University Quantum Mechanics II · Emission and Absorption of Radiation · 13.9
Rate equations, population inversion and the laser
Everything this unit built (A, the B coefficients, the line shape) goes into two coupled rate equations. They show why no lamp can invert a pair of levels, why ruby must excite about half its chromium before it gains at all while Nd:YAG needs under a ten-thousandth, and how a lifetime and a linewidth predict a laser's gain.
Build the model
Connect the measurement to the mechanism.
Once light is incoherent, phases average away and only populations are left, so the unit's coefficients become bookkeeping: level 2 fills at W₁₂N₁ and empties at (W₂₁ + A)N₂, with g₁W₁₂ = g₂W₂₁ inherited from g₁B₁₂ = g₂B₂₁. That relation settles the two-level question. Light that lifts atoms pushes them back down at the same rate per state, and spontaneous emission only adds to the downward traffic, so optical pumping can at most bring N₂/g₂ up to N₁/g₁: the line bleaches to transparency and never amplifies.
Inversion needs a pump acting on a different transition from the one that lases. In a three-level scheme such as ruby the lower laser level is the ground state, and about half the ions must be lifted before any gain appears; in a four-level scheme such as Nd:YAG the lower level sits far enough above the ground, and drains fast enough, to stay empty, and the first excited ion already counts. Once inverted, a beam grows as e(γz) with γ = σ(ν)[N₂ − (g₂/g₁)N₁] and σ(ν) = λ²A g(ν)/(8π), so a lifetime and a linewidth predict the gain; a laser oscillates when round-trip gain meets round-trip loss.
The model stops where phase matters. A coherent π pulse inverts two levels completely, if only for a moment; the coherence and modes of laser light lie outside a description that discarded phase at its first step; and its A came from detailed balance with Planck's law, not from a quantised field.
- Simple definition
- Population inversion on a transition means its upper level holds more atoms per quantum state than its lower level, N₂/g₂ > N₁/g₁; only then does stimulated emission outrun absorption, so light at that frequency is amplified.
- Example
- Neon's 632.8 nm line has g₂ = 3 and g₁ = 5. With 2.0 × 10¹⁵ m⁻³ atoms up and 3.0 × 10¹⁵ m⁻³ down, that is 6.7 × 10¹⁴ against 6.0 × 10¹⁴ per state: an inversion, although the upper level holds fewer atoms.
Valid once incoherent light has destroyed the phases: topic 13.02's golden-rule rates up and down, and 13.03's detailed-balance A, with N₁ + N₂ fixed.
N₁, N₂ in m⁻³; W₁₂, W₂₁ and A in s⁻¹. Each W is B times the spectrum ρ folded through the line shape, Bρ at line centre for broadband light, so g₁B₁₂ = g₂B₂₁ makes g₁W₁₂ = g₂W₂₁ for any spectrum
The last factor is below 1 at any intensity, so the upper level's population per state only approaches the lower's: the line bleaches but never inverts.
Steady state of the rate equation; the pump light sets both W₁₂ and W₂₁ = (g₁/g₂)W₁₂, in s⁻¹
The lower laser level is the ground. Inversion needs Wₚτ > g₂/g₁, so with equal weights more than half the atoms must be lifted first: ruby's burden.
Wₚ: pump rate per ground-state atom, s⁻¹; τ = 1/A of the upper laser level, s; the pump level drains into it at once
Positive for any pump rate, so threshold costs only ΔNₜₕ/N of the atoms: about 7 × 10⁻⁵ in a typical Nd:YAG rod, against about one half in ruby.
Same Wₚ and τ. If level 2 decays only into level 1, of lifetime τ₁, then N₁ = (τ₁/τ)N₂ and inversion needs g₂τ₁ < g₁τ
A beam grows as I₀e(γz). The first bracket is the gain cross-section σ(ν), in m², which reaches its ceiling λ²/(2π) for a line of purely natural width.
