University Quantum Mechanics II · Emission and Absorption of Radiation · 13.3

Einstein's A and B coefficients and spontaneous emission

An excited atom in a dark box still emits, and nothing in the semiclassical theory built so far says why. This lesson gets the spontaneous rate the way Einstein did, by demanding that atoms and thermal light agree on a temperature, turns it into a number from one dipole matrix element, and keeps count of what was borrowed on the way.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Put two atomic levels, a below b, in a cavity at temperature T. The previous topic supplied the rates light can drive: absorption at B_ab ρ(ω₀) per lower atom and stimulated emission at B_ba ρ(ω₀) per upper atom, both proportional to the spectral energy density and both zero in the dark. Those two alone cannot hold Boltzmann's populations at every temperature, so Einstein added a third process, a downward rate A that does not depend on the light at all.

Demanding that the steady spectrum be Planck's then fixes two relations between atomic constants, gₐ B_ab = g_b B_ba and A = (ħω₀³/π²c³) B_ba, and with the dipole B the spontaneous rate becomes A = ω₀³|p_ab|²/(3πε₀ħc³), 6.27 × 10⁸ s⁻¹ for hydrogen's Lyman α. The argument borrows as much as it derives: Planck's law and the Boltzmann factor are inputs, and a semiclassical atom, whose excited eigenstate is stationary when no field acts, cannot produce A by itself. Quantum electrodynamics can, by the golden rule summed over the empty modes of a quantised field, and it returns the same A; that calculation is quoted here.

Simple definition: Einstein's A coefficient is the rate at which an excited atom decays with no light present; the B coefficients turn the spectral energy density at the transition frequency into absorption and stimulated-emission rates.
Example: Sodium's 3p₃/₂ level lives 16.2 ns in the dark, so A = 6.17 × 10⁷ s⁻¹. In a 3000 K cavity at 589 nm, stimulated emission adds only A/(e^8.14 − 1) = 1.80 × 10⁴ s⁻¹, about 3400 times less.

Rate equation for the upper leveldN_b/dt = −A N_b − B_ba ρ(ω₀) N_b + B_ab ρ(ω₀) Nₐ

At ρ = 0 only −A N_b survives, so the upper level empties as e^(−At) in the dark. The two B terms are 13.02's golden-rule rates; A is the new one.

N in m⁻³; A in s⁻¹; ρ(ω) in J s m⁻³, energy per volume per rad s⁻¹; B in m³ J⁻¹ s⁻². Einstein's postulate, not derived.

First Einstein relationgₐ B_ab = g_b B_ba

Per sublevel, absorption and stimulated emission are equally strong. For the sodium D2 line g_b/gₐ = 4/2, so B_ab = 2B_ba.

gₐ, g_b: numbers of sublevels in the lower level a and the upper level b. Derived here from the T → ∞ limit.

Second Einstein relationA = (ħω₀³/(π²c³)) B_ba

It holds for any radiative line, dipole or not, so a measured lifetime fixes B_ba: 16.2 ns at 589 nm gives B_ba = 4.76 × 10²¹ m³ J⁻¹ s⁻².

ω₀ = (E_b − Eₐ)/ħ in rad s⁻¹; ħω₀³/(π²c³) in J s m⁻³. Derived here by matching Planck's law, which is quoted.

Spontaneous rate of an electric-dipole lineA = ω₀³ |p_ab|² / (3πε₀ħc³)

As (4α/3)ω₀³|r_ab|²/c² it is first order in α, and A/ω₀ ≈ 0.10α³ for Lyman α: 6.27 × 10⁸ s⁻¹, a 1.60 ns lifetime, as measured.

|p_ab|² = e²(|x_ab|² + |y_ab|² + |z_ab|²) in C² m², one upper sublevel summed over lower ones. Derived here.

Stimulated emission in photons per modeB_ba ρ(ω₀) = A n̄, with n̄ = π²c³ρ(ω₀)/(ħω₀³)

Stimulated emission wins only when n̄ exceeds 1, above kT = ħω₀/ln 2. At 500 nm and 300 K n̄ is 2 × 10⁻⁴²; the 21 cm line in the 2.725 K background has n̄ = 39.

n̄ is dimensionless, the mean photon number per mode near ω₀; thermal light has n̄ = 1/(e^(ħω₀/kT) − 1). Derived here.

