University Quantum Mechanics II · Time-Dependent Perturbation Theory · 12.4
Harmonic perturbations: absorption and stimulated emission
Clocks, spectrometers and masers all do one thing: shake a pair of levels with an oscillating field and count how many systems changed state. Here the first-order amplitude becomes that count. You find the half of the cosine that does the work, price the half you throw away, and see emission come out as absorption run backwards.
Build the model
Connect the measurement to the mechanism.
A prescribed field V cos ωt acts on levels a and b, with ω₀ = (Eb − Eₐ)/ħ, and the first-order integral of UQMII-12.03 turns it into a transition probability. The organising move is to read the cosine as two exponentials, (V/2)e(−iωt) and (V/2)e(iωt). Each acts like a constant perturbation of strength V/2 across a gap shifted by ∓ħω, so the amplitude has two terms, with denominators ω₀ − ω and ω₀ + ω. A term of this kind can never exceed t, nor 2 divided by its denominator.
Near resonance the first therefore grows as t while the second stays below about 1/ω₀, and once the drive has run many cycles the second can be dropped. That is the rotating-wave approximation, and it leaves P = (|Vab|/ħ)² sin²(δt/2)/δ² with δ = ω − ω₀. Start in b and the other exponential resonates. Because V is Hermitian and cos ωt is real, the probability down equals the probability up, so absorption and stimulated emission are one calculation, and whether a sample absorbs or amplifies depends only on which level holds more population.
The model has three edges. It is first order, so P must stay small. The rotating-wave step fails far from resonance, returning at ω = 0 a quarter of the true static answer. And the drive is a classical function of time, so an undriven upper level never decays: spontaneous emission, slower than ω₀ by order α³ in an atom, needs the quantised field of Unit 13.
- Simple definition
- A harmonic perturbation H′ = V cos ωt drives a pair of levels with a first-order transition probability that peaks where ħω matches their spacing, and is the same whether the system starts in the lower level (absorption) or the upper one (stimulated emission).
- Example
- A caesium atom driven on its 9 192 631 770 Hz clock line for 10.0 ms, with |Vab|/ħ = 2π × 5.00 Hz = 31.42 rad s⁻¹, changes state with probability (|Vab|t/2ħ)² = 0.0247, whichever clock state it started in.
Each half acts as a constant perturbation of strength V/2, so every first-order probability for a harmonic drive carries |Vab|²/4.
V is a time-independent Hermitian operator in J, ω the drive's angular frequency in rad s⁻¹, t in s
Derived from UQMII-12.03's integral. A term (exp[ixt] − 1)/x has size 2|sin(xt/2)|/|x|, at most t and at most 2/|x|: only the ω₀ − ω term can grow.
ω₀ = (Eb − Eₐ)/ħ in rad s⁻¹ and Vba = ⟨b|V|a⟩ in J. Each bracketed term is a time in s, so cb is a pure number
Derived by dropping the ω₀ + ω term. Valid for |δ| ≪ ω₀, ω₀t ≫ 1 and P ≪ 1, which together require ω₁ ≪ ω₀; it is Unit 3's exact Rabi formula to lowest order in ω₁.
P is dimensionless; ω₁ = |Vab|/ħ and δ in rad s⁻¹. On resonance P = (ω₁t/2)²
Derived: starting in b, the e(iωt) half resonates instead. Both sides vanish when V = 0, so nothing here empties an undriven upper level.
State by state, at first order, for any real f(t) in H′ = V f(t); it uses |Vab| = |Vba|, which Hermiticity guarantees
Derived. The interference term, linear in the dropped amplitude, dominates: the error ripples at 2ω₀ under the envelope 1/(ω₀t) + 1/(ω₀t)², so ω₀t ≥ 101 guarantees 1%.
Exactly at ω = ω₀, drive switched on at a crest of cos ωt. ω₀t is a pure number, in radians
Derived from the equality above. Equal populations are transparent, a thermal pair absorbs in proportion to tanh(ħω₀/2kT), and an inverted pair amplifies the drive.
pₐ and pb are the initial occupations, a diagonal density operator; Δpb is the dimensionless gain in b's occupation
Read the cosine as two rotating halves
Quote UQMII-12.03: with cₐ(0) = 1, the first-order amplitude is cb(t) = −(i/ħ)∫₀ᵗ H′ba(t′) exp(iω₀t′) dt′, with ω₀ = (Eb − Eₐ)/ħ positive. For H′ = V cos ωt, write cos ωt = [exp(iωt) + exp(−iωt)]/2. The integrand then holds (Vba/2) exp[i(ω₀ + ω)t′] and (Vba/2) exp[i(ω₀ − ω)t′], each integrating to (exp[ixt] − 1)/(ix), and the i cancels the −i in front. So cb(t) = −(Vba/2ħ) times the sum of (exp[i(ω₀ + ω)t] − 1)/(ω₀ + ω) and (exp[i(ω₀ − ω)t] − 1)/(ω₀ − ω). Dimensions: each term is a time and Vba/ħ a rate, so cb is a pure number. It is the constant-perturbation result twice over, with V halved and the gap ħω₀ replaced by ħ(ω₀ ± ω). The exp(−iωt) half acts as if level b sat at Eb − ħω, so it resonates when ħω bridges the gap. The condition Eb = Eₐ + ħω comes out of a phase, with no photon in sight.
