University Physics IV · The Hydrogen Atom · 10.9
Dipole Selection Rules & Hydrogen Spectra
Hydrogen has bound states everywhere, and its spectrum still shows only certain lines. This is the rule that decides which pairs of states a single photon can join — where it comes from, why it says nothing whatever about Δn, and what becomes of the states it strands.
Build the model
Connect the measurement to the mechanism.
Nothing in the Coulomb solution forbids any pair of hydrogen levels from being joined; the forbidding comes from the photon. Treat the light wave as a small time-dependent perturbation and Fermi's golden rule makes the rate proportional to |⟨f| e(ik⋅r) ε⋅p |i⟩|². Across an atom the phase in that exponential is tiny — Lyman-α is 121.6 nm long against a Bohr radius of 0.0529 nm, so k a₀ = 2.7 × 10⁻³ — and setting the exponential to 1 leaves the electric-dipole operator d = −e r.
Everything follows from that one vector. It is odd under parity, so the two states must have opposite parity, which in hydrogen means lf + lᵢ odd; and it is a rank-one spherical tensor built from Y₁q, so the angular integral of three spherical harmonics survives only for |Δl| ≤ 1 and Δm = q ∈ (0, ±1). Parity removes Δl = 0, leaving Δl = ±1 exactly — the one unit of angular momentum the photon carries off, paid out of the electron's orbit.
The radial integral enforces nothing at all, so Δn is free. The cost is that these are the rules of a leading term only: everything they forbid still happens, through the terms the expansion dropped, at rates smaller by roughly (k a₀)² ≈ 10⁻⁵ per order — which is why 2s survives 0.12 s where 2p lasts 1.60 ns.
- Simple definition
- An electric-dipole selection rule is a condition on two states' quantum numbers that must hold for the matrix element ⟨f|r|i⟩ to be non-zero, and so for one photon to connect them at leading order in the atom's size over the wavelength.
- Example
- 3p → 1s has Δl = −1, so it is allowed: it radiates Lyman-β at 102.6 nm with A = 1.67 × 10⁸ s⁻¹. 3s → 1s has Δl = 0, so the same matrix element is exactly zero and that photon is never emitted.
Turns an integral into a stopwatch: for 2p → 1s it gives 6.27 × 10⁸ s⁻¹, a lifetime of 1.60 ns.
ω in rad s⁻¹, the matrix element in C m, A in s⁻¹; the level's lifetime is τ = 1/ΣAfi
The atom sits in a uniform field. Each further multipole order costs a factor (k a₀)² ≈ 10⁻⁵.
λ = 121.6 nm against a₀ = 0.0529 nm for Lyman-α, so k a₀ = ω a₀/c = 2.7 × 10⁻³
Kills every Δl = 0 transition — s → s, p → p, d → d — with no integral to evaluate.
r is odd under r → −r, so the integrand must be even overall for the integral to survive
With parity already excluding lf = lᵢ, this leaves Δl = ±1 and Δm = 0, ±1 exactly.
q = 0 for z (π light), q = ±1 for x ± iy (σ± light); r is a rank-one tensor built from Y₁q
8p → 1s is as allowed as 2p → 1s. Only the radial overlap, never a rule, makes the high lines weak.
In the coupled basis: Δj = 0, ±1, with j = 0 → j = 0 forbidden because the photon still carries one unit
Carries no selection rule; it is why Lyman rates fall 1 : 0.27 : 0.11 across 2p, 3p and 4p → 1s.
