
Physics V course sequence · Interactive lesson
Line profiles and Doppler-free spectroscopy
Every optical line you measure is a homogeneous Lorentzian smeared over a distribution of velocities. Learn to take that convolution apart — fit the Voigt, read the natural width off the wings — and then get rid of the Doppler smear outright with a counterpropagating beam.
UPV-13.09 · Lesson 9 of 9 in this unit. Position is not a completion score.Build the model
One profile, two mechanisms — separate them before you quote a linewidth.
The key idea
A line profile is the distribution of absorption strength over frequency, and Doppler-free spectroscopy is any scheme that reports the homogeneous width of that profile by making the whole sample respond as a single velocity class.
Example. Sodium's D2 line at 589.0 nm has a 9.8 MHz natural width from its 16.2 ns lifetime, but a 300 K cell shows 1.32 GHz; a saturated-absorption dip in that same cell returns about 15 MHz at I = 3 Iₛₐₜ.
The complete physical model and assumptions
An optical line carries two widths of different kinds. Spontaneous decay truncates the excited-state amplitude as exp(−Γt/2), so its Fourier transform is a Lorentzian of FWHM Γ = 1/τ that every atom carries identically — homogeneous, and beyond reach. The first-order Doppler shift instead relabels atoms: one moving at vz is resonant at ν₀(1 + vz/c), so a Maxwell–Boltzmann spread over vz widens the ensemble into a Gaussian of FWHM (ν₀/c)√(8 ln2 kBT/M) while broadening no single atom at all. What a spectrometer records is the convolution, the Voigt profile, and in a 300 K sodium cell the Gaussian is 134 times the Lorentzian: it owns the core and essentially all of the width, while the Lorentzian owns the wings past about 2.2 GHz. That asymmetry is the opening. You cannot select inside a homogeneous width, but you can address one velocity class at a time — a strong pump and a counterpropagating probe from one laser coincide only at line centre, leaving a narrow Lamb dip in the pedestal, while counterpropagating two-photon absorption cancels the first-order shift for every class at once. The cost is that the lineshape is no longer a rate-equation result but the steady state of the optical Bloch equations, and what you recover is not Γ but (Γ/2)[1 + √(1 + I/Iₛₐₜ)], floored further by transit time, laser jitter and collisions.
1Two widths, and only one is a property of the atom
Spontaneous decay truncates the excited-state amplitude as exp(−Γt/2), and the Fourier transform of a truncated exponential is a Lorentzian of FWHM Γ = 1/τ in angular frequency, or 1/2πτ in Hz. Every atom in the sample carries that profile identically, which is what homogeneous means: there is no subset of atoms with a narrower line. The Doppler width is a different kind of object. An atom with axial velocity vz is resonant with a laser tuned to ν₀(1 + vz/c), so the Maxwell–Boltzmann distribution over vz relabels the resonance frequency atom by atom, producing a Gaussian of FWHM (ν₀/c)√(8 ln2 kBT/M). Nothing about any individual atom has been broadened; the ensemble is simply a set of differently tuned absorbers. For sodium at 300 K the two numbers are 9.8 MHz and 1.32 GHz. The whole of Doppler-free spectroscopy exists because you can address an inhomogeneous distribution one class at a time and can never do the same to Γ.
2Convolve once, with the Faddeeva function, not with an FFT
The measured profile is the Lorentzian convolved with the Gaussian: every velocity class contributes a full Γ-wide line centred on its own shifted frequency. That convolution is the Voigt function, and it has a closed form in terms of the Faddeeva function w(z) = exp(−z²)erfc(−iz), namely V(ν) = Re[w(z)]/(σ√(2π)) with z = (ν − ν₀ + iγ)/(σ√2). In Python that is scipy.special.wofz, evaluated pointwise at whatever frequencies you like. Resist the temptation to FFT-convolve two sampled arrays instead. With γ/σ ≈ 1/114 you need a grid fine enough to resolve a 10 MHz core across a ±5 GHz span — some 2000 points before the core is sampled even twice, and of order 10⁴ to resolve its shape — and the periodic wrap-around of a discrete convolution contaminates exactly the far wings you were trying to measure. Use wofz for the model; keep the FFT for the data.