The same ratio with ρ per unit frequencyA/B_ba(ν) = 8πhν³/c³

Most tables and older texts use this form. Pairing a B(ν) with a ρ(ω) is a silent factor-of-2π error, so check which density a B multiplies first.

ν in Hz, h = 2πħ. This B multiplies ρ_ν = 2πρ_ω, so B(ν) = B(ω)/(2π) and the rate Bρ is unchanged. Derived by change of variable.

An excited atom in the dark has nothing to make it decay

The previous topic, in one line: in incoherent, isotropic light of spectral energy density ρ(ω), a lower-level atom is excited at B_ab ρ(ω₀) and an upper-level atom is driven down at B_ba ρ(ω₀), with B_ba = π|p_ab|²/(3ε₀ħ²) from the golden rule. Both rates are proportional to ρ. Switch the light off and the perturbation H′ = −p⋅E(t) vanishes, so the upper state |b⟩, an eigenstate of the atomic Hamiltonian, only gathers the phase e^{(−iE_b t/ħ)} and stays put forever. Its dipole expectation value ⟨b|p|b⟩ does not oscillate, and vanishes for a state of definite parity, so even a classical picture finds nothing to radiate. Yet hydrogen's 2p level decays in 1.60 ns and sodium's 3p in 16.2 ns inside a dark, cold vacuum chamber. The semiclassical theory is missing a process, and thermodynamics can supply its rate without saying what the process is.

Detailed balance in a cavity demands a third process

Fill a cavity at temperature T with Nₐ atoms in level a and N_b in level b, with gₐ and g_b sublevels. Equilibrium statistical mechanics, quoted here, gives N_b/Nₐ = (g_b/gₐ) e^{(−ħω₀/kT)} and a radiation field with Planck's spectrum. With only the two driven processes, a steady state needs N_b B_ba ρ = Nₐ B_ab ρ, so N_b/Nₐ = B_ab/B_ba: a constant fixed by the atom and blind to T, which Boltzmann forbids. Einstein's repair is a downward rate A that runs whether light is present or not. Balance becomes N_b(A + B_ba ρ) = Nₐ B_ab ρ, and solving for the field the atoms can coexist with gives ρ(ω₀) = (A/B_ba) / [(gₐ B_ab/(g_b B_ba)) e^(ħω₀/kT) − 1]. That must be Planck's law at every temperature. Planck's law is an input with a measured pedigree: the cosmic microwave background follows it at 2.725 K to within about 50 parts per million of its peak.

Two limits fix the two Einstein relations

Planck's law, quoted, is ρ(ω) = (ħω³/π²c³)/(e^(ħω/kT) − 1). Let T → ∞: Planck's ρ grows without bound while the exponential tends to 1, so the bracket in the balance result must vanish there, which forces gₐ B_ab = g_b B_ba. At any finite T the two expressions then have the same shape, and matching numerators gives A/B_ba = ħω₀³/(π²c³). Neither relation mentions T, the cavity or the atom's structure, and 13.02's golden-rule B's already obey the first, because summed over sublevels |⟨a|p|b⟩|² = |⟨b|p|a⟩|². The −1 in Planck's denominator is stimulated emission: delete B_ba from the balance and the same algebra returns Wien's law, ρ ∝ ω³e^(−ħω/kT), only 1% low at ħω/kT = 4.6 but badly wrong once ħω ≲ kT. Einstein read stimulated emission out of that −1 in 1916 and 1917, nearly four decades before the first maser.