γ in m⁻¹; λ in m, A in s⁻¹, line shape g(ν) in s (topic 13.07), N in m⁻³. In a host of index n, λ → λ/n (stated)
Unity round-trip gain, R₁R₂e(2(γ − α)L) = 1. Dividing by σ gives the inversion ΔNₜₕ = γₜₕ/σ that the pump has to sustain.
γₜₕ and the internal loss α in m⁻¹; L the gain length in m; mirror reflectances R₁ and R₂ dimensionless
From amplitudes to rate equations
Topic 13.02 found that incoherent light makes a transition probability grow linearly in time, because frequency components with random phases add in probability rather than in amplitude. That licenses rate equations: once phases are lost, an atom is described only by the level it occupies, and a population N₂, in atoms per m³, changes as dN₂/dt = W₁₂N₁ − W₂₁N₂ − AN₂, with dN₁/dt its negative. Each stimulated rate is B times the spectrum ρ folded through the line shape, which is Bρ at line centre for broadband light; both directions fold the same spectrum through the same line, so 13.03's g₁B₁₂ = g₂B₂₁ turns into g₁W₁₂ = g₂W₂₁ for whatever spectrum is present: per quantum state, up and down are equally likely. Test the model on its own origin. In blackbody light W₂₁ = n̄A, with n̄ = 1/(e(hν/kT) − 1) photons per mode, and the steady state is N₂/N₁ = (g₂/g₁)e(−hν/kT); for ruby's 694.3 nm line at 300 K, hν/kT = 69 and the ratio is 10⁻³⁰. The model fails when the light is coherent, its spectrum narrower than the Rabi frequency it drives, and nothing else scrambles the phases faster: then amplitudes add, and Unit 3's Rabi oscillation replaces a steady rate.
Why light alone cannot invert two levels
Pump the pair with light on the transition itself. In steady state W₁₂N₁ = (W₂₁ + A)N₂, and with g₁W₁₂ = g₂W₂₁ this becomes N₂/g₂ = (N₁/g₁) × W₂₁/(W₂₁ + A). The factor is below one at every intensity, so the upper level's population per state trails the lower's and approaches it only when W₂₁ ≫ A. With equal weights the inversion is ΔN/N = −1/(1 + 2W₂₁τ), τ = 1/A: the absorption coefficient falls by the factor 1 + 2W₂₁τ, the medium bleaches, and a transmitted beam stops losing energy but gains none. Spontaneous emission is not the real obstacle. Set A = 0 and the light still drives atoms down per state exactly as fast as it drives them up, so it can only equalise. In the language of 13.03, inversion is a negative temperature, N₂/N₁ = (g₂/g₁)e(−hν/kT) with T < 0, and pumping on the line itself can only push the effective temperature towards +∞, where the per-state populations are equal.
Three levels: pump one transition, lase on another
Break the symmetry with a third level. In ruby, Cr³⁺ ions in sapphire absorb green and violet light into broad bands (level 3) that relax without radiating, far faster than anything else here, into the ²E level (level 2), which lives τ = 3.0 ms and emits the 694.3 nm line to the ground (level 1). The reverse of the pump transition no longer matters, because level 3 is always nearly empty. With Wₚ the pump rate per ground-state ion and stimulated emission on 2 → 1 still negligible, the steady state is N₂ = WₚτN₁, and N₁ + N₂ = N gives [N₂ − (g₂/g₁)N₁]/N = (Wₚτ − g₂/g₁)/(1 + Wₚτ). With equal weights, transparency sits at Wₚτ = 1: every ground-state ion must be excited once per lifetime, and half the ions must sit in level 2 before the crystal stops absorbing its own line. For ruby's 1.6 × 10²⁵ ions per m³ that is 2.7 × 10²⁷ excitations per m³ per second just to reach zero gain. Limit check: as Wₚτ → ∞ the inversion tends to N, every ion up.
