University Quantum Mechanics II · Time-Dependent Perturbation Theory · 12.2
The interaction picture and the Dyson series
Nearly every transition amplitude and rate in the rest of this course is a term of one series. This topic builds it: move the solved motion into the operators, turn what is left into an integral equation, and iterate it, keeping track of which kick came first, because that order is where the physics hides.
Build the model
Connect the measurement to the mechanism.
When H = H₀ + H′(t) and H₀ is solved, most of a state's motion is the fast, known phase e(−iEₙt/ħ) of each eigenstate, and transitions are the slow drift underneath it. The interaction picture separates the two by undoing the known part exactly, |ψI(t)⟩ = e(iH₀t/ħ)|ψ(t)⟩. The H₀ terms then cancel from the Schrödinger equation, leaving iħ d|ψI⟩/dt = H′I(t)|ψI⟩ with the perturbation dressed as H′I = e(iH₀t/ħ)H′e(−iH₀t/ħ), and the components of |ψI⟩ are the coefficients cₙ of the previous topic.
Nothing has been approximated; what the change buys is an integral equation that can be fed into itself. Each pass adds one more action of H′I at an earlier time, and the result is the Dyson series: first order is one kick, second order is two kicks with the later one standing to the left. Only when H′I commutes with itself at all pairs of times does the series collapse to the ordinary exponential of ∫H′I dt, which is why Unit 3 chose a frame turning at the drive frequency, where the Hamiltonian left over was static.
The approximation lives entirely in cutting the series off. A cut series is unitary only to the order kept and fails once the phase or amplitude that H′ has built up nears 1, which even a tiny H′ manages given time; for unbounded perturbations the convergence of the full series is assumed, not proved.
- Simple definition
- The interaction picture is the frame that turns with the solved Hamiltonian H₀, so a state changes only under the dressed perturbation H′I(t); the Dyson series expands that change in the number of times H′I acts, each later action written to the left.
- Example
- An electron spin in 0.350 T under a resonant 0.100 mT rotating field: H₀ turns the relative phase of |↑⟩ and |↓⟩ at ω₀ = 6.16 × 10¹⁰ rad s⁻¹, but the interaction-picture state moves only at ω₁ = 1.76 × 10⁷ rad s⁻¹. After 10.0 ns H₀ has turned that phase by 616 rad; ψI has tipped by 0.176 rad.
Derived as an exact unitary change of frame. Its components ⟨n|ψI⟩ are the coefficients cₙ(t), so every probability |cₙ|² is the same in both pictures.
|ψ(t)⟩ the Schrödinger-picture state; H₀ the solved Hamiltonian in J; t in s; ħ = 1.055 × 10⁻³⁴ J s
Derived by differentiating the definition: the H₀ terms cancel exactly. The state moves only while the perturbation acts, and at H′ = 0 it is frozen.
H′(t) the perturbation in J; H′I(t) the same operator carried along by H₀, also in J
Derived. Projecting the equation of motion onto ⟨m| returns iħ ċₘ = Σₙ H′ₘₙ e(iωₘₙt) cₙ exactly: the coefficient equations are this picture written in components.
Eₘ, Eₙ eigenvalues of H₀ in J; ωₘₙ in rad s⁻¹; the phase is the only trace H₀ leaves
Derived by integrating the equation of motion. The unknown sits under the integral, so it can be substituted into itself: each pass adds one factor of H′I/iħ at an earlier time.
t′ in s runs over earlier times only; the initial condition is built in, and the equation is still exact
Derived by iteration. With ‖H′‖ ≤ V the nth term is at most (Vt/ħ)ⁿ/n!, so for bounded H′, as on any finite set of levels, the series converges; for unbounded H′ that is assumed.
|ψI(t)⟩ = UI(t)|ψI(0)⟩; 0 ≤ t″ ≤ t′ ≤ t with the later operator on the left; every term is dimensionless
A name for the Dyson series, not a new way to sum it. Dropping T wrongly loses ½(1/iħ)² ∫∫ [H′I(t′), H′I(t″)] over t″ ≤ t′ at second order.