From B to a number: the rate of a dipole line

Insert 13.02's B_ba = π|p_ab|²/(3ε₀ħ²) into A = (ħω₀³/π²c³)B_ba to get A = ω₀³|p_ab|²/(3πε₀ħc³), derived here from a derived B and a quoted Planck law. Dimensions: s⁻³ × C² m² ÷ (C² J⁻¹ m⁻¹ × J s × m³ s⁻³) = s⁻¹. With e²/(4πε₀) = αħc it reads A = (4α/3)ω₀³|r_ab|²/c², first order in α. For hydrogen ħω₀ ~ α²mc² and |r_ab| ~ a₀ = ħ/(αmc), so A/ω₀ ~ α³: the line is sharp, and Lyman α oscillates about 4 × 10⁶ optical cycles before the atom decays. With |⟨1s|z|2p₀⟩| = 0.745 a₀, which 13.04 builds from the radial functions, A = 6.27 × 10⁸ s⁻¹ and τ = 1.60 ns, the measured lifetime. A classical cross-check: Larmor's power for a dipole of amplitude 2|p_ab|, (2|p_ab|)²ω₀⁴/(12πε₀c³), equals ħω₀A exactly.

Spontaneous or stimulated is set by the photons per mode

Divide the two downward rates: B_ba ρ/A = π²c³ρ/(ħω₀³) = n̄. The factor ω₀²/(π²c³) is the mode count behind Planck's law, quoted: radiation modes per unit volume per unit angular frequency, both polarisations included, so n̄ is the mean number of photons in each mode near ω₀, and spontaneous emission runs at the rate one extra photon per mode would stimulate. Thermal light has n̄ = 1/(e^(ħω₀/kT) − 1), so the two rates are equal at ħω₀ = kT ln 2. Visible lines sit deep on the spontaneous side: at 500 nm and 300 K, A/(B_ba ρ) = e^95.9 ≈ 4.5 × 10⁴¹, and even radiation at the Sun's 5800 K leaves it at 142. Radio lines sit on the other side: the 21 cm line in the 2.725 K microwave background has n̄ = 39. Because A/B ∝ ω₀³, the gap widens fast at short wavelengths, which is why masers came before lasers and X-ray lasers remain hard.

What quantum electrodynamics adds, and where A stops

The semiclassical theory has no A because its field is a fixed classical function. Quantise the field, which this course does not do, and each mode becomes an oscillator whose emission matrix element carries √(n + 1): the rate into a mode holding n photons goes as n + 1, stimulated plus spontaneous. The golden rule of Unit 12, summed over every empty mode with density ω²/(π²c³), returns exactly A = ω₀³|p_ab|²/(3πε₀ħc³); that result is quoted, not derived. “Emission stimulated by the vacuum” is a slogan rather than a mechanism: in one standard analysis vacuum fluctuations and the atom's reaction to its own field each supply half of A. The quantised picture also marks the limit of Einstein's value. A counts free-space modes, so it changes when the modes do: Hulet, Hilfer and Kleppner (1985) lengthened a Rydberg state's lifetime at least twenty-fold by placing the atoms between plates that removed the modes it decays into.

Change one variable at a time

Make the relationship visible.

Interactive model

Wavelength, log₁₀(λ / 1 m)−6.2

Temperature, log₁₀(T / 1 K)3.8

Degeneracy ratio g_b/gₐ2.0

At the default, 631 nm in a 6300 K cavity, near the Sun's surface temperature, 97% of decays are spontaneous. Now set λ to 10^-0.7 m, near the 21 cm line, and T to 10^0.4 K, the microwave background: stimulated emission outruns A over thirty-fold. Changing g_b/gₐ moves only the dashed curve.

PHOTONS PER MODE n̄ = B_baρ/A0.028

SPONTANEOUS SHARE OF DECAYS97.3 %

ABSORPTION PER LOWER ATOM ÷ A0.055

CROSSOVER log₁₀(T / 1 K)4.52

Live interpretationPHOTONS PER MODE n̄ = B_baρ/A: 0.028. SPONTANEOUS SHARE OF DECAYS: 97.3 %. ABSORPTION PER LOWER ATOM ÷ A: 0.055. CROSSOVER log₁₀(T / 1 K): 4.52

Catch the common trap

Explain before calculating.