One term grows, the other is capped
Each term has the form (exp[ixt] − 1)/x, of size 2|sin(xt/2)|/|x|, which never exceeds t, its x → 0 limit, nor 2/|x|. Near resonance x = ω₀ − ω is small, so that term climbs toward t, while the other has x = ω₀ + ω ≈ 2ω₀ and stays below 1/ω₀ however long the drive runs. Their ratio on resonance is at most 1/(ω₀t), about 2 × 10⁻⁹ for a 10 ms drive of the 9.19 GHz caesium line. Dropping the capped term is the rotating-wave approximation: P(a → b) = (|Vab|²/ħ²) sin²(δt/2)/δ², δ = ω − ω₀, even in δ. Limits check it. On resonance P = (ω₁t/2)² with ω₁ = |Vab|/ħ, quadratic in t as every coherent start is. As t → 0, P → (ω₁t/2)² at any δ, since a brief drive cannot resolve frequency. As ω₁ → 0 at fixed δ, Unit 3's exact (ω₁²/Ω²) sin²(Ωt/2), with Ω = √(ω₁² + δ²), reduces to exactly this. The line shape is UQMII-12.03's: first zeros at δ = ±2π/t, full width at half height 5.57/t.
What the dropped term costs
Small is not harmless, because amplitudes add before they are squared. On resonance, with the drive switched on at a crest, the two terms combine to Pfull/PRWA = 1 + sin(2ω₀t)/(ω₀t) + sin²(ω₀t)/(ω₀t)². The last term is the dropped amplitude squared. The middle one is its interference with the kept amplitude, linear in 1/(ω₀t), and it dominates: the error ripples at 2ω₀ under the envelope 1/(ω₀t) + 1/(ω₀t)², so ω₀t ≥ 101, about 16 cycles, guarantees 1%. Off resonance the damage grows, since the kept term shrinks and the dropped one does not. At ω = 0 the two denominators are equal, the terms are equal, and |cb|² = 4|Vab|² sin²(ω₀t/2)/(ħω₀)², exactly UQMII-12.03's switched-on constant perturbation, while the rotating-wave formula returns a quarter of it. So the approximation belongs near resonance, |δ| ≪ ω₀, after many cycles. For a spin in a linear coil the dropped term is Unit 3's counter-rotating field; its Bloch–Siegert shift, ω₁²/4ω₀, is second order in ω₁ and quoted there.
Start in b: stimulated emission is the same integral
Now let cb(0) = 1. The amplitude for a is −(i/ħ)∫₀ᵗ Vab cos(ωt′) exp(−iω₀t′) dt′, since the Bohr frequency for b → a is −ω₀. The exp(iωt) half now gives exp[i(ω − ω₀)t′] and resonates, and the exp(−iωt) half becomes the capped term. Worked through, the bracket is minus the complex conjugate of the one for a → b, and Vab = Vba* because V is Hermitian, so |cₐ|² = |cb|² term by term, before any approximation. The reason is general: for H′ = V f(t) with f real, the two directions need ∫ f(t′) exp(±iω₀t′) dt′, which are complex conjugates. Energy balance sits in the phase: going down, the system hands ħω₀ to the drive, which a prescribed field absorbs without being tracked. The equality is between states, not levels: with gₐ states in a and gb in b, averaging over initial states and summing over final ones gives gₐ P̄(a → b) = gb P̄(b → a). For a two-state system it holds to all orders, since any 2 × 2 unitary has |Uab| = |Uba|.
Populations decide whether a sample absorbs or amplifies
A real sample is a mixture. For a diagonal density operator with occupations pₐ and pb, and P(a → b) = P(b → a) = P, the occupation of b rises by pₐP − pbP = (pₐ − pb)P, and each system takes ħω₀(pₐ − pb)P from the drive on average. Equal populations are transparent however strong the drive. An inverted pair, with pb above pₐ, gives energy to the drive, which is how a maser or laser amplifies. In equilibrium pb/pₐ = exp(−ħω₀/kT), so an isolated pair has pₐ − pb = tanh(ħω₀/2kT). At 300 K that is 7.4 × 10⁻⁴ for the 9.19 GHz caesium line, 1.9 × 10⁻³ for the 23.87 GHz ammonia inversion line and 3.2 × 10⁻⁵ for protons at 400 MHz: of every 100 000 transitions an NMR magnet drives, all but about 3 are cancelled by one going the other way. That, and not only weak coupling, is why radio-frequency absorption is faint, and why the first maser, in 1954, sorted ammonia molecules into the upper level before they reached its cavity.
