University Quantum Mechanics II · Emission and Absorption of Radiation · 13.2

Absorption and stimulated emission of incoherent light

Almost none of the light an atom meets is a laser. This topic takes Unit 12's single-frequency result to the rate at which an atom really absorbs and emits in lamp, star or thermal light, shows which averages, over frequency, direction and polarisation, produce it, and tests when a rate is the wrong answer.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Light from a lamp, a star or a hot cavity is a band of components whose phases nothing controls. Each component gives Unit 12's first-order probability (E₀²|p_ba⋅n̂|²/ħ²) sin²[(ω₀ − ω)t/2]/(ω₀ − ω)², and because the phases are random the cross terms between components average away: probabilities add, not amplitudes. Writing each component's energy density (ε₀/2)E₀² as ρ(ω)dω and summing, the sin² kernel, t²/4 high and 2π/t wide, keeps an area πt/2, so a spectrum flat across it gives a probability growing linearly in t, a rate π|p_ab⋅n̂|²ρ(ω₀)/(ε₀ħ²).

Light from every direction with every polarisation replaces |p⋅n̂|² by |p|²/3; a single unpolarised beam, in general, does not. Started in the upper level, the other rotating term resonates with the same |V|, so stimulated emission runs at the same rate and incoherent light can at most equalise two populations. All of this is derived here, with the light a prescribed classical field.

It stops working when the phases stop being random over the times that matter: the rate needs t longer than 2π/Δω and shorter than 1/R, a window open only while the Rabi frequency ω₁ that the same energy would drive coherently is small beside the bandwidth, since R = πω₁²/(2Δω). A single-frequency laser fails that test and Rabi-oscillates, and an excited atom in the dark, which this model never lets decay, is topic 13.03's problem.

Simple definition: In incoherent light, absorption and stimulated emission are constant rates, each proportional to the squared transition dipole and to the light's spectral energy density at the transition frequency, and equal to each other for non-degenerate levels.
Example: Levels with |p_ab| = 2.54 × 10⁻²⁹ C m (3.0 e a₀) in isotropic light of ρ(ω₀) = 4.9 × 10⁻¹⁸ J s m⁻³, roughly a 3000 K cavity at 2.00 eV, absorb at π|p_ab|²ρ(ω₀)/(3ε₀ħ²) = 3.4 × 10⁴ s⁻¹, and an atom in the upper level is stimulated down at the same rate.

One frequency component (quoted from Unit 12)P(a → b) = (E₀²|p_ba⋅n̂|²/ħ²) sin²[(ω₀ − ω)t/2]/(ω₀ − ω)²

The rotating-wave harmonic result with V_ba = −E₀ p_ba⋅n̂. On resonance it grows as t², so a single frequency on its own never yields a rate.

E₀ in V m⁻¹; p_ba = q⟨b|r|a⟩ in C m; n̂ the unit polarisation; ω, ω₀ in rad s⁻¹; ħ in J s; P dimensionless.

Energy density of a component, and of a band(ε₀/2)E₀² → ρ(ω) dω

Electric and magnetic parts each average ε₀E₀²/4. Per unit ordinary frequency the density is 2π times larger; for a beam it is I_λ λ²/(2πc²) from the irradiance per unit wavelength.

ε₀ = 8.854 × 10⁻¹² F m⁻¹; ρ(ω) is energy per unit volume per unit angular frequency, in J s m⁻³.

Rate for light of fixed polarisation n̂R(a → b) = π|p_ab⋅n̂|²ρ(ω₀)/(ε₀ħ²)

Random phases add probabilities, and the sin² kernel, t²/4 high and 2π/t wide, has area πt/2: summed over a broad band, t² becomes t.

R in s⁻¹: |p|²/(ε₀ħ²) is in m³ J⁻¹ s⁻², ρ in J s m⁻³. Derived here for a spectrum flat across 2π/t.