T puts every product in latest-left order; it may be dropped only if [H′I(t₁), H′I(t₂)] = 0 for all t₁, t₂
Take the solved motion out of the state
Quantum Mechanics I evolved states with U(t) = e(−iHt/ħ) for a time-independent H. With H = H₀ + H′(t), most of the motion is still H₀'s: an electron spin in 0.350 T turns its relative phase at ω₀ = gμB B₀/ħ = 6.16 × 10¹⁰ rad s⁻¹ whatever else acts; with the measured g = 2.00232, ω₀/2π = 9.81 GHz, the X-band frequency at which ESR spectrometers work near that field. So undo exactly that motion: |ψI(t)⟩ = e(iH₀t/ħ)|ψ(t)⟩. Apply iħ d/dt by the product rule. The exponential contributes −H₀|ψI⟩, the Schrödinger equation contributes e(iH₀t/ħ)(H₀ + H′)|ψ⟩, and because H₀ commutes with its own exponential the two H₀ terms cancel, leaving iħ d|ψI⟩/dt = H′I(t)|ψI⟩. Nothing was approximated. Operators now carry H₀, AI(t) = e(iH₀t/ħ)A e(−iH₀t/ħ), and states carry H′, so ⟨ψI|AI|ψI⟩ = ⟨ψ|A|ψ⟩ at every time: the picture sits halfway between Schrödinger's and Heisenberg's. Limiting case: at H′ = 0 the state is frozen.
The components are the coefficients you already had
Expand the Schrödinger state as the previous topic did, |ψ(t)⟩ = Σₙ cₙ(t) e(−iEₙt/ħ)|n⟩. Then ⟨n|ψI⟩ = e(iEₙt/ħ)⟨n|ψ⟩ = cₙ(t): the interaction-picture state is the list of slow coefficients, and |cₙ|² is one probability in both pictures because the frames differ by a phase per level. Sandwich the dressed perturbation between eigenstates: ⟨m|H′I|n⟩ = e(iEₘt/ħ)⟨m|H′|n⟩e(−iEₙt/ħ) = e(iωₘₙt)H′ₘₙ. Projecting the equation of motion onto ⟨m| returns iħ ċₘ = Σₙ H′ₘₙ e(iωₘₙt) cₙ, with nothing added or lost. That phase is H₀'s only surviving trace, and it does real work. A static transverse field on the spin becomes, in this picture, an operator turning at ω₀ = 6.16 × 10¹⁰ rad s⁻¹ while the cₙ barely move, so its pushes cancel turn by turn: a perturbation drives a large transition only when its own time dependence cancels e(iωₘₙt).
Feed the integral equation into itself
Integrate from 0 to t: |ψI(t)⟩ = |ψI(0)⟩ + (1/iħ)∫₀ᵗ H′I(t′)|ψI(t′)⟩ dt′. It is exact and carries the initial condition, but the unknown appears on both sides. Replace |ψI(t′)⟩ under the integral by the whole right-hand side at t′, whose integral runs over t″ from 0 to t′, and repeat. Each pass adds one factor of H′I/iħ at an earlier time, which builds the Dyson series with 0 ≤ t″ ≤ t′ ≤ t and the later operator always on the left, because it acts later. Starting in |i⟩, the components are cₘ⁽¹⁾ = (1/iħ)∫₀ᵗ e(iωₘᵢt′)H′ₘᵢ(t′) dt′ and cₘ⁽²⁾ = (1/iħ)² Σₖ ∫₀ᵗ dt′ ∫₀(t′) dt″ e(iωₘₖt′)H′ₘₖ(t′) e(iωₖᵢt″)H′ₖᵢ(t″). Read cₘ⁽²⁾ from the right: at t″ the system goes i → k, carries k's phase, and at t′ goes k → m; k runs over every state, i and m included. Dimensions: each factor H′ dt/ħ is J × s over J s, a pure number.