Units of length; the r² Jacobian is explicit in the R form. ⟨1s|r|2p⟩ = (128√6/243) a₀ = 1.290 a₀
One small number licenses the dipole approximation
Light is a travelling wave, so the perturbation the atom feels is E₀ ε cos(k⋅r − ωt), and Fermi's golden rule makes the transition rate proportional to |⟨f| e(ik⋅r) ε⋅p |i⟩|². The exponential is the awkward factor, and one number disposes of it. For Lyman-α the wavelength is 121.6 nm while the electron cloud is about a₀ = 0.0529 nm across, so the phase changes by only k a₀ = ω a₀/c = 2.7 × 10⁻³ from one side of the atom to the other. More generally, a transition of order one Rydberg gives k a₀ ≈ α/2 = 3.6 × 10⁻³. Set e(ik⋅r) = 1 and the field is uniform over the whole atom; what survives, after the standard commutator step that converts ε⋅p into r, is the electric-dipole operator d = −e r. Every rule in this topic is a statement about that single vector. The price is stated in the same breath: the next term in the expansion, i k⋅r, is smaller by 10⁻³, and the rates it produces are smaller by (k a₀)² ≈ 10⁻⁵. Nothing is being forbidden here — it is being postponed.
Parity does half the work before any integral
Under the parity operation r → −r a hydrogen eigenfunction picks up a sign fixed entirely by l: P ψₙₗₘ = (−1)l ψₙₗₘ, because the radial factor is untouched and Yₗₘ(π − θ, φ + π) = (−1)l Yₗₘ(θ, φ). The dipole operator is itself odd, since r → −r by construction. So the integrand of ⟨f|r|i⟩ carries the overall sign (−1)(l_f) × (−1) × (−1)(lᵢ), and the integral of an odd function over all space is zero. The matrix element can survive only if lf + lᵢ is odd, that is, only if Δl is odd. That one line — Laporte's rule — kills every Δl = 0 transition at once: s → s, p → p, d → d, and with them 2s → 1s. It has done half the work of the topic without evaluating a single integral, and it costs nothing beyond noticing that a central potential is invariant under inversion, so parity is a good quantum number in the first place.
The angular integral gives Δl = ±1 and Δm = 0, ±1
Parity says Δl is odd; it does not say Δl = ±1. For that, write the components of r in the same language as the wavefunctions: z = r √(4π/3) Y₁₀, and x ± iy = ∓ r √(8π/3) Y_(1±1). Each component is r times a Y₁q with q = 0 or ±1, so the angular part of ⟨f|r|i⟩ is ∫ Y*_(lf mf) Y₁q Y_(lᵢ mᵢ) dΩ — an integral of three spherical harmonics. It vanishes unless lf, lᵢ and 1 satisfy the triangle inequality |lᵢ − 1| ≤ lf ≤ lᵢ + 1, and unless mf = mᵢ + q. The triangle rule permits lf = lᵢ − 1, lᵢ or lᵢ + 1; parity has already removed the middle case. What is left is Δl = ±1 and Δm = 0, ±1, exactly. Read physically, the photon is a spin-one object of odd parity, and the atom must hand over precisely the one unit of angular momentum it carries away. The value of q is the polarisation — q = 0 is light linearly polarised along z, q = ±1 is circular — which is why the components of a Zeeman-split line are polarised differently.
Nothing restricts Δn; the radial integral only sets strength
The radial factor ∫₀^∞ R_(nf lf)(r) r R_(nᵢ lᵢ)(r) r² dr is where every quantum number except l has been sent, and it enforces nothing. It is tempting to hope orthogonality will kill some of them, but radial functions are orthogonal only within the same l, and Δl = ±1 guarantees the two belong to different l — different radial equations, no orthogonality to invoke. The integral is therefore generically non-zero for any Δn, including Δn = 0: in hydrogen 3p → 3s radiates nothing only because the two are degenerate, while in sodium, where screening lifts that degeneracy, the very same Δn = 0, Δl = −1 transition is the yellow D lines at 589 nm. What the radial integral does control is strength. It shrinks as the two radial extents diverge, going as nᵢ(−3/2) once nᵢ is large, and A goes as ω³ times its square. Across the Lyman series the rates run 6.27 × 10⁸, 1.67 × 10⁸ and 6.82 × 10⁷ s⁻¹ for 2p, 3p and 4p → 1s — a ratio of 1 : 0.27 : 0.11. Weak and forbidden are different words, and only one of them is a selection rule.