Misconception check

The commonest slip is to treat gₐ B_ab = g_b B_ba and A = (ħω₀³/π²c³)B_ba as facts about thermal radiation, and then refuse to use them for a laser, a lamp or starlight, which are far from equilibrium. They are facts about the atom. The dipole formula shows it: A and both B's are built from ω₀ and a single matrix element p_ab, and the light enters only through the ρ(ω₀) that multiplies B. The cavity at temperature T was a device for discovering relations between three constants that never knew the cavity was there. Two things do limit them, and neither is the light's spectrum: the light must be broadband compared with the line, or a coherent drive gives Rabi oscillation instead of rates, and A is the free-space value, which a cavity that reshapes the modes can change.

Calcium's 422.7 nm line joins the ground level 4s² ¹S₀ (gₐ = 1) to 4s4p ¹P₁ (g_b = 3). In broadband, isotropic, unpolarised light each ground-state atom absorbs at 1.5 × 10³ s⁻¹. In the same light, at what rate is each excited atom driven down by stimulated emission?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTreat the light in a 9000 K stellar atmosphere as Planck radiation. For hydrogen's Balmer-α line at 656.3 nm, what fraction of the downward transitions on that line are spontaneous? Compare a 12.0 GHz molecular line in a 295 K cavity, and find the temperature at which Balmer-α's two downward rates would be equal.

The downward rates per upper atom are A and B_ba ρ(ω₀) = A n̄, with n̄ = 1/(e^x − 1) and x = ħω₀/kT. The spontaneous share is A/(A + A n̄) = 1 − e^(−x).
Balmer α: ħω₀ = 1239.84 eV nm ÷ 656.3 nm = 1.889 eV, and kT = 8.617 × 10⁻⁵ eV K⁻¹ × 9000 K = 0.7755 eV, so x = 2.436 and the spontaneous share is 1 − e^(−2.436) = 0.912.
Microwave line: x = hν/kT = (6.626 × 10⁻³⁴ J s × 1.20 × 10¹⁰ Hz) ÷ (1.381 × 10⁻²³ J K⁻¹ × 295 K) = 1.95 × 10⁻³, so n̄ = 512 and the spontaneous share is 1 − e^(−x) ≈ x = 0.195%.
Equal rates need n̄ = 1, so e^x = 2 and x = ln 2: T = 1.889 eV ÷ (8.617 × 10⁻⁵ eV K⁻¹ × 0.6931) = 3.16 × 10⁴ K.
Limits: x ≫ 1 recovers the dark box, where every decay is spontaneous, and x ≪ 1 gives n̄ ≈ kT/ħω₀, the classical equipartition count for one mode.

Answer91% of Balmer-α decays at 9000 K are spontaneous, against 0.20% for the 12.0 GHz line at 295 K, where stimulated emission is 512 times faster. Balmer α's two rates would match at 3.2 × 10⁴ K.

MediumSodium's 3p ²P₃/₂ level (g_b = 4) decays only to the 3s ²S₁/₂ ground level (gₐ = 2), emitting at 589.0 nm, and its measured lifetime is 16.2 ns. Find the dipole matrix element |p_ab| in units of e a₀, then B_ba and B_ab for ρ per unit angular frequency.

One decay channel, so A = 1/τ = 1/(16.2 × 10⁻⁹ s) = 6.17 × 10⁷ s⁻¹. The angular frequency is ω₀ = 2πc/λ = 2π × 2.998 × 10⁸ m s⁻¹ ÷ 589.0 × 10⁻⁹ m = 3.198 × 10¹⁵ rad s⁻¹.
Invert the dipole formula: |p_ab|² = 3πε₀ħc³A/ω₀³ = 3π × 8.854 × 10⁻¹² × 1.0546 × 10⁻³⁴ × (2.998 × 10⁸)³ × 6.17 × 10⁷ ÷ (3.198 × 10¹⁵)³ = 4.475 × 10⁻⁵⁸ C² m², so |p_ab| = 2.12 × 10⁻²⁹ C m.
With e a₀ = 1.602 × 10⁻¹⁹ C × 5.292 × 10⁻¹¹ m = 8.48 × 10⁻³⁰ C m, |p_ab| = 2.50 e a₀, over three times hydrogen's Lyman-α value of 0.745 e a₀, because sodium's valence electron sits much further out.
Second relation: B_ba = π²c³A/(ħω₀³) = 2.659 × 10²⁶ m³ s⁻³ × 6.17 × 10⁷ s⁻¹ ÷ (1.0546 × 10⁻³⁴ J s × 3.271 × 10⁴⁶ s⁻³) = 4.76 × 10²¹ m³ J⁻¹ s⁻². Units: B times ρ in J s m⁻³ is s⁻¹, a rate.
First relation: B_ab = (g_b/gₐ)B_ba = 2 × 4.76 × 10²¹ = 9.52 × 10²¹ m³ J⁻¹ s⁻². Cross-check with 13.02's B = π|p_ab|²/(3ε₀ħ²) = π × 4.475 × 10⁻⁵⁸ ÷ (3 × 8.854 × 10⁻¹² × 1.112 × 10⁻⁶⁸) = 4.76 × 10²¹, the same.

