
Physics V course sequence · Interactive lesson
Einstein coefficients and oscillator strengths
Three processes — absorption, spontaneous emission, stimulated emission — and only one independent number among them. This is where you learn to move between a lifetime, an oscillator strength and a measured absorption depth, and to say which of the three a spectrum actually pins down.
UPV-13.03 · Lesson 3 of 9 in this unit. Position is not a completion score.Build the model
One measured lifetime can fix all three rates of a transition.
The key idea
The Einstein coefficients are the three rate constants of a transition — A₂₁ for spontaneous emission, B₁₂ and B₂₁ for absorption and stimulated emission — and detailed balance against the Planck spectrum leaves only one of them free.
Example. Sodium's D₂ line at 589.16 nm has A₂₁ = 6.16 × 10⁷ s⁻¹, so the 3p level lives 16.2 ns; that one number gives B₂₁ = A₂₁c³/8πhν³ = 7.6 × 10²⁰ m³ J⁻¹ s⁻² and, with g₁ = 2 and g₂ = 4, B₁₂ = 1.5 × 10²¹ in the same units.
The complete physical model and assumptions
Einstein's 1917 argument is a thermodynamic bootstrap. Put a two-level atom in a blackbody cavity, insist its populations settle to the Boltzmann ratio while the field stays Planckian, and the algebra refuses to close unless a third process exists: absorption and spontaneous emission alone reproduce Wien's law, not Planck's. Adding stimulated emission repairs it and, in the same stroke, fixes g₁B₁₂ = g₂B₂₁ and A₂₁/B₂₁ = 8πhν³/c³. What the argument cannot supply is the size of any one coefficient — equilibrium relates them, it does not compute them. Quantum mechanics supplies the missing scale through the dipole matrix element ⟨f|r|i⟩, and spectroscopy repackages it as the dimensionless oscillator strength f, the transition measured against one classical electron on a spring. That form keeps the bookkeeping honest: Thomas–Reiche–Kuhn says the f values out of a level, continuum included, sum to the electron count, so absorption strength is conserved rather than invented, and Beer–Lambert then turns a line's area — not its depth — into a column density. The cost is that all of it is a rate equation for populations, holding only where dephasing is fast enough that the density matrix's coherences never build up. Drive the atom with a narrow, intense laser instead of a broadband field and the populations Rabi-flop rather than relax, and no Einstein B describes what happens.
1Einstein's bootstrap: why stimulated emission is forced
Put one atom in a cavity at temperature T. Absorption drains level 1 at B₁₂uνN₁ and spontaneous emission refills it at A₂₁N₂. In steady state those two alone give uν = (A₂₁/B₁₂)(g₂/g₁)e(−hν/kBT), which is Wien's law — right in the far tail, catastrophically wrong at low frequency, where Planck's spectrum grows as ν². Einstein's repair was a third channel proportional to the field but driven out of the upper level, B₂₁uνN₂. Steady state now gives uν = A₂₁ / [B₁₂(g₁/g₂)e(hν/kBT) − B₂₁], and this matches (8πhν³/c³)/(e(hν/kBT) − 1) only if two conditions hold at once: g₁B₁₂ = g₂B₂₁, and A₂₁/B₂₁ = 8πhν³/c³. Nothing about the atom was assumed beyond two levels and a Boltzmann ratio, so the relations bind every atom — and for exactly that reason they cannot deliver a single number. Equilibrium constrains ratios; it does not compute matrix elements.