Reading the rules off hydrogen's observed series
Point the rules at hydrogen and the observed series fall out. The Lyman series is np → 1s and nothing else, because 1s has l = 0 and only l = 1 is one step away; a 3d atom therefore cannot reach the ground state directly and must cascade 3d → 2p → 1s. The Balmer series is richer, because n = 2 offers both parities: ns → 2p, nd → 2p and np → 2s are all allowed. So Balmer-α, the red line at 656.3 nm, is not one transition but a blend of 3s → 2p, 3p → 2s and 3d → 2p, degenerate in the pure Coulomb solution and separated only at the 0.01 nm level by fine structure. Now switch on a magnetic field and the Δm rule becomes visible. Ignore spin for a moment: the emitted photon energy shifts by (mᵢ − mf) μB B, and Δm is restricted to 0 and ±1, so however many sublevels the two levels carry, the line splits into exactly three — the normal Zeeman pattern. Balmer-α's fifteen separate dipole channels produce only three frequencies.
Past the dipole term: the metastable 2s state
The 2s state is the rules' own test case. It cannot reach 1s by an electric-dipole photon, since both states have l = 0. Electric quadrupole is closed too: that operator goes as Y₂q, and its angular integral between two spherically symmetric s states vanishes. Magnetic dipole is nearly closed, because the operator L + 2S does not touch the radial functions and ⟨1s|2s⟩ = 0; only a relativistic correction cracks it open, at about 10⁻⁶ s⁻¹. What remains is two photons emitted at once, sharing the 10.2 eV in any proportion, at a rate of 8.23 s⁻¹ — a lifetime of 0.121 s against 1.60 ns for 2p, a factor of 7.6 × 10⁷. That factor is what a selection rule is worth. It is also observable twice over: 2s hydrogen in a nebula lives long enough to radiate a smooth two-photon continuum instead of a line, and a stray electric field that mixes a little 2p into 2s quenches the state at once.
Change one variable at a time
Make the relationship visible.
Start at 3d with the final shell at n = 2: one arrow lands on 2p, the Balmer-α transition. Drop the final shell to n = 1 and both arrows vanish, because n = 1 has no p state for 3d to reach. Then set nᵢ = 2, lᵢ = 0, nf = 1 — the metastable 2s, with no dipole route at all.
ALLOWED DIPOLE CHANNELS1
PHOTON ENERGY1.89 eV
VACUUM WAVELENGTH656.5 nm
Δn1
Live interpretationALLOWED DIPOLE CHANNELS: 1. PHOTON ENERGY: 1.89 eV. VACUUM WAVELENGTH: 656.5 nm. Δn: 1
Catch the common trap
Explain before calculating.
Hydrogen's 2p state decays to 1s in 1.60 ns, but the 2s state, which sits at essentially the same energy, survives 0.12 s — longer by a factor of 7.6 × 10⁷. What accounts for the difference?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyDecide which of 4d → 2p, 3p → 1s and 2s → 1s can occur by emitting a single electric-dipole photon, then give the wavelength of the allowed one that ends on the ground state. Take the hydrogen Rydberg energy as 13.60 eV and hc = 1240 eV nm.
- Read l off the letters: s is l = 0, p is l = 1, d is l = 2. The rule to apply is Δl = ±1; Δn is not restricted at all.
- 4d → 2p: Δl = 1 − 2 = −1. Allowed. That Δn = −2 is irrelevant — no rule mentions n.
- 3p → 1s: Δl = 0 − 1 = −1. Allowed. This is Lyman-β.
- 2s → 1s: Δl = 0 − 0 = 0. Forbidden. The photon must carry away one unit of angular momentum, and neither state has any orbital angular momentum to give it.
- Wavelength of 3p → 1s: ΔE = 13.60 × (1/1² − 1/3²) = 13.60 × 8/9 = 12.09 eV, so λ = 1240/12.09 = 102.6 nm.
Answer4d → 2p and 3p → 1s are allowed; 2s → 1s is dipole-forbidden. Lyman-β sits at 102.6 nm, in the vacuum ultraviolet.
