
Advanced Quantum Mechanics course sequence · Interactive lesson
The Einstein relations and Planck's law as outputs
Einstein could fix his A and B coefficients only by demanding equilibrium with thermal radiation and matching its classical high-temperature limit. With the quantised field we can run his argument backwards: the n̄ and n̄ + 1 of the photon factors fix A/B and the degeneracy rule outright, and atoms in thermal equilibrium then hand us Planck's spectrum. This is where the vacuum's 1 pays for itself.
UAQM-12.03 · Lesson 3 of 9 in this unit. Position is not a completion score.Build the model
Run Einstein's 1917 argument backwards: the photon's n̄ + 1 makes Planck's law an output.
The key idea
The Einstein relations, A/B₂₁ = ħω³/(π²c³) and g₁B₁₂ = g₂B₂₁, fix the absorption and stimulated-emission coefficients B from the spontaneous rate A; with the quantised field they follow from the photon factors n̄ and n̄ + 1, and Planck's law follows from them.
Example. In black-body radiation at 300 K a line at 10.0 µm sees n̄ = 1/(e4.796 − 1) = 8.3 × 10⁻³ photons per mode, so a non-degenerate molecule there is stimulated to emit, and absorbs, at 0.83% of its spontaneous rate A, whatever its dipole moment.
The complete physical model and assumptions
Topic 12.01 showed that the quantised coupling carries √n for absorption and √(n + 1) for emission, and topic 12.02 summed the empty-mode golden rule into A = ω³|d|²/(3πε₀ħc³). Now fill the modes. If the field is isotropic, unpolarised and broadband across the line, its mean occupation n̄ at the line frequency factors out of the same mode sum, so an atom emits at A(n̄ + 1) and absorbs at A n̄. Everything Einstein postulated in 1917 then follows by bookkeeping, with no temperature in sight: rewriting n̄ as an energy density defines B coefficients that are equal for non-degenerate levels, obey g₁B₁₂ = g₂B₂₁ once magnetic substates are counted, and stand to A in the universal ratio ħω³/(π²c³). Thermodynamics enters only at the end. If atoms held at Boltzmann populations are to stay steady under these rates, the field must hold n̄ = 1/(e(ħω/kT) − 1) photons per mode, which is Planck's law: an output, where Quantum Mechanics II took it as an input and Einstein needed equilibrium and its classical limit to fix A/B. The vacuum's 1 does the work, since without it no finite temperature balances. The picture is a rate picture, and it stops where its inputs do: a narrow coherent drive, a directed beam or a cavity breaks the broadband, isotropic, free-space assumptions, and Unit 14 recovers these rates as the population part of a master equation.
1Mean photon numbers enter as n̄ and n̄ + 1
SI units throughout, with k for Boltzmann's constant; no four-vector is needed, but where one appears the metric is (+, −, −, −). Quoted from topic 12.01: in the dipole and rotating-wave approximations, emission takes |2; n⟩ to |1; n + 1⟩ in one mode with squared matrix element (ħω/(2ε₀V))|ê*⋅d₁₂|²(n + 1), and absorption takes |1; n⟩ to |2; n − 1⟩ with n in place of n + 1. Summed over final field states these become ⟨aa†⟩ = n̄ + 1 and ⟨a†a⟩ = n̄, for any state of the field: derived. Let the field be isotropic and unpolarised, so every mode near ω₀ holds the same n̄ whatever its direction and polarisation, and broadband, so n̄ barely changes across the line. Then n̄(ω₀) leaves the mode sum, and what remains is topic 12.02's sum, A = ω₀³|d₁₂|²/(3πε₀ħc³), quoted. For non-degenerate levels each upper atom emits at A(n̄ + 1) and each lower atom absorbs at A n̄: derived. Stimulated and spontaneous emission are one matrix element, its n̄ and its 1, and n̄ → 0 returns the vacuum rate A.