2The ν³ ratio decides between masers and lasers
Divide the stimulated emission rate by the spontaneous one: B₂₁uν/A₂₁ = uνc³/8πhν³, and for a thermal field that is exactly n̄ = 1/(e(hν/kBT) − 1), the mean photon number per mode. Which channel wins is therefore a question about the field, not the atom. At 300 K on the sodium D₂ line hν/kBT = 81.4, so n̄ ≈ 4 × 10⁻³⁶: in a warm room essentially every optical photon an atom emits is spontaneous, and an optical amplifier must manufacture its own mode occupation by pumping a population inversion. At 1.42 GHz the same temperature gives hν/kBT = 2.27 × 10⁻⁴ and n̄ ≈ 4.4 × 10³, so stimulated emission beats spontaneous by four thousand to one with nobody doing anything. That is why the maser preceded the laser by seven years. The 21 cm hydrogen line, with a spontaneous rate of 2.9 × 10⁻¹⁵ s⁻¹ — a lifetime of eleven million years — shows the same imbalance in space: even the 2.7 K cosmic background makes its stimulated rate some forty times the spontaneous one.
3The oscillator strength is the dimensionless currency
Einstein stopped at ratios; first-order time-dependent perturbation theory supplies the scale. For an electric-dipole transition the spontaneous rate is A₂₁ = ω³Σf|⟨f|d|i⟩|²/3πε₀ħc³, with d = −er, summed over the lower sublevels and averaged over the g₂ upper ones. Alongside A, spectroscopists tabulate the oscillator strength f₁₂ = (2mₑω₂₁/3ħ)Σf|⟨f|r|i⟩|², which measures the transition against one classical electron bound on a spring at the same frequency. It is dimensionless, usually between 10⁻³ and 1, and for exact eigenfunctions it comes out the same whether you evaluate the matrix element in the length gauge or the velocity gauge — so the gap between those two values is a standard diagnostic of an approximate wavefunction. The bridge back to a rate is A₂₁ = (2πe²ν²/ε₀mₑc³)(g₁/g₂)f₁₂. For Lyman-α, |⟨2p₀|z|1s⟩| = 0.7449a₀, the three 2p sublevels give Σ|⟨f|r|i⟩|² = 1.665a₀², f = 0.416, and A₂₁ = 6.26 × 10⁸ s⁻¹ — a 1.60 ns lifetime.
4Thomas–Reiche–Kuhn: absorption strength is conserved
How do you know a computed set of oscillator strengths is not simply too big? Take the double commutator [x,[H, x]] for H = p²/2m + V(r). Since [H, x] = −iħp/m, the whole object collapses to the pure number ħ²/m, with no reference to V at all. Sandwich it in the ground state and expand over a complete set: 2Σₙ(Eₙ − E₀)|⟨n|x|0⟩|² = ħ²/m. Divide through by ħ²/m and every term becomes an oscillator strength, giving Σₙ fₙ₀ = 1 for one electron and Nₑ for Nₑ of them. A strong line has to be paid for somewhere. The sum runs over the continuum as well, and in hydrogen that is no footnote: the entire bound Lyman series sums to 0.565, so 43.5% of the sum rule sits in photoionisation above 13.6 eV. Use it as a test on any numerical dipole calculation — add your computed f values, subtract from Nₑ, and if what is left for the continuum comes out negative, the calculation is wrong.
5Beer–Lambert: from a measured depth to a column density
A spectrum reaches the atoms only through the optical depth. Write σ(ν) = (e²/4ε₀mₑc) f₁₂ φ(ν) with ∫φ dν = 1; then I(ν) = I₀ exp[−σ(ν)nL], and the frequency-integrated optical depth, ∫σ(ν)nL dν = (e²/4ε₀mₑc) f₁₂N, depends on f and the column density N = nL alone, never on the broadening. Depth is a different story. A Doppler profile has ΔνD = (ν₀/c)√(2kBT/M) and peak φ(ν₀) = 1/(√π ΔνD), so heating the gas at fixed N flattens the line while its area holds. For Na D₂ at 350 K, ΔνD = 854 MHz and σ₀ = 1.12 × 10⁻¹⁵ m²; 20% transmission through a 7.5 cm cell means τ₀ = ln 5 = 1.61 and n = 1.9 × 10¹⁶ m⁻³. A resolved laser scan reads τ₀ this way for as long as the centre still transmits measurably. A spectrometer that cannot resolve the line sees only its equivalent width, and that tracks N linearly only while τ₀ ≲ 1: push further and the centre saturates, the equivalent width barely grows, and only the damping wings — growing as √N — still count atoms. Reading that flat middle of the curve of growth as if the line were thin is the standard way to underestimate a column density tenfold.
