golden rule turns every question about the spectrum into a question about one number: ⟨n′ l′ m′| r |n l m⟩. The move that makes that number tractable is algebraic, not analytic.

The vector r is a rank-one spherical tensor, so the Wigner-Eckart theorem factors the element into a Clebsch-Gordan coefficient — pure geometry, fixed by rotational symmetry, carrying all the m dependence — times a reduced matrix element that holds the radial physics and every metre. The selection rules then fall out of the two factors for free: the triangle rule allows Δl = 0, ±1 with m′ = m + q; parity kills Δl = 0 because r is odd; spin is untouched because r is the identity on the spin factor, so Δmₛ = 0; and nothing whatever constrains Δn, because the radial integral never vanishes by symmetry. What the model costs is the term you discarded.

"Forbidden" means only that the electric-dipole amplitude vanishes: restore the (k⋅r) correction and magnetic-dipole and electric-quadrupole channels reappear, suppressed by (k a₀)² ≈ 7×10⁻⁶, which is how the 21 cm line radiates at all. Hydrogen's 2s is stricter than that. Because 2s → 1s is l = 0 → l′ = 0 and one photon always carries at least one unit of angular momentum, every single-photon multipole is shut, not merely suppressed; the level empties by two-photon E1⋅E1 emission at 8.23 s⁻¹ and lives 0.12 s against 1.6 ns for 2p.

Simple definition: The electric-dipole matrix element ⟨n′ l′ m′| r |n l m⟩ is the amplitude that decides whether a spectral line exists and how fast it radiates; Wigner-Eckart splits it into a Clebsch-Gordan factor set by rotational geometry and one reduced radial integral set by the states.
Example: For 2p → 1s the whole 3 × 1 × 3 array of amplitudes reduces to one radial number, ⟨1s|r|2p⟩ = 2⁷√6/3⁵ a₀ = 1.290 a₀ = 6.83×10⁻¹¹ m, which gives A = 6.3×10⁸ s⁻¹ and a lifetime of 1.6 ns.

Dipole operator and its matrix elementd = −e r, d_fi = −e ⟨n′ l′ m′| r |n l m⟩

One complex number per pair of states. Every rate, intensity and lifetime in the spectrum is built from it.

e = 1.602×10⁻¹⁹ C and r in m, so d is in C m; e a₀ = 8.48×10⁻³⁰ C m = 2.54 D.

Position as a rank-one spherical tensorr¹₀ = z, r¹_(±1) = ∓(x ± iy)/√2

Rewriting r this way is the step that lets Wigner-Eckart act: q = 0 is π light, q = ±1 is σ light.

The three components carry photon projection q = +1, 0, −1 on the quantisation axis; each has units of length.

Wigner-Eckart factorisation⟨n′l′m′| r¹_q |n l m⟩ = ⟨l m1 q | l′ m′⟩ ⟨n′l′‖r‖n l⟩ / √(2l′+1)

Collapses the 3(2l+1)(2l′+1) integrals of a level pair to one radial number times a table of coefficients.

The Clebsch-Gordan factor is a pure number from rotations; the reduced element carries the metres and all n dependence.

The radial integral you actually computeR_(n′l′, nl) = ∫₀^∞ R_(n′l′)(r) · r · R_{nl}(r) r² dr

Never zero by symmetry, so Δn is unrestricted; it only sets how bright an already-allowed line is.

Units of metres. For 1s ← 2p it equals 2⁷√6/3⁵ a₀ = 1.290 a₀ = 6.83×10⁻¹¹ m.

Electric-dipole selection rulesΔl = ±1, Δmₗ = 0, ±1, Δmₛ = 0, Δn unrestricted

Decides which of the 2n² substates can reach which, before a single integral is attempted.

The triangle rule gives Δl = 0, ±1 and shuts l = l′ = 0 outright; parity, requiring l + l′ odd, removes the rest of Δl = 0.

Spontaneous rate and lifetimeA = ω³ |d_fi|² / (3π ε₀ ħ c³), τ = 1 / ΣA

The ω³ is why an allowed ultraviolet line fires in nanoseconds. It is not the whole story for a forbidden one: against Lyman-α the 21 cm line loses 5×10¹⁸ to ω³ and a further 4×10⁴ to its magnetic-dipole matrix element.

