University Physics V · Atomic Spectroscopy · 13.2

Atom–Field Coupling & the Dipole Matrix Element

Light does not push an atom; the operator you get by putting the vector potential inside the momentum does. Build that operator from minimal coupling, throw away everything the atom is too small to feel, and one number is left — ⟨f|d⋅ε|i⟩ — that fixes every rate in the spectrum.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

An atom in a light field is not a new problem; it is the old Hamiltonian with p replaced by p + eA. Expand that square in the Coulomb gauge and all of atom–light physics falls out of two extra terms: a cross term (e/m)A⋅p linear in the field, and an A² term quadratic in it. Because an optical wavelength is some thousand times an atomic diameter, exp(ik⋅r) equals 1 across the atom to within k a₀ ≈ 3 × 10⁻³, and once it is set to one the A² term is r-independent — proportional to the identity on the atomic Hilbert space, so it shifts a phase and moves no population.

That leaves (e/m)A⋅p as the only operator connecting two atomic states. A Göppert-Mayer gauge transformation, generated by exp(−i q r⋅A/ħ), rewrites the velocity form as −d⋅E, and the identity ⟨f|p|i⟩ = i m ω_fi ⟨f|r|i⟩ makes the two agree whenever |i⟩ and |f⟩ are exact eigenkets of H₀. First-order perturbation theory with the rotating-wave approximation then supplies everything: a Rabi frequency Ω = ⟨f|d⋅ε|i⟩E₀/ħ, a resonance narrowing as 2π/t, and a rate proportional to |⟨f|d|i⟩|².

The cost is spontaneous emission. A classical field is a c-number: switch the drive off and the interaction is identically zero, so an excited eigenket lives forever. Hydrogen 2p's 1.60 ns lifetime has to be imported, not derived.

Simple definition: The electric-dipole matrix element ⟨f|d⋅ε|i⟩ is the single complex number, with units of charge times length, through which the interaction −d⋅E lets light polarised along ε carry an atom from |i⟩ to |f⟩.
Example: For hydrogen 1s → 2p(m = 0) with ε along z, ⟨2p₀|z|1s⟩ = 0.7449 a₀, so d_fi = 6.32 × 10⁻³⁰ C m (1.89 D); a field of 1.00 × 10⁵ V m⁻¹ then flops the population at Ω = 5.99 × 10⁹ rad s⁻¹.

Minimal-coupling interactionH = H₀ + (e/m)A⋅p + e²A²/2m

Names the operator. Every rate in this unit is one of its matrix elements between H₀ eigenkets.

Coulomb gauge ∇⋅A = 0, electron charge −e; A in V s m⁻¹, p in kg m s⁻¹, H in J

Dipole-approximation parameterexp(i k⋅r) = 1 + i k⋅r − ½(k⋅r)² + … , k a = 2π a/λ

Keeping only the 1 is E1; the next term carries M1 and E2, down by (k a)² ≈ 7 × 10⁻⁶ in rate.

k = 2π/λ in m⁻¹, a the orbital radius in m; Lyman-α gives k a₀ = 2.7 × 10⁻³

Göppert-Mayer gauge transformation|ψ′⟩ = exp(−i q r⋅A/ħ)|ψ⟩ ⟹ H₁ = −d⋅E, with d = q r

Trades A⋅p for a length-gauge operator whose parity and rank-1 tensor character are visible.

q = −e for an electron, so d = −e r; d in C m, E in V m⁻¹, product in J

Velocity–length identity⟨f|p|i⟩ = i m ω_fi ⟨f|r|i⟩, ω_fi = (E_f − Eᵢ)/ħ

Converts either gauge's matrix element into the other's, so a gap between them is basis error.

Follows from [H₀, r] = −iħp/m; exact only between exact eigenkets of H₀

Rabi frequency and first-order amplitudeΩ = ⟨f|d⋅ε|i⟩E₀/ħ, |c_f(t)|² = Ω² sin²(δt/2)/δ², δ = ω_fi − ω

On resonance it grows as Ω²t²/4, and first order expires the moment that approaches 1.

E₀ the peak field in V m⁻¹; Ω, δ, ω in rad s⁻¹ and t in s, so |c_f|² is a probability

Golden-rule rate in a broadband fieldW = π|⟨f|d|i⟩|² ρ(ω_fi)/(3ε₀ħ²) = B ρ(ω_fi)

Turns the t² growth into a constant rate once the bandwidth destroys coherence, giving B.