6Where the rate equations stop: coherences and saturation
Everything above is a rate equation for populations, and populations are only the diagonal of the density matrix. The optical Bloch equations also carry the coherence ρ₁₂, and eliminating it is legitimate only when it relaxes far faster than the populations do — dephasing rate γ much larger than the Rabi frequency Ω = |d⋅E|/ħ. A broadband thermal field satisfies that easily, its correlation time being femtoseconds; a narrowband laser does not, and the atom then Rabi-flops at Ω instead of relaxing towards a steady state, a regime in which no B coefficient appears at all. Even inside the rate-equation regime the two-level system has a hard ceiling: setting dN₂/dt = 0 under strong driving brings N₂/g₂ up to N₁/g₁ at best — N₂/(N₁ + N₂) → ½ for non-degenerate levels — so a two-level atom can be bleached but never inverted. For Na D₂ the saturation intensity Iₛₐₜ = πhc/3λ³τ is 6.3 mW cm⁻². Real lasers need a third or fourth level precisely because Einstein's own equations forbid the two-level shortcut.
Key equations
Read the units, assumptions and limits before substituting.
Detailed balance against Planck
g₁B₁₂ = g₂B₂₁ · A₂₁/B₂₁ = 8πhν³/c³
Measure one lifetime and all three coefficients follow, provided the upper level has no other decay channel.
g are the level degeneracies and B is defined on the spectral energy density uν (J s m⁻³), so B is in m³ J⁻¹ s⁻²; drop B₂₁ and the same algebra returns Wien's law, not Planck's.Spontaneous rate from the dipole element
A₂₁ = ω³ Σf |⟨f|d|i⟩|² / (3πε₀ħc³)
The ω³ is why a 589 nm line lives 16 ns while the same dipole at 1.42 GHz would live some 24 years.
d = −e r; sum |⟨f|d|i⟩|² over the lower sublevels and average over the g₂ upper ones; ω = 2πν.Oscillator strength
f₁₂ = (2mₑω₂₁/3ħ) Σf |⟨f|r|i⟩|²
f = 1 is one classical electron on a spring. Lyman-α gives 0.416, the Na D₂ line 0.641.
Dimensionless. The 1/3 averages over polarisation; the sum runs over the g₂ upper sublevels reached.From f to the Einstein A
A₂₁ = (2πe²ν²/ε₀mₑc³)(g₁/g₂) f₁₂
Na D₂: f = 0.641 with g₁/g₂ = ½ gives A₂₁ = 6.16 × 10⁷ s⁻¹, that is τ = 16.2 ns.
ν in Hz; the g₁/g₂ factor turns an absorption f into an emission rate out of the upper level.Thomas–Reiche–Kuhn sum rule
Σₙ fₙ₀ = Nₑ
Hydrogen's whole bound Lyman series reaches only 0.565 — the other 0.435 is photoionisation.
Sum over every final state, bound and continuum; Nₑ is the number of electrons in the atom.Beer–Lambert and the integrated cross-section
I(ν) = I₀ exp[−σ(ν)nL] · ∫σ dν = (e²/4ε₀mₑc) f₁₂
A line's area measures column density; its depth measures the broadening as well.
σ in m², n in m⁻³, L in m. The constant is 2.654 × 10⁻⁶ m² Hz, the classical electron value.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Hold N at 1 and sweep the Doppler FWHM from 0.4 to 3 GHz: the dip flattens from 98% to 41%, but the area ∫τ dν never moves — it is fixed by f × N alone. Then set the FWHM back to 1.4 GHz and raise N to 6: the centre absorption pins at 99.9% while the area keeps climbing.
