University Quantum Mechanics II · Time-Dependent Perturbation Theory · 12.9
Second-order transitions through virtual intermediate states
Selection rules shut the one-photon door between states of equal parity; this lesson opens the two-photon one. You build the amplitude from two steps through states the atom never occupies, see why the energy mismatch sits in the denominator and the intensity enters squared, and drive hydrogen's 1s → 2s line at 243 nm.
Build the model
Connect the measurement to the mechanism.
When the perturbation cannot connect i to f directly, Vfi = 0 and the first-order amplitude vanishes at every time, so the leading term of the Dyson series is second order: the perturbation acts once to take i to some state m, the system carries m's own phase until it acts again, and the amplitudes of all such paths add. For a harmonic drive H′ = V e(−iωt) + V† e(iωt), doing the inner time integral leaves, apart from a bounded switch-on term, an amplitude of the first-order form, with Vfi replaced by Mfi = Σₘ Vfm Vₘᵢ/(Eᵢ + ħω − Eₘ) and ω by 2ω. The golden rule therefore carries over, with energy conserved at Ef = Eᵢ + 2ħω and nowhere else.
The level at Eᵢ + ħω is virtual: no eigenstate lives there, and each real state m merely follows the drive with an amplitude of order Vₘᵢ/(Eᵢ + ħω − Eₘ) that oscillates and never grows. Every link carries one power of the field amplitude E₀, so the rate goes as E₀⁴, that is as I². This is how two 243 nm photons drive hydrogen's parity-forbidden 1s → 2s line.
The treatment fails when a denominator shrinks to the size of its coupling or of the intermediate level's width: m then fills for real, and that step must be solved exactly. It also assumes a prescribed classical field, so spontaneous two-photon emission, which needs the quantised field, is quoted from quantum electrodynamics rather than derived.
- Simple definition
- A second-order transition is one whose amplitude comes from the perturbation acting twice, i → m → f, summed over every intermediate state m, each path weighted by 1/(Eᵢ + ħω − Eₘ), the distance of m from the virtual level that one photon reaches.
- Example
- Hydrogen 1s → 2s is parity-forbidden for one photon, but two 243.1 nm photons of 5.10 eV each drive it through the p states, chiefly 2p, which lies 5.10 eV above the virtual level at E₁ₛ + ħω = −8.50 eV, so 2p is passed through virtually and never filled.
When Vfi = 0 this is the leading amplitude. Every path i → m → f with both links non-zero contributes, and the paths add as amplitudes, before anything is squared.
H′fm(t) = ⟨f|H′(t)|m⟩ in J; ωfm = (Ef − Eₘ)/ħ in rad s⁻¹; t″ ≤ t′, so the i → m step comes first
It replaces Vfi in every first-order formula. Each path is weighted by how far m sits from the virtual level Eᵢ + ħω, and levels on opposite sides of it have denominators of opposite sign.
Vₘᵢ = ⟨m|V|i⟩ in J, the e(−iωt) amplitude of H′; the sum runs over every m, continuum included; Mfi in J
Energy is conserved between i and f only. The rule inherits first order's window: t long enough for the sinc peak to sit inside the band, short enough that i is not depleted.
W in s⁻¹, ρ in J⁻¹; if H′ = F cos ωt, each V is F/2, so Mfi = ¼ Σₘ Ffm Fₘᵢ/Δₘ with Δₘ = Eᵢ + ħω − Eₘ
The virtual population oscillates with period h/|Δₘ| and never exceeds 4|Vₘᵢ|²/Δₘ²: m is polarised by the drive, never filled.
dimensionless; Vₘᵢ and Δₘ in J, t in s; trustworthy while |Vₘᵢ| ≪ |Δₘ|
Double the intensity and a two-photon rate quadruples; on a log–log plot of signal against power the slope is 2, the experimental fingerprint of a two-photon process.