Time ordering, and when an exponential is enough
Square ∫₀ᵗ H′I dt and you cover the triangle t″ ≤ t′ twice, once with H′I(t′)H′I(t″) and once with the product reversed. If H′I at different times commuted the two would agree, the ordered double integral would be half the square, every term would be (1/n!)[(1/iħ)∫H′I dt]ⁿ, and the series would sum to exp[−(i/ħ)∫₀ᵗ H′I dt]. In general the Dyson term exceeds half the square by ½(1/iħ)² ∫∫ [H′I(t′), H′I(t″)] over t″ ≤ t′. Test it on a rotating drive, H′ = (ħω₁/2)[e(−iωt)σ₊ + e(iωt)σ₋] on H₀ = (ħω₀/2)σz, with σ₊ = |↑⟩⟨↓| from Quantum Mechanics I. Dressing multiplies σ₊ by e(iω₀t), so H′I = (ħω₁/2)[e(−iΔt)σ₊ + e(iΔt)σ₋] with Δ = ω − ω₀, and since [σ₊, σ₋] = σz, [H′I(t₁), H′I(t₂)] = −(i/2)ħ²ω₁² sin(Δ(t₁ − t₂)) σz. It vanishes at every pair of times only when Δ = 0.
The rotating frame was this move at the drive frequency
Unit 3 undid e(−iωσz t/2), turning with the drive rather than with H₀. The same differentiation leaves HR = (ħ/2)[(ω₀ − ω)σz + ω₁σₓ], static: the detuning that the interaction picture turns into the phase e(−iΔt) is kept there as a constant energy instead. A static HR commutes with itself, its evolution is an ordinary exponential, and that is the whole reason Unit 3's Rabi formula, P = (ω₁²/Ω²) sin²(Ωt/2) with Ω = √(ω₁² + Δ²), came out exact; it is quoted here, not rederived. Which part of H you call H₀ is a choice that changes what looks static, never the physics. On resonance the two frames coincide and H′I = (ħω₁/2)σₓ commutes with itself even if ω₁ varies in time. Off resonance, dropping T is a real error: at Δ = ω₁ and ω₁t = 2π the ordinary exponential predicts no flip at all, sin²(sin π) = 0, while the full series gives ½ sin²(√2 π) = 0.46.
Truncation, unitarity and convergence
UI is exactly unitary; a series cut at order n is unitary only to order n, and that is a check to run every time. When H′ has no diagonal element, the probability that leaves |i⟩ at second order, −2Re cᵢ⁽²⁾, must equal Σₘ |cₘ⁽¹⁾|²; for the detuned drive both are (ω₁/Δ)² sin²(Δt/2). The same cᵢ⁽²⁾ also holds a secular term, −iω₁²t/(4Δ) for the lower level, the first term of the phase e(−iω₁²t/(4Δ)): the drive shifts the level by ħω₁²/(4Δ), the light shift of atomic physics. The ordinary exponential has no such term: at second order, dropping T discards −i(ω₁²/(4Δ))(t − sin(Δt)/Δ), this secular term plus a bounded ripple. A term growing as t spoils the cut series once ω₁²t/(4Δ) nears 1, however small ω₁/Δ is. As for the whole series: with ‖H′‖ ≤ V the nth term is at most (Vt/ħ)ⁿ/n!, the 1/n! coming from the ordered region, whose volume is tⁿ/n!, so for bounded H′, as on any finite set of levels, it converges at every t. For an unbounded perturbation such as eE₀z no such bound exists, and convergence is assumed, not proved.
Change one variable at a time
Make the relationship visible.
Set Δ to zero: H′I then commutes with itself and the two curves are one. At Δ = ω₁ they agree only at first; by ω₁t = 2π the ordinary exponential is back at zero while the full series sits near 0.46, because it keeps the commutator that time ordering carries.
FULL SERIES P0.458
ORDINARY EXPONENTIAL P0.000
ERROR FROM DROPPING T-0.458
CEILING ω₁²/Ω²0.500
Live interpretationFULL SERIES P: 0.458. ORDINARY EXPONENTIAL P: 0.000. ERROR FROM DROPPING T: −0.458. CEILING ω₁²/Ω²: 0.500
Catch the common trap
Explain before calculating.