Live results
- PEAK OPTICAL DEPTH τ₀
- 1.14
- CENTRE ABSORPTION
- 68.0 %
- LINE AREA ∫τ dν
- 1.70 GHz
- UPPER-LEVEL LIFETIME 1/A
- 16.3 ns
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Oscillator strength f: 0.64; Column density N: 1.0 ×10¹⁵ m⁻²; Doppler FWHM: 1.4 GHz
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Students read A₂₁/B₂₁ = 8πhν³/c³ as an equilibrium result and stop applying it the moment the field is not a blackbody. It is not. A and B are set by the atom's dipole matrix element alone; the cavity was scaffolding, used once to relate coefficients that were already fixed before anyone mentioned a cavity. The ratio holds inside a laser beam, in a 20 K interstellar cloud, and in the vacuum. The second slip is numerical. Writing B₂₁ = 7.6 × 10²⁰ beside A₂₁ = 6.2 × 10⁷ looks like stimulated emission winning by 10¹³, but the two carry different units — s⁻¹ against m³ J⁻¹ s⁻² — and only B₂₁uν is a rate. Their ratio is the photon occupation n̄ = 1/(e(hν/kBT) − 1): at 300 K that is 4 × 10⁻³⁶ on the sodium D₂ line and 4.4 × 10³ at 1.42 GHz.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
The sodium D₂ line at 589.16 nm has a measured upper-state lifetime of 16.2 ns, with degeneracies g₁ = 2 for 3s ²S₁/₂ and g₂ = 4 for 3p ²P₃/₂. Find A₂₁, B₂₁ and B₁₂ on the spectral-energy-density convention.
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.
- The 3p level decays by this one channel only, so A₂₁ = 1/τ = 1/(16.2 × 10⁻⁹ s) = 6.17 × 10⁷ s⁻¹.
- Frequency: ν = c/λ = 2.998 × 10⁸ / (589.16 × 10⁻⁹) = 5.088 × 10¹⁴ Hz.
- Detailed balance gives A₂₁/B₂₁ = 8πhν³/c³ = 8π(6.626 × 10⁻³⁴)(5.088 × 10¹⁴)³/(2.998 × 10⁸)³ = 8.14 × 10⁻¹⁴ J s m⁻³.
- So B₂₁ = A₂₁ ÷ (8πhν³/c³) = 6.17 × 10⁷ / 8.14 × 10⁻¹⁴ = 7.58 × 10²⁰ m³ J⁻¹ s⁻².
- The degeneracy relation g₁B₁₂ = g₂B₂₁ gives B₁₂ = (4/2)(7.58 × 10²⁰) = 1.52 × 10²¹ m³ J⁻¹ s⁻².
Answer. A₂₁ = 6.17 × 10⁷ s⁻¹, B₂₁ = 7.58 × 10²⁰ m³ J⁻¹ s⁻², B₁₂ = 1.52 × 10²¹ m³ J⁻¹ s⁻². B₁₂ is the larger only because the upper level carries twice the sublevels.
MediumExample 2
Hydrogen's Lyman-α transition has |⟨2p, m = 0|z|1s⟩| = 0.7449 a₀ and λ = 121.57 nm. Find the absorption oscillator strength f₁₂ and the 2p lifetime, then say what share of hydrogen's sum rule this single line uses.
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.
- The three 2p sublevels are equivalent, so Σf|⟨f|r|1s⟩|² = 3|⟨2p₀|z|1s⟩|² = 3(0.7449)²a₀² = 1.665a₀² = 1.665 × (5.292 × 10⁻¹¹ m)² = 4.662 × 10⁻²¹ m².
- Angular frequency: ω = 2πc/λ = 2π(2.998 × 10⁸)/(1.2157 × 10⁻⁷) = 1.5495 × 10¹⁶ rad s⁻¹.