Isotropic, unpolarised light: the one-thirdR(a → b) = π|p_ab|²ρ(ω₀)/(3ε₀ħ²)

Needs every direction and polarisation. One unpolarised beam along k̂ gives (|p|² − |p⋅k̂|²)/2 instead, anywhere from 0 to |p|²/2 depending on its angle to p.

|p_ab|² = |pₓ|² + |p_y|² + |p_z|², summed over final and averaged over initial substates. Derived: |p⋅n̂|² averages to |p|²/3.

Stimulated emissionR(b → a) = R(a → b), for non-degenerate levels a and b

Two levels in such light obey dN_b/dt = R(Nₐ − N_b), so populations at most equalise. With degeneracies, gₐ R(a → b) = g_b R(b → a), topic 13.03's relation.

Both in s⁻¹. Derived: from b the e^(+iωt) part of cos ωt resonates, and |V_ab| = |V_ba| because V is Hermitian.

When a rate exists: the Rabi-frequency testR = πω₁²/(2Δω), with ω₁ = E_eq|p_ab⋅n̂|/ħ and (ε₀/2)E_eq² = ρΔω

The window 2π/Δω ≪ t ≪ 1/R is open only if ω₁ ≪ Δω. Narrow the band at fixed energy and the rate gives way to Unit 3's Rabi oscillation.

Flat band of width Δω in rad s⁻¹; ω₁ in rad s⁻¹ is the Rabi frequency of one wave carrying the band's energy.

Random phases make probabilities add

Light from a lamp or a hot cavity is a superposition E(t) = Σₖ Eₖ n̂ cos(ωₖt + φₖ) whose phases φₖ are set by independent emitters in the source, so they bear no fixed relation to one another. First-order amplitudes are linear in the perturbation, so the amplitude to reach b is a sum, c_b = Σₖ cₖ, each term carrying its own factor e^(−iφₖ). Squaring gives Σₖ|cₖ|² plus cross terms cₖcₗ* proportional to e^{(iφₗ − iφₖ)}, and averaged over independent phases uniform on 0 to 2π every cross term vanishes. The mean probability is therefore the sum of the probabilities each component would produce alone. With N equal components the difference is stark: in phase they give N²|c|², with random phases N|c|². This one step is where incoherence enters the physics; nothing else in the derivation tells a lamp from a laser. The average is over an ensemble, many atoms or many realisations of the light, and one atom in one realisation fluctuates about it, which is why the unit's lab averages hundreds of runs.

The frequency integral turns t² into t

Each component contributes Unit 12's (Eₖ²|p_ba⋅n̂|²/ħ²) sin²(Δₖt/2)/Δₖ², with Δₖ = ω₀ − ωₖ. Its mean energy density is (ε₀/2)Eₖ², electric and magnetic parts together, and a continuous spectrum writes that as ρ(ω)dω, so Eₖ² becomes (2/ε₀)ρ(ω)dω and P(t) = (2|p_ba⋅n̂|²/(ε₀ħ²)) ∫ρ(ω) sin²[(ω₀ − ω)t/2]/(ω₀ − ω)² dω. The kernel peaks at t²/4, has its first zeros at ω₀ − ω = ±2π/t and encloses an area πt/2, from ∫ sin²u/u² du = π over all u. If ρ is flat across the peak it leaves the integral at ω₀, and P = π|p_ba⋅n̂|²ρ(ω₀)t/(ε₀ħ²): linear in t, a constant rate. Flatness needs t ≫ 2π/Δω for a source of bandwidth Δω. At t = 1 ns the first zeros lie 1 GHz either side of resonance, so a thermal continuum some 10¹³ Hz wide is flat after picoseconds, while light only 1 GHz wide needs many nanoseconds. The units close: |p|²/(ε₀ħ²) is in m³ J⁻¹ s⁻², and multiplying by ρ in J s m⁻³ leaves s⁻¹.