What a prescribed drive cannot do
Set V = 0 and every probability here is zero for all time: |b⟩ is an eigenstate of H₀ and stays put. A classical drive only stimulates; it never makes an isolated excited state decay. Hydrogen in its 2p state does decay to the ground state, with a measured lifetime of 1.60 ns, because the electromagnetic field is itself quantised, a result quoted from quantum electrodynamics and taken up in Unit 13. The quoted rate, A = (2/3)⁸α⁵mc²/ħ = 6.27 × 10⁸ s⁻¹, sits beside ω₀ = (3/8)α²mc²/ħ = 1.55 × 10¹⁶ rad s⁻¹, so A/ω₀ = 0.104α³ = 4.0 × 10⁻⁸. Being third order in α is what lets a driven two-level model hold for times short compared with 1/A = 1.6 ns. The same quoted theory makes A grow as ω₀³, so at 24 GHz an ammonia molecule waits months, and this lesson's model is essentially exact there. So the lesson derives the amplitudes, the rotating-wave result, its error and the up-down equality; the decay rate, the Bloch–Siegert shift and the examples' matrix elements are quoted.
Change one variable at a time
Make the relationship visible.
At 2 cycles and ω = 0.8ω₀ the faint term alone is under 0.01, yet solid and dashed differ by 0.11: the dropped term acts through interference. Set 1.5 cycles and ω = 0 to see the static factor of 4, then add cycles: the curves merge near ω₀, while at ω = 0 they shrink together but keep their factor of 4. The b → a readout always matches a → b.
BOTH TERMS, a → b0.683
ROTATING-WAVE TERM ONLY0.573
COUNTER-ROTATING TERM ONLY0.0071
BOTH TERMS, b → a0.683
Live interpretationBOTH TERMS, a → b: 0.683. ROTATING-WAVE TERM ONLY: 0.573. COUNTER-ROTATING TERM ONLY: 0.0071. BOTH TERMS, b → a: 0.683
Catch the common trap
Explain before calculating.
A molecule starts in the upper level b of a pair separated by ħω₀ and is driven by H′ = V cos ωt with ω = ω₀ for a time t, where |Vab|t/ħ = 0.20 and ω₀t ≈ 10⁶. To first order, what is the probability of then finding it in the lower level a?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyCaesium atoms in the clock state |F = 3, mF = 0⟩, level a, cross a microwave field for t = 10.0 ms. It acts as H′ = V cos ωt with |Vab|/ħ = 2π × 5.00 Hz = 31.42 rad s⁻¹, and level b, |F = 4, mF = 0⟩, lies 9 192 631 770 Hz higher. Find the transition probability on resonance and 50.0 Hz off it, and for atoms that enter in b.
- Check the premises: ω₀t = 2π × 9.19263 × 10⁹ Hz × 0.0100 s = 5.78 × 10⁸, so the capped counter-rotating term is at most 1/(ω₀t) = 1.7 × 10⁻⁹ of the resonant one. The rotating-wave approximation is safe.
- On resonance P = (ω₁t/2)² with ω₁ = |Vab|/ħ: ω₁t/2 = 31.42 rad s⁻¹ × 0.0100 s/2 = 0.1571, so P = 0.0247, small enough for first order.
- At 50.0 Hz detuning, δ = 2π × 50.0 Hz = 314.2 rad s⁻¹ and δt/2 = 1.571 rad = π/2, so sin²(δt/2) = 1 and P = (ω₁/δ)² = (5.00/50.0)² = 0.0100.
- Units and limits: ω₁/δ is a ratio of rates and ω₁t/2 an angle, so P is a pure number. The first zero of the line is at δ/2π = 1/t = 100 Hz, so 50.0 Hz lies inside the central peak, as the drop from 0.0247 to 0.0100 suggests.
- Atoms entering in b are driven down with the same probabilities, 0.0247 and 0.0100. Each downward transition hands hν₀ = 6.626 × 10⁻³⁴ J s × 9.19263 × 10⁹ Hz = 6.09 × 10⁻²⁴ J to the field.
AnswerP = 0.0247 on resonance and 0.0100 at 50.0 Hz detuning, the same for a → b and b → a; each downward transition gives 6.09 × 10⁻²⁴ J to the drive.
