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Advanced Quantum Mechanics course sequence · Interactive lesson

Emission depends on the modes: the Purcell factor

Until now a lifetime has looked like a property of the atom, as fixed as its mass. Put the same atom in a cavity and the lifetime moves by orders of magnitude, up or down, while the atom is untouched. This topic shows why, by how much, and where the golden rule behind it runs out.

UAQM-12.08 · Lesson 8 of 9 in this unit. Position is not a completion score.
Starting model: Emission depends on the modes: the Purcell factorExcited-state population Pₑ(t) over 200 ns for an emitter at the antinode of a resonant cavity mode, with g/2π = 5.0 MHz, κ/2π = 60 MHz and Γ₀/2π = 0.4 MHz. Solid: exact solution for the atom and the leaking mode. Dashed: golden-rule decay at Γ₀ + 4g²/κ. Faint: free space alone. Dashed vertical line: cavity memory 2/κ. Now 2g/κ = 0.17.Pₑ11/e0100200 ns2/κsolid: exact, dashed: golden rulefaint: free space only2g/κ = 0.17
Start with a picture. Change the model in the lab below.
1

Build the model

The same dipole decays faster or slower depending on the modes around it.

The key idea

The Purcell factor FP is the rate at which an emitter radiates into one resonant cavity mode divided by its free-space rate Γ₀: FP = (3/(4π²))(λ³/V)Q for a dipole aligned with the mode's field at its maximum.

Example. A mode with Q = 2.0 × 10³ and V = λ³ has FP = 0.0760 × 2000 = 152, so a resonant, aligned emitter whose free-space lifetime is 20 ns puts a photon into that mode in about 20 ns/152 = 0.13 ns.

The complete physical model and assumptions

Topic 12.02 made spontaneous emission a sum over modes, and nothing in that sum required the modes to be those of empty space. The rate takes two things from the field: how strongly each mode couples, which grows with the vacuum field per photon √(ħω/(2ε₀V)) as the volume shrinks, and how many modes sit at the atomic frequency. A cavity trades free space's smooth ω² density for a single mode whose leakage spreads it into a Lorentzian of width κ = ω/Q, so on resonance it can offer far more density than free space, and off resonance or below a cutoff far less. Put that density into the golden rule and the rate into the mode is 4g²/κ, which against free space is the Purcell factor (3/(4π²))(λ³/V)Q: a lifetime belongs to the atom and its surroundings together. The result leans on topic 12.06's Markov step, which needs the cavity to forget faster than the atom decays. Because Γcav/κ = (2g/κ)², raising Q eventually breaks that step: once g passes about κ/4 the photon is re-absorbed before it leaks out, decay turns into Unit 11's reversible vacuum Rabi oscillation, and there is no longer a rate for the golden rule to give.

1Keep the golden rule, change the modes

SI units throughout, with every rate and linewidth an angular frequency in s⁻¹; nothing here is relativistic, so no metric signature enters. Topic 12.02 summed the golden rule, Γ = 2π Σ|gₖ|² δ(ωₖ − ω₀), over free-space modes and found Γ₀ = ω³d²/(3πε₀ħc³). Nothing in that sum belonged to the atom alone: it needed the coupling to each mode, ħgₖ = d⋅Evac, and the number of modes at ω₀. A cavity changes both. One mode of volume V, with the dipole along its field at the maximum, has ħg = d√(ħω/(2ε₀V)), Unit 11's Jaynes-Cummings coupling. Its mirrors leak, so its energy rings down as e(−κt) with κ = ωc/Q, and its spectrum is the Lorentzian ρc(ω), normalised to one mode. If the emitter's own line is narrower than κ, ρc is smooth across it and the golden rule gives Γcav = 2πg²ρc(ω₀). At resonance ρc = 2/(πκ), so Γcav = 4g²/κ, derived. Dimensions: s⁻² over s⁻¹ is a rate. The rate grows as κ falls, because the same one mode is squeezed into a narrower band.