Directions and polarisations: where the one-third comes from

The rate so far belongs to light polarised along n̂, through |p_ab⋅n̂|², with p_ab = q⟨a|r|b⟩ a vector whose components may be complex. Radiation filling a hot cavity arrives from every direction with every polarisation, so average n̂ over the sphere: the mean of nᵢnⱼ is δᵢⱼ/3, the mean of |p⋅n̂|² is |p|²/3, and R = π|p_ab|²ρ(ω₀)/(3ε₀ħ²). An unpolarised beam alone does not supply the one-third. Travelling along k̂ it holds only the two transverse polarisations, whose average is (|p|² − |p⋅k̂|²)/2: for p along z, a beam along x gives |p|²/2 and a beam along z gives nothing. Averaging (1 − cos²θ)/2 over beam directions gives (1 − 1/3)/2 = 1/3 again, and randomly oriented atoms in a polarised beam reach it the same way. When the lower level is spherically symmetric and every upper sublevel counts, the total cannot depend on n̂: for hydrogen 1s → 2p each sublevel has |⟨100|r|21m⟩|² = 0.555 a₀², quoted from topic 13.04, and any one polarisation collects 0.555 a₀², a third of the 1.665 a₀² total.

Stimulated emission runs at the same rate

Start the atom in b instead. Of the two rotating parts of cos ωt it is now e^(+iωt) that resonates, carrying V_ab, which equals V_ba* because V is Hermitian and so has the same modulus. The kernel, the frequency integral and the direction average then repeat line for line, so R(b → a) = R(a → b) for non-degenerate levels, derived rather than assumed. Each downward transition returns ħω₀ to the light, so a gas absorbs net power in proportion to Nₐ − N_b and amplifies only if N_b exceeds Nₐ. For two levels, dN_b/dt = R(Nₐ − N_b) gives P_b(t) = ½(1 − e^(−2Rt)) from P_b(0) = 0: incoherent light at most equalises the populations, whereas a coherent π pulse inverts them completely. With degenerate levels, sum over final substates and average over initial ones; that gives gₐ R(a → b) = g_b R(b → a), which topic 13.03 recovers from detailed balance. That the emitted energy joins the very mode of the stimulating light is a statement about photons, quoted from quantum electrodynamics; with a classical field the matching picture is a dipole radiating in step with the wave that drives it.

When a rate is the wrong description

The linear growth needs t ≫ 2π/Δω, so that the kernel is narrower than the spectrum, and Rt ≪ 1, so that first order holds. Give the band a total energy density U = ρΔω and let ω₁ = E_eq|p⋅n̂|/ħ be the Rabi frequency of one coherent wave with the same energy, (ε₀/2)E_eq² = U. Then R = πω₁²/(2Δω), and the window 2π/Δω ≪ t ≪ 1/R is open only while ω₁ ≪ Δω. A 1 nm slice of lamp light at 600 nm spans 8.3 × 10¹¹ Hz, or Δω = 5.2 × 10¹² s⁻¹; even at 10 W m⁻² it gives ω₁ ≈ 2 × 10⁷ s⁻¹ for |p| = 2.5 × 10⁻²⁹ C m, so a rate is right. A single-frequency laser, Δω → 0 at fixed U, fails the test: amplitudes add, first order gives ω₁²t²/4, and the exact answer is Unit 3's sin²(ω₁t/2). In between, with Δω comparable to ω₁, neither picture holds and the amplitude equations must be integrated, as the unit's lab does. A laser narrower than an atomic line already broadened by decay or collisions gives a rate again, with the laser's energy density times the atom's normalised line shape in place of ρ, the form topics 13.07 and 13.09 use.

Checks: dimensions, order in α, and hydrogen

With q = e and e²/(ε₀ħc) = 4πα, the isotropic rate reads R = (4π²α/3)|⟨a|r|b⟩|² cρ(ω₀)/ħ. Here cρ/ħ is in m⁻² s⁻¹ and |⟨a|r|b⟩|² in m², so the units close a second way. At fixed spectral density the rate is first order in α: one factor of e² from one interaction with the field, the mark of a one-photon process, where Unit 12's two-photon rates carry α² and the intensity squared. Two limits: ρ → 0 stops transitions in both directions, so this semiclassical atom never decays by itself and the spontaneous rate must come from topic 13.03; and spreading the same total energy over a wider band lowers ρ(ω₀), and the rate, as 1/Δω. The measured test is hydrogen 1s → 2p. Its only atomic input, 0.555 a₀² per polarisation from topic 13.04, is the matrix element that reproduces the measured 1.60 ns lifetime of 2p, and fed through this rate it gives an integrated absorption strength 0.416 times a classical electron's, the tabulated Lyman-α oscillator strength, as the medium example shows.