3The core belongs to the Gaussian, the wings to the Lorentzian
Far from line centre the Gaussian falls as exp(−x²/2σ²)/(σ√(2π)) and the Lorentzian only as γ/πx², so however small γ is, the Lorentzian eventually wins. For sodium at 300 K, with σ = 559 MHz and γ = 4.9 MHz, the crossing sits at 3.92σ — about 2.19 GHz from centre, where the profile has dropped to 1.1 × 10⁻³ of its peak. Two consequences follow for fitting. The Olivero–Longbothum width fV ≈ 0.5346 fL + √(0.2166 fL² + fG²) says a 9.8 MHz Lorentzian widens a 1317 MHz Gaussian to 1322 MHz, a change of 0.4 per cent: a Voigt fitted to the core alone cannot determine γ, because the fit trades it against σ and against your baseline. To pull γ out you either need three or four decades of dynamic range in the wings, or you fix fG from a measured temperature and let the fit carry fL alone.
4Saturated absorption: burn a hole, then look through it
Split one laser into a strong pump and a weak counterpropagating probe. At detuning δ the pump is resonant with atoms moving at vz = +δ/k and the probe with atoms at vz = −δ/k, so for any δ ≠ 0 the two beams talk to different halves of the velocity distribution and the probe reports the undisturbed Doppler profile. At δ = 0 both address the same class, vz = 0. The pump has already driven a fraction (S/2)/(1+S) of that class into the excited state, so the probe finds fewer atoms to absorb and more to stimulate emission: the population difference falls to 1/(1+S) of its unpumped value and the probe absorption dips. The hole is Γ√(1+S) wide, but the probe brings its own Γ-wide response, and tuning the laser sweeps hole and probe class through vz in opposite directions, so the dip traced out in detuning is (Γ/2)[1 + √(1+S)] wide — Γ at low power — and its depth, 1 − 1/√(1+S), creeps towards the full pedestal as the pump is raised. Watch for crossover resonances: when one lower level feeds two upper levels separated by Δ, the single class vz = Δ/2k is pumped on one transition and probed on the other, giving a dip midway between the two real ones — often the strongest feature in the trace, and belonging to no level pair at all.
5Two photons from opposite sides cancel the shift for everyone
Saturated absorption throws away all but one velocity class. Two-photon spectroscopy keeps them all. Let an atom absorb one photon from each of two counterpropagating beams: the resonance condition is (ω₁ + ω₂) − (k₁ + k₂)⋅v = ωeg, and with k₂ = −k₁ the velocity term vanishes identically, for every v. Every atom in the cell contributes to one line at ωeg/2, so there is no pedestal to sit in — only a weak broad background from photon pairs taken out of the same beam. What survives is second order: time dilation shifts the resonance by −ν₀v²/2c², which for a 1 km s⁻¹ hydrogen beam is −13.7 kHz on the 1S–2S line, ten thousand natural widths, a systematic to be modelled rather than averaged away. The selection rules change too: two dipole steps means no net parity change, so ΔL = 0 or ±2, and the one-photon-forbidden 1S–2S transition becomes the whole point.