I = ½ε₀cE₀² in W m⁻²; each V ∝ E₀, and e²E₀² = 8παħI, so |Mfi|² ∝ α²I² where a one-photon |Vfi|² ∝ αI
Bohr levels with the reduced mass. The nearest intermediate state, 2p, sits 5.10 eV from the virtual level, as far as the photon energy itself, so no denominator comes close to zero.
RH = R_∞/(1 + mₑ/mₚ) = 1.096 776 × 10⁷ m⁻¹; each photon carries ħω = 5.099 eV, half the 10.199 eV interval
First order is silent, so follow two steps
Expand the state as Σₙ cₙ(t) e(−iEₙt/ħ)|n⟩ and iterate the exact coefficient equations twice. If Vfi = 0, as for an electric-dipole perturbation between two states of equal parity, then cf⁽¹⁾ = 0 at every t, and the first term that can move population is cf⁽²⁾ = (iħ)⁻² Σₘ ∫₀ᵗ dt′ e(iωfm t′) H′fm(t′) ∫₀(t′) dt″ e(iωₘᵢ t″) H′ₘᵢ(t″). Read it right to left, as a history: at t″ the perturbation takes i to m, the system then carries m's own phase, and at the later time t′ the perturbation takes m to f. The nested limits enforce t″ ≤ t′, the time ordering of the Dyson series, which matters because H′ at two different times need not commute. A path contributes only if both links exist, Vₘᵢ ≠ 0 and Vfm ≠ 0, so for a dipole perturbation between two s states every usable m is a p state. The paths are summed as amplitudes, so they can reinforce or cancel before anything is squared.
The inner integral produces the energy denominator
Take H′ = V e(−iωt) + V† e(iωt) and keep the term that absorbs at both steps, the only one that can reach Ef = Eᵢ + 2ħω. The inner integral is [e(i(ωₘᵢ − ω)t′) − 1]/(i(ωₘᵢ − ω)). The −1 is the switch-on transient: carried into the outer integral it oscillates at ωfm − ω, which on two-photon resonance equals Δₘ/ħ, far from zero, so it stays bounded. The first piece joins e(iωfm t′) to make e(i(ωfi − 2ω)t′), and because ħ(ωₘᵢ − ω) = −(Eᵢ + ħω − Eₘ), the result is cf⁽²⁾(t) = (1/iħ) Mfi ∫₀ᵗ e(i(ωfi − 2ω)t′) dt′. That is the first-order harmonic amplitude with Vfi → Mfi and ω → 2ω, so everything proved at first order carries over: the sinc² peak, now at 2ħω = Ef − Eᵢ, and, summed over a band of final states, W = (2π/ħ)|Mfi|²ρ(Ef). Compare Unit 6's second-order energy, Σ|Vₘₙ|²/(Eₙ − Eₘ): the denominator is the same idea with one photon's energy added to the initial state. Dimensions check: energy²/energy is an energy, as Vfi was.
Virtual means small, oscillating and never filled
Energy is conserved between i and f, not at the halfway point. At first order the intermediate amplitude is cₘ(t) = Vₘᵢ[e(−iΔₘt/ħ) − 1]/Δₘ with Δₘ = Eᵢ + ħω − Eₘ, so |cₘ|² = (4|Vₘᵢ|²/Δₘ²) sin²(Δₘt/2ħ): it rises, falls back to zero every h/|Δₘ|, and never exceeds 4|Vₘᵢ|²/Δₘ². With |Vₘᵢ| = 1.0 meV and Δₘ = 0.50 eV the ceiling is 1.6 × 10⁻⁵ and the period 8.3 fs. No eigenstate lives at Eᵢ + ħω; the virtual level is only the energy of the atom plus the one photon absorbed so far, and the real levels respond to it the way a driven oscillator follows a force far from its resonance, with a small amplitude locked to the drive. That is also why the transition needs no intermediate level near Eᵢ + ħω at all. The familiar phrase that m is borrowed for a time of order ħ/|Δₘ| is a mnemonic for this oscillation, whose period h/|Δₘ| is 2π times that time, not an extra law of nature.