Change one variable at a time

Make the relationship visible.

Interactive model

Band half-width γ4.0 ω₁

Time marker ω₁t3.0

At γ = 4ω₁ and ω₁t = 3 the single wave has almost inverted the atom while the band has moved about a quarter of it, and the solid curve never crosses ½. Widen the band and the rate falls as 1/γ; narrow it to γ = 0.5ω₁ and the faint curve follows the Rabi curve early on, so the rate equation fails.

INCOHERENT RATE R = ω₁²/(2γ)0.125 ω₁

BAND, RATE EQUATION P_b0.264

Rt, FIRST ORDER NEEDS ≪ 10.375

ONE WAVE, RABI P_b0.995

Live interpretationINCOHERENT RATE R = ω₁²/(2γ): 0.125 ω₁. BAND, RATE EQUATION P_b: 0.264. Rt, FIRST ORDER NEEDS ≪ 1: 0.375. ONE WAVE, RABI P_b: 0.995

Catch the common trap

Explain before calculating.

Misconception check

A resonant laser is not just bright incoherent light. It is tempting to write a laser's spectral density as ρ = I/(cΔω), with Δω its linewidth, and read off a rate π|p⋅n̂|²ρ/(ε₀ħ²) proportional to the intensity I. That rate rested on two things a single-frequency wave lacks: random phases, which let probabilities rather than amplitudes add, and a spectrum wider than the sin² kernel, which turned t² into t. For one coherent wave the amplitudes add, first order gives ω₁²t²/4, and the exact answer is Unit 3's oscillation sin²(ω₁t/2), with ω₁ proportional to √I, not to I. The formula even refutes itself there: rewritten as R = πω₁²/(2Δω), it exceeds ω₁ once Δω falls below πω₁/2, a rate claiming to empty the lower level faster than the atom can complete one Rabi cycle. And the end states differ: a coherent π pulse inverts the atom completely, while incoherent light can never push P_b past ½.

Atoms whose a → b transition dipole p_ab points along z are lit by a collimated, unpolarised beam travelling along x, with spectral energy density ρ(ω₀) at the transition. How does the absorption rate compare with the isotropic result π|p_ab|²ρ(ω₀)/(3ε₀ħ²) for the same ρ(ω₀)?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA Rydberg atom has two non-degenerate levels a and b, 15.0 GHz apart, with |p_ab| = 1.00 × 10³ e a₀. It sits in the isotropic, unpolarised thermal radiation of a 300 K room. Find the absorption rate a → b and the stimulated-emission rate b → a, and the time for the absorption probability of an atom starting in a to reach 1%.

