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Advanced Quantum Mechanics course sequence · Interactive lesson

The emitted line: a Lorentzian from the photon amplitudes

This is where the line in every spectrum stops being a quoted shape. We let the decaying atom fill the field mode by mode, read the line off the photon amplitudes, and get its shape, width, centre and normalisation from one calculation, then watch it form in time and find where it stops being a Lorentzian.

UAQM-12.07 · Lesson 7 of 9 in this unit. Position is not a completion score.
Starting model: The emitted line: a Lorentzian from the photon amplitudesPhoton spectrum Γ₂ dP/dω against detuning (ω − ω₀′)/Γ₂ from the shifted line centre, for an upper level decaying at Γ₂ to a lower level decaying at Γ₁ = 0.00 Γ₂. Solid: the photon in the field at Γ₂t = 2.0, area 0.865. Dashed: the t → ∞ Lorentzian, area 1, whose full width 1.00 Γ₂ is marked by the bar at half height.spectrum Γ₂ dP/dωdashed: t → ∞solid: at Γ₂t = 2.0Γ₁ = 0.00 Γ₂area 0.865bar: FWHM 1.00 Γ₂−50+5(ω − ω₀′)/Γ₂
Start with a picture. Change the model in the lab below.
1

Build the model

Read the line shape off the photon the atom leaves behind, not off an assumed decay.

The key idea

The natural emission line is the distribution over frequency of the photon a decaying atom leaves in the field; computed from the photon amplitudes, it is a normalised Lorentzian of full width Γ centred on the radiatively shifted frequency ω₀′.

Example. Hydrogen's 2p level lives 1.596 ns, so Γ = 6.27 × 10⁸ s⁻¹ and the 121.6 nm Lyman-α line is a Lorentzian of full width Γ/2π = 99.7 MHz: half its photons fall within ±49.9 MHz of the centre, yet 3.2% lie more than ten widths out.

The complete physical model and assumptions

The line an atom emits is the probability distribution of the photon it leaves in the field, so the honest way to find its shape is to look at the photon amplitudes themselves. In the one-excitation sector each mode amplitude is driven only by the atom, iċₖ = gₖ* cₑ e(i(ωₖ − ω₀)t), and once topic 12.06's Markov step has made cₑ = e(−(Γ/2 + iΔ)t), that equation integrates in one line. The long-time probability in mode k is |gₖ|²/[(ωₖ − ω₀′)² + Γ²/4]: a Lorentzian whose width is the same Γ that empties the upper level, and whose centre is the shifted frequency ω₀′ = ω₀ + Δ, because the atom radiates at the frequency it actually oscillates at. Summed over modes, with the coupling-weighted mode density G(ω) frozen at its resonant value Γ/2π, the field holds exactly 1 − e(−Γt) at every instant: what the atom has lost. Quantum Mechanics II reached the same curve by Fourier transforming an exponential it assumed; here nothing is assumed after the Markov step, and the same calculation shows the line forming through broad, fringed spectra, and the widths of two decaying levels adding. The model stops in the far wings, where G(ω) grows as a power of ω and vanishes below zero frequency: the Lorentzian, good to a few parts in 10⁸ near the centre of Lyman-α, is wrong by per cent a nanometre away, where fewer than two photons in a million lie.

1Integrating the photon amplitudes against the decaying atom

SI units with ħ explicit; nothing here is relativistic, so no metric signature enters. Quoted from topic 12.05: in the one-excitation sector |ψ(t)⟩ = cₑ(t) e(−iω₀t)|e, 0⟩ + Σₖ cₖ(t) e(−iωₖt)|g, 1ₖ⟩, and in the rotating-wave approximation iċₖ = gₖ* cₑ e(i(ωₖ − ω₀)t) exactly, with ħgₖ = ⟨e, 0|V|g, 1ₖ⟩. Quoted from topic 12.06: the Markov step gives cₑ(t) = e(−(Γ/2 + iΔ)t). Insert it, with no further approximation, and integrate from cₖ(0) = 0: cₖ(t) = gₖ* [1 − e(i(ωₖ − ω₀′)t − Γt/2)]/(ωₖ − ω₀′ + iΓ/2), with ω₀′ = ω₀ + Δ. Dimensions: gₖ and the denominator are both s⁻¹, so cₖ is a pure number. Two features are fixed already. The centre is ω₀′, not ω₀, because the phase e(−iΔt) of cₑ beats against e(i(ωₖ − ω₀)t) in the integrand: the atom radiates at the frequency it actually oscillates at. The width is Γ because the envelope e(−Γt/2) switches the source off, and a source lasting about 2/Γ cannot radiate a narrower line.