2From photons per mode to energy density: the B coefficients
Einstein wrote the induced rates as B times the field's spectral energy density; take it here as u(ω), the energy per unit volume per unit angular frequency. Unit 11's mode count, ω²/(π²c³) per unit volume per unit ω with both polarisations (quoted), times ħω per photon gives u(ω) = ħω³n̄/(π²c³). Setting A n̄ = B u defines B₁₂ = B₂₁ = π²c³A/(ħω³) = π|d₁₂|²/(3ε₀ħ²), so A/B₂₁ = ħω³/(π²c³): derived. Dimensions: ħω³/c³ is J s × s⁻³ ÷ (m³ s⁻³) = J s m⁻³, which is s⁻¹ ÷ (m³ J⁻¹ s⁻²), the units of A/B. No matrix element survives, because the ratio only converts photons per mode into energy density: A/B is the u(ω) of a field with one photon per mode, where stimulated and spontaneous emission are equal. Per unit frequency ν, uν = 2πu(ω) turns it into 8πhν³/c³. Quantum Mechanics II's semiclassical golden rule with a random classical field found this same B and an A of zero: a classical field carries the n̄ but not the 1. For Lyman-α, ħω³/(π²c³) = 1.475 × 10⁻¹² J s m⁻³.
3Degenerate levels: g₁B₁₂ = g₂B₂₁
Give the lower level g₁ magnetic substates and the upper level g₂. In an isotropic, unpolarised field the direction and polarisation sums weight every Cartesian component of d alike, so each pair of substates contributes in proportion to |⟨1, m₁|d|2, m₂⟩|². By the Wigner-Eckart theorem (quoted), the sum of these over m₁ is the same for every m₂, namely S/g₂, and the sum over m₂ is the same for every m₁, namely S/g₁, where the line strength S is the sum over both. So every upper substate decays at A = ω³S/(3πε₀ħc³g₂), every lower substate absorbs at (g₂/g₁)A n̄, and B₁₂ = (g₂/g₁)B₂₁: derived, which is g₁B₁₂ = g₂B₂₁ = πS/(3ε₀ħ²), with A/B₂₁ unchanged. For Lyman-α, g₁ = 2 and g₂ = 6 with spin, or 1 and 3 without; the ratio is 3 either way, because d does not act on spin, and B₁₂ = 1.27 × 10²¹ m³ J⁻¹ s⁻². With one substate in each level the relation returns B₁₂ = B₂₁.
4Steady Boltzmann populations force Planck's law
Now hold the atoms at temperature T, by collisions or contact with the walls, so that N₂/N₁ = (g₂/g₁)e(−ħω/kT): the canonical ensemble, quoted. For radiation on this line to leave that ratio steady, the downward flow N₂A(n̄ + 1) must equal the upward flow N₁(g₂/g₁)A n̄. The g's cancel against the Boltzmann factor and A cancels outright, leaving (n̄ + 1)/n̄ = e(ħω/kT), so n̄ = 1/(e(ħω/kT) − 1): derived. Times ħω³/(π²c³) it is Planck's spectrum. Nothing about the atom survives, so every line of every atom demands the same n̄, and the field must be Planckian wherever matter can absorb. Quantum Mechanics II ran this the other way, Planck in and A/B out, and Einstein in 1917 needed equilibrium itself, with Wien's displacement law and the classical high-temperature limit, to fix both relations; here they came from the photon factors and Planck comes out. The step assumes Boltzmann populations and balance line by line, which for one line is just steady state. Cross-check: e(ħω/kT) is also p(n)/p(n + 1) for a mode in its own canonical ensemble, which gives the same n̄. Drop the 1 and N₂n̄ = N₁(g₂/g₁)n̄ balances at finite T only if n̄ = 0.
5Limits and measured numbers
Limits first. For ħω ≪ kT, n̄ = kT/(ħω) − ½ + …, so u → ω²kT/(π²c³), the classical Rayleigh-Jeans law with kT per mode, and stimulated emission outruns spontaneous by n̄ ≫ 1. For ħω ≫ kT, n̄ ≈ e(−ħω/kT), Wien's law, and spontaneous emission wins. The two are equal at n̄ = 1, where ħω = kT ln 2: at 300 K that is a wavelength of 69.2 µm, so at every shorter wavelength a room-temperature body emits mostly spontaneously, with stimulated emission under 1% of it at 10 µm, and equality at 500 nm would need 41 500 K. Then measured numbers. Integrating u over ω with ∫x³/(eˣ − 1)dx = π⁴/15 (quoted) gives U = π²k⁴T⁴/(15ħ³c³), and the flux cU/4 through a surface gives σ = π²k⁴/(60ħ³c²) = 5.670 × 10⁻⁸ W m⁻² K⁻⁴, the Stefan-Boltzmann constant that absolute radiometry measures. Per unit frequency the spectrum peaks where hν/kT = 3(1 − e(−hν/kT)), at hν = 2.821kT, which is 160.2 GHz at the 2.7255 K of the cosmic microwave background, where COBE's FIRAS instrument found the spectrum Planckian to within 50 parts per million of its peak.