Each photon brings one power of the intensity
For an electric-dipole drive H′ = eE₀z cos ωt, each rotating part is V = eE₀z/2, so every link of a path carries one power of E₀: Mfi ∝ E₀² and W ∝ E₀⁴. With I = ½ε₀cE₀², that is W ∝ I², against W ∝ I for one photon. Written with the fine-structure constant, e²E₀² = 8παħI, so a one-photon rate goes as αI and a two-photon rate as α²I²: each absorbed photon costs one more factor of αI. The ratio of successive orders is of order |V/Δ|², roughly (E₀/Eₐₜ)² for atoms, with Eₐₜ = e/(4πε₀a₀²) = 5.14 × 10¹¹ V m⁻¹, the field whose work across a₀ is the Hartree energy, 27.2 eV. That is why two-photon absorption needs a laser, and why a measured slope of 2 on a log–log plot of signal against power is the standard test that a signal is two-photon. At fixed average power, pulses help, since for rectangular pulses of length τ repeated at rate ν the time average of I² is ⟨I⟩²/(τν); a tight focus helps too, confining the excitation to where I² is largest.
Hydrogen 1s → 2s at 243 nm
Both states have l = 0 and even parity while the dipole operator is odd, so ⟨2s|z|1s⟩ = 0 by Unit 5's parity rule and one photon cannot do it. Two can: each link changes parity, so the intermediate states are the p states, bound and continuum. Energy conservation between the ends gives 2ħω = E₂ₛ − E₁ₛ = (3/4)RH hc = 10.199 eV, so ħω = 5.099 eV and λ = 243.1 nm (derived). The virtual level sits at E₁ₛ + ħω = −8.50 eV, below every p state, so every denominator is negative and large: −5.10 eV for 2p, −6.99 eV for 3p and at most −8.50 eV for the continuum. Counter-propagating photons carry Doppler shifts +ωv/c and −ωv/c that cancel, so the line is Doppler-free, and 2s decays almost entirely by two-photon emission, at 8.23 s⁻¹ from quantum electrodynamics (quoted), a natural width of 1.3 Hz. Measured at Garching, the 1S–2S frequency is 2 466 061 413 187 kHz, good to 4 parts in 10¹⁵ (quoted). Bohr's value is 23.0 GHz lower; fine structure from Unit 8 adds 30.1 GHz, order α² relative to the interval, and the Lamb shifts, quoted from QED, remove 7.1 GHz, leaving a residual of order 10 MHz.
When a denominator nears zero, solve that step exactly
Mfi rests on |Vₘᵢ| ≪ |Δₘ|. Tune the photon toward a real level and Δₘ shrinks: the ceiling 4|Vₘᵢ|²/Δₘ² approaches 1, and Vfm Vₘᵢ/Δₘ diverges although no probability can. The intermediate level then fills for real and absorption becomes stepwise, so the resonant step has to be treated exactly. For one intermediate level with equal couplings V, the rotating-frame Hamiltonian is a constant 3 × 3 matrix; (|i⟩ − |f⟩)/√2 decouples from m, and (|i⟩ + |f⟩)/√2 Rabi-oscillates with m at coupling √2 V and detuning Δ. That gives a peak Pₘ = 4V²/(Δ² + 8V²) and, for |Δ| ≫ V, Pf = sin²(V²t/ħΔ), whose early-time form is second order's (Mt/ħ)². If m has a width Γ, the denominator becomes Eᵢ + ħω − Eₘ + iΓ/2 (quoted), which caps the enhancement. Strong fields bring a second limit: sums of the same kind with f = i, together with their counter-rotating partners with denominators Eᵢ − ħω − Eₘ, shift each level by an amount proportional to I, the light shift, which unequal shifts of i and f turn into a detuning that precision two-photon measurements must correct for.
Change one variable at a time
Make the relationship visible.
At Δ/h = 15 MHz, Pₘ stays below 4% while Pf tracks the dashed (Mt/ħ)² until depletion bends it. Double V, four times the intensity: M quadruples and the dashed curve is sixteen times higher at early t. Then drag Δ/h to 2 MHz: m fills, the dashed curve shoots off, and only the exact solution holds.