2Watching the line form

Take the modulus squared and write it with real functions: |cₖ(t)|² = |gₖ|² [1 − 2e(−Γt/2) cos(δₖt) + e(−Γt)]/(δₖ² + Γ²/4), with δₖ = ωₖ − ω₀′. Early-time limit: for Γt ≪ 1 the bracket becomes 2(1 − cos δₖt), giving |gₖ|² sin²(δₖt/2)/(δₖ/2)², the broad sinc² of Quantum Mechanics II's golden rule, growing as t² at resonance. Late-time limit: the bracket tends to 1 and the Lorentzian is left. In between, the centre fills as (1 − e(−Γt/2))², the square of an amplitude built at Γ/2, so at t = 1/Γ it stands at 15% of its final height although the photon is already in the field with probability 0.63; at 2/Γ it reaches 40%, at 8/Γ 96%. Off resonance the fringes, 2π/t apart, overshoot the final curve by up to a factor (1 + e(−Γt/2))², and that excess flows back through the atom to be re-emitted nearer ω₀′. Single-mode probabilities are not monotonic in time; only their sum is.

3Summing over modes: the population lost is found in the field

To sum over box modes, both polarisations and all directions, write Σₖ |gₖ|² F(ωₖ) = ∫ dω G(ω) F(ω), with G(ω) = Σₖ |gₖ|² δ(ω − ωₖ), the coupling-weighted mode density in s⁻¹; topic 12.06's rate is Γ = 2πG(ω₀). G changes on the scale ω₀ and the line on the scale Γ, tens of millions of times smaller for an optical line, so set G = Γ/2π across the line. Then ∫ dδ/(δ² + Γ²/4) = 2π/Γ and ∫ dδ cos(δt)/(δ² + Γ²/4) = (2π/Γ) e(−Γt/2) give Σₖ |cₖ(t)|² = 1 − 2e(−Γt) + e(−Γt) = 1 − e(−Γt), derived: at every instant the field holds exactly the population the atom has lost, the check that no probability leaks out of the one-excitation sector. At long times dP/dω = (Γ/2π)/[(ω − ω₀′)² + Γ²/4], in s, with area 1, peak 2/(πΓ), and half height where (ω − ω₀′)² = Γ²/4, so the full width is Γ. Half the photons lie inside that width and 3.2% beyond ten widths: the 1/δ² tails are heavy.

4What one calculation buys over Quantum Mechanics II

Quantum Mechanics II assumed an amplitude e(−Γt/2) e(−iω₀t), Fourier transformed it, took Γ from the Einstein A coefficient and put the centre at the unperturbed ω₀. Nothing in that recipe guaranteed that the line held one photon, that its width was the decay rate rather than merely resembling it, or where its centre sat. Here all three follow from one G(ω): Γ = 2πG(ω₀) both empties the level and sets the width, and the centre carries Δ = P∫ dω G(ω)/(ω₀ − ω), the principal value topic 12.06 found divergent and left for Unit 13. The line centre is where a radiative shift becomes observable. In a real atom the ground level shifts too, through the counter-rotating terms the rotating-wave approximation dropped, so a measured centre is the difference of two shifted levels. Measured check: sodium's D2 upper level has a lifetime of 16.25 ns, measured by time-resolved fluorescence, so Γ/2π = 9.79 MHz, and the width measured directly on the line in laser-cooled sodium agrees with it to a few tenths of a per cent.

5Two decaying levels: the widths add

For a line from level 2 to a lower level 1 that itself decays, add amplitudes c₁ₖ, atom in 1 with photon k, and c₀ₖₗ, atom in 0 with photons k and l. The Markov step applied to level 1 removes the l photons and leaves ċ₁ₖ = −i gₖ* c₂ e(i(ωₖ − ω₂₁)t) − (Γ₁/2 + iΔ₁) c₁ₖ, with c₂ = e(−(Γ₂/2 + iΔ₂)t). Its solution is a difference of two exponentials, and the probability that the first photon went into mode k is |c₁ₖ|² plus what has already passed on, Γ₁∫|c₁ₖ|² dt. At long times that is |gₖ|² [(Γ₁ + Γ₂)/Γ₂]/[(ωₖ − ω₂₁′)² + (Γ₁ + Γ₂)²/4], derived in outline: summed with G = Γ₂/2π it has area 1, full width Γ₁ + Γ₂, and centre ω₂₁′ = ω₂₁ + Δ₂ − Δ₁, the difference of the shifted levels. Limits: Γ₁ → 0 returns the two-level line, and Γ₂ → 0 leaves a width Γ₁. If level 2 has other decay channels the line's area is its branching ratio and its width is unchanged. The figure's second slider is this result.

