University Quantum Mechanics II · Emission and Absorption of Radiation · 13.1
The electric-dipole Hamiltonian and its validity
Every rate in this unit starts from one line, H′ = −qE₀ ε̂⋅r cos ωt, and this is where it is earned. Start from the minimal coupling you already trust, spend the one small number an atom offers, its size over the wavelength, and keep a ledger of what you threw away, so that when a line comes out forbidden you know which term to revisit.
Build the model
Connect the measurement to the mechanism.
Light reaches an atom through minimal coupling. With the field kept classical and the Coulomb gauge chosen, H = p²/2m + V(r) − (q/m)A⋅p + q²A²/2m, and a plane wave carries the factor e(ik⋅r) inside A. The atom confines r to about a₀ = 5.29 × 10⁻¹¹ m, so the phase k⋅r barely moves across it: for visible light k a₀ is below 10⁻³, and for every line of hydrogen k a₀ = ħω/(αmc²) is below α/2.
Expanding e(ik⋅r) = 1 + ik⋅r + … is an expansion in the fine-structure constant, and keeping the 1 makes the field uniform across the atom. A then depends on time alone, a gauge change moves the whole coupling into a scalar potential, and H′ = −qE₀ ε̂⋅r cos ωt = −d⋅E(t): the static Stark coupling, set oscillating. The A⋅p form is not a rival.
The commutator [H₀, r] = −iħp/m makes its resonant matrix element the length form's times ωfi/ω, which is 1 wherever the golden rule looks. What is dropped is O(α) in amplitude and O(α²) in rate: the ik⋅r term and the wave's magnetic force, both restored in 13.08, while A², a multiple of the identity once A is uniform, moves no population. The model stops working where those terms stop being small, for X-rays on outer electrons and in heavy ions where Zα nears 1; where the dipole matrix element vanishes, so the dropped terms become the whole answer; and it never makes an excited atom radiate in the dark, which needs the quantised field, whose result 13.03 quotes.
- Simple definition
- The electric-dipole Hamiltonian H′ = −d⋅E(t), with d = qr, is the coupling of an atom to a light wave whose field is treated as uniform across the atom: the leading term of an expansion in k a, the atom's size times the wave number, which is of order α.
- Example
- For hydrogen in its own Lyman-α light at 121.6 nm, k a₀ = 2π(0.0529 nm)/(121.6 nm) = 2.7 × 10⁻³: the field differs by at most 0.55% between opposite sides of the atom, and the dropped terms change a rate by about (k a₀)² = 7.5 × 10⁻⁶.
Opening Unit 3's (p − qA)²/2m leaves one term linear in the field, which drives transitions, and A², which is second order in E₀.
q in C, m in kg, p in kg m s⁻¹, A in V s m⁻¹; ∇⋅A = 0 makes p⋅A = A⋅p. Every term is in J.
Hydrogen's lines have ħω below α²mc²/2, so k a₀ is below α/2: 3α/8 = 2.74 × 10⁻³ for Lyman-α, as the measured 121.6 nm line gives.
k in m⁻¹, a₀ = 5.29 × 10⁻¹¹ m, ħω in J; dimensionless. At 500 nm, k a₀ = 6.6 × 10⁻⁴.
With e(ik⋅r) set to 1, the gauge function χ = −r⋅A(t) removes A and leaves the potential −E(t)⋅r: the static Stark coupling, oscillating.
E₀ in V m⁻¹, r in m, ε̂ the unit polarisation, H′ in J. For an electron q = −e, so H′ = eE₀ ε̂⋅r cos ωt.
Sandwich [H₀, r] = −iħp/m between ⟨f| and |i⟩: a derivative integral becomes a position integral, and the velocity form reduces to the length form.
ωfi = (Ef − Eᵢ)/ħ in s⁻¹; kg × s⁻¹ × m = kg m s⁻¹. |i⟩ and |f⟩ must be eigenstates of H₀ = p²/2m + V(r).