2The Purcell factor is a ratio of mode densities

Divide 4g²/κ by Γ₀ and the dipole cancels: FP = 6πc³Q/(ω³V) = (3/(4π²))(λ³/V)Q, derived; λ³/V and Q are pure numbers, as a ratio of rates needs. Read it as two factors. Per unit volume the cavity offers a peak density 2Q/(πωV) where free space offers ω²/(π²c³), a ratio of 2πQc³/(ω³V). The remaining 3 is orientation: averaged over directions and both polarisations, |ε̂⋅d̂|² is 1/3, so a free-space dipole collects a third of each mode's g², while the aligned cavity dipole collects all of it. Now the free-field limit. In a box far larger than λ the modes lie closer together than κ, their Lorentzians merge into the smooth ω² density, and summing 2πg²ρc(ω₀) over all of them, with each g² averaged over the box and over orientation, returns exactly Γ₀ whatever Q is. Enhancement therefore comes from isolating one mode in a small volume, not from mirrors as such. For hydrogen's 2p, Γ₀ = 6.27 × 10⁸ s⁻¹ matches the measured 1.60 ns lifetime. A mode at 121.6 nm with V = λ³ and Q = 10³, if one could be built, gives FP = 76 and a lifetime near 21 ps, with Γcav/κ = 3 × 10⁻³, safely a rate. At Q = 10⁴ the formula promises FP = 760 and 2.1 ps, but there Γcav/κ = 0.31 and g = 0.28κ is past κ/4, so the decay oscillates and no such lifetime exists.

3Detuned, displaced or tilted, the emitter loses the peak

The emitter samples ρc at its own frequency, so a detuning Δ = ω₀ − ωc gives Γcav(Δ) = (4g²/κ)/(1 + 4Δ²/κ²), derived: half a linewidth off, the rate halves. And g² is the square of d⋅E at the emitter, so a dipole tilted by θ from the local field, at a point where the field is a fraction of its peak, carries a further |E(r)/Eₘₐₓ|² cos²θ. FP is a ceiling. What the total rate does depends on what else the emitter can radiate into. An open cavity, such as two small mirrors, leaves most directions untouched, so the emitter keeps about Γ₀ through the sides and Γ = Γ₀ + Γcav: the cavity adds a channel and takes none away. A closed cavity offers nothing else, and an emitter detuned by many linewidths sees only the Lorentzian's wing, g²κ/Δ², which can lie far below Γ₀: emission is inhibited. The fraction of photons that enter the mode, β = Γcav/Γ, is what a single-photon source is built to push towards 1.

4Below a cutoff there is nothing to emit into

Suppression can be total. Between perfectly conducting plates a distance L apart, the part of E parallel to the plates must vanish on both, so a mode carrying such a field needs kz = nπ/L with n ≥ 1, and ω ≥ πc/L. With L below λ/2 the transition lies under that cutoff, a dipole parallel to the plates has no mode at all, and Γ∥ = 0, derived. A perpendicular dipole couples to the n = 0 mode, whose field is normal to the plates and uniform across the gap: with Evac² = ħω/(2ε₀AL) over an area A and Aω/(2πc²) modes per unit ω, the golden rule gives Γ⊥ = (3λ/(4L))Γ₀, derived, 1.5Γ₀ just under the cutoff. Both enhancement and inhibition have been measured, and the results are quoted. In 1983 sodium Rydberg atoms in a superconducting cavity at 340 GHz with Q near 10⁶ emitted about 500 times faster than in free space; the Purcell formula gives that with V ≈ 1.5 × 10² λ³, a mode about five wavelengths across. In 1985 caesium atoms in circular Rydberg states, between plates just under half a wavelength apart, lived at least 20 times longer.

5The golden rule predicts its own failure

Topic 12.05's equations hold unchanged with one leaking mode in place of the free-space continuum. Eliminating the photon amplitude leaves, at resonance, the kernel g² e(−κτ/2), which remembers for 2/κ, and topic 12.06's Markov step needs that memory short beside the decay time 1/Γcav. Since Γcav/κ = 4g²/κ² = (2g/κ)², the rate is self-consistent only while 2g ≪ κ. Neglecting Γ₀, differentiating once turns the memory integral into c̈ₑ + (κ/2)ċₑ + g²cₑ = 0, with roots s = −κ/4 ± √(κ²/16 − g²), derived. For g ≪ κ/4 the slow root is −2g²/κ, and the population falls at 4g²/κ: the golden rule recovered. Past g = κ/4 the roots turn complex and cₑ oscillates; as κ → 0 it becomes cos(gt), so Pₑ = cos²(gt), Unit 11's vacuum Rabi oscillation at 2g, the one-mode limit. The photon is now re-absorbed before it escapes, and the excitation, shared between atom and mode, drains at κ/2 however large g is. Lowering κ further makes emission slower, not faster.