ρ the spectral energy density in J m⁻³ (rad s⁻¹)⁻¹; the 3 is the isotropic polarisation average

Put the field inside the momentum, not into a force

Start from the gauge-invariant substitution p → p − qA, which for an electron of charge q = −e reads p → p + eA. The atom's Hamiltonian becomes H = (p + eA)²/2m + V(r). Choose the Coulomb gauge, ∇⋅A = 0: then A commutes with p, the two cross terms are equal, and the scalar potential of a free radiation field vanishes. Expanding gives H = H₀ + (e/m)A⋅p + e²A²/2m, with H₀ = p²/2m + V(r) the bare atomic Hamiltonian whose eigenkets |n l m⟩ are the basis you will work in. Nothing has been approximated yet. The whole of what follows is the business of deciding which of those two added operators can move an electron from one |n l m⟩ to another, and how big its matrix element is.

The dipole approximation expands in ka — and it kills A²

A plane wave carries the factor exp(ik⋅r) across the atom. Compare the two lengths: at hydrogen's Lyman-α line, λ = 121.6 nm, so k = 2π/λ = 5.17 × 10⁷ m⁻¹, and with a₀ = 5.29 × 10⁻¹¹ m the phase varies by k a₀ = 2.7 × 10⁻³ over one Bohr radius. Set exp(ik⋅r) → 1 and two things happen at once. The retained operator is the electric-dipole (E1) one; the discarded i k⋅r term supplies the magnetic-dipole and electric-quadrupole channels, smaller by k a in amplitude and (k a)² ≈ 7 × 10⁻⁶ in rate. And A(t) is now independent of r, so e²A²/2m is a multiple of the identity on the atomic Hilbert space: ⟨f|A²|i⟩ ∝ ⟨f|i⟩ = 0 for f ≠ i. It contributes a common time-dependent phase and drives nothing.

Göppert-Mayer moves the same physics into the length gauge

The velocity form (e/m)A⋅p is awkward: the operator is p, whose parity and angular-momentum content are less transparent than r's, and A depends on gauge. Apply the unitary transformation |ψ′⟩ = exp(−i q r⋅A(t)/ħ)|ψ⟩. Because the generator carries its own time dependence, the transformed Hamiltonian gains iħ(∂U/∂t)U† = q r⋅(∂A/∂t) = −q r⋅E, while U(p + eA)U† = p strips A out of the kinetic term altogether. What is left is H₁ = −d⋅E with d = q r = −e r. The A² term is absorbed in the same step. Two warnings. The transformation is unitary, so all observables agree, but |ψ⟩ and |ψ′⟩ are different kets while the field is on. And the gauge you compute in must be the gauge you interpret in.

One commutator ties the two forms together

The bridge is [H₀, r] = [p²/2m, r] = −iħp/m, which holds because V(r) commutes with r. Sandwich it between exact eigenkets: ⟨f|[H₀, r]|i⟩ = (E_f − Eᵢ)⟨f|r|i⟩, so ⟨f|p|i⟩ = (i m/ħ)(E_f − Eᵢ)⟨f|r|i⟩ = i m ω_fi ⟨f|r|i⟩. For hydrogen 1s → 2p₀ with ω_fi = 1.55 × 10¹⁶ rad s⁻¹ and ⟨2p₀|z|1s⟩ = 0.7449 a₀ = 3.94 × 10⁻¹¹ m, that gives |⟨2p₀|p_z|1s⟩| = 5.56 × 10⁻²⁵ kg m s⁻¹, about 0.28 of ħ/a₀. The identity used H₀|i⟩ = Eᵢ|i⟩ twice, so it is exact only for exact eigenkets. Run a Hartree-Fock or configuration-interaction calculation and the two gauges disagree; the size of the gap is a standard measure of how good the wavefunctions are.

First order, and dropping the counter-rotating term

Write the drive as H₁(t) = −d⋅ε E₀ cos ωt = −(d⋅ε E₀/2)[exp(iωt) + exp(−iωt)] and put it through first-order time-dependent perturbation theory. The amplitude c_f(t) = (1/iħ)∫₀ᵗ ⟨f|H₁|i⟩ exp(iω_fi t′) dt′ collects two denominators: δ = ω_fi − ω, which is small near resonance, and ω_fi + ω, which is not. Dropping the second is the rotating-wave approximation, and at Lyman-α, a detuning of 8 × 10⁹ rad s⁻¹ against ω = 1.55 × 10¹⁶ rad s⁻¹ makes the discarded amplitude smaller by δ/2ω = 3 × 10⁻⁷. What survives is |c_f(t)|² = Ω² sin²(δt/2)/δ², a resonance whose peak grows as Ω²t²/4 and whose width shrinks as 2π/t — energy conservation emerging from the clock, not assumed.