Exactly 1 on resonance, where the golden rule's delta function puts every rate. Off resonance the full amplitudes differ only by a gauge term that vanishes once A = 0.
H′ᵥ = −(q/m)A(t)⋅p and H′ₗ = −qE(t)⋅r, in J; res marks the e(−iωt) part, which drives absorption.
The same order as the ik⋅r term: both change amplitudes by O(α) and rates by O(α²), and 13.08 restores them as magnetic-dipole and quadrupole lines.
B₀ = E₀/c for a plane wave, c = 3.00 × 10⁸ m s⁻¹; a hydrogen electron has v ≈ αc = 2.19 × 10⁶ m s⁻¹.
Minimal coupling, with the light kept classical
The starting point is quoted from Unit 3 (3.06): a charge in potentials has H = (p − qA)²/2m + qφ + V(r). Describe the wave in the Coulomb gauge, φ = 0 and A = (E₀/ω) ε̂ sin(k⋅r − ωt), which gives E = −∂A/∂t = E₀ ε̂ cos(k⋅r − ωt) and a magnetic field of size E₀/c. Because ε̂⋅k = 0, ∇⋅A = 0, so p⋅A = A⋅p acting on any ψ, and the square opens into H₀ − (q/m)A⋅p + q²A²/2m (derived). The units close: (q/m)Ap is C kg⁻¹ × V s m⁻¹ × kg m s⁻¹ = C V = J. The field is a prescribed function of time, not an operator. That is the semiclassical model of the whole unit, and it is why the spontaneous rate must later be imported rather than derived.
The one small number: atomic size over wavelength
Across the atom the phase k⋅r changes by about k a, where a is the extent of the wavefunctions. For hydrogen a₀ = 0.0529 nm, and 500 nm light has a₀/λ = 1.06 × 10⁻⁴, k a₀ = 6.6 × 10⁻⁴. The size is set by α. Since ħc/a₀ = αmc² = 3.73 keV, k a₀ = ħω/(αmc²), and a photon between hydrogen's bound levels carries less than α²mc²/2 = 13.6 eV, so k a₀ is below α/2 = 3.6 × 10⁻³. For Lyman-α, ħω = (3/8)α²mc² gives k a₀ = 3α/8 = 2.737 × 10⁻³, against 2πa₀/(121.567 nm) = 2.735 × 10⁻³ from the measured line; the 0.05% gap is the reduced mass. So e(ik⋅r) = 1 + ik⋅r − (k⋅r)²/2 + … is an expansion in α, and keeping the 1 is its leading order. The same algebra says where it ends: k a₀ = 1 at λ = 2πa₀ = 0.33 nm, a 3.73 keV X-ray.
A uniform field is a scalar potential
Keep only the 1, with the nucleus as origin. Then A(t) = −(E₀/ω) ε̂ sin ωt is the same at every point of the atom and B = ∇ × A = 0: the wave's magnetic field is gone at this order. A vector potential that depends only on t is pure gauge. The gauge change of Unit 3 (3.07) with χ = −r⋅A(t) gives A′ = A + ∇χ = 0 and φ′ = −∂χ/∂t = r⋅∂A/∂t = −r⋅E(t), and multiplies ψ by e(−iqr⋅A(t)/ħ). The whole coupling is now potential energy, H′ = −qE(t)⋅r = −d⋅E(t) with d = qr, and for E(t) = E₀ ε̂ cos ωt it is H′ = −qE₀ ε̂⋅r cos ωt, in joules (derived). As ω → 0 it becomes Unit 9's Stark Hamiltonian. A state of definite parity has ⟨n|d|n⟩ = 0, so H′ acts through the transition dipole ⟨f|d|i⟩ between states of opposite parity, which does not depend on the choice of origin because ⟨f|i⟩ = 0.