6Where the Lorentzian stops: the far wings

Before G was frozen the long-time line read dP/dω = G(ω)/[(ω − ω₀′)² + Γ²/4]. With Unit 11's A⋅p coupling, |gₖ|² ∝ |ε⋅⟨e|p|g⟩|²/ωₖ and the mode density grows as ωₖ², so G(ω) = (Γ/2π)(ω/ω₀) for ω ≥ 0; written with E⋅d it is (Γ/2π)(ω/ω₀)³. The two agree on resonance and part in the wings, a sign that the two-level truncation, not only the Lorentzian, fails there; Lamb met the same ambiguity in 1952 fitting hydrogen line shapes. Either way the error is of order |ω − ω₀|/ω₀: 4 × 10⁻⁸ one width from the centre of Lyman-α, about a per cent at 1 nm. And G vanishes for ω ≤ 0, while the Lorentzian puts Γ/(2πω₀) = 6.4 × 10⁻⁹ of its photons at negative frequency; that lower edge is what makes decay non-exponential at very late times (quoted, Khalfin 1958). Measured check: damped Lyman-α absorbers in quasar spectra are fitted with this Lorentzian and Γ = 6.27 × 10⁸ s⁻¹ several ångströms from the centre, some 10⁵ widths out.

2

Key equations

Read the units, assumptions and limits before substituting.

Photon amplitude from the decaying atom

cₖ(t) = gₖ* [1 − e(i(ωₖ − ω₀′)t − Γt/2)] / (ωₖ − ω₀′ + iΓ/2)

Derived: iċₖ = gₖ* cₑ e(i(ωₖ − ω₀)t) integrated from cₖ(0) = 0 with cₑ = e(−(Γ/2 + iΔ)t). Width and centre both sit in one pole, at ωₖ = ω₀′ − iΓ/2.

gₖ the mode coupling, Γ the decay rate, ωₖ and ω₀′ = ω₀ + Δ angular frequencies, all in s⁻¹; cₖ is dimensionless

Time-resolved photon spectrum

|cₖ(t)|² = |gₖ|² [1 − 2e(−Γt/2) cos(δₖt) + e(−Γt)] / (δₖ² + Γ²/4), with δₖ = ωₖ − ω₀′

Derived: the amplitude's squared modulus, in real form. Its fringes lie 2π/t apart, so a line emitted for a time t is no sharper than about 2π/t and takes its final shape only once Γt ≫ 1.

δₖ and Γ in s⁻¹, t in s; for Γt ≪ 1 it becomes |gₖ|² sin²(δₖt/2)/(δₖ/2)², the golden rule's start

The emitted line at long times

dP/dω = (Γ/2π) / [(ω − ω₀′)² + Γ²/4]

Derived, not assumed: shape, width and centre come from one amplitude. The width is the same Γ that empties the level, so a lifetime τ gives a full width of 1/(2πτ) in hertz.

P per unit angular frequency, in s; full width Γ in s⁻¹, or Γ/2π in Hz; peak 2/(πΓ); takes G(ω) ≈ G(ω₀) = Γ/2π

The field holds what the atom lost

Σₖ |cₖ(t)|² = 1 − e(−Γt) = 1 − |cₑ(t)|²

Derived from ∫ dδ cos(δt)/(δ² + Γ²/4) = (2π/Γ) e(−Γt/2). With G flat the Markov solution keeps the norm at every instant, so the finished line has area exactly 1.

Σₖ |gₖ|² F(ωₖ) → ∫ dω G(ω) F(ω), with G(ω) = Σₖ |gₖ|² δ(ω − ωₖ) in s⁻¹; both sides dimensionless

Two decaying levels: the widths add

dP/dω = [(Γ₁ + Γ₂)/2π] / [(ω − ω₂₁′)² + (Γ₁ + Γ₂)²/4]

Derived in outline: the lower level damps the amplitude c₁ₖ at Γ₁/2. The limit Γ₁ → 0, a stable lower level, returns the line above; Γ₂ → 0 does not make it sharper than Γ₁.

Γ₂ and Γ₁ total decay rates of the upper and lower level, in s⁻¹; ω₂₁′ = ω₂₁ + Δ₂ − Δ₁; area 1 if 2 decays only to 1

The line before G is frozen

dP/dω = G(ω) / [(ω − ω₀′)² + Γ²/4], with G(ω) = (Γ/2π)(ω/ω₀) for ω ≥ 0 and 0 below

Derived from Unit 11's mode sum. One width from Lyman-α's centre the Lorentzian is off by Γ/ω₀ = 4 × 10⁻⁸ (A⋅p); 1 nm out, by 0.8% (A⋅p) or 2.4% (E⋅d), and 1.6 × 10⁻⁶ of photons lie beyond.

A⋅p coupling: |gₖ|² ∝ 1/ωₖ times a mode density ∝ ωₖ²; written with E⋅d, G ∝ (ω/ω₀)³; G in s⁻¹
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

2.0
0.5 to 8; steps of 0.5.
0.00
0 to 3; steps of 0.25.