M/h = V²/(hΔ)0.150 MHz
PEAK Pₘ, EXACT0.037
Pf AT 1 μs, EXACT0.620
(Mt/ħ)² AT 1 μs0.89
Live interpretationM/h = V²/(hΔ): 0.150 MHz. PEAK Pₘ, EXACT: 0.037. Pf AT 1 μs, EXACT: 0.620. (Mt/ħ)² AT 1 μs: 0.89
Catch the common trap
Explain before calculating.
A two-photon transition i → f between states of equal parity, so that Vfi = 0, runs through one intermediate level m. Take Eᵢ = 0, Eₘ = 2.30 eV, Ef = 4.00 eV, light with ħω = 2.00 eV, and |Vₘᵢ| = |Vfm| = 3.0 meV, where V is the amplitude of the e(−iωt) part of H′. What is |Mfi|?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA dye absorbs at 400 nm and has no level near 800 nm, but two 800 nm photons can still excite it. In the same focal spot, compare the time-averaged two-photon excitation from a continuous-wave laser of 1.0 W with that from a pulsed laser of 20 mW average power emitting 100 fs pulses at 80 MHz, treating the pulses as rectangular. Then say what halving the pulsed laser's power does.
- Energy bookkeeping: ħω = hc/λ = 1.550 eV at 800 nm, so two photons deliver 3.10 eV, the 400 nm transition. No level sits near 1.55 eV, so the intermediate states are virtual and the process is second order.
- Each link V ∝ E₀, so Mfi ∝ E₀² and W ∝ E₀⁴ ∝ I². With the focus fixed, I ∝ P, so the time-averaged rate follows ⟨P²⟩, not ⟨P⟩².
- Continuous wave: ⟨P²⟩ = (1.0 W)² = 1.0 W².
- Pulsed: the duty cycle is τν = (100 × 10⁻¹⁵ s)(80 × 10⁶ s⁻¹) = 8.0 × 10⁻⁶, so the power during a pulse is P̄/(τν) = 20 mW/(8.0 × 10⁻⁶) = 2.5 kW, present for that fraction of the time: ⟨P²⟩ = τν(P̄/τν)² = P̄²/(τν) = (0.020 W)²/(8.0 × 10⁻⁶) = 50 W².
- Ratio 50/1.0: the pulsed beam excites 50 times more at one fiftieth of the average power. Halving its power to 10 mW divides ⟨P²⟩, and the signal, by 4.
AnswerThe pulsed laser gives about 50 times the two-photon signal (50 W² against 1.0 W²); halving its average power cuts the signal to a quarter. Real pulse shapes change the 50 by a factor of order one.
MediumLevels i (E = 0) and f (E = 3.20 eV) have equal parity, so Vfi = 0. Two levels couple to both: m at 1.20 eV with Vₘᵢ = 5.0 meV and Vfm = 2.0 meV, and n at 2.40 eV with Vₙᵢ = 3.0 meV and Vfn = 4.0 meV, all real and positive e(−iωt) amplitudes. Final states form a band of density 50 eV⁻¹ around 3.20 eV. On two-photon resonance, find Mfi and the rate, then compare with the path through m alone and with the intensity tripled.
- Two-photon resonance: 2ħω = 3.20 eV, so ħω = 1.60 eV and the virtual level sits at 1.60 eV, between the two real levels.
- Denominators Eᵢ + ħω − E: for m, 1.60 − 1.20 = +0.40 eV; for n, 1.60 − 2.40 = −0.80 eV. Both are at least 80 times the couplings, and the intermediate populations stay below 4(5.0 meV/0.40 eV)² = 6.3 × 10⁻⁴, so second order applies.
- Paths: via m, (2.0 meV)(5.0 meV)/(0.40 eV) = +25 μeV; via n, (4.0 meV)(3.0 meV)/(−0.80 eV) = −15 μeV. The level below the virtual level and the one above it enter with opposite signs, so Mfi = +10 μeV.