Four levels: keep the lower laser level empty
Lift the lower laser level off the ground and the condition changes character. In Nd:YAG, 808 nm diode light pumps Nd³⁺ into a band that relaxes to ⁴F₃/₂ (level 2, τ = 230 μs); the 1064 nm line runs to ⁴I₁₁/₂ (level 1), which drains to the ground manifold (level 0) thousands of times faster than level 2 decays. If every decay of level 2 lands in level 1, whose lifetime is τ₁, the steady state has N₁ = (τ₁/τ)N₂, so ΔN = N₂(1 − g₂τ₁/(g₁τ)) is positive whenever g₂τ₁ < g₁τ, which for equal weights says an ion leaves level 1 faster than it leaves level 2, and ΔN/N → Wₚτ/(1 + Wₚτ) as τ₁ → 0: any pump inverts. Thermal filling of level 1 is small because it lies 2111 cm⁻¹ above the ground, ten times kT/hc = 208.5 cm⁻¹ at 300 K, so it holds e(−10.1) = 4 × 10⁻⁵ as many ions per state as the lowest ground level. Pump power goes as N₂/τ, so holding an inversion ΔNₜₕ costs ΔNₜₕ/τ excitations per m³ per second here, against (N + ΔNₜₕ)/(2τ) in a three-level medium with equal weights: a ratio (N + ΔNₜₕ)/(2ΔNₜₕ).
Gain is stimulated emission outrunning absorption
Send a narrow-band beam of intensity I and frequency ν along z. Each upper atom is stimulated at W₂₁ = σ(ν)I/(hν), each lower atom absorbs at W₁₂ = (g₂/g₁)W₂₁, and each event adds or removes one photon hν; spontaneous photons spread over 4π and barely touch the beam. So dI/dz = σ(ν)[N₂ − (g₂/g₁)N₁]I = γ(ν)I. To find σ, give the beam the spectral energy density (I/c)δ(ν′ − ν) and fold it through topic 13.07's line shape g(ν′). Referred to energy per unit frequency, ρ(ν) = 2πρ(ω), 13.03's A/B = ħω³/(π²c³) becomes A/B = 8πhν³/c³, so W₂₁ = B(I/c)g(ν) and σ(ν) = (c²/8πν²)A g(ν) = λ²A g(ν)/(8π). Random atomic orientations, or the sum over m sublevels, supply the same 1/3 that isotropic light did. Dimensions: m² × s⁻¹ × s = m². Limit: a line of purely natural width has g(ν₀) = 4/A, so σ = λ²/(2π), 5.5 × 10⁻¹⁴ m² at 589 nm against an atomic area near 10⁻¹⁹ m². It holds neither α nor the dipole moment, because a stronger line broadens itself in proportion. Room-temperature solid-state lines are 10⁸ to 10¹⁰ times broader than their natural width, and σ falls in proportion.
Threshold, clamping, and what rate equations leave out
Put a gain medium of length L between mirrors of reflectance R₁ and R₂. A round trip multiplies the intensity by R₁R₂e(2(γ − α)L), α the internal loss, and oscillation starts when that factor reaches one: γₜₕ = α + ln[1/(R₁R₂)]/(2L), stated here as gain matching loss. A 0.10 m Nd:YAG rod with R₁ = 1, R₂ = 0.95 and negligible α needs γₜₕ = 0.26 m⁻¹; with the tabulated σ = 2.8 × 10⁻²³ m² that is ΔNₜₕ = γₜₕ/σ = 9.2 × 10²¹ m⁻³, only 7 × 10⁻⁵ of the 1.38 × 10²⁶ Nd ions per m³ in 1% doped YAG. The oscillation is seeded by spontaneous photons that happen to leave along the axis. Above threshold, stimulated emission by the growing field holds the inversion at ΔNₜₕ, so every extra excitation the pump supplies is turned into laser photons; that clamping is stated, not derived here. Rate equations cannot say which frequencies oscillate, why the output is coherent or how narrow its line is. Those belong to the cavity modes, spaced by c/(2L) in an empty cavity, and to the phase of the field, which this description gave up at its first step, and they are left to optics.
Change one variable at a time
Make the relationship visible.