Try this. Set Γ₁/Γ₂ = 0 and Γ₂t = 0.5: the early line is low and broad. Step the time up: fringes 2π/t apart appear, overshoot the dashed curve and sink back, while the centre fills as (1 − e(−Γ₂t/2))². Then raise Γ₁/Γ₂: the final line widens to Γ₁ + Γ₂, its area still 1.

Live results

PHOTON IN THE FIELD
0.865
FINAL FWHM Γ₁ + Γ₂
1.00 Γ₂
FINAL PEAK Γ₂ dP/dω
0.637
FRINGE SPACING 2π/t
3.14 Γ₂

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

Interactive Advanced Quantum Mechanics modelPhoton spectrum Γ₂ dP/dω against detuning (ω − ω₀′)/Γ₂ from the shifted line centre, for an upper level decaying at Γ₂ to a lower level decaying at Γ₁ = 0.00 Γ₂. Solid: the photon in the field at Γ₂t = 2.0, area 0.865. Dashed: the t → ∞ Lorentzian, area 1, whose full width 1.00 Γ₂ is marked by the bar at half height.spectrum Γ₂ dP/dωdashed: t → ∞solid: at Γ₂t = 2.0Γ₁ = 0.00 Γ₂area 0.865bar: FWHM 1.00 Γ₂−50+5(ω − ω₀′)/Γ₂
Photon spectrum Γ₂ dP/dω against detuning (ω − ω₀′)/Γ₂ from the shifted line centre, for an upper level decaying at Γ₂ to a lower level decaying at Γ₁ = 0.00 Γ₂. Solid: the photon in the field at Γ₂t = 2.0, area 0.865. Dashed: the t → ∞ Lorentzian, area 1, whose full width 1.00 Γ₂ is marked by the bar at half height.

Predict → change one variable → check

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

Saved baseline. Elapsed time Γ₂t: 2.0; Lower-level width Γ₁/Γ₂: 0.00

Baseline 2.0; choose 0.5 to 8, 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

A long-lived upper level does not guarantee a sharp line. It is tempting to read the natural width as 1/(2πτ) of the level the photon leaves, so that hydrogen's 3s level, which lives 158 ns, should emit its 3s → 2p line 1.0 MHz wide. The line is 101 MHz wide. The photon's frequency is set by the energies of both states, and the lower one, 2p, lives only 1.60 ns. In the amplitude picture the amplitude c₁ₖ, atom in 2p and photon in mode k, is fed for as long as 3s survives but damped at Γ₁/2 by the 2p decay, and its long-time weight is a Lorentzian of full width Γ₁ + Γ₂ = (6.27 + 0.063) × 10⁸ s⁻¹. The upper lifetime sets how long the emission lasts, not how sharp the line is; a narrow line needs both levels long-lived, which is why optical clocks run on transitions from the ground state to a metastable level.

5

Practice & worked examples

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

EasyExample 1

Sodium's D2 line at 589.16 nm comes from the 3p₃/₂ level, whose measured lifetime is 16.25 ns and whose only decay is to the ground state. Treating it as a two-level emitter, find Γ, the natural full width in hertz and in wavelength, the line's quality factor, and the fraction of photons within one full width of the centre on either side.

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 ⁸⁷Rb atom is put in 5p₃/₂ at t = 0; that level has a measured lifetime of 26.24 ns and decays only to 5s. Using the time-resolved photon spectrum, find at t = 52.48 ns the probability that the photon is in the field, the spectral density at the line centre in MHz⁻¹ and as a fraction of its final value, and the spectral density 9.53 MHz from the centre compared with the final Lorentzian there.

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

Hydrogen's Lyman-α line (λ₀ = 121.567 nm; 2p decays to the stable 1s at Γ = 6.266 × 10⁸ s⁻¹) is examined 1.00 nm either side of the centre, as in a strong absorber's damping wings. How many full widths out is that? By what fraction does the true spectral density differ from the Lorentzian there for A⋅p coupling (G ∝ ω) and E⋅d coupling (G ∝ ω³)? What fraction of photons lies beyond ±1.00 nm, and what fraction does the Lorentzian put at negative frequency?

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.

6

Quick check

Select an answer and explain it before checking.

A two-level atom, prepared in |e⟩ at t = 0, decays at rate Γ. At t = 1/Γ, what is the spectral density of the photon already in the field at the shifted line centre ω₀′, as a fraction of its t → ∞ value?
7

Review & continue

The natural emission line is the distribution over frequency of the photon a decaying atom leaves in the field; computed from the photon amplitudes, it is a normalised Lorentzian of full width Γ centred on the radiatively shifted frequency ω₀′.

  • 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.