Answer|p_ab| = 2.12 × 10⁻²⁹ C m = 2.50 e a₀; B_ba = 4.76 × 10²¹ m³ J⁻¹ s⁻² and B_ab = 9.52 × 10²¹ m³ J⁻¹ s⁻².

HardA rubidium vapour cell is lit by a broadband, unpolarised beam whose spectral irradiance is flat at 0.80 W m⁻² nm⁻¹ across the 780.2 nm D2 line. The 5p ²P₃/₂ level (g_b = 4, lifetime 26.2 ns) decays only to 5s ²S₁/₂ (gₐ = 2). Find n̄, the absorption rate per ground-state atom, and the steady N_b/Nₐ with only this light and spontaneous emission acting. What cavity temperature gives the same rates?

A beam of irradiance I carries energy density I/c, so per unit wavelength u_λ = 0.80 × 10⁹ W m⁻³ ÷ 2.998 × 10⁸ m s⁻¹ = 2.67 J m⁻⁴. For a J = ½ ground level the unpolarised-beam rate equals the isotropic one at the same energy density, so 13.02's B applies.
Change variable: ρ(ω) = u_λ|dλ/dω| = u_λλ²/(2πc) = 2.67 J m⁻⁴ × (780.2 × 10⁻⁹ m)² ÷ (2π × 2.998 × 10⁸ m s⁻¹) = 8.62 × 10⁻²² J s m⁻³.
With ω₀ = 2πc/λ = 2.414 × 10¹⁵ rad s⁻¹, one photon per mode corresponds to ħω₀³/(π²c³) = 5.58 × 10⁻¹⁵ J s m⁻³, so n̄ = 8.62 × 10⁻²² ÷ 5.58 × 10⁻¹⁵ = 1.55 × 10⁻⁷.
A = 1/(26.2 ns) = 3.82 × 10⁷ s⁻¹. Absorption per ground atom: B_ab ρ = (g_b/gₐ)A n̄ = 2 × 3.82 × 10⁷ s⁻¹ × 1.55 × 10⁻⁷ = 11.8 s⁻¹. Stimulated emission per excited atom, averaged over the four upper sublevels, is half that, 5.9 s⁻¹, negligible beside A.
Set dN_b/dt = 0: N_b/Nₐ = B_ab ρ/(A + B_ba ρ) = (g_b/gₐ)n̄/(1 + n̄) = 3.09 × 10⁻⁷. Limiting case: as n̄ → ∞ the ratio climbs only to g_b/gₐ = 2, equal population per sublevel and never an inversion.
Thermal light with the same n̄ has e^x = 1 + 1/n̄, so x = ln(6.47 × 10⁶) = 15.68 and T = hc/(λkx) = 0.014388 m K ÷ (780.2 × 10⁻⁹ m × 15.68) = 1.18 × 10³ K. The beam is not thermal; this only labels its strength at one frequency.

Answern̄ = 1.55 × 10⁻⁷; absorption 11.8 s⁻¹ per ground-state atom; N_b/Nₐ = 3.1 × 10⁻⁷; a cavity at about 1.18 × 10³ K would drive the same rates.