Start at τ₁/τ = 0 with equal weights and drag the threshold down: the four-level circle follows ΔNₜₕ/N itself while the three-level one stays above g₂/g₁, so at 0.01 the pump-power ratio passes 50. Raise τ₁/τ to watch that advantage shrink. Nothing lifts the faint two-level curve above zero.
3-LEVEL THRESHOLD Wₚτ1.11
4-LEVEL THRESHOLD Wₚτ0.053
3-LEVEL N₂/N AT THRESHOLD52.5 %
PUMP POWER 3-LEVEL / 4-LEVEL10.5
Live interpretation3-LEVEL THRESHOLD Wₚτ: 1.11. 4-LEVEL THRESHOLD Wₚτ: 0.053. 3-LEVEL N₂/N AT THRESHOLD: 52.5 %. PUMP POWER 3-LEVEL / 4-LEVEL: 10.5
Catch the common trap
Explain before calculating.
A three-level medium with non-degenerate levels is pumped from the ground state at a rate Wₚ per ground-state atom into a band that drains at once into the upper laser level, which decays only back to the ground with lifetime τ. Below threshold, with Wₚτ = 1.5, what is the inversion (N₂ − N₁)/N?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyBroadband light pumps sodium's D2 line, 3s ²S₁/₂ → 3p ²P₃/₂ at 589.0 nm, with g₁ = 2, g₂ = 4 and a measured A = 6.16 × 10⁷ s⁻¹. What stimulated-emission rate W₂₁ puts exactly half the atoms in 3p? Is the vapour then amplifying, and how bright would isotropic light have to be?
- Steady state of the two-level rate equation: N₂/g₂ = (N₁/g₁)W₂₁/(W₂₁ + A), so N₂/N₁ = 2W₂₁/(W₂₁ + A). Setting N₂/N₁ = 1 gives W₂₁ = A = 6.16 × 10⁷ s⁻¹, and each 3s atom absorbs at W₁₂ = (g₂/g₁)W₂₁ = 1.23 × 10⁸ s⁻¹.
- Gain test, per state: N₂ − (g₂/g₁)N₁ = 0.5N − 2 × 0.5N = −0.50N. It is negative, so the vapour still absorbs with as many atoms up as down.
- Unpumped, N₁ = N and the same quantity is −2N, so the absorption coefficient has fallen to a quarter: the line is saturating, not inverting.
- Limit: as W₂₁ → ∞, N₂/N₁ → g₂/g₁ = 2, two-thirds of the atoms up, and N₂ − 2N₁ → 0. Transparency is the ceiling.
- For isotropic light W₂₁/A is the photon number per mode, n̄ = 1/(e(hν/kT) − 1), from 13.03 with Planck's law quoted. n̄ = 1 means hν/kT = ln 2; with hν = 2.105 eV, T = 2.105 eV/(8.617 × 10⁻⁵ eV K⁻¹ × 0.693) = 3.52 × 10⁴ K.
AnswerW₂₁ = 6.2 × 10⁷ s⁻¹ (W₁₂ = 1.2 × 10⁸ s⁻¹); N₂ − 2N₁ = −0.50N, so the vapour still absorbs, at a quarter of its unpumped rate. Isotropic light would need a brightness temperature of 3.5 × 10⁴ K.
MediumRuby's R₁ line at 694.3 nm has a Lorentzian room-temperature width of 330 GHz, the ²E level lives 3.0 ms, and sapphire's index is 1.76. Treating the decay as purely radiative, A = 1/τ, predict the peak gain cross-section and compare it with the tabulated 2.5 × 10⁻²⁴ m².
- Peak of a normalised Lorentzian (topic 13.07): g(ν₀) = 2/(πΔν) = 2/(π × 3.30 × 10¹¹ Hz) = 1.93 × 10⁻¹² s, and A = 1/(3.0 × 10⁻³ s) = 333 s⁻¹.
