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Physics V course sequence · Interactive lesson

Atom–field coupling and 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.

UPV-13.02 · Lesson 2 of 9 in this unit. Position is not a completion score.
Starting model: Atom–field coupling and the dipole matrix elementExcitation probability against laser detuning after a drive lasting t. Solid: the exact two-level RWA result. Dashed: first-order perturbation theory. A matrix element of 0.75 e a₀ in a field of 0.50 × 10⁵ V m⁻¹ gives Rabi frequency 3.01 × 10⁹ rad s⁻¹ and pulse area 1.66 rad, so the exact peak is 0.544 against first order's 0.687.excitation probability PΩ t = 1.66 rad1.00.50−80+8detuning δ = ωfi − ω / 10⁹ rad s⁻¹
Start with a picture. Change the model in the lab below.
1

Build the model

One matrix element sets every rate the light can drive.

The key idea

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 dfi = 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.

The complete physical model and assumptions

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 ωfi ⟨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.

1Put 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.

2The 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.

3Göppert-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.

4One 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⟩ = (Ef − Eᵢ)⟨f|r|i⟩, so ⟨f|p|i⟩ = (i m/ħ)(Ef − Eᵢ)⟨f|r|i⟩ = i m ωfi ⟨f|r|i⟩. For hydrogen 1s → 2p₀ with ωfi = 1.55 × 10¹⁶ rad s⁻¹ and ⟨2p₀|z|1s⟩ = 0.7449 a₀ = 3.94 × 10⁻¹¹ m, that gives |⟨2p₀|pz|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.

5First 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 cf(t) = (1/iħ)∫₀ᵗ ⟨f|H₁|i⟩ exp(iωfi t′) dt′ collects two denominators: δ = ωfi − ω, which is small near resonance, and ωfi + 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 |cf(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.

6What 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.

2

Key equations

Read the units, assumptions and limits before substituting.

Minimal-coupling interaction

H = 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 parameter

exp(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 ωfi ⟨f|r|i⟩, ωfi = (Ef − 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 amplitude

Ω = ⟨f|d⋅ε|i⟩E₀/ħ, |cf(t)|² = Ω² sin²(δt/2)/δ², δ = ωfi − 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 |cf|² is a probability

Golden-rule rate in a broadband field

W = π|⟨f|d|i⟩|² ρ(ωfi)/(3ε₀ħ²) = B ρ(ωfi)

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
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

0.75 e a₀
0.2 to 1.5 e a₀; steps of 0.05.
0.50 10⁵ V m⁻¹
0.4 to 1 10⁵ V m⁻¹; steps of 0.05.
0.55 ns
0.3 to 1.2 ns; steps of 0.05.

Try this. 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.

Live results

RABI 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

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Physics V modelExcitation probability against laser detuning after a drive lasting t. Solid: the exact two-level RWA result. Dashed: first-order perturbation theory. A matrix element of 0.75 e a₀ in a field of 0.50 × 10⁵ V m⁻¹ gives Rabi frequency 3.01 × 10⁹ rad s⁻¹ and pulse area 1.66 rad, so the exact peak is 0.544 against first order's 0.687.excitation probability PΩ t = 1.66 rad1.00.50−80+8detuning δ = ωfi − ω / 10⁹ rad s⁻¹
Excitation probability against laser detuning after a drive lasting t. Solid: the exact two-level RWA result. Dashed: first-order perturbation theory. A matrix element of 0.75 e a₀ in a field of 0.50 × 10⁵ V m⁻¹ gives Rabi frequency 3.01 × 10⁹ rad s⁻¹ and pulse area 1.66 rad, so the exact peak is 0.544 against first order's 0.687.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Matrix element ⟨f|d⋅ε|i⟩: 0.75 e a₀; Field amplitude E₀: 0.50 10⁵ V m⁻¹; Drive duration t: 0.55 ns

Baseline 0.75; choose 0.2 to 1.5, in steps of 0.05.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

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 = ωfi, 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.

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

The 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 4 steps revealed.

MediumExample 2

A 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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 5 steps revealed.

HardExample 3

For 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|pz|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.

Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.

0 of 6 steps revealed.

6

Quick check

Select an answer and explain it before checking.

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?
7

Review & continue

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⟩.

  • Explain the relationship without looking at the equation.
  • Check a sign, a unit and a model assumption in your working.
  • Return to the lab or worked example if a step is still unclear.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.