6What sets the floor once the Doppler width is gone
At this level the rate equations run out: the pump's coherence ρeg matters, and the honest model is the optical Bloch equations, ρ̇ = −(i/ħ)[H, ρ] + relaxation, with H carrying the Rabi frequency Ω = d⋅E/ħ. Their steady state is ρₑₑ = (S/2)/(1 + S + 4δ²/Γ²) with S = 2Ω²/Γ², which is where Γ√(1+S) comes from; push Ω past Γ and the same equations give the Autler–Townes splitting that no rate equation contains. Then count the remaining floors. Transit time: an atom crossing a waist w is driven for only w/v, giving roughly 0.4 v/w — 0.21 MHz for sodium at 526 m s⁻¹ through a 1 mm waist, but 400 kHz for a 1 km s⁻¹ hydrogen beam. Pressure broadening runs at tens of MHz per torr. And the laser's own linewidth convolves in directly, which is why a 9.8 MHz dip needs a laser locked well below it.
Key equations
Read the units, assumptions and limits before substituting.
Natural Lorentzian from spontaneous decay
L(ω) = (Γ/2π) / [(ω − ω₀)² + Γ²/4]
Homogeneous: every atom carries it, so no cooling and no velocity selection removes it. Na 3p, τ = 16.2 ns, gives 9.8 MHz.
Γ = 1/τ in rad s⁻¹; the FWHM is Γ in ω, or Δν = 1/2πτ in HzDoppler Gaussian FWHM
ΔνD = (ν₀/c) √(8 ln2 · kB T / M)
Inhomogeneous: a label on velocity class, not a property of any atom. Na at 300 K gives 1.32 GHz, 134 times the natural width.
M is the atom mass in kg and T in K; the Gaussian σ is ΔνD/2.3548Voigt profile via the Faddeeva function
V(ν) = Re[w(z)] / (σ√(2π)), z = (ν − ν₀ + iγ)/(σ√2)
One line of Python — wofz(z).real/(σ*√(2*π)) — with no grid, no wrap-around, and the wings exact.
γ is the Lorentzian HWHM in Hz, σ the Gaussian standard deviationVoigt FWHM, Olivero–Longbothum
fV ≈ 0.5346 fL + √(0.2166 fL² + fG²)
Invert it to get fL from a fitted fV and a temperature-fixed fG: the deconvolution you actually perform.
all three are FWHMs in one unit; the approximation is good to 0.02 per centSaturation and power broadening
ρₑₑ = (S/2) / (1 + S + 4δ²/Γ²), FWHM = Γ√(1 + S)
This is the hole the pump burns in the velocity distribution. The weak probe's own Γ-wide response is swept across it, so the Lamb dip is (Γ/2)[1 + √(1 + S)] wide, and its depth 1 − 1/√(1 + S) tends to the full pedestal, not to half.
S = I/Iₛₐₜ and δ = ωL − ω₀; Iₛₐₜ = πhc/3λ³τ is 6.3 mW cm⁻² for Na D2Two-photon Doppler cancellation
(k₁ + k₂)⋅v = 0 for k₂ = −k₁ · Δν/ν₀ = −v²/2c²
Kills the pedestal outright, but leaves −13.7 kHz on hydrogen 1S–2S for a 1 km s⁻¹ beam.
first order cancels for every velocity class; the residual term is time dilationInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Raise the saturation parameter from 0.25 to 8: the dip deepens from 0.11 to 0.67 of the pedestal, but its width grows as (Γ/2)[1 + √(1+S)], from 10 to 20 MHz — power broadening is what a deep dip costs. Then sweep the Doppler width; the pedestal flattens and the dip does not move.