MediumProtons (γ/2π = 42.577 MHz T⁻¹) in the Earth's field, B₀ = 50.0 μT, receive a weak pulse 2B₁ cos(ωt) x̂ from a linear coil, with B₁ = 0.100 μT, exactly on resonance, switched on at a crest and lasting t = 5.00 ms. Find the flip probability in the rotating-wave approximation, then with both first-order terms, and the pulse length beyond which the envelope of the rotating-wave error guarantees it below 1%.
- Rates: ω₀ = γB₀ = 2π × 42.577 × 10⁶ × 5.00 × 10⁻⁵ = 1.3376 × 10⁴ rad s⁻¹, or 2128.9 Hz. The coupling is V = −2γB₁Sₓ, so |Vab| = 2γB₁(ħ/2) = ħγB₁ and ω₁ = γB₁ = 26.75 rad s⁻¹.
- Rotating wave: PRWA = (ω₁t/2)² = (26.75 × 5.00 × 10⁻³/2)² = 0.06688² = 4.47 × 10⁻³.
- Here ω₀t = 66.88, only 10.6 cycles. With both terms, Pfull/PRWA = 1 + sin(2ω₀t)/(ω₀t) + sin²(ω₀t)/(ω₀t)² = 1 + 0.9709/66.88 + 0.6198/4473 = 1 + 0.01452 + 0.00014 = 1.0147.
- So Pfull = 1.0147 × 4.47 × 10⁻³ = 4.54 × 10⁻³, 1.5% above the rotating-wave value, nearly all of it from the interference term rather than the dropped amplitude squared.
- Guarantee: the error never exceeds its envelope 1/(ω₀t) + 1/(ω₀t)². Setting this equal to 0.01 gives ω₀t = 101.0, so t ≥ 101.0/(1.3376 × 10⁴ rad s⁻¹) = 7.55 ms, about 16 cycles.
- Check first order at that length: PRWA = (26.75 × 7.55 × 10⁻³/2)² = 0.0102, still ≪ 1.
AnswerPRWA = 4.47 × 10⁻³; with both terms P = 4.54 × 10⁻³, 1.5% higher; pulses of at least 7.55 ms keep the rotating-wave error below 1%.
HardAmmonia molecules on the 23.870 GHz inversion line cross a cavity in t = 60.0 μs, where a field E₀ cos ωt with E₀ = 0.150 V m⁻¹ is exactly resonant; H′ = −dzE₀ cos ωt with |⟨a|dz|b⟩| = 3.70 × 10⁻³⁰ C m. The beam carries 1.00 × 10¹³ molecules per second in this pair of levels. Find the power exchanged with the field for a thermal beam at 300 K and for a beam sorted into the upper level, and check the premises.
- Coupling: |Vab| = E₀|⟨a|dz|b⟩| = 0.150 V m⁻¹ × 3.70 × 10⁻³⁰ C m = 5.55 × 10⁻³¹ J, so ω₁ = |Vab|/ħ = 5.263 × 10³ rad s⁻¹ and ω₁t/2 = 0.1579. Per molecule P = 0.1579² = 0.0249, in either direction.
- Premises: ω₀t = 2π × 2.3870 × 10¹⁰ Hz × 6.00 × 10⁻⁵ s = 9.0 × 10⁶, so the dropped term is at most 1/(ω₀t) = 1.1 × 10⁻⁷ of the kept one; and P ≪ 1, with Rabi's sin²(ω₁t/2) = 0.0247 only 0.8% lower.
- Thermal beam: hν₀ = 6.626 × 10⁻³⁴ J s × 2.3870 × 10¹⁰ Hz = 1.582 × 10⁻²³ J, and hν₀/kT = 1.582 × 10⁻²³/(1.381 × 10⁻²³ × 300) = 3.82 × 10⁻³, so pₐ − pb = tanh(1.91 × 10⁻³) = 1.91 × 10⁻³.
- Net power taken from the field: N hν₀(pₐ − pb)P = 1.00 × 10¹³ s⁻¹ × 1.582 × 10⁻²³ J × 1.91 × 10⁻³ × 0.0249 = 7.5 × 10⁻¹⁵ W.
- Sorted beam, pb = 1: every transition is stimulated emission, and the field gains N hν₀P = 1.00 × 10¹³ s⁻¹ × 1.582 × 10⁻²³ J × 0.0249 = 3.94 × 10⁻¹² W, 524 times the thermal figure and of opposite sign.
- Spontaneous emission, from Unit 13's formula A = ω₀³|dab|²/(3πε₀ħc³) quoted here, is 2 × 10⁻⁷ s⁻¹, so its chance in 60 μs is 1 × 10⁻¹¹: leaving it out is safe at 24 GHz, as it would not be on an optical line.
AnswerP = 0.0249 per molecule either way; the thermal beam absorbs a net 7.5 × 10⁻¹⁵ W, while the upper-level beam delivers 3.94 × 10⁻¹² W to the field, 524 times more and of opposite sign.