Spectral density, from Planck's law (quoted): ρ(ω) = (ħω³/(π²c³))/(e^(ħω/kT) − 1). Here hν/kT = 2.40 × 10⁻³, so the Rayleigh-Jeans form ρ = ω²kT/(π²c³) is good to 0.1%. With ω = 2π × 15.0 GHz = 9.42 × 10¹⁰ s⁻¹, ρ(ω₀) = (8.88 × 10²¹ s⁻²)(4.14 × 10⁻²¹ J)/(2.66 × 10²⁶ m³ s⁻³) = 1.38 × 10⁻²⁵ J s m⁻³.
|p_ab| = 1.00 × 10³ × (1.602 × 10⁻¹⁹ C)(5.292 × 10⁻¹¹ m) = 8.48 × 10⁻²⁷ C m. The room's radiation arrives from every direction with every polarisation, so the isotropic form applies: R = π|p_ab|²ρ(ω₀)/(3ε₀ħ²).
Units first: |p_ab|²/(ε₀ħ²) = 7.30 × 10²⁶ m³ J⁻¹ s⁻², and multiplying by ρ in J s m⁻³ leaves s⁻¹. Then R = (π/3)(7.30 × 10²⁶)(1.38 × 10⁻²⁵) s⁻¹ = 1.06 × 10² s⁻¹.
The levels are non-degenerate, so an atom in b is stimulated down at the same 1.06 × 10² s⁻¹. From a, the probability grows as Rt and reaches 1% at t = 0.01/R = 95 μs; near 15 GHz the thermal ρ ∝ ω² changes only on the scale of ω₀ itself, so it is flat across the kernel after a few nanoseconds and the rate holds from then on.
Check: topic 13.03's spontaneous rate for this pair, quoted, is 0.25 s⁻¹, smaller by 417, the thermal occupation kT/hν. At room temperature the radiation, not spontaneous decay, drives transitions this far into the microwave.

AnswerR(a → b) = R(b → a) = 1.1 × 10² s⁻¹; the absorption probability reaches 1% after about 95 μs.

MediumA hydrogen atom in 1s sits in a collimated, unpolarised ultraviolet beam whose spectral irradiance is flat near Lyman-α, λ = 121.6 nm, at I_λ = 5.0 × 10⁻² W m⁻² nm⁻¹. Using |⟨100|r|21m⟩|² = (2¹⁵/3¹⁰) a₀² for each m, quoted from topic 13.04, find the rate of excitation to 2p, check it against the tabulated oscillator strength f = 0.416, and find the stimulated rate back down per 2p atom. Ignore spin.

Convert the spectrum. A beam's energy density is irradiance over c, so per unit wavelength it is I_λ/c, and per unit angular frequency ρ = (I_λ/c)|dλ/dω| = I_λ λ²/(2πc²). With I_λ = 5.0 × 10⁷ W m⁻³ and λ² = 1.479 × 10⁻¹⁴ m², ρ(ω₀) = (7.39 × 10⁻⁷ W m⁻¹)/(5.647 × 10¹⁷ m² s⁻²) = 1.31 × 10⁻²⁴ J s m⁻³.
Choose the average. A beam is not isotropic, but 1s is spherically symmetric, so the rate summed over the three 2p sublevels cannot depend on n̂. Take n̂ = ẑ: only m = 0 contributes, and Σₘ|⟨21m|n̂⋅r|100⟩|² = (2¹⁵/3¹⁰) a₀² = 0.555 a₀², a third of the 1.665 a₀² summed over all m and all three components.
Rate for fixed polarisation: R = π e²(0.555 a₀²)ρ(ω₀)/(ε₀ħ²). With 0.555 e²a₀² = 3.99 × 10⁻⁵⁹ C² m² and π/(ε₀ħ²) = 3.19 × 10⁷⁹ m C⁻² J⁻¹ s⁻², R = (3.19 × 10⁷⁹)(3.99 × 10⁻⁵⁹)(1.31 × 10⁻²⁴) s⁻¹ = 1.67 × 10⁻³ s⁻¹, one excitation about every 10 minutes.
Check against the table. The rate is also the integrated cross-section times the photon flux per unit frequency. The first is πrₑ c f = (2.654 × 10⁻⁶ m² s⁻¹)(0.416) = 1.104 × 10⁻⁶ m² s⁻¹, quoted; the second is (I_λ λ²/c)/(hν) = (2.466 × 10⁻¹⁵ W m⁻² Hz⁻¹)/(1.634 × 10⁻¹⁸ J) = 1.51 × 10³ m⁻² s⁻¹ Hz⁻¹. Their product is 1.67 × 10⁻³ s⁻¹: the routes agree.
Stimulated emission. With ẑ polarisation a 2p atom in m = 0 is driven down at 1.67 × 10⁻³ s⁻¹ and one in m = ±1 not at all, so the average per 2p atom is a third, 5.6 × 10⁻⁴ s⁻¹: gₐ R(a → b) = g_b R(b → a) with g = 1 for 1s and g = 3 for 2p. Against the quoted spontaneous rate of 6.27 × 10⁸ s⁻¹ it is negligible.

