University Physics V · Atomic Spectroscopy · 13.9

Line Profiles & 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.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

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 v_z is resonant at ν₀(1 + v_z/c), so a Maxwell–Boltzmann spread over v_z widens the ensemble into a Gaussian of FWHM (ν₀/c)√(8 ln2 k_BT/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√(1 + I/Iₛₐₜ), floored further by transit time, laser jitter and collisions.

Simple definition: 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 20 MHz at I = 3 Iₛₐₜ.

Natural Lorentzian from spontaneous decayL(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 Hz

Doppler Gaussian FWHMΔν_D = (ν₀/c) √(8 ln2 · k_B 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 Δν_D/2.3548

Voigt profile via the Faddeeva functionV(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 deviation

Voigt FWHM, Olivero–Longbothumf_V ≈ 0.5346 f_L + √(0.2166 f_L² + f_G²)

Invert it to get f_L from a fitted f_V and a temperature-fixed f_G: the deconvolution you actually perform.

all three are FWHMs in one unit; the approximation is good to 0.02 per cent

Saturation and power broadeningρₑₑ = (S/2) / (1 + S + 4δ²/Γ²), FWHM = Γ√(1 + S)

The Lamb dip's width and depth both come from here; the depth saturates at half the pedestal however hard you pump.

S = I/Iₛₐₜ and δ = ω_L − ω₀; Iₛₐₜ = πhc/3λ³τ is 6.3 mW cm⁻² for Na D2

Two-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 dilation

Two 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 v_z is resonant with a laser tuned to ν₀(1 + v_z/c), so the Maxwell–Boltzmann distribution over v_z relabels the resonance frequency atom by atom, producing a Gaussian of FWHM (ν₀/c)√(8 ln2 k_BT/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.

Convolve 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 — of order 10⁴ points before the core is sampled even twice — 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.

The core belongs to the Gaussian, the wings to the Lorentzian

Far from line centre the Gaussian falls as exp(−x²/2σ²) 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 4.6 × 10⁻⁴ of its peak. Two consequences follow for fitting. The Olivero–Longbothum width f_V ≈ 0.5346 f_L + √(0.2166 f_L² + f_G²) 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 f_G from a measured temperature and let the fit carry f_L alone.

Saturated 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 v_z = +δ/k and the probe with atoms at v_z = −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, v_z = 0. The pump has already driven a fraction (S/2)/(1+S) of that class into the excited state, so the probe finds fewer ground-state atoms and its absorption dips. The dip's width is the homogeneous width of the hole, Γ√(1+S), and its depth saturates at half the pedestal. Watch for crossover resonances: when one lower level feeds two upper levels separated by Δ, the single class v_z = 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.

Two 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 = ω_eg, and with k₂ = −k₁ the velocity term vanishes identically, for every v. Every atom in the cell contributes to one line at ω_eg/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.

What sets the floor once the Doppler width is gone

At this level the rate equations run out: the pump's coherence ρ_eg 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.

Change one variable at a time

Make the relationship visible.

Interactive model

Doppler FWHM of the pedestal1300 MHz

Saturation parameter S = I/Iₛₐₜ2.00

Natural width Γ/2π9.8 MHz

Raise the saturation parameter from 0.25 to 8: the dip deepens towards half the pedestal, but its width grows as √(1+S), from 11 to 29 MHz — power broadening is what a deep dip costs. Then sweep the Doppler width; the pedestal flattens and the dip does not move.

LAMB DIP FWHM Γ√(1+S)17.0 MHz

DIP DEPTH S/2(1+S)0.33 of pedestal

DOPPLER / DIP WIDTH77 ×

VELOCITY CLASS Δv10.0 m/s

Live interpretationLAMB DIP FWHM Γ√(1+S): 17.0 MHz. DIP DEPTH S/2(1+S): 0.33 of pedestal. DOPPLER / DIP WIDTH: 77 ×. VELOCITY CLASS Δv: 10.0 m/s

Catch the common trap

Explain before calculating.

Misconception check

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 v_z = 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 it is twice Γ before anything else is added. Second, transit time, laser jitter and collisions each contribute their own Lorentzian, and Lorentzian widths add linearly, not in quadrature, so they accumulate faster than the Gaussian intuition expects. 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.

A saturated-absorption spectrum of the sodium D2 line is taken in a 300 K cell, where the Doppler width is 1.32 GHz and the natural width is Γ/2π = 9.8 MHz. The pump runs at I = 3 Iₛₐₜ, and transit-time, collisional and laser-jitter contributions are all negligible. How wide should the Lamb dip be?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySodium'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.

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 k_BT/M) = √(5.545 × 1.381 × 10⁻²³ × 300 / 3.818 × 10⁻²⁶) = √(6.016 × 10⁵) = 775.7 m s⁻¹, so Δν_D = 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: exp(−x²/2σ²) beats γ/πx² only out to 3.92σ, about 2.19 GHz from centre, where the profile is 4.6 × 10⁻⁴ of peak — beyond that the line is pure natural wing.

AnswerΔνₙₐₜ = 9.8 MHz, Δν_D = 1.32 GHz, Δν_Voigt = 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.

MediumA 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.

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-broadened homogeneous width: Γ√(1+S)/2π = 9.82 × √3.03 = 9.82 × 1.740 = 17.1 MHz.
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. It is a Lorentzian, so it adds linearly: 17.1 + 0.21 = 17.3 MHz.
For a 12.0 MHz dip the power-broadened part must be 12.0 − 0.21 = 11.79 MHz, so √(1+S) = 11.79/9.82 = 1.201, S = 0.440, I = 0.440 × 6.28 = 2.77 mW cm⁻².
Pump power P = I × area = 2.77 × 0.03142 = 0.087 mW, a factor of 4.6 below the original.

AnswerS = 2.03, so the dip is 17.1 MHz from power broadening plus 0.21 MHz of transit time, near 17.3 MHz. Cutting the pump to 0.087 mW gives 12.0 MHz; the natural 9.8 MHz arrives only as the power goes to zero and the dip vanishes with it.

HardHydrogen'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.

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.

Answer0.65 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.