Keep A⋅p instead: the commutator makes them agree
Skip the gauge change and the dipole-limit coupling is H′ᵥ = −(q/m)A(t)⋅p = (qE₀/mω) sin ωt ε̂⋅p, plus q²A(t)²/2m, which now multiplies the identity: zero between different states, and the same phase on every state. Since V(r) commutes with r, [H₀, r] = [p²/2m, r] = −iħp/m, and between eigenstates (Ef − Eᵢ)⟨f|r|i⟩ = −(iħ/m)⟨f|p|i⟩, so ⟨f|p|i⟩ = imωfi⟨f|r|i⟩ (derived). The part of ⟨f|H′ᵥ|i⟩ that varies as e(−iωt), the one that drives absorption, is then −½qE₀(ωfi/ω)⟨f|ε̂⋅r|i⟩: the length form's −½qE₀⟨f|ε̂⋅r|i⟩ times ωfi/ω, which is 1 on resonance. Hydrogen checks it. ∂ψ₁ₛ/∂z = −(cos θ/a₀)ψ₁ₛ gives ⟨2p₀|pz|1s⟩ = (iħ/a₀)⟨2p₀|cos θ|1s⟩ = 0.279iħ/a₀, while imωfi times the 0.745a₀ that 13.04 computes is i(3/8)(0.745)ħ/a₀ = 0.279iħ/a₀.
Off resonance the forms differ by a gauge term
Off resonance the factor ωfi/ω looks like a contradiction. It is not. Write Unit 12's first-order amplitude in the velocity form, cf = (1/iħ)∫₀ᵗ(−iqωfi)A(t′)⋅rfi e(iωfi t′)dt′ with rfi = ⟨f|r|i⟩, notice that iωfi e(iωfi t′) is the time derivative of e(iωfi t′), and integrate by parts. Since −dA/dt′ = E, the integral becomes the length-form amplitude exactly, plus a boundary term (iq/ħ)A(t)⋅rfi e(iωfi t) when A(0) = 0 (derived). That term is the first-order piece of the gauge phase: the length gauge multiplied ψ by e(−iqr⋅A/ħ), so the velocity-gauge state is e(iqr⋅A/ħ) ≈ 1 + iqr⋅A/ħ times the length-gauge one, and while A ≠ 0 the amplitude in |f⟩ means different things in the two gauges. It stays bounded, of size qE₀|rfi|/ħω, while a resonant amplitude grows in proportion to t, and it vanishes for any pulse that ends with A = 0. So probabilities measured after the light is off agree at every detuning.
What was dropped, and where the model fails
The wave's magnetic field, acting on the orbit and on the spin, is smaller than the electric force by v/c, and a bound hydrogen electron moves at about αc = 2.19 × 10⁶ m s⁻¹. The field's change across the atom, the ik⋅r term, is k a, at most of order α for the atom's own lines. The two are the same order: the ik⋅r term splits into the magnetic-dipole and electric-quadrupole couplings that 13.08 restores. Each changes an amplitude by O(α) and a rate by O(α²), about 10⁻⁵, below the precision of most measured lifetimes. The model fails where these stop being small: a 0.1 nm X-ray on hydrogen has k a₀ = 3.3, and in a hydrogen-like ion k a = (3/8)Zα for Lyman-α, 0.25 in U⁹¹⁺. It fails where ⟨f|ε̂⋅r|i⟩ = 0, so the dropped terms, or two dipole photons, carry the whole rate (13.05, 13.08). And with the field classical, an excited atom in the dark feels no H′: spontaneous emission is quoted from quantum electrodynamics, or reached by Einstein's detailed balance in 13.03.
Change one variable at a time
Make the relationship visible.
Start in the visible, where the wave lies on the dashed line, then drag λ down: the k a₀ bar passes α at 45.6 nm, the hartree, and reaches 1 at 0.33 nm, where the field reverses inside the atom. Sweep ωt to see that the dipole form keeps the oscillation in time.