6When the emitter's line is the broader one

Sampling ρc at one frequency assumed the emitter's line, of full width Γₑ, much narrower than κ. A free atom meets that easily, with Γ₀/2π of a few MHz against cavity widths of tens of MHz or more. An emitter in a solid at room temperature, its line broadened by phonons, may not. For a spread of emitter frequencies with a normalised Lorentzian distribution ρₑ, the rate is 2πg²∫ρc ρₑ dω, and a convolution of two Lorentzians is a Lorentzian whose widths add, as topic 12.07 found for two decaying levels. So Γcav = 4g²/(κ + Γₑ) at resonance, derived for that case; that a dephasing-broadened line gives the same is quoted. In FP the cavity's Q becomes an effective ω/(κ + Γₑ). A mode at 637 nm with Q = 1.0 × 10⁴ has κ/2π = 47 GHz; an emitter line 1.0 THz wide drops the effective Q to 449 and FP by a factor of 22, and better mirrors then buy almost nothing. The limit Γₑ → 0 returns 4g²/κ.

2

Key equations

Read the units, assumptions and limits before substituting.

Density of one leaking cavity mode

ρc(ω) = (κ/2π)/[(ω − ωc)² + κ²/4]

Derived: the power spectrum of a mode whose amplitude rings down as e(−κt/2). It stands in for free space's ω²V/(π²c³) and peaks at 2/(πκ) = 2Q/(πωc).

ρc in s per mode; ω, ωc and κ in s⁻¹; κ = ωc/Q is the rate at which the mode's energy leaks; ∫ρc dω = 1.

Emission rate into the mode at resonance

Γcav = 2πg²ρc(ω₀) = 4g²/κ

Derived at ω₀ = ωc for an aligned dipole whose line is narrower than κ. g/2π = 5 MHz and κ/2π = 50 MHz give Γcav = 1.26 × 10⁷ s⁻¹, a lifetime of 80 ns.

ħg = dEvac with Evac = √(ħω/(2ε₀V)) at the field maximum (Unit 11); g, κ in s⁻¹; d in C m; V in m³.

Purcell factor

FP = Γcav/Γ₀ = (3/(4π²))(λ³/V)Q

Derived: d² cancels, so only the cavity's geometry and Q remain. Q = 1.0 × 10⁴ in a mode of V = 10λ³ gives FP = 76; the same Q in V = λ³ gives 760.

Γ₀ = ω³d²/(3πε₀ħc³) from topic 12.02; λ = 2πc/ω in m; V in m³; Q and FP are dimensionless.

Detuned emitter

Γcav(Δ) = (4g²/κ)/(1 + 4Δ²/κ²)

Derived: a narrow emitter samples the Lorentzian at its own frequency. Half a linewidth away, Δ = κ/2, the rate halves; at Δ = κ it is a fifth.

Δ = ω₀ − ωc in s⁻¹. Off the antinode, or tilted by θ from the local field, multiply by |E(r)/Eₘₐₓ|² cos²θ.

When the golden-rule rate is valid

Γcav/κ = (2g/κ)² ≪ 1

Derived: the exact decay stays overdamped only while g is below κ/4 (for Γ₀ ≪ κ). Past it Pₑ oscillates at 2√(g² − κ²/16), which tends to Unit 11's vacuum Rabi frequency 2g, and no rate exists.

Γcav, g and κ in s⁻¹. The kernel g² e(−κτ/2) remembers for 2/κ, which must be short beside 1/Γcav.

Parallel dipole between conducting plates

Γ∥ = 0 when ω₀ lies below the cutoff πc/L

Derived: E parallel to the plates must vanish on both, which needs kz = nπ/L with n ≥ 1. Quoted: in 1985 such plates lengthened a Rydberg lifetime at least twentyfold.

L is the plate spacing in m, so the condition reads L below λ/2; perfect conductors, dipole parallel to the plates.
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

5.0 MHz
0 to 20 MHz; steps of 0.5.
60 MHz
2 to 80 MHz; steps of 2.
0.4 MHz
0.2 to 6 MHz; steps of 0.2.

Try this. At the default the dashed golden-rule curve tracks the solid exact one, five times faster than the faint free-space decay. Lower κ/2π: the golden rule keeps speeding up as 1/κ, but below κ/2π ≈ 20 MHz, where g = κ/4, the exact curve oscillates instead. Set g to zero and all three merge.