Live results
- LAMB DIP FWHM (Γ/2)[1+√(1+S)]
- 13.4 MHz
- DIP DEPTH 1−1/√(1+S)
- 0.42 of pedestal
- DOPPLER / DIP WIDTH
- 97 ×
- VELOCITY CLASS Δv
- 10.0 m/s
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. Doppler FWHM of the pedestal: 1300 MHz; Saturation parameter S = I/Iₛₐₜ: 2.00; Natural width Γ/2π: 9.8 MHz
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The dip is not the natural line. Students read "Doppler-free" as "linewidth-free" and quote Γ/2π = 1/2πτ for whatever the saturated-absorption trace shows. Three things stand in the way. First, the pump does not select vz = 0; it burns a hole of homogeneous width Γ√(1 + I/Iₛₐₜ) into the velocity distribution, so the dip is already power-broadened — at I = 3 Iₛₐₜ the hole is twice Γ and the dip one and a half times Γ before anything else is added. Second, collisions and fast laser frequency noise each contribute their own Lorentzian, and Lorentzian widths add linearly, not in quadrature, so they accumulate faster than the Gaussian intuition expects; transit time through a Gaussian beam adds a Gaussian of its own on top. Third, some of the strongest peaks in a real trace are crossover resonances sitting midway between two transitions, corresponding to no level pair at all. Doppler-free means the inhomogeneous label has been removed, not that the homogeneous width has.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Sodium's D2 line sits at 589.0 nm and the 3p state lives 16.2 ns. For a vapour cell at 300 K (M = 22.99 u), find the natural FWHM, the Doppler FWHM and the Voigt FWHM, and say which mechanism sets the width and which sets the wings.
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.
- Natural width: the amplitude decays as exp(−t/2τ), so the intensity profile is a Lorentzian of FWHM Δνₙₐₜ = 1/(2πτ) = 1/(2π × 16.2 × 10⁻⁹ s) = 9.82 MHz.
- Doppler width: ν₀ = c/λ = 2.998 × 10⁸ / 589.0 × 10⁻⁹ = 5.090 × 10¹⁴ Hz, and M = 22.99 × 1.6605 × 10⁻²⁷ = 3.818 × 10⁻²⁶ kg.
- √(8 ln2 kBT/M) = √(5.545 × 1.381 × 10⁻²³ × 300 / 3.818 × 10⁻²⁶) = √(6.016 × 10⁵) = 775.7 m s⁻¹, so ΔνD = ν₀ × 775.7/c = 1.317 GHz — 134 times the natural width.
- Voigt FWHM from Olivero–Longbothum: 0.5346 × 9.82 + √(0.2166 × 9.82² + 1316.9²) = 5.25 + 1316.91 = 1322 MHz.
- So the Lorentzian widens the line by 0.4 per cent and is invisible in the core. The wings are another matter: the Gaussian term exp(−x²/2σ²)/(σ√(2π)) beats γ/πx² only out to 3.92σ, about 2.19 GHz from centre, where the profile is 1.1 × 10⁻³ of peak — beyond that the line is pure natural wing.
Answer. Δνₙₐₜ = 9.8 MHz, ΔνD = 1.32 GHz, ΔνVoigt = 1.32 GHz. The Gaussian sets the core and effectively all of the width; the Lorentzian sets the wings beyond about 2.2 GHz from centre.
MediumExample 2
A sodium saturated-absorption cell is pumped by a beam of 2.0 mm diameter carrying 0.40 mW. Take Iₛₐₜ = πhc/(3λ³τ) for the D2 line, with λ = 589.0 nm, τ = 16.2 ns and Γ/2π = 9.82 MHz. Find the saturation parameter and the Lamb-dip width, add the transit-time contribution for a mean speed of 526 m s⁻¹, and find the pump power that would bring the dip to 12.0 MHz.
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.
- Iₛₐₜ = πhc/(3λ³τ) = (π × 6.626 × 10⁻³⁴ × 2.998 × 10⁸) / (3 × (5.890 × 10⁻⁷)³ × 1.62 × 10⁻⁸) = 6.24 × 10⁻²⁵ / 9.93 × 10⁻²⁷ = 62.8 W m⁻² = 6.28 mW cm⁻².
- Beam area = π(0.10 cm)² = 0.03142 cm², so I = 0.40 / 0.03142 = 12.73 mW cm⁻² and S = I/Iₛₐₜ = 2.03.