6Where the rate picture stops
Each step leaned on the golden rule, quoted from Quantum Mechanics II and justified for a continuum by the Markov step of topics 12.05 and 12.06, and on a field isotropic, unpolarised and broadband across the line. Remove each in turn. A collimated, polarised beam drives each pair of substates at a rate set by |ê⋅d|² for its own direction and polarisation; only when the substates are equally populated does the average restore the isotropic B. A source narrower than the line no longer samples n̄ at ω₀ alone, and the rate becomes an overlap of its spectrum with the line shape of topic 12.07; once the Rabi frequency exceeds both the line width and the source's bandwidth, the populations oscillate coherently instead of relaxing, and no B describes them. A cavity changes the mode density, so A and A/B change with it (topic 12.08). And Boltzmann populations were an input: a laser medium with N₂/g₂ above N₁/g₁ obeys the same A and B but no longer forces Planck's n̄. Unit 14 writes these rates as the population part of a master equation that also evolves coherences.
Key equations
Read the units, assumptions and limits before substituting.
Emission in an occupied field
W↓ = A [n̄(ω₀) + 1]
Derived: topic 12.01's √(n + 1), squared and summed over modes. A n̄ is stimulated emission and A is spontaneous: one matrix element, not two processes.
W↓ and A in s⁻¹; n̄ is the mean photon number per mode at the line, the same for every direction and polarisation.Absorption from the lower level
W↑ = (g₂/g₁) A n̄(ω₀)
Derived from √n and Wigner-Eckart sums: a lower atom absorbs faster when the upper level has more substates to receive it.
Per atom in level 1, in s⁻¹; g₁ and g₂ count magnetic substates. For g₁ = g₂ it equals the stimulated-emission rate A n̄.Photons per mode as energy density
u(ω) = ħω³ n̄(ω)/(π²c³)
Quoted mode density times ħω per photon. It is the only bridge between Einstein's B, defined through u, and the field's n̄.
u per unit volume and angular frequency, in J m⁻³ s; ω²/(π²c³) modes per unit volume per unit ω, both polarisations.Einstein's A/B ratio
A/B₂₁ = ħω³/(π²c³)
Derived, with no matrix element left: A/B is the spectral energy density of one photon per mode, where stimulated emission equals spontaneous.
B in m³ J⁻¹ s⁻²; B₂₁ = π|d₁₂|²/(3ε₀ħ²) for non-degenerate levels. Per unit frequency ν the ratio is 8πhν³/c³.Degeneracy relation
g₁B₁₂ = g₂B₂₁ = πS/(3ε₀ħ²)
Derived for an isotropic, unpolarised field. B₁₂ = B₂₁ only when g₁ = g₂; for hydrogen's Lyman-α line B₁₂ = 3B₂₁.
S = Σ|⟨1, m₁|d|2, m₂⟩|² over all substates, in C² m²; B₁₂ governs absorption and B₂₁ stimulated emission.Planck's occupation, as an output
n̄ = 1/(e(ħω/kT) − 1)
Derived: A, d and the g's all cancel, so every line of every atom forces the same n̄. Times ħω³/(π²c³) it is Planck's spectrum.
k = 1.380649 × 10⁻²³ J K⁻¹; from N₂A(n̄ + 1) = N₁(g₂/g₁)A n̄ with Boltzmann populations N₂/N₁ = (g₂/g₁)e(−ħω/kT).Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Slide n̄ to the crossing: the net rate falls to about zero at Planck's occupation and the field temperature reads close to 1. Raise ħω/kT and the crossing moves toward zero, Wien's regime. Set the vacuum term to 0: both lines start at the origin and meet nowhere else, so no field balances a finite temperature.