- Rate: W = (2π/ħ)|Mfi|²ρ = (2π/6.582 × 10⁻¹⁶ eV s)(1.0 × 10⁻⁵ eV)²(50 eV⁻¹) = 4.8 × 10⁷ s⁻¹, valid for times short against 1/W = 21 ns and long against ħ over the band's width.
- Comparisons: m alone gives 25 μeV and a rate 6.25 times larger, 3.0 × 10⁸ s⁻¹, while adding the paths' squares instead of their amplitudes would overestimate by (25² + 15²)/10² = 8.5. Tripling the intensity triples each M, so W rises ninefold, to 4.3 × 10⁸ s⁻¹.
AnswerMfi = +10 μeV and W ≈ 4.8 × 10⁷ s⁻¹; the path through m alone would give 3.0 × 10⁸ s⁻¹, and tripling the intensity gives 4.3 × 10⁸ s⁻¹.
HardHydrogen atoms at rest in 1s are lit for 5.0 μs by linearly polarised light on the 1s → 2s two-photon resonance, of intensity 1.0 × 10⁸ W m⁻². Keeping only the 2p₀ path, with ⟨2p₀|z|1s⟩ = 0.745 a₀ and ⟨2s|z|2p₀⟩ = −3.00 a₀, find the peak 2p population and P(2s). Then correct P(2s) using the full sum over p states, 7.85 in atomic units against 11.92 for 2p alone.
- Field: E₀ = √(2I/ε₀c) = √(2 × 1.0 × 10⁸/(8.854 × 10⁻¹² × 2.998 × 10⁸)) V m⁻¹ = 2.74 × 10⁵ V m⁻¹. For H′ = eE₀z cos ωt the absorbing part is V = eE₀z/2, so the 1s → 2p link is eE₀(0.745 a₀)/2 = 5.41 × 10⁻⁶ eV and the 2p → 2s link is eE₀(−3.00 a₀)/2 = −2.18 × 10⁻⁵ eV.
- Denominator: ħω = 5.099 eV, so E₁ₛ + ħω − E₂ₚ = −13.598 + 5.099 + 3.400 = −5.099 eV. The 2p population never exceeds 4(5.41 × 10⁻⁶/5.099)² = 4.5 × 10⁻¹², and its oscillation period is h/5.1 eV = 8.1 × 10⁻¹⁶ s: 2p is virtual.
- 2p path: M = (−2.18 × 10⁻⁵ eV)(5.41 × 10⁻⁶ eV)/(−5.099 eV) = +2.31 × 10⁻¹¹ eV, so M/ħ = 3.51 × 10⁴ s⁻¹. Dimensions: energy times energy over energy is an energy, as a matrix element must be.
- 2s is a single level and the light is on resonance, so c₂ₛ = Mt/(iħ) and P = (Mt/ħ)² = (3.51 × 10⁴ s⁻¹ × 5.0 × 10⁻⁶ s)² = (0.176)² = 3.1 × 10⁻².
- Full sum: in atomic units the 2p term is (−3.00)(0.745)/(−0.1875) = 11.92, while a numerical sum over every p state and the continuum gives 7.85 (quoted); the 3p path alone contributes −2.05, opposite in sign. Scaling M by 7.85/11.92 = 0.659 gives M/ħ = 2.31 × 10⁴ s⁻¹, Mt/ħ = 0.116 and P = 1.34 × 10⁻².
- Checks: P ≪ 1, so second order holds, and P ∝ I², so doubling I would give 5.4 × 10⁻². Light shifts of 1s and 2s have been ignored, and so has ionisation of 2s by a third photon, which at this intensity runs at about 7.6 × 10⁴ s⁻¹ (quoted, from the 2s photoionisation cross-section) and would lower P by about a sixth.
AnswerP(2s) ≈ 1.3 × 10⁻² after 5.0 μs; the 2p path alone gives 3.1 × 10⁻², 2.3 times too large, and the 2p population never exceeds 4.5 × 10⁻¹².