Live results

2g/κ
0.17
PURCELL FACTOR 4g²/(κΓ₀)
4.2
GOLDEN-RULE LIFETIME
77.0 ns
Pₑ OSCILLATION FREQUENCY
0.0 MHz

Results follow this lesson’s model and assumptions. Displayed values are rounded.

Interactive Advanced Quantum Mechanics modelExcited-state population Pₑ(t) over 200 ns for an emitter at the antinode of a resonant cavity mode, with g/2π = 5.0 MHz, κ/2π = 60 MHz and Γ₀/2π = 0.4 MHz. Solid: exact solution for the atom and the leaking mode. Dashed: golden-rule decay at Γ₀ + 4g²/κ. Faint: free space alone. Dashed vertical line: cavity memory 2/κ. Now 2g/κ = 0.17.Pₑ11/e0100200 ns2/κsolid: exact, dashed: golden rulefaint: free space only2g/κ = 0.17
Excited-state population Pₑ(t) over 200 ns for an emitter at the antinode of a resonant cavity mode, with g/2π = 5.0 MHz, κ/2π = 60 MHz and Γ₀/2π = 0.4 MHz. Solid: exact solution for the atom and the leaking mode. Dashed: golden-rule decay at Γ₀ + 4g²/κ. Faint: free space alone. Dashed vertical line: cavity memory 2/κ. Now 2g/κ = 0.17.

Predict → change one variable → check

Plan a change before running it. Every other input stays at the saved baseline.

Saved baseline. Coupling g/2π: 5.0 MHz; Cavity linewidth κ/2π: 60 MHz; Free-space rate Γ₀/2π: 0.4 MHz

Baseline 5.0; choose 0 to 20, in steps of 0.5.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

Catch the common trap

FP = (3/(4π²))(λ³/V)Q looks like a dial with no end stop: double Q, double the rate, forever. The formula carries its own expiry date. Q enters only through κ = ω/Q, and 4g²/κ is a golden-rule rate, valid only while the photon leaves the cavity faster than the atom can put it there, Γcav ≪ κ, which is 2g ≪ κ. Raise Q far enough and the predicted Γcav overtakes κ. The photon is then re-absorbed before it escapes, Pₑ oscillates at the vacuum Rabi frequency 2g, and the excitation drains out at about (κ + Γ₀)/2, which falls as Q rises: past that point a better cavity makes emission slower. Nor is the enhancement stimulated emission. The mode holds no photons; what a small V buys is a larger vacuum field per photon, √(ħω/(2ε₀V)).

5

Practice & worked examples

3 authored examples. Make an attempt, then reveal the reasoning one step at a time.

EasyExample 1

An emitter radiating at λ = 850 nm, with free-space lifetime τ₀ = 30.0 ns, sits with its dipole along the field at the antinode of a resonant mode with Q = 4.0 × 10³ and V = 2.5λ³. The cavity is open, so the other modes keep the free-space rate. Find the Purcell factor and the new lifetime, and check that the golden rule applies.

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.

MediumExample 2

A mode at λ = 780 nm has Q = 1.2 × 10⁴ and V = 3.0λ³. An emitter is placed where the mode's field amplitude is 0.70 of its maximum, with its dipole at 30° to the local field, and its transition frequency lies 20 GHz above the cavity resonance. Find Γcav/Γ₀, and the fraction of emitted photons that enter the mode if the sides keep Γ₀.

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.

HardExample 3

A two-level emitter at λ = 780 nm with free-space lifetime τ₀ = 26.2 ns sits, aligned, at a field maximum of the TEM₀₀ mode of a Fabry-Perot cavity of length L = 100 μm and waist w₀ = 20 μm, with mode volume V = πw₀²L/4. The mirrors give Q = 1.0 × 10⁸. Find g, κ and the golden-rule rate 4g²/κ, decide whether that rate describes the emission, and find the Q below which the decay stops oscillating.

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.

6

Quick check

Select an answer and explain it before checking.

An emitter at the antinode of a resonant mode has g/2π = 5.0 MHz in a cavity with κ/2π = 50 MHz, and its decay into other modes is negligible. New mirrors raise Q a hundredfold with the mode volume unchanged. What happens to the emission?
7

Review & continue

The Purcell factor FP is the rate at which an emitter radiates into one resonant cavity mode divided by its free-space rate Γ₀: FP = (3/(4π²))(λ³/V)Q for a dipole aligned with the mode's field at its maximum.

  • 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.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.