- Power broadening: the pump burns a hole Γ√(1+S)/2π = 9.82 × √3.03 = 9.82 × 1.740 = 17.1 MHz wide. The weak probe's own 9.82 MHz response is swept across it in the opposite direction, so the Lamb dip is (9.82/2) × (1 + 1.740) = 4.91 × 2.740 = 13.5 MHz wide.
- Transit time for a Gaussian beam of waist w = 1.0 mm: Δνₜₜ ≈ 0.4 v̄/w = 0.4 × 526 / 1.0 × 10⁻³ = 0.21 MHz. A Gaussian beam gives a Gaussian transit profile, so it combines with the Lorentzian dip through Olivero–Longbothum: 0.5346 × 13.45 + √(0.2166 × 13.45² + 0.21²) = 13.45 MHz, a change of 0.004 MHz.
- Transit time is negligible, so a 12.0 MHz dip needs 4.91 × [1 + √(1+S)] = 12.0: √(1+S) = 12.0/4.91 − 1 = 1.444, S = 1.085, I = 1.085 × 6.28 = 6.81 mW cm⁻².
- Pump power P = I × area = 6.81 × 0.03142 = 0.214 mW, a factor of 1.9 below the original.
Answer. S = 2.03, so the pump burns a 17.1 MHz hole and the probe sees a 13.5 MHz Lamb dip, to which the 0.21 MHz Gaussian transit width adds almost nothing. Cutting the pump to 0.214 mW gives 12.0 MHz; the natural 9.8 MHz arrives only as the power goes to zero and the dip vanishes with it.
HardExample 3
Hydrogen's 1S–2S transition at ν₀ = 2.4661 × 10¹⁵ Hz is driven by two counterpropagating 243 nm photons. The 2S state decays by two-photon emission at 8.2 s⁻¹. For a thermal beam at v = 1.0 km s⁻¹ crossing a laser waist w = 1.0 mm, find the natural width as seen in laser frequency, the second-order Doppler shift and the transit-time width, and say what actually limits the measurement.
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.
- First order: the atom takes one photon from each beam, and with k₂ = −k₁ the total shift (k₁ + k₂)⋅v = 0 for every velocity class — no pedestal, no velocity selection, and the whole beam feeds one line.
- Natural width: Δνₙₐₜ = Γ/2π = 8.2/(2π) = 1.31 Hz on the transition. The laser supplies half the energy per photon, so scanning the laser the resonance is only 0.65 Hz wide.
- Second-order Doppler is time dilation and does not cancel: Δν/ν₀ = −v²/2c² = −(1.0 × 10³)²/(2 × 8.988 × 10¹⁶) = −5.56 × 10⁻¹², so Δν = −2.4661 × 10¹⁵ × 5.56 × 10⁻¹² = −1.37 × 10⁴ Hz, a 13.7 kHz redshift — ten thousand natural widths.
- Transit time: Δνₜₜ ≈ 0.4 v/w = 0.4 × 1.0 × 10³ / 1.0 × 10⁻³ = 4.0 × 10⁵ Hz = 400 kHz, twenty-nine times the second-order Doppler shift and 3 × 10⁵ natural widths.
- Transit time is the floor. Reaching 1 kHz needs w/v ≥ 4 × 10⁻⁴ s: even a 5.8 K cryogenic beam at about 350 m s⁻¹ would need a 0.14 m waist, which is why the real experiment sends the atoms along the standing-wave axis of a build-up cavity and selects slow atoms by delayed detection — which also cuts the second-order Doppler shift by a factor of eight.
Answer. 0.65 Hz natural width in laser frequency, a −13.7 kHz second-order Doppler shift, and 400 kHz of transit-time broadening. Transit time dominates by a factor of twenty-nine, so the interaction geometry, not the Doppler effect, is what limits this measurement.
Quick check
Select an answer and explain it before checking.
Review & continue
A line profile is the distribution of absorption strength over frequency, and Doppler-free spectroscopy is any scheme that reports the homogeneous width of that profile by making the whole sample respond as a single velocity class.
- 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.