WAVELENGTH λ501.19 nm
PHOTON ENERGY ħω2.5 eV
k a₀ IN UNITS OF 10⁻³0.663
log₁₀ (k a₀)², SHARE OF A RATE-6.4
Live interpretationWAVELENGTH λ: 501.19 nm. PHOTON ENERGY ħω: 2.5 eV. k a₀ IN UNITS OF 10⁻³: 0.663. log₁₀ (k a₀)², SHARE OF A RATE: −6.4
Catch the common trap
Explain before calculating.
In the long-wavelength limit the velocity-form Hamiltonian keeps q²A(t)²/2m. For an electron in an 800 nm pulse at 10¹⁴ W cm⁻² its cycle average is about 6.0 eV, comparable with hydrogen's 10.2 eV spacing between n = 1 and n = 2. Why does it still drive no transitions between atomic states?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasySodium's yellow line, 589.0 nm, comes from the 3p → 3s transition, whose wavefunctions extend to about a = 0.30 nm from the nucleus. Find k a, the largest fractional change of the field across the atom's diameter, and the rough size of the dropped terms in a rate. Repeat for an 8.05 keV Cu Kα X-ray, λ = 0.154 nm, meeting the same outer electron, and decide in each case whether H′ = −qE₀ ε̂⋅r cos ωt can be used.
- Sodium line: k = 2π/λ = 2π/(589.0 × 10⁻⁹ m) = 1.067 × 10⁷ m⁻¹, so k a = (1.067 × 10⁷ m⁻¹)(0.30 × 10⁻⁹ m) = 3.2 × 10⁻³, dimensionless as it must be.
- Across the diameter the phase kx runs from −k a to +k a, and cos(kx − ωt) differs between the two sides by 2 sin(k a) sin ωt, at most 2k a = 6.4 × 10⁻³: the field is uniform to 0.64%.
- The first dropped term, ik⋅r, has the opposite parity to the dipole term, so for a given pair of states they do not interfere, and the leading change to a dipole rate is of order (k a)² = 1.0 × 10⁻⁵. The dipole form is excellent, as k a of order α led us to expect.
- X-ray: k = 2π/(0.154 × 10⁻⁹ m) = 4.08 × 10¹⁰ m⁻¹ and k a = 12. The 0.60 nm diameter holds 0.60/0.154 = 3.9 wavelengths, so the field reverses several times inside the atom and e(ik⋅r) cannot be replaced by 1.
AnswerSodium D line: k a = 3.2 × 10⁻³, field uniform to 0.64%, rate error about 10⁻⁵, so the dipole form holds. Cu Kα on the outer electron: k a ≈ 12, and it fails.
MediumAn electron in a well of width L = 1.20 nm with infinitely high walls, ψₙ = √(2/L) sin(nπx/L), is driven on its 1 → 2 transition by light polarised along x. Find ħω₂₁, the resonant wavelength and kL; evaluate ⟨2|x|1⟩ and ⟨2|p|1⟩ by direct integration and confirm ⟨2|p|1⟩ = imω₂₁⟨2|x|1⟩; then give the ratio of the velocity-form to the length-form resonant amplitude if the light is tuned 3.0% above resonance.
- E₁ = h²/(8mL²) = (6.626 × 10⁻³⁴ J s)²/(8 × 9.109 × 10⁻³¹ kg × (1.20 × 10⁻⁹ m)²) = 4.18 × 10⁻²⁰ J = 0.261 eV. So ħω₂₁ = 3E₁ = 0.783 eV, ω₂₁ = 1.19 × 10¹⁵ s⁻¹, λ = (1239.8 eV nm)/(0.7834 eV) = 1583 nm, and kL = 2π(1.20)/1583 = 4.8 × 10⁻³: the dipole form is safe.
- ⟨2|x|1⟩ = (2/L)∫₀ᴸ x sin(2πx/L) sin(πx/L) dx = −16L/(9π²) = −0.216 nm. Measuring x from the centre instead gives the same number, because the shift adds (L/2)⟨2|1⟩ = 0.