A student replaces the Dyson second-order term (1/iħ)² ∫₀ᵗ dt′ ∫₀(t′) dt″ H′I(t′) H′I(t″) by ½[(1/iħ) ∫₀ᵗ H′I dt′]², the second term of an ordinary exponential. What is the Dyson term minus the student's?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron spin (g = 2.00232) sits in B₀ = 0.350 T along z, prepared in |↑⟩. At t = 0 the field is raised by δB = 2.00 μT, still along z. Find the exact interaction-picture amplitude c↑(t), the time at which the first-order truncation of the Dyson series overstates |c↑|² by 1.0%, and what the second-order truncation gives then.
- H′ = (ħδ/2)σz with δ = gμB δB/ħ = 1.7609 × 10¹¹ rad s⁻¹ T⁻¹ × 2.00 × 10⁻⁶ T = 3.522 × 10⁵ rad s⁻¹, the same form as H₀ = (ħω₀/2)σz. The measured g exceeds 2 by α/π at leading order, a result quoted from quantum electrodynamics.
- H′ commutes with H₀, so H′I(t) = e(iH₀t/ħ)H′e(−iH₀t/ħ) = H′: constant, hence commuting with itself at all times. Time ordering does nothing, the nth Dyson term is (1/n!)(−iH′t/ħ)ⁿ, and the series sums to UI = e(−iH′t/ħ).
- On |↑⟩, where σz = +1, c↑(t) = e(−iδt/2) exactly, and c↓ stays zero at every order because a diagonal H′ has no matrix element between the levels. |c↑|² = 1 at all times.
- First-order truncation: c↑ ≈ 1 − iδt/2, so |c↑|² = 1 + (δt/2)². That reaches 1.010 when δt/2 = 0.100, at t = 0.200/δ = 0.200/(3.522 × 10⁵ rad s⁻¹) = 5.68 × 10⁻⁷ s.
- Second order at the same moment: 1 − iφ − φ²/2 with φ = 0.100 gives |c↑|² = (1 − φ²/2)² + φ² = 1 + φ⁴/4 = 1.000025. Keeping more orders pushes the norm error to higher powers of δt: φ² at first order, φ⁴ at second.
- Check: δB/B₀ = 5.7 × 10⁻⁶, yet first order is 1% wrong after 0.57 μs. The number that governs truncation is the phase δt/2 that H′ has built up, not H′ beside H₀; meanwhile H₀ has turned the Schrödinger-picture phase by ω₀t/2 = 1.75 × 10⁴ rad, which this picture never has to follow.
Answerc↑(t) = e(−iδt/2) with δ = 3.52 × 10⁵ rad s⁻¹. First order overstates |c↑|² by 1.0% at t = 0.568 μs, when the second-order truncation is off by only 2.5 × 10⁻⁵.
MediumAn electron spin in B₀ = 0.350 T is in |↓⟩ long before a field rotating about z at exactly ω₀ is pulsed with amplitude B₁(t) = Bₚₖ exp(−t²/(2τ²)), τ = 4.00 ns, centred on t = 0. Show that the Dyson series sums to an ordinary exponential, find the Bₚₖ that flips the spin completely, and compare with the first term, and with the terms through third order, of the series.
- With ω = ω₀ the dressing factor e(iω₀t) cancels the drive's e(−iωt), so H′I(t) = (ħω₁(t)/2)(σ₊ + σ₋) = (ħω₁(t)/2)σₓ, where ω₁(t) = gμB B₁(t)/ħ. Every H′I(t) is a multiple of σₓ, so all of them commute, whatever the pulse shape.
- The series is then exp[−(i/ħ)∫H′I dt] = exp(−iAσₓ/2) = cos(A/2) − i sin(A/2)σₓ, with pulse area A = ∫ω₁ dt over the whole pulse, so P(↓ → ↑) = sin²(A/2) exactly.
- A full flip needs A = π. The Gaussian integral gives A = ωₚₖ τ√(2π), so ωₚₖ = √(π/2)/τ = 1.2533/(4.00 × 10⁻⁹ s) = 3.133 × 10⁸ rad s⁻¹, and Bₚₖ = ħωₚₖ/(gμB) = 3.133 × 10⁸ rad s⁻¹ ÷ 1.7609 × 10¹¹ rad s⁻¹ T⁻¹ = 1.78 × 10⁻³ T.