Live results
- PLANCK n̄ AT THIS T
- 0.986
- NET UP-RATE PER ATOM
- 0.341 A
- FIELD TEMPERATURE ÷ T
- 1.73
- STIMULATED SHARE OF EMISSION
- 0.667
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. Level spacing ħω/kT: 0.70; Field photons per mode n̄: 2.00; Vacuum term (1 quantised, 0 classical): 1
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Because Einstein reached A/B = ħω³/(π²c³) and g₁B₁₂ = g₂B₂₁ by demanding equilibrium with black-body radiation, it is tempting to read them as facts about thermal equilibrium, which would fail for a lamp, a laser or an inverted population. They are not. Here they follow from the matrix elements √n and √(n + 1) and the free-space mode sum alone, with no temperature anywhere, so they hold for any isotropic, broadband field and any populations; that is why laser gain is computed with the same B that governs absorption. Temperature entered only at the last step, to ask which n̄ a gas at Boltzmann populations tolerates, and the answer is Planck's. What does limit the relations is geometry, bandwidth and the density of modes: a beam acting on oriented atoms, a source narrower than the line or a cavity changes them, and a non-thermal spectrum does not.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A non-degenerate transition at λ = 2.00 µm with A = 3.0 × 10⁶ s⁻¹ sits in the black-body radiation of a furnace at 1200 K. Find the mean photon number per mode at the line, the stimulated-emission and absorption rates per atom, and the temperature at which stimulated emission would match spontaneous emission.
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.
- Photon and thermal energies: ħω = hc/λ = 1239.84 eV nm ÷ 2000 nm = 0.6199 eV, and kT = 8.617 × 10⁻⁵ eV K⁻¹ × 1200 K = 0.1034 eV, so ħω/kT = 5.995.
- Planck's occupation, derived in this lesson: n̄ = 1/(e5.995 − 1) = 1/(401.4 − 1) = 2.50 × 10⁻³. This is deep in Wien's regime, where n̄ ≈ e(−ħω/kT) = 2.49 × 10⁻³.
- Stimulated emission per upper atom runs at A n̄ = 3.0 × 10⁶ s⁻¹ × 2.50 × 10⁻³ = 7.5 × 10³ s⁻¹, and absorption per lower atom at the same rate, since the levels are non-degenerate. Spontaneous emission, at 3.0 × 10⁶ s⁻¹, is 1/n̄ = 400 times faster.
- The two kinds of emission are equal when n̄ = 1, that is when e(ħω/kT) = 2: T = ħω/(k ln 2) = 0.6199 eV ÷ (8.617 × 10⁻⁵ eV K⁻¹ × 0.6931) = 1.04 × 10⁴ K.
- Checks: n̄ is dimensionless, as a photon count must be; as T → 0 only spontaneous emission survives, and for kT ≫ ħω the occupation grows as kT/(ħω), the classical limit.
Answer. n̄ = 2.50 × 10⁻³; stimulated emission and absorption each run at 7.5 × 10³ s⁻¹ per atom against 3.0 × 10⁶ s⁻¹ spontaneous; they would match at 1.04 × 10⁴ K.
MediumExample 2
Hydrogen's Lyman-α line, 121.57 nm, joins 1s (g₁ = 2 with spin) to 2p (g₂ = 6), and each 2p substate decays at A = 6.26 × 10⁸ s⁻¹ (topic 12.02; the measured 2p lifetime of 1.60 ns agrees). Find B₂₁ and B₁₂ per unit angular frequency, check B₂₁ against Quantum Mechanics II's matrix element, convert B₂₁ to per unit frequency, and find both induced rates in isotropic black-body radiation at 20 000 K.
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.
- ω = 2πc/λ = 1.5494 × 10¹⁶ s⁻¹, so ħω³/(π²c³) = 1.0546 × 10⁻³⁴ J s × 3.720 × 10⁴⁸ s⁻³ ÷ (9.870 × 2.694 × 10²⁵ m³ s⁻³) = 1.475 × 10⁻¹² J s m⁻³.
- B₂₁ = A ÷ (ħω³/(π²c³)) = 6.26 × 10⁸ s⁻¹ ÷ 1.475 × 10⁻¹² J s m⁻³ = 4.24 × 10²⁰ m³ J⁻¹ s⁻², and B₁₂ = (g₂/g₁)B₂₁ = 3 × 4.24 × 10²⁰ = 1.27 × 10²¹ m³ J⁻¹ s⁻².
- Cross-check: for a 2p substate, |d|² summed over components is (2¹⁵/3¹⁰)e²a₀² = 0.5549 × (8.4784 × 10⁻³⁰ C m)² = 3.989 × 10⁻⁵⁹ C² m², and π|d|²/(3ε₀ħ²) = 4.24 × 10²⁰ m³ J⁻¹ s⁻², the semiclassical B, agreeing as it must.