- ⟨2|p|1⟩ = −iħ(2/L)(π/L)∫₀ᴸ sin(2πx/L) cos(πx/L) dx. With u = πx/L the integral is (L/π)∫₀π sin 2u cos u du = (L/π)(4/3), so ⟨2|p|1⟩ = −8iħ/(3L) = −2.34 × 10⁻²⁵ i kg m s⁻¹.
- Check: imω₂₁⟨2|x|1⟩ = i(9.109 × 10⁻³¹ kg)(1.190 × 10¹⁵ s⁻¹)(−2.162 × 10⁻¹⁰ m) = −2.34 × 10⁻²⁵ i kg m s⁻¹, the same. Symbolically m(3π²ħ/2mL²)(16L/9π²) = 8ħ/3L exactly, with units kg s⁻¹ m = kg m s⁻¹.
- Tuned 3.0% high, ω = 1.030ω₂₁, and the resonant parts differ by ω₂₁/ω = 1/1.030 = 0.971: the velocity form's resonant amplitude is 2.9% smaller, its probability 5.7% smaller. Keep the antiresonant parts too and the full amplitudes differ only by the gauge term (iq/ħ)A⋅rfi e(iωfi t), which is zero once the light is off.
Answerħω₂₁ = 0.783 eV (λ = 1583 nm, kL = 4.8 × 10⁻³); ⟨2|x|1⟩ = −0.216 nm and ⟨2|p|1⟩ = −2.34 × 10⁻²⁵ i kg m s⁻¹ = imω₂₁⟨2|x|1⟩; 3.0% above resonance the resonant-part ratio is 0.971.
HardA hydrogen-like ion of nuclear charge Z emits Lyman-α (2p → 1s). Scale hydrogen's nonrelativistic results: orbit size a₀/Z, energies Z² times hydrogen's. Show that k a = (3/8)Zα; estimate the Z at which the dropped terms' share of a rate, about (k a)², reaches 1%; then evaluate k a, (k a)² and the orbital v/c for U⁹¹⁺, and compare the nonrelativistic photon energy with the Lyman-α lines measured near 97.6 and 102.2 keV.
- With a = a₀/Z and ħω = (3/4)Z²(13.606 eV) = (3/8)Z²α²mc², k a = ħω a/(ħc) = (3/8)Z²α²mc² a₀/(Zħc). Since a₀ = ħ/(αmc), a₀/(ħc) = 1/(αmc²), so k a = (3/8)Zα. It is dimensionless, and at Z = 1 it gives 2.74 × 10⁻³, what the measured 121.6 nm line gives.
- (k a)² = 0.010 needs k a = 0.10, so Z = 0.10 × 8/(3α) = 0.10 × 8 × 137.04/3 = 36.5. From about Z = 37, hydrogen-like rubidium, the dropped terms are a percent-level correction; for hydrogen they are (2.74 × 10⁻³)² = 7.5 × 10⁻⁶.
- U⁹¹⁺, Z = 92: k a = 0.375 × 92/137.04 = 0.252 and (k a)² = 0.063, a 6% effect on rates, which must now be computed with the full e(ik⋅r).
- The magnetic term is not small either: the 1s electron moves at v/c ≈ Zα = 92/137.04 = 0.67, so the wave's magnetic force on it is up to two-thirds of the electric force.
- The same number condemns the nonrelativistic H₀. Its Lyman-α energy, (3/4)(92²)(13.606 eV) = 86.4 keV, falls 12% and 15% short of the measured 97.6 and 102.2 keV fine-structure pair. Corrections of order (Zα)² = 0.45 are no longer small; Dirac's equation, quoted, comes within half a percent, and the finite size of the nucleus and quantum electrodynamics, both quoted, supply the rest.
Answerk a = (3/8)Zα; (k a)² reaches 1% near Z = 37. For U⁹¹⁺, k a = 0.25, (k a)² = 0.063 and v/c ≈ 0.67: the dipole form and the nonrelativistic Hamiltonian fail together, 86.4 keV against 97.6 and 102.2 keV measured.