- The Dyson terms for c↑ are the Taylor terms of −i sin(A/2). The first alone gives |c↑|² = (A/2)² = (π/2)² = 2.47, an impossible probability. Through third order, A/2 − (A/2)³/6 = 1.5708 − 0.6460 = 0.9248, so P = 0.855; adding fifth order gives 1.0045, so P = 1.009.
- Checks: [ωₚₖ τ] = s⁻¹ × s, a pure number. A constant ω₁ gives A = ω₁t and P = sin²(ω₁t/2), Unit 3's resonant Rabi result. Halving Bₚₖ gives A = π/2 and P = 0.500 exactly. Off resonance H′I gains the phases e(∓iΔt), stops commuting, and the area alone no longer fixes the answer.
AnswerBₚₖ = 1.78 mT, a pulse 9.4 ns wide at half maximum. The summed series gives P = 1 exactly; its first term alone predicts 2.47 and its terms through third order 0.855, so at A = π the series converges but is within 1% only from fifth order on.
HardAn electron spin starts in |↓⟩ under a rotating drive with ω₁ = 2π × 0.400 MHz, tuned Δ = ω − ω₀ = 2π × 8.00 MHz above resonance. Evaluate the second-order Dyson amplitude c↓⁽²⁾(t) with its time ordering, check probability conservation at order ω₁², read off the level shift it implies and compare it with the exact rotating-frame value, and find when the truncated survival probability is 0.010 too large.
- The only nonzero matrix elements are ⟨↑|H′I|↓⟩ = (ħω₁/2)e(−iΔt) and its conjugate, so the intermediate state is |↑⟩ and c↓⁽²⁾ = (1/iħ)² ∫₀ᵗ dt′ (ħω₁/2)e(iΔt′) ∫₀(t′) dt″ (ħω₁/2)e(−iΔt″), with t″ ≤ t′: up at t″, back down at t′.
- The inner integral is (1 − e(−iΔt′))/(iΔ), so c↓⁽²⁾ = −(ω₁²/4) ∫₀ᵗ (e(iΔt′) − 1)/(iΔ) dt′ = −iω₁²t/(4Δ) − (ω₁²/(4Δ²))(1 − e(iΔt)). As Δ → 0 the two pieces combine to −(ω₁t)²/8, the second-order term of cos(ω₁t/2), as resonance requires.
- Conservation: 2Re c↓⁽²⁾ = −(ω₁²/(2Δ²))(1 − cos Δt) = −(ω₁/Δ)² sin²(Δt/2), which cancels the first-order |c↑⁽¹⁾|² = (ω₁/Δ)² sin²(Δt/2) exactly. Here (ω₁/Δ)² = 0.00250, so at most about 0.25% ever leaves |↓⟩.
- The term −iω₁²t/(4Δ) grows without bound: it is the first term of e(−iδt) with δ = ω₁²/(4Δ) = 2π × (0.400 MHz)²/(4 × 8.00 MHz) = 2π × 5.000 kHz, so the drive raises |↓⟩ by ħδ. The exact rotating-frame value, quoted from Unit 3, is (Ω − Δ)/2 = 2π × (√64.16 − 8.00)/2 MHz = 2π × 4.997 kHz: second order is high by (ω₁/Δ)²/4 = 0.06%.
- Truncation: |1 + c↓⁽²⁾|² ≈ 1 − (ω₁/Δ)² sin²(Δt/2) + (δt)², while the exact survival probability has no (δt)² piece. The excess reaches 0.010 when δt = 0.100, at t = 0.100/(3.142 × 10⁴ rad s⁻¹) = 3.18 μs, about 25 detuning periods in. Integrating the exact equations numerically gives a phase of c↓ of −0.0999 rad there, against −δt = −0.100 rad.
Answerc↓⁽²⁾ = −iω₁²t/(4Δ) − (ω₁²/(4Δ²))(1 − e(iΔt)). It implies a shift δ/2π = 5.000 kHz against the exact 4.997 kHz, and the truncated survival probability is 0.010 too large by t = 3.18 μs, although ω₁/Δ is only 0.05.