AnswerR(1s → 2p) = 1.7 × 10⁻³ s⁻¹, about one excitation every 10 minutes, confirmed by f = 0.416; the stimulated rate back down is 5.6 × 10⁻⁴ s⁻¹ per 2p atom.

HardThe 15.0 GHz Rydberg transition with |p_ab| = 1.00 × 10³ e a₀ is driven by microwave noise polarised along p_ab, with a Lorentzian spectrum ρ(ω) = (U/π)γ/[(ω − ω₀)² + γ²] of half-width γ = 2π × 10.0 MHz and the energy density U of a coherent wave of amplitude 1.0 × 10⁻² V m⁻¹. Find P_b(t) to first order in closed form, the rate, and P_b at the time a coherent wave of that amplitude completes a π pulse. Ignore spontaneous decay.

Set up. With the field along p_ab, P_b(t) = (2|p_ab|²/(ε₀ħ²)) ∫ρ(ω) sin²(xt/2)/x² dω, with x = ω₀ − ω. Put ω₁² = 2U|p_ab|²/(ε₀ħ²), the squared Rabi frequency of the coherent wave with (ε₀/2)E₀² = U. Then P_b(t) = (ω₁²γ/π) ∫ sin²(xt/2)/[x²(x² + γ²)] dx over all x.
Partial fractions, 1/[x²(x² + γ²)] = (1/γ²)[1/x² − 1/(x² + γ²)], with sin²(xt/2) = (1 − cos xt)/2 and the standard integrals ∫(1 − cos xt)/x² dx = πt and ∫(1 − cos xt)/(x² + γ²) dx = (π/γ)(1 − e^(−γt)), give P_b(t) = (ω₁²/(2γ²))(γt − 1 + e^(−γt)).
Limits. For γt ≪ 1, expanding e^(−γt) to second order gives ω₁²t²/4, the coherent first-order result: over less than a coherence time the noise is a coherent wave. For γt ≫ 1, P_b → Rt − R/γ with R = ω₁²/(2γ), equal to π|p_ab|²ρ(ω₀)/(ε₀ħ²) since ρ(ω₀) = U/(πγ). The rate sets in after a delay of one coherence time, 1/γ.
Numbers. ω₁ = E₀|p_ab|/ħ = (1.0 × 10⁻² V m⁻¹)(8.48 × 10⁻²⁷ C m)/(1.055 × 10⁻³⁴ J s) = 8.04 × 10⁵ s⁻¹ and γ = 6.28 × 10⁷ s⁻¹, so γ/ω₁ = 78 and R = ω₁²/(2γ) = 5.14 × 10³ s⁻¹. At t = 1/γ = 15.9 ns, P_b = 3.0 × 10⁻⁵ against Rt = 8.2 × 10⁻⁵; by t = 10/γ, P_b has reached 0.90 of Rt.
At the coherent π-pulse time t = π/ω₁ = 3.91 μs, the coherent wave has put the atom in b with certainty, while the noise gives P_b = 0.0200, and the rate equation with stimulated emission, ½(1 − e^(−2Rt)), gives 0.0197. The window 1/γ ≪ t ≪ 1/R runs from 16 ns to 0.19 ms.
The rate description needs γ ≫ ω₁. At this field it fails once the noise is narrowed to γ ≈ ω₁, a half-width of about 2π × 128 kHz; below that the drive acts coherently and the atom Rabi-oscillates.

AnswerP_b(t) = (ω₁²/(2γ²))(γt − 1 + e^(−γt)) with R = ω₁²/(2γ) = 5.1 × 10³ s⁻¹; at t = 3.91 μs the noise gives P_b = 0.020, against certainty for the coherent π pulse.