- In the crystal λ → λ₀/n, so σ = λ₀²A g(ν₀)/(8πn²) = (6.943 × 10⁻⁷ m)² × 333 s⁻¹ × 1.93 × 10⁻¹² s/(8π × 1.76²) = 3.98 × 10⁻²⁴ m². Units: m² × s⁻¹ × s = m².
- That is 60% above the tabulated value. But 3.0 ms is the lifetime of the whole ²E level, which is two sublevels 29 cm⁻¹ apart, the measured R₁ to R₂ spacing, and only the lower one, Ē, emits R₁.
- Equal weights and a Boltzmann factor e(−29 × 1.4388/300) = 0.870 give Ē a share 1/1.870 = 53.5% of the ²E ions. If both sublevels radiate at similar rates, the cross-section per ²E ion is 0.535 × 3.98 × 10⁻²⁴ m² = 2.13 × 10⁻²⁴ m².
- Ceiling check: a natural-width line would give λ₀²/(2πn²) = 2.48 × 10⁻¹⁴ m², 6.2 × 10⁹ times more, which is exactly the ratio of 330 GHz to the natural width A/(2π) = 53 Hz.
Answerσ ≈ 4.0 × 10⁻²⁴ m² from A and the line shape, or 2.1 × 10⁻²⁴ m² once only Ē is counted: within 20% of the tabulated 2.5 × 10⁻²⁴ m².
HardA ruby rod 0.10 m long, with N = 1.6 × 10²⁵ Cr³⁺ ions per m³, σ = 2.5 × 10⁻²⁴ m² and τ = 3.0 ms, sits between mirrors of reflectance 1.00 and 0.90 with no other loss. Treat the levels as non-degenerate and pump at 550 nm. Find the threshold inversion and the absorbed pump power per unit volume at threshold, then repeat for a four-level medium with the same σ, τ, N, cavity and pump photons.
- Threshold gain: γₜₕ = ln[1/(1.00 × 0.90)]/(2 × 0.10 m) = 0.1054/0.20 m = 0.527 m⁻¹, so ΔNₜₕ = γₜₕ/σ = 0.527 m⁻¹/(2.5 × 10⁻²⁴ m²) = 2.11 × 10²³ m⁻³, 1.3% of N.
- Three-level: N₂ − N₁ = ΔNₜₕ with N₁ + N₂ = N gives N₂ = (N + ΔNₜₕ)/2 = 8.11 × 10²⁴ m⁻³, which needs Wₚτ = N₂/N₁ = 1.027.
- In steady state each ion leaving level 2, N₂/τ per m³ per second, was put there by one absorbed pump photon of hνₚ = hc/(550 nm) = 3.61 × 10⁻¹⁹ J, so P₃ = hνₚN₂/τ = 3.61 × 10⁻¹⁹ J × 8.11 × 10²⁴ m⁻³/(3.0 × 10⁻³ s) = 9.8 × 10⁸ W m⁻³.
- Four-level: level 1 is empty, so N₂ = ΔNₜₕ = 2.11 × 10²³ m⁻³, reached at Wₚτ = 0.013, and P₄ = hνₚΔNₜₕ/τ = 2.5 × 10⁷ W m⁻³.
- Ratio P₃/P₄ = (N + ΔNₜₕ)/(2ΔNₜₕ) = 38. Limits: as the loss falls, ΔNₜₕ → 0 and the four-level cost vanishes while the three-level cost stays at hνₚN/(2τ); only if ΔNₜₕ approached N would the two schemes cost the same.
- Scale check: 9.8 × 10⁸ W m⁻³ is about 1 kW per cm³ absorbed continuously in a rod of a few cm³, which is why ruby lasers, from Maiman's in 1960 onwards, are almost always pumped in pulses by flashlamps.
AnswerΔNₜₕ = 2.1 × 10²³ m⁻³. Three-level: P ≈ 9.8 × 10⁸ W m⁻³, about 1 kW cm⁻³; four-level: P ≈ 2.5 × 10⁷ W m⁻³, a factor of 38 less.