ω in rad s⁻¹, d_fi in C m, A in s⁻¹. For 2p → 1s, A = 6.3×10⁸ s⁻¹ and τ = 1.6 ns.

Why the amplitude, not the energy, decides

The Bohr frequency condition ħω = Eᵢ − E_f says where a line would sit. Whether it appears at all is a separate question, answered by Fermi's golden rule: the rate carries |⟨f| H′ |i⟩|², so a vanishing matrix element means no line no matter how good the energy match. The coupling to a plane-wave field is (e/mₑ) A⋅p with A carrying e^(ik⋅r), and across an atom that exponential barely moves: for Lyman-α, k a₀ = 2π(0.0529 nm)/(121.6 nm) = 2.7×10⁻³. Set e^(ik⋅r) ≈ 1 and the interaction becomes −d⋅E with d = −e r. Name the operator and its home: r acts on the radial and angular factors of the state space H_radial ⊗ Hₛₚₕₑᵣₑ ⊗ Hₛₚᵢₙ, and is the identity on the spin factor. Everything that follows is a statement about that one operator.

r is a rank-one spherical tensor

Cartesian components x, y, z mix messily under rotation. Recombine them as r¹₀ = z and r¹_(±1) = ∓(x ± iy)/√2 and the three components transform among themselves exactly as the l = 1 spherical harmonics do — that is what rank one means. Two facts follow immediately. First, the label q on r¹_q is the angular-momentum projection the photon carries along the quantisation axis, so σ⁺ light drives q = +1 and light linearly polarised along z drives q = 0. Second, r is odd under the parity operator: Π r Π† = −r, whereas H_Coulomb commutes with Π, so every hydrogen eigenstate has a definite parity (−1)^l. Rank and parity are independent pieces of information about the same operator, and each one cuts the spectrum differently.

Wigner-Eckart: geometry out, dynamics in one number

For any rank-k tensor, ⟨n′l′m′| T^k_q |n l m⟩ = ⟨l m; k q | l′ m′⟩ ⟨n′l′‖T^k‖n l⟩ / √(2l′+1). The Clebsch-Gordan coefficient depends only on the angular labels and is fixed by rotational symmetry; the reduced element depends on n, n′, l, l′ and nothing else — no m, no m′, no q. Count what that saves for 2p → 1s: three initial substates × one final state × three components of r is nine integrals, and all nine are one number, 1.290 a₀, times coefficients. Three of them are non-zero, each with |⟨1 m; 1 −m|0 0⟩|² = 1/3, and summing |⟨l′m′|C⁽¹⁾_q|lm⟩|² over m, m′ and q recovers the standard angular line strength |⟨l′‖C⁽¹⁾‖l⟩|² = max(l, l′), here equal to 1.

Two independent cuts: the triangle rule and parity

The Clebsch-Gordan factor is non-zero only when |l − 1| ≤ l′ ≤ l + 1 and m′ = m + q, giving Δl = 0, ±1 and Δmₗ = 0, ±1. That is the whole content of "the photon carries one unit of angular momentum", and it does not by itself remove Δl = 0 — though it does kill l = l′ = 0 outright, since coupling 0 to 1 can only return 1. Parity removes the rest, and it acts on the other factor: inserting Π†Π around an odd operator gives ⟨n′l′‖r‖nl⟩ = (−1)^(l+l′+1) ⟨n′l′‖r‖nl⟩, so the reduced element vanishes unless l + l′ is odd. Combine the two and only Δl = ±1 survives. The cuts are genuinely different: 3p → 2p (l = 1 → 1) passes the triangle rule and dies on parity, while 4f → 2s (l = 3 → 0) passes parity and dies on the triangle rule. Spin needs no argument at all — r is the identity on the spin factor, so ⟨mₛ′|mₛ⟩ = δ and Δmₛ = 0.

From rules to the spectrum you actually see

Apply them to hydrogen and the Grotrian diagram draws itself. The Lyman series is np → 1s only: Lyman-α at 121.6 nm is 2p → 1s, and the 2s level, sitting at the same energy, has no line at all. Balmer-α at 656.3 nm is not one transition but three — 3s → 2p, 3p → 2s and 3d → 2p, every one of them Δl = ±1 — while 3s → 2s and 3d → 2s are dead. In the pure Coulomb Hamiltonian all three components coincide, because energy depends on n alone; fine structure spreads them over roughly 0.33 cm⁻¹, which is what a Fabry-Pérot etalon resolves and a school grating does not. Note the limits of the rules: they say allowed or forbidden, never bright or faint. Intensity needs the radial integral, the ω³ factor and the population of the upper level.