- f₁₂ = (2mₑω/3ħ)Σf|⟨f|r|i⟩|² = [2(9.109 × 10⁻³¹)(1.5495 × 10¹⁶)]/[3(1.0546 × 10⁻³⁴)] × 4.662 × 10⁻²¹ = (8.923 × 10¹⁹ m⁻²)(4.662 × 10⁻²¹ m²) = 0.416.
- Convert to a rate with ν = 2.466 × 10¹⁵ Hz, g₁ = 2, g₂ = 6: the prefactor 2πe²ν²/ε₀mₑc³ is 4.514 × 10⁹ s⁻¹, so A₂₁ = 4.514 × 10⁹ × (1/3) × 0.416 = 6.26 × 10⁸ s⁻¹.
- Lifetime τ = 1/A₂₁ = 1.60 ns. Against Σf = 1 for hydrogen's single electron, this one line spends 41.6% of the whole sum rule.
Answer. f₁₂ = 0.416, A₂₁ = 6.26 × 10⁸ s⁻¹, τ(2p) = 1.60 ns. Lyman-α alone uses 41.6% of the sum rule; the rest of the bound series adds only 0.149, and photoionisation carries the remaining 0.435.
HardExample 3
A 7.5 cm sodium cell at 350 K is scanned with a narrowband laser on the D₂ line (589.16 nm, f₁₂ = 0.641). Transmission at line centre is 20%. Find the Doppler width, the peak cross-section, the number density and the column density, and give one reason the answer is only good to a factor of about two.
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.
- Doppler 1/e half-width: ΔνD = (ν₀/c)√(2kBT/M) with M = 22.99 u = 3.818 × 10⁻²⁶ kg. Here √(2 × 1.381 × 10⁻²³ × 350 / 3.818 × 10⁻²⁶) = 503 m s⁻¹ and ν₀/c = 1/λ = 1.697 × 10⁶ m⁻¹, so ΔνD = 8.54 × 10⁸ Hz, an FWHM of 2√(ln2)ΔνD = 1.42 GHz.
- Peak of the normalised Gaussian profile: φ(ν₀) = 1/(√π ΔνD) = 1/(1.7725 × 8.54 × 10⁸) = 6.61 × 10⁻¹⁰ s.
- Line-centre cross-section: σ₀ = (e²/4ε₀mₑc) f₁₂ φ(ν₀) = (2.654 × 10⁻⁶ m² Hz)(0.641)(6.61 × 10⁻¹⁰ s) = 1.12 × 10⁻¹⁵ m².
- Optical depth from Beer–Lambert: I/I₀ = e(−τ₀) = 0.20, so τ₀ = −ln 0.20 = 1.609.
- τ₀ = σ₀nL gives n = 1.609/[(1.124 × 10⁻¹⁵ m²)(0.075 m)] = 1.9 × 10¹⁶ m⁻³, that is 1.9 × 10¹⁰ cm⁻³, and the column density is N = nL = 1.4 × 10¹⁵ m⁻².
- Caveat: the D₂ ground-state hyperfine splitting is 1.77 GHz, comparable with the 1.42 GHz Doppler FWHM, so the true profile is two overlapping Doppler lines sharing the oscillator strength. Treating it as one line at the full f overstates σ₀ and so understates n.
Answer. ΔνD = 854 MHz (FWHM 1.42 GHz), σ₀ = 1.12 × 10⁻¹⁵ m², n = 1.9 × 10¹⁶ m⁻³, N = 1.4 × 10¹⁵ m⁻². Unresolved hyperfine structure of comparable width makes this a factor-of-two estimate, not a measurement.
Quick check
Select an answer and explain it before checking.
Review & continue
The Einstein coefficients are the three rate constants of a transition — A₂₁ for spontaneous emission, B₁₂ and B₂₁ for absorption and stimulated emission — and detailed balance against the Planck spectrum leaves only one of them free.
- 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.