- Per unit frequency, uν = 2πu(ω) makes B₂₁ smaller by 2π, giving 6.75 × 10¹⁹ m³ J⁻¹ s⁻², and A/B₂₁ becomes 8πhν³/c³ = 9.27 × 10⁻¹² J s m⁻³.
- At 20 000 K, kT = 1.7235 eV against ħω = 10.1986 eV, so ħω/kT = 5.917 and n̄ = 1/(e5.917 − 1) = 2.70 × 10⁻³. Each 1s atom absorbs at (g₂/g₁)A n̄ = 3 × 6.26 × 10⁸ s⁻¹ × 2.70 × 10⁻³ = 5.07 × 10⁶ s⁻¹; each 2p atom is stimulated to emit at A n̄ = 1.69 × 10⁶ s⁻¹.
- Check: balancing these flows gives N(2p)/N(1s) = 3n̄/(n̄ + 1) = 8.08 × 10⁻³, exactly the Boltzmann value 3e(−5.917), as the derivation guarantees.
Answer. B₂₁ = 4.24 × 10²⁰ m³ J⁻¹ s⁻² (6.75 × 10¹⁹ per unit frequency) and B₁₂ = 1.27 × 10²¹ m³ J⁻¹ s⁻²; at 20 000 K, n̄ = 2.70 × 10⁻³, absorption is 5.07 × 10⁶ s⁻¹ per 1s atom and stimulated emission 1.69 × 10⁶ s⁻¹ per 2p atom.
HardExample 3
Non-degenerate molecules with a transition at 10.0 µm and A = 0.50 s⁻¹ are held at 300 K by collisions with a buffer gas, inside an enclosure whose walls, at 600 K, fill it with black-body radiation. Find the net rate per molecule at which the radiation excites them and the power each absorbs. Then switch the collisions off: find the relaxation time, the populations reached and their excitation temperature.
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.
- ħω = hc/λ = 0.12398 eV = 1.986 × 10⁻²⁰ J. For the radiation ħω/kT = 0.12398 ÷ 0.05170 = 2.398, so n̄ = 1/(e2.398 − 1) = 1/(11.001 − 1) = 0.1000. For the gas ħω/kT = 4.796, so N₂/N₁ = e(−4.796) = 8.26 × 10⁻³, giving fractions N₁ = 0.99180 and N₂ = 0.00820.
- Net upward rate per molecule: A[N₁n̄ − N₂(n̄ + 1)] = 0.50 s⁻¹ × (0.099173 − 0.009015) = 0.50 s⁻¹ × 0.090158 = 0.045 s⁻¹. Most of it is absorption; stimulated and spontaneous emission from the few excited molecules take back about a tenth of it.
- Each net upward transition takes ħω from the field, so each molecule absorbs 0.04508 s⁻¹ × 1.986 × 10⁻²⁰ J = 9.0 × 10⁻²² W. It is positive because radiation hotter than the gas heats it; with the two temperatures swapped the same formula gives −0.038 s⁻¹, radiative cooling, and with them equal it gives zero.
- Without collisions, dN₂/dt = A n̄ N₁ − A(n̄ + 1)N₂ with N₁ + N₂ = 1 becomes dN₂/dt = A n̄ − A(2n̄ + 1)N₂, so the populations relax at A(2n̄ + 1) = 0.50 s⁻¹ × 1.200 = 0.600 s⁻¹, a time constant of 1.67 s.
- The steady state is N₂ = n̄/(2n̄ + 1) = 0.0833, so N₂/N₁ = n̄/(n̄ + 1) = 0.0909 = e(−2.398): the molecules take the radiation's 600 K as their excitation temperature, the Planck balance read in reverse.
- Checks: rates come out in s⁻¹ and power in W; as n̄ → 0 the relaxation rate falls to A, pure spontaneous decay, so stimulated emission and absorption speed the approach by the factor 2n̄ + 1 = 1.20.
Answer. Net excitation 0.045 s⁻¹ per molecule, absorbing 9.0 × 10⁻²² W; without collisions the populations relax at 0.600 s⁻¹ (1.67 s) to N₂/N₁ = 0.0909, an excitation temperature of 600 K.
Quick check
Select an answer and explain it before checking.
Review & continue
The Einstein relations, A/B₂₁ = ħω³/(π²c³) and g₁B₁₂ = g₂B₂₁, fix the absorption and stimulated-emission coefficients B from the spontaneous rate A; with the quantised field they follow from the photon factors n̄ and n̄ + 1, and Planck's law follows from them.
- 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.