What forbidden costs: metastability and its cure

The 2s level has only 1s beneath it, and 2s → 1s is Δl = 0, so no electric-dipole channel exists. The next multipole does not rescue it either: with l = l′ = 0 the transition is 0 → 0, and one photon of any multipolarity carries at least one unit of angular momentum, so M1 and E2 are shut as well. What is left is two-photon E1⋅E1 emission, at about 8.23 s⁻¹ — a lifetime of 0.12 s against 1.6 ns for 2p, a factor of 7.5×10⁷. A different escape route runs the 21 cm line: the 1s hyperfine transition F = 1 → 0 also has Δl = 0, but the magnetic dipole couples to spin and F does change, so M1 is genuinely open — at A = 2.85×10⁻¹⁵ s⁻¹, τ ≈ 11 Myr, visible only because a galaxy supplies enough atoms to make an absurdly slow rate into a bright line. And because the 2s rule rests on a symmetry, breaking the symmetry repeals it: a static field of a few volts per centimetre mixes the near-degenerate 2s and 2p states, the mixed state inherits 2p's dipole amplitude, and the metastability is quenched.

Change one variable at a time

Make the relationship visible.

Interactive model

Initial orbital l1

Final orbital l′ (cursor)2

Set l = 2 and walk l′ from 0 to 6: only l′ = 1 and l′ = 3 raise a bar, and l′ = l + 1 is always the taller because S = max(l, l′). Then set l = 0, where the window collapses to l′ = 1 alone and the open circle disappears — which is why 2s has no dipole route down to 1s.

Δl = l′ − l1

TRIANGLE (1 = pass)1

PARITY (1 = pass)1

STRENGTH S = max(l, l′)2

Live interpretationΔl = l′ − l: 1. TRIANGLE (1 = pass): 1. PARITY (1 = pass): 1. STRENGTH S = max(l, l′): 2

Catch the common trap

Explain before calculating.

Misconception check

The standard mnemonic — "the photon carries one unit of angular momentum, so l must change by one" — reaches the right rule for the wrong reason, and then fails. Spin-1 gives the triangle rule |l − 1| ≤ l′ ≤ l + 1, and for every l ≥ 1 that permits Δl = 0: 3p → 2p is perfectly legal as far as angular momentum is concerned. What kills it is parity, an independent statement about the reduced matrix element, since r is odd and both p states carry parity (−1)¹. A student running on the mnemonic cannot say why electric-quadrupole transitions have Δl = 0, ±2 rather than 0, ±1, ±2. The second half of the error is reading "forbidden" as "impossible". It means only that the leading multipole amplitude vanishes; 2s → 1s still empties, by two photons, at 8.2 s⁻¹ rather than the 6.3×10⁸ s⁻¹ its dipole-allowed neighbour 2p manages.

In hydrogen, 3d → 2p is an allowed electric-dipole line and 3p → 2p is not. Which statement identifies correctly what forbids 3p → 2p?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA hydrogen atom sits in the 3d state. Decide which of 3d → 2p, 3d → 2s and 3d → 1s are electric-dipole allowed, and give the wavelength of the allowed n = 3 → 2 line. Use Eₙ = −13.6 eV/n² and hc = 1240 eV nm.

Read off the orbital labels: 3d has l = 2, 2p has l = 1, and both 2s and 1s have l = 0.
Triangle rule first. With l = 2 and a rank-one operator, only l′ = 1, 2, 3 are reachable, so 3d → 2s and 3d → 1s (both l′ = 0, Δl = −2) are out before parity is consulted. 3d → 2p has l′ = 1 and survives.
Parity next: l + l′ must be odd. For 3d → 2p, 2 + 1 = 3, odd, so the reduced element is not forced to zero. Nothing constrains Δn, so dropping two principal levels is no obstacle.
Photon energy: E₃ − E₂ = −13.6/9 + 13.6/4 = −1.511 + 3.400 = 1.889 eV, so λ = 1240/1.889 = 656.4 nm.

AnswerOnly 3d → 2p is allowed (Δl = −1). 3d → 2s and 3d → 1s are Δl = −2 and fail the triangle rule. The surviving line is Balmer-α, computed at 656.4 nm and observed at 656.3 nm.

MediumHydrogen atoms excited to 2p sit in a field B = 0.50 T along z. Treating the orbital Zeeman shift alone, ΔE = μ_B B mₗ with μ_B = 9.274×10⁻²⁴ J T⁻¹, find how many Lyman-α components appear, the polarisation of each, and their frequency spacing.

The lower state is 1s, so m′ = 0. Wigner-Eckart makes ⟨1s 0| r¹_q |2p m⟩ vanish unless m′ = m + q, hence q = −m. Each of m = +1, 0, −1 therefore has exactly one open channel: three components, not nine.
Identify the polarisations. q = 0 is the z-component of r, radiated as π light — linearly polarised along B and invisible looking straight down the field. q = ±1 are ∓(x ± iy)/√2, the σ components, seen as circular along B and linear across it.
Only the upper level splits, since 1s has m = 0 alone. The shift per unit of mₗ is μ_B B = 9.274×10⁻²⁴ × 0.50 = 4.637×10⁻²⁴ J, which is 2.89×10⁻⁵ eV.
Frequency spacing: Δν = μ_B B / h = 4.637×10⁻²⁴ / 6.626×10⁻³⁴ = 7.00×10⁹ Hz.
In wavelength at λ = 121.6 nm: Δλ = λ² Δν / c = (1.216×10⁻⁷)² × 7.00×10⁹ / 2.998×10⁸ = 3.45×10⁻¹³ m = 0.35 pm.

AnswerThree components — σ (q = +1), π (q = 0) and σ (q = −1) — evenly spaced by 7.00 GHz, or 0.35 pm at 121.6 nm. Spin, untouched by r, adds the real fine structure this idealisation ignores.

HardThe radial integral for 1s ← 2p is ⟨1s|r|2p⟩ = 2⁷√6/3⁵ a₀. Compute the spontaneous emission rate A for 2p → 1s and the 2p lifetime, then say why the 2s level alongside it lives some 10⁸ times longer. Take ΔE = 10.20 eV, a₀ = 5.292×10⁻¹¹ m, ħ = 1.0546×10⁻³⁴ J s, ε₀ = 8.854×10⁻¹² F m⁻¹.

Radial part: 2⁷√6/3⁵ = 128 × 2.4495 / 243 = 1.290, so ⟨1s|r|2p⟩ = 1.290 a₀ = 6.828×10⁻¹¹ m.
Angular part from Wigner-Eckart: for each m, Σ_q |⟨1s 0| r¹_q |2p m⟩|² = ⅓ |⟨1s|r|2p⟩|², since |⟨1 m; 1 −m|0 0⟩|² = ⅓. So |d_fi|² = e² (6.828×10⁻¹¹)²/3 = 2.566×10⁻³⁸ × 1.554×10⁻²¹ = 3.99×10⁻⁵⁹ C² m².
Frequency: ω = ΔE/ħ = (10.20 × 1.602×10⁻¹⁹)/1.0546×10⁻³⁴ = 1.549×10¹⁶ rad s⁻¹, so ω³ = 3.72×10⁴⁸ s⁻³.
Denominator of A: 3π ε₀ ħ c³ = 9.425 × 8.854×10⁻¹² × 1.0546×10⁻³⁴ × 2.695×10²⁵ = 2.371×10⁻¹⁹.
Rate: A = ω³|d_fi|²/(3πε₀ħc³) = (3.72×10⁴⁸ × 3.99×10⁻⁵⁹)/2.371×10⁻¹⁹ = 6.26×10⁸ s⁻¹, and τ = 1/A = 1.60×10⁻⁹ s.
The 2s level has only 1s below it, and 2s → 1s is l = 0 → l′ = 0: no dipole channel, and no single-photon channel of any multipolarity, since the photon must carry at least one unit of angular momentum. It empties by two-photon emission at about 8.23 s⁻¹, τ = 0.12 s.

AnswerA = 6.3×10⁸ s⁻¹ and τ(2p) = 1.6 ns, against τ(2s) = 0.12 s — a factor of 7.5×10⁷ bought by one unit of orbital angular momentum.