MediumIn the n = 3 shell an atom can sit in 3s, 3p or 3d, and in n = 2 in 2s or 2p. Which of the six level pairs are electric-dipole allowed, and how many distinct (l, m) → (l′, m′) channels does the Balmer-α line at 656.3 nm actually contain? Ignore spin.
- Apply Δl = ±1 to the six pairs. Allowed: 3s → 2p (Δl = +1), 3p → 2s (Δl = −1), 3d → 2p (Δl = −1). Blocked: 3s → 2s and 3p → 2p by Δl = 0, and 3d → 2s by Δl = −2.
- Three of the six survive. Now count sublevels: 3s has 1 value of m, 3p has 3, 3d has 5; 2s has 1 and 2p has 3.
- 3s → 2p: the single m = 0 reaches m′ = 0 and ±1, all inside Δm = 0, ±1. That is 3 channels.
- 3p → 2s: each of m = 0, ±1 reaches the single m′ = 0, and |Δm| ≤ 1 in every case. Another 3 channels.
- 3d → 2p: pair each m in (−2, −1, 0, 1, 2) with m′ in (−1, 0, 1) keeping |m − m′| ≤ 1, giving 1 + 2 + 3 + 2 + 1 = 9 channels.
- Total 3 + 3 + 9 = 15. With no field they are degenerate and blend into one line; in a magnetic field the shift depends only on Δm, so those 15 channels collapse onto just three frequencies.
AnswerThree of the six level pairs are allowed — 3s → 2p, 3p → 2s and 3d → 2p — carrying 15 dipole channels in all, which a magnetic field resolves into only three frequencies.
HardThe 1s–2p radial integral is ∫₀^∞ R₁₀ r R₂₁ r² dr = (128√6/243) a₀ = 1.290 a₀, and for the m = 0 sublevel |⟨1s|r|2p⟩|² is one third of its square. Use A = ω³ e² |⟨r⟩|² / (3π ε₀ ℏ c³) to find the 2p decay rate and lifetime. Take a₀ = 5.29 × 10⁻¹¹ m, ΔE = 10.20 eV, e = 1.602 × 10⁻¹⁹ C, ℏ = 1.055 × 10⁻³⁴ J s, ε₀ = 8.854 × 10⁻¹² F m⁻¹ and c³ = 2.694 × 10²⁵ m³ s⁻³.
- Matrix element: 1.290 × 5.29 × 10⁻¹¹ = 6.82 × 10⁻¹¹ m, whose square is 4.66 × 10⁻²¹ m², and one third of that is 1.55 × 10⁻²¹ m².
- Angular frequency: ω = ΔE/ℏ = (10.20 × 1.602 × 10⁻¹⁹)/(1.055 × 10⁻³⁴) = 1.634 × 10⁻¹⁸/1.055 × 10⁻³⁴ = 1.549 × 10¹⁶ rad s⁻¹, so ω³ = 3.72 × 10⁴⁸ s⁻³.
- Numerator: 3.72 × 10⁴⁸ × (1.602 × 10⁻¹⁹)² × 1.55 × 10⁻²¹ = 3.72 × 10⁴⁸ × 2.567 × 10⁻³⁸ × 1.55 × 10⁻²¹ = 1.48 × 10⁻¹⁰.
- Denominator: 3π ε₀ ℏ c³ = 9.425 × 8.854 × 10⁻¹² × 1.055 × 10⁻³⁴ × 2.694 × 10²⁵ = 2.37 × 10⁻¹⁹.
- A = 1.48 × 10⁻¹⁰ / 2.37 × 10⁻¹⁹ = 6.2 × 10⁸ s⁻¹, and τ = 1/A = 1.6 × 10⁻⁹ s.
- Set that beside 2s, whose dipole rate is exactly zero: it goes by two photons at 8.23 s⁻¹, a lifetime of 0.121 s. The selection rule, not the energy, is the whole of the difference.
AnswerA(2p → 1s) ≈ 6.2 × 10⁸ s⁻¹ and τ ≈ 1.6 ns, against tabulated values of 6.265 × 10⁸ s⁻¹ and 1.596 ns. The 2s state, blocked by Δl = 0, lives 0.121 s instead.