What a classical field cannot do

Turn the drive off and −d⋅E is identically zero, so an excited eigenket of H₀ is stationary and hydrogen 2p never decays. That is not a small blemish: the observed rate is A = 6.26 × 10⁸ s⁻¹, a 1.60 ns lifetime, and this model gives zero. The reason is structural — the field is a c-number, and a c-number has no vacuum fluctuation for the atom to emit into. Two repairs exist. Einstein's detailed balance against the Planck law fixes A/B = ħω³/π²c³ on the energy-density convention, taking A from the stimulated rate you already have; or quantise the field, where |0⟩ still supplies a non-zero ⟨1|a†|0⟩. Both give the same answer. Neither is contained in the classical calculation.

Change one variable at a time

Make the relationship visible.

Interactive model

Matrix element ⟨f|d⋅ε|i⟩0.75 e a₀

Field amplitude E₀0.50 10⁵ V m⁻¹

Drive duration t0.55 ns

Raise the matrix element at fixed t: the first-order peak climbs as its square, while the solid curve also broadens, because once Ω passes 2π/t the Rabi frequency sets the width. Then drive the pulse area past π, where the dashed curve runs above 1 while the exact one has turned back — first order expiring, not the atom.

RABI FREQUENCY Ω3.01 10⁹ rad s⁻¹

PULSE AREA Ω t1.66 rad

EXACT P AT δ = 00.544

FIRST ORDER AT δ = 00.687

Live interpretationRABI FREQUENCY Ω: 3.01 10⁹ rad s⁻¹. PULSE AREA Ω t: 1.66 rad. EXACT P AT δ = 0: 0.544. FIRST ORDER AT δ = 0: 0.687

Catch the common trap

Explain before calculating.

Misconception check

The dipole approximation freezes the field's variation in space, not in time. Read exp(ik⋅r) → 1 as "the field is uniform, so the atom just sits in a static field" and you will compute a Stark shift and wonder where the transition went; the field is still E₀ cos ωt, and it is that time dependence, resonant at ω = ω_fi, that moves population. The second half of the same error is hearing "electric-dipole forbidden" as "impossible". A vanishing ⟨f|d|i⟩ only removes the leading term of an expansion whose next term is down by k a ≈ 3 × 10⁻³ in amplitude, so an M1 or E2 channel at the same frequency still runs at roughly (k a)² ≈ 10⁻⁵ of an allowed rate. Hydrogen 2s is a sharper case, and not an instance of that estimate: 2s → 1s is E1-forbidden by parity, carries no E2 amplitude at all because a rank-2 tensor cannot connect two l = 0 states, and has no non-relativistic M1 amplitude either, since ⟨1s|L + gₛ S|2s⟩ ∝ ⟨1s|2s⟩ = 0. What is left is two-photon E1⋅E1 emission at 8.23 s⁻¹, so 2s lives 0.12 s rather than 1.60 ns — long, but not forever.

Working from the same approximate variational wavefunctions, a student computes the hydrogen 1s → 2p transition rate twice: once from the velocity form (e/m)A⋅p, once from the length form −d⋅E. The two answers differ by 30%. What has gone wrong?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe sodium D₂ line sits at 589.0 nm, and the 3s valence orbital has a mean radius of about 0.25 nm. Evaluate the dipole-approximation expansion parameter, say which term of exp(ik⋅r) you are keeping, and estimate the fractional rate at which the next multipole enters.

Photon wavevector: k = 2π/λ = 6.2832 / (5.890 × 10⁻⁷ m) = 1.067 × 10⁷ m⁻¹.
Expansion parameter over the orbital: k a = 1.067 × 10⁷ × 2.5 × 10⁻¹⁰ = 2.67 × 10⁻³. Over one orbital radius the field's phase turns by 2.67 × 10⁻³ rad, about 0.15°, so the wave is flat across the atom to a part in four hundred.
exp(ik⋅r) = 1 + i k⋅r − ½(k⋅r)² + … . Keeping the 1 gives the electric-dipole operator; the i k⋅r term is the magnetic-dipole and electric-quadrupole piece, smaller in amplitude by k a.
Rates go as the square of an amplitude, so those channels enter at about (k a)² = (2.67 × 10⁻³)² = 7.1 × 10⁻⁶ of an allowed E1 rate.

Answerk = 1.07 × 10⁷ m⁻¹ and k a = 2.67 × 10⁻³; keeping the leading 1 is the E1 approximation, and M1/E2 enter the rate at roughly 7 × 10⁻⁶ of it.

MediumA laser tuned to hydrogen's Lyman-α line (121.6 nm, ω = 1.55 × 10¹⁶ rad s⁻¹) drives 1s → 2p(m = 0) with polarisation along z, for which ⟨2p₀|z|1s⟩ = 0.7449 a₀. The peak field amplitude is E₀ = 1.00 × 10⁵ V m⁻¹. Find the dipole matrix element and the Rabi frequency, give the π-pulse duration, test the rotating-wave approximation, and compare Ω with the 2p decay rate Γ = 6.26 × 10⁸ s⁻¹.

Matrix element: d_fi = e⟨2p₀|z|1s⟩ = 1.602 × 10⁻¹⁹ C × 0.7449 × 5.292 × 10⁻¹¹ m = 6.32 × 10⁻³⁰ C m, which is 1.89 D.
Rabi frequency: Ω = d_fi E₀/ħ = (6.32 × 10⁻³⁰ × 1.00 × 10⁵) / 1.0546 × 10⁻³⁴ = 5.99 × 10⁹ rad s⁻¹.
A π pulse is the area Ωt = π, so t = π/Ω = 3.1416 / 5.99 × 10⁹ = 5.25 × 10⁻¹⁰ s = 525 ps.
RWA check: Ω/ω = 5.99 × 10⁹ / 1.55 × 10¹⁶ = 3.9 × 10⁻⁷. The counter-rotating term's leading effect is the Bloch–Siegert shift Ω²/4ω = 3.59 × 10¹⁹ / 6.20 × 10¹⁶ = 5.8 × 10² rad s⁻¹, a millionth of Γ, so dropping it costs nothing.
Coherence check: Ω/Γ = 5.99 × 10⁹ / 6.26 × 10⁸ = 9.6, so the Rabi phase advances 9.6 rad — about one and a half full flops — within one 1.60 ns lifetime. Coherent flopping, not a rate equation.

Answerd_fi = 6.32 × 10⁻³⁰ C m (1.89 D), Ω = 5.99 × 10⁹ rad s⁻¹, π pulse of 525 ps. Ω ≈ 9.6 Γ and Ω/ω ≈ 4 × 10⁻⁷, so the RWA is excellent and the flopping outruns decay.

HardFor the same 1s → 2p₀ transition, the level separation is 10.20 eV and ⟨2p₀|z|1s⟩ = 0.7449 a₀. Obtain the velocity-form matrix element ⟨2p₀|p_z|1s⟩ from the length-form one, then show that the velocity and length couplings to a field of amplitude E₀ are the same size, and say what changes if the wavefunctions are only approximate.

Transition frequency: ω_fi = ΔE/ħ = (10.20 × 1.602 × 10⁻¹⁹ J) / 1.0546 × 10⁻³⁴ = 1.634 × 10⁻¹⁸ / 1.0546 × 10⁻³⁴ = 1.5495 × 10¹⁶ rad s⁻¹.
Identity: [H₀, r] = −iħp/m, so between eigenkets ⟨f|p|i⟩ = (i m/ħ)(E_f − Eᵢ)⟨f|r|i⟩ = i m ω_fi ⟨f|r|i⟩.
Numbers: ⟨2p₀|z|1s⟩ = 0.7449 × 5.292 × 10⁻¹¹ = 3.942 × 10⁻¹¹ m, so |⟨2p₀|p_z|1s⟩| = 9.109 × 10⁻³¹ × 1.5495 × 10¹⁶ × 3.942 × 10⁻¹¹ = 5.56 × 10⁻²⁵ kg m s⁻¹, or 0.28 ħ/a₀.
Velocity form: A = ε A₀ cos ωt with A₀ = E₀/ω, so the coupling is (e/m)A₀⟨f|p_z|i⟩ = (e/m)(E₀/ω)(i m ω_fi ⟨f|z|i⟩), which on resonance is i e E₀⟨f|z|i⟩.
Length form: −d⋅E with d = −e r and E = ε E₀ sin ωt gives e E₀⟨f|z|i⟩. Same magnitude; the factor i is just the quarter-cycle phase between A and E, so the two rates are identical.
Step 2 used H₀|i⟩ = Eᵢ|i⟩ and H₀|f⟩ = E_f|f⟩. With approximate kets that fails, the two forms separate, and the gap is a diagnostic of the basis rather than physics.

Answer|⟨2p₀|p_z|1s⟩| = 5.56 × 10⁻²⁵ kg m s⁻¹ ≈ 0.28 ħ/a₀. Both gauges give the coupling e E₀⟨2p₀|z|1s⟩, so the rates agree exactly — for exact eigenkets